Differential Equations I MATB44H3F. Version September 15, 2011-1949

Differential Equations I MATB44H3F Version September 15, 211-1949

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Contents 1 Introduction 1 11 Preliminaries 1 12 Sample Application of Differential Equations 2 2 First Order Ordinary Differential Equations 5 21 Separable Equations 5 22 Exact Differential Equations 7 23 Integrating Factors 11 24 Linear First Order Equations 14 25 Substitutions 17 251 Bernoulli Equation 17 252 Homogeneous Equations 19 253 Substitution to Reduce Second Order Equations to First Order 2 3 Applications and Examples of First Order ode s 25 31 Orthogonal Trajectories 25 32 Exponential Growth and Decay 27 33 Population Growth 28 34 Predator-Prey Models 29 35 Newton s Law of Cooling 3 36 Water Tanks 31 37 Motion of Objects Falling Under Gravity with Air Resistance 34 38 Escape Velocity 36 39 Planetary Motion 37 31 Particle Moving on a Curve 39 iii

iv CONTENTS 4 Linear Differential Equations 45 41 Homogeneous Linear Equations 47 411 Linear Differential Equations with Constant Coefficients 52 42 Nonhomogeneous Linear Equations 54 5 Second Order Linear Equations 57 51 Reduction of Order 57 52 Undetermined Coefficients 6 521 Shortcuts for Undetermined Coefficients 64 53 Variation of Parameters 66 6 Applications of Second Order Differential Equations 71 61 Motion of Object Hanging from a Spring 71 62 Electrical Circuits 75 7 Higher Order Linear Differential Equations 79 71 Undetermined Coefficients 79 72 Variation of Parameters 8 73 Substitutions: Euler s Equation 82 8 Power Series Solutions to Linear Differential Equations 85 81 Introduction 85 82 Background Knowledge Concerning Power Series 88 83 Analytic Equations 89 84 Power Series Solutions: Levels of Success 91 85 Level 1: Finding a finite number of coefficients 91 86 Level 2: Finding the recursion relation 94 87 Solutions Near a Singular Point 97 88 Functions Defined via Differential Equations 111 881 Chebyshev Equation 111 882 Legendre Equation 113 883 Airy Equation 115 884 Laguerre s Equation 115 885 Bessel Equation 116 9 Linear Systems 121 91 Preliminaries 121 92 Computing e T 123 93 The 2 2 Case in Detail 129 94 The Non-Homogeneous Case 133

CONTENTS v 95 Phase Portraits 135 951 Real Distinct Eigenvalues 137 952 Complex Eigenvalues 139 953 Repeated Real Roots 141 1 Existence and Uniqueness Theorems 145 11 Picard s Method 145 12 Existence and Uniqueness Theorem for First Order ODE s 15 13 Existence and Uniqueness Theorem for Linear First Order ODE s 155 14 Existence and Uniqueness Theorem for Linear Systems 156 11 Numerical Approximations 163 111 Euler s Method 163 1111 Error Bounds 165 112 Improved Euler s Method 166 113 Runge-Kutta Methods 167

vi CONTENTS

Chapter 1 Introduction 11 Preliminaries Definition (Differential equation) A differential equation (de) is an equation involving a function and its derivatives Differential equations are called partial differential equations (pde) or ordinary differential equations (ode) according to whether or not they contain partial derivatives The order of a differential equation is the highest order derivative occurring A solution (or particular solution) of a differential equation of order n consists of a function defined and n times differentiable on a domain D having the property that the functional equation obtained by substituting the function and its n derivatives into the differential equation holds for every point in D Example 11 An example of a differential equation of order 4, 2, and 1 is given respectively by ( ) 3 dy + d4 y dx dx 4 + y = 2 sin(x) + cos3 (x), 2 z x 2 + 2 z y 2 =, yy = 1 1

2 CHAPTER 1 INTRODUCTION Example 12 The function y = sin(x) is a solution of ( ) 3 dy + d4 y dx dx 4 + y = 2 sin(x) + cos3 (x) on domain R; the function z = e x cos(y) is a solution of 2 z x 2 + 2 z y 2 = on domain R 2 ; the function y = 2 x is a solution of yy = 2 on domain (, ) Although it is possible for a de to have a unique solution, eg, y = is the solution to (y ) 2 + y 2 =, or no solution at all, eg, (y ) 2 + y 2 = 1 has no solution, most de s have infinitely many solutions Example 13 The function y = 4x + C on domain ( C/4, ) is a solution of yy = 2 for any constant C Note that different solutions can have different domains solutions to a de is call its general solution The set of all 12 Sample Application of Differential Equations A typical application of differential equations proceeds along these lines: Real World Situation Mathematical Model Solution of Mathematical Model Interpretation of Solution

12 SAMPLE APPLICATION OF DIFFERENTIAL EQUATIONS 3 Sometimes in attempting to solve a de, we might perform an irreversible step This might introduce extra solutions If we can get a short list which contains all solutions, we can then test out each one and throw out the invalid ones The ultimate test is this: does it satisfy the equation? Here is a sample application of differential equations Example 14 The half-life of radium is 16 years, ie, it takes 16 years for half of any quantity to decay If a sample initially contains 5 g, how long will it be until it contains 45 g? Solution Let x(t) be the amount of radium present at time t in years The rate at which the sample decays is proportional to the size of the sample at that time Therefore we know that dx/dt = kx This differential equation is our mathematical model Using techniques we will study in this course (see 32, Chapter 3), we will discover that the general solution of this equation is given by the equation x = Ae kt, for some constant A We are told that x = 5 when t = and so substituting gives A = 5 Thus x = 5e kt Solving for t gives t = ln(x/5) /k With x(16) = 25, we have 25 = 5e 16k Therefore, ( ) 1 16k = ln = ln(2), 2 giving us k = ln(2) /16 When x = 45, we have t = ln(x/5) k = ln(45/5) ln(8/1) = 16 = 16 ln(1/8) ln(2) /16 ln(2) ln(2) 16 15 16 152 2432 693 Therefore, it will be approximately 2432 years until the sample contains 45 g of radium Additional conditions required of the solution (x() = 5 in the above example) are called boundary conditions and a differential equation together with boundary conditions is called a boundary-value problem (BVP) Boundary conditions come in many forms For example, y(6) = y(22); y (7) = 3y(); y(9) = 5 are all examples of boundary conditions Boundary-value problems, like the one in the example, where the boundary condition consists of specifying the value of the solution at some point are also called initial-value problems (IVP) Example 15 An analogy from algebra is the equation y = y + 2 (11)

4 CHAPTER 1 INTRODUCTION To solve for y, we proceed as y 2 = y, (y 2) 2 = y, (irreversible step) y 2 4y + 4 = y, y 2 5y + 4 =, (y 1) (y 4) = Thus, the set y {1, 4} contains all the solutions We quickly see that y = 4 satisfies Equation (11) because while y = 1 does not because 4 = 4 + 2 = 4 = 2 + 2 = 4 = 4, 1 = 1 + 2 = 1 = 3 So we accept y = 4 and reject y = 1

Chapter 2 First Order Ordinary Differential Equations The complexity of solving de s increases with the order We begin with first order de s 21 Separable Equations A first order ode has the form F (x, y, y ) = In theory, at least, the methods of algebra can be used to write it in the form y = G(x, y) If G(x, y) can be factored to give G(x, y) = M(x) N(y),then the equation is called separable To solve the separable equation y = M(x) N(y), we rewrite it in the form f(y)y = g(x) Integrating both sides gives f(y)y dx = g(x) dx, f(y) dy = f(y) dy dx dx Example 21 Solve 2xy + 6x + ( x 2 4 ) y = We use the notation dy/dx = G(x, y) and dy = G(x, y) dx interchangeably 5

6CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Solution Rearranging, we have ( x 2 4 ) y = 2xy 6x, y = 2xy 6x, y + 3 = 2x x 2 4, x ±2 ln( y + 3 ) = ln ( x 2 4 ) + C, ln( y + 3 ) + ln ( x 2 4 ) = C, where C is an arbitrary constant Then (y + 3) ( x 2 4 ) = A, (y + 3) ( x 2 4 ) = A, y + 3 = A x 2 4, where A is a constant (equal to ±e C ) and x ±2 Also y = 3 is a solution (corresponding to A = ) and the domain for that solution is R Example 22 Solve the ivp sin(x) dx + y dy =, where y() = 1 Solution Note: sin(x) dx + y dy = is an alternate notation meaning the same as sin(x) + y dy/dx = We have y dy = sin(x) dx, y dy = sin(x) dx, y 2 2 = cos(x) + C 1, y = 2 cos(x) + C 2, where C 1 is an arbitrary constant and C 2 = 2C 1 Considering y() = 1, we have 1 = 2 + C 2 = 1 = 2 + C 2 = C 2 = 1 Therefore, y = 2 cos(x) 1 on the domain ( π/3, π/3), since we need cos(x) 1/2 and cos(±π/3) = 1/2

22 EXACT DIFFERENTIAL EQUATIONS 7 An alternate method to solving the problem is y 1 y 2 y dy = sin(x) dx, y dy = x sin(x) dx, 2 12 = cos(x) cos(), 2 y 2 2 1 = cos(x) 1, 2 y 2 2 = cos(x) 1 2, y = 2 cos(x) 1, giving us the same result as with the first method Example 23 Solve y 4 y + y + x 2 + 1 = Solution We have ( y 4 + 1 ) y = x 2 1, y 5 5 + y = x3 3 x + C, where C is an arbitrary constant This is an implicit solution which we cannot easily solve explicitly for y in terms of x 22 Exact Differential Equations Using algebra, any first order equation can be written in the form F (x, y) dx + G(x, y) dy = for some functions F (x, y), G(x, y) Definition An expression of the form F (x, y) dx + G(x, y) dy is called a (first-order) differential form A differentical form F (x, y) dx + G(x, y) dy is called exact if there exists a function g(x, y) such that dg = F dx + G dy If ω = F dx+g dy is an exact differential form, then ω = is called an exact differential equation Its solution is g = C, where ω = dg Recall the following useful theorem from MATB42:

8CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 24 If F and G are functions that are continuously differentiable throughout a simply connected region, then F dx + G dy is exact if and only if G/ x = F/ y Proof Proof is given in MATB42 Example 25 Consider ( 3x 2 y 2 + x 2) dx + ( 2x 3 y + y 2) dy = Let Then note that ω = ( 3x 2 y 2 + x 2) dx + ( 2x 3 y + y 2) dy } {{ } } {{ } F G G x = 6x2 y = F y By Theorem 24, ω = dg for some g To find g, we know that g x = 3x2 y 2 + x 2, g y = 2x3 y + y 2 (21a) (21b) Integrating Equation (21a) with respect to x gives us g = x 3 y 2 + x3 3 So differentiating that with respect to y gives us Eq (21b) {}}{ g y + h(y) (22) = 2x 3 y + dh dy, 2x 3 y + y 2 = 2x 3 y + dh dy, dh dy = y2, h(y) = y3 3 + C

22 EXACT DIFFERENTIAL EQUATIONS 9 for some arbitrary constant C Therefore, Equation (22) becomes g = x 3 y 2 + x3 3 + y3 3 + C Note that according to our differential equation, we have ) d (x 3 y 2 + x3 3 + y3 3 + C = which implies x 3 y 2 + x3 3 + y3 3 + C = C for some arbitrary constant C Letting D = C C, which is still an arbitrary constant, the solution is x 3 y 2 + x3 3 + y3 3 = D Example 26 Solve ( 3x 2 + 2xy 2) dx + ( 2x 2 y ) dy =, where y(2) = 3 Solution We have (3x 2 + 2xy 2) dx = x 3 + x 2 y 2 + C for some arbitrary constant C Since C is arbitrary, we equivalently have x 3 + x 2 y 2 = C With the initial condition in mind, we have 8 + 4 9 = C = C = 44 Therefore, x 3 + x 2 y 2 = 44 and it follows that y = ± 44 x 3 x 2 But with the restriction that y(2) = 3, the only solution is y = 44 x 3 x 2 on the domain ( 3 44, 3 44 ) \ {} Let ω = F dx + G dy Let y = s(x) be the solution of the de ω =, ie, F + Gs (x) = Let y = s(x ) and let γ be the piece of the graph of y = s(x) from (x, y ) to (x, y) Figure 21 shows this idea Since y = s(x) is a solution to ω =, we must have ω = along γ Therefore, ω = This can be seen γ

1CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS y (x,y) γ 2 y =s(x) (x,y) γ 1 γ (x,y ) x Figure 21: The graph of y = s(x) with γ connecting (x, y ) to (x, y) by parameterizing γ by γ(x) = (x, s(x)), thereby giving us ω = γ x x F dx + Gs (x) dx = dx = x x This much holds for any ω Now suppose that ω is exact Then the integral is independent of the path Therefore = ω = γ F dx + G dy + γ 1 F dx + G dy γ 2 = y x G(x, y) dy + F (x, y) dx y x We can now solve Example 26 with this new method Solution (Alternate solution to Example 26) We simply have = 4 2 2 2 y dy + x 3 2 ( 3x 2 + 2xy 2) dx = 4y 2 4 ( 3) 2 + x 3 + x 2 y 2 2 3 2 2 y 2 = 4y 2 36 + x 3 + x 2 y 2 8 4y 2, finally giving us x 3 + x 2 y 2 = 44, which agrees with our previous answer Remark Separable equations are actually a special case of exact equations, that is, f(y)y = g(x) = g(x) dx + f(y) dy = = x f(y) = = y ( g(x))

23 INTEGRATING FACTORS 11 So the equation is exact 23 Integrating Factors Consider the equation ω = Even if ω is not exact, there may be a function I(x, y) such that Iω is exact So ω = can be solved by multiplying both sides by I The function I is called an integrating factor for the equation ω = Example 27 Solve y/x 2 + 1 + y /x = Solution We have We see that x ( y ) x 2 + 1 dx + 1 dy = x ( ) 1 = 1x 1 x 2 x 2 = ( y ) y x 2 + 1 So the equation is not exact Multiplying by x 2 gives us ( y + x 2 ) dx + x dy =, ) d (xy + x3 =, 3 xy + x3 3 = C for some arbitrary constant C Solving for y finally gives us y = C x x3 3 There is, in general, no algorithm for finding integrating factors But the following may suggest where to look It is important to be able to recognize common exact forms: x dy + y dx = d(xy), x dy y dx y ) x 2 = d(, ( x ( x dx + y dy ln x 2 + y 2) ) x 2 + y 2 = d, 2 x dy d dx ( x 2 + y 2 = d tan 1( y )), x x a 1 y b 1 (ay dx + bx dy) = d ( x a y b)

12CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Example 28 Solve ( x 2 y 2 + y ) dx + ( 2x 3 y x ) dy = Solution Expanding, we have x 2 y 2 dx + 2x 3 y dy + y dx x dy = Here, a = 1 and b = 2 Thus, we wish to use d ( xy 2) = y 2 dx + 2xy dy This suggests dividing the original equation by x 2 which gives y 2 dx + 2xy dy + y dx x dy x 2 = Therefore, xy 2 + y = C, x, x where C is an arbitrary constant Additionally, y = on the domain R is a solution to the original equation Example 29 Solve y dx x dy ( x 2 + y 2) dx = Solution We have y dx x dy x 2 + y 2 dx =, unless x = and y = Now, it follows that tan 1( y ) x = C, x tan 1( y ) = C x, x tan 1( y ) = D x, (D = C) x y = tan(d x), x y = x tan(d x), where C is an arbitrary constant and the domain is D x (2n + 1) π 2, x (2n + 1) π 2 for any integer n Also, since the derivation of the solution is based on the assumption that x, it is unclear whether or not should be in the domain, ie, does y = x tan(d x) satisfy the equation when x =? We have y xy

23 INTEGRATING FACTORS 13 ( x 2 + y 2) = If x = and y = x tan(d x), then y = and the equation is satisfied Thus, is in the domain Proposition 21 Let ω = dg Then for any function P : R R, P (g) is exact Proof Let Q = P (t) dy Then d(q(g)) = P (g) dg = P (g)ω To make use of Proposition 21, we can group together some terms of ω to get an expression you recognize as having an integrating factor and multiply the equation by that The equation will now look like dg + h = If we can find an integrating factor for h, it will not necessarily help, since multiplying by it might mess up the part that is already exact But if we can find one of the form P (g), then it will work Example 211 Solve ( x yx 2) dy + y dx = Solution Expanding, we have y dx + x dy yx 2 dy = } {{ } d(xy) Therefore, we can multiply teh equation by any function of xy without disturbing the exactness of its first two terms Making the last term into a function of y alone will make it exact So we multiply by (xy) 2, giving us y dx + x dy x 2 y 2 1 1 dy = = ln( y ) = C, y xy where C is an arbitrary constant Note that y = on the domain R is also a solution Given M dx + N dy =, ( ) we want to find I such that IM dx + IN dy is exact If so, then x (IN) = } {{ } y (IM) } {{ } I xn+in x I ym+im y

14CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS If we can find any particular solution I(x, y) of the pde I x N + IN x = I y M + IM y, ( ) then we can use it as an integrating factor to find the general solution of ( ) Unfortunately, ( ) is usually even harder to solve than ( ), but as we shall see, there are times when it is easier Example 212 We could look for an I having only x s and no y s? For example, consider I y = Then I x N + IN x = IM y implies I x I = M y N x N This works if (M y N x ) /N happens to be a function of x alone Then I = e R My Nx N dx Similarly, we can also reverse the role of x and y If (N x M y ) /M happens to be a function of y alone, then e R Nx My M dy works 24 Linear First Order Equations A first order linear equation (n = 1) looks like y + P (x)y = Q(x) An integrating factor can always be found by the following method Consider dy + P (x)y dx = Q(x) dx, (P (x)y Q(x)) dx + } {{ } dy }{{} = M(x,y) N(x,y) We use the de for the integrating factor I(x, y) The equation IM dx + IN dy is exact if I x N + IN x = I y M + IM y

24 LINEAR FIRST ORDER EQUATIONS 15 In our case, I x + = I y (P (x)y Q(x)) + IP (x) ( ) We need only one solution, so we look for one of the form I(x), ie, with I y = Then ( ) becomes di = IP (x) dx This is separable So di = P (x) dx, I ln( I ) = P (x) dx + C, I = e R P (x) dx, e x > I = e R P (x) dx We conclude that e R P (x) dx is an integrating factor for y + P (x)y = Q(x) Example 213 Solve y (1/x) y = x 3, where x > Solution Here P (x) = 1/x Then I = e R P (x) dx = e R 1 x dx = e ln( x )dx = 1 x = 1 x, where x > Our differential equation is x dy y dx x = x 3 dx Multiplying by the integrating factor 1/x gives us Then x dy y dx x 2 = x 2 dx y x = x3 3 + C, y = x3 3 + Cx on the domain (, ), where C is an arbitrary constant (x > is given)

16CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS In general, given y + P (x)y = Q(x), multiply by e R P (x) dx to obtain e R P (x) dx y + e R P (x) dx P (x)y = Q(x)e R P (x) dx } {{ } d(ye R P (x) dx )/dx Therefore, ye R P (x) dx = Q(x)e R P (x) dx dx + C, y = e R P (x) dx Q(x)e R P (x) dx dx + Ce R P (x) dx, where C is an arbitrary constant Example 214 Solve xy + 2y = 4x 2 Solution What should P (x) be? To find it, we put the equation in standard form, giving us y + 2 x y = 4x Therefore, P (x) = 2/x Immediately, we have I = e R (2/x)dx = e ln(x2 ) = x 2 Multiplying the equation by x 2 gives us x 2 y + 2xy = 4x 3, x 2 y = x 4 + C, y = x 2 + C x 2, where C is an arbitrary constant and x Example 215 Solve e y dy + dx + 2x dy = Solution This equation is linear with x as a function of y So what we have is dx dy + 2x = e y,

25 SUBSTITUTIONS 17 where I = e R 2 dy = e 2y Therefore, 2y dx e dy + 2xe2y = e y, xe 2y = e y + C, where C is an arbitrary constant We could solve explicitly for y, but it is messy The domain is not easy to determine 25 Substitutions In many cases, equations can be put into one of the standard forms discussed above (separable, linear, etc) by a substitution Example 216 Solve y 2y = 5 Solution This is a first order linear equation for y Let u = y Then the equation becomes u 2u = 5 The integration factor is then I = e R 2 dx = e 2x Thus, u e 2x 2ue 2x = 5e 2x, ue 2x = 5 2 e 2x + C, where C is an arbitrary constant But u = y, so y = 5 2 x + C 2 e2x + C 1 = 5 2 x + C 1e 2x + C 2 on the domain R, where C 1 and C 2 are arbitrary constants We now look at standard substitutions 251 Bernoulli Equation The Bernoulli equation is given by dy dx + P (x)y = Q(x)yn Let z = y 1 n Then dz dy = (1 n) y n dx dx,

18CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS giving us which is linear in z n dy y dx + P (x)y1 n = Q(x), 1 dz + P (x)z = Q(x), 1 n dx dz + (1 n) P (x)z = (1 n) Q(x), dx Example 217 Solve y + xy = xy 3 Solution Here, we have n = 3 Let z = y 2 If y, then Therefore, our equation becomes dz dy = 2y 3 dx dx y3 z 2 + xy = xy3, z 2 + xy 2 = x, z 2xy = 2x We can readily see that I = e R 2x dx = e x2 Thus, e x2 z 2xe x2 = 2xe x2, e x2 z = e x2 + C, z = 1 + Ce x2, where C is an arbitrary constant But z = y 2 So 1 y = ± 1 + Ce x 2 The domain is R, C > 1, x > ln( C), C 1 An additional solution is y = on R

25 SUBSTITUTIONS 19 252 Homogeneous Equations Definition (Homogeneous function of degree n) A function F (x, y) is called homogeneous of degree n if F (λx, λy) = λ n F (x, y) For a polynomial, homogeneous says that all of the terms have the same degree Example 218 The following are homogeneous functions of various degrees: 3x 6 + 5x 4 y 2 homogeneous of degree 6, 3x 6 + 5x 3 y 2 not homogeneous, x x 2 + y 2 homogeneous of degree 2, ( y sin homogeneous of degree, x) 1 x + y homogeneous of degree 1 If F is homogeneous of degree n and G is homogeneous of degree k, then F/G is homogeneous of degree n k Proposition 219 If F is homogeneous of degree, then F is a function of y/x Proof We have F (λx, λy) = F (x, y) for all λ Let λ = 1/x Then F (x, y) = F (1, y/x) Example 22 Here are some examples of writing a homogeneous function of degree as a function of y/x 5x2 + y 2 ( y ) 2, = 5 + x x y 3 + x 2 y x 2 y + x 3 = (y/x)3 + (y/x) (y/x) + 1 Consider M(x, y) dx + N(x, y) dy = Suppose M and N are both homogeneous and of the same degree Then dy dx = M N,

2CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS This suggests that v = y/x (or equivalently, y = vx) might help In fact, write M(x, y) ( y ) N(x, y) = R x Then Therefore, dy ( y = R = R(v) }{{} dx x) v+x dv dx x dv = R(v) v, dx dv R(v) v = dx x, which is separable We conclude that if M and N are homogeneous of the same degree, setting y = vx will give a separable equation in v and x Example 221 Solve xy 2 dy = ( x 3 + y 3) dx Solution Let y = vx Then dy = v dx + x dv, and our equation becomes xv 2 x 2 (v dx + x dv) = ( x 3 + v 3 x 2) dx, x 3 v 3 dx + x 4 v 2 dv = x 3 dx + v 3 x 3 dx Therefore, x = or v 2 dv = dx/x So we have v 3 3 = ln( x ) + C = ln( x ) + ln( A ) = ln( Ax ) = ln(ax) } {{ } C where the sign of A is the opposite of the sign of x Therefore, the general solution is y = x (3 ln(ax)) 1/3, where A is a nonzero constant Every A > yields a solution on the domain (, ); every A < yields a solution on (, ) In addition, there is the solution y = on the domain R 253 Substitution to Reduce Second Order Equations to First Order A second order de has the form F (y, y, y, x) =

25 SUBSTITUTIONS 21 If it is independent of y, namely, F (y, y, x) =, then it is really just a first order equation for y as we saw in earlier examples Consider now the case where it is independent of x, namely, F (y, y, y) = Substitute v = dy/dx for x, ie, eliminate x between the equations and v = dy/dx Then Therefore, ( d 2 y F dx 2, dy ) dx, y = d 2 y dx 2 = dv dx = dv dy dy dx = dv dy v ( d 2 y F dx 2, dy ) ( ) dv dx, y = F v, v, y = dy This is a first order equation in v and y Example 222 Solve y = 4 (y ) 3/2 y Solution Let v = dy/dx Then d 2 y dx 2 = dv dy v and our equation becomes dv dy v = 4v3/2 y, dv v = 4y dy, v, 2 v = 2y 2 + C 1, v = y 2 + C 2,

22CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS where C 1 is an arbitrary constant and C 2 = C 1 /2 But v = dy/dx, so we have dy dx = ( y 2 ) 2 + C 2, dy dx = (y 2 + C 2 ) 2, dy x = (y 2 + C 2 ) 2 = 1 2C 3/2 2 1 ( tan 1 ( y C2 ) + C 2y y 2 +C 2 ) + C 3, C 2 >, 3y + C 3 3, C 2 =, y 2 y 2 +C 2 + C 3, C 2 < 1 2( C 2) 3/2 Next consider second order linear equations That is, P (x)y + Q(x)y + R(x)y = We can eliminate y by letting y = e v Then y = e v v and y = e v (v ) 2 + e v v The equation then becomes ( P (x) e v (v ) 2 + e v v ) + Q(x)e v v + R(x)e v =, which is a first order equation in v Example 223 Solve x 2 y + ( x x 2) y e 2x y = Solution Let y = e v Then the equation becomes x 2 e v (v ) 2 + x 2 e v v + ( x x 2) e v v e 2x e v = Write z = v Then x 2 z + x 2 z 2 + (1 x) xz = e 2x Now we are on our own there is no standard technique for this case Suppose we try u = xy Then z = u/x and z = u x 2 + 1 x u Then it follows that xu + u 2 xu = e 2x This is a bit simpler, but it is still nonstandard We can see that letting u = se x

25 SUBSTITUTIONS 23 will give us some cancellation Thus, u = s e x + se x and our equation becomes xs e x + xse x + s 2 e 2x xse x = e 2x, xs + s 2 e x = e x, s xs = e x ( 1 s 2), 1 s 2 = ex x, ( ) 1 2 ln 1 + s e x 1 s = x dx Working our way back through the subsitutions we find that s = zxe x so our solution becomes ( 1 2 ln 1 + zxe x ) e x 1 zxe x = x dx Using algebra, we could solve this equation for z in terms of x and then integrate the result to get v which then determines y = e v as a function of x The algebraic solution for z is messy and we will not be able to find a closed form expression for the antidervative v, so we will abandon the calculation at this point In practical applications one would generally use power series techniques on equations of this form and calculate as many terms of the Taylor series of the solution as are needed to give the desired accuracy Next consider equations of the form (a 1 x + b 1 y + c 1 ) dx + (a 2 x + b 2 y + c 2 ) dy = If c 1 = c 2 =, the equation is homogeneous of degree 1 If not, try letting x = x h and ȳ = y k We try to choose h and k to make the equation homogeneous Since h and k are constants, we have d x = dx and dȳ = dy Then our equation becomes (a 1 x + a 1 h + b 1 ȳ + b 1 k + c 1 ) d x + (a 2 x + a 2 h + b 2 ȳ + b 2 k + c 2 ) dȳ = We want a 1 h + b 1 k = c 1 and a 2 h + b 2 k = c 2 We can always solve for h and k, unless a 1 b 1 a 2 b 2 =

24CHAPTER 2 FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS So suppose a 1 b 1 a 2 b 2 = Then (a 2, b 2 ) = m (a 1, b 1 ) Let z = a 1 x + b 1 y Then dz = a 1 dx + b 1 dy If b 1, we have dy = dz a 1 dx, b 1 (z + c 1 ) dx + (mz + c 2 ) dz a 1 dx =, b ( 1 z + c 1 + a ) ( ) 1 mz + c2 dx + dx =, b 1 b 1 b 1 dx = mz + c 2 z + c 1 + a 1 /b 1 dz This is a separable equation If b 1 = but b 2, we use z = a 2 x + b 2 y instead Finally, if both b 1 = and b 2 =, then the original equation is separable

Chapter 3 Applications and Examples of First Order Ordinary Differential Equations 31 Orthogonal Trajectories An equation of the form f(x, y, C) = determines a family of curves, one for every value of the constant C Figure 31 shows curves with several values of C Differentiating f(x, y, C) = with respect to x gives a second equation y C=9 C=4 θ=π/4 C=1 θ= x Figure 31: Graphs of x 2 + y 2 C = with various values of C 25

26CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S g(x, y, C, dy/dx) =, which is explicitly given by Example 31 We have g = f f dy (x, y, C) + (x, y, C) x y dx x 2 + y 3 + Cx 4 = implies 2x + 3y 2 dy dx + 4Cx2 = We can eliminate C between the equations f = and g =, that is, Therefore, f = = C = x2 + y 2 x 4 g = = 2x 3y 2 dy dx + y 3 4x2 = x This yields a first order de, where the solution includes the initial family of curves So just as a first order de usually determines a one-parameter family of curves (its solutions), a one-parameter family of curves determines a de A coordinate system (Cartesian coordinates, polar coordinates, etc) consists of two families of curves which always intersect at right angles Two families of curves which always intersect at right angles are called orthogonal trajectories of each other Given a one-parameter family of curves, we wish to find another family of curves which always intersects it at right angles Example 32 Find orthogonal trajectories of y = C sin(x) Solution Figure 32 shows the plot of y = C sin(x) with several values of C We have C = y/ sin(x), so it follows that dy dx = }{{} C cos(x) = y cos(x) sin(x) = y cot(x) y/ sin(x) Two curves meet at right angles if the product of their slopes is 1, ie, m new = 1/m old So orthogonal trajectories satisfy the equation dy dx = 1 y cot(x) = tan(x) y

32 EXPONENTIAL GROWTH AND DECAY 27 5 3 1-3 p -2 p - p 2 p 3 p -1 x -3-5 Figure 32: A plot of the family y = C sin(x) (in black) and its orthogonal trajectories (in gray) This is a separable equation, and we quickly have y dy = tan(x), y 2 = ln( cos(x) ) + C, 2 y = ± 2 ln( cos(x) ) + C 1, where C 1 = 2C is also an arbitrary constant 32 Exponential Growth and Decay There are many situations where the rate of change of some quantity x is proportional to the amount of that quantity, ie, dx/dt = kx for some constant k The general solution is x = Ae kt 1 Radium gradually changes into uranium If x(t) represents the amount of radium present at time t, then dx/dt = kx In this case, k < 2 Bank interest The amount of interest you receive on your bank account is proportional to your balance In this case, k > Actually, you do not do quite this well at the bank because they compound only daily rather than continuously, eg, if we measure this time in years, our model predicts that x = x e kt, where x is your initial investment and k is the interest rate Actually, you get x = x (1 + k n t ) n,

28CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S where n is the number of interest periods throughout the year Typically, n = 365 (daily interest), but the difference is small when n = 365 Note that lim n (1 + kt/n) n = e kt 3 Population growth of rabbits The number of rabbits born at any time is roughly proportional to the number of rabbits present If x(t) is the number of rabbits at time t, then dx/dt = kx with k > Obviously, this model is not accurate as it ignores deaths Example 33 The half-life of radium is 16 years, ie, it takes 16 years for half of any quantity to decay If a sample initially contains 5 g, how long will it be until it contains 45 g? Solution Let x(t) be the amount of radium present at time t in years Then we know that dx/dt = kx, so x = x e kt With x() = 5, we quickly have x = 5e kt Solving for t gives t = ln(x/5) /k With x(16) = 25, we have 25 = 5e 16k Therefore, ( ) 1 16k = ln = ln(2), 2 giving us k = ln(2) /16 When x = 45, we have t = ln(x/5) k = ln(45/5) ln(8/1) = 16 = 16 ln(1/8) ln(2) /16 ln(2) ln(2) 16 15 16 152 2432 693 Therefore, it will be approximately 2432 years until the sample contains 45 g of radium 33 Population Growth Earlier, we discussed population growth where we considered only births We now consider deaths also Take, as our model, the following Let N(t) be the number of people at time t Assume that the land is intrinsically capable of supporting L people and that the rate of increase is proportional to both N and L N Then dn dt = kn (L N),

34 PREDATOR-PREY MODELS 29 and the solution is N = where k is the proportionality constant L 1 + (L/N 1) e klt, 34 Predator-Prey Models Example 34 (Predator-Prey) Consider a land populated by foxes and rabbits, where the foxes prey upon the rabbits Let x(t) and y(t) be the number of rabbits and foxes, respectively, at time t In the absence of predators, at any time, the number of rabbits would grow at a rate proportional to the number of rabbits at that time However, the presence of predators also causes the number of rabbits to decline in proportion to the number of encounters between a fox and a rabbit, which is proportional to the product x(t)y(t) Therefore, dx/dt = ax bxy for some positive constants a and b For the foxes, the presence of other foxes represents competition for food, so the number declines proportionally to the number of foxes but grows proportionally to the number of encounters Therefore dy/dt = cx + dxy for some positive constants c and d The system is our mathematical model dx = ax bxy, dt dy = cy + dxy dt If we want to find the function y(x), which gives the way that the number of foxes and rabbits are related, we begin by dividing to get the differential equation dy cy + dxy = dx ax bxy with a, b, c, d, x(t), y(t) positive This equation is separable and can be rewritten as (a by) dy y = ( c + dx) dx x Integrating gives a ln(y) by = c ln(x) + dx + C,

3CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S or equivalently y a e by = kx c e dx (31) for some positive constant k = e C The graph of a typical solution is shown in Figure 33 d c b a Figure 33: A typical solution of the Predator-Prey model with a = 94, b = 158, c = 684, d = 13, and k = 754 Beginning at a point such as a, where there are few rabbits and few foxes, the fox population does not initially increase much due to the lack of food, but with so few predators, the number of rabbits multiplies rapidly After a while, the point b is reached, at which time the large food supply causes the rapid increase in the number of foxes, which in turn curtails the growth of the rabbits By the time point c is reached, the large number of predators causes the number of rabbits to decrease Eventually, point d is reached, where the number of rabbits has declined to the point where the lack of food causes the fox population to decrease, eventually returning the situation to point a 35 Newton s Law of Cooling Let T and T s be the temperature of an object and its surroundings, respectively Let T and T s be initial temperatures Then Newton s Law of Cooling states

36 WATER TANKS 31 that dt dt = k (T s T ), k > As T changes, the object gives or takes heat from its surroundings We now have two cases Case 1: T s is constant This occurs either because the heat given off is transported elsewhere or because the surroundings are so large that the contribution is negligible In such a case, we have which is linear, and the solution is dt dt + kt = kt s, T = T e kt + T s ( 1 e kt ) Case 2: The system is closed All heat lost by the object is gained by its surroundings We need to find T s as a function of T We have Therefore, change in temperature = ( dt dt + k 1 + wc w s c s heat gained weight specific heat capacity T T = T s T s, wc w s c s T s = T s + wc (T T ), w s c ) ( s ) T = k T s + wc w s c s T This is linear with different constants than the previous case The solution is T = T + 36 Water Tanks ( ) Ts + wc w sc s T ( wc k(1+ 1 + wc 1 e )t) wscs w sc s Example 35 A tank contains a salt water solution consisting initially of 2 kg of salt dissolved into 1 l of water Fresh water is being poured into the tank at a rate of 3 l /min and the solution (kept uniform by stirring) is flowing out at 2 l /min Figure 34 shows this setup Find the amount of salt in the tank after 5 minutes

32CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S litre litre Figure 34: Fresh water is being poured into the tank as the well-mixed solution is flowing out Solution Let Q(t) be the amount (in kilograms) of salt in the tank at time t (in minutes) The volume of water at time t is The concentration at time t is given by kg per litre Then dq dt 1 + 3t 2t = 1 + t amount of salt volume = Q 1 + t = (rate at which salt is leaving) = Q 1 + t 2 = 2Q 1 + t Thus, the solution to the problem is the solution of dq dt = 2Q 1 + t evaluated at t = 5 We see that it is simply a separable equation

36 WATER TANKS 33 To solve it, we have where C is a constant Thus, dq Q = 2 dt 1 + t, dq Q = 2 dt 1 + t, ln( Q ) = 2 ln( 1 + t ), ( ln( Q ) = ln 1 + t 2) + C, ( ln( Q ) = ln 1 + t 2) + C, ( = ln A 1 + t 2), Q = A 1 + t 2 where A = e C But Q (we cannot have a negative amount of salt) and t (we do not visit the past), so we remove absolute value signs, giving us Q = A (1 + t) 2 Initially, ie, at t =, we know that the tank contains 2 kg of salt Thus, the initial condition is Q() = 2, and we have Q(2) = = A A 2 = 2 = = 2 = A = 2 (1 + ) 1 Our equation therefore becomes Q(t) = 2 (1 + t) 2 Figure 35 shows a plot of the solution Evaluating at t = 5 gives Q(5) = 2 15 2 = 8 225 = 8 9 889 Therefore, after 5 minutes, the tank will contain approximately 889 kg of salt Additional conditions desired of the solution (Q() = 2) in the above example) are called boundary conditions and a differential equations together with boundary conditions is called a boundary-value problem (BVP)

34CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S Q 2 15 1 5 1 2 3 4 5 6 t Figure 35: The plot of the amount of salt Q(t) in the tank at time t shows that salt leaves slower as time moves on A = A (1+t) is the general solution to the DE dq 2 dt = 2dt 1+t Q = 2 (1+t) is the 2 solution to the boundary value problem dq dt = 2dt 1+t ; Q() = 2 Boundaryvalue problems like this one where the boundary conditions are initial values of various quantities are also called initial-value problems (IVP) 37 Motion of Objects Falling Under Gravity with Air Resistance Let x(t) be the height at time t, measured positively on the downward direction If we consider only gravity, then a = d2 x dt 2 is a constant, denoted g, the acceleration due to gravity Note that F = ma = mg Air resistance encountered depends on the shape of the object and other things, but under most circumstances, the most significant effect is a force opposing the motion which is proportional to a power of the velocity So }{{} F = mg kv n ma and d 2 x dt 2 = g k m ( ) n dx, dt

37 MOTION OF OBJECTS FALLING UNDER GRAVITY WITH AIR RESISTANCE35 which is a second order de, but there is no x term So it is first order in x Therefore, dv dt = g k m vn This is not easy to solve, so we will make the simplifying approximation that n = 1 (if v is small, there is not much difference between v and v n ) Therefore, we have dv dt = g k m v, dv dt + k m v = g, The integrating factor is I = e R k m dt = e kt/m Therefore, ( dv dt + k ) m v e kt/m = ge kt/m, e kt/m v = gm k ekt/m + C, v = mg k + Ce kt/m, where C is an arbitrary constant Note that So we have Note that x x dx } {{ } x x v() = v = v = mg k = Thus, we finally have t }{{} v = mg ( k + dx/dt v dt = mg k t m k x = x + mg k t + m k ( + C = C = v mb k v mg k ( v mg k v mg k ) e kt/m ) ( ) e kt/m 1 ) ( 1 e kt/m)

36CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S 38 Escape Velocity Let x(t) be the height of an object at time t measured positively upward (from the centre of the earth) Newton s Law of Gravitation states that }{{} F = kmm x 2, ma where m is the mass of the object, M is the mass of the Earth, and k > Note that x = km x 2 We define the constant g known as the acceleration due to gravity by x = g when x = R, where R = radius of the earth So k = gr 2 /M Therefore x = gr 2 /x 2 Since t does not appear, letting v = dx/dt so that d 2 x dt 2 will reduce it to first order Thus, = dv dx dx dt = dv dx v v dv dx = gr2 x 2, v dv = gr2 x 2 dx, v dv = gr2 x 2 dx, v 2 2 = gr2 x + C, for some arbitrary constant C Note that v decreases as x increases, and if v is not sufficiently large, eventually v = at some height x max So C = gr 2 /x max and ( 1 v 2 = 2R 2 g x 1 ) x max Suppose we leave the surface of the Earth starting with velocity V How far

39 PLANETARY MOTION 37 will we get? (v = V when x = R) We have ( 1 V 2 = 2R 2 g R 1 ), x max V 2 2R 2 g = 1 R 1, x max 1 = 1 x max R V 2 2R 2 g = 2Rg V 2 2R 2, g x max = 2R2 g 2Rg V 2 That is, if V 2 < 2Rg, we will get as far as 2R 2 g 2Rg V 2 and then fall back To escape, we need V 2Rg Thus, the escape velocity is 2Rg 69 mi /s, or roughly 25 mph 39 Planetary Motion Newton s law of gravitation states that }{{} F = km r 2 r ma Suppose we let a r = k/r 2 and a = Then a r = a x cos(θ) + a y sin(θ), a θ = a x sin(θ) + a y cos(θ) Let x = r cos(θ) and y = r sin(θ) Then x = r cos(θ) r sin(θ)θ, a x = x = r cos(θ) r sin(θ)θ r sin(θ)θ r cos(θ) (θ ) 2 r sin(θ)θ, y = r sin(θ) + r cos(θ)θ, a y = y = r sin(θ) + r cos(θ)θ + r cos(θ)θ r sin(θ) (θ ) 2 + r cos(θ)θ = r sin(θ) + 2r θ cos(θ) (θ ) 2 r sin(θ) + θ r cos(θ)

38CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S Now a r becomes a r = r cos 2 (θ) 2r θ sin(θ) cos(θ) (θ ) 2 r cos 2 (θ) θ sin(θ) cos(θ) + r sin 2 (θ) + 2r θ sin(θ) cos(θ) (θ ) 2 r sin 2 (θ) + θ r sin(θ) cos(θ) = r (θ ) 2 r and a θ becomes a θ = r sin(θ) cos(θ) + 2r θ sin 2 (θ) + (θ ) 2 r sin(θ) cos(θ) + θ r sin 2 (θ) + r sin(θ) cos(θ) + 2r θ cos 2 (θ) (θ ) 2 r sin(θ) cos(θ) + θ r cos 2 (θ) = 2r θ + θ r Note that Equation ( ) implies that r (θ ) 2 r = k r 2, 2r θ + θ r = ( ) ( ) 2rr θ + θ } {{ r } 2 = = r 2 θ = h d(r 2 θ ) We want the path of the planet, so want an equation involving r and θ Therefore, we eliminate t between Equation ( ) and r 2 θ = h Thus, r = dr dθ θ = dr h dθ r 2, r = d2 r h dθ 2 θ r 2 2 dr h dθ r 2 r = d2 r h h dθ 2 r 2 r 2 2 dr h dr h ( dθ r 3 dθ r 2 ( ) ) 2 = h2 2 dr r 2 r 3 1 d 2 r dθ r 2 dθ 2 Note that Z θ Z r(θ) Z θ r (θ) 2 A = r dr dθ = dθ, θ θ 2 so da/dθ = r(θ) 2 /2 So in our case, A = h/2 Therefore, A = ht/2, which is Kepler s Second Law (area swept out is proportional to time)

31 PARTICLE MOVING ON A CURVE 39 Let u = 1/r Then du dθ = 1 dr r 2 dθ, d 2 u dθ 2 = 2 ( ) 2 dr r 3 1 d 2 r dθ r 2 dθ 2 = r r2 h 2 = r h 2 u 2 Therefore, Equation ( ) implies that Therefore, ( ) 2 h 2 u 2 d2 u h dθ 2 r 2 r = k r 2, h 2 u 2 d2 u dθ 2 u3 h 2 = ku 2, u, d 2 u dθ 2 + u = k h 2 where Therefore, }{{} u = B cos(θ δ) + k h 2 = C 1 sin(θ) + C 2 cos(θ) + k h 2, 1/r r = 1 k/h 2 + B cos(θ δ) = 1 k/h 2 + C 1 sin(θ) + C 2 cos(θ) k h { 2 r+c1x+c2y }} { k h 2 r + C 1r sin(θ) + C 2 r cos(θ) = 1, k h 2 r = 1 C 1x C 2 x, k 2 ( x 2 h 4 + y 2) = (1 C 1 x C 2 y) 2, which is the equation of an ellipse 31 Particle Moving on a Curve Let x(t) and y(t) denote the x and y coordinates of a point at time t, respectively Pick a reference point on the curve and let s(t) be the arc length of the piece of the curve between the reference point and the particle Figure 36 illustrates this idea Then

4CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S Reference point Particle Figure 36: A particle moving on a curve v = ds dt, a = a = d2 s dt 2 The vector v is in the tangent direction The goal is to find components a in the tangential and normal directions Let θ be the angle between the tangent line and the x axis Then cos(θ) = dx ds, dy sin(θ) = ds In the standard basis, we have a = (a x, a y ), where a x = d2 x dt 2, a y = d2 y dt 2 To get components in T, n basis, apply the matrix representing the change of basis from T, n to the standard basis, ie, rotation by θ, ie, ie, cos(θ) sin(θ) sin(θ) cos(θ) a x a y = a x cos(θ) + a y sin(θ) a x sin(θ) + a y cos(θ) dx a T = a x cos(θ) + a y sin(θ) = a x ds + a dy y ( ) ds dx = a x dt + a dy dt y dt ds = a xv x + a y v y v and a n = a x sin(θ) + a y cos(θ) = a yv x a x v y v Also, we have v = vx 2 + vy, 2 so, d 2 s dt 2 = dv dt = 2v x dv x dt + 2v y dv y dt 2 vx 2 + vy 2 = v xa x + v y a y v = a T

