Problem 1. MATH 241-2 FA 214 HOMEWORK 1 SOUTIONS Note that u E (x) 1 ( x) is an equilibrium distribution for the homogeneous pde that satisfies the given boundary conditions. We therefore want to find solutions v u u E v t 2 v x + 2 e 2t sin 5x, v(, t) v(, t), v(x, ) 1 (x ). The eigenfunctions of the related homogeneous problem satisfy d 2 X + λx, X() X(). dx2 We therefore know that these eigenfunctions are sin nx, corresponding to λ n 2. We therefore wish to write v(x, t) b n (t) sin nx. Plugging v into the pde yields b n(t) sin nx n 2 b n (t) sin nx + e 2t sin 5x. We therefore have that b n (t) solves the ode b 5(t) + 25b 5 (t) e 2t b n(t) + n 2 b n (t) if n 5. Moreover, using the initial condition, we have that and so 1 (x ) b n () sin nx, b n () 2 2 (x ) sin nx dx 2 n. When n 5, the homogeneous ode has solutions ce n2t. We therefore have that b n (t) 2 n e n2t. When n 5, the ode will have a specific solution 1 23 e 2t. The general solution is therefore 1 23 e 2t + ce 25t, and we see that b 5 (t) 1 23 e 2t ( 1 + 2 23 5 )e 25t. Since we have determined b n (t), we know v(x, t), and thus u(x, t). 1
Problem 2. (a) The eigenfunctions of the related homogeneous problem satisfy: d 2 X λx, X() X(). dx2 We therefore have that the eigenfunctions are sin nx, corresponding to λ ( ) n 2. We therefore wish to write u(x, t) b n (t) sin nx. Plugging u into the pde yields b n(t) sin nx ( cn ) 2 bn (t) sin nx + q n (t) sin nx where q n (t) sin nx Q(x, t), so that q n(t) 2 Q(x, t) sin nx dx. We therefore have that b n (t) solves the ode ( cn ) 2 b n(t) + bn (t) q n (t). We can obtain two boundary conditions using the two initial conditions. We have that We also have that f(x) u(x, ) The general solution to the ode is c 1 cos cnt b n () sin nx, so that b n() 2 u t (x, ) b n() sin nx, so that b n(). + c 2 sin cnt + cn Using the boundary conditions, we have that c 1 2 therefore determined b n (t), and thus we have determined u. t f(x) sin nx dx. cn(t τ) q n (τ) sin dτ (see 8.5.19). nx f(x) sin dx and c 2. We have (b) When Q(x, t) g(x) cos ωt, we have that q n (t) a n cos ωt, where a n sin nx is the Fourier sine series of g(x). Once again writing u(x, t) b n(t) sin nx, we have that b n(t) + ( cn In this case, for generic ω, we see that that ) 2 bn (t) a n cos ωt, b n () 2 b n (t) c 1 cos cnt a n ω 2 ( cn f(x) sin nx dx, b n(). ) 2 cos ωt is a particular solution for the ode, so + c 2 sin cnt + a n ω 2 ( cn 2 ) 2 cos ωt,
where c 1 and c 2 are obtained as in part (a). This formula, however, is not defined when ω cn for some n. It is for these values that resonance occurs. In this case, we have a specific (nonperiodic) solution an cnt cn t sin. For the n with ω, we can once again determine b 2cn n(t) using this specific solution. Problem 3. (a) We first consider the corresponding eigenvalue problem 2 u + λu, u(a, θ). When considering a solution u f(r)g(θ), we obtain the system of odes g + µg, g( ) g(), g ( ) g ( ) ( r d dr r df dr) + (λr 2 µ)f, f(a), f() < The solutions to the θ-valued equation are linear combinations of cos mθ and sin mθ, corresponding to µ m 2. The eigenvalues