u(r, θ) = 1 + 3a r n=1

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1 Mth 45 / AMCS 55. etuck Assignment 8 ue Tuesdy, Apil, 6 Topics fo this week Convegence of Fouie seies; Lplce s eqution nd hmonic functions: bsic popeties, computions on ectngles nd cubes Fouie!, Poisson s fomul fo the disk Eighth Homewok Assignment - due Tuesdy, Apil Reding: Red sections 6.4, 7., 9., 9. nd. of the text Be peped to discuss the following poblems in clss: Pge 75 poblems, 6 Pge 83 poblem 3 Pge 33 poblems, 3 Pge 4 poblems 3,5 Pge 63 poblems, 5 Pge 75, poblem. Fo u outside the cicle nd u bounded t infinity, we need to use the Fouie seies with the n tems. Since the boundy dt on is u, θ + sin 3θ, which is ledy tigonometic polynomil, the seies will hve just two nonzeo tems: u, θ + 3 sin θ. Pge 75, poblem 6. solution of u : Since u, u, π, we need only the n sin nπθ tems in the u, θ B n n sin nθ. n And to hve u, θ π sin θ sin θ, we should tke B π, B nd the othe B n. So the solution is u, θ π sin θ sin θ. Pge 83, poblem 3. This is like pge 75, poblem, except the coefficient x is function the thn constnt nd we e in n dimensions now. But the poof poceeds the sme wy: As usul with uniqueness poofs, let u nd u be ny two solutions of the poblem, nd then thei diffeence v u u stisfies v in, v n + u on bdy

2 Wite v v nd ecll the identity v v v v + v nd clculte: v v d vol v v v d vol bd bd v v n dσ v d vol v v n dσ v d vol v dσ v d vol bd the lst equlity becuse of the boundy condition. Since x > the only wy fo the lst line to be zeo is fo both integnds to be zeo, in which cse v is constnt becuse v thoughout, nd the constnt is zeo becuse v on the boundy of. Thus vx so the two solutions u nd u of the oiginl poblem must be the sme, poving uniqueness. Pge 33, poblems, 3. Let s do these togethe nd in n dimensions. Wite x x, x,... x n. We ll clculte fo wht exponent p the function ux, t c t x x p stisfies the wve eqution pehps except on the light cone emnting fom the oigin, whee it will not be defined if p <. Clculte: so we get nd Theefoe u t t c u u x i px i c t x x p u x i u t pc c nd u t pc tc t x x p pc t x x p + 4pp x i c t x x p pc c t x x p + 4pp c 4 t c t x x p n i p c t x x p + 4pp c 4 t c n x i i pc n + c t x x p + 4pp c c t x xc t x x p pc n + + p c t x x p c t x x p nd this will be zeo fo ll x nd t if eithe p duh o n + + p, in othe wods if p n nd this specilizes to the esults of poblem nd 3 fo n 3 nd espectively. Fo n this only gives the constnt solution, but note tht in this cse ux, t ln c t x woks.

3 3 Pge 4, poblem 3. We wnt to solve the thee-dimensionl wve eqution with initil dt ux, y, z, nd u t x, y, z, y. Accoding to fomul 3 in section 9., the solution is ux, t 4πc y ds t whee S is the sphee of dius ct centeed t x. Now we don t hve to do n ctul sufce integl hee, since it s cle tht the vege vlue of y on ny sphee centeed t the point x, y, z is y, nd the e of sphee of dius ct is 4πc t. So the integl is equl to 4πc t y. So S ux, y, z, t 4πc t y 4πc t ty. Pge 4, poblem 5. By the pinciple of cuslity, if the initil dt vnish outside the sphee of dius R, then the solution ux, t vnishes outside the sphee of dius R + ct. Pge 63, poblem. Homogeneous Neumnn conditions implies tht the eigenfunctions of the Lplcin on the sque tke the fom cos mx cos ny fo m, n. So the genel solution of the wve eqution u tt c u with these boundy conditions is ux, t A + B t + m n m,n, cos mx cos nya mn cos m + n ct + B mn sin m + n ct The initil condition ux, mens tht ll the A mn, nd u t x, sin x + cos x mens tht B, B 4 nd ll the othe B mn e zeo. So. ux, t t + cos x sin t 4 Pge 63, poblem 5. Since u + x u is homogeneous line second-ode odiny diffeentil eqution, its solutions e detemined by the choice of initil dt vlue of u nd of u t one point, so the dimension of the solution spce is. b The genel solution of u + π u is l u c cos πt l + c sin πt l nd ll of these solutions stisfy peiodic boundy conditions on the intevl l, l, so the dimension of the eigenspce is. c The solution of the Neumnn poblem on the disk o on ny domin is unique up to dding constnt. So the hmonic functions e the constnt functions on the disk, which compise vecto spce of dimension. d The Neumnn solutions on the sque e of the fom cos mπx cos nπy fo m, n,,,..., nd the coesponding eigenvlues e λ mn m + n π. Thee e fou wys to get 5π, nmely

