x x2 2 B A ) v(0, t) = 0 and v(l, t) = 0. L 2. This is a familiar heat equation initial/boundary-value problem and has solution

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1 Hints to homewok d. The poblem is u t ku xx + k ux fx u t A u t B. It has a souce tem and inhomogeneous bounday conditions but none of them depend on t. So as in example 3 of the notes we should find the equilibium tempeatue distibution u eq x and subtact that fom u. The function u eq x satisfies: ku eq + k u eq A u eq B. so u eq c + c 2 x 2 x2. To have u eq A we need c A and to have u eq B we need to solve A + c B fo c 2. So c B A and u eq x A B A x x2 2. Next let vx t ux t u eq x. Since ku eq xx k we ll have and since u eq A and u eq B we ll have Finally we have v t u t ku xx + k ku u eq xx kv xx v t and v t. vx ux u eq x fx A 2 B A x + x2 2. This is a familia heat equation initial/bounday-value poblem and has solution whee Finally c n 2 with these same constants c n. vx t c n e kn2 π 2 t/ 2 nπx sin [ fx A 2 B A ux t A B A x x2 2 + ] x + x2 sin 2 c n e kn2 π 2 t/ 2 sin The poblem is u t k u + Q u f ua t T

2 2 togethe with the implicit condition that u t is finite. Once again this is a poblem with an equilibium solution u eq. We find it by solving k u eq + Q u eq finite u eq a T. To solve the diffeential equation move the Q to the ight and unwind the deivatives and multiplications that ae applied to u eq on the left as follows: Fist multiply both sides by /k and integate fom to to get u eq k ρqρ dρ. Next divide by and integate this time fom to a to avoid the / singulaity at be caeful to change the sign since the is the lowe limit and add a constant to match the bounday condition: a u eq T + kσ σ ρqρ dρ dσ. Now as usual let v t u t u eq. Since k/u eq Q we ll have v t u t k u + Q k u u eq k v and v will be finite and va. The initial conditions fo v will be v u u eq f a kσ σ ρqρ dρ dσ. The v poblem is a familia heat equation poblem on the disk and has solution 2 a 2 J z n 2 vx t zn c n e kz2 n t/a2 J a whee z n is the nth positive zeo of the Bessel function J x and a [ a σ c n f ρqρ dρ dσ Finally u t with these same constants c n. a kσ σ kσ ρqρ dρ dσ + ] zn J d. a zn c n e kz2 n t/a2 J a 8.3.c. This time the poblem is u ku xx + Qx t ux fx u t At u x t. No equilibium solution this time so we ll just take something simple to get id of the bounday values: u bd x t At so that u bd is zeo when x and u bd is equal to At when x. We have u bd t A t and u bd xx

3 3 so the function vx t ux t u bd x t satisfies v t u t u bd t ku xx + Qx t A t ku u bd xx + Qx t A t kv xx + Qx t A t togethe with the bounday conditions v t and v x t and the initial condition vx fx A. Since the x pat actually the non-t pat of the diffeential equation is simply v xx we seek to expand vx t as a seies of eigenfunctions of X + λx X X and with time-dependent coefficients as follows: 2 vx t a n t sin πx n and we know the eigenvalues ae n π 2 / 2. We substitute this into the diffeential equation fo v to get a nt + kn π 2 n 2 a n t sin 2 πx Qx t A t. We have to expand the ight side in a simila way as a sinish seies in x with t-dependent coefficients so let n t 2 [Qx t A 2 t] sin πx Then the equation becomes a nt + kn π 2 n 2 a n t sin and so we have to solve the diffeential equations 2 πx a n + kn π 2 2 a n n t n n t sin 2 πx fo a n t. The integating facto fo this fist-ode linea equation is e kn+ 2 2 π 2 t/ 2 so we have t a n t c n e kn+ 2 2 π 2 t/ 2 + e kn+ 2 2 π 2 t/ 2 e kn+ 2 2 π 2 τ/ 2 n τ dτ. We detemine the constants c n using the initial conditions fo v since we have vx 2 c n sin πx fx A. Expand the ight side as a Fouie seies and lean that c n 2 [fx A 2 ] sin πx

4 4 Put all of this togethe to get whee in which and ux t At + n a n t sin 2 πx t a n t c n e kn+ 2 2 π 2 t/ 2 + e kn+ 2 2 π 2 t/ 2 e kn+ 2 2 π 2 τ/ 2 n τ dτ n t 2 c n 2 [Qx t A 2 t] sin πx dx [fx A 2 ] sin πx This time the poblem is u ku xx + Qx ux fx u t u t. Even though we expect an equilibium solution the text diects us to use eigenfunction expansion so we wite ux t a n t sin and we know the eigenvalues ae n 2 π 2 / 2. We substitute this into the diffeential equation fo u to get a nt + kn2 π 2 2 a n t sin Qx. We expand the ight side as a sine seies in x and let n 2 Then the equation becomes a nt + kn2 π 2 2 Qx sin a n t sin and so we have to solve the diffeential equations a n + kn2 π 2 2 a n n n sin fo a n t. By integating factos o undetemined coefficients we get that the solution is a n t c n 2 n kn 2 π 2 e kn2 π 2 t/ n kn 2 π 2 we wite it this way so that a n c n. We detemine the constants c n using the initial conditions fo u since we have ux c n sin fx.