31 PARTICLE MOVING ON A CURVE 41 But this was clear to start with because it is the component of a in the tangential direction that affects ds/dt We now have y = dy dx = dy/dt dx/dt = v y v x, y = v x dvy dx v y dvx dx v 2 x = v x dvy dt dt dx v y dvx dt dt dx vx 2 = v xa y v y a x v 3 x = a nv vx 3, where a n = y v 3 x/v and v x = dx dt = dx ds ds dt = dx ds v, so Let ( ) a n = y dx 3 ds v 3 = y v 2 v (ds/dx) 3 = y (1 + (y ) 2) 3/2 v2 (1 + (y ) 2) 3/2 R = y (32) so that a n = v 2 /R The quantity R is called the radius of curvature It is the radius of the circle which mostly closely fits the curve at that point This is shown in Figure 37 If the curve is a circle, then R is a constant equal to the R Figure 37: The radius of curvature is the circle whose edge best matches a given curve at a particular point radius Conversely, suppose that R is a constant Then Ry = ( 1 + (y ) 2) 3/2

42CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S Let z = y Then and so Rz = ( 1 + z 2) 3/2 dx = R dz (1 + z 2 ) 3/2, R Rz x = dz = (1 + z 2 3/2 ) + A, 1 + z 2 R x A = (1/z)2 + 1, ( ) 2 1 + 1 = z ( ) 2 1 = z R 2 (x A) 2, R 2 (x A) 2 1 = R2 (x A) (x A) 2, z 2 (x A) 2 = R 2 (x A) 2, x A }{{} z = ± dy/dx R 2 (x A) 2 2 Therefore, y = ± x A R 2 (x A) 2 dx = ± R 2 (x A) 2 + B, (y B) 2 = R 2 (x A) 2, (x A) 2 + (y B) 2 = R 2 Suppose now that the only external force F acting on the particle is gravity, ie, F = (, mg) Then in terms of the T, n basis, the components of F are cos(θ) sin(θ) sin(θ) cos(θ) mg = mg sin(θ) mg cos(θ) = mg dy ds mg dx ds

31 PARTICLE MOVING ON A CURVE 43 Equating the tangential components gives us m d2 s dt 2 = mg dy ds, ds dt = v, d 2 s dt 2 = dv dt = ds dt = dv ds v, dv dy v = g ds ds, v dv = g dy, v 2 2 = g(y y ) = g(y y), ds dt = v = ± 2g(y y) To proceed, we must know y(s), which depends on the curve Therefore, ds dt = ± 2g(y y(s)), t = ± ds 2g(y y(s)) If the ma n ever becomes greater than the normal component of F, the particle will fly off the curve This will happen when m y v 3 x v = mg dx ds y v 2 x = g, dx dt = g dt ds = g v x v, v x = dx dt = dx ds ds dt = dx ds v Therefore, ( ) 2 dx g = y v 2 = y v 2 ds (ds/dx) 2 = y v 2 1 + (y ) 2 = y 2g(y y) 1 + (y ) 2 = 2y (y y ) = 1 + (y ) 2 Example 36 A particle is placed at the point (1, 1) on the curve y = x 3 and released It slides down the curve under the influence of gravity Determine whether or not it will ever fly off the curve, and if so, where

44CHAPTER 3 APPLICATIONS AND EXAMPLES OF FIRST ORDER ODE S Solution First note that Therefore, y = x 3, y = 3x 2, y = 6x, y = 1 2y (y y ) = 1 (y ) 2, 12x ( x 3 1 ) = 1 + 9x 4, 12x 4 12x = 1 + 9x 4, 3x 4 12x 1 = Solving for the relevant value of x gives us x 83, which corresponds to y 57 Therefore, the particle will fly off the curve at ( 83, 57) Figure 38 shows the graph 8 6 4 2 H1,1L H-83,-57L -2-1 1 2-2 x -4-6 -8 Figure 38: A particle released from (1, 1), under the influence of gravity, will fly off the curve at approximately ( 83, 57)

Chapter 4 Linear Differential Equations Although we have dealt with a lot of manageable cases, first order equations are difficult in general But second order equations are much worse The most manageable case is linear equations We begin with the general theory of linear differential equations Specific techniques for solving these equations will be given later Definition (nth order linear differential equation) An nth order linear differential equation is an equation of the form d n y dx n + P 1(x) dn 1 y dx n 1 + P 2(x) dn 2 y dx n 2 + + P n(x)y = Q(x) An nth order linear equation can be written as a linear system (see Chapter 9) as follows Let y 1 (x) = y(x), y 2 (x) = dy dx, y 3(x) = d2 y dx 2,, y n(x) = dn 1 y dx n 1 Then dy 1 dx = y 2, dy 2 dx = y 3,, dy n 1 dx = y n If the coefficient of d n y/dx n is not 1, we can divide the equation through this coefficient to write it in normal form A linear equation is in normal form when the coefficient of y (n) is 1 45

46 CHAPTER 4 LINEAR DIFFERENTIAL EQUATIONS and we have Therefore, where dy n dx = dn y dx n = P 1(x) dn 1 y dx n 1 P n(x)y + Q(x) A(x) = = P 1 (x)y n P n (x)y 1 + Q(x) dy dx = A(x)Y + B(x), 1 1 1 P n (x) P n 1 (x) P n 2 (x) P 1 (x) and B(x) = Q(x) Theorem 41 If P 1 (x),, P n (x), Q(x) are continuous throughout some interval I and x is an interior point of I, then for any numbers W,, W n 1, there exists a unique solution throughout I of satisfying d n y dx n + P 1(x) dn 1y dx n 1 + + P n(x)y = Q(x) y(x ) = w, y (x ) = w 1,, y (n 1) (x ) = w n 1 Proof The proof is given in Chapter 1

41 HOMOGENEOUS LINEAR EQUATIONS 47 41 Homogeneous Linear Equations To begin, consider the case where Q(x) called the homogeneous case, where y (n) + P 1 (x)y (n 1) + P 2 (x)y (n 2) + + P n (x)y = (41) Proposition 42 1 If r(x) is a solution of Equation (41), then so is ar(x) for all r R 2 If r(x) and s(x) are solutions of Equation (41), then so is r(x) +s(x) Proof 1 Substituting y = ar into the lhs gives ( ) ar (n) + P 1 (x)ar (n 1) + + P n (x)ar = a r (n) + + P n (x)r But since r is a solution, we have Therefore, ar is a solution ( ) a r (n) + + P n (x)r = a 2 Substituting y = r + s in the lhs gives r (n) + s (n) +P 1 (x) ( r (n 1) + + s (n 1)) + +P n (x) (r + s) = ( r (n) + + P n (x)r +s (n) + + P n (x)s = + = ) Therefore, r + s is a solution Corollary 43 The set of solutions to Equation (41) forms a vector space

48 CHAPTER 4 LINEAR DIFFERENTIAL EQUATIONS Proof Let V be the set of solutions to Equation (41) and let y 1, y 2,, y n V Further, let x R Set W (y 1, y 2,, y n ) (x) = the Wronskian of y 1, y 2,, y n Then two equal rows y 1 (x) y 2 (x) y n (x) y 1(x) y 2(x) y n(x), y (n 1) 1 (x) y (n 1) 2 (x) y n (n 1) (x) two equal rows { }} { { }} { y 1(x) y n(x) y 1 (x) y n (x) dw dx = y 1(x) y n(x) y 1 (x) y n(x) + y 1 (x) y n(x) + y (n 1) 1 (x) y n (n 1) (x) y (n 1) 1 (x) y n (n 1) (x) y 1 (x) y n (x) y 1 (x) y n (x) + y (n 2) 1 (x) y n (n 2) = (x) y (n 2) 1 (x) y (n 2) n (x) y (n) 1 (x) y n (n) (x) y (n) 1 (x) y n (n) (x) y 1 (x) y n (x) = y (n 2) 1 (x) y n (n 2) (x) n k=1 P k(x)y (n k) 1 n k=1 P k(x)y n (n k) y 1 (x) y n (x) n = k=1 P k(x) y (n 2) 1 (x) y n (n 2) (x) = P 1(x)W y (n) 1 (x) y n (n) (x) Therefore, W = Ae R P 1(x) dx This is Abel s formula There are two cases: A = = W (x), A = W (x) x for all x

41 HOMOGENEOUS LINEAR EQUATIONS 49 Theorem 44 Let V be the set of solutions to Equation (41) and suppose that y 1, y 2,, y n V Then y 1, y 2,, y n are linearly independent if and only if W Proof Let V be the set of solutions to Equation (41) and suppose that y 1, y 2,, y n V Suppose that y 1, y 2,, y n are linearly dependent Then there exists constants c 1, c 2,, c n such that c 1 y 1 (x) + c 2 y 2 (x) + + c n y n (x) =, c 1 y 1(x) + c 2 y 2(x) + + c n y n(x) =, c 1 y (n 1) 1 (x) + c 2 y (n 1) 2 (x) + + c n y (n 1) n (x) = Since there exists a nonzero solution for (c 1, c 2,, c n ) of the equation, we have c 1 c 2 M c n =, so it follows that M = = W Now suppose that W = Select an x R Since W (x), we have W (x ) = Thus, there exists constants c 1, c 2,, c n not all zero such that c 1 c 2 M c n, where M = y 1 (x ) y n (x ) y (n 1) 1 (x ) y n (n 1) (x ),

5 CHAPTER 4 LINEAR DIFFERENTIAL EQUATIONS that is, c 1 y 1 (x ) + c 2 y 2 (x ) + + c n y n (x ) =, c 1 y 1(x ) + c 2 y 2(x ) + + c n y n(x ) =, c 1 y (n 1) 1 (x ) + c 2 y (n 1) 2 (x ) + + c n y (n 1) n (x ) = Let y(x) = c 1 y 1 (x) + c 2 y 2 (x) + + c n y n (x) Since y 1, y 2,, y n are solutions to Equation (41), so is f, and f satisfies f(x ) =, f (x ) =, f (x ) =,, f (n 1) (x ) = But y = also satisfies these conditions, and by the uniqueness part of Theorem 41, it is the only function satisfying them Therefore, f, that is, c 1 y 1 (x ) + c 2 y 2 (x ) + + c n y n (x ) Therefore, y 1, y 2,, y n are linearly dependent Corollary 45 Let V be the set of solutions to Equation (41) Then we have dim(v ) = n Proof Let V be the set of solutions to Equation (41) Pick a point x R Then by Theorem 41, given any numbers w 1, w 2,, w n, there exists a unique solution of Equation (41) satisfying y (j 1) (x ) for j = 1, 2,, n Let y 1 be such a solution with (w 1, w 2,, w n ) = (,,, ), let y 2 be such a solution with (w 1, w 2,, w n ) = (, 1,, ), etc, finally letting y n be such a solution with (w 1, w 2,, w n ) = (,,, 1) We claim that y 1, y 2,, y n are linearly independent Suppose that c 1 y 1 (x) + c 2 y 2 (x) + + c n y n (x) Then differentiating gives c 1 y (j) 1 (x) + c 2y (j) 2 (x) + + c ny (j) n (x)

41 HOMOGENEOUS LINEAR EQUATIONS 51 for all j In particular, c 1 y (j) 1 (x ) + c 2 y (j) 2 (x ) + + c n y (j) n (x ) for all j Therefore, y 1, y 2,, y n are linearly independent and dim(v )) n Conversely, let y 1,, y n be linearly independent Then W (y 1,, y n ) (x) for all x Suppose that y V and let M = y 1 (x ) y 2 (x ) y n (x ) y 1(x ) y 2(x ) y n(x ) y (n 1) 1 (x ) y (n 1) 2 (x ) y n (n 1) (x ) Therefore, M = W (x ) = So there exists constants (c 1, c 2,, c n ) such that c 1 y(x ) c 2 M = y (x ), y (n 1) (x ) c n that is, y (j) (x ) = c 1 y (j) 1 + c 2 y (j) 2 + + c 1 y (j) 1 for i n 1 Let f(x) = c 1 y 1 (x) + c 2 y 2 (x) + + c n y n (x) Then f is a solution to Equation (41) and y (j) (x ) = f (j) (x ) for j =,, n 1 By uniqueness, y = f, ie, y(x) c 1 y 1 (x) + c 2 y 2 (x) + + c n y n (x) This means that y is a linear combination of y 1,, y n So y 1,, y n forms a basis for V Therefore, dim(v ) = n We conclude that, to solve Equation (41), we need to find n linearly independent solutions y 1,, y n of Equation (41) Then the general solution to Equation (41) is c 1 y 1 + c 2 y 2 + + c n y n Example 46 Solve y + 5y + 6y = Solution It is easy to check that y = e 3x and y = e 2x each satisfies y + 5y + 6y = Therefore, the general solution is c 1 e 3x + c 2 e 2x, where c 1 and c 2 are arbitrary constants

52 CHAPTER 4 LINEAR DIFFERENTIAL EQUATIONS 411 Linear Differential Equations with Constant Coefficients The next problem concerns how to find the n linearly independent solutions For n > 1, there is no good method in general There is, however, a technique which works in the special cases where all the coefficients are constants Consider the constant coefficient linear equation a y (n) + a 1 y (n 1) + + a n 1 y + a n y = (42) We first look for solutions of the form y = e λx If y = e λx is a solution, then Now since e λx, we have a λ n e λx + a 1 λ n 1 e λx + + a n e λx = a λ n + a 1 λ n 1 + + a n = Let p(λ) = a λ n + a 1 λ n 1 + + a n (43) If Equation (43) has n distinct real roots r 1, r 2,, r n, then we have n linearly independent solutions e r1x, e r2x,, e rnx We would like one solution for each root like this Suppose that Equation (43) has a repeated real root r Then we do not have enough solutions of the form e λx But consider y = xe rx Then y = re rx + rxe rx = (rx + 1) e rx, y = re rx + r (rx + 1) e rx = ( r 2 x + 2r ) e rx, y = r 2 e rx + r ( r 2 x + 2r ) e rx = ( r 3 x + 3r 2) e rx, y (n) = ( r n x + nr n 1) e rx

41 HOMOGENEOUS LINEAR EQUATIONS 53 Substituting y = xe rx in Equation (42) gives us ( n 1 ( n ) a k r n k x + (n k) r n k) = a k r n k k= + k= ( n 1 xe rx ) a k (n k) r n k 1 k= = p(r)xe rx + p (r)e rx e rx Since r is a repeated root of p, both p(r) = and p (r) =, that is, p(λ) = (λ r) 2 q(λ) = p (λ) = 2 (λ r) q(λ) + (λ r) 2 q (λ) = p (r) = Therefore, xe rx is also a solution of Equation (42) What about complex roots? They come in pairs, ie, r 1 = a + ib and r 2 = a ib, so e (a+ib)x and e (a ib)x satisfy Equation (42) But we want real solutions, not complex ones But Let If z 1 and z 2 are solutions, then e (a+ib)x = e ax (cos(bx) + i sin(bx)), e (a ib)x = e ax (cos(bx) i sin(bx)) z 1 = e ax (cos(bx) + i sin(bx)), z 2 = e ax (cos(bx) i sin(bx)) z 1 + z 2 2 = e ax cos(bx), z 1 z 2 2 = e ax sin(bx) are solutions So the two roots a + ib and a ib give two solutions e ax cos(bx) and e ax sin(bx) For repeated complex roots, use xe ax cos(bx), xe ax sin(bx), etc So in all cases, we get n solutions The expression c 1 e ax cos(bx) + e ax c 2 sin(bx) can be rewritten as follows Let A = c 2 1 + c2 2 Then ( ) θ = cos 1 c 1, c 2 1 + c 2 2

54 CHAPTER 4 LINEAR DIFFERENTIAL EQUATIONS that is, cos(θ) = c 1, sin(θ) = 1 cos 2 c (θ) = 2 c 2 1 + c 2 2 c 2 1 + c 2 2 Therefore, c 1 e ax cos(bx) + e ax c 2 sin(bx) = Ae ax ( c 1 A cos(bx) + c 2 A sin(bx) ) = Ae ax (cos(θ) cos(bx) + sin(θ) sin(bx)) = Ae ax (cos(bx θ)) Example 47 Solve y (5) 4y (4) + 5 + 6 36y + 4y = Solution We have λ 5 4λ 4 + 5λ 3 + 6λ 2 36λ + 4 =, (λ 2) 2 (λ + 2) ( λ 2 2λ + 5 ) = Therefore, λ = 2, 2, 2, 1 ± 2i and y = c 1 e 2x + c 2 xe 2x + c 3 e 2x + c 4 e x cos(2x) + c 5 sin(2x), where c 1, c 2, c 3, c 4, c 5 are arbitrary constants 42 Nonhomogeneous Linear Equations A nonhomogeneous linear equation is an equation of the form y (n) + p 1 (x)y (n 1) + p 2 (x)y (n 2) + + p n (x)y = q(x) (44) Let Equation (41) (p 47) be the corresponding homogeneous equation If u and v are solutions of Equation (44), then u v is a solution of Equation (41) Conversely, if u is a solution of Equation (44) and v is a solution of Equation (41), then u + v is a solution of Equation (44) Let y 1, y 2,, y n be n linearly independent solutions of Equation (41) and let y p be a solution of Equation (44) If y is any solution to Equation (44), then y y p is a solution to Equation (41), so y y p = c 1 y 1 + c 2 y 2 + + c n y n

42 NONHOMOGENEOUS LINEAR EQUATIONS 55 Therefore, y = c 1 y 1 + c 2 y 2 + + c n y n + y p is the general solution to Equation (44) Example 48 Solve y y = x Solution The corresponding homogeneous equation is y y = follows that λ 2 1 = λ = ±1 Therefore, It then y 1 = e x, y 2 = e x By inspection, y p = x solves the given equation Therefore, the general solution is y = c 1 e x + c 2 e x x, where c 1 and c 2 are arbitrary constants We will consider more systematically how to find y p later

56 CHAPTER 4 LINEAR DIFFERENTIAL EQUATIONS

Chapter 5 Second Order Linear Equations A second order linear equation is an equation of the form y + P (x)y + Q(x)y = R(x) (51) To find the general solution of Equation (51), we need Two solutions: y 1 and y 2 of the corresponding homogeneous equation y + P (x)y + Q(x) = One solution: y p of Equation (51) Then the general solution is y = c 1 y 1 + c 2 y 2 + y p, where c 1 and c 2 are arbitrary constants Except in the case where P (x) and Q(x) are constants (considered earlier on p 52), there is no general method of finding y 1, but 1 given y 1, there exists a method of finding y 2 2 given y 1 and y 2, there exists a method of finding y p We will discuss (1) first 51 Reduction of Order As mentioned earlier, we can find a second solution from the first develop the tools we need We first 57

58 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS Lemma 51 Let f(x) be a twice differentiable function on a closed bounded interval J If {x J : f(x) = } is infinite, then there exists an x J such that f(x ) = and f (x ) = Proof Let S = {x J : f(x) = } If S is infinite, then by the Bolzano- Weierstrass Theorem, S has an accumulation point x, ie, there exists an x S such that every interval about x contains another point of S Therefore, by Rolle s Theorem, every interval about x contains a point where f is Therefore, by continuity, f (x ) = Corollary 52 Let y 1 (x) be a solution of y + P (x)y + Q(x)y = on a closed bounded interval J If y 1 is not the zero solution, then y 1 (x) = for at most finitely many x in J Proof Let V be the set of solutions of y + P (x)y + Q(x)y = on J Then V is a two-dimensional vector space So if y 1, then there exists a y 2 V such that y 1 and y 2 are linearly independent Therefore, W (y 1, y 2 ), ie, W (y 1, y 2 ) is never zero (the Wronskian is either always or never ) But if y 1 (x) = for infinitely many x in J, by Lemma 51, there exists an x J such that y 1 (x ) = and y 1(x ) = Therefore, W (y 1, y 2 )(x ) = y 1 (x ) y 2 (x ) y 1(x ) y 2(x ) = y 2 (x ) y 2(x ) = This is a contradiction Therefore, y 1 (x) = for at most finitely many x in J Let y 1 a solution to y + P (x)y + Q(x)y = on a closed bounded interval J We want to find a linearly independent second solution y 2 We break J into a finite collection of subintervals so that y 1 (x) on the interior of each subinterval We will describe how to find the second solution y 2 Note that f is necessarily but certainly not sufficiently continuous if f is differentiable

51 REDUCTION OF ORDER 59 on the interior of each interval This will define y 2 throughout J except for a few isolated points Consider a subinterval I on which y 1 (x) is never zero y 2 (x) < throughout I We have y 1 y 2 W (y 1, y 2 ) = y 1 y 2 So y 1 (x) > or If y 2 is also a solution to y + P (x)y + Q(x)y =, then W = Ae R P (x) dx for some constant A (Albel s formula) Suppose we choose A = 1 Then y 1 y 2 y 1y 2 = e R P (x) dx, where y 1 is known This is a linear first order equation for y 2 In standard form, we have R P (x) dx y 2 y 1 y 2 = e, y 1 y 1 where y 1 throughout I The integrating factor is R y e 1 y 1 = e ln( y1 )+C = C y 1, where C = ±1, depending on the sign of y 1 We use 1/y 1 Therefore, finally giving us y 2 R P (x) dx y y 1 y1 2 y 2 = e y1 2, y 2 e R P (x) dx = y 1 y 2 1 R e P (x) dx y 2 = y 1 y1 2 dx dx, (52a) This defines y 2 on J except at a few isolated points where y 1 = At those points, we have y 1 y 2 y 1y 2 = e R P (x) dx, so R P (x) dx y 2 = e y 1 (52b) This defines y 2 are the boundary points of the subintervals Different A s correspond to different y 2 s

6 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS Example 53 Observe that y = x is a solution of y x + 2 x y + x + 2 x 2 y = and solve y x + 2 x y + x + 2 x 2 y = xe x Solution Using the fact that y 1 = x, we have y 2 = x e R ( x+2 x )dx x 2 e x e 2 ln( x ) = x x 2 R e (1+ 2 x)dx dx = x x 2 e x x 2 dx = x x 2 dx = x e x+2 ln( x ) dx = x x 2 e x dx = xe x dx Therefore, W = x xe x 1 e x + xe x = xex + x 2 e x xe x = x 2 e x From this, we have v 1 = y 2R W = xex xe x x 2 e x = e x = v 1 = e x, v 2 = y 1R W = xxex x 2 e x = 1 = v 2 = x Therefore, y p = v 1 y 1 + v 2 y 2 = xe x + x 2 e x, and the general solution is y = C 1 x + C 2 xe x xe x + x 2 e x = C 1 x + C 2xe x + x 2 e x, x, where C 1, C 2, and C 2 are arbitrary constants 52 Undetermined Coefficients Suppose that y 1 and y 2 are solutions of y + P (x)y + Q(x)y = The general method for finding y p is called variation of parameters We will first consider another method, the method of undetermined coefficients, which does not always work, but it is simpler than variation of parameters in cases where it does work