4 4 m, n 3, 4, m, n 4, 3, m, n 5, nd m, n, 5. So the dimension of this spce is 4. e The spce of ll solutions of the wve eqution is the set of functions ux, t of the fom ux, t fx + ct + gx ct fo bity functions f nd g of one vible, so this spce is of infinite dimension. Wite up solutions of the following to hnd in: Pge 75 poblems 4, 5, Pge 83 poblem 5 Pge 33 poblems, 5 Pge 4 poblems 4, 6 Pge 63 poblems, 3 Pge 75, poblem 4. The poblem is u outside the cicle with u, θ hθ nd u bounded s, so we know tht whee A n π u, θ A + n n A n cos nθ + B n sin nθ n π π hθ cos nθ dθ nd B n π π π hθ sin nθ dθ. We jm this ll togethe, chnging the vible of integtion in the coefficients to ϕ, to wite: u, θ π π π π π π π π π hϕ hϕ hϕ π n π π hϕ dϕ + π n hϕ cos nϕ dϕ cos nθ + hϕ sin nϕ dϕ sin nθ n π π π + n cos nϕ cos nθ + sin nϕ sin nθ dϕ πn n π + n cosnθ nϕ dϕ πn + n n n n e inθϕ + e inθϕ dϕ

5 5 Now the quntity in the bckets is + the sum of two geometic seies; it is e iθϕ n e iθϕ n n n e iθϕ eiθϕ + e iθϕ eiθϕ + eiθϕ e iθϕ + + eiθϕ e iθϕ e iθϕ + e θϕ e iθϕ e eθϕ + cosθ ϕ + Theefoe which is fomul 9. ux, t π π hϕ π cosθ ϕ + dϕ Pge 75, poblem 5. We stt by witing the genel solution fo the nnulus, which hs ll the n tems s well s the n tems: u, θ A + C ln + n A n cos nθ + B n sin θ + n C n cos nθ + n sin nθ. n Now fo ech pt of the poblem we hve to mtch the boundy conditions. The boundy conditions e u, θ nd u, θ sin θ cos θ. So we should hve the constnt tem A the coefficient of the ln tem is zeo becuse its deivtive t is not zeo nd then we hve to choose A nd C so tht fo we will hve A + C cos θ cos θ nd, since the noml deivtive on the oute boundy is u, θ n n A n cos nθ + B n sin nθ n n C n cos nθ + n sin nθ n fo we need 4A 4 C cos θ. So A nd C stisfy the line system 4 4 A C Theefoe A C so A 34 nd C 8 7. We conclude tht u, θ 34 cos θ 8 7 cos θ.

