(2) f(x) = o(g(x)) as x ( little o ) if lim. f(x)


 Delphia Hampton
 2 years ago
 Views:
Transcription
1 1. Introduction To perturbation Theory & Asymptotic Expansions Example Consider x = ε coshx (1.1) For ε 0 we cannot solve this in closed form. (Note: ε = 0 x = ) The equation defines a function x : ( ε, ε) R (some range of ε either side of 0) We might look for a solution of the form x = x 0 +εx 1 +ε x + and by subsititung this into equation (1.1) we have x 0 + εx 1 + ε x + = ε cosh(x 0 + εx 1 + ) Now for ε = 0, x 0 = and so for a suitably small ε + εx 1 ε cosh( + εx 1 + ) x 1 cosh() x(ε) = + ε cosh(ε) + For example, if we set ε = 10 we get x = where the exact solution is x = Landau or Order Notation. Definition Let f and g be real functions defined on some open set containing zero, ie: 0 D R. We say: (1) f(x) = O(g(x)) as x ( Big O ) if K > 0 and ε > 0 such that( ε, ε) D and x ( ε, ε) f(x) < K g(x) f(x) () f(x) = o(g(x)) as x ( little o ) if lim x 0 g(x) = 0 (3) f(x) g(x) as x (asymptotically equivalent) if lim x 0 f(x) g(x) = 1 Remark (1) Could define these for x x 0 or x () Abuse of notation: for say, sin x = x + O(x ), O(x ) should be an equivalence class, ie: sin x x O(x ) f(x) Lemma If lim m < then f(x) = O(g(x)) x 0 g(x) Proof. Suppose f(x) g(x) m < ε and x < δ then f(x) g(x) m f(x) < ε g(x) < m + ε f(x) < ( m + ε) g(x) But this is just the definition of O with m + ε for K Example (i) x = o(x) as x since x x 0 as x (ii) 3x = O(x ) since 3x 5x 0 as x 0 1
2 (iii) x = o( x 1 ) as x 0 x (iv) 1 + x = o(1) since x/(1 + x ) 0 as x 0 1 x (v) 1 + x = O(x) sin x x cosx 1 (vi) sinx = x + o(x) as x 0 since lim = lim = 0 x 0 x x 0 1 (vii) sinx = x + O(x 3 sin x x cos x 1 ) since lim x 0 x = lim 3 x 0 3x sin x cos x = lim x 0 6x = lim x 0 6 = 1 6 (viii) sinx x x3 3! since lim = 1 (ix) sinx = x x3 3! + O(x4 ) sin x x 0 x x3 3! The definition of f (x) in o notation is: f(x + h) = f(x) + f (x)h + o(h) as h 0 The Taylor series for f(x + h) is given by f(x + h) = f(x) + f (x)h + + f(n) (x)h n n! + o(h) If the Taylor series is a convergent power series then o(h n ) can be replaced by O(h n+1 ). N N Any convergent power series a n x n = a n x n + O(x n+1 ) n=0 Examples: 1 + x = ( x) + 1 ( 1 )( x)! + O(x 3 ) = 1 x x + O(x3 ) ln(1 + x) = x x + O(x3 ) x sin( 1 x ) = O(x ) It is important to note that this is big O with no limit as x sin( 1 x ) 1 x though x sin( 1 x ) x n=0 = sin( 1 x ), and this has no limit. 1.. The Fundamental Theorem of Perturbation Theory. Theorem If A 0 +A 1 ε+ +A N ε N = O(ε N+1 ) then A 0 = A 1 = A N = 0 Proof. Suppose A 0 + A 1 ε + + A N ε N = O(ε N+1 ) but not all A k are zero. Let A M be the first nonzero. Consider A M ε M + A M+1 ε M A N ε N+1 ε N+1 = A M + A M+1 ε + + as ε ε N+1 M Then we have a contradiction with big O.
3 Perturbation Theory of Algebraic Equations. Example Consider x 3x++ε = 0. Assume the roots have the following expansion: x 0 + εx 1 + ε x + O(ε 3 ) then by substitution we get (x 0 + εx 1 + ε x + ) 3(x 0 + εx 1 + ε x + ) + ε + = O(ε 3 ) = (x 0 + 3x 0 + ) + ε(x 0 x 1 3x 1 + 1) + ε (x 1 + x 0 x 1 3x ) = O(ε 3 ) Terms in ε 0 : x 0 3x 0 + = 0 x 0 = 1 or x 0 = Terms in ε 1 : x 0 x 1 3x = 0 if x 0 = 1 then x 1 = 1 otherwise if x 0 = then x 1 = 1 Terms in ε : x 1 = x 0x 3x = 0 if x 0 = 1 then x 1 = 1, and so x = 1 otherwise if x 0 = then x 1 = 1, and so x = 1 x = 1 + ε + ε + O(ε 3 ) or x = ε ε + O(ε 3 ) We can solve x 3x + + ε = 0 directly to get x = 3 ± 1 4ε Now 1 4ε = 1 ε ε +O(x 3 ) and substituting this into 3 ± 1 4ε we get x = 3 ± (1 ε ε ) which is the same answer as above. Example 1.3. (Singular Perturbation).. Consider εx x+1 = 0 and again assume there is an expansion x 0 +εx 1 +ε x + O(ε 3 ). We get for terms in ε 0 : x = 0 x 0 = 1 terms in ε 1 x 0 x 1 = 0 x 1 = 1 8 terms in ε x 0 x 1 x = 6 x = 1 8 and so we have x = 1 + ε 8 + ε 8 and this gives us one of the roots, but where is the second? The exact solution is given by: and the other root should be x = 1 ± 1 ε ε x = ε 1 + O(ε)
4 4 Last time in example 1.3. we did not find the other root of εx x + 1 = 0 using the expansion of form x 0 + εx 1 + ε x + O(ε 3 ). If instead we try ω = εx, then ω ε = w ε + 1 = 0 ω ω + ε = 0 this, we can assume in the usual way has an expansion of the form: ω 0 + εω 1 + ε ω O(ε 3 ), and: Terms in ε 0 : ω0 ω 0 = 0 ω 0 = 0 or Terms in ε 1 : ω ε = x = { 1 + O(ε) ε 1 + O(ε) ω 0 ω 1 ω = 0 ω 1 = 1 or 1 Example (Non Regular Expansions). Consider x x( + ε) + 1 = 0. assume x = x 0 + εx 1 + ε x + O(ε 3 ) ε 0 : x 0 x = 0 x 0 = 1 twice ε 1 : (1+ε) (1+εx 1 )(+ε)+1 = O(ε ) εx 1 εx 1 +1 = O(ε ) 1 = O(ε ) This contradicts the assumption that there was a regular expansion. The exact roots are: x = +ε± 4ε ε = 1 ( + ε ± ε 4 ε) = 1 ( + ε ± ε 4 ε) = 1 ± ε 1 + O(ε) Try x = x 0 + ε 1 x1 + ε x + ε 0 : x 0 = 1 as before. ε 1 : (1 + ε 1 x 1 ) (1 + ε 1 )( + ε) + 1 = O(ε 3 ) ε 1 x 1 ε 1 x 1 = O(ε) 0 = 0 ε 1 : (1 + ε 1 x 1 + ε(x ) (1 + ε 1 x 1 + εx )( + ε) + 1 = O(ε 3 ) which gives x 1 = 1 x 1 = ±1 so x = 1 ± ε 1 + O(ε 3 ) 1.4. Perturbation Theory of Odes. Example (Regular Problem). Consider the following ODE: ẋ + x = εx, x(0) = 1 We try an expansion of the form: x(t) = x 0 (t) + εx 1 (t) + O(ε ) which leads to: ε 0 : { ẋ0 + x 0 = 0 x x 0 (0) = 1 0 (t) = e t ε 1 : and so { ẋ1 + x 1 = x 0 = e t x 1 (0) = 0 (no ε in x(0) = 1) x 1(t) = e t e t x = e t + ε(e t e t + O(ε )