52 UNDETERMINED COEFFICIENTS 61 The method of undetermined coefficients has two requirements, that 1 we have constant coefficients, ie, P (x) = p and Q(x) = q for some constants p and q 2 R(x) is nice Example 54 Solve y y = e 2x Solution We have λ 2 1 = So y 1 = e x and y 2 = e x Since R(x) = e 2x, we look for a solution of the form y = Ae 2x Then y = 2Ae 2x, y = 4Ae 2x Then according to the differential equation, we have Therefore, the general solution is 4Ae 2x Ae } {{ 2x = e } 2x 1 = A = 3 3Ae 2x y = c 1 e x + c 2 e x + 1 3 e2x, where c 1 and c 2 are arbitrary constants Undetermined coefficients works when P (x) and Q(x) are constants and R(x) is a sum of the terms of the forms shown in Table 51 Table 51: The various forms of y p ought to be given a particular R(x) R(x) Form of y p Cx n Ce rx Ce rx cos(kx) Ce rx sin(kx) A x n + A 1 x n 1 + + A n 1 x + A n Ae rx Ae rx cos(kx) + Be rx sin(kx) The success of undetermined coefficients depends on the derivative of R(x) having some basic form as R(x) There is an easy way to deal with a complication: whenever a term in the trial y p is already part of the general solution to the homogeneous equation, multiply it by x For example, if we have y y = e x, as before, we have y 1 = e x and y 2 = e 2x Thus, we try y p = Axe 2x and find A

62 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS Example 55 Solve y + 3y + 2y = e x 3 Solution We have λ 2 + 3λ + 2 = Factoring gives (λ + 2) (λ + 1) =, so λ { 1, 2}, and we have y 1 = e x and y 2 = e 2x Let y = Ae x + B Then y = Ae x and y = Ae x, giving us y + 3y + 2y = Ae x + 3Ae x + 2Ae x + 2B = 6Ae x + 2B Comparing coefficients with e x 3, we have 6A = 1 = A = 1 6, 2B = 3 = B = 3 2 Therefore, y p = e x /6 3/2, and we have y = c 1 e x + c 2 e 2x + ex 6 3 2, where c 1 and c 2 are arbitrary constants Example 56 Solve y + y = sin(x) Solution We have λ 2 + 1 =, so λ = ±i, and it follows that y 1 = cos(x) and y 2 = sin(x) Let y = Ax sin(x) + Bx cos(x) Then y = A sin(x) + Ax cos(x) + B cos(x) Bx sin(x) = (A Bx) sin(x) + (Ax + B) cos(x), y = B sin(x) + (A Bx) cos(x) + A cos(x) (Ax + B) sin(x) = ( Ax 2B) sin(x) + (2A Bx) cos(x) Therefore, y + y = ( Ax 2B) sin(x) + (2A Bx) cos(x) + Ax sin(x) + Bx cos(x) Comparing coefficients with sin(x), we see that 2B = 1 = B = 1 2, 2A = = A =

52 UNDETERMINED COEFFICIENTS 63 Therefore, y p = x cos(x)/2, and the general solution is y = c 1 sin(x) + c 2 cos(x) x cos(x), 2 where c 1 and c 2 are arbitrary constants Example 57 Solve y 4y = x 2 Solution We have λ 2 4 =, so λ = ±2, and it follows that y 1 = e 2x and y 2 = e 2x Let y = Ax 2 +Bx+C Then y = 2Ax+B and y = 2A Therefore, y 4y = 2A 4Ax 2 4B 4C = 4Ax 2 + x + (2A 4B 4C) Comparing coefficients with x 2, we immediately see that 4A = 1 = A = 1 4, 4B = = B =, 2A 4C = = C = A 2 = 1/4 2 = 1 8 Therefore, y p = x 2 /4 1/8, and the general solution is y = c 1 e 2x + c 2 e 2x x2 4 1 8, where c 1 and c 2 are arbitrary constants Example 58 Solve y y = x Solution We have λ 2 λ =, so λ {1, }, and it follows that y 1 = e x and y 2 = 1 Let y = Ax 2 + Bx Then y = 2Ax + B and y = 2A Therefore, y y = 2A 2Ax B Comparing coefficients with x, we immediately see that 2A = 1 = A = 1 2, 2A B = = B = 1 Therefore, y p = x 2 /2 x, and the general solution is y = c 1 e x + c 2 x2 2 x,

64 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS where c 1 and c 2 are arbitrary constants 521 Shortcuts for Undetermined Coefficients Suppose we have y + ay + by = R(x) Let p(λ) = λ 2 + aλ + b We now consider two cases for R(x) Case I: R(x) = ce αx It follows that y p = Ae αx, so y = αae αx and y = αae αx Then y + ay + by = ce αx ( α 2 + aα + b ) Ae αx = p(α)ae αx Therefore, A = c/p(α), unless p(α) = c What if p(α) =? Then e αx is a solution to the homogeneous equation Let y = Axe αx Then Thus, y = Ae αx + Aαxe αx, y = Aαe αx + Aαe αx + Aα 2 xe αx = 2Aαe αx + Aα 2 xe αx y + y + y = A ( α 2 x + aαx + b ) + 2α + a e αx = A (p(α)xe αx + p (α)e αx ) = Ap (α)e αx Therefore, A = c/p (α), unless p (α) = too If p(α) = and p (α) =, then e αx and xe αx both solve the homogeneous equation Letting y = Ax 2 e αx, we have A = c/p (α) Case II: R(x) = ce rx cos(bx) or ce rx sin(bx) We use complex numbers Set α = r + ib Then ce αx = ce rx cos(bx) + ice rx sin(bx) Let z = y + iw Consider the equation z + az + bz = ce αx (53)

52 UNDETERMINED COEFFICIENTS 65 The real part and imaginary parts are y + ay + by = ce rx cos(bx), w + aw + bw = ce rx sin(bx), (54a) (54b) respectively The solutions of Equation (53,54a,54b) are then respectively z = ceαx p(α), ( ) ce αx y = Re, w = Im p(α) ( ce αx p(α) ), Example 59 Solve y + 8y + 12y = 3 cos(2x) Solution We have p(λ) = λ 2 + 8λ + 12 = (λ + 2) (λ + 6) { 2, 6} and it follows that y 1 = e 2x and y 2 = e 6x So Therefore, λ z = 3e2ix p(2i) = = 3e2ix 8 + 16i = 3 8 3e 2ix (2i) 2 + 16i + 12 = 1 1 + 2i e2ix = 3 8 3e 2ix 4 + 16i + 12 1 2i (1 2i) (1 + 2i) e2ix = 3 8 1 2i 1 + 2i e2ix = 3 (1 2i) e2ix 4 = 3 (1 2i) (cos(2x) + i sin(2x)) 4 = 3 cos(2x) + 2 sin(2x) + i ( 2 cos(2x) + sin(2x)) 4 Therefore, Re(z) = 3 3 (cos(2x) + 2 sin(2x)), Im(z) = ( 2 cos(x) + sin(2x)) 4 4 Hence, the solution is y = c 1 e 2x + c 2 e 6x + 3 (cos(2x) + 2 sin(2x)), 4 where c 1 and c 2 are arbitrary constants We conclude that there is no shortcut when R(x) is a polynomial or any multiple of a polynomial

66 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS 53 Variation of Parameters Consider the equation y + P (x)y + Q(x)y = R(x), (I) where P (x) and Q(x) are not necessarily constant Let y 1 and y 2 be linearly independent solutions of the auxiliary equation y + P (x)y + Q(x)y = (H) So for any constants C 1 and C 2, C 1 y 1 + C 2 y 2 satisfies Equation (H) The idea of variation of parameters is to look for functions v 1 (x) and v 2 (x) such that v 1 y 1 + v 2 y 2 satisfies Equation (I) Let y 1 and y 2 If we want y to be a solution to Equation (I), then substitution into Equation (I) will give one condition on v 1 and v 2 Since there is only one condition to be satisfied and two functions to be found, there is lots of choice, ie, there are many pairs v 1 and v 2 which will do So we will arbitrarily decide only to look for v 1 and v 2 which also satisfy v 1y 1 + v 2y 2 = We will see that this satisfaction condition simplifies things So consider y = v 1 y 1 + v 2 y 2, y = v 1 y 1 + v 2 y 2 + v 1y 1 + v 2y 2 = v 1 y 1 + v 2 y 2, y = v 1 y 1 + v 2 y 2 + v 1y 1 + v 2y 2 To solve Equation (I) means that Therefore, ( v 1 y 1 + v 2 y 2 + v 1y 1 + v 2y 2 + P (x)v 1 y 1 +P (x)v 2 y 2 + Q(x)v 1 y 1 + Q(x)v 2 y 2 v 1 (y 1 + P (x)y 1 + Q(x)y 1 ) +v 2 (y 2 + P (x)y 2 + Q(x)y 2 ) +v 1y 1 + v 2y 2 v 1y 1 + v 2y 2 = R(x), v 1y 1 + v 2y 2 =, ) = R(x), = R(x)

53 VARIATION OF PARAMETERS 67 and it follows that where v 1 = y 2R W, v 2 = y 1R W, y 1 y 2 W = y 1 y 2, ie, it is never zero regardless of x, since y 1 and y 2 are linearly independent Example 51 Solve y 5y + 6y = x 2 e 3x Solution The auxiliary equation is y 5y + 6y = λ 2 5λ + 6 = = λ = 3, 2 Therefore, the solutions are y 1 = e 3x and y 2 = e 2x Now, Therefore, y 1 y 2 W = y 1 y 2 = v 1 = y 2R W e 3x 3e 3x e 2x 2e 2x = e2x x 2 e 3x e 5x = 2e5x 3e 5x = e 5x = x 2 = v 1 = x3 3 We can use any function v 1 with the right v 1, so choose the constant of integration to be For v 2, we have v 2 = y 1R W = e3x x 2 e 3x e 5x = x 2 e x, so v 2 = x 2 e x dx } {{ } by parts = x 2 e x + 2 = x 2 e x + 2xe x 2 e x dx xe x dx } {{ } by parts = x 2 e x + 2xe x 2e x = ( x 2 2x + 2 ) e x Finally, y p = x3 3 e3x ( x 2 2x + 2 ) ( ) x e x e 2x = e 3x 3 3 x2 + 2x 2,

68 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS and the general solution is ( ) x y = C 1 e 3x + C 2 e 2x + e 2x 3 3 x2 + 2x 2 ( ) x = e 3x 3 + C 2 e 2x 3 x2 + 2x 2 + C 1 ( x 3 = 3 x2 + 2x + C 1 ) e 3x + C 2 e 2x, where C 1 and C 2 are arbitrary constants Example 511 Solve y + 3y + 2y = sin(e x ) Solution We immediately have λ 2 + 3λ + 2 =, which gives λ { 1, 2} Therefore, y 1 = e x and y 2 = e 2x Now, Therefore, W = e x e x e 2x 2e 2x = 2e 3x + e 3x = e 3x v 1 = e 2x sin(e x ) e 3x = e x sin(e x ) = v 1 = cos(e x ) For v 2, we have Therefore, v 2 = e x sin(e x ) e 3x = e 2x sin(e x ) v 2 = e 2x sin(e x ) dx } {{ } = substitute t = e x ( t cos(t) + = t sin(t) dt } {{ } ) cos(t) dt by parts = ( t cos(t) + sin(t)) = e x cos(e x ) sin(e x )

53 VARIATION OF PARAMETERS 69 Finally, y p = v 1 y 1 + v 2 y 2 = e x cos(e x ) + e 2x (e x cos(e x ) sin(e x )) = e x cos(e x ) + e x cos(e x ) e 2x sin(e x ) = e 2x sin(e x ), and the general solution is y = C 1 e x + C 2 e 2x e 2x sin(e x ), where C 1 and C 2 are arbitrary constants

7 CHAPTER 5 SECOND ORDER LINEAR EQUATIONS

Chapter 6 Applications of Second Order Differential Equations We now look specifically at two applications of first order de s We will see that they turn out to be analogs to each other 61 Motion of Object Hanging from a Spring Figure 61 shows an object hanging from a spring with displacement d d Figure 61: An object hanging from a spring with displacement d The force acting is gravity, spring force, air resistance, and any other external 71

72CHAPTER 6 APPLICATIONS OF SECOND ORDER DIFFERENTIAL EQUATIONS forces Hooke s Law states that F spring = kd, (61) where F is the spring force, d is the displacement from the spring to the natural length, and k > is the spring constant As we lower the object, the spring force increases There is an equilibrium position at which the spring and gravity cancel Let x be the distance of the object from this equilibrium position measured positively downwards Let s be the displacement of the spring from the natural length at equilibrium Then it follows that ks = mg If the object is located at x, then the forces are 1 gravity 2 spring 3 air resistance 4 external Considering just gravity and the spring force, we have F gravity + F spring = mg k (x + s) = kx Suppose air resistance is proportional to velocity Then F air = rv, r > In other words, the resistance opposes motion, so the force has the opposite sign of velocity We will suppose that any external forces present depend only on time but not on position, ie, F external = F (t) Then we have so it follows that F } total = kx rv + F (t), {{ } ma Therefore, mλ 2 + rλ + k = and we have λ = r ± r 2 4mk 2m m d2 x dt 2 + r dx + kx = F (t) (62) dt = r ( r ) 2 2m ± k 2m m Intuitively, when x =, they balance Changing x creates a force in the opposite direction attempting to move the object towards the equilibrium position

61 MOTION OF OBJECT HANGING FROM A SPRING 73 We now have three cases we need to consider Case 1: 4mk > r 2 Let α = r/2m, let ω = k/m, and let Then λ = α ± iβ and β = k ( r ) 2 m = ω2 α 2m 2 x = e αt (C 1 cos(βt) + C 2 sin(βt)) = Ae αt cos(βt θ), (63) where C 1 and C 2 are arbitrary constants Figure 62a shows how this solution qualitatively looks like Damped vibrations is the common case The oscillations die out in time, where the period is 2π/β If r =, then α = In such a case, there is no resistance, and it oscillates forever Case 2: 4mk = r 2 In such a case, we simply have x = e r 2m t (C 1 + C 2 t), (64) where C 1 and C 2 are arbitrary constants Figure 62b shows how this solution qualitatively looks like Case 3: 4mk < r 2 Let a = r/2m and Then the solution is b = ( r ) k 2m m x = C 1 e (a b)t + C 2 e (a+b)t, (65) where C 1 and C 2 are arbitrary constants This is the overdamped case The resistance is so great that the object does not oscillate (imagine everything immersed in molasses) It will just gradually return to the equilibrium position Figure 62c shows how this solution qualitatively looks like Consider Equation (62) again We now consider special cases of F (t), where F cos(ωt), F (t) = F sin(ωt)

74CHAPTER 6 APPLICATIONS OF SECOND ORDER DIFFERENTIAL EQUATIONS t t (a) 4mk > r 2 (b) 4mk = r 2 (c) 4mk < r 2 t Figure 62: The various cases for a spring-mass system Use undetermined coefficients to get the solution of the form x p = B cos(ωt) + C sin(ωt) For F (t) = F cos(ωt), we get x p = where F (k mω 2 ) 2 2 (k mω) cos(ωt) + rω sin(ωt) = A cos(ωt δ), + (rω) A = F (k mω 2 ) 2 + (rω) 2, cos(δ) = sin(δ) = k mω 2 (k mω 2 ) 2 + (rω) 2, rω (k mω 2 ) 2 + (rω) 2 For F (t) = F sin(ωt), we get x p = A sin(ωt δ), where A and δ is as above

62 ELECTRICAL CIRCUITS 75 In any solution e αt ( ) + A sin(ωt δ), the first term eventually dies out, leaving A sin(ωt δ) as the steady state solution Here we consider a phenomenon known as resonance The above assumes that cos(ωt) and sin(ωt) are not part of the solution to the homogeneous equation Consider now m d2 x dt 2 + kx = F cos(ωt), where ω = k/m (the external force frequency coincides with the natural frequency) Then the solution is x p = F 2 km t sin(ωt) Another phenomenon we now consider is beats Consider r = with ω near but not equal to k/m Let ω = k/m Then the solution, with x() = = x (), is where ε = (ω ω) /2 F x = m ( ω 2 ω 2 cos(ωt ωt) ) ( ) ( ) 2F (ω ω) t (ω + ω) t = ( ω + ω) ( ω ω) sin sin 2 2 F sin(εt) sin(ωt), 2ωεm 62 Electrical Circuits Consider the electrical circuit shown in Figure 63 Let Q(t) be the charge in the capacitor at time t (Coulombs) Then dt/dt is called the current, denoted I The battery produces a voltage (potential difference) resulting in current I when the switch is closed The resistance R results in a voltage drop of RI The coil of wire (inductor) produces a magnetic field resisting change in the current The voltage drop created is L (di/dt) The The amplitude of the vibrations is which increases with time F 2 km t,

76CHAPTER 6 APPLICATIONS OF SECOND ORDER DIFFERENTIAL EQUATIONS resistor battery I amperes inductor switch + capacitor Figure 63: An electrical circuit, where resistance is measured in Ohms, capacitance is measured in Farads, and the inductance is measured in Henrys capacitor produces a voltage drop of Q/C Unless R is too large, the capacitor will create sine and cosine solutions and, thus, an alternating flow of current Kirkhoff s Law states that the sum of the voltage changes around a circuit is zero, so E(t) + RI + L di dt + Q C, so we have L d2 Q dt 2 + R dq dt + Q C = E(t) (66) The solution is the same as in the case of a spring The analogs are shown in Table 61 Table 61: Analogous terms between a spring-mass system and an electrical circuit Circuit Spring Charge Q Inductance I Resistance R Capacitance inverse 1/C Voltage generated by battery E(t) Amperes I Displacement x Mass m Friction r Spring constant k External force F (t) Velocity v An example of this application is choosing a station on an old-fashioned radio The radio has a capacitor with two charged metal bars When the user turns the tuning dial, it changes the distance between the bars, which in

62 ELECTRICAL CIRCUITS 77 turn changes the capacitance C This changes the frequency of the solution of the homogeneous equation When it agrees with the frequency E(t) of some incoming signal, resonance results so that the amplitude of the current produced from this signal is much greater than any other

78CHAPTER 6 APPLICATIONS OF SECOND ORDER DIFFERENTIAL EQUATIONS

Chapter 7 Higher Order Linear Differential Equations In this section we generalize some of the techniques for solving second order linear equations discussed in Chapter 5 so that they apply to linear equations of higher order Recall from Chapter 4 that an nth order linear differential equation has the form y (n) + P 1 (x)y (n 1) + P 2 (x)y (n 2) + + P n (x)y = Q(x) (71) The general solution is y = C 1 y 1 + C 2 y 2 + + C n y n + y p, where y 1, y 2,, y n are linearly independent solutions to the auxiliary homogeneous equation and y p is a solution of Equation (71) So given y 1, y 2,, y n, how do we find y p? We again consider the two methods we have looked at in 52 and 53 (p 6 and p 66, respectively) 71 Undetermined Coefficients If P 1 (x), P 2 (x),, P n (x) are constants and R(x) is nice, we can use undetermined coefficients Example 71 Solve y (4) y = 2e 2e + 3e x 79

8CHAPTER 7 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Solution We have Therefore, p(λ) = λ 4 1 = (λ 1) (λ + 1) ( λ 2 + 1 ) y 1 = e x, y 2 = e x, y 3 = cos(x), y 4 = sin(x) Let y p = Ae 3x + Bxe 3x Then by the short cut method ( 521, p 64), we have Therefore, the general solution is A = 2 p(2) = 2 16 1 = 2 15, B = 3 p (1) = 3 4 1 3 = 3 4 y = C 1 e x + C 2 e x + C 3 cos(x) + C 4 sin(x) + 2 15 e2x + 3 4 xex 72 Variation of Parameters Suppose that y 1, y 2,, y n are solutions of the auxiliary homogeneous equation We look for solutions of Equation (71) of the form y = v 1 y 1 + v 2 y 2 + + v n y n, where v 1, v 2,, v n are functions of x One condition on v 1, v 2,, v n is given by substituting into Equation (71) We choose n 1 conditions to be v 1y 1 + v 2y 2 + + v ny n =, (1) v 1y 1 + v 1y 2 + + v ny n =, (2) v 1y (n 2) 1 + v 2y (n 2) 2 + + v ny (n 2) n = (n 1)

72 VARIATION OF PARAMETERS 81 If conditions (1),, (n 1) are satisfied, then Therefore, y = y = y (n 1) = y (n) = n (v i y i ) = i=1 n (v i y i) = i=1 n i=1 n i=1 v i y (n 1) i + v i y (n) i + y (n) + P 1 (x)y (n 1) + + P n (x)y = n v i y i + i=1 n i=1 n i=1 = = = v i y i + n v iy i = i=1 n v iy i = i=1 n v iy (n 2) = i=1 v iy (n 1) i n i=1 + P 1 (x) n i=1 + i=1 v i y (n) i + n i=1 v iy (n 1) i ( n v i i=1 n i=1 n i=1 v i y (n 1) i y (n) i n v i y i, i=1 n i=1 v i y i, v i y (n 1) i, v iy (n 1) i n n v iy (n 1) i + v i i=1 n v iy (n 1) i i=1 Therefore, Equation (71) becomes the first condition n i=1 + + P n (x) + P 1 (x)y (n 1) i + +P n (x)y i n v i y i i=1 v iy (n 1) i y = R(x) (n) )

82CHAPTER 7 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS Conditions (1),, (n 1) can be written as M v 1 v 2 v n = R(x), where M = Therefore, M = W y 1 y n y 1 y n y (n 1) 1 y n (n 1) 73 Substitutions: Euler s Equation Consider the equation x n y (n) + a 1 x n 1 y (n 1) + + a n 1 y + a n y = R(x), (72) where a 1, a 2,, a n are constants Consider x > Let x = e u so that u = ln(x) (for x <, we must use u = ln( x)) Then dy dx = dy du du dx = dy 1 du x, d 2 y dx 2 = d2 y 1 du 2 x 2 dy ( 1 d 2 du x 2 = y du 2 dy ) 1 du x 2, d 3 ( ) ( y dx 3 = 3 ydx 3 d2 y 1 d 2 d du 2 x 3 2 y du 2 dy ) 1 du x 3 ( d 3 y = du 3 3 d2 y du 2 + 2 dy ) 1 du x 3, d n ( y d n dx n = y du n + C d n 1 ) y 1 du n 1 + + C dy 1 n du x n

