General Theory of Differential Equations Sections 2.8, , 4.1


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1 A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, , 4.1 Dr. John Ehrke Department of Mathematics Fall 2012
2 Questions Of Existence and Uniqueness Part of the difficulty in working with differential equations is understanding what constitutes a solution and knowing when a solution even exists. In this lecture, we will try to address questions of existence and uniqueness as they relate to solutions of linear differential equations. This will allow us to build up a general theory supporting our study of differential equations throughout the semester. We will begin with a small example to illustrate what can go wrong. Example Solve the differential equation dy ( y ) dx = 2. x Solution: This equation is separable and so we proceed as follows: Slide 2/27 Dr. John Ehrke Lecture 3 Fall 2012 dy y = 2 dx = ln y = 2 ln x + c x = e ln y 2 ln x+c = e = y = e c (e ln x) 2 = y = A x 2
3 Specifying a Solution The solutions of this equation are parabolas of the form y = A x 2, but we have a couple of questions to still be resolved. Are there some initial conditions for which no solutions exist? If a solution does exist, over what interval(s) are they uniquely defined? In answering the second question, consider what would happen if I asked you to solve the IVP corresponding to y(0) = 0. This is bad for two reasons: There is no rate of change defined at the origin, since f (x, y) is undefined there. A solution could come in on one branch and leave on a completely different branch, i.e. a solution of the IVP corresponding to y(1) = 2 is the parabola y = 2x 2, but this parabola passes through the origin and so cannot be extended uniquely for negative x. The solution is unique only from [0, ). Even for the first question above, no solution exists for any initial value along the yaxis other than the origin, because all the parabolas must pass through the origin. This example highlights just a few of the things that can go wrong when solving differential equations. Slide 3/27 Dr. John Ehrke Lecture 3 Fall 2012
4 Existence Uniqueness Theorem At this point one might wonder if there is any way to predict the behavior of solutions of a given differential equation without solving the ODE. The answer to this question is yes, and is summarized by the existence uniqueness theorem (EUT). First Order Existence Uniqueness Theorem Suppose that both the function f (x, y) and its partial derivative f / y are continuous on some rectangle R in the xyplane containing the point (a, b) in its interior. Then, for some open interval I R containing the point a, the initial value problem dy = f (x, y), y(a) = b (1) dx has one and only one solution that is defined on the interval I. We will not do a complete proof of this result, but will introduce the machinery behind the proof and a make a few remarks. Our text does not have a complete proof of this result but many undergraduate texts include such a proof in the appendices. If you are interested in reading through the complete proof let me know and I would be happy to provide the proof for you. Slide 4/27 Dr. John Ehrke Lecture 3 Fall 2012
5 Method of Successive Approximations The approach we employ to establish the existence of solutions is called the method of successive approximations and utilizes a series of approximations called Picard iterates. This method is based on the fact that a function y(x) satisfies the initial value problem, (1) on the open interval I containing x = a if and only if it satisfies the integral equation y(x) = b + x a f (t, y(t)) dt (2) for all x I. In particular, if y(x) satisfies (2), then clearly y(a) = b, and differentiating both sides gives y (x) = f (x, y(x)) as desired. It is important to note that we traded solving a differential equation for an integral equation. This is almost always the case when one studies differential equations, as we will see. Slide 5/27 Dr. John Ehrke Lecture 3 Fall 2012
