PHYSICS 116C Homework 4 Solutions

Transcription

1 PHYSICS 116C Homework 4 Solutions 1. ( Simple hrmonic oscilltor. Clerly the eqution is of the Sturm-Liouville (SL form with λ = n 2, A(x = 1, B(x =, w(x = 1. Legendre s eqution. Clerly the eqution is of the SL form with λ = n(n+1, A(x = 1 x 2, B(x =, w(x = 1. Bessel s eqution. This is not obviously in the SL form. To mke it so, multiply by f(x nd choose f(x such tht the coefficient of y is the derivtive of the coefficient of y, i.e. or which hs solution f(x = 1/x nd so xy +y + [c which is of the SL for with d ( x 2 f(x = xf(x dx f (x = f(x x 2nx ν2 x ] y =, λ = c 2 n, A(x = x, B(x = ν 2 /x, w(x = x. Hermite s eqution. Proceeding s for Bessel s eqution we multiply by f(x nd require which hs solution nd so which is of the SL form with df dx = 2xf(x f(x = exp( x 2, e x2 y 2xe x2 y +2ne x2 y =, λ = 2n, A(x = e x2, B(x =, w(x = e x2. Lguerre s eqution. Proceeding s for Bessel s eqution we multiply by f(x nd require or which hs solution f = e x nd so which is of the SL form with d (xf = (1 xf(x dx f = f, xe x y +(1 xe x y +ne x y =, λ = n, A(x = xe x, B(x =, w(x = e x. 1

2 (b The equtions for y 1 nd y 2 re d [ ] A(xy dx 1 +[λ1 w(x+b(x]y 1 = (1 d [ ] A(xy dx 2 +[λ2 w(x+b(x]y 2 =. (2 Multiply Eq. (1 by y 2 nd Eq. (2 by y 1, subtrct, nd integrte from to b. This gives ( d [ ] y 2 A(xy d [ ] b dx 1 y1 A(xy dx 2 dx+(λ 1 λ 2 w(xy 1 (xy 2 (xdx =. (3 The integrnd in the first term cn be written s d [ ] y 2 A(xy d [ ] dx 1 y1 A(xy dx 2 Eq. (3 therefore becomes [ A(x(y 1 y 2 y 2y 1 ] b +(λ 1 λ 2 = d [ A(x(y dx 1 y 2 y 2y 1 ] ( A(xy 1y 2 A(xy 2y 1 = d [ A(x(y dx 1 y 2 y 2y 1 ]. Hence, for λ 1 λ 2 we obtin the orthogonlity condition provided the boundry term vnishes, i.e. w(xy 1 (xy 2 (xdx =. w(xy 1 (xy 2 (xdx =, (4 [ A(x(y 1 y 2 y 2y 1 ] b =. (5 (c Simple hrmonic oscilltor. =,b = 2π. The y n re sinnx nd cosnx which re periodicwithperiod2π. A(x = 1ndso the contributions from the two limits in Eq. (5 cncel. The orthogonlity condition, Eq.(4, is just the fmilir orthogonlity of sines nd cosines over the intervl from to 2π. Legendre s eqution. = 1,b = 1. The y n re Legendre polynomils, P n (x. Eq. (5 is stisfied becuse A(x = 1 x 2 vnishes t x = ±1. The orthogonlity condition, Eq. (4, is 1 1 P n (xp m (xdx = (n m, which ws shown in clss. Bessel s eqution. =,b = 1. The y n re Bessel functions J ν (c n x. In Eq. (5, the contribution t x = vnishes becuse A(x = x nd the contribution t x = 1 vnishes becuse c n is zero of the Bessel function nd hence y(1 =. The orthogonlity condition, Eq. (4, is 1 which ws shown in clss. xj ν (c n xj ν (c m xdx = (n m, 2