5 5 For t [0, ] the constants in O definition can be independent of t Sometimes we want only ε > 0 versions of o, O with one sided limits lim We use a series 1, ε 1 3, ε, ε or in general: ϕ0 (ε), ϕ 1 (ε)... ϕ1 +1(ε) = (ε)) is what is needed. o(ϕ1 Example 1.4. (Singular Model Equation). Consider εẋ + x = 1, x(0) = 0, and suppose x = x 0 + εx 1 O(ε ) Then ε(ẋ 0 + εx 1 ) + x 0 + εx 1 = 1 + O(ε ) ε o +. ε 0 : x 0 = 1 but x(0) = 0 cannot solve the initial condition. We can rescale time, ie: t = ετ which gives τ = t dt ε and dε = ε, dx dτ = dx dt dt dτ = εẋ dx + x = 1, x(0) = 0 dτ Now use x = x 0 + εx 1 O(ε ) ε 0 : dx 0 dτ + x 0 = 1, x 0 (0) = 0 x 0 = 1 e τ = 1 e t ε ε 1 : dx 1 dτ + x 1 = 1, x 1 (0) = 0 x 1 = 0 (similarly for ε, ε 3 etc...) Therefore the solution is: x = 1 e tε Example ( Singular In The Domain ). Consider ẋ + εx = 1, x(0) = o, t > 0 and assume x = x 0 + εx 1 + ε x O(ε ) ε 0 : ẋ 0 = 1, x 0 (0) = 0 x 0 = t ε 1 : (t + εx 1 ) + ε(t + εx 1 ) = 1 + O(ε ) 1 + εẋ 1 + ε(t + εtx 1 ) = 1 + O(ε ) εẋ 1 + εt = O(ε ) ẋ 1 + t = 0 x 1 = t3 3 (If we carry on we find the ε term t 5 ) The solution is: This is not regular for t [0, ) x = t ε t3 3 + O(ε3 ) Example (Damped Harmonic Motion with Small Damping (ε > 0)). Consider ẍ + εẋ+x = 0, x(0) = 0, ẋ(0) = 1 and assume x = x 0 + εx 1 + ε x O(ε 3 ) ε 0 : ẍ 0 + x 0 = 0, x 0 (0) = 0, ẋ 0 = 1 x 0 = sin t ε 1 : ẍ 1 + ẋ 0 + x 1 = 0 ẍ 1 + x 1 = cost, x 1 (0) = 0, ẋ 1 (0) = 0 and we should get: x 1 = t sin t whereby the solution is: x = sint 1 εt sin t
6 6 This again is not uniform for t [0, ) The exact solution is: x = e ε t sin( 1 = ε 4 )t, the expansion above is only good for small t Asymptotic Expansions. Definition A sequence of functions {ϕ n }, n = 0, 1,,... is called an asymptotic series as x x 0 if (ie: ϕ n+1 (x) = o(ϕ n (x))) ϕ n+1 (x) lim = 0 x x 0 ϕ n (x) Note: We could have x 0 = or a onesided limit x x + 0 Examples: (i) x 1,1, x,x, 1 x 3,... x 0 + (ii) 1, 1 x, 1 x ln(x), 1, x x 3 ln(x) (iii) tanx, (x π), (sin x) 3, x π Taylor s Theorem: f(x) = N f (n) (x)(x x 0 ) n + n! n=0 remainder { }} { R N (x), f C N+1 [x 0 r, x 0 + r], r > 0 There are remainder formulas to bound R N (x), for example, Cauchy: R N (x) M Nr N+1 (n + 1)!, M N > 0, R N = O((x x 0 ) N+1 ) as x x 0. It is important to remember that Taylor R N 0 as N In fact we know N plenty of power series that do not converge, eg: ( 1) n n!x n diverges for all x. n=0 Another famous nonconvergent series is Stirling s formula for n!: lnn! = n lnn + 1 ln(πn) n n 3 + O( 1 n 4 ) Definition Let {ϕ n } be an asymptotic sequence as x x 0. N The sum a n ϕ n (x) is called an asymptotic expansion of f with N terms if f(x) n=0 N a n ϕ n (x) = o(ϕ N (x)). n=0 The coefficients a n are called the coefficients of the asymptotic expansion. is called an asymptotic series. Note: Some people use the stronger definition: O(ϕ N (x)) Notation f(x) a n ϕ n (x) as x x 0 n=0 a n ϕ n (x) n=0
7 7 Clearly any Taylor series is an asymptotic series. Example: Find an asymptotic expansion for f(x) = (1 + xt) 1 = 1 xt + x t + f(x) = Now, it can be shown that 0 0 t n e t dt = n! and hence 0 e t 1 + xt dt (1 xt + x t + )e t dt f(x) = 1 x +!x 3!x 3 + n!x n which diverges by the ratio test = n x x 0. (n 1)!x n 1 It could still be be an asymptotic expansion however; we d need to check N f(x) ( 1) n n!x n = o(x N ) This is a special case of: n=0 Lemma (Watson s Lemma). Let f be a function with convergent power series and radius of convergence R, and f(t) = O(e αt ) as t (for some α > 0) then: 0 In last example we had f(x) = e au du (looks like Watson s) u + u 0 e at f(t)dt 0 n=0 Example (Incomplete Gamma Function).. γ(a, x) = = x 0 N ( 1) n n=0 n! f n (o) a n+1 e t 1 + x dt as x 0+ let xt = u, a = 1/x, then t a 1 e e t dt = x 0 t n+a 1 dt = x 0 N t a 1 ( t) n dt n! N n=0 n=0 ( 1) n n!(n + a) xn+a Note: Power series under integral is convergent, hence uniformly convergent. We have a convergent power series for γ(a, x)in x Example E i (x) = x e t t dt (exponential integral)
8 8 E 1 (x) = e tx 1 0 t dt (Cauchy principal value), it turns out that E 1 (x) = E i ( x) E i (x) = γ + ln(x) + x + x! + x3 3 3! + O(x4 ) ( N where γ = e x 1 ) ln(x)dx = lim n k ln(n) k=0. ODEs In The Plane We consider systems of ODEs of the form: { ẋ = u(x, y) ẏ = v(x, y) where x(0) = x 0, y(0) = y 0. (Note: A nd order ODE can be expressed as a coupled system of 1 st order ODEs since if ẍ + f(x)ẋ + g(x) = 0 then if we say ẋ = y it follows that ẏ = ẍ so we get { ẋ = y ẏ = f(x) g(x) Where in this case it turns out that u(x, y) = y, v(x, y) = f(x) g(x)).1. Linear Plane autonomous Systems. { ẋ = ax + by the 1D case is easy: ẋ = ax, x(t) = x(0)eat ẏ = cx + dy Example { ẋ = y Consider 3x = y ( x If we say x = then we can write y) x = We solve these two ODEs to get ( ) 3 0 x 0 x(t) = x(0)e 3t, y(t) = y(0)e t (which may be written as) ( ) ( ) e 3t 0 x(0) x(t) = 0 e t y(0) We can construct a Phase plot with solutions being curves in the plane called trajectories or orbits. We could eliminate t as follows: ( x ) 3 y = y(0), x(0) 0 x(0) y x Figure.1.1:
9 9 Example.1.. { ẋ = x Similar to the above example consider ẏ = y We solve in both cases to get x(t) = x(0)e t, y(t) = y(0)e t and eliminate t such that: Noting that x = ( x ) 1 y = y(0) x(0) ( ) 1 0 x we end up with a phase plot which looks like this 0 1 Figure.1.: Example.1.3 (Simple Harmonic Motion: ẍ + ẋ = 0). { ( ) ẋ = y 0 1 x = x ẏ = x 1 0 x(t) = Acost + B sin t, x(0) = A ẋ(t) = Asin t + B cost, ẋ(0) = B ( ) ( ) ( ) x(0)cos t + y(0)sint cost sin t x(0) ẋ = = x(0)sin t + y(0)cost sint cost y(0) } {{ } rotation matrix The orbits are circles. y x Figure.1.3: Example.1.4 (Damped Harmonic Motion: ẍ + bẋ + x = 0). { ( ) ẋ = y 0 1 ẋ = x ẏ = x by 1 b Try x(t) = e λt giving the characteristic polynomial λ + bλ + 1 λ = b b ± 4 1
10 10 If b is small then b 4 and so y 1 < 0 such that λ = b ± i 1 b 4 = α + iβ x(t) = e αt (Acos(βt) + B sin(βt)) y x x Figure.1.4: Theorem.1.5. Let A R be a real matrix with eigenvalues λ 1, λ then: (i) If λ 1 λ are real then there exists an invertible matrix P such that ( ) P 1 λ1 0 AP = 0 λ (ii) If λ 1 = λ then either A is diagonal, A = λi or A is not diagonal and there is a P such that ( ) P 1 λ 1 AP = 0 λ (iii) If λ 1 = α + iβ, λ = α iβ, β 0 then there is a P such that ( ) P 1 α β AP = β α How does this help? Put x = A x and let y = P 1 x then x = P y, y = P 1 x, P x = PA x and so y = P 1 AP y This allows us to generalise the work we did above. Example.1.6. For case 1 in the theorem, consider u = P 1 AP u then u 1 = λ 1 u 1, u = λ u (since P 1 AP u is just a matrix of eigenvectors acting on u) Then u i (t) = u i (0)e λit and so ( ) ( ) e λ 1t 0 u1 (0) u(t) = 0 e λt u (0) ( )( ) ( e λ 1t 0 u1 (0) e λ 1t 0 Now x = P u so x(t) = P 0 e λt = P u (0) 0 e λt u 1 and u are related by eliminating t: ( u 1 (t) = u 1 (0)e λ1t u1 u 1 (0) ) 1 λ 1 = e t, then u = u (0)( u1 u 1 (0) ) ) λ λ 1. ( ) P 1 x1 (0) x (0)
