The formulas for derivatives are particularly useful because they reduce ODEs to algebraic expressions. Consider the following ODE d 2 dx + p d

Transcription

1 Solving ODEs using Fourier Transforms The formulas for derivatives are particularly useful because they reduce ODEs to algebraic expressions. Consider the following ODE d 2 dx + p d 2 dx + q f (x) R(x) x where p and q are constants. Transform both sides of the equations FT d 2 f (x) df (x) + p + qf (x) dx 2 dx ( (ik)2 + p(ik) + q) f (k) FT (R(x)) R(k) where ~ denotes the Fourier transform. We then have R(k) f (k) k 2 + ipk + q so that the solution (particular) is f (x) 1 e ikx f (k)dk 1 e ikx R(k) k 2 + ipk + q dk This very formal solution to the problem is called the integral representation of the solution. In general, we need complex integration techniques to evaluate these integrals. We will see how later. Solving ODEs with the Laplace Transform We now have all the necessary tools to solve ODEs using the Laplace transform. We consider an ODE with constant coefficients Now if d 2 y 2 + a dy L(y(t)) Y (s) + by r(t) we then have L d 2 y + a dy 2 + by L(r(t)) R(s) s 2 Y (s) sy() y'() + a sy (s) y() [ ] + by (s) R(s) which is now an algebraic equation. The solution is (s + a)y() + y'() R(s) Y (s) + s 2 + as + b s 2 + as + b Examples: y(t) L 1 (Y (s)) Page 1

2 (1) Consider the homogeneous ODE d 2 y + 4 dy 2 + 8y with initial conditions y() 2, y'(). We have Now or and or Y (s) 2s + 8 2(s + 2) s 2 + 4s + 8 (s + 2) (s + 2) s L(e at f (t)) F(s a) and L(cosωt) s 2 + ω 2 L(e 2t cos2t) L(cos2(t + 2)) (s + 2) (s + 2) ω L(e at f (t)) F(s a) and L(sinωt) s 2 + ω 2 L(e 2t sin2t) L(sin2(t + 2)) 2 (s + 2) Therefore, 2s + 8 2(s + 2) Y (s) s 2 + 4s + 8 (s + 2) (s + 2) L(e 2t cos2t) + 2L(e 2t sin2t) L(y(t)) L(2e 2t (cos2t + sin2t)) y(t) 2e 2t (cos2t + sin2t) (2) Consider the nonhomogeneous ODE 2 d 2 q + 5q 1sinωt 2 with initial conditions q() i() dq(). This ODE arises from a series circuit with an AC voltage source, a capacitor and an inductance. We have Q(s) (s + a)q() + i() s L(5sinωt) s L(5sinωt) s ω (s 2 + ω 2 )(s ) or using the partial fraction rules when a quadratic remains 5ω Q(s) (s 2 + ω 2 )(s ) A s + A 1 2 s 2 + ω + B s + B s Now for this form Page 2

3 or Therefore, P(s) B 1 (a + ib) + B 2 Q(s) / (s a) 2 + b 2 5ω A 1 (iω) + A 2 s ω B 1 (5i) + B 2 s 2 + ω 2 siω s5i sa+ib 5ω ω A, A 5ω ω 2 5ω ω 2 25 B 1, B 2 5ω ω ω Q(s) 25 ω 2 s 2 + ω ω 2 s or 5 q(t) [ sinωt sin5t ] 25 ω 2 Clearly, this form of the solution is valid only if ω 5 since the amplitude becomes unbounded. This corresponds to resonance in the circuit and we have an unbounded amplitude because there is no damping in this circuit (no resistors). If ω 5, we go back to Q(s) 5ω Q(s) (s ) A s + A (s ) + B s + B s (s ) 2 which using the same procedure gives q(t) 25[ sin5t 5t cos5t] unbounded as t gets large Green Function and Convolution Consider the special equation shown below: d 2 dx + p d 2 dx + q G(x x') δ(x x') x In this equation the inhomogeneous function is a delta-function, x' is an arbitrary point, and we denote this special solution as G(x). It is called the Green function. We now show that the solution of the equation d 2 dx + p d 2 dx + q f (x) R(x) x is easily expressed in terms of the Green function. Page 3

4 We now show that the solution of the inhomogeneous ODE can be written as the expression f (x) G(x x')r(x')dx' Substituting into the ODE we get d 2 dx + p d 2 dx + q G(x x')r(x')dx' as it should! d 2 dx + p d 2 dx + q G(x x')r(x')dx' δ(x x')r(x')dx' R(x) So the Green function representation of the solution is valid. This means if we solve for G(x-x') for a particular differential operator, then we have solved for all inhomogeneous solutions of the ODE given by that particular differential operator. As we saw earlier, an integral of the form above involving G(xx') is called a convolution, and is denoted by the symbol (*). f (x) G(x x')r(x')dx' (G * R) x The Fourier transform of a convolution is always the product of the transforms(same as for Laplace transforms) FT ((G * R) x ) 1 1 dxe ikx G(x x')r(x')dx' 1 1 dxe ik(x x ') G(x x')e ikx ' R(x')dx' 1 1 e ikx ' R(x')dx' dxe ik(x x ') G(x x') FT (G)F(R) In Linear Response Theory, the inhomogeneity function R(x) is called the input to, the solution f(x) is called the output from, the system, while the Green function is called the response function, since it describes how the system responds to the input. Examples: Consider Newton's 2nd law, Page 4