3 Hermite s eqution. =,b =. The y n re Hermite polynomils H n (x. Eq. (5 is stisfied becuse A(x = exp( x 2 vnishes for x ±. The orthogonlity condition, Eq. (4, is e x2 H n (xh m (xdx = (n m, which ws stted in clss. Lguerre s eqution. =,b =. The y n re Lguerre polynomils. L n (x. Eq. (5 is stisfied becuse A(x = xe x vnishes t x = nd. The orthogonlity condition, Eq. (4, is which ws stted in clss. e x L n (xl m (xdx = (n m, (d Assuming completeness of the y n we cn expnd function f(x in terms of them s f(x = m m y m (x. (6 Multiplying by y n (x, integrting from to b, nd using the orthogonlity reltion, Eq. (4, gives w(xy n (xf(xdx = n w(xyn(xdx, 2 which gives where n = 1 I n I n = w(xy n (xf(xdx. (7 w(xy 2 n(xdx. (8 (e For the simple hrmonic oscilltor eqution, =,b = 2π, nd the solutions re cosnx nd sinnx. Hence Eq. (7 is just the fmilir Fourier series f(x = m ( m cosmx+b m sinmx. I n in Eq. (8 is or nd hence Eq. (7 corresponds to 2π 2π cos 2 nxdx = π sin 2 nxdx = π, n = 1 π b n = 1 π 2π 2π f(xcosnxdx f(xsinnxdx, 3

4 which re indeed the coefficients in Fourier series. 2. ( Using the divergence theorem FdV = R B F ds with F = u u gives ( u 2 u+ u 2 dv = R B u u ds = B u ds, (9 n where we used (ua = u A+A u, nd / n is the derivtive of u in the direction of the (outwrd norml to the surfce. (b Consider two solutions of Lplce s eqution u 1 nd u 2 which hve the sme vlues on the boundry B. Denote the difference u 1 u 2 by u. This stisfies Lplce s eqution, 2 u =, nd hs the vlue u = on the boundry. Hence inserting these properties into Eq. (9 gives u 2 dv =. (1 R Hence u =, so u is constnt, in the volume V, but since u = on the boundry S we must hve u = everywhere in V. Hence u 1 = u 2 nd so the solution to Lplce s eqution whose vlue is specified on the boundry is unique. (c Repet the rgument of the previous section but with / n = on the surfce. This gin yields u 2 dv =, (11 R showing tht u = so u is constnt. Since we do not know the vlue of u u 1 u 2 t ny point, ll we cn sy is tht the solution is determined up to n rbitrry constnt. 3. ( In one dimension, the diffusion eqution is t = u D 2 x 2. (12 Using we hve u(x,t = 1 exp ( x2 t1/2 4Dt t = 1 2t (13 x2 u+ u, (14 4Dt2 where the first term comes from differentiting 1/t 1/2 nd the second from differentiting the 1/t in the exponentil. We lso hve x = x 2Dt u, 4

5 nd differentiting gin gives x 2 = 1 ( x 2 2Dt u+ u. 2Dt We see tht this is just D 1 times Eq. (14, nd hence the diffusion eqution, Eq. (12, hs solution given by by u(x,t in Eq. (13. (b In sphericl polr coordintes in three dimensions, the diffusion eqution is t = D 1 ( r 2 r 2, (15 r r where we ssumed sphericl symmetry. Tking u(r,t = 1 exp ( r2 t3/2 4Dt (16 we hve We lso hve r = r 2Dt u, nd tking the pproprite second derivtive gives t = 3 r2 u+ u, (17 2t 4Dt2 ( r 2 ( = 3r2 r 2 r r 2Dt u+r2 u. 2Dt Thisisjustr 2 /D timeseq.(17, ndhencethediffusionequtioninsphericlpolrs, Eq.(15, hs solution given by u(r,t in Eq. ( The wve eqution in two dimensions is x u y 2 = 1 v 2 t 2. (18 Tking where k 2 = k 2 x +k 2 y, we hve nd similrly u(x,y,t = sin(k x xsin(k y ysin(kvt, (19 y 2 = k2 yu, x 2 = k2 xu, t 2 = k2 v 2. Hence the wve eqution, Eq. (18 hs solution given by Eq. (19. 5