11 11 y x λ 1 > λ > 0 λ 1 < λ < 0 Figure.1.5: For λ 1 λ, distinct eigenvectors v 1, v, A v i = λ i v i Half lines through the origin in eigendirections are trajectories. y eigen directions x e.g. λ 1 > 0 > λ Figure.1.6:
12 1.. Phase Space Plots.. Eigenvalues real, different, same sign Eigenvalues real, different, same sign Node: Source Eigenvalues real, different, opposite sign Node: Sink Eigenvalues real, equal, λ > 0, A = λi Saddle Eigenvalues real, equal λ < 0, A = λi Node: Source Eigenvalues real, equal λ > 0, A λi Node: Sink Eigenvalues real, equal λ < 0, A λi Degenerate Source Eigenvalues complex, λ 1 = α+iβ, λ = α  iβ α < 0, β 0 Degenerate Sink Eigenvalues complex, λ 1 = α+iβ, λ = α  iβ α > 0, β 0 Stable Spiral Eigenvalues purely imaginary, λ 1 = iβ, λ  iβ, β 0 Unstable Spiral Figure..1: Ellipse
13 13 { ẋ = 3x + y Example..1. consider ẏ = x y ( 0 The critical points of this system are at, and the Jacobian is given by 0) ( ) 3 1 ( ) 3 λ 1 we find the eigenvalues by setting A I = = 0, ie: λ λ + 5λ + 4 = 0 λ = 4, 1 this corresponds to a node (sink) As for the associated eigenvectors; ( ) ( A 4I = v = is in the null space (hence an eigenvector) ) ( ) ( 1 1 A I = v = is the other eigenvector 1 ) We can get further information to help in curve sketching by considering isoclines: ẏ ẋ = dy x y = dx 3x + y Figure..:.3. Linear Systems. A linear system for which the eigenvalues are wholly imaginary (λ = ±β) is called called a centre. The characteristic equation, in general of A R = λ (trace A)λ + deta. In this case (for λ imaginary), λ + β = 0, trace A = 0, deta > 0. Consider a simple case: { ẋ = y ẏ = cx eliminate t to get ẏ ẋ = dy dx = cx y y dy = cxdx y + cx = const This is the equation for an ellipse. To determine (ẋ ) the direction of the arrows, set y = 0 then on x axis ẋ = 0, ẏ = cx and is a vector in the direction of solutions, and we that for x positive ẋ is ẏ positive.
14 14 The Oscillatory nature of these graphs don t reflect any real life situations. x(t) Figure.3.1:.4. Linear Approximations.. Consider ẋ = u(x, y), ẏ = v(x, y), Critical points occur when u = v = 0. Let (x 0, y 0 ) be critical points, put ξ = x x 0, η = y y 0 and Taylor expand about (x 0, y 0 ) to get (near equilibrium point) u(x, y) = u(x 0, y 0 ) + ξ u x +η u (x0, y 0) y (x0, y +O(ξ + η ) as (ξ, η) (0, 0) 0) v(x, y) = v(x 0, y 0 ) + ξ v x +η v (x0, y 0) y (x0, y +O(ξ + η ) as (ξ, η) (0, 0) 0) and so ( u u = x v) v x y(t) u ( y ξ v + O(ξ η) + η ) y We can now make the approximation ( ) u u ) ξ = x y ξ η v v + O(ξ η (x0, y x y 0)( + η ) which is a linear system. Example.4.1 (PreditorPrey).. We wish to model the dynamics between predators and prey. Without considering external & environmental variables, as the number of predators increases, we expect the population growth rate of the prey should lessen; this then, should result in a slow down in the growth rate of predators (as there is more competition for fewer prey). let x be a population of prey (eg: rabbits), y be a population of predators (eg: foxes) with x, y > 0 (note: this model relies on large x and y such that we are able to talk about derivatives etc... (since x, y are integers!) { ẋ = x(a αy) A simple model is (a, c, α, γ > 0). ẏ = y( c + γx) Then u(x, y) = x(a αy), v(x, y) = y( c + γx) and so for an equilibrium, u = v = 0, x(a αy), y( c + γx) = 0 so for u = 0, either x = 0 or a αy = 0 (y = a α )
15 15 for v = 0, either y = 0 or c + γx = 0 (x = c γ ) Therefore the criticals are at (0, 0), ( c γ, a α ) More specifically if we put a = 1, α = 1, c = 3 4, γ = 1 4 { then: ẋ = x(1 y ) ẏ = y( x 4 ) with critical points at (0, 0), (3, 0) Near (0, 0): ( ) ( ( ) ξ 1 0 ξ = η 0 4) 3 η The corresponding eigenvalues being: ( 1 λ 1 = 1, v 1 =, λ 0) = 3 ( 0 4, v = 1) This is a saddle. Near (3, ): ( ) ξ = η with eigenvalues: λ 1, = ±i ( ) ( ξ 1 0 η) 3 ie: ξ + η = const, so ellipses. Now dy dx = y( x ) x(1 y ) 3 4 lnx + lny y x 4 = const. It is possible, albeit tricky to show this is a closed curve. x(t) y(t) Figure.4.1: Example.4. (Circular Pendulum).. We consider a circular pendulum given by nondimensional units { }} { ẍ + sin x = 0 where x is an angle. x m mg Figure.4.: Note that for small angles x, x sin x and so we get ẍ + x = 0 (simple harmonic
16 16 motion) We shall solve ẍ + sin x = 0 qualititively since it can t easily be solved analytically. Let ẋ = y = u, ẏ = sin x = v The critical points are at ẋ = ẏ = 0 y = 0, x = nπ, n Z u u ( ) x y 0 1 v v = ( 1) n at y = 0, x = nπ 0 x y ( ) 0 1 ξ Near the critical point we consider ( 1) 0)( n+1 η The characteristic equation is λ 1 ( 1) n+1 λ = 0 = λ + ( 1) n = 0 If n even λ + 1 = 0 λ = ±i, if n odd λ 1 = 0 λ = ±1 ( ( ) 1 1 For n odd we get eigenvectors v 1 =, v 1) = which is a saddle. 1 For n even we get a centre. The centres correspond to small swings The saddles correspond to swings just large enough that they stop at the top Everywhere else corresponds to big swings, where with no damping, the pendulem doesn t stop. Figure.4.3:.5. NonLinear Operators.. ẍ+x = 0 represents simple harmonic motion (SHM) with solution x(t) = Acos(t)+ B sin(t), a centre. Consider ẍ + βẋ + x = 0 making the substitution ẋ = y. The system of ODEs is { ẋ = y given by, and the roots of the resulting char poly are ẏ = β x λ = β ± i 1 β and so for 0 < β < 1 we get a stable spiral Example.5.1 (Stiff Spring System).. In a simple spring, force and hence acceleration is proportional to the extension of the spring (Hooke s Law), instead we think of a stiff spring with force proportional to x + βx 3. For small x this behaves like x, for large x, like x 3. { ẋ = y ẍ + x + βx 3 = 0, ẏ = x βx 3, u = y, v = x βx 3 The critical points are at u = v = 0 y = 0, x + βx 3 = 0 x = 0 or 1 + βx = 0 (no real solutions).