5 m d 2 x(t) F(t) force function 2 This is a 2nd-order inhomogeneous ODE with p q and R F. The Green function satisfies this equation with the force function replaced by a delta-function. A delta-function force is called an impulsive force. Thus, the Green function also describes the response of a mechanical system to an impulsive driving force. Now let us consider a damped, driven oscillator x(t) + 2β x(t) + ω 2 x(t) R(t) The solution will be of the form x(t) G(t t ')R(t ') ' where d 2 G(t t ') dg(t t ') + 2β + ω 2 2 G(t t ') δ(t t ') Let the Fourier transform of G(t-t') be given by FT (G(t t ')) G(ω) 1 e iω (t t ') G(t t ') d(t t ') Now take the Fourier transform of the Green function ODE FT d 2 G(t t ') dg(t t ') + 2β + ω 2 2 G(t t ') FT (δ(t t ')) or ω iβω + ω 1 G(ω) G(ω) 1 1 (ω 2 ω 2 ) + 2iβω Finally we compute the Green function itself by taking the inverse Fourier transform G(t) 1 e iωt 1 1 G(ω)dω e iωt (ω 2 ω 2 ) + 2iβω dω Complex integration(later) gives the result where G(t) 1 ω 1 e βt sin(ω 1 t)h (t) Page 5

6 ω 1 ω 2 β 2 and H (t) 1 t> step function t< Finally, the solution of the original equation is x(t) 1 e β (t t ') sin(ω 1 (t t '))H (t t ')R(t ') ' ω 1 t 1 e β (t t ') sin(ω 1 (t t '))R(t ') ' ω 1 Note the interesting feature that the integral over t' takes into account the effects of all driving forces occurring in the past (t' < t) due to the presence of the step function. It contains no effect due to the driving force in the future (t' > t), because these forces have not yet occurred. Hence, the result is explicitly consistent with the physical requirement of causality. Is the answer correct? Let us choose a driving force where we know the answer. In Physics 7 we discussed sinusoidal driving forces; so we choose R(t) e iωt Direct integration of the x(t) equation then gives the result e iωt x(t) (ω 2 Ω 2 ) + 2iβΩ which corresponds to the resonance amplitude formula of the damped, driven oscillator and agrees with the earlier result. More details of these integration techniques later. Examples: Consider M d 2 x + α dx + Kx d(t) 2 Assume the Fourier transform of d(t) is D(ω) and that X(ω) Substituting we get 1 e iωt x(t) x(t) 1 dωe iωt X(ω) Page 6

7 Now defining we have x(t) 1 ( Mω 2 + iαω + K)X(ω) D(ω) X(ω) dωe iωt D(ω) / M ω 2 iαω / M K / M ω ± iα 2M ± K M α 2 4M 2 D(ω) / M ω 2 iαω / M K / M 1 D(ω) / M dω (ω ω + )(ω ω ) eiωt where x(t) is the response of the oscillator to the forcing function d(t). Response to a Delta-Function Impulse Consider This gives d(t) I δ(t) D(ω) I 1 x(t) 1 I dω / M (ω ω + )(ω ω ) eiωt Using complex integration techniques which we learn later we get t < x(t) t > x(t) I KM α M sin e αt K M α 2 4M t 2 The solution looks like Page 7

8 We note that the solution obeys causality, i.e., there is no motion until after the impulse is applied. A delta-function impulse implies that x() and v x () Notice, however, that we have a solution with no unknown constants. We started with a second-order equation. Normally, that means we need to use 2 boundary conditions to completely define the solution. Somehow we have already imposed them. Where? It turns out that the choice of integration path in the complex integration was the place! Effectively, we have done the following: causality x() δ function impulse v x () I / M Fourier methods fail where the response diverges(unstable systems). In this case we must use the Laplace transform. Page 8

9 Example - Unstable Electric Circuit C o + c _ i(t) + + o -R r o - l + L o This circuit has a negative resistance (a theorist s circuit) and is therefore unstable. We let s (t) t < sinω t t > For t <, all voltages and i(t). We have for t > the integral-diffeq s (t) c (t) + r (t) + l (t) 1 di i(t) R C i(t) + L Taking the derivative wrt t we have d 2 i L R di i(t) d s C which is a second-order non-homogeneous diffeq where i(t) dependent variable and d s / is the driving term. Neither s nor d s / have a valid Fourier transform. Thus, we use Laplace transforms. We get s 2 L I(s) sr I(s) + 1 C I(s) s s (s) where I(s) L(i(t) and s (s) L( s (t)) The initial conditions (due to causality) are Page 9

10 Now i() and s (s) di t ω (s + iω )(s iω ) which gives I(s) ω s (s + iω )(s iω )(L s 2 R s + 1 ω s ) (s + iω )(s iω )L (s s + )(s s ) C where s ± R 2L ± i 1 R 2 L C Note that R R negative resistance Real(s ± ) >. Inverting(using complex integration methods) we have i(t) 1 ds i ω se st (s + iω )(s iω )L (s s + )(s s ) L For simplicity we choose ω 1, L 1 so that i(t) 1 ds i se st (s + i)(s i)(s 1+ 2i)(s 1 2i) L For t < we get i(t). For t > we close to the left, we get i(t) cos ( t +.15π ) et cos( 2t +. ) The first term is a result of the driving voltage at ω 1. It persists at a constant amplitude for all t >. The second term is the characteristic response of the circuit (determined solely by R, L,C ). It grows exponentially in time because of R < (an unstable circuit). It looks like(where we have enhanced the initial interference terms by replacing e t with e.1t. 4L 2 Page 1

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