6 Picard Iterates (Demonstrating Existence) To attempt to solve (2), we begin with the initial function y 0 (x) b, and then define iteratively a sequence y 1, y 2, y 3,... of functions that we hope will converge to the solution. Specifically, we let y 1 (x) = b + y 2 (x) = b + y n+1 (x) = b +. x a x a x a f (t, y 0 (t)) dt f (t, y 1 (t)) dt f (t, y n (t)) dt Suppose we know that each of these functions {y n (x)} 0 is defined on some open interval (the same for each n) containing x = a, and that the limit y(x) = lim y n (x) exists at each point of this interval. n Slide 6/27 Dr. John Ehrke Lecture 3 Fall 2012
7 Continuing... Continuing from the previous slide, it follows that y(x) = lim y n+1 (x) n [ = lim b + n = b + lim = b + = b + x a x n a x a x a f ] f (t, y n (t)) dt f (t, y n (t)) dt ( t, lim n y n (t) f (t, y(t)) dt This solution method works provided that: ) dt we can validate the interchange of limit operations above. we can show there exists only this one solution. Slide 7/27 Dr. John Ehrke Lecture 3 Fall 2012
8 Demonstrating Uniqueness If we assume at this point that we have established the existence of a solution via successive approximations (there still is some work to be done to make that argument solid), then we only need to show that the hypotheses of the theorem lead to a unique solution. When proving uniqueness you generally assume there exists another solution to the IVP different from y(x), say φ(x) and look at the difference y(x) φ(x) and try to reach a contradiction by showing this difference must be zero. In this case, that is done by proving that y(x) φ(x) A x 0 y(t) φ(t) dt. To do this requires the use of some heavy duty analysis and the determination of a Lipschitz condition. A Lipschitz condition for two functions exists if there is a positive constant A such that f (x, y 1 ) f (x, y 2) A y 1 y 2. Arguments of this type are generally considered in a first semester ODE course in graduate school. We will not pursue this problem beyond this point. Slide 8/27 Dr. John Ehrke Lecture 3 Fall 2012
9 Applying the Existence Uniqueness Theorem Example Consider the first order ODE, xy = y 1. Identify for what values of (a, b), the initial value problem xy = y 1, y(a) = b has a unique solution, no solution, or more than one solution. Solution: Consider the point(s), (0, b) for all b 1. At these points, x = 0, and so f (x, y) = (y 1)/x fails to be continuous at x = 0 and so by the EUT there can be no solution to the IVP, dy dx = y 1, y(0) = b. x Consider the point, (0, 1). When we consider the point (0, 1), f (x, y) = 0/0 which is indeterminate. So the EUT says nothing about this point in regards to existence. We may have solutions, or we may not have solutions. In the case, a solution exists might it be unique? We will check the partial derivative, ( ) y 1 = 1 y x x which is discontinuous at x = 0. This means, that solutions to the IVP y(0) = a when they exist are not guaranteed to be unique and in particular we see this for (0, 1). Slide 9/27 Dr. John Ehrke Lecture 3 Fall 2012
10 Working With Complex Numbers To achieve our goals this semester we will need to learn how to handle complex numbers and operations involving complex variables. A complex number is of the form z = a + bi where a, b R. The complex conjugate of z, denoted z is given by z = a bi. Remember, that multiplying a number by its complex conjugate turns the number to a real number. In particular, z z = a 2 + b 2. The absolute value of z, denoted z, involves conjugation since Moreover, using z and z we obtain z = z z = a 2 + b 2. a = R(z) = z + z and b = I(z) = z z. 2 2i Recall that when working with complex numbers you will often need to multiply or divide by i. Recall that i = 1 and so i 2 = 1, i 3 = i and i 4 = 1. Slide 10/27 Dr. John Ehrke Lecture 3 Fall 2012