17 17 The only critical is at (0, 0) u u ( ) x y 0 1 v v = 1 3βx = 0 x y This is a centre λ = ±i ( ) 0 1 evaluated at (0, 0) 1 0 Example.5. (Soft Spring).. We could change the sign before β and simulate a soft spring ie: ẍ + x βx 3 = 0 The critical points in this case lie at (0, 0), ( ±1 β, 0) and the jacobian is given by ( ) βx 0 for x close to ±1 β we have a source; which, in physical terms, means we are actually adding energy to the system. 3. Limit Cycles Orbits, that is, trajectories of a system of ODEs cannot cross. x(0) x(0) = x(t) x(t) Figure 3.0.1: ( ) x(t) If x = and x(0) = x(t) (for T 0) then y(t) x(0) = x(t). when this happens we have a periodic orbit, eg a clock, oscillator, cycle in the economy, biology etc... Suppposing we have such a cycle what can be said about what happens around it? It could be the case that both outside and inside the orbit we spiral towards it; but then again, something else could happen instead. The equilibrium point at the centre doesn t give us this information. Figure 3.0.: Example (Cooked up). Consider a system of the contrived form: { ẋ = y x(x + y 1) 1 ẏ = x y(x + y 1)
18 18 we see that by construction, when x, y satisfy x +y = 1 we obtain simple harmonic motion. There is only one equilibriums point and that occurs at (0,0). The Jacobian at this point is given by f x g x f y g y (x,y)=(0,0) = ( ) The characteristic equation is λ λ + = 0 which has roots λ = 1 ± i. This is an unstable spiral. If instead however, we look at this in polar coordinates Differentiating implicitely we get: r = x + y, tan θ = y x rṙ = xẋ + yẏ, θ = xẏ yẋ x + y If we multiply (1) and () by x & y respectively we get; xẋ = x( y x(x + y 1)), yẏ = y(x y(x + y 1)) xẋ + yẏ = (x + y )(r 1) rṙ = r (r 1) ṙ = r(r 1) This reveals that there is also an equilibrium point at r = 1 (ie; for r = 1 we stay on the circle). Furthermore for r > 1, ṙ < 0 which is a stable spiral, whilst for r < 1, ṙ > 0 which is the unstable spiral we found above. r < 0 r > 0 Figure 3.0.3: Now if we multiply (1) and () by y & x respectively and subtract we get: xẏ yẋ = x xy(r 1) + y + xy(r 1) = x + y = r θ = r r = 1 The equilibrium point correctly told us that locally we have an unstable spiral but it failed to illuminate the behaviour as we move further out. We shall establish a couple of results that allow us determine when & where there there exist no closed orbits. First recall Theorem (Divergence Theorem/Divergence theorem in the plane). Let C be a closed curve, let A R be the region it encloses, and u, v be functions with continuous derivatives. Then A u x + v dx dy = y C u dy v dy = where (x(s), y(s)) is a parameterisation of C C u dy ds vdx ds ds ds
19 Theorem (Bendixon s Negative criterion). Consider the system with u and v continuously differentiable. Let A R be a region of the plane for which u x + v y Then there is no closed orbit contained within A. 19 { ẋ = u(x, y) ẏ = v(x, y) does not change sign. Proof. Suppose for contradiction there exists a closed orbit in A. Then this orbit forms a closed curve C in A. A A x(0)= x(t) Figure 3.0.4: Let A A be the region enclosed by C (ie A = C). Then by the divergence theorem T 0 (xẏ yẋ)dt = T 0 u dy dt vdx dt dt dt = but u x + v y is either > 0 or < 0 x, y A hence C u dy v dy = Example (returned to cooked example).. { ẋ = u = y x(x + y 1) ẏ = v = x y(x + y 1) u x = 3x y + 1, v y = x 3y + 1 and so u x + v y = 4x 4y + = 4(x + y ) + A u x + v y dxdy = 0 A 0 For values of x, y such that x + y > 1 this fails to be always positive or always negative; and so we cannot say there exists no closed orbit. (Though we can be confident there is no such orbit for x + y = r < 1 ) Bendixon s Criterion is not much use for answering the question Is there a closed orbit between r = 1 or r = 1? Example (Damped Harmonic Motion).. For ẍ + βẋ + x = 0 we have ẋ = u = y, ẏ = v = βy x This is a stable spiral at (0,0) and u x + v = 0 β which is constant y Hence there is no sign change for u x + v y and so no closed orbits.
20 0 Example (General Damped Oscillator).. This system is characterised by ẍ+f(x)ẋ +g(x) = 0 which with the usual substitution ẋ = y yields } {{ } damping f(x)>0 { ẋ = y ẏ = f(x)y g(x) Now u x + v = 0 f(x) is always negative and so general damped systems of this y form have no closed orbits Theorem (PoincaréBendixon Theorem). Given a region A R and an orbit of a system of ODEs C which remains in A t 0 then C must approach either a limit cycle or equilibrium. Remark (1) orbit = trajectory = solution curve () limit cycle = closed orbit that nearby orbits approach (3) We can use this result with time running backwards, ie: orbits come from unstable equilibrium or closed orbits. 3.. energy (brief).. { ẋ = y consider an oscillator of the form characterising ẍ + f(x) = 0 ẏ = f(x) Since there is no damping we expect energy to be conserved. Consider ε = ẋ y ẋ + F(x) = + F(x) where can be considered kinetic energy, F(x) the potential energy. Then dε(x, y) dt = yẏ + f (x) = (ẍ + F (x))ẋ Set F = f then dε = 0 along solution curves. dt So ε is constant on a solution (x(t), y(t)). we call this a first integral Example 3..1 (Duffing s Equation).. This is the hard spring system we met earlier ẍ + ω x + εx 3 = 0 f(x) = ω x + εx 3 F(x) = ω x + εx4 + some constant we need not worry 4 about in this context of constant solution curves. Then ε(x, y) = y + ω x + εx4 4 for ε > 0. As ε(x, y) is constant then the solutions are bounded for all t (closed curves). We can say that x + y max(, ω) (y + ω x + εx4 4 ) = max(, )ε. Ie; ω solutions stay in the circle. For constant ε we can check that for y 0 there are two solutions for x.
