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1 AMS B Perturbation Methods Lecture 3 Copyright by Hongyun Wang, UCSC Recap Iterative method for finding asymptotic series requirement on the iteration formula to make it work Singular perturbation use scaling to get rid of singularity Divergence of asymptotic series Next we discuss asymptotic expansions of functions, such as solutions of ODEs. Asymptotic expansion of ƒ(x, ϵ) Definition: Suppose φ i ( ε) N a i i= { } is an asymptotic sequence. ( x)φ i ( ε) is called a straightforward asymptotic series of ƒ(x, ϵ) as ε if at each fixed x. N f ( x, ε) a i ( x)φ i ( ε) = o φ N ε i= ( ) as ε Initial value problems (IVP) of ODE Example: (nondimensionalization and identifying the ε) Consider the -D motion of an object launched upward from the surface of the earth. Let v: initial velocity τ: time h(τ): height (from the surface of the earth) at time τ R = 637 km: radius of the earth The gravitational force is f = G mm ( R + h) - -

2 AMS B Perturbation Methods m: mass of the object M: mass of the Earth The gravitational acceleration at height h is a = f m = G M ( R + h) The gravitational acceleration at h = is given G M g = 9.8 m/s R ==> a = g R ( R + h) = g + h R The governing equation of the object is d h dτ = g + h R =, h ( ) = v h Non-dimensionalization: We select time and length scales: time = v g length = v g We introduce non-dimensional time and height: τ t = ( v g), y = h v g ( g) y ==> τ = ( v g)t, h = v ==> d dτ = ( v g) d dt d dτ = v d ( g) dt - -

3 ==> dh dτ = v d h dτ = v ( g) dy AMS B Perturbation Methods dτ = v ( g) d y h R = v g R ( g) g dt = v ( v ) dy dτ = v y = v R g y The non-dimensional equation: d y dt = + v gr y Identifying the ε: Let ε = v gr. ( g) y g dt d ( v ) When v is not very large, we have ϵ dy dt = g d y dt To make it more precise, let us consider the escape velocity. Escape velocity (view from physics): The gravitational potential is G mm R + h The kinetic energy is mv If the object can escape form the gravity of the Earth, then (by conservation of energy) we have mv G mm R = mv ==> v G M R = G M R R = gr ==> v ( esc ) = gr h=

4 AMS B Perturbation Methods v ( esc ) = gr.8 km/s Escape velocity (view from differential equations): (Skip the derivation in lecture) We re-write the dimensional equation as dv dt = g + h R Multiply both sides by v, we get vdv = g vdt = g dh + h R + h R Let H be the maximum height the object can reach (that is, when h(t) = H, we have v(t) = ). Integrating both sides yields v H vdv = g dh + h R h=h ==> v = gr + h R h= = gr + H R Setting H =, we obtain the escape velocity = gr v ( esc ) ==> v ( esc ) = gr Based on the expression of escape velocity v ( esc ) = gr, we conclude ε = v gr = v v esc - 4 -

5 AMS B Perturbation Methods That is, ε = v gr << corresponds to v << v ( esc). The non-dimensional IVP: d y dt = + ε y =, y ( ) = y We seek an expansion of the form: y( t ) = a ( t )+ εa ( t )+! Initial conditions: a ( )+ εa ( )+!= ==> a() =, a() = a ( )+ ε a ( )+!= ==> a () =, a () = Substituting into equation ( ) a ( t )+ ε a ( t )+!= ( + εa ( t )+!) = εa t ==> a ( t )+ + ε a t a ( t ) +!= ε : a ( t )+ = =, a ( ) = a ==> a ( ) = t +t ε : ( t ) a t a = =, a ( ) = a ==> t a = a ( t ) = t +t ==> a t = t 4 + t

6 AMS B Perturbation Methods ==> y( t )~ t +t + ε t 4 + t 3 3 Let us see a more meaningful example. Example: (oscillation of a pendulum) (Draw a pendulum with length L, angle θ) Governing equation (Newton s second law) ml θ = mgsin θ ==> θ = g L sin ( θ ) = ε, θ ( ) = θ This set of initial conditions corresponds to the case of displacing a pendulum and then letting it go. Nondimensionalization: time = t new = t old L g g L ==> dθ = dθ dt new = g dt old dt new dt old L dθ dt new ==> d θ dt = g old L d θ dt new = ε, θ ( ) = θ = sin θ θ This is a singular perturbation. At ϵ =, there is no oscillation. Scaling (so that the scaled displacement remains O() as ε ) Let = θ ( t ) y t ε - 6 -

7 AMS B Perturbation Methods ==> =, y ( ) = y = ε sin ε y y Goal: To find an asymptotic expansion of y(t, ϵ). We notice that ε sin ( ε y ) is a function of ε only. ε sin ε y = ( ε ε y ε y ) 3 = y ε y 3 3! ( + ε y ) 5 5! 3! + ε4 y 5 5!! So we seek an expansion of the form: y( t ) = a ( t )+ ε a ( t )+! Initial conditions: a ( )+ ε a ( )+!= ==> a() =, a() =. a ( )+ ε a ( )+!= ==> a () =, a () =, Substituting it into the equation, a + ε a = ( a + ε a )+ 6 ε 3 a ==> a + a + ε a + a 6 a 3 +!=! ε : a + a = =, a ( ) = a Result #: The solution of a + a = =, a ( ) = a is a(t) = cos(t)