11 Polar Representation of a Complex Number z = a + bi = r cos θ + i(r sin θ) = r (cos θ + i sin θ) = re iθ Euler s Formula: e iθ = cos θ + i sin θ The form z = re iθ is called the polar form of the complex number a + bi. Under this description, a is called the real part, b the imaginary part when z is written in rectangular form. In polar form, r is called the modulus of z (denoted z ), and θ is called the argument of z (denoted arg(z)). Slide 11/27 Dr. John Ehrke Lecture 3 Fall 2012
12 The Complex Exponential Typically, when we call something an exponential in mathematics, we have in mind two very specific properties that it must satisfy. An exponential should satisfy the exponential rule. That is, if e is an exponential, then e x e y = e x+y. An exponential should satisfy the differential equation dy dt = ay. This says that an exponential has the property that it is selfreplicating under differentiation (and integration). Question? Does our definition of the complex exponential, e iθ as described by Euler s formula define an exponential as stated above? e iθ = cos θ + i sin θ is called a complexvalued function of a real variable Slide 12/27 Dr. John Ehrke Lecture 3 Fall 2012
13 Power Series Representations The final motivation for our description of cos θ + i sin θ as being the complex exponential is provided by our knowledge of the Taylor series for the exponential, which is given by Inserting x = iθ we obtain e x = 1 + x 1! + x2 2! + x3 3! + x4 4! +. e iθ = 1 + iθ 1! + (iθ)2 + (iθ)3 + (iθ)4 + 2! 3! 4! ) = (1 θ2 2! + θ4 4! + + i (θ θ3 3! + ). This points to the fact that the real part of e iθ should be cos θ and the imaginary part is the Taylor series for sin θ. Slide 13/27 Dr. John Ehrke Lecture 3 Fall 2012
14 The Art of Complexifying To see how truly amazing and wondrous operations with complex variables can be we will demonstrate two examples which take advantage of the art of complexifying (yes that s a word). Example Obtain a solution to the integral e x cos x dx by complexifying the integral. Solution: Looking at the integrand, we can rewrite it as the real part of a complex variable, e x cos(x) = R ( e x e ix). By substituting this in for the real integrand we complexify the integral, ( ) ( ) e e x cos(x) dx = R e ( 1+i)x ( 1+i)x dx = R. 1 + i Taking the real part of this expression gives ( ) ( ) 1 1 i R 1 + i e( 1+i)x = R e x (cos x + i sin x) = e x (sin x cos x). 2 2 Slide 14/27 Dr. John Ehrke Lecture 3 Fall 2012
15 Constant Coefficient Equations In this example, we will introduce a technique that we will lean on later in the semester called exponential response. Example Obtain a general solution to the first order linear equation with constant coefficients given by x (t) + 2x(t) = cos(t). Solution: Complexifying the ODE leads to z (t) + 2z(t) = e it. We change to z(t) to represent the fact that we expect the solution to such an equation to be complex. We guess that the solution to such an equation is of the form z(t) = Ae it (i.e. matches the exponential response term up to a constant) where A is complex valued. Substituting our guess into the ODE gives A = 1/(2 + i) so z(t) = i eit R(z(t)) = x(t) = 2 5 cos t + 1 sin t. 5 Slide 15/27 Dr. John Ehrke Lecture 3 Fall 2012
16 The Exponential Response Theorem This method is so useful that we will establish the general case, that is, solve x + kx = Ae rt. The results of this are summarized in the theorem below: Exponential Response Theorem The exponential response to the first order linear constant coefficient ODE x + kx = Ae rt is given by as long as r + k 0. x(t) = A r + k ert (3) This formula is valid for complex values of r, but depending on the specific nature of the original oscillation you might have to obtain the real or imaginary part of (3). Slide 16/27 Dr. John Ehrke Lecture 3 Fall 2012