21 1 4. Lindstedt s Method Example 4.0. (Duffing s Equation). ẍ + ω x + εx 3 = 0, 0 < ε << 1 We know the solutions are periodic for y(0) = ẋ(0) 0. The solutions will resemble slightly square ellipses. Figure 4.0.1: consider a straightforward expansion x = x 0 + εx 1 + ε x O(ε 3 ) Terms in ε 0 : ẍ 0 + ω x 0 = 0 x 0 = a cosωt We may suppose without loss of generality that the initial conditions for this system are x(0) = a, and ẋ(0) = 0 since we are just picking out a particular solution curve; we still get all of them. Terms in ε 1 : ẍ 1 + ω x 1 + x 3 0 = 0 ẍ 1 + ω x 1 = a 3 cos 3 (ωt) To deal with cos 3 (ωt) we use the identity that cos 3 (ωt) = 3cos(ωt) 4 + cos(3ωt) 4 and so ẍ 1 + ω x 1 = a 3( 3cos(ωt) + cos(3ωt) ) 4 4 since cos(ωt) appears in the homogeneous solution we would need to introduce a t sin(ωt) term (secular term), and this is at odds with what we already know about this system; that it is bounded as t The trick to get rid of the secular term is to introduce another series: τ = Ωt where Ω = Ω 0 + εω 1 + Example (Lindstedt s Method). We try again with Duffings equation (prefixing ε with a minus sign this time) using the above idea and expecting to see a solution that has periodic orbits for ε small enough. Without loss of generality, we rescale time (setting ω = 1) such that we try to solve ẍ + x εx 3 = 0 We define τ = Ωt where Ω = Ω 0 + εω 1 + and so x(t) becomes x(τ). coupling this with the standard expansion we use for x we have x(τ) = x 0 ((Ω 0 + εω 1 )t) + εx 1 ((Ω 0 + εω 1 )t) We want x i (τ) = x i (τ + π) (ie; period of π for i = 1,, 3,...). For ε = 0 we have Ω 0 = 1 (had we not set ω = 1 earlier then we d have instead
22 Ω 0 = ω; actually we could choose what we want for Ω 0 and depending on how difficult we wnat to make things, this choice determines Ω 1 later. It makes sense to keep things simple and choose Ω 0 such that for ε = 0 we have x(τ) = x(ωt)) Now dτ dx = Ω so dt dt = dx dτ dτ dt = Ωdx dτ (so ẋ = Ωx ) and d x dt = Ω d x dτ and so returning to the ODE we have Ω x εx 3 = 0 ( =Ω0 1 +εω 1 + ) (x 0 + εx 1 + ) + (x 0 + εx 1 + ) ε(x 0 + εx 1 + ) 3 = 0 Terms in ε 0 : x 0 + x 0 = 0 As before we will assume initial conditions x(0) = a, ẋ(0) = 0 to pick a critical point on some curve. We know there exists a solution with these properties by assuming ellipsoidal shaped orbits; and by varying a we get all the curves. Since there is no ε in initial conditions we get Then the solution for x 0 is Terms in ε 1 : x 0 (0) = a, x 1 (0) = 0, x 0(0) = x 1(0) = 0 x 0 = a cosτ x 1 + Ω 1x 0 + x 1 x 3 0 = 0 x 1 + x 1 = Ω 1 ( a cos(τ)) + (a cos(τ)) 3 = 0 x 1 + x 1 = (Ω 1 a a3 )cos(τ) a3 cos(3τ) The whole point of doing this was to eliminate the cos(τ) term which would have introduced a secular term and so we set Ω 1 a a = 0 so it now remains to solve Ω 1 = 3 8 a3 x 1 + x 1 = 1 4 a3 cos(3τ) We try x 1 = Acos(3τ) 9Acos(3τ) + Acos(3τ) = 1 4 a3 cos(3τ) A = a3 3 The general solution for x 1 before applying boundary conditions is x 1 = α cos(τ) + β sin(τ) a3 3 cos(3τ) Now x 1 (0) = 0 α = a3 3 and x 1 (0) = 0 β = 0 giving Thus x 1 = a3 a3 cos(τ) 3 3 cos(3τ) x(t) = a cos(ωt) + a3 (cos(ωt) cos(3ωt)) 3 Where Ω = a +
23 3 5. Method of Multiple scales In the last section we used Lindstedt s method to take account of varying frequencies; now we develop a more general method for situations with two time scales. An example of where this is relevant is the damped circular pendulem ẍ + βẋ + sinx = 0 There are two things going on here; there is an oscillation which is captured by one time scale; and a slow loss of energy due to the damping term captured by another time scale. By considering two scales we are able to capture different features of the system. t T Figure 5.0.: Consider small oscillations and small energy ẍ + εẋ + sin x = 0 (D.H.M) If we did a standard expansion we d end up with a solution x(t, ε) = sin t + ε t sin t + and this is not a uniform expansion for the solution. The exact solution for this system is: 1 ) x = e εt sin (t 1 ε 4 where x(0) = 1 ẋ(0) = 0 1 ε The Method.. (1) Introduce a new variable T = εt, and think of t as fast time, and T as slow time. () Treat t and T as independent variables for the function x(t, t, ε). Using the chain rule we have dx dt = x t t t + x T T t + x ε ε t = x t + ε x T d x dt = d ( x dt t + ε x ) = ( x T t t + ε x ) 1 + ( x T T t + ε x ) ε T = x t + ε x x + ε t T T (3) Try an expansion x = x 0 (t, T) + εx 1 (t, T) + where of course T = εt =0
24 4 (4) Use the extra freedom of x depending on T to kill off any secular terms. 5.. Application of Multiple scales to D.H.M. Example we consider ẍ + εẋ + x = 0 with boundary conditions x(0) = 0, ẋ(0) = 1, this becomes x t + ε x t T + ε x ( x T + ε t + ε x ) + x = 0 T Now let X = x 0 + εx 1 + ε x + to get ( ) ( +ε +ε t t T T (x 0 +εx 1 +ε x + )+ε t +ε T Terms in ε 0 : this is a PDE with solution +x 0 + εx 1 + ε x + = 0 x 0 t + x 0 = 0 x 0 = A 0 (T)cost + B 0 (T)sint ) (x 0 +εx 1 +ε x + ) where A 0 (T), B 0 (T) are some functions of T (as opposed to just constants) Using boundary conditions x(0) = 0, ẋ(0) = 1 then Terms in ε 1 : x 0 (0) = A 0 (T)cost + B 0 (T)sin t = 0 A 0 (0) = 0 ẋ 0 (0) = A 0 (T)sint + B 0 (T)cost = 1 B 0 (0) = 1 x 1 t + x 0 t T + x 0 t + x 1 = 0 Now x 0 = A 0 (T)sin t + B 0 (T)cost and x 0 t t T = da 0(T) sin t + db 0(T) cost dt dt and so it remains to solve x ( 1 t + x 1 = da 0(T) sin t + db ) 0(T) cost + A 0 (T)sint B 0 (T)cost dt dt The RHS contains terms in sin(t), cos(t) which will induce secular terms. we therefore choose A 0 (T) and B 0 (T) such that they go away. In other words: and so da 0 dt + A 0 = 0 A 0 = A 0 (0)e 1 T = 0 db 0 dt + B 0 = 0 B 0 = B 0 (0)e 1 T = e 1 T x 0 (t) = e 1 T sin t = e εt sint to get the x 1 term we would need higher order terms to fully specify it. Notice that with the exception of constants, the first order part captures most of the features of the exact solution. In general we would have multiple time scales: T 0 = t, T 1 = εt,...,t n = ε n t If we consider a series x 0 (T 0, T 1,...,T n ) + εx 1 (T 0, T 1,...,T n ) +... then d dt = t=t 0 + ε T 1 + ε T +