8 AMS B Perturbation Methods (See Appendix A for derivation.) ε : =, a ( ) = a + a = 6 cos3 t a Result #: The solution of =, a ( ) = a + a = 6 cos3 t a a ( t ) = t 6 sin ( t )+ (See Appendix B for derivation.) ( ) 9 cos ( t ) cos 3t is Combining results # & #, we arrive at the expansion: ( ) t y( t,ε)~cos( t )+ ε 6 sin ( t )+ 9 cos ( t ) cos 3t Remark: As t, we have y(t, ϵ), which is inconsistent with the periodic behavior of the physical solution. ==> This asymptotic approximation is invalid for large t. It is valid only for short time. For large t, we will use two methods: *) multi-scale expansion *) strained variable expansion Now we find the period of oscillation. Let T(ε) be the period of oscillation. As ε, we have y(t, ϵ) = cos(t) ==> T() = π ==> T ε = π( + ε α +! ) - 8 -

9 AMS B Perturbation Methods Note: Again, T(ε) should be a function of ε. Goal: Find the value of α. By definition, T(ε) satisfies ( ) = y T ε Differentiating with respect to t, we have ( ) y ( t,ε)~ sin( t )+ ε 6 sin ( t )+ t 6 cos ( t )+ 9 sin ( t )+3sin 3t Substituting T( ε) = π + ε απ +! into the expansion of y (t, ϵ) and keeping only terms up to O(ε ), we obtain ε απ + ε π 6 +!= ==> α = 6 ==> T ε = π + ε 6 +! Appendix A Derivation of Result #: The solution of a + a = =, a ( ) = a Solution using characteristic equation: The characteristic equation for a + a = is is a(t) = cos(t). λ + = It has two roots λ = i and λ = i where i = A general solution of a + a = is - 9 -

10 a ( t ) = α exp it AMS B Perturbation Methods + α exp it ==> a ( t ) = c cos( t )+ c sin t Using the initial conditions a() = and a () =, we obtain a(t) = cos(t). Solution using Laplace transform: Laplace transform of function f(t) is a function of s defined as L f ( t ) e st f t F s s dt Laplace transform of derivatives L f t = s L f t f L f t = s L f t f = s L f ( t ) s f ( ) f Let A(s) = L[a(t)]. Taking the Laplace transform of a + a = and using a() = and a () =, we have Recall s A( s) sa ( ) a ( )+ A s ==> A s s + ==> A( s) = s s + = s = L sin at = a s + a, a L s + a = sin at L cos at = s s + a, s L s + a = cos at Taking the inverse Laplace transform, we obtain = L A( s) a t = s L s = cos t + - -

11 AMS B Perturbation Methods Appendix B Derivation of Result #: The solution of =, a ( ) = a + a = 6 cos3 t a a ( t ) = t 6 sin ( t )+ ( ) 9 cos ( t ) cos 3t is To solve Result #, we first need to express cos 3 (t) in terms of cos(3t) and cos(t). exp cos( t ) ( it )+ exp( it ) = cos 3 t = 8 ==> cos 3 t ( exp( it )+ exp( it )) 3 = ( exp( 3it )+3exp( it )+3exp( it )+ exp( 3it )) 8 = ( cos( 3t )+3cos( t )) 4 = 4 ( cos( 3t )+3cos( t )) Solution using method of undetermined coefficients: ) Write the ODE as a + a = cos( 3t ) cos ( t ) ) Use the method of undetermined coefficients to find a particular solution of a + a = cos( 3t ). 4 Substituting p ( t ) = c cos( 3t )+ c sin( 3t ) into equation - -

12 AMS B Perturbation Methods ==> p ( t ) = cos( 3t ) 9 3) Use the method of undetermined coefficients to find a particular solution of a + a = 3 4 cos ( t ). Substituting p ( t ) = c t cos( t )+ c t sin( t ) into equation ==> p ( t ) = t 6 sin ( t ) Think about why we use p ( t ) = c t cos( t )+ c t sin( t ). 4) A general solution of a + a = cos( 3t ) cos ( t ) is a ( t ) = c cos( t )+ c sin( t )+ t 6 sin ( t ) cos( 3t ) 9 Using the initial conditions a() = and a () =, we obtain a ( t ) = t 6 sin ( t )+ ( ) 9 cos ( t ) cos 3t Solution using Laplace transform: Let A(s) = L[a(t)]. Taking the Laplace transform of a + a = 4 conditions a() = and a () =, we have s A( s) sa ( ) a ( )+ A s = ==> A( s) ( s +) = 4 s s s s + ==> A( s) = ( s + 9) s s + cos( 3t )+ 3 4 cos ( t ), and using initial 4 s s s s + 4 s s + = s 9 ( s +) s ( s + 9 ) s s + - -

13 AMS B Perturbation Methods Recall that L s s + a = cos at Differentiating with respect to a, we obtain L s ( s + a ) a s ==> L s + a = t sin( at ) = t sin( at ) a Taking the inverse Laplace transform yields = L a t s 9 ( s +) s ( s + 9 ) s s + = ( 9 cos ( t ) cos( 3t ))+ t 6 sin t - 3 -