of the r-calued equation are λ r(f ) 2 + m2 f 2 dr r a f. 2 r dr We therefore have that f satisfies the order m Bessel equation, so that f(r) J m ( λr). Choose λ m n to be the n th smallest number so that J m ( λ mn a). We then have that the eigenvalues are φ mn cos mθj m ( λ mn r) and ψ mn sin mθj m ( λ mn r). We therefore wish to write u(r, θ) m1 A mn φ mn + Plugging this into the original pde yields Q(r, θ) λ mn A mn φ mn + The coefficients are m1 A mn 1 λ m n B mn 1 λ m n m1 m1 B mn ψ mn. λ mn B mn ψ mn. Q(r, θ) cos mθj m( λ mn r)r dr dθ cos2 mθj m ( λ mn r) 2 r dr dθ Q(r, θ) sin mθj m( λ mn r)r dr dθ sin2 mθj m ( λ mn r) 2 r dr dθ 3
(c) We will first begin by solving 2 u E, u E (a, θ) f(θ). A separable solution u(r, θ) R(r)g(θ) to aplace s equation will satisfy the ode g + λg, g( ) g(), g ( ) g ( ) ( ) r d dr r dr dr λr, R() < The θ-valued equation has solutions cos mθ and sin mθ, corresponding to λ n 2. The corresponding r-valued equation is then solved by R(r) r n. We therefore have u E (r, θ) A n r n cos nθ + B n r n sin nθ. We can solve for the coefficients using the boundary condition: A 1 A n 2 a n B n 2 a n f(θ) dθ f(θ) cos nθ dθ when n 1 f(θ) sin nθ dθ We have therefore determined u E. It remains to solve 2 u Q(r, θ), u(a, θ). However, we have already solved this in part (a). The solution to the given pde is then u E + u. Problem 4. The eigenfunctions of the related homogeneous problem satisfy: d 2 X λx, X() X(). dx2 We therefore have that the eigenfunctions are sin nx, corresponding to λ n 2. We therefore wish to write u(x, t) b n (y) sin nx. Plugging u into the pde yields b n(y) sin nx + n 2 b n (y) sin x e 2y sin x. We therefore have that b n solves the ode b n(y) n 2 b n (y) if n 1 b 1(y) b 1 (y) e 2y 4
Moreover, since u(x, ) b n () sin nx, we have that b n(). Additionally, since f(x) u(x, ) b n () sin nx, we have that b n() 2 f(x) sin nx dx. The homogeneous ode b n(y) n 2 b n (y) has solutions sinh ny and cosh ny. For n 1, we can now determine b n by evaluating the boundary conditions. We see that b n (y) B n sinh y, where B n 2 sinh n f(x) sin nx dx. To determine b 1,we still need to determine a specific solution for the given ode. There should be a particular solution of the form ce 2y. By plugging this function into the ode, we see that 1 3 e2y is a specific solution. The general solution to the ode is then 1 3 e2y + A sinh y + B sinh( y) (we have chosen a basis for the the homogeneous solutions that are nicer with respect to the boundary conditions). Plugging in the boundary conditions, we see that we get b 1 (y) when B 1 3 sinh and A 1 ( 2 f(x) sin nx dx 1 ) sinh 3 e 2. Since we have determined the b n (y), he know u(x, y). Problem 5. (i) (ii) [ ] 4x F F (x 2 + 9) 2 F[e 4(x 2)2 ] e 2iω F[e 4x2 ] 1 e 2iω e ω2 16 16 [ ( )] [ ] d 2 2 iωf 13 [ dx x 2 + 9 x 2 + 9 iωf 6 x 2 + 9 ] 1 3 iωe 3 ω (iii) F [f(x)] 3 sin 2ω ω 5