25 5 Example 5.. (Van Der Pol s Equation). Consider ẍ + ε(x 1)ẋ + x = 0 Immediately we see that if x > 1 we have damping, x < 1 we have negative damping; so it would seem there is a tendency to head towards x = 1. Perhaps we could use energy methods; anyhow... ( ) + ε t t T + (x 0 + εx 1 + ) ( +ε((x 0 + 1) + ε ) (x 0 + εx 1 + ) + x 0 + εx 1 + = 0 T Terms in ε 0 : x 0 t + x 0 = 0 x 0 = A(T)cost + B(T)sint We can write this in complex form x 0 = A 0 (T)e it + A 0 (T)e it (noting that z + z = Re(z)) We can justify this step since in general, if z = a + ib then (a + ib)e it + (a ib)e it = a(e it + e it ) + ib(e it e it ) = a cost b sint Terms in ε 1 : x 1 t + x 1 + x 0 t T + (x 0 1) x 0 t x 1 t + x 1 + (A ie it A ie it ) + = 0 [ (Ae it + Ae it ) 1 ] (iae it iae it ) = 0 where A = A (T) = A T If we wish to find A(T) as opposed to x 1 we need only consider secular terms. So we need to equate coefficients of e it (and e t ) to zero and kill them off. considering terms in e it we have: ia ia ia A + ia A = 0 A A + A A = 0 Similarly, if we consider e it we get the conjugate of this expression; ie; whatever A kills off the e it terms also kills off the e it terms. We should now use the polar form of A, as A A = A 3, and if we write A = A e iϕ with ϕ = arg(a) and A = 1 a then Now da dt = 1 da dt eiϕ + 1 dϕ aie iϕ dt which dividing through by e iϕ gives x 0 = a cos(t + ϕ) and we now wish to find a and so it remains to solve a e iϕ + iae iϕ ϕ 1 aeiϕ + a 4 eiϕ a e iϕ = 0 a = 1 a a3 iaϕ = 0 We now equate real and imaginary parts: ϕ = 0 ϕ is constant
The Phase Plane. Phase portraits; type and stability classifications of equilibrium solutions of systems of differential equations
The Phase Plane Phase portraits; type and stability classifications of equilibrium solutions of systems of differential equations Phase Portraits of Linear Systems Consider a systems of linear differential
More informationLecture L19  Vibration, Normal Modes, Natural Frequencies, Instability
S. Widnall 16.07 Dynamics Fall 2009 Version 1.0 Lecture L19  Vibration, Normal Modes, Natural Frequencies, Instability Vibration, Instability An important class of problems in dynamics concerns the free
More informationCALCULUS 2. 0 Repetition. tutorials 2015/ Find limits of the following sequences or prove that they are divergent.
CALCULUS tutorials 5/6 Repetition. Find limits of the following sequences or prove that they are divergent. a n = n( ) n, a n = n 3 7 n 5 n +, a n = ( n n 4n + 7 ), a n = n3 5n + 3 4n 7 3n, 3 ( ) 3n 6n
More informationcorrectchoice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:
Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that
More informationProblem 1 (10 pts) Find the radius of convergence and interval of convergence of the series
1 Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series a n n=1 n(x + 2) n 5 n 1. n(x + 2)n Solution: Do the ratio test for the absolute convergence. Let a n =. Then,
More informationSOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve
SOLUTIONS Problem. Find the critical points of the function f(x, y = 2x 3 3x 2 y 2x 2 3y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Partial derivatives
More informationThis makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5
1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy,
More information3.2 Sources, Sinks, Saddles, and Spirals
3.2. Sources, Sinks, Saddles, and Spirals 6 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay 00 C By 0 C Cy D 0. These are linear equations with constant coefficients
More information2 Integrating Both Sides
2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation
More information1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives
TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of
More informationEXAM. Practice Questions for Exam #2. Math 3350, Spring April 3, 2004 ANSWERS
EXAM Practice Questions for Exam #2 Math 3350, Spring 2004 April 3, 2004 ANSWERS i Problem 1. Find the general solution. A. D 3 (D 2)(D 3) 2 y = 0. The characteristic polynomial is λ 3 (λ 2)(λ 3) 2. Thus,
More information4 Linear Dierential Equations
Dierential Equations (part 2): Linear Dierential Equations (by Evan Dummit, 2012, v. 1.00) Contents 4 Linear Dierential Equations 1 4.1 Terminology.................................................. 1 4.2
More informationDefinition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =
Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a
More informationMATH 381 HOMEWORK 2 SOLUTIONS
MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e
More informationLectures 56: Taylor Series
Math 1d Instructor: Padraic Bartlett Lectures 5: Taylor Series Weeks 5 Caltech 213 1 Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular,
More informationExample 1: Competing Species
Local Linear Analysis of Nonlinear Autonomous DEs Local linear analysis is the process by which we analyze a nonlinear system of differential equations about its equilibrium solutions (also known as critical
More informationTaylor Polynomials and Taylor Series Math 126
Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will
More informationChapter 8. Nonlinear systems. 8.1 Linearization, critical points, and equilibria
Chapter 8 Nonlinear systems 8. Linearization, critical points, and equilibria Note: lecture, 6. 6. in [EP], 9. 9.3 in [BD] Except for a few brief detours in chapter, we considered mostly linear equations.
More informationEigenvalues and Eigenvectors
Chapter 6 Eigenvalues and Eigenvectors 6. Introduction to Eigenvalues Linear equations Ax D b come from steady state problems. Eigenvalues have their greatest importance in dynamic problems. The solution
More informationSequences and Series
Sequences and Series Consider the following sum: 2 + 4 + 8 + 6 + + 2 i + The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite
More informationTaylor and Maclaurin Series
Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions
More informationPROBLEM SET. Practice Problems for Exam #2. Math 2350, Fall Nov. 7, 2004 Corrected Nov. 10 ANSWERS
PROBLEM SET Practice Problems for Exam #2 Math 2350, Fall 2004 Nov. 7, 2004 Corrected Nov. 10 ANSWERS i Problem 1. Consider the function f(x, y) = xy 2 sin(x 2 y). Find the partial derivatives f x, f y,
More informationPractice Final Math 122 Spring 12 Instructor: Jeff Lang
Practice Final Math Spring Instructor: Jeff Lang. Find the limit of the sequence a n = ln (n 5) ln (3n + 8). A) ln ( ) 3 B) ln C) ln ( ) 3 D) does not exist. Find the limit of the sequence a n = (ln n)6
More informationSimple Harmonic Motion
5 Simple Harmonic Motion Note: this section is not part of the syllabus for PHYS26. You should already be familiar with simple harmonic motion from your first year course PH115 Oscillations and Waves.
More information3. INNER PRODUCT SPACES
. INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.