17 Second Order Linear Equations The general form of a second order differential equation is given by y = f (t, y, y ). In this lecture, we will consider the second order, linear homogeneous ODE with constant coefficients, given by ay + by + cy = 0 (4) We are going to make some assumptions about solutions to (4) which we will come back and verify later. 1 A general solution (4) is of the form y(x) = c 1 y 1 + c 2 y 2. 2 y 1 and y 2 are linearly independent solutions of (4). (y 1 cy 2, for some constant c 0) 3 Initial conditions involve both y and y which help to determine c 1 and c 2 above. 4 The solutions, y 1 and y 2 are called the fundamental set of solutions for the equation (4). Slide 17/27 Dr. John Ehrke Lecture 3 Fall 2012
18 Making a guess... A natural question to ask at this point is, What is the nature of the solutions y 1 and y 2? This question is best answered by making a guess (as we saw in the last lecture) that such solutions are exponential in nature. So, to try and solve the equation ay + by + cy = 0, (5) we guess that y(t) = e rt is a solution. Substituting into (2) gives ar 2 e rt + bre rt + ce rt = 0 = (r 2 + pr + q)e rt = 0 where p = b/a and q = c/a and a 0. This implies that y(t) = e rt is a solution if and only if r 2 + pr + q = 0. Slide 18/27 Dr. John Ehrke Lecture 3 Fall 2012
19 The Characteristic Equation So we have determined that y = e rt if and only if r 2 + pr + q = 0. This equation is called the characteristic equation for the ODE, (2). There are three cases to consider, each of which leads to different solution behaviors. 1 distinct real roots r 1, r 2 y 1 = e r 1t, y 2 = e r 2t 2 complex conjugate pair r = a ± bi y 1 = e (a+bi)t, y 2 = e (a bi)t 3 repeated real root, r gives you y 1 = e rt, but how do we find y 2? We will explore the physical interpretations of these cases, as well as solution techniques in the next lecture, but for the remainder of this lecture, we will discuss the general theory which underpins these techniques. The questions we need to answer are: 1 Why is c 1 y 1 + c 2 y 2 a solution to the equation, (2)? 2 Why are all solutions of this form? Slide 19/27 Dr. John Ehrke Lecture 3 Fall 2012
20 Superposition Principle Superposition Principle If y 1 and y 2 are solutions to a linear homogeneous ODE, then y = c 1 y 1 + c 2 y 2 is also a solution. To make our lives easier in manipulating differential equations we often want to think about differential equations as linear operators acting on y. We define the linear operator L corresponding to the ODE y + py + qy = 0 by y + py + qy = 0 = D 2 y + pdy + qy = 0 = (D 2 + pd + q)y = 0 = Ly = 0 We denote by D, the differential operator, D = d/dy. L = D 2 + pd + q is a linear operator if it satisfies the following properties: L(u 1 + u 2 ) = L(u 1 ) + L(u 2 ) L(αu) = αl(u), for some constant α. Slide 20/27 Dr. John Ehrke Lecture 3 Fall 2012
21 Linear Operators Under the definition of a linear operator on the previous slide, we show that L = D 2 + pd + q is linear. L(u 1 + u 2) = (D 2 + pd + q)(u 1 + u 2) = D 2 (u 1 + u 2) + pd(u 1 + u 2) + q(u 1 + u 2) = u 1 + u 2 + p(u 1 + u 2) + qu 1 + qu 2 = (u 1 + pu 1 + qu 1 ) + (u 2 + pu 2 + qu 2) = Lu 1 + Lu 2 We claim it is obvious that L(αu) = αl(u) by properties of the derivative. We are now prepared to answer our first question: If we suppose that y 1 and y 2 are solutions to y + py + qy = 0, does that automatically mean c 1 y 1 + c 2y 2 is a solution? L(c 1 y 1 + c 2y 2) = L(c 1 y 1 ) + L(c 2y 2) = c 1 L(y 1 ) + c 2L(y 2) = 0 (since y 1 solves the ODE, then L(y 1 ) = 0) Thus, c 1 y 1 + c 2y 2 solves y + py + qy = 0 if y 1 and y 2 do. Slide 21/27 Dr. John Ehrke Lecture 3 Fall 2012
22 Two solutions are better than one... In answering the second of our questions for this lecture, Why do all solutions look like c 1 y 1 + c 2y 2? we will show that this solution form is necessary to satisfy IVPs. Theorem The set of solutions {c 1 y 1 + c 2y 2} is enough to satisfy any initial values imposed on the equation y + py + qy = 0. This means there does not exist some initial condition y(x 0) = a, y (x 0) = b such that a solution not of the form c 1 y 1 + c 2y 2 is the only solution that satisfies this IVP. Proof: Let y(x 0) = a, y (x 0) = b. Then applying these initial conditions to the solution c 1 y 2 + c 2y 2 we obtain the system of linear equations with unknowns c 1, c 2 given by c 1 y 1 (x 0) + c 2y 2(x 0) = a c 2y 1(x 0) + c 2y 2(x 0) = b This system of equations is solvable if and only if the determinant, y 1 y 2 y 1 y 0. 2 Slide 22/27 Dr. John Ehrke Lecture 3 Fall 2012