Problem 6. et U(ω, y) be the Fourier transform of u(x, y) with respect to x. Then [ ] [ ] u F F k 2 u t x + c u 2 x U t kω2 U ciωu subject to U(ω, ) F (ω), where F (ω) F[f]. We therefore have that U(ω, t) F (ω)e (kω2 +ciω)t. Note that so that We therefore have that u(x, t) f(x) [ ] F 1 e kω2 t [ ] F 1 e ciωt e kω2 t (x+ct) 2 kt e 4kt. x 2 kt e 4kt, kt e (x+ct) 2 4kt. Problem 7. et U(ω, y) be the Fourier transform of u(x, y) with respect to x. Then F [ ] 2 u x + 2 u 2 U 2 y 2 y 2 ω2 U, subject to U(x, ) F 1 (ω) and U(x, H) F 2 (ω), where F i is the Fourier transform of f i. Note that the ode has solutions sinh ωy and sinh ω(h y) (we could use other bases for the space of solutions, such as cosh ωy and sinh ωy, but the chosen basis is more convenient). We therefore have that U(ω, y) A(ω) sinh ωy + B(ω) sinh ω(h y) for some functions A and B. By using the boundary conditions, we conclude that We then have that u(x, y) U(ωy) F 2(ω) sinh ωy + F 1(ω) sinh ω(h y). sinh ωh sinh ωh ( F2 (ω) sinh ωy + F ) 1(ω) sinh ω(h y) e iωx dx. sinh ωh sinh ωh 6
Problem 8. et U(ω 1, ω 2, t) be the Fourier transform of u(x, y) with respect to both x and y. Then [ ] [ ] u 2 u F F k 1 t x + k 2 u 2 2 y 2 U t k 1ω1U 2 k 2 ω2u 2 subject to U(ω 1, ω 2, ) F (ω 1, ω 2 ), where F F[f]. It follows that U(ω 1, ω 2, t) F (ω 1, ω 2 )e (k 1ω 2 1 +k 2ω 2 2 )t. Note that It follows that u(x, y, t) [ ] F 1 e (k 1ω1 2+k 2ω2 2)t e (k 1ω 2 1 +k 2ω 2 2 )t e i(ω 1x+ω 2 y) dx dy e k 1ω 2 1 t e ixω 1 dx k1 k 2 t 1 t 2 e x2 k 1 t 1 y2 k 2 t 2. k1 k 2 t 1 t 2 f(x, y) e x2 k 1 t 1 y2 k 2 t 2. e k 2ω 2 2 t e ixω 2 dx Prolbem 9. Note that T is an equilibrium solution to the pde, satisfying the given boundary condition. We may therefore examine u t k r r ( r u ) + Q(r), u(r, ) f(r), u(a, t), u(, t) <. r If u(r, t) R(r)T (t) is a solution to the related homogeneous pde, then R and T satisfy T + λkt ( ) r d dr r dr dr + λr 2 R, R(a), R() <. Since the eigenvalues of the r-valued ode are non-negative, the r-valued equation is the order Bessel equation. We therefore have that R(r) J ( λ n r), where λ n is the n th smallest zero of J (r). We will therefore write u(r, t) b n(t)j ( λ n r). Plugging this into the pde yields b n(t)j ( λ n r) λ n kb n (t)j ( λ n r) + q n J ( λ n r). Here, q n are the coefficients for the generalized Fourier series of Q with respect to J ( λ n r): q n Q(r)J ( λ n r)r dr J 2 (. λ n r)r dr 7
We therefore have that b n (t) solves the ode b n(t) + λ n kb n (t) q n. Note that the solutions to the associated homogeneous ode are ce λnkt. We also have a specific solution of kλ n. We therefore have that b n (t) c n e λnkt + qn kλ n for some value of c n. To determine c n, we will use the initial condition Using orthogonality, we see that f(r) u(r, ) b n () b n ()J ( λ n r). f(r)j ( λ n r)r dr J 2 (, λ n r)r dr and so c n b n () f(r)j ( λ n r)r dr J 2 ( q n λ n r)r dr. We therefore know b n (t), and therefore know u(r, t). q n Problem 1. To prove that u(x, t) c(ω)e iωx e kω2t dω is real, it suffices to show that u is equal to u. Now u(x, t) u(x, t) c(ω)e iωx e kω2 t dω c( ω)e iωx e kω2t dω c(ω)e iωx e kω2t dω c(ω)e iωx e kω2t dω (use a change of coordinates from ω to ω) 8
Problem 11. If f(x) then f(x β) F (ω)e iωx dω, F (ω)e iω(x β) dω In other words, e iωβ F (ω) is the Fourier transform of f(x β). e iωβ F (ω)e iωx dω. 9