6 6 b This time, u, θ nd u, θ cos θ. So we e going to need both A nd C s well s A nd C. Since the constnt tem t is just A, we hve A. But the constnt tem t is A + C ln, so fo this to be zeo we need C / ln. Next, fo the cos θ tems we need A + C t nd 4A + 4 C t. So this time, A nd C stisfy the line system A 4 4 C Theefoe A C so A 3 nd C 8 5. We conclude tht u, θ ln ln + 8 cos θ cos θ. 3 5 Pge 75, poblem. As usul with uniqueness poofs, let u nd u be ny two solutions of the poblem, nd then thei diffeence v u u stisfies v in, v n + u on bdy Wite v v nd ecll the identity v v v v + v nd clculte: v v d vol bd bd v v v d vol v v n dσ v d vol v v n dσ v d vol v dσ v d vol bd the lst equlity becuse of the boundy condition. Since > the only wy fo the lst line to be zeo is fo both integnds to be zeo, in which cse v is constnt becuse v thoughout, nd the constnt is zeo becuse v on the boundy of. Thus vx, y, z so the two solutions u nd u of the oiginl poblem must be the sme, poving uniqueness. Pge 83, poblem 5 Let u be hmonic function in which is unique up to dding constnt tht stisfies du dn hx on bd nd let v be ny function on. Using the divegence theoem nd the identity g f g f +

7 7 g f we cn conclude: Eu + v Eu + Eu + u + v u + v dx Eu bd u v + v v dx u v dx v u dx + Eu + v v dx Eu bd hu + hv ds bd v u n ds v v dx hv ds + v v dx to get the next-to-lst lines we used tht u is solution of the Neumnn poblem. This shows tht the enegy functionl is minimized when u is solution of the Neumnn poblem, nd tht Eu + v Eu exctly when v is constnt. Pge 33, poblem. We might s well do this one in n spce dimensions, so suppose tht x x, x,... x n nd k k, k,..., k n. If ux, t fk x ct, then nd If u tt c u, then u x i k i f k x ct u x i nd k i f k x ct nd u t cf k x ct c c k kf k x ct u t c f k x c t. so eithe f is line nd k is bity, o else k k nd f is bity. This sques with wht we ledy know bout the n cse Pge 33, poblem 5. Cuslity mens tht fo some time t >, the vlues of the solution u t points in the disk centeed t x x, y with dius depend only on the initil dt in the disk centeed t x with dius + ct. Equivlently, if the initil dt position nd velocity e identiclly zeo in the disk of dius + ct centeed t x, then the solution is identiclly zeo t time t in the disk of dius centeed t x. To do this, we ll show tht the enegy Et u t + c u dx dy whee is the disk of dius in the t t plne centeed t x, t x, y, t is no lge thn the enegy E u t + c u dx dy whee is the disk of dius + ct in the t plne centeed t x, x, y,. So if the enegy E, then the enegy Et must lso be zeo, since the enegy is clely non-negtive function of t, which in tun will imply tht the solution is identiclly zeo in the disk t time t.

8 8 To pove this enegy inequlity, we e going to pply the divegence theoem to egion detemined by the vlues x, y, t nd. We ll let t t + so tht the cone in xyt-spce with vetex c x, t nd bse in the t plne being the disk centeed t x, with dius ct + ct hs the initil disk nd the time t disk s coss-sections. We ll need to use the vecto clculus identity: with v u t to get the enegy identity v u v u + v u u t + c u t u t u tt + c u u t u t u tt c u t u + c u t u c u t u if u stisfies the wve eqution u tt c u. Note tht the gdients, divegences nd Lplcien e tken in -dimensionl xy-spce. A consequence of this is tht the thee-dimensionl divegence of the thee-dimensionl vecto in xyt-spce: V c u t u x, c u t u y, u t + c u c u t u, u t + c u is zeo. So the sufce integl of V n dσ ove the boundy of ny solid egion in xyt-spce is zeo. The solid egion to which we e going to pply the peceding to is constucted s in the fist pgph bove: Stting fom the point x, t ssume t >, we conside the cone with vetex x, t nd with bse in the t plne being the disk centeed t x, nd with dius ct. So the ltel sufce of the cone is given by the eqution which we cn ewite letting x x, y s x x + y y c t t x x c t t. The solid we e going to conside is fustum F of this cone between the two pllel plnes t nd t t whee < t < t The sufce of F hs thee pts: the top, which is the disk of dius ct t centeed t x, t in the plne t t, the bottom, which is the bse of the cone, nd the side, which is the gph of the eqution of the cone fo < t < t. We obseve tht since the outwd-pointing noml fom bdf on the top t t disk is the unit vecto,, pointing in the positive t diection, the pt of the sufce integl of V n dσ ove the top sufce is pecisely the enegy Et. Likewise, the outwd pointing noml fom bdf on the bottom t disk is the unit vecto,, pointing in the negtive t divesion, so tht the pt of the sufce integl of V n dσ ove the bottom sufce is the negtive of the enegy t time t, i.e., it is E. Fo the integl ove the side, we need the outwd pointing noml, which is the nomlized gdient of the defining eqution: in othe wods, it is the nomlized vesion of x x c t t, N x x, c t t Now we clculte: N V c t tu t + c u u t x x u