23 The Wronskian Given y 1, y 2, the Wronskian of y 1, y 2, denoted W(y 1, y 2 ) is given by W(y 1, y 2 ) = y 1 y 2 y 1 y = y 1y 2 y 2 y 1 2 The Wronskian explains why linearly independent solutions are required since if y 2 = cy 1, then W(y 1, y 2 ) = y 1 y 2 y 1 y = y 1 cy 1 2 y 1 cy = cy 1y 1 cy 1 y 1 = 0 1 In this case, because W(y 1, y 2 ) = 0, then the resulting system of equations from the initial conditions is not solvable. This leads to a very important theorem, Abel s Theorem If y 1, y 2 are solutions to the ODE y + py + qy = 0, then either 1 W(y 1, y 2) 0 for all x 2 or W(y 1, y 2) 0 for all x Slide 23/27 Dr. John Ehrke Lecture 3 Fall 2012
24 Abel s Theorem Proof: To prove Abel s theorem we note that y 1 and y 2 satisfy y 1 + p(t)y 1 + q(t)y 1 = 0 (6) y 2 + p(t)y 2 + q(t)y 2 = 0 (7) If we multiply the first equation by y 2, the second by y 1 and add the results, we obtain (y 1 y 2 y 1 y 2) + p(t)(y 1 y 2 y 1y 2) = 0. (8) Next, we let W(t) = W(y 1, y 2)(t) and observe that W (t) = y 1 y 2 y 1 y 2. Then we can write (3) in the form W + p(t)w = 0 (9) which is a first order linear equation in W, and has solution ( ) W(t) = c exp p(t) dt where c is a constant. The value of c depends on the fundamental solutions, y 1 and y 2. However, since the exponential function is never 0, W(t) is not 0 unless c = 0, in which case W(t) 0 for all t, which completes the proof. (10) Slide 24/27 Dr. John Ehrke Lecture 3 Fall 2012
25 Normalized Solutions Not all solutions y 1 and y 2 are created equal. There are in fact, best solution, called normalized solutions which we will introduce below. Normalized Solutions Two solutions Y 1 and Y 2 of the second order linear ODE, y + py + qy = 0 are called normalized if they satisfy the initial conditions, Y 1 (0) = 1, Y 1 (0) = 0 Y 2 (0) = 0, Y 2 (0) = 1 Example Using the above definition, find the normalized solutions for y + y = 0 and y y = 0. Compare your results. Remember to use the characteristic equation to determine your fundamental set of solutions. Slide 25/27 Dr. John Ehrke Lecture 3 Fall 2012
26 Why do we care about normalized solutions? If Y 1 and Y 2 are normalized at x = 0, then the solution to the IVP y + py + qy = 0 with initial conditions y(0) = a, y (0) = b is y(x) = ay 1 + by 2. In example, we can simply read the solutions off straight from the ODE. This is highly desired by those in the engineering fields, and is generally a nicer version of linearly independent solutions than what we have been working with thus far. The general solution y(x) = c 1 Y 1 + c 2 Y 2 is called the normalized general solution. Slide 26/27 Dr. John Ehrke Lecture 3 Fall 2012
27 Wrapping up the discussion... We are now ready to finish answering the second of our two questions, Why all solutions look like c 1 y 1 + c 2y 2? First, we state the second order existence uniqueness theorem. Second Order Existence Uniqueness Theorem Given the second order linear differential equation y + p(x)y + q(x)y = 0, (11) where p and q are continuous for all x in a rectangle about the initial point x 0, there is exactly one solution satisfying the initial conditions, y(x 0) = a and y (x 0) = b. How does this help answer our question? Well, given a solution u(x) satisfying u(0) = a, u (0) = b, then by the existence uniqueness theorem above, there can only be one such solution, and we know that ay 1 + by 2 is a solution, where Y 1 and Y 2 are normalized solutions of (11). This says that u(x) is of the form c 1 y 1 + c 2y 2 for some fundamental set of solutions, y 1, y 2, which is what we needed to show. Slide 27/27 Dr. John Ehrke Lecture 3 Fall 2012
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