M412 Assignment 9 Solutions

Transcription

1 M4 Assignment 9 Solutions. [ pts] Use separation of variables to show that solutions to the quarter-plane problem can be written in the form u t u xx ; t>,<x< u(t, ) u(t, + ) bounded u(,x)f(x), <x<, u(t, x) e ωt sin ωxdω, for some appropriate constant. Solution. Looking for solutions u(t, x) T(t)X(x), we find which gives the eigenvalue problem T (t) T (t) X (x) X(x) λx, X (x)+λx(x) X() X(x) <. x We conclude that the eigenvalues are all positive real numbers λ>, with associated eigenfunctions X λ (x) sin λx. In this way, the most general solution is u(t, x) A(λ)e λt sin λxdλ, where A(λ) is to be determined from the initial condition. In order to put this in the form stated in the problem, make the change of variable ω λ, which gives dλ ωdω and consequently u(t, x) A(ω )e ωt sin ωx(ω)dω. Upon choosing ωa(ω ), we arrive at the claimed relationship.. [ pts] Show that the coefficient from Problem satisfies π ( is called the Fourier sine transform of f.) Solution. First, the initial condition u(,x) f(x) gives f(x) f(x)sinωxdx. sinωxdx.

2 For any real number k>, multiply both sides of this last expression by sin kx and integrate x from to. Wehave f(x)sinkxdx sin kx N sinωxdωdx sin kx N sinωxdωdx sin kx sin ωxdxdω, where in the second step we have switched the order of integration. (In general, this step requires some assumptions on, but since is what we are searching for, we assume the steps are valid, keeping in mind that in principle they should later be verified. Though we won t carry out the verification here.) We now employ the trigonometric relationship to write this last integral as sin A sin B cos(a B) cos(a + B), Upon integration over x, this last expression becomes N cos[(k ω)x] cos[(k + ω)x]dxdω. [ sin[(k ω)n] (k ω) sin[(k + ω)n] ] dω. (k + ω) For the integration involving sin[(k + ω)n], (k + ω) we observe that with k>we never have the issue of division by, and hence the Riemann Lebesgue lemma gives that the it as N of this integral is. For the remaining term, we make the change of variable, z k ω, fromwhichwehave sin[(k ω)n] dω (k ω) L L N k k L k L + k L C(k z) sin[zn] ( dz) C(k z) sin[zn] dz ( ) sin[zn] C(k z) C(k) dz C(k) sin[zn] dz. The Riemann Lebesgue lemma asserts that the first of these two integrals goes to in the it as N. For the second, we make the change of variable y zn, from which we observe, k L C(k) sin[zn] dz L C(k) N N (k L)N C(k) sin[y] y N sin y y dy C(k)π, N dy

3 where in this final equality we have used the integration sin y dy π. y Solving for C(k), we conclude C(k) f(x)sinkxdx, π which is precisely the relationship we set out to establish, with the independent variable denoted k rather than ω. 3. [5 pts] Haberman.3.3. Solution. First recall the relationship (e iθ e iθ, where denote complex conjugate. (That is, (α + iβ α iβ. This relationship is straightforward to prove: (e iθ (cosθ+isin θ (cosθ isin θ) e iθ.) The Fourier transform of f(x) is F(ω) e iωx f(x)dx. The complex conjugate of this is F (ω) e iωx f(x)dx F ( ω). If you re uncertain about bringing the complex conjugate inside the integration, consider it in the following way. ( + F (ω) e iωx f(x)dx ( + (cos ωx + i sinωx)f(x)dx ( ( 4. [5 pts] Haberman.3.7. Solution. In this case, compute directly, F [F (ω)] cos ωxf(x)dx + i cos ωxf(x)dx i e iωx f(x)dx. e iωx e α ω dω e (ix+α)ω dω + ix + α ix α α x α α α + x. sin ωxf(x)dx sin ωxf(x)dx e (ix α)ω dω (ix α) (ix + α) (ix + α)(ix α) 3

4 5. In this problem, we will combine three problems from Haberman to solve the PDE 5a. [5 pts] Haberman.3.8. Solution. Recalling that F (ω) we differentiate with respect to ω to find u t ku xxx u(,x)f(x). e iωx f(x)dx, df dω ixe iωx f(x)dx. The claimed relationship follows from multiplication on both sides of this last expression by i. 5b. [ pts] Haberman.4.6. Proceed here by taking a Fourier transform of the equation for y(x) and using (5a). Solution. Taking the Fourier transform of d y xy, dx we have F[ d y ] F[xy]. dx According to (5a), this gives ω ŷ + iŷ (ω), which can be solved by separation of variables for ŷ(ω) Ce i 3 ω3, for some constant of integration C. Inverting, we have y(x) e iωx Ce i 3 ω3 dω. In order to put this in the form Haberman recommends, use Euler s formula to get y(x) C C C e iωx i 3 ω3 dω cos[ωx + 3 ω3 ] i sin[ωx + 3 ω3 ]dω cos[ωx + 3 ω3 ]dω, where in the last equality we have observed the integration properties of even and odd functions. Finally, according to Haberman, we have the initial condition y() π 4 cos[ 3 ω3 ]dω,

5 which provides C. We conclude y(x) cos[ωx + π 3 ω3 ]dω, typically referred to as the Airy function and denoted Ai(x). 5c. [ pts] Haberman.4.7, Parts (a), (b), and (c). In Part (a), Haberman is only asking that you show u(t, x) F [ˆf(ω)e ikω3t ]. In Part (b), you will write this as a convolution, while in Part (c) you will need to compute F [e ikω3t ] in terms of the Airy function Ai(x) from Part (5b). In this last calculation, you will want to make the change of variables z /3 ω. You can check your final answer in the back of Haberman. Solution. First, taking a Fourier transform of the equation and its initial conditions, we find û t iω 3 kû û(,ω)ˆf(ω), which is solved by û(t, ω) ˆf(ω)e ikω3t. By the Fourier Inversion Theorem, this gives u(t, x) F [ˆf(ω)e ikω3t ]. For Part (b), we note that by the Convolution Theorem, u(t, x) F [ˆf(ω)] F [e ikω3t ]f(x) F [e ikω3t ]. For Part (c), the key is to recognize how F [e ikω3t ] can be written in terms of the Airy equation. We have F [e ikω3t ] e iωx+ikω3t dω, in which we make the suggested change of variables z /3 ω, giving e i x /3 z i 3 z3 F dz [e ikω3 ] /3 x cos[ /3 z + /3 3 z3 ]dz x Ai( ), / /3 5

6 where Ai( ) is as in Haberman.4.6. We conclude x u(t, x) f(x) Ai( / ) /3 / / Ai( y x )f(y)dy. /3 Ai( (x y) )f(y)dy /3 6