9 9 Let x x ct t. Then we cn fcto t t out nd get: N V c u t + c u x x cu t c c u x x t cu t u + u t c x x u c x x u c x x u + c u c x x u u + c u The sum of the lst two tems in the pentheses is non-negtive, since x x, so x x / is unit vecto, which mens tht the lst tem is c u cos θ whee θ is the ngle between u nd x x. So the whole expession dds up to something non-negtive, which mens tht the integl of N V dσ ove the side of the cone is non-negtive. Summing up the thee pts, we hve Et E + something non-negtive which implies tht E Et, which is just wht we needed to show. Theefoe the enegy nd in tun u will be zeo in the disk if it is zeo in the disk. Pge 4, poblem 4. The Lplcin in spheicl coodintes is u u ρρ + ρ u ρ + ρ u θθ + cot θu θ + sin θ u ϕϕ But since the initil dt fo this poblem depend only on ρ, nmely ux, nd u t x, ρ, we expect the solution to depend only on ρ nd t. Thus we seek the solution uρ, t of u tt c u ρρ + ρ u ρ uρ, t u t ρ, t ρ. But then, following the deivtion of Kichhoff s fomul we know tht the function vρ, t ρuρ, t stisfies v tt c v ρρ vx, v t x, ρ 3. But now, by d Alembet s fomul, vρ, t c c ρ+ct ρct s 3 ds 8c ρ 3 ct + ρc 3 t 3 ρ 3 t + ρc t 3 ρ + ct 4 ρ ct 4 Theefoe uρ, t ρ vρ, t ρ t + c t 3 x t + y t + z t + c t 3 Pge 4, poblem 6. S is the sphee with cente t x nd dius R fo convenience, let s put the point x on the z-xis t z nd ssume tht >. We wnt the sufce e of the pt of S tht lies inside the sphee Q of dius ρ centeed t the oigin. Fist, we ll ssume tht > ρ, so the point x is outside the sphee Q. It s esy to see tht if R < ρ, then the sphees S nd

10 Q will not intesect. No will they intesect if R > + ρ. So the sufce e is zeo in both these situtions. If ρ < R < + ρ, then the intesection of the two sphees will be cicle pllel to the xy-plne. Wht we need to find is the z-coodinte of the cicle of intesection. We cn estict ou ttention to the points in the yz-plne, nd find the intesection of the two cicles S yz centeed t,, with dius R nd Q yz centeed t,, with dius ρ. So fo some vlue of the ngles θ nd ϕ, the coodintes of the point well, points, since thee e two of them of intesection e nd, ρ cos θ, ρ sin θ s seen fom Q yz, R cos ϕ, R sin ϕ s seen fom S yz. These e the coodintes of the sme point, so we hve the equtions ρ cos θ R cos ϕ nd ρ sin θ R sin ϕ. Sque the equtions nd dd them togethe to get o ρ + R R cos ϕ R cos ϕ + R ρ Now the z-coodinte of the point of intesection ws R cos ϕ, so we conclude tht we e inteested in the e of the pt of S between the pllel plnes z R nd z + R ρ. Now, emkble fct bout sphees tht you should pove fo youself if you ve neve done it is tht the sufce e of the pt of sphee between two pllel plnes tht intesect o e tngent to it is equl to the ltel sufce e of cylinde with the sme dius s the sphee nd height equl to the distnce between the pllel plnes. So the sufce e we e inteested in is πr + R ρ R πr R R + ρ πr ρ R Now this fomul is vlid fo ny x outside the sphee Q with x eplcing the pmete. So fo x outside of Q we hve SA fo R < x ρ πr ρ x R x fo x ρ < R < x + ρ fo R > x + ρ Next, if < ρ, so tht the point x is inside the sphee Q of dius ρ centeed t the oigin. Then fo R ρ the entie sphee of dius R centeed t x is contined in the sphee Q, so the sufce e is 4πR. Fo ρ R ρ +, we hve the sufce e of the pt of the sphee of dius R centeed t x between the plnes z R nd z + R ρ s befoe. so in this cse: SA 4πR πr ρ x R x fo R < ρ x fo ρ x < R < x + ρ fo R > x + ρ