5 5 Expand the ight side as a Fouie seies and lean that Put all of this togethe to get whee in which and a n t c n 2 ux t fx sin a n t sin c n 2 n kn 2 π 2 e kn2 π 2 t/ n kn 2 π 2 n 2 c n 2 Qx sin fx sin dx As t a n t tends to the constant 2 n /kn 2 π 2 so we have lim ux t t 2 kn 2 π 2 n sin. This function is clealy zeo at both ends of the inteval and its second deivative is by the definition of n as we would expect. n nπx k sin k Qx The poblem is u t u xx + e 2t sin 5x u t uπ t ux. et u bd x x/π and vx t ux t u bd x. Since u bd is linea in x v t u t u xx + e 2t sin 5x v xx + e 2t sin 5x and v t vπ t and vx u bd x x/π. Seek vx t in the fom vx t a n t sin nx. The diffeential equation fo v becomes a n + n 2 a n sin nx e 2t sin 5x which tells us that a n + n 2 a n unless n 5 in which case a a 5 e 2t. Theefoe a n t c n e n2 t if n 5

6 6 and a 5 t c 5 e 25t e 2t via the method of undetemined coefficients we subtacted /23 fom c 5 so that a 5 c 5 just as a n c n fo all the othe values of n. We detemine the coefficients fom the initial condition vx a n sin nx c n sin nx x π. Thus c n 2 π π x π sin nx dx 2 π [ x π cos nx n π + nπ π ] cos nx dx 2 nπ. We conclude that and so vx t 23 e 2t e 25t sin 5x ux t x π + 23 e 2t e 25t sin 5x 2 2t nπ e n sin nx 2 2t nπ e n sin nx b. The poblem is u tt c 2 u xx + gx cos ωt u t u t ux fx u t x. Since the bounday values ae homogeneous we launch into ou eigenfunction expansion letting whee a n t must satisfy whee ux t a n t sin a n + c2 n 2 π 2 2 a n n cos ωt n 2 gx sin ae the Fouie sine coefficients of gx and whee to match the initial data. a 2 fx sin dx dx and a We ll use the method of undetemined coefficients to solve the diffeential equation fo a n t. If ω cnπ/ then we guess a n A cos ωt and get that A n c 2 n 2 π 2 2 ω 2

7 7 so a n t 2 n c n c 2 n 2 π 2 2 ω 2 cos cnπt + d n sin cnπt 2 n + c 2 n 2 π 2 2 cos ωt. ω2 If ω cnπ/ fo all n then thee is no esonance and we will have d n fo all n and c n a n so can wite the solution of the poblem as whee ux t n 2 [ 2 n c n c 2 n 2 π 2 2 ω 2 cos gx sin cnπt + dx and c n 2 2 ] n c 2 n 2 π 2 2 cos ωt sin ω2 fx sin If ω cnπ/ fo some positive intege N then we will have esonance and we have to solve sepaately fo a N t fom all the othes the othe a n s emain the same as calculated above. The function a N t satisfies which we ll wite as a N + c2 N 2 π 2 2 a N N cos a N + ω 2 a N N cos ωt cnπt fo simplicity with N a N and a N as befoe. Fo ou undetemined coefficients this time we ll have to guess a N t At sin ωt which gives us a N + ω 2 a N 2Aω cos ωt Aω 2 t sin ωt + Aω 2 t sin ωt 2Aω cos ωt N cos ωt so A N /2ω and so a N t c N cos ωt + d N sin ωt + Nt sin ωt. 2ω Since a N c N we detemine c N the same way as all the othe c n s. And a N ωd N is supposed to be zeo so d N as well. Theefoe in the case of esonance whee ω cnπ/ we have ux t Nt Nπx [ 2 n cnπt 2 ] n nπx sin ωt sin + c n 2ω c 2 n 2 π 2 2 ω 2 cos + c 2 n 2 π 2 2 cos ωt sin ω2 whee n 2 gx sin dx and c n 2 fx sin The exta tem with t sin ωt will make the solution oscillate moe and moe violently as t gows The Fouie tansfom of fx { if x > a if x < a is assuming a > F ω fxe iωx dx a e iωx dx eiωx 2π 2π a 2πiω a a eiωa e iωa 2πiω sin ωa πω.

8 The invese Fouie tansfom of F ω e ω α is povided α > fx eα ixω α ix F ωe iωx dω e αω e iωx dω + e α ixω dω + ω ω α ix + α + ix 2α α 2 + x 2 + e α ixω α ix e αω e iωx dω e α ixω dx the limiting values of the exponentials at ± ae zeo because e a x goes to zeo as x goes to ± and e iωx stays bounded. ω ω.3.8. The Fouie tansfom of xfx is F[xfx]ω xfxe iωx 2π The deivative with espect to ω of the Fouie tansfom of f is d dω F[fx]ω d fxe iωx dx dω 2π 2π fx d dω eiωx dx i xfxe iωx dx 2π we assume we can exchange the deivative and the integal this will be tue povided fx is smooth and goes to zeo quickly enough as x goes to ±. Compaing these two equations shows that d F[fx] if[xfx] dω o multiplying both sides by i F[xfx] i d dω F[fx].