73 SUBSTITUTIONS: EULER S EQUATION 83 for some constants C 1, C 2,, C n Therefore, Equation (72) becomes d n y du n + b d n 1 y 1 du n 1 + + b dy n 1 du + b ny = R(e u ) for some constants b 1, b 2,, b n Example 72 Solve x 2 y + xy y = x 3 for x > Solution Let x = e u so that u = ln(x) Then dy dx = dy du du dx = dy 1 du x, d 2 y dx 2 = d2 y 1 du 2 x 2 dy 1 du x 2 = ( d 2 y du 2 dy ) 1 du x 2 Therefore, ( d 2 y du 2 dy ) + dy du du y = e3u, d 2 u du 2 y = e3u Considering the homogeneous case, we have λ 2 1 = λ = ±1 Therefore, y 1 = e u and y 2 = e u Using undetermined coefficients, let y p = Ae 3u Then y = 3Ae 3u and y = 9Ae 3u Substitution gives 9Ae 3u Ae 3u } {{ } 8Ae 3u = e 3u We can immediately see that A = 1/8 Therefore, the general solution is y = C 1 e u + C 2 e u + 1 8 e3u = C 1 e ln(x) + C 2 e ln(x) + 1 ln(x) e3 8 = C 1 x + C 2 x + x3 8, where C 1 and C 2 are arbitrary constants We can also consider a direct method Solving d 2 y dx 2 + p dy du + qy =

84CHAPTER 7 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS involves looking for solutions of the form y = e λu = (e u ) λ and finding λ Since x = e u, we could directly look for solutions of the form x λ and find the right λ With this in mind and considering Example 72 again, we have Therefore, as before y = x λ, y = λx λ 1, y = λ (λ 1) x λ 2 x 2 y + xy y = = λ (λ 1) x λ + λx λ x λ = = λ (λ 1) + λ 1 = = λ 2 1 = = λ = ±1,

Chapter 8 Power Series Solutions to Linear Differential Equations 81 Introduction Consider the equation y (n) + P 1 (x)y (n 1) + + P n (x)y = R(x) (81) Given solutions y 1,, y n to the homogeneous equation y (n) + P 1 (x)y + + P n y =, we know how to solve Equation (81), but so far there is no method of finding y 1,, y n unless P 1,, P n are constants In general, there is no method of finding y 1,, y n exactly, but we can find the power series approximation Example 81 Solve y xy 2y = Solution Let y = a + a 1 x + a 2 x 2 + + a n x n + = a n x n 85

86CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Then it follows that y = a 1 + 2a 2 x + + na n x n 1 + (n + 1) a n+1 x n + = na n x n 1 = (k + 1) a k+1 x k, n=1 k= where k = n 1 Relabeling, we have y = By the same procedure, we have (n + 1) a n+1 x n y = n (n + 1) a n+1 x n 1 = (n + 1) (n + 2) a n+2 x n Now considering y xy 2y =, we have { y }} { (n + 1) (n + 2) a n+2 x n x { y }} { na n x n 1 2 { y }} { a n x n =, (n + 1) (n + 2) a n+2 na n 2a n x n = This gives us the recurrence relation for all n Therefore, (n + 1) (n + 2) a n+2 na n 2a n =, (n + 1) (n + 2) a n+2 (n + 2) a n =, (n + 1) a n+2 = a n, a n+2 = a n n + 1 a 2 = a 1, a 4 = a 2 3 = a 1 3, a 6 = a 4 5 = a 1 3 5,, a 2n = a 3 = a 1 2, a 5 = a 3 4 = a 1 2 4, a 7 = a 5 6 = a 1 2 4 6,, a 2n+1 = n k=1 a (2k 1), a 1 n k=1 2k

81 INTRODUCTION 87 Therefore, y = a + a 1 x + a 2 x 2 + a 3 x 3 + + a n x n + a n k=1 (2k + 1)x2n = a + a 1 x + a x 2 + a 1 2 x3 + + a 1 n k=1 2k x2n+1 + = a (1 + x 2 x 2n ) ( ) + + n k=1 (2k 1) + + a 1 x + x3 2 + + x2n+1 n k=1 2k + x 2n = a n k=1 (2k 1) + a x 2n+1 1 2 n n! That is, we have y = C 1 y 1 + C 2 y 2, where C 1 = a and C 2 = a 1 such that y 1 = x 2n n k=1 (2k 1), y 2 = x 2n+1 2 n n! Can we recognize y 1 and y 2 as elementary functions? In general, no But in this case, we have y 2 = x 2n+1 2 n n! = x x 2n 2 n n! = x ( x 2 ) n 2 n n! ( x 2 /2 ) n = x n! = xe x2 /2 Having one solution in closed form, we can try to find another (see 51, p 57) by R e P (x) dx y 1 = y 2 y 2 2 = xe x2 /2 e x 2 /2 x 2 e x2 = xe x2 /2 e R P (x) dx x 2 e x2 dx = xe x 2 /2 dx 1 x 2 e x2 /2 dx, which is not an elementary function

88CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS 82 Background Knowledge Concerning Power Series Theorem 82 Consider the power series a n (z p) n 1 There exists an R such that R < called the radius of convergence such that a n (z p) n converges for z p < R and diverges for z p > R The radius of convergence is given by if the limit exists (ratio test) R = lim n a n a n+1 (82) 2 The function f(z) = a nz n defined for z p < R can be differentiated and integrated term-wise within its radius of convergence Proof Proof is given in MATA37 or MATB43 Remark The series may or may not converge when z p = R Using (2) of Theorem 82, f (z) exists for z p < R with f (z) = na nz n 1 Similarly, f (z) = f (z) = f(z) = a n (z p) n, na n (z p) n 1, n (n 1) a n (z p) n 2, z = p = a = f(p), z = p = a 1 = f (p), z = p = a 2 = f (p), 2 f (n) (z) = n! a n, z = p = a n = f (n) (p) n!

83 ANALYTIC EQUATIONS 89 Conversely, if f is infinitely differentiable at p, we define a n = f (n) (p)/n! Then a n (z p) n is called the Taylor series of f about p From the above, if any power series converges to f near p, it must be its Taylor series But 1 the Taylor series might not converge 2 if it does converge, it does not have to converge to f Definition (Analytic function, ordinary/singular point) A function f is called analytic at p if a n (z p) n converges to f(z) in some neighbourhood of p (ie, if there exists an r > such that it converges for z p < r) If f is analytic at p, then p is called an ordinary point of f If f is not analytic at p, then p is called a singular point of f Theorem 83 (Theorem from complex analysis) Let f(z) be analytic at p Then the radius of convergence of the Taylor series of f at p equals the distance from p to the closest singularity of f in the complex plane Proof Proof is given in MATC34 Example 84 Consider f(x) = 1/ ( x 2 + 1 ) The singularities are ±i Therefore, the radius of convergence of the Taylor series about the origin is 1, with the series given by f(x) = ( 1) n x 2n The radius of convergence of the Taylor series about x = 1 is 2 83 Analytic Equations Consider y + P (x)y + Q(x)y = If both P (x) and Q(x) are analytic at p, then p is called an ordinary point of the de Otherwise, p is called a singular

9CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS point or a singularity of the de When looking at a de, do not forget to write it in the standard form For example, x 2 y + 5y + xy = has a singularity at x =, since the standard form is ( ) 5 y + x 2 y + ( ) 1 y = x Theorem 85 Let x = p be an ordinary point of y + P (x)y + Q(x)y = Let R be the distance from p to the closest singular point of the de in the complex plane Then the de has two series y 1 (x) = a nx n and y 2 (x) = b nx n which converge to linearly independent solutions to the de on the interval x p < R Remark The content of the theorem is that the solutions are analytic within the specified interval The fact that solutions exist on domains containing at least this interval is a consequence of theorems in Chapter 1 Proof The proof is given in Chapter 1 Example 86 Consider ( x 2 + 1 ) y + 2xy + 2y = Dividing by x 2 + 1, we have y + 3x x 2 + 1 y + 2 x 2 + 1 y = The singularities are ±i If p =, then r = i = 1 Therefore, analytic solutions about x = exist for x < 1 If p = 1, then r = i 1 = 2 Therefore, there exist two linearly independent series of the form a n (x 1) n converging to the solutions for at least x 2 < 2 Note that the existence and uniqueness theorem (see Chapter 1) guarantees solutions from to (since x 2 + 1 has no real zeros), but only in a finite subinterval can we guarantee that the solutions are expressible as convergent power series Example 87 Consider y xy 2y = It has no singularities Therefore, a series solution converging on R exists about any point

84 POWER SERIES SOLUTIONS: LEVELS OF SUCCESS 91 84 Power Series Solutions: Levels of Success When using power series techniques to solve linear differential equations, ideally one would like to have a closed form solution for the final answer, but most often one has to settle for much less There are various levels of success, listed below, and one would like to get as far down the list as one can For simplicity, we will assume that we have chosen to expand about the point x =, although the same considerations apply to x = c for any c Suppose is an ordinary point of the equation L(x)y +M(x)y +N(x)y = and let y = a nx n be a solution One might hope to: 1 find the coefficients a n for a few small values of n; (For example, find the coefficients as far as a 5 which thereby determines the 5th degree Taylor approximation to y(x)); 2 find a recursion formula which for any n gives a n+2 in terms of a n 1 and a n ; 3 find a general formula a n in terms of n; 4 recognize the resulting series to get a closed form formula for y(x) In general, level (1) can always be achieved If L(x), M(x), and N(x) are polynomials, then level (2) can be achieved Having achieved level (2), it is sometimes possible to get to level (3) but not always Cases like the one in Example 81 where one can achieve level (4) are rare and usually beyond expectation 85 Level 1: Finding a finite number of coefficients Suppose c is an ordinary point of Equation (83) One approach is to finding power series solutions is as follows Expand P (x) and Q(x) into Taylor series P (x) = p n (x c) n, Q(x) = q n (x c) n and set y = a n (x c) n solves the equation Then y = na n (x c) n 1, y = n (n 1) a n x n 2,

92CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS and Equation (83) becomes P (x) y { }} { {( }} { ) {( }} { ) n (n 1) a n x n 2 + p n (x c) n na n (x c) n 1 ( ) ( ) + q n (x c) n a n (x c) n = } {{ } Q(x) y } {{ } y Since p n s and q n s are known (as many as desired can be computed from derivatives of P (x) and Q(x)), in theory we can inductively find the a n s one after another In the special case where P (x) and Q(x) are polynomials, we can get a recursion formula by writing a n+2 in terms of a n and a n 1 (level 2) and, under favourable circumstances, we can find the general formula for a n (level 3) as in Example 81 But in general, computation gets impossibly messy after a while and, although a few a n s can be found, the general formula cannot But if you only want a few a n s, there is an easier way We have a n = y (n) (c)/n!, so and so on a 2 = y (c) = P (x)y (c) + Q(c)y(c) 2! 2 a 3 = y (c) = 1 6 a 4 =, = a 1P (c) a Q(c), 2 = 1 3! 6 ( P (c)y (c) P (c)y Q (c)y(c) Q(c)y (c)) ( P (c)a 1 + a 1P (c) 2 ) + a P (c)q(c) a Q (c) a 1 Q(c), 2 Example 88 Compute the Taylor expansion about as far as degree 4 for the solution of y e 7x y + xy =, which satisfies y() = 2 and y () = 1 Solution Rearranging the equation, we have y = e 5x y xy Therefore y = e 5x y + 5e 5x y xy y y (4) = e 5x y + 5e 5x y + 5e 5x y xy y y = e 5x y + 1e 5x y xy 2y

85 LEVEL 1: FINDING A FINITE NUMBER OF COEFFICIENTS 93 Substituting gives y () = e y () = 1 y () = e y () + 5e y () y() = 1 + 5 2 = 4 y (4) () = e y () + 1e y () 2y () = 4 + 1 2 = 12 Therefore, y = 2 + x + x2 2! + 4x3 + 12x4 + 3! 4! = 2 + x + 1 2 x2 + 2 3 x3 1 2 x4 + Example 89 Find the series solution for y + (x + 3) y 2y =, where y() = 1 and y () = 2 Solution Rearranging the equation, we have y = (x + 3) y + 2y Then Therefore, y () = ( + 3) y () + 2y() = 3 2 + 2 1 = 4, y = y (x + 3) y + 2y = (x + 3) y + y, y () = 3y () + y () = ( 3) ( 4) + 2 1 = 12 + 2 = 14, y (4) = = 34 y = 1 + 2x 4x2 2! + 14x3 3! 34x4 4! + = 1 + 2x 2x 2 + 7 3 x3 17 12 x4 + In this case, since the coefficients were polynomials, it would have been possible to achieve level 2, (although the question asked only for a level 1 solution) However even had we worked out the recursion formula (level 2) it would have been too messy to use it to find a general formula for a n

94CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS 86 Level 2: Finding the recursion relation Example 81 Consider ( x 2 5x + 6 ) y 5y 2y =, where y() = 1 and y () = 1 Find the series solution as far as x 3 and determine the lower bound on its radius of convergence from the recursion formula Solution We have 2y = 2a n x n, 5y = 5na n x n 1 = 5 (n + 1) a n+1 x n = 5 (n + 1) a n+1 x n, 6y = 6n (n 1) a n x n 2 = 6 (n + 2) (n + 1) a n+2 x n, 5xy = 5n (n 1) a n x n 1 = 5 (n + 1) na n+1 x n, x 2 y = n (n 1) a n x n Therefore, n (n 1) a n 5 (n + 1) na n+1 +6 (n + 2) (n + 1) a n+2 xn = 5 (n + 1) a n+1 2a n 6 (n + 2) (n + 1) a n+2 5 (n + 1) (n + 1) a n + ( n 2 n 2 ) a n x n, so it follows that 6 (n + 2) (n + 1) a n+2 5 (n + 1) 2 a n+1 + (n + 1) (n 2) a n =, a n+2 = 5 6 n + 1 n + 2 a n+1 1 6 n 2 n + 2 a n With a = y() = 1 and a 1 = y () = 1, we have a 2 = 5 6 1 2 a 1 1 6 2 2 a = 5 12 + 2 12 = 7 12, a 3 = 5 6 2 3 1 6 1 3 a 1 = 5 9 7 12 + 1 6 1 3 = 35 18 + 6 18 = 41 18 Therefore, y = 1 + x + 7 12 x2 + 41 18 x3 +

86 LEVEL 2: FINDING THE RECURSION RELATION 95 Notice that although we need to write our equation in standard form (in this case dividing by x 2 5x + 6 to determine that is an ordinary point of the equation), it is not necessary to use standard form when computing the recursion relation Instead we want to use a form in which our coefficients are polynomials Although our recursion relation is too messy for us to achieve find a general formula for a n (level 3) there are nevertheless some advantages in computing it In particular, we can now proceed to use it to find the radius of convergence To determine the radius of convergence, we use the ratio test R = 1/ L, where L = lim n (a n+1 /a n ), provided that the limit exists It is hard to prove that the limit exists, but assuming that taking the limit gives a n+2 = 5 a n+1 6 n + 1 n + 2 1 6 n 2 n + 2 a n, a n+1 L = 5 6 1 1 6 1 L = 5 6 1 6 1 L So 6L 2 = 5L 1, and solving for L gives L = 1/2 or L = 1/3 So R = 2 or R = 3 The worst case is when R = 2 Therefore, the radius of convergence is at least 2 Theorem 83 also gives this immediately Example 811 Consider ( x 2 5x + 6 ) y 5y 2y = Find two linearly independent series solutions y 1 and y 2 as far as x 3 Solution Let y 1 (x) be the solution satisfying y 1 () = 1; y 1() = and let y 2 (x) be the solution satisfying y 2 () = ; y 2() = 1 Then y 1 (x) and y 2 (x) will be linearly independent solutions We already computed the recursion relation in Example 81 For y 1 (x), we have a = y() = 1 and a 1 = y () = Thus, a 2 = 5 6 1 2 a 1 1 ( 2 ) a = 1 6 2 6, a 3 = 5 6 2 3 a 2 1 6 1 3 a 1 = 5 54 Therefore, y 1 = 1 + x2 6 + 5 54 x3 +

96CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS For y 2, we have a = y() = and a 1 = y () = 1, so Therefore, With a 2 = 5 6 1 2 a 1 2 2 a = 5 12, a 3 = 5 6 2 ( 3 a 2 1 3 ) a 1 = 5 6 2 3 y 2 = x + 5 12 x2 + 61 18 x3 + y = 1 + x + 7 12 x2 + 41 18 x3 +, 5 12 + 1 6 3 = 25 18 + 6 18 = 31 18 Looking at the initial values, the solution y(x) of Example 81 ought to be given by y = y 1 +y 2, and comparing are answers we see that they are consistent with this equation Summarizing: Let p be a regular point of the equation y + P (x)y + Q(x)y =, (83) Let D be the distance from p to the closest singularities of P or Q in the complex plane Then Equation (83) has two linearly independent solutions expressible as a convergent power series a n (x p) n with x p < D Theorem 85 guarantees that the radius of convergence is at least D, but it might be larger Given a = y(p) and a 1 = y (p), to find a n, we have a n = y (n) (p)/n! We can find as many a n s as desired, but in general we cannot find the formula Since we have y = P (x)y Q(x)y, it can be differentiated to give y (p) and higher derivatives can be found by further differentiation Substituting the values at p gives formulas for the a n s In the special case where P (x) and Q(x) are polynomials, we can instead find a recurrence relation From the recurrence relation there is a chance to get the formula for a n (if so, then there is a chance to actually recognize the series) Even without the formula, we can determine the radius of convergence from the recurrence relation (even if P (x) and Q(x) are not polynomials, we could at least try to get a recurrence relation by expanding them into Taylor series,but in practice this is not likely to be successful)

87 SOLUTIONS NEAR A SINGULAR POINT 97 87 Solutions Near a Singular Point A singularity a of y + P (x)y + Q(x)y = is called a regular singularity if (x a) P (x) and (x a) 2 Q(x) are analytic at x = a To simplify notation, we assume that a = (we can always change the variable by v = x a to move the singularity to the origin) We assume that x > For x <, we substitute t = x and apply below to t Our primary goal in this section is to consider the Method of Frobenius We look for solutions of the form y = x α (a + a 1 x + + a n x n + ) = a x α + a 1 x α+1 + + a n x α+n + for some α, where a Then y = a 1 αx α 1 + a 1 (α + 1) x α + + a n (α + n) x α+n 1 +, y = a 1 α (α 1) x α 2 + a 1 α (α + 1) x α 1 + + a n (α + n) (α + n 1) x α+n 2 +, xp (x) = p + p 1 x + + p n x n +, x 2 Q(x) = q + q 1 x + + q n x n + So if x, then a α (α 1) x α 2 + a 1 (α + 1) αx α 1 + + a n (α + n) (α + n 1) x α+n 2 + +x α 2 (a α + a 1 (α + 1) x + ) (p + p 1 x + + p n x n + ) +x α 2 (a + a 1 x + + a n x n ) (q + q 1 x + + q n x n + ) = We have a α (α 1) + a αp + a q =, () a 1 (α + 1) α + a 1 (α + 1) p + a αp 1 + a q 1 + a 1 q =, (1) Note that Equation () implies (n) α (α 1) + αp + q =, α 2 + (p 1) α + q = (84)

98CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Equation (84) is known as an indicial equation Solving for α gives us two solutions Then Equation (1) gives a 1 in terms of a, Equation (2) gives a 2 in terms of a 1, etc We will get the solution y = a x α ( ) Other α gives a second solution y = b x β ( ) The plan is to find a n from Equation (n) for n 1 Equation (n) looks like ( ) a n + ( ) a n 1 + + ( ) a =, where the coefficients are in terms of α, p,, p n, q,, q n There is one possible problem: what happens if the coefficient of a n is zero? Note that so we have y + P (x)y + Q(x)y = y + xp (x) y x + x2 Q(x) y x 2, y + xp (x) y x + x2 Q(x) y x 2 = a n (α + n) (α + n 1) x α+n 2 In Equation (n), the coefficient of a n is + ( p + p 1 x + p 2 x 2 + ) a n (α + n) x α+n 2 + ( q + q 1 x + q 2 x 2 + ) a n x α+n 2 (α + n) (α + n 1) + p (α + n) + q Suppose that (α + n) (α + n 1) + p (α + n) + q = Let β = α + n Then β (β 1) + p β + q = is the indicial equation again, ie, β is the other solution to the indicial equation We conclude that if two solutions of the indicial equation do not differ by an integer, the method of Frobenius provides two solutions If the two solutions do differ by an integer, then the larger one will give a solution by this method, but the smaller one may fail Let F (λ) = λ (λ 1) + p λ + q Suppose the solutions to F (λ) = are r This includes the case where the solutions are complex, ie, in this case, we get a complex solution z = x a+ib (c + c 1 x + ), so y 1 = Re(x) and y 2 = Im(z) give two real solutions

87 SOLUTIONS NEAR A SINGULAR POINT 99 and r + k, where k Z + Given f(x), set L(f)(x) = f (x) + P (x)f (x) + Q(x)f(x) Let w = a nx n, where a n is to be chosen For any given α, we have L(x α w) = c n x α 2+n, where c n = F (α + n)a n + ( ) a n 1 + + ( ) a Chose a = 1 Then for n 1, we can always solve c n = for a n in terms of its predecessors unless F (α + n) = Provided that α s 1, s 2,, we have F (α + n) for any n, so we can solve it Suppose that we call the solution a n (α), ie, a (α) = 1 for any α and a n is chosen to make c n = for this α Set W (α, x) a n(α)x n Then L(x α w(α, x)) = c n x α 2+n = c x α 2 But since c n = for n by the choice of a n (α), we have L(x α w(α, x)) = F (α)x α 2 Since F (s) =, y 1 (x) = x s w(s, x) is the solution to the de We now need a second solution Unfortunately, we cannot set α = r since our definition of a n (α) depends on α s (positive integer) Instead, set L(x α w(α, x)) = F (α)x α 2 and differentiated with respect to α Note that for g(α, x), we have ( ) g x (Lg) = L x since differentiation with respect to x commutes with differentiation with respect to α (cross derivatives equal) Therefore, ( L x α ln(x)w(α, x) + x α w ) = F (α)x α 2 + F (α)x α 2 ln(x) α

1CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS But note that ( ( )) k w { }} { {}}{ α = s = L ln(x)x s w(s, x) + x s α = F (s) x s 2 + F (s) x s 2 ln(x) ( ( ) ) α=s w = L ln(x)y 1 + α x s = kx s 2 α=s If k =, then is the second solution to our de If k >, consider L ( x r ( ) w y 2 = ln(x)y 1 + α α=s a n x n ) = L ( x r x s c n x r+n 2 ) where a n s are to be chosen and c n = F (r + n)a n + We have c = F (r) =, where r is the root of the indicial equation Select a = 1 and inductively choose a 1,, a n such that c 1,, c n = With this choice, we have L ( x r a n x n ) = c n x r+n 2 = c k x s 2 + n=k n=k+1 where c k is independent of the choice of a k Set A = c k /k Then L ( x r ( w )) a n x n A ln(x)y 1 (x) + x s α = α=s a + a 1 x + + a k 1 x k 1 x r ( + a L k A w ) α x k α=s = + a n x n A ln(x)y 1 (x) n=k+1, c n (r)x r+n 2, n=k+1 n=k+1 c n x r+n 2, c n x r+n 2 The choice of a k does not matter we can choose any Then inductively solve for a k+1, a k+2, in terms of predecessor a s to make c n = for n > k This gives a solution to the de of the form y 2 = x r a n x n A ln(x)y 1 (x)