More informationExam 1 Sample Question SOLUTIONS. y = 2x
Exam Sample Question SOLUTIONS. Eliminate the parameter to find a Cartesian equation for the curve: x e t, y e t. SOLUTION: You might look at the coordinates and notice that If you don t see it, we can
More informationSystem of First Order Differential Equations
CHAPTER System of First Order Differential Equations In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown functions
More information5 Indefinite integral
5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse
More information1 3 4 = 8i + 20j 13k. x + w. y + w
) Find the point of intersection of the lines x = t +, y = 3t + 4, z = 4t + 5, and x = 6s + 3, y = 5s +, z = 4s + 9, and then find the plane containing these two lines. Solution. Solve the system of equations
More informationHooke s Law. Spring. Simple Harmonic Motion. Energy. 12/9/09 Physics 201, UWMadison 1
Hooke s Law Spring Simple Harmonic Motion Energy 12/9/09 Physics 201, UWMadison 1 relaxed position F X = kx > 0 F X = 0 x apple 0 x=0 x > 0 x=0 F X =  kx < 0 x 12/9/09 Physics 201, UWMadison 2 We know
More informationREVIEW EXERCISES DAVID J LOWRY
REVIEW EXERCISES DAVID J LOWRY Contents 1. Introduction 1 2. Elementary Functions 1 2.1. Factoring and Solving Quadratics 1 2.2. Polynomial Inequalities 3 2.3. Rational Functions 4 2.4. Exponentials and
More informationSECONDORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS
L SECONDORDER LINEAR HOOGENEOUS DIFFERENTIAL EQUATIONS SECONDORDER LINEAR HOOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS A secondorder linear differential equation is one of the form d
More information1. Firstorder Ordinary Differential Equations
Advanced Engineering Mathematics 1. Firstorder ODEs 1 1. Firstorder Ordinary Differential Equations 1.1 Basic concept and ideas 1.2 Geometrical meaning of direction fields 1.3 Separable differential
More informationMath 150, Fall 2009 Solutions to Practice Final Exam [1] The equation of the tangent line to the curve. cosh y = x + sin y + cos y
Math 150, Fall 2009 Solutions to Practice Final Exam [1] The equation of the tangent line to the curve at the point (0, 0) is cosh y = x + sin y + cos y Answer : y = x Justification: The equation of the
More informationCalculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum
Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum UNIT I: The Hyperbolic Functions basic calculus concepts, including techniques for curve sketching, exponential and logarithmic
More informationSecondOrder Linear Differential Equations
SecondOrder Linear Differential Equations A secondorder linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1
More informationNumerical Methods for Differential Equations
Numerical Methods for Differential Equations Course objectives and preliminaries Gustaf Söderlind and Carmen Arévalo Numerical Analysis, Lund University Textbooks: A First Course in the Numerical Analysis
More information1 Mathematical Induction
Extra Credit Homework Problems Note: these problems are of varying difficulty, so you might want to assign different point values for the different problems. I have suggested the point values each problem
More informationSome Notes on Taylor Polynomials and Taylor Series
Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited
More informationSolutions to Linear Algebra Practice Problems
Solutions to Linear Algebra Practice Problems. Find all solutions to the following systems of linear equations. (a) x x + x 5 x x x + x + x 5 (b) x + x + x x + x + x x + x + 8x Answer: (a) We create the
More informationMATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.
MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin
More informationMATH 461: Fourier Series and Boundary Value Problems
MATH 461: Fourier Series and Boundary Value Problems Chapter III: Fourier Series Greg Fasshauer Department of Applied Mathematics Illinois Institute of Technology Fall 2015 fasshauer@iit.edu MATH 461 Chapter
More informationDynamics. Figure 1: Dynamics used to generate an exemplar of the letter A. To generate
Dynamics Any physical system, such as neurons or muscles, will not respond instantaneously in time but will have a timevarying response termed the dynamics. The dynamics of neurons are an inevitable constraint
More informationPhysics 53. Oscillations. You've got to be very careful if you don't know where you're going, because you might not get there.
Physics 53 Oscillations You've got to be very careful if you don't know where you're going, because you might not get there. Yogi Berra Overview Many natural phenomena exhibit motion in which particles
More informationPURE MATHEMATICS AM 27
AM Syllabus (015): Pure Mathematics AM SYLLABUS (015) PURE MATHEMATICS AM 7 SYLLABUS 1 AM Syllabus (015): Pure Mathematics Pure Mathematics AM 7 Syllabus (Available in September) Paper I(3hrs)+Paper II(3hrs)
More informationPURE MATHEMATICS AM 27
AM SYLLABUS (013) PURE MATHEMATICS AM 7 SYLLABUS 1 Pure Mathematics AM 7 Syllabus (Available in September) Paper I(3hrs)+Paper II(3hrs) 1. AIMS To prepare students for further studies in Mathematics and
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationExamination paper for Solutions to Matematikk 4M and 4N
Department of Mathematical Sciences Examination paper for Solutions to Matematikk 4M and 4N Academic contact during examination: Trygve K. Karper Phone: 99 63 9 5 Examination date:. mai 04 Examination
More informationMath 2280  Assignment 6
Math 2280  Assignment 6 Dylan Zwick Spring 2014 Section 3.81, 3, 5, 8, 13 Section 4.11, 2, 13, 15, 22 Section 4.21, 10, 19, 28 1 Section 3.8  Endpoint Problems and Eigenvalues 3.8.1 For the eigenvalue
More informationReview Solutions MAT V1102. 1. (a) If u = 4 x, then du = dx. Hence, substitution implies 1. dx = du = 2 u + C = 2 4 x + C.
Review Solutions MAT V. (a) If u 4 x, then du dx. Hence, substitution implies dx du u + C 4 x + C. 4 x u (b) If u e t + e t, then du (e t e t )dt. Thus, by substitution, we have e t e t dt e t + e t u
More information4. Complex integration: Cauchy integral theorem and Cauchy integral formulas. Definite integral of a complexvalued function of a real variable
4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complexvalued function of a real variable Consider a complex valued function f(t) of a real variable
More informationThe Geometric Series
The Geometric Series Professor Jeff Stuart Pacific Lutheran University c 8 The Geometric Series Finite Number of Summands The geometric series is a sum in which each summand is obtained as a common multiple
More informationPeriodic Motion or Oscillations. Physics 232 Lecture 01 1
Periodic Motion or Oscillations Physics 3 Lecture 01 1 Periodic Motion Periodic Motion is motion that repeats about a point of stable equilibrium Stable Equilibrium Unstable Equilibrium A necessary requirement
More informationMath 201 Lecture 23: Power Series Method for Equations with Polynomial
Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients Mar. 07, 2012 Many examples here are taken from the textbook. The first number in () refers to the problem number in
More informationAn important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate.
Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set
More information6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives
6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise
More information299ReviewProblemSolutions.nb 1. Review Problems. Final Exam: Wednesday, 12/16/2009 1:30PM. Mathematica 6.0 Initializations
99ReviewProblemSolutions.nb Review Problems Final Exam: Wednesday, /6/009 :30PM Mathematica 6.0 Initializations R.) Put x@td = t  and y@td = t . Sketch on the axes below the curve traced out by 8x@tD,
More informationLecture 31: Second order homogeneous equations II
Lecture 31: Second order homogeneous equations II Nathan Pflueger 21 November 2011 1 Introduction This lecture gives a complete description of all the solutions to any differential equation of the form
More informationEIGENVALUES AND EIGENVECTORS
Chapter 6 EIGENVALUES AND EIGENVECTORS 61 Motivation We motivate the chapter on eigenvalues b discussing the equation ax + hx + b = c, where not all of a, h, b are zero The expression ax + hx + b is called
More informationWASSCE / WAEC ELECTIVE / FURTHER MATHEMATICS SYLLABUS
Visit this link to read the introductory text for this syllabus. 1. Circular Measure Lengths of Arcs of circles and Radians Perimeters of Sectors and Segments measure in radians 2. Trigonometry (i) Sine,
More informationAgain, the limit must be the same whichever direction we approach from; but now there is an infinity of possible directions.
Chapter 4 Complex Analysis 4.1 Complex Differentiation Recall the definition of differentiation for a real function f(x): f f(x + δx) f(x) (x) = lim. δx 0 δx In this definition, it is important that the
More information2 FirstOrder Equations: Method of Characteristics
2 FirstOrder Equations: Method of Characteristics In this section, we describe a general technique for solving firstorder equations. We begin with linear equations and work our way through the semilinear,
More informationDerive 5: The Easiest... Just Got Better!