11 b Accoding to Kichhoff s fomul, to find ux, t, we need to set R ct in the sufce e fomuls bove, multiply them by A, nd divide by 4πc t. So we get: If x > ρ: ux, t fo ct < x ρ A ρ x ct 4c x fo x ρ < ct < x + ρ fo ct > x + ρ nd If x < ρ: ux, t At A ρ x ct 4c x fo ct < x ρ fo x ρ < ct < x + ρ fo ct > x + ρ c In the figue, the pofile fo t.5 is ed, fo t is blue nd fo t is geen. d

12 In the figue, the pth fo.5 is ed, fo is blue nd fo is geen. nd e Fist of ll, if x < ρ nd v c, then x + tv x + t v < ρ + ct x + tv t v x > ct ρ so fo ll t the vlue of ux + tv, t is detemined by the middle complicted line in the fomul fo u given in pt b. So the poblem is to clculte: lim t ux At + tv, t lim ρ x + tv ct tø t 4c x + tv lim t A 4c v lim t A 4c A 4c A 4c A 4c A 4c x ρ x + tv ct x + tv + ct t + v x + tv + ct ρ x + tv c t x + tv + ct lim ρ x + tv x + tv t v t x + tv + ct lim ρ x + tx v t x + tv + t v x lim + x v t ρ t x t + v + v ρ x v v

13 3 Pge 63, poblem. The iichlet boundy conditions imply tht the eigenfunctions of the Lplcin on the sque tke the fom sin mx sin ny fo m, n >. So the genel solution of the wve eqution u tt c u with these boundy conditions is ux, t sin mπx nπy m π sin A mn cos b + n π m π b ct + B mn sin + n π b ct m n The initil conditions u t x, y, implies tht ll the B mn e zeo. The condition mens tht The denominto is sin mπx nd the numeto is xy xb y, sin mπx Now So likewise ux, y, xy xb y xy xb y, sin mπx nπy sin A mn b sin mπx sin nπy mπx nπy, sin sin b b sin nπy mπx nπy, sin sin b b sin nπy b b b sin mπx nπy sin dy dx b b 4 xy xb y sin mπx x x sin mπx x x sin mπx dx mπx x x cos mπ mπx m x sin π b 3 mπx m 3 cos π3 yb y sin nπy b Letting m k nd n l we cn wite fo the coefficient of sin kπx ux, t m n sin lπy b A kl dy 4b 3 n 3 π b 3 k 3 l 3 π 6 nd conclude + mπ + m π 4 3 m 3 π 3 dx b nπy sin dy dx b yb y sin nπy b x cos mπx sin mπx if m is odd if m is even if n is odd if n is even 6 3 b 3 k πx l πy k 3 l 3 sin sin cos π6 b dx dy dx m π + n π b ct

14 4 Pge 63, poblem 3. Since u t k u + γu, with iichlet boundy conditions, we know tht ech tem in the seies solution of the poblem will hve fcto of the fom e λkt, whee λ is n eigenvlue of + γ on the cube, in othe wods, k γ u + k + λ u. The eigenfunctions e ϕ lmn x, y, z sin lπx mπy sin s usul, but the eigenvlue coesponding to ϕ lmn is sin nπz fo l, m.n,, 3,... λ lmn l π + m π + n π γ k. We need the smllest eigenvlue, nmely λ to be non-negtive, so this mens tht γ 3kπ.

Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.

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