87 SOLUTIONS NEAR A SINGULAR POINT 11 We conclude that if is a regular singularity, then there are always two linearly independent solutions (converging near ) y 1 = x s a n x n, y 2 = x r b n x n + B(ln(x))y 1, where r and s are the two solutions to the indicial equation We have B =, unless s = r + k for some k {, 1, 2, } Example 812 Solve 4xy + 2y y = Solution Let y = a nx α+n Then y = (α + n) a n x α+n 1, y = (α + n) (α + n 1) a n x α+n 2 Our differential equation now becomes 4 (α + n) (α + n 1) a n x α+n 1 + 2 (α + n) a n x α+n 1 a n x α+n =, Note that 4 (α + n) (α + n 1) + 2 (α + n) a n x α+n 1 a n 1 x α+n 1 = n=1 4 (α + n) (α + n 1) + 2 (α + n) = 2 (α + n) (2α + 2n 2 + 1) = 2 (α + n) (2α + 2n 1) Therefore, 2α (2α 1) a x α 1 + 2 (α + n) (2α + 2n 1) a n 1 x α+n 1 n=1 Therefore, the indicial equation is 2α (α 1) α {, 1/2} and the recursion relation is 2 (α + n) (2α + 2n 1) a n a n 1 = for n 1 For α =, we have 2n (2n 1) a n a n 1 = = a n = 1 2n (2n 1) a n 1

12CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Thus, a n = a 1 = 1 2 1 a, a 2 = 1 4 3 a 1 = 1 4! a, 1 (2n) (2n 1) a 1 n 1 = 2n (2n 1) (2n 2) = 1 (2n)! a Therefore, ( y = x a 1 + x ) 2! + x2 4! + + xn (2n)! + ( = a 1 + x ) 2! + x2 4! + + xn (2n)! + = a cosh ( x ) Thus, For α = 1/2, we have ( ) 1 2 2 + n 1 (1 + 2n 1) b n = b n 1 = b n = (2n + 1) (2n) b n 1 b 1 = 1 3 2 b, b 2 = 1 5 4 b 1 = 1 5! b, b n = 1 (2n + 1)! b

87 SOLUTIONS NEAR A SINGULAR POINT 13 Therefore, ( y = x 1/2 b 1 + x 3! + x2 5! + + x n ) (2n + 1)! + ( = b x 1/2 + x3/2 + x5/2 + + x(2n+1)/2 3! 5! (2n + 1)! + = b sinh ( x ) ) In general, however, it is possible to only calculate a few low terms We cannot even get a general formula for a n, let alone recognize the series As mentioned, the above methods work near a by using the above technique after the substitution v = x a We can also get solutions near infinity, ie, for large x, we substitute t = 1/x and look at t = Then Then dy dx = dy dt dt dx = dy ( 1x ) dx 2 = 1 dy x 2 dt d 2 y dx 2 = 2 dy x 3 dt = 1 d 2 y dt x 2 dt 2 dx = 2 dy x 3 dt 1 d 2 y x 2 dt 2 d 2 y dy + P (x) dx2 dx + Q(x)y = = t4 d2 y dy + 2t3 dt2 dt P dy = t2 dt, ( 1x ) 2 = 2t 3 dy dt + t4 d2 y dt 2 ( ) 1 t 2 dy t dt + Q ( 1 t ) y = Example 813 (Legendre s equation) Find the behaviour of the solutions for large x of ( 1 x 2) y 2xy + n (n + 1) y =, where n is a constant Solution Let t = 1/x so that x = 1/t Then our differential equation becomes (1 1t ) ( 2 2t 3 dy ) dt + t4 d2 y dt 2 2 ( t 2 dy ) + n (n + 1) y =, t dt ( t 4 t 2) d 2 y dt 2 + ( 2t 3 2t + 2y ) dy + n (n + 1) y =, dt d 2 y dt 2 + 2t t 2 1 dy n (n + 1) + dt t 4 t 2 y = We see that we have a regular singularity at t = Therefore, there exists a solution y = t α a nt n near t =, or equivalently, since t = 1/x, y = 1 x α 1 x n

14CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS for large x Example 814 Find the solutions of y + xy + y/x = for positive x near Solution Let Then y = x α a n x n = a n x n+α y = a n (n + α) x n+α 1, y = a n (n + α) (n + α 1) x n+α 2, xy = a n x n+α = a n 2 x n+α 2, n=2 y x = a n x n+α 1 = a n 1 x n+α 2 n=1 Note that Indicial equation { }} { a α (α 1) =, () a 1 α (α + 1) α + a =, (1) a n (n + α) (n + α 1) + a n 1 + a n 2 = (n) Note that Equation () implies that α =, 1 The roots differ by an integer The largest always gives the solution For α = 1, Equation (1) implies that 2a 1 + a =, so a 1 = a /2 Equation (n) then becomes a n (n + 1) n + a n 1 + a n 2 = = a n = a n 1 a n 2 n (n + 1) Equation (2) implies that ( 6a 2 = a a 1 = a 1 + 1 ) = a 2 2 = a 2 = a 12 ;

87 SOLUTIONS NEAR A SINGULAR POINT 15 Equation (3) implies that 12a 3 = a ( 1 2 + 1 12 ) = a 7 12 = a 3 = a 7 144 Therefore, ( y 1 = a x 1 1 2 x 1 12 x2 + 7 ) 144 x3 + As for the second solution, we have y = x ( b + b 1 x + b 2 x 2 + ) + C ln(x)y 1, y = b 1 + 2b 2 x + + C x y 1 + C ln(x)y 1, y = 2b 2 + 6b 3 x + C x 2 y 1 + C x y 1 + C x y 1 + C ln(x)y 1 Therefore, our differential equation becomes y + xy + y x =, ( 2b2 + 6b 3 x + 24b 4 x 2 + ) C x 2 y 1 + 2C x y 1 + C ln(x)y 1 + ( b 1 x + 2b 2 x 2 + ) + Cy 1 + C ln(x)xy ( 1 ) b + x + b 1 + b 2 x + + C ln(x) y 1 =, x ( 2b2 + 6b 3 x + 24b 4 x 2 + ) C x 2 y 1 + 2C x y 1 + ( b 1 x + 2b 2 x 2 + ) ( ) b + Cy 1 + x + b 1 + b 2 x + + C ln(x)y 1 + C ln(x)xy 1 + C ln(x) y =, 1 } {{ x } ( 2b2 + 6b 3 x + 24b 4 x 2 + ) C ( 1 1 x 2 x 1 12 x2 + 7 + 2C ( 1 x + 1 x 4 x2 + 7 ) 36 x3 + ( x 1 2 x2 1 12 x3 + 7 144 x4 + + ( b 1 x + 2b 2 x 2 + ) + C ( ) b + x + b 1 + b 2 x + ) 144 x3 + ) = Looking at the coefficient of x 1, we have b C + 2C = C = b Note that b 1 is arbitrary, so choose b = Then the coefficient of x implies that 2b 2 + C/2 2C = b 2 = (3/4) b ; the coefficient of x 1 implies that

16CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS 6b 3 + C/12 + C/2 + C + b 2 = b 3 = (7/18) b Therefore, y 2 = b (1 3 4 x2 + 7 ) ( 18 x2 + ln(x) x 1 2 x2 1 12 x3 + 7 ) 144 x4 + Example 815 Solve x 2 y xy + 1e x y = Solution Dividing out by x 2, we have y y x + 1ex x 2 y = We have a singularity at x = Note that xp (x) = 1 and x 2 Q(x) = 1e x = 1 xn /n! Therefore, y = y = a n x α+n, (α + n) a n x α+n 1, y = y x 2 = a n x α+n 2, y x = (α + n) a n x α+n 2, (α + n) (α + n 1) a n x α+n 2 With these components in place, our differential equation becomes (α + n) (α + n 1) a n x α+n 2 (α + n) a n x α+n 2 +1 ( x n n! ) ( ) a n x α+n 2 = We now have α (α 1) a αa + 1a = Therefore, F (α) = α (α 1) α + 1 = α = 1 ± 3i For 1 + 3i, we have (α + 1) αa 1 (α + 1) a 1 + 1 (a + a 1 ) =, F (α + 1)a 1 + 1a 1 = Therefore, 1 a 1 = F (α + 1) a 1 = F (2 + 3i) a

87 SOLUTIONS NEAR A SINGULAR POINT 17 and Therefore, F (2 + 3i) = (2 + 3i) 2 2 (2 + 3i) + 1 = 4 + 12i 9 4 6i + 1 = 1 + 6i a 1 = 1 1 + 6i a 1 (1 6i) = (1 + 6i) (1 6i) a = 1 6i 1 + 36 a = 1 6i a 37 Also note that and we have Therefore, and we have But so ( a ) (α + 2) (α + 1) a 2 (α + 2) a 2 + 1 2! + a 1 + a 2 =, 1 6i F (α + 2)a 2 + 1a 1 + 5a = F (α + 2)a 2 a + 175 37 37 a 75 + 6i = F (α + 2)a 2 + a 37 a 2 = 75 + 6i 37F (α + 2) a ( ) ) 1 6i z = x 1+3i a (1 x + 37 x 1+3i = xx 3i = xe 3i ln(x) = x cos (3 ln(x)) + i sin (3 ln(x)), z = a x cos(3 ln(x)) + a i sin(3 ln(x)) ( + a 1 37 x2 cos(3 ln(x)) 6 37 x2 sin(3 ln(x)) + 6 ) sin(3 ln(x)) 37 + 6 37 ix2 cos(3 ln(x)) 1 37 ix2 sin(3 ln(x)) +

18CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Therefore, Re(z) = a x cos(3 ( ln(x)) +x 2 1 37 6 cos(3 ln(x)) 37 +x 3 cos(3 ln(x)) + sin(3 ln(x)) ) sin(3 ln(x)) + Example 816 Solve xy + (1 x) y y = Solution Note that and p = lim xp (x) = lim x 1 x = 1 x x x ( q = lim x 2 Q(x) = lim x 2 1 ) = x x x Therefore, is a regular singular point The indicial equation is Therefore, α = is a double root and y 1 α (α 1) + p α + q =, x n, y 2 = α (α 1) + α =, α 2 α + α =, α 2 = b n x n + C ln(x)y 1, x > To find y 1, we have y = a n x n, y = na n x n 1 = (n + 1) a n+1 x n, y = (n + 1) nx n 1

87 SOLUTIONS NEAR A SINGULAR POINT 19 Therefore, our differential equation becomes xy + (1 x) y y =, (n + 1) na n+1 + (n + 1) a n+1 na n a n x n = From this, we have Therefore, So we have (n + 1) na n+1 + (n + 1) a n+1 na n a n =, (n + 1) 2 a n+1 = (n + 1) a n, a n+1 = a n n + 1 a 1 = a 1, a 2 = a 1 2 = a 2, a 3 = a 2 2 = a 3!,, a n = a n! y = a x n n! = a e x Immediately, y 1 = e x To find y 1 (see 51, p 57), we have R e P (x) dx R e y 2 = y 1, dx = e x 1 x x dx dx y 2 1 R e = e x ( 1 x 1)dx e 2x = e x e x /x e 2x dx = ex e 2x e dx = e x (ln(x) x) e 2x e x x dx, dx which is not an elementary function Example 817 Solve ( x 2 + 2x ) y 2 ( x 2 + 2x 1 ) y + ( x 2 + 2x 2 ) y = Solution Let y = a nx α+n Then y = (α + n) a n x α+n 1, y = (α + n) (α + n 1) a n x α+n 2

11CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS The de then becomes (α + n) (α + n 1) a n x α+n + 2 (α + n) (α + n 1) a n x α+n 1 2 (α + n) a n x α+n+1 4 (α + n) a n x α+n =, +2 (α + n) a n x α+n 1 + a n x α+n+2 + 2 a n x α+n+1 2 a n x α+n (α + n 1) (α + n 2) a n 1 x α+n 1 n=1 +2 (α + n) (α + n 1) a n x α+n 1 2 (α + n 2) a n 2 x α+n 1 n=2 =, 4 (α + n 1) a n 1 x α+n 1 + 2 (α + n) a n x α+n 1 n=1 + a n 3 x α+n 1 + 2 a n 2 x α+n 1 2 a n 1 x α+n 1 n=3 n=2 2 (α + n) (α + n 1) + 2 (α + n) a n + (α + n 1) (α + n 2) 4 (α + n 1) 2 a n 1 x α+n 1 =, + 2 (α + n 2) + 2 a n 2 + a n 3 2 (α + n)2 a n + (α + n) 2 7 (α + n) + 4 a n 1 = + 2 (α + n) + 6 a n 2 + a n 3 Note that 2α 2 = α =, giving us a double root Therefore we get the recursion relation Substituting gives n=1 2n 2 a n + ( n 2 7n + 4 ) a n 1 + ( 2n + 6) a n 2 + a n 3 = 2a 1 2a = = a 1 = a, 8a 2 6a 1 + 2a = = 8a 2 = 6a 1 2a = (6 2) a = 4a = a = 1 2, 18a 3 + ( 8) a 2 + () a 1 + a 2 = = 18a 3 = 8 1 2 1 = 3 = a 3 = 1 6, 32a 4 8a 3 ( 2) a 2 + () a 1 + a 2 = = 32a 3 = 8 1 6 2 1 2 1 = 4 3 = a 3 = 1 24

88 FUNCTIONS DEFINED VIA DIFFERENTIAL EQUATIONS 111 If we our intuition leads us to guess that perhaps a n = 1/n! be can use induction to check that this guess is in fact correct Explicitly, suppose by induction that a k = 1/k! for all k < n Then the recursion relation yields 2n 2 a n = n2 7n + 4 (n 1)! + 2n 6 (n 2)! 1 (n 3)! = n2 + 7n 4 + (2n 6)(n 1) (n 1)(n 2) (n 1)! = n2 + 7n 4 + 2n 2 8n + 6 n 2 + 3n 2 (n 1)! = 2n (n 1)! and so a n = 1/n!, completing the induction This gives the solution y 1 = xn /n! = e x (If we had guessed initially that this might be the solution, we could easily have checked it by direct subsitution into the differential equation) Using reduction of order, one then finds that the solutions are y 1 = e x, y 2 = xe x + 2 ln(x) y 1 88 Functions Defined via Differential Equations 881 Chebyshev Equation The Chebyshev equation is given by ( 1 x 2 ) y xy + λ 2 y = (85)

112CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS of which is an ordinary point The series solution is guaranteed to converge for x < 1 We have y = y = y = a n x n, na n x n 1, n (n 1) a n x n 2 = and our differential equation becomes (n + 2) (n + 1) a n+2 x n, (n + 2) (n + 1) an+2 n (n 1) a n na n + λ 2 a n x n = From this, we see that Therefore, a n+2 = n (n 1) + n λ2 n 2 λ 2 a n = (n + 2) (n + 1) (n + 2) (n + 1) a n a = 1, a 1 =, a 2 = λ2 2, a 3 =, ( ) ( a 4 = λ2 4 λ 2 4 3 = λ 2), a 5 =, ( ) ( 4! 16 λ 2 4 λ 2) ( λ 2) a 6 =, 6! ) ) (4 (n 1) 2 λ (4 2 (n 2) 2 ( λ λ 2 2) a 2n = a 1 (2n)!

88 FUNCTIONS DEFINED VIA DIFFERENTIAL EQUATIONS 113 Also, ( ) 1 λ 2 a =, a 1 = 1, a 2 =, a 3 = a 1, ( ) ( 3 2 9 λ 2 1 λ 2) a 4 =, a 5 = a 1, 5! ) ((2n 1) 2 λ 2 (1 λ 2) a 2n+1 = a 1 (2n + 1)! Note that if λ is an integer, then one of these solutions terminates, ie, a k = beyond some point In this case, one solution is a polynomial known as a Chebyshev polynomial 882 Legendre Equation The Legendre equation is given by ( 1 x 2 ) y 2xy + λ (λ + 1) y = (86) of which is an ordinary point The series solution is guaranteed to converge for x < 1 Expressed as a power series, the equation becomes = (n + 2) (n + 1) a n+2 n (n 1) a n 2na n + λ (λ + 1) a n x n = From this, we have a n+2 = n (n 1) + 2n λ (λ + 1) a n = (n + 2) (n + 1) n (n + 1) λ (λ + 1) a n (n + 2) (n + 1) If λ is an integer, then a k = for k λ gives the λth Legendre polynomial Instead, expanding around x = 1 gives p(x) x 1 x 2

114CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Then x p = lim (x 1) P (x) = lim x 1 x 1 x + 1 = 1 2, q = lim x 1 (x 1) 2 Q(x) = lim x 1 (x 1) 2 λ 2 x 2 1 = Therefore, x = 1 is a regular singular point The indicial equation is α (α 1) + p α + q =, α (α 1) + 1 2 α + q =, α 2 α + 1 2 α =, α 2 1 2 α = Therefore, α = or α = 1/2, so the solutions look like for x > 1 y 1 = a n (x 1) n, y 2 = (x 1) 1/2 b n (x 1) n For a Legendre equation, around x = 1, we have ( p = lim (x 1) 2x ) x 1 1 x 2 = 1, q = lim x 1 (x 1) 2 λ (λ + 1) x 2 1 = Therefore, x = 1 is a regular singular point The indicial equation is α (α 1) + p α + q =, α (α 1) + α =, α 2 = Therefore, α = is a double root and the solutions look like for x > 1 y 1 = a n (x 1) n, y 2 = b n (x 1) n + C ln(x 1) y 1

88 FUNCTIONS DEFINED VIA DIFFERENTIAL EQUATIONS 115 883 Airy Equation The Airy equation is given by y xy = (87) of which is an ordinary point The solution converges for all x Expressed as a power series, we have (n + 2) (n + 1) a n+2 a n 1 x n = From this, we see that therefore, a n+2 = a n 1 (n + 2) (n + 1), a 3 = a 3 2, a 6 = a 6 5,, a 3n = a 4 = a 1 4 3,, a 3n+1 = a 2 3 5 6 8 9 (3n 1) (3n), a 1 3 4 6 7 (3n) (3n + 1), a 2 = a 1 2 =, = a 2 = a 5 = a 8 = 884 Laguerre s Equation Laguerre s equation is given by of which is a regular singular point Note that xy (1 x) y + λy = (88) ( ) (1 x) p = lim x = 1, x x q = lim x x 2 λ x = Applicable in the context of the hydrogen atom

116CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Therefore, x = is a regular singular point The indicial equation is α (α 1) α =, α 2 α α =, α 2 2α =, α (α 2) = Therefore, α = or α = 2, so the solutions look like for x > y 1 = x 2 a n x n, y 2 = b n x n + C ln(x)y 1 885 Bessel Equation The Bessel equation is given by We have The indicial equation is then x 2 y + xy + ( x 2 λ 2) y = (89) p = lim x x x x 2 = 1, q = lim x 2 x2 λ 2 x x 2 = λ α (α 1) + α λ 2 =, α 2 λ 2 =, (α + λ) (α λ) = Therefore, α = ±λ If λ k/2 for some integer k, then we get y 1 = x λ a n x n, y 2 = x λ a n x n

88 FUNCTIONS DEFINED VIA DIFFERENTIAL EQUATIONS 117 for x > We encounter trouble if λ = k/2 Consider λ = 1/2 Then For y 1, we have y 1 = x 1/2 a n x n, y 2 = x 1/2 b n x n + C ln(x)y 1 y = a n x n+1/2, ( y = n + 1 ) a n x n 1/2, 2 ( y = n + 1 ) ( n 1 ) a n x n 3/2 2 2 } {{ } n 2 1/4 Therefore, the equation x 2 y + xy + ( x 2 1/4 ) y = becomes From this, we see that Therefore, ( n 2 1 ) ( a n + n + 1 ) a n + a n 2 1 4 2 4 a n x n+1/2 = ( n 2 1 4 + n + 1 2 1 ) a n + a n 2 =, 4 a n = a n 2 n 2 + n = a n 2 n (n + 1) a 2 = a 3 2, a 4 = a 2 5 4 = a 5!, a 6 = a 7!,, a 2n = ( 1) n a (2n + 1)! Therefore y 1 = ( 1) n x 2n+1/2 (2n + 1)! = 1 x ( 1) n x 2n+1 (2n + 1)! = sin(x) x

118CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS Example 818 Identify y = x 3n (3n)! = 1 + x3 3! + x6 6! + x9 9! + Solution We seek to solve y = y From this, we see that Therefore, λ = 1 or λ 3 1 =, (λ 1) ( λ 2 + λ + 1 ) = λ = 1 ± 1 4 2 Therefore, the general solution is = 1 2 ± 3 2 i ( ) ( ) 3 3 y = C 1 e x + C 2 e x/2 sin 2 x + C 3 e x/2 cos 2 x, where C 1, C 2, C 3 are arbitrary constants To find C 1, C 2, C 3, we note that Therefore, y() = 1, y () =, y () = C 1 + C 2 + C 3 = 1, 3 C 1 + 2 C 2 1 2 C 3 =, 3 C 1 2 C 2 1 2 C 3 =

88 FUNCTIONS DEFINED VIA DIFFERENTIAL EQUATIONS 119 Solving this system, we have 1 1 1 1 3 2 1 2 1 3 2 1 2 1 1 1 1 2 3 1 3 2 1 3 1 1 1 3 2 3 2 3 2 1 3 2 1 1 1 3 1 3 1 2 3 1 1 1 3 3 2 2 1 3 2 1 1 3 1 1 2 3 Therefore, C 1 = 1/3, C 2 =, and C 3 = 2/3, so ( ) y = 1 3 ex + 2 3 3 e x/2 cos 2 x

12CHAPTER 8 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS

Chapter 9 Linear Systems 91 Preliminaries Consider where A = a c dy dx = AY + B, b, Y = d y 1 y 2 such that dy 1 dx = ay 1 + by 2, dy 2 dx = cy 1 + dy 2 We will first consider the homogeneous case where B = Only in the case when the entries of A are constants is there an algorithmic method of finding the solution If A a is a constant, then we know that y = Ce ax is a solution to dy/dx = ay We will define a matrix e A such that the general solution of dy/dx = AY is Y = Ce xa, where C = C 1, C 2,, C n is a vector of arbitrary constants We define e T for an arbitrary n n matrix T and apply T = xa Indeed, we know that Therefore, let us analogously define e t = 1 + t + t2 2! + t3 3! + t4 4! + + tn n! + e T I + T + T2 2! + T3 3! 121 + + Tn n! + (91)