Liverpool John Moores University, 115 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de Technologie Supérieure, Canada Email; mbeaudin@seg.etsmtl.ca 1. Introduction Engineering
More informationSolutions for Review Problems
olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector
More informationLinear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices
MATH 30 Differential Equations Spring 006 Linear algebra and the geometry of quadratic equations Similarity transformations and orthogonal matrices First, some things to recall from linear algebra Two
More informationSection 10.7 Parametric Equations
299 Section 10.7 Parametric Equations Objective 1: Defining and Graphing Parametric Equations. Recall when we defined the x (rcos(θ), rsin(θ)) and ycoordinates on a circle of radius r as a function of
More informationFourier Series. A Fourier series is an infinite series of the form. a + b n cos(nωx) +
Fourier Series A Fourier series is an infinite series of the form a b n cos(nωx) c n sin(nωx). Virtually any periodic function that arises in applications can be represented as the sum of a Fourier series.
More informationGeneral Theory of Differential Equations Sections 2.8, 3.13.2, 4.1
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, 3.13.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions
More information2 Complex Functions and the CauchyRiemann Equations
2 Complex Functions and the CauchyRiemann Equations 2.1 Complex functions In onevariable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)
More informationIntroduction to Algebraic Geometry. Bézout s Theorem and Inflection Points
Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a
More information1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style
Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with
More informationtegrals as General & Particular Solutions
tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx
More informationSection 1.1. Introduction to R n
The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to
More informationA First Course in Elementary Differential Equations. Marcel B. Finan Arkansas Tech University c All Rights Reserved
A First Course in Elementary Differential Equations Marcel B. Finan Arkansas Tech University c All Rights Reserved 1 Contents 1 Basic Terminology 4 2 Qualitative Analysis: Direction Field of y = f(t, y)
More informationChapter 1 Vectors, lines, and planes
Simplify the following vector expressions: 1. a (a + b). (a + b) (a b) 3. (a b) (a + b) Chapter 1 Vectors, lines, planes 1. Recall that cross product distributes over addition, so a (a + b) = a a + a b.
More informationProblems for Quiz 14
Problems for Quiz 14 Math 3. Spring, 7. 1. Consider the initial value problem (IVP defined by the partial differential equation (PDE u t = u xx u x + u, < x < 1, t > (1 with boundary conditions and initial
More informationSimple harmonic motion
PH122 Dynamics Page 1 Simple harmonic motion 02 February 2011 10:10 Force opposes the displacement in A We assume the spring is linear k is the spring constant. Sometimes called stiffness constant Newton's
More informationNumerical Methods for Differential Equations
Numerical Methods for Differential Equations Chapter 1: Initial value problems in ODEs Gustaf Söderlind and Carmen Arévalo Numerical Analysis, Lund University Textbooks: A First Course in the Numerical
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationMatrix Methods for Linear Systems of Differential Equations
Matrix Methods for Linear Systems of Differential Equations We now present an application of matrix methods to linear systems of differential equations. We shall follow the development given in Chapter
More informationCalculus. Contents. Paul Sutcliffe. Office: CM212a.
Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical
More information4.3 Lagrange Approximation
206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average
More informationPractice Problems for Midterm 2
Practice Problems for Midterm () For each of the following, find and sketch the domain, find the range (unless otherwise indicated), and evaluate the function at the given point P : (a) f(x, y) = + 4 y,
More informationLecture 3: Solving Equations Using Fixed Point Iterations
cs41: introduction to numerical analysis 09/14/10 Lecture 3: Solving Equations Using Fixed Point Iterations Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mark Cowlishaw, Nathanael Fillmore Our problem,
More informationThe Fourth International DERIVETI92/89 Conference Liverpool, U.K., 1215 July 2000. Derive 5: The Easiest... Just Got Better!
The Fourth International DERIVETI9/89 Conference Liverpool, U.K., 5 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de technologie supérieure 00, rue NotreDame Ouest Montréal
More informationCOLLEGE ALGEBRA. Paul Dawkins
COLLEGE ALGEBRA Paul Dawkins Table of Contents Preface... iii Outline... iv Preliminaries... Introduction... Integer Exponents... Rational Exponents... 9 Real Exponents...5 Radicals...6 Polynomials...5
More informationCalculating Areas Section 6.1
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 Calculating Areas Section 6.1 Dr. John Ehrke Department of Mathematics Fall 2012 Measuring Area By Slicing We first defined
More informationwww.mathsbox.org.uk ab = c a If the coefficients a,b and c are real then either α and β are real or α and β are complex conjugates
Further Pure Summary Notes. Roots of Quadratic Equations For a quadratic equation ax + bx + c = 0 with roots α and β Sum of the roots Product of roots a + b = b a ab = c a If the coefficients a,b and c
More informationBlue Pelican Calculus First Semester
Blue Pelican Calculus First Semester Teacher Version 1.01 Copyright 20112013 by Charles E. Cook; Refugio, Tx Edited by Jacob Cobb (All rights reserved) Calculus AP Syllabus (First Semester) Unit 1: Function
More informationD f = (2, ) (x + 1)(x 3) (b) g(x) = x 1 solution: We need the thing inside the root to be greater than or equal to 0. So we set up a sign table.
. Find the domains of the following functions: (a) f(x) = ln(x ) We need x > 0, or x >. Thus D f = (, ) (x + )(x 3) (b) g(x) = x We need the thing inside the root to be greater than or equal to 0. So we
More informationk=1 k2, and therefore f(m + 1) = f(m) + (m + 1) 2 =
Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 +2 2 + +n 2 = 1 n(n+1)(2n+1) for all n N. 6 Put f(n) = n(n + 1)(2n + 1)/6. Then f(1) = 1, i.e the theorem
More information18.03 Final Exam Review
8.3 Final Exam Review Chris Kottke Contents Definitions & Terminology. Single Equations............................................ Systems of Equations......................................... 3.3 Initial
More informationAdvanced Higher Mathematics Course Assessment Specification (C747 77)
Advanced Higher Mathematics Course Assessment Specification (C747 77) Valid from August 2015 This edition: April 2016, version 2.4 This specification may be reproduced in whole or in part for educational
More informationHigher Order Linear Differential Equations with Constant Coefficients
Higher Order Linear Differential Equations with Constant Coefficients Part I. Homogeneous Equations: Characteristic Roots Objectives: Solve nth order homogeneous linear equations where a n,, a 1, a 0
More informationArea Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by
MATH 42, Fall 29 Examples from Section, Tue, 27 Oct 29 1 The First Hour Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x
More information1. Periodic Fourier series. The Fourier expansion of a 2πperiodic function f is:
CONVERGENCE OF FOURIER SERIES 1. Periodic Fourier series. The Fourier expansion of a 2πperiodic function f is: with coefficients given by: a n = 1 π f(x) a 0 2 + a n cos(nx) + b n sin(nx), n 1 f(x) cos(nx)dx
More informationComplex Numbers and the Complex Exponential
Complex Numbers and the Complex Exponential Frank R. Kschischang The Edward S. Rogers Sr. Department of Electrical and Computer Engineering University of Toronto September 5, 2005 Numbers and Equations
More informationG.A. Pavliotis. Department of Mathematics. Imperial College London
EE1 MATHEMATICS NUMERICAL METHODS G.A. Pavliotis Department of Mathematics Imperial College London 1. Numerical solution of nonlinear equations (iterative processes). 2. Numerical evaluation of integrals.
More informationInfinite series, improper integrals, and Taylor series
Chapter Infinite series, improper integrals, and Taylor series. Introduction This chapter has several important and challenging goals. The first of these is to understand how concepts that were discussed
More informationFINAL EXAM SOLUTIONS Math 21a, Spring 03
INAL EXAM SOLUIONS Math 21a, Spring 3 Name: Start by printing your name in the above box and check your section in the box to the left. MW1 Ken Chung MW1 Weiyang Qiu MW11 Oliver Knill h1 Mark Lucianovic
More information