122 CHAPTER 9 LINEAR SYSTEMS We will consider systematically how to compute e T later, but first we consider an ad hoc example Example 91 Solve dy dx = 1 1 1 Y Solution First, we write where A = 1 1 1 N = 1 = I + N, so that NI = IN = N and N 2 = Note that Binomial exp { }} { A n = (I + N) n ( ) n = I n + I n 1 N + 1 = I + nn = 1 n 1 ( ) n I n 1 N 2 + + 2 ( ) n IN n 1 + N n n 1 Therefore, with A n properly defined, Equation (91) gives e Ax 1 1 = I + x + 1 = = 1 2 1 x 2 2 + + 1 n 1 x n n! + 1 + x + x 2 2! + + xn n! + + x + 2 x2 2! + 3 x3 3! + + n xn n! ex 1 + x + x2 2! + + xn n! + ( ) x 1 + x + x2 2! + + xn n! + e x xe x = e x e x Therefore, the solution is y 1 y 2 = e x xe x e x C 1 C 2 = C 1 e x + C 2 xe x C 2 e x,

92 COMPUTING E T 123 that is, y 1 = C 1 e x + C 2 xe x, y 2 = C 2 e x Is this in fact the solution of the equation? We have d dx eax = d ) (I + Ax + A 2 x2 xn + + An dx 2! n! + = A + A 2 x + + A n x n 1 = Ae Ax (n 1)! + So if Y = Ce Ax, then dy dx = ACeAx = AY, as required Therefore, Y = Ce Ax is indeed the solution 92 Computing e T The easiest case to do is a diagonal matrix If λ 1 λ 2 D =, λ n then e D = e λ1 e λ2 e λn First, we consider the effect of changing bases Suppose T is diagonalizable, that is, there exists a P such that P 1 TP is diagonalizable Therefore, T = PDP 1 and T 2 = ( PDP 1) ( PDP 1) = PD 2 P 1

124 CHAPTER 9 LINEAR SYSTEMS So in general, we have T n = PD n P 1 It then follows that e T = I + T + T2 + + Tn 2! n! = P (I + D + D2 + + Dn 2! n! = Pe D P 1 So if T is diagonalizable, we can compute e T ) + P 1 Example 92 Compute e T, where T = 4 1 3 2 Solution First note that 4 λ 1 3 2 λ =, (4 λ) (2 λ) 3 =, 8 6λ + λ 2 3 =, λ 2 6λ + 5 =, (λ 5) (λ 1) = Therefore, λ = 1 or λ = 5 For λ = 1, we have We see that b = 3a, so 4 1 1 3 2 1 3 1 3 1 a b = a b a b 3a + b 3a + b a 3a = a = = = 1 3,,

92 COMPUTING E T 125 For λ = 5, we have 4 5 1 3 2 5 1 1 3 3 a b a b a + b 3 3b = = =,, We see that a = b, so a b = a a = a 1 1 So let Therefore, So P 1 = 1 P P = 1 1 3 1 1 1 3 1 1 1 3 1 PTP = 1 4 = 1 1 1 4 3 1 = 1 1 4 2 = 1 4 4 1 3 2 1 5 3 5 1 1 3 1 1 1 3 1 What is P? For each eigenvalue λ j, we find its corresponding eigenvector v j Then for P = v 1, v 2,, v n, which is an n n matrix, we have λ 1 λ 2 A = P λ n P 1

126 CHAPTER 9 LINEAR SYSTEMS For Y = AY, we have constant { }} { Y = e Ax C = e PDP 1x C = Pe xd P 1 C = P e λ1x e λ2x = v 1, v 2,, v n e λ1x e λnx e λ2x C e λnx = C 1 e λ1x v 1 + C 2 e λ2x v 2 + + C n e λnx v n, C where C = P 1 C Example 93 Solve Solution First note that Y = 4 1 3 2 4 λ 1 3 2 λ Y =, (4 λ) (2 λ) 3 =, 8 6λ + λ 2 3 =, λ 2 6λ + 5 =, (λ 5) (λ 1) = Therefore, λ = 1 or λ = 5 For λ = 1, we have 4 1 1 3 2 1 3 1 3 1 a b a b 3a + b 3a + b = = =,,

92 COMPUTING E T 127 We see that b = 3a, so a b = a 3a = a 1 3 For λ = 5, we have 4 5 1 3 2 5 1 1 3 3 a b a b a + b 3 3b = = =,, We see that a = b, so a b = a a = a 1 1 Therefore, P = 1 1 3 1, D = 1 5 and we have Y = = 1 1 3 1 e x C 1 e x e 5x e 5x = C 2 3e x e 5x = C 1 e x 1 + C 2 e 5x 1, 3 1 C 1 e x + C 2 e 5x 3C 1 e x + C 2 e 5x C 1 C 2 that is, y 1 = C 1 e x + C 2 e 5x, y 2 = 3C 1 e x + C 2 e 5x If T does not have n distinct eigenvalues, computing e T requires greater knowledge of linear algebra Property 94 1 e PAP 1 = Pe A P 1 2 If AB = BA, then e A+B = e A e B = e B e A MATB24

128 CHAPTER 9 LINEAR SYSTEMS 3 If D = λ 1 λ 2, λ n then e D = e λ1 e λ2 e λn 4 If A n+1 =, then e A = I + A + A2 2! + + An n! Proof We have already proved (1), and (3) and (4) are trivial What remains is (2) So note that Also ) e A e B = (I + A + A2 + + (I An + B + B2 2! n! 2! = I + A + B + 1 ( A 2 + 2AB + B 2) + 2 ) + + Bn n! e A+B (A + B)2 (A + B)3 = I + A + B + + + 2! 3! = I + A + B + 1 ( A 2 + AB + BA + B 2) +, 2 which is the same as the above if AB = BA Theorem 95 Given T over C, there exists matrices P, D, and N with D diagonal, N k+1 = for some k, and DN = ND such that T = P (D + N) P 1

93 THE 2 2 CASE IN DETAIL 129 Proof Proof is given in MATB24 Corollary 96 We have e T = Pe D e N P 1 (e D and e N can be computed as above) Proof Proof is given in MATB24 The steps involved in finding P, N, and D are essentially the same as those involved in putting T into Jordan Canonical form (over C) 93 The 2 2 Case in Detail Theorem 97 (Cayley-Hamilton) Let p(λ) = A λi Then p(a) = Proof Proof is given in MATB24 The eigenvalues of A are the solutions of p(λ) = We now consider three cases for the solutions of p(λ) = Case 1: p(λ) has two distinct real roots λ 1 and λ 2 This is the diagonalizable case considered earlier We find eigenvectors v and w for the eigenvalues λ 1 and λ 2 Let v 1 w 1 P = v 2 w 2 Then so P 1 TP = e T = P λ 1 λ 2 e λ1 e λ2, P 1 The solution of Y = TY would be e λ1x Y = P e λ2x C 1 C 2

13 CHAPTER 9 LINEAR SYSTEMS in this case Case 2: p(λ) has a double root r Let S = T ri so that T = S + ri Since SI = IS, we have e T = e S e ri = e s e r e r Then p(t) =, (T ri) 2 =, S 2 = Therefore, e S = I + S = I + T ri = T + (1 r) I and e T = (T + (1 r) I) e r e r Case 3: p(λ) has complex roots r ± iq, where q Let S = T ri so that T = S + ri Then e T = e S e r e r and p(λ) = λ 2 2rλ + ( r 2 + q 2) = (λ r) 2 + q 2 We then have p(t) =, (T ri) 2 + (qi) 2 =, S 2 + (qi) 2 =, S 2 + q 2 I =

93 THE 2 2 CASE IN DETAIL 131 Therefore, S 2 q 2 I and it follows that e S = I + S + S2 2! + S3 3! + = I + S q2 I 2! = I q2 S 3! + q4 I 4! (1 q2 2! + q4 4! q6 6! + = I cos(q) + S q sin(q) + q4 S q6 I + ) 5! 6! + S (1 q2 3! + q4 5! ) Therefore, e T = ( cos(q)i + ( T ri q ) ) e r sin(q) e r Example 98 (Double root) Solve Y = 3 9 Y 4 9 Solution We have p(λ) =, 3 λ 9 4 9 λ =, ( 3 λ) (9 λ) + 36 =, 27 6λ + λ 2 + 36 =, λ 2 6λ + 9 =, (λ 3) 2 = Therefore, λ = 3 is a double root and we have Y = e Ax C For A, the eigenvalue

132 CHAPTER 9 LINEAR SYSTEMS is 3 Therefore, for T, the eigenvalue is r = 3x Hence, e r Y = (T + (1 r) I) e r e 3x = (Ax + (1 3x) I) e 3x C ( ) 3x 9x 1 3x e 3x = + 4x 9x 1 3x e 3x C 1 6x 9x e 3x = 4x 1 + 6x e 3x C e 3x 6xe 3x 9xe 3x C 1 = 4xe 3x e 3x + 6xe 3x C 2 ( C 1 e 3x 6xe 3x) 9C 2 xe 3x = ( 4C 1 xe 3x + C 2 e 3x + 6xe 3x) = C 1 e 3x 1 + C 1 xe 3x 6 + C 2 e 3x 9 + C 2 xe 3x 4 1 1 9 6 Example 99 (Complex roots) Solve Y = 1 4 1 1 Y Solution We have p(λ) =, 1 λ 4 1 1 λ =, ( 1 λ) ( 1 λ) + 4 =, (λ + 1) 2 + 4 = Therefore, λ = 1 ± 2i For A, the roots are 1 ± 2i, so for T, the roots are ( 1 ± 2i) x, ie, r = x and q = 2x Note that T ri q = 1 2x 4x x = 2 1 2

94 THE NON-HOMOGENEOUS CASE 133 Thus, Y = e Ax C ( ) 2 e x = cos(2x)i + sin(2x) 1 2 e x C cos(2x) 2 sin(2x) e x = 1 2 sin(2x) cos(2x) e x C e x cos(2x) 2e x sin(2x) = 1 1 2 C 1e x sin(2x) + C 2 e x cos(2x) C 1 C 2 2 e x sin(2x) e x cos(2x) C1 e x cos(2x) + 2C 2 e x sin(2x) = 94 The Non-Homogeneous Case The general solution of Y = AY + B is Y = Y h C + Y p, where Y h C is the general solution of Y = AY Therefore, Y h C = AY hc for all C, giving us Y h = AY h, (92a) and Y p is a particular solution to Y = AY + B, giving us Y p = AY p + B (92b) To find Y p, we use variation of parameters Thus, we have Y p = Y h V, (92c) where V = v 1 v 2

134 CHAPTER 9 LINEAR SYSTEMS To find V, we have Y p = Y h V, Y p = Y h V + Y h V, }{{} }{{} (92b) (92a) AY p + B = A Y h V +Y } {{ } h V = AY p + Y h V (92c) Therefore, Y h V = B V = Y 1 h B Example 91 Solve Y = 4 1 3 2 Y + e 5x e 2x Solution As before, we have λ 1 = 1 and λ 2 = 5 Thus, V 1 = 1 3, V 2 = 11 Therefore, P = 1 1 3 1, D = 1 5, and the solution of Y = AY is Y = Pe Dx C = Y h C So we have Y h = Pe Dx 1 1 e x e x e 5x = 3 1 e 5x 3e x e 5x, Y 1 h = e Dx P 1 = 1 e x 1 1 P e 5x = 1 3 1 4 e x 3e 5x e x e 5x It follows that V = Y 1 h B = 1 4 1 4 e4x e x e x 3e 5x e x e 5x e 5x e 2x = 1 4 e 4x e x 3 + e 3x, V = 1 4 3x 1 3 e 3x

95 PHASE PORTRAITS 135 and e x e 5x 1 4 e4x e x Y p = Y h V = 1 4 3e x e 5x 3x 1 3 e 3x = 1 1 4 e5x e 2x + 3xe 5x 1 3 e2x 4 3 4 e5x + 3e 2x + 9xe 5x 1 3 e2x 1 = e 5x 16 1 3 + e 2x 4 + xe 5x 4 3 3 9 16 4 4 + e 2x 1 12 1 12 Therefore, Y = Y h C + Y p = C 1 e x 1 3 + e 2x 1 4 3 4 + C 2 e 5x 1 1 + xe 5x 3 4 9 4 + e 5x 1 16 3 16 + e 2x 1 12 1 12, where C 1 and C 2 are arbitrary constants 95 Phase Portraits: Qualitative and Pictorial Descriptions of Solutions of Two-Dimensional Systems Let V : R 2 R 2 be a vector field Imagine, for example, that V(x, y) represents the velocity of a river at the point (x, y) We wish to get the description of the path that a leaf dropped in the river at the point (x, y ) will follow For example, Figure 91 shows the vector field of V(x, y) = ( y, x 2) Let γ(t) = (x(t), y(t)) be such a path At any time, the leaf will go in the direction that the river is flowing at the point at which it is presently located, ie, for all t, we have γ (t) = V(x(t), y(t)) If V(F, G), then dx dt = F (x, y), } {{ } y dy dt = G(x, y) } {{ } x 2 t We are assuming here that V depends only on the position (x, y) and not also upon time

y 136 CHAPTER 9 LINEAR SYSTEMS 4 4 3 2 2 1 y -1-2 -2-4 -3-4 -2 2 4 (a) The vector field plot of `y, x 2 x -3-2 -1 1 2 3 4 x (b) The trajectories of `y, x 2 Figure 91: The vector field plot of ( y, x 2) and its trajectories In general, it will be impossible to solve this system exactly, but we want to be able to get the overall shape of the solution curves, eg, we can see that in Figure 91, no matter where the leaf is dropped, it will head towards (, ) as t Before considering the general case, let us look at the linear case where we can solve it exactly, ie, V = (ax + by, cx + dy) with dx dt = ax + by, dy dt = cx + dy, or X = AX, where X = x y, A = a c b d Recall from Theorem 112 (p 157) that if all the entries of A are continuous, then for any point (x, y ), there is a unique solution of X = AX satisfying x(t ) = x and y(t ) = y, ie, there exists a unique solution through each point; in particular, the solution curves do not cross The above case can be solved explicitly, where X = e At is a solution passing through (x, y ) at time t = We will consider only cases x y

95 PHASE PORTRAITS 137 where A 951 Real Distinct Eigenvalues Let λ 1 and λ 2 be distinct eigenvalues of A and let v and w be their corresponding eigenvectors Let P = (v, w) Then P 1 λ 1 AP = λ 2 } {{ } D Therefore, At = P (Dt) P 1 and we have X = e At x y = Pe Dt P 1 = (v, w) C 1 e λ1t C 2 e λ2t x y = (v, w) e λ1t e λ2t = C 1 e λ1t v + C 2 e λ2t w C 1 C 2 Different C 1 and C 2 values give various solution curves Note that C 1 = 1 and C 2 = implies that X = e λ1t v If λ 1 <, then the arrows point toward the origin as shown in Figure 92a in which contains a stable node Note that ( ) X = C 1 e λ1t v + C 2 e λ2t w = e λ2t C 1 e (λ1 λ2)t v + C 2 w The coefficient of v goes to as t, ie, as t, X (, ), approaching along a curve whose tangent is w As t, X = e λ1t ( C 1 v + C 2 e (λ2 λ1)t w ), ie, the curves get closer and closer to being parallel to v as t We have the degenerate case when λ 1 < λ 2 =, in which case X = C 1 e λ1t v+ C 2 w The case when λ 1 < < λ 2 gives us the phase portrait shown in Figure 92b in which contains a saddle point This occurs when A < The case when < λ 1 < λ 2 gives us the phase portrait shown in Figure 92c in which contains an unstable node We have ( ) X = C 1 e λ1t v + C 2 e λ2t w = e λ1t C 1 v + C 2 e (λ2 λ1)t w Therefore, as t, X approaches parallel to w asymptotically Note that in all cases, the origin itself is a fixed point, ie, at the origin, x = and y =, so anything dropped at the origin stays there Such a point

138 CHAPTER 9 LINEAR SYSTEMS (a) λ 1 < λ 2 < (b) λ 1 < < λ 2 (c) < λ 1 < λ 2 Figure 92: The cases for λ 1 and λ 2

95 PHASE PORTRAITS 139 is called an equilibrium point; in a stable node, if it is disturbed, it will come back in an unstable node; if perturbed slightly, it will leave the vicinity of the origin 952 Complex Eigenvalues Complex eigenvalues come in the form λ = α ± βi, where β In such a case, we have X = C 1 Re ( e λt v ) + C 2 Im ( e λt v ), where v = p + iq is an eigenvector for λ Then Therefore, e λt v = e αt e βit (p + iq) = e αt (cos(βt) + i sin(βt)) (p + iq) = e αt (cos(βt)p sin(βt)q + i cos(βt)q + i sin(βt)p) X = e αt (C 1 cos(βt)p C 1 sin(βt)q) + C 1 cos(βt)q + C 2 sin(βt)q = e αt k 1 cos(βt) + k 2 sin(βt) k 3 cos(βt) + k 4 sin(βt) ) = e αt (cos(βt) k 1 k 3 + sin(βt) k 2 k 4 Note that tr(a) = 2α So α = = tr(a) =, α > = tr(a) <, α < = tr(a) > Consider first α = To consider the axes of the ellipses, first note that X = p 1 q 1 } p 2 q 2 {{ } P C 1 C 2 } C 2 C 1 {{ } C cos(βt) sin(βt) Recall that the trace of a matrix is the sum of the elements on the main diagonal

14 CHAPTER 9 LINEAR SYSTEMS Except in the degenerate case, where p and q are linearly dependent, we have Therefore, cos(βt) + sin(βt) cos(βt) sin(βt) cos(βt) sin(βt) = C 1 P 1 X = X t ( P 1) t ( C 1 ) t C 1 P 1 X, cos 2 (βt) + sin 2 (βt) = X t ( P 1) t ( C 1 ) t C 1 P 1 X, 1 = X t ( P 1) t ( C 1 ) t C 1 P 1 X Note that C = C t, so CC t C 1 C 2 = C 2 C 1 C 2 = C 2 1 + C 2 2 C 2 1 + C 2 2 = ( C 2 1 + C 2 2) I C 1 C 2 C 2 C 1 Therefore, C 1 C = C1 2 +, C2 2 ( C 1 ) t = C 1 C = C1 2 +, C2 2 ( C 1 ) t C 1 = ( ( C 1) 2 C 2 = 1 + C2) 2 I (C1 2 + = I C2 2 )2 C1 2 + C2 2 Therefore, 1 = X t ( P 1) t ( C 1 ) t C 1 P 1 X = 1 X t ( P 1) t P 1 X C1 2 + C2 2 Let T = ( P 1) t P 1 Then X t TX = C 2 1 + C 2 2 and T = T t (T is symmetric) Therefore, the eigenvectors of T are mutually orthogonal and form the axes of the ellipses Figure 93 shows a stable spiral and an unstable spiral

95 PHASE PORTRAITS 141 (a) α < (b) α > Figure 93: The cases for α, where we have a stable spiral when α > and an unstable spiral when α > 953 Repeated Real Roots We have N = A λi, where N 2 = and A = N + λi So e At = e Nt+λtI = e Nt e λti = (I + Nt) e λt e λt Therefore, X = e At C 1 C 2 = (I + Nt) e λt e λt C 1 C 2 = e λt (I + Nt) C 1 C 2 Note that N 2 = N 2 = N = Therefore, N = n 1 n 2 αn 1 αn 2 Also, N 2 = tr(n) = n 1 + αn 2 = Let V = 1α Then Nv = n 1 + αn 2 α (n 1 + αn 2 ) =

142 CHAPTER 9 LINEAR SYSTEMS So Av = (N + λi) v = λv, ie, v is an eigenvector for λ Therefore, X = e λt C 1 (I + Nt) = e λt C 1 + (n 1 C 1 + n 2 C 2 ) t C 2 C 2 + α (n 1 + n 2 C 2 ) t ( ) = e λt C 1 1 + (n 1 C 1 + n 2 C 2 ) t C 2 α ( ) = e λt C 1 + (n 1 C 1 + n 2 C 2 ) tv C 2 If λ <, we have as t If λ >, then x y x y as t What is the limit of the slope? In other words, what line is approached asymptotically? We have ie, it approaches v Similarly, y lim t x = lim C 2 + (n 1 C 1 + n 2 C 2 ) tv 2 = v 2, t C 1 + (n 1 C 1 + n 2 C 2 ) tv 1 v 1 y lim t x = v 2, v 1 ie, it also approaches v as t Figure 94 illustrates the situation We encounter the degenerate case when N = This does not work, but then A = λi, so X = e At C 1 C 2 = e λt e λt C 1 C 2 = e λt C 1 C 2, which is just a straight line through Figure 95 illustrates this situation C 1 C 2

95 PHASE PORTRAITS 143 (a) λ < (b) λ > Figure 94: The cases for λ, where we have a stable node when λ < and an unstable node when λ > (a) λ < (b) λ > Figure 95: The degenerate cases for λ when N =, where we have a stable node when λ < and an unstable node when λ >

144 CHAPTER 9 LINEAR SYSTEMS

Chapter 1 Existence and Uniqueness Theorems 11 Picard s Method Consider the general first order ivp dy dx = f(x, y), y(x ) = y ( ) Picard s Method is a technique for approximating the solution to Equation ( ) If y(x) is a solution of Equation ( ), then x dy dt dt = x y x y dy = y y, x f(t, y(t)) dt, giving us x y = y + f(t, y(t)) dt x Conversely, if y satisfies ( ), then differentiating gives ( ) dy = + f(x, y(x)) = f(x, y) dx 145

146 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS and y(x ) = y + = y Therefore, the ivp ( ) is equivalent to the integral equation ( ) If p(x) is a function, define a new function P (p) by x (P (p)) (x) = y + f(t, p(t)) dt x So a solution to ( ) is a function y(x) such that P (y) = y Picard s Method is as follows Start with a function y (x) for which y (x ) = y Let y 1 = P (y ), y 2 = P (y 1 ),, y n = P (y n 1 ),, and so on Under the right conditions (discussed later), the function y n will converge to a solution y Intuitively, we let y = lim n y n Since y n+1 = P (y n ), taking limits as n gives y = P (y) Later, we will attempt to make this mathematically precise Example 11 Solve y = y, where y() = 1 Solution We begin with y (x) 1 With x =, y = 1, f(x, y) = y, we have x ( ) 1 y 1 (x) = y + x, y (t) dt and and = 1 + f x x ( 1) dt = 1 x, x y 2 (x) = y + f(t, y 1 (t)) dt x y 3 (x) = 1 + So in general, we have Therefore, we finally have = 1 + x x (1 t) dt = 1 x + x2 2, ( ) 1 t + t2 dt = 1 x + x2 2 2 x3 3! y n (x) = 1 x + x2 2! + + ( 1)n x n n! y(x) = lim n y n(x) = ( 1) n x n n!

11 PICARD S METHOD 147 In this case, we recognize the solution as y = e x Example 12 Solve y = x + y 2, where y(1) = 1 Solution Let y (x) 1 With x = 1, y = 1, and f(x, y) = x + y 2, we have and y 2 (x) = 1 + y 1 (x) = 1 + = 1 + x 1 ( t + 1 2 ) x dt = 1 + (t 1) + 2 dt (t 1) 2 2 = 1 + 2 (x 1) + x 1 t + ( + 2 (t 1) (x 1)2 2 1 + 2 (t 1) + 1 x 1 = 1 + (x 1)2 2 ) 2 (t 1)2 dt 2 + 2 (x 1) = 1 + 2 (x 1) + 5 2 (x 1)2 + 5 3 (x 1)3 + 1 4 (x 1)4 + 1 2 (x 1)5 Picard s Method is not convenient for actually finding solutions, but it is useful for proving that solutions exist under the right conditions Theorem 13 If f : X R is continuous, where X R n is closed and bounded (compact), then f(x) is closed and bounded (compact) In particular, there exists an M such that f(x) M for all x X Proof Proof is given in MATB43 Corollary 14 Let R be a (closed) rectangle Let f(x, y) be such that f/ y exists and is continuous throughout R Then there exists an M such that f(x, y 2 ) f(x, y 1 ) M y 2 y 1 } {{ } Lipschitz condition with respect to y for any points (x, y 1 ) and (x, y 2 ) in R

148 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS Proof Let R be a (closed) rectangle Let f(x, y) be such that f/ y exists and is continuous throughout R By Theorem 13, there exists an M such that f y (r) M for all r R By the Mean Value Theorem, there exists a c (y 1, y 2 ) such that f(x, y 2 ) f(x, y 1 ) = f y (x, c)(y 2 y 1 ) Since R is a rectangle, we have (x, c) R, so f (x, c) y M Therefore, f(x, y 2 ) f(x, y 1 ) M y 2 y 1 Theorem 15 Let f be a function Then 1 f is differentiable at x implies that f is continuous at x 2 f is continuous on B implies that f is integrable on B, ie, B f dv exists Proof Proof is given in MATB43 Definition (Uniform convergence) A sequence of functions {f n } defined on B is said to converge uniformly to a function f if for all ɛ >, there exists an N such that f(x) f n (x) < ɛ for all n N The point is that the same N works for all x s If each x has an N x that worked for it but there was no N working for all x s at once, then it would converge, but not uniformly

11 PICARD S METHOD 149 Theorem 16 Suppose that {f n } is a sequence of functions that converges uniformly to f on B if f n is integrable for all n, then f is integrable and B f(x) dv = lim f n (x) dv n B Proof Proof is given in MATB43 Theorem 17 Let {f n } be a sequence of functions defined on B Suppose that there exists a positive convergent series n=1 a n such that f n (x) f n 1 (x) a n for all n Then f(x) = lim n f n (x) exists for all x B and {f n } converges uniformly to f Proof Let {f n } be a sequence of functions defined on B Suppose that there exists a positive convergent series n=1 a n such that f n (x) f n 1 (x) a n for all n Let Therefore and f n (x) = f (x) + (f 1 (x) f (x)) + + (f n (x) f n 1 (x)) n = f (x) + (f k (x) f k 1 (x)) k=1 lim f n(x) n } {{ } f(x) f(x) f n (x) = = f (x) + k=n+1 (f k (x) f k 1 (x)) k=1 (f k (x) f k 1 (x)) k=n+1 a k < ɛ for n sufficiently large, since n=1 a n converges Therefore, f n converges uniformly to f

15 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS 12 Existence and Uniqueness Theorem for First Order ODE s Theorem 18 (Existence and uniqueness theorem for first order ode s ) Let R be a rectangle and let f(x, y) be continuous throughout R and satisfy the Lipschitz Condition with respect to y throughout R Let (x, y ) be an interior point of R Then there exists an interval containing x on which there exists a unique function y(x) satisfying y = f(x, y) and y(x ) = y To prove the existence statement in Theorem 18, we will need the following lemma Lemma 19 Let I = x α, x + α and let p(x) satisfy the following: 1 p(x) is continuous on I 2 p(x) y a for all x I Then x x f(t, p(t)) dt exists for all x I and q(x) = y + x x f(t, p(t)) dt also satisfies (1) and (2) Proof Let I = x α, x + α and let p(x) satisfy the following: 1 p(x) is continuous on I 2 p(x) y a for all x I Immediately, (2) implies that (t, p(t)) R for t I We have I R f R t (t, p(t)) x so f is continuous, hence integrable, ie, x f(t, p(t)) dt exists for all x I Since q(x) is differentiable on I, it is also continuous on I Also, since

12 EXISTENCE AND UNIQUENESS THEOREM FOR FIRST ORDER ODE S151 f(t, p(t)) M, it follows that x q(x) y = f(t, p(t)) dt M x x Mα a x Proof (Proof of existence statement (Theorem 18)) Let R be a rectangle and let f(x, y) be continuous throughout R that satisfy the Lipschitz Condition with respect to y throughout R Let (x, y ) be an interior point of R Since f is continuous, there exists an M such that f(r) M for all r R Let a be the distance from (x, y ) to the boundary of R Since (x, y ) is an interior point, we have a > Let α = min(a, a/m) We will show that y = f(x, y), where y(x ) = y, has a unique solution on the domain I = x α, x + α Inductively define functions (1) and (2) of Lemma 19 by y (x) y for all x I and x y n (x) = y + f(t, y n 1 (t)) dt x By Lemma 19, the existence of y n 1 satisfying (1) and (2) guarantees the existence of y n By hypothesis, there exists an A such that f(x, v) f(x, w) A v w, whenever (x, v), (x, w) R As in the proof of Lemma 19, we have for all x I Likewise, we have y 1 (x) y (x) = y 1 (x) y M x x x y 2 (x) y 1 (x) = f(t, y 1 (t)) f(t, y (t)) dt x x f(t, y 1 (t)) f(t, y (t)) dt x x A y 1 (t) y (t) dt x x AM t x dt x MA x x 2, 2

152 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS and x y 3 (x) y 2 (x) = f(t, y 2 (t)) f(t, y 1 (t)) dt x x f(t, y 2 (t)) f(t, y 1 (t)) dt x ( ) x AMA t x 2 dt 2 x MA 2 x x 3 3! Continuing, we get y n (x) y n 1 (x) MAn 1 α n n! We have y n (x) = y (x) + (y 1 (x) y (x)) + (y 2 (x) y 1 (x)) + + (y n (x) y n 1 (x)) n = y (x) + (y k (x) y k 1 (x)) Since and k=1 converges, we must have y k (x) y k 1 (x) MAk 1 α k k=1 k! MA k 1 α k = M ( e Aα 1 ), k! A lim y n(x) = y (x) + n (y k (x) y k 1 (x)) converging for all x I Therefore, y(x) = lim n y n (x) exists for all x I and y n converges uniformly to y k=1 Note that f(x, y n (x)) f(x, y n 1 (x)) A y n (x) y n 1 (x)

12 EXISTENCE AND UNIQUENESS THEOREM FOR FIRST ORDER ODE S153 implies that f(x, y n (x)) converges uniformly to f(x, y(x)) With we then have y(x) = lim y n(x) = lim n = y + x x y n (x) = y + f(y, y n 1 (t)) dt, x x ( n ( x y + f(t, y n 1 (t)) dt x lim f(t, y n 1(t)) n ) dt = y + x ) x f(t, y(t)) dt implying that y satisfies y (x) = f(x, y) and y(x ) = y To prove the uniqueness statement in Theorem 18, we will need the following lemma Lemma 11 If y(x) is a solution to the de with y(x ) = y, then y(x) y a for all x I In particular, (x, y(x)) R for all x I Proof Suppose there exist an x I such that y(x) y > a We now consider two cases Suppose x > x Let s inf({t : t > x and y(t) y > a}) So s < x x + α x + a By continuity, we have a = y(s) y = y(s) y(x ) = y (c) (s x ) = f(c, y(c)) (s x ) } {{ } M(s x ) for some c (x, s) However, s I = s x α = α M (s x ) Mα a Therefore, M (s x ) = Ma = s x = a = s = x + a, which is a contradiction because we have s < x + a The case with x < x is similarly shown Proof (Proof of uniqueness statement (Theorem 18)) Suppose y(x) and z(x)

154 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS both satisfy the given de and initial condition Let σ(x) = (y(x) z(x)) 2 Then Therefore for all x I, σ (x) = 2 (y(x) z(x)) (y (x) z (x)) = 2 (y(x) z(x)) (f(x, y(x)) f(x, z(x))) σ (x) 2 y(x) z(x) A y(x) z(x) = 2A (y(x) z(x)) 2 = 2Aσ(x) Therefore, 2Aσ(x) σ (x) 2Aσ(x), and we have σ (x) 2Aσ(x) = σ (x) 2Aσ(x) if x x Similarly, = e 2Ax (σ (x) 2Aσ(x)), σ (x) = d ( e 2Ax σ(x) ) = e 2Ax σ(x) decreasing on I = e 2Ax σ(x) e 2Ax σ(x ) = σ(x) σ (x) 2Aσ(x) = σ (x) + 2Aσ(x) = e 2Ax (σ (x) + 2Aσ(x)), σ (x) = d ( e 2Ax σ(x) ) = e 2Ax σ(x) increasing on I = e 2Ax σ(x) e 2Ax σ(x ) = σ(x) if x x Therefore, σ(x) for all x I But obviously, σ(x) for all x from its definition Therefore, σ(x), ie, y(x) z(x) Therefore, the solution is unique dx dx

13 EXISTENCE AND UNIQUENESS THEOREM FOR LINEAR FIRST ORDER ODE S155 13 Existence and Uniqueness Theorem for Linear First Order ODE s Here, we consider the special case of a linear de dy = a(x)y + b(x) dx Theorem 111 (Existence and uniqueness theorem for linear first order de s) Let a(x) and b(x) be continuous throughout an interval I Let x be an interior point of I and let y be arbitrary Then there exists a unique function y(x) with y(x ) = y satisfying dy/dx = a(x)y + b(x) throughout I Proof Let a(x) and b(x) be continuous throughout an interval I Let x be an interior point of I and let y be arbitrary Let A = max({ a(x) : x I}) Inductively define functions on I by y (x) = y and x y n+1 (x) = y + (a(t)y n (t) + b(t)) dt x Then x y + (a(t)y n 1 (t) + b(t)) dt y n (x) y n 1 (x) = x x y (a(t)y n 2 (t) + b(t)) dt x x = a(t) (y n 1 (t) y n 2 (t)) dt x Assuming by induction that y n 1 (x) y n 2 (x) A n 2 x x n (n 1)!, Assuming by induction that y n(t) is continuous, a(t)y n(t) + b(t) is continuous throughout I and thus integrable, so the definition of y n+1 (x) makes sense Then y n+1 (x) is also differentiable, ie, y n+1 (x) = a(x)yn(x) + b(x), and thus continuous, so the induction can proceed

156 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS we then have x a(t) (y n 1 (t) y n 2 (t)) dt x thereby completing the induction Therefore, x x ( AA n 2 t x n 1 (n 1)! A n 1 x x n, n! y n (x) y n 1 (x) An 1 α n, n! ) dt where α is the width of I The rest of the proof is as before 14 Existence and Uniqueness Theorem for Linear Systems A system of differential equations consists of n equations involving n functions and their derivatives A solution of the system consists of n functions having the property that the n functional equations obtained by substituting these functions are their derivatives into the system of equations, and they hold for every point in some domain D A linear system of differential equations has the form dy1 dx = a 11(x)y 1 + a 12 (x)y 2 + + a 1n (x)y n + b 1 (x), dy 2 dx = a 21(x)y 1 + a 22 (x)y 2 + + a 2n (x)y n + b 2 (x), dy n dx = a n1(x)y 1 + a n2 (x)y 2 + + a nn (x)y n + b n (x) We can write the system as dy dx = AY + B, The point is that y 1, y 2,, y n and their derivatives appears linearly

14 EXISTENCE AND UNIQUENESS THEOREM FOR LINEAR SYSTEMS157 where A = a 11 (x) a 1n (x) a 21 (x) a 2n (x) a n1 (x) a nn (x), Y = y 1 (x) y 2 (x) y n (x), B = b 1 (x) b 2 (x) b n (x) Theorem 112 (Existence and uniqueness theorem for linear systems of n de s) Let A(x) be a matrix of functions, each continuous throughout an interval I and let B(x) be an n-dimensional vector of functions, each continuous throughout I Let x be an interior point of I and let Y be an arbitrary n-dimensional vector Then there exists a unique vector of functions Y(x) with Y(x ) = Y satisfying throughout I dy dx = A(x)Y + B(x) Proof Let A(x) be a matrix of functions, each continuous throughout an interval I and let B(x) be an n-dimensional vector of functions, each continuous throughout I Let x be an interior point of I and let Y be an arbitrary n-dimensional vector Let Λ = max({ A(x) : x I}) Inductively define Y (x) = Y and x Y n+1 (x) + Y + (A(t)Y n (t) + B(t)) dt x } {{ } vector obtained by integrating componentwise

158 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS Then x Y n (x) Y n 1 (x) = A(t) (Y n 1 (t) Y n 2 (t)) dt x ( ) x ΛΛ n 1 t x n 1 dt (n 1)! x = Λ n x x n n! Λ n α n, where α is the width of I before The rest of the proof is essentially the same as Recall Theorem 85 (p 9) restated here as follows Theorem 113 (Theorem 85) Let x = p be an ordinary point of y + P (x)y + Q(x)y = Let R be the distance from p to the closest singular point of the de in the complex plane Then the de has two series y 1 (x) = a nx n and y 2 (x) = b nx n which converge to linearly independent solutions to the de on the interval x p < R Proof Let x = p be an ordinary point of y + P (x)y + Q(x)y = Let R be the distance from p to the closest singular point of the de in the complex plane We first claim that if f(z) is analytic at p and R is the radius of convergence of f, then for r < R, there exists an M (depending on r) such that 1 f (n) (p) M n! r n for all n N To prove this claim, since converges absolutely, let M = 1 n f n (p)r n = f(r) 1 n! f n (p) r n ( )

14 EXISTENCE AND UNIQUENESS THEOREM FOR LINEAR SYSTEMS159 Then 1 n! f (n) (p)r n M, so Equation ( ) holds Therefore, our claim holds Let w(z) be a solution of y + P (x)y + Q(x)y = Being a solution to the de, w is twice differentiable Furthermore, differentiating the de gives a formula for w in terms of w, w, w Continuing, w is n times differentiable for all n, so w has a Taylor series We now wish to show that the Taylor series of w converges to w for z p < R Since z p < R, there exists an a such that z p < a < R As above, find constants M such that for all n N Let Note that so A(z) = M 1 P (n) (p) M n! a n, 1 Q (n) (p) N n! a n M A(z) = 1 z p, B(z) = a ( 1 + z p a N 1 z p a ( ) 2 ( ) ) n z p z p + + + +, a a A(p) = M, A (p) = M a, A (p) = 2 M a 2, A (n) (p) = n! M P a n (n) (p) Similarly, Consider B (n) (p) = n! N Q a n (n) (p) y = A(z)y + B(z)y ( )

16 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS Let v(z) be a solution of Equation ( ) satisfying v(p) = w(p) and v (p) = w (p) In general, w (n) (p) = C 1 w (p) + C 2 w(p), where C 1 and C 2 are arbitrary constants depending on P and Q and their derivatives at p, eg, w = P (p)w (p) + Q(p)w(p), w = P (p)w (p) + P (p)w (p) + Q (p)w(p) + Q(p)w (p) = P (p)w (p) + P (p) (P (p)w (p) + Q(p)w(p)) + Q (p)w(p) + Q(p)w (p) = ( P 2 (p) + P (p) + Q(p) ) w (p) + (P (p)q(p) + Q (p)) w(p) Similar formulas hold v (n) (p) involving derivatives of A and B at p Using P (n) (p) A (n) (p) and Q (n) (p) B (n) (p) gives w (n) (p) v (n) (p) The Taylor series of W about p is a n (z p) n, where a n = w (n) (p)/n!; the Taylor series of v about p is b n (z p) n, where b n = v (n) (p)/n! Since we showed that a n b n, the first converges anywhere the second does So we need to show that the Taylor series of w(x) has a radius of convergence equal to a (this implies that the Taylor series for w converges for n pk, but a < R was arbitrary) We have Let u = (z p) /a Then Then we have v M = 1 z p v + a v = dv 1 dv du a du N 1 z p a v = a d2 v du 2 du dz = 1 a 2 d 2 v du 2 1 d 2 v a 2 du 2 = M 1 dv 1 u a du + N 1 u v, (1 u) d2 v dv = am du2 du + a2 Nv Write v = γ nu n Using its first radius of convergence, we have ( ) n z p v = γ n u n = γ n = γ n a a n (z p)n = b n (z p) n, v

14 EXISTENCE AND UNIQUENESS THEOREM FOR LINEAR SYSTEMS161 which implies that γ n = b n a n > Note that Now we have dv du = nγ n u n 1 = (n + 1) γ n+1 u n, n=1 } {{ } adjust indices d 2 v du 2 = (n + 1) nγ n+1 u n 1 = (n + 2) (n + 1) γ n+2 u n n=1 } {{ } adjust indices (1 u) (n + 2) (n + 1) γ n+2 u n = am (n + 1) γ n+1 u n + a 2 Nγ n u n, (n + 2) (n + 1) γ n+2 u n (n + 2) (n + 1) γ n+2 u n+1 (n + 2) (n + 1) γ n+2 u n = (n + 1) nγ n+1 u n Therefore, (n + 2) (n + 1) γ n+2 u n = (n + 1) (n + am) γn+1 + a 2 Nγ n u n, so it follows that (n + 2) (n + 1) γ n+2 = (n + 1) (am + n) γ n+1 + a 2 Nγ n, for all n N Since γ n /γ n+1 < 1, we have Therefore, γ n+2 = am + n n + 2 γ n+1 + a2 Nγ n n + 2, γ n+2 = n + am γ n+1 n + 2 + a2 N γ n n + 2 lim n γ n a 2 N = n + 2 γ n+1 γ n+2 lim = 1 + = 1 n γ n+1 γ n+1 So by the ratio test (Equation (82) of Theorem 82, p 88), the radius conver-

162 CHAPTER 1 EXISTENCE AND UNIQUENESS THEOREMS gence of is a γ n u n γ n = (z p) n = a n ( ) n z p b n = 1 a The point of Theorem 85 is to state that solutions are analytic (we already know solutions exists from early theorems) The theorem gives only a lower bound on the interval The actual radius of convergence may be larger

Chapter 11 Numerical Approximations 111 Euler s Method Consider y = f(x, y), where y(x ) = y The goal is to find y(x end ) Figure 111 shows this idea We pick a large N and divide x, x end into segments of actual value y N y y 1 y 2 approximate value x x +h = x 1 x +2h = x 2 x end = x N Figure 111: Graphical representation of Euler s method length h = (x end x ) /N Then y(x 1 ) y(x ) + y (x ) (x 1 x ) } {{ } h 163

164 CHAPTER 11 NUMERICAL APPROXIMATIONS is the linear/tangent line approximation of y(x 1 ) Set y 1 = y + hf(x, y ) y 1 f(x 1 ), y 2 = y 1 + hf(x 1, y 1 ) y 2 f(x 2 ), y 3 = y 2 + hf(x 2, y 2 ) y 3 f(x 3 ), y answer = y N 1 + hf(x N 1, y N 1 ) y answer f(x end ) Example 111 Consider y = x 2y, where y() = 1 Approximate y(3) using a step size of h = 1 Solution First note that x = With a step size of h = 1, we have Therefore, Therefore, y(3) 54 x 1 = 1, x 2 = 2, x 3 = 3 y 1 = y + hf(x, y ) = 1 + 1 ( 2) = 1 2 = 8, y 2 = 8 + hf(1, 8) = 8 + h(1 16) = 8 1 15 = 8 15 = 65, y 3 = 65 + hf(2, 65) = 65 + h (2 13) = 65 1 11 = 65 11 = 54 What about the actual value? To solve the de, we have y + 2y = x, e 2x y + 2e 2x y = xe 2x, ye 2x = xe 2x dx = 1 2 xe2x = 1 2 xe2x 1 4 e2x + C, 1 2 e2x dx

111 EULER S METHOD 165 where C is an arbitrary constant Therefore y = 1 2 x 1 4 + Ce 2x Note that y() = 1 = 1 4 + C = 1 = C = 5 4, so the general solution is y = 1 2 x 1 4 + 5 4 e 2x Therefore, y(3) = 15 25 + 5 4 e 6 1 + 6861 5861, so y(3) = 5861 is the (approximated) actual value 1111 Error Bounds The Taylor second degree polynomial of a function y with step size h is given by y(a + h) = y(a) + hy (a) + h 2 y (z) 2 for some z a, h, where h 2 y (z)/2 is the local truncation error, ie, the error in each step We have h 2 y (z) M 2 2 h2 if y (z) M on a, h Therefore, dividing h by, say, 3 reduces the local truncation error by 9 However, errors also accumulate due to y n+1 being calculated using the approximation of y n instead of the exact value of f(x n ) Reducing h increases N, since N = (x end x ) /h, so this effect wipes out part of the local gain from the smaller h The net effect is that (overall error global truncation error) Mh, eg, dividing h by 3 divides the overall error only by 3, not 9

166 CHAPTER 11 NUMERICAL APPROXIMATIONS 112 Improved Euler s Method To improve upon Euler s method, we consider y n+1 = y n h, where, instead of using f (x n ), it is better to use the average of the left and right edges, so = f (x n ) + f (x n+1 2 = f(x n, y n ) + f(x n+1, y n+1 ) 2 Figure 112 illustrates this idea However, we have one problem in that we do x n x n+1 Figure 112: A graphical representation of the improved Euler s method not know y n+1 (that is what we are working out at this stage) Instead, we fill into the inside y n+1 from the original Euler s method, ie, 1 2 (f(x n, y n ) + f(x n+1, y n + hf(x n, y n ))), ie, y n+1 = y n + h 2 f(x n, y n ) + f(x n+1, y n + hf(x n, y n )) This improves the local truncation error from h 2 to h 3

113 RUNGE-KUTTA METHODS 167 113 Runge-Kutta Methods We again modify Euler s method and consider y n+1 = y n h At each step, for, we use some average of the values over the interval y n, y n+1 The most common one is x n + h/2 Figure 113 illustrates this idea Set x n x n + h/2 x n + 1 = x n + h Figure 113: A graphical representation of the Runge-Kutta Methods k n1 = f(x n, y n ), k n2 = f (x n + 12 h, y n + 12 ) hk n1, k n3 = f (x n + 12 h, y n + 12 ) hk n2, = f (x n + 12 h, y n + 12 ) hk n3 k n4 Use y n+1 = y n + h 6 (k n 1 + 2k n2 + 2k n3 + 3k n4 ) This gives a local truncation error proportional to h 5

168 CHAPTER 11 NUMERICAL APPROXIMATIONS