6 5. We hve to find the solution of Lplce s eqution in the rectngulr region x, y b which stisfies the boundry conditions u(,y = u(,y = u(x, =, nd u(x,b = u sin(πx/. We look for seprble solution of Lplce s eqution, i.e. u(x, y = X(xY(y. Substituting nd dividing by u gives X X + Y Y =. Since the first term is only function of x nd the second term only function of y ech term must be constnt in order for the eqution to be true for ll x nd y. Setting the first term to equl k 2 we hve X +k 2 X =, Y k 2 Y =. A solution which stisfies the boundry conditions u(,y = u(x, = is X(x = sin(kx,y(y = sinh(ky. To stisfy X( = we need k = 2πn/ (n = 1,2,3,, nd so we form the liner combintion. u(x,y = C n sin sinh ( nπy. WedeterminethecoefficientsC n byrequiringthtthelstboundrycondition,u(x,b = u sin(πx/, is stisfied. By inspection we hve C n = (n 1, nd C 1 = u /sinh(πb/, so u(x,y = u sinh( πb sin ( πx sinh ( πy. Note: One could work out ll the Fourier coefficients explicitly nd show tht they re ll zero except for C 1, but this is unnecessry work. If you cn see the solution by inspection, like here, then you cn use this informtion (unless you re expressly told to go through the derivtion. 6. We look for seprble solution of Lplce s eqution, i.e. u(x, y = X(xY(y. Substituting nd dividing by u gives X X + Y Y =. Since the first term is only function of x nd the second term only function of y ech term must be constnt in order for the eqution to be true for ll x nd y. Setting the first term to equl k 2 we hve X +k 2 X =, Y k 2 Y =. A solution which stisfies the boundry conditions u(,y = u(x, = is X(x = sin(kx,y(y = sinh(ky. To stisfy X( = we need k = 2πn/ (n = 1,2,3,, nd so we form the liner combintion. u(x,y = C n sin sinh ( nπy. To stisfy the boundry condition t y = b we need [ ( ] nπb u x( x = C n sinh sin. 6

7 In other words we hve to determine the sine Fourier series for u x( x in the intervl x nd the n-th Fourier coefficient is C n sinh(nπb/, i.e. ( nπb 2 C n sinh = u x( x sin dx. Integrting by prts twice gives ( { nπb 2 [ ] C n sinh = u x( xcos + ( 2xcos nπ = 2u ( nπx ( 2xcos dx nπ = 2u { [ ] ( 2xsin ( 2 sin nπ nπ = 4u ( nπx (nπ 2 sin dx = 4u [ ] (nπ 2 cos = 4u nπ (nπ 2 nπ [1 ( 1n ] (n even, = 8 2 (nπ 3 u (n odd. Hence the solution is (writing n = 2m 1 with m = 1,2,3, u(x,y = u 2 ( 8 π 3 m=1 1 (2m 1 3 } dx } dx ( (2m 1πy sinh ( (2m 1πx ( sin. (2m 1πb sinh 7. We strt with [ ] sin (2n 1πx L T(x,y = 4T e (2n 1πy/L. (2n 1π Since sin[(2n 1πx/L] = Im[exp(i(2n 1πx/L] we cn write where T(x,y = 4T π Im ( z 2n 1, (2 2n 1 [ ] π(ix y z = exp = exp( yπ/l(cos(πx/l+isin(πx/l. L Usingtheseriesforln(1+z = z z 2 /2+z 3 /3, whichlsoimpliesln(1 z = z z 2 /2 z 3 /3, we see tht the series in Eq. (2 is just tht for ln(1+z ln(1 z = ln[(1+z/(1 z]. Hence T(x,y = 2T [ ( ] 1+z π Im ln. (21 1 z 7

8 For complex number w = r iθ we hve Imlnw = Imln(re iθ = Im(lnr + iθ = θ = rg(w = tn 1 (Imw/Rew. Letting z = s+it we hve 1+z 1 z = 1+s+it 1 s it = (1+s+it(1 s+it (1 s it(1 s+it = 1 s2 t 2 +2it (1 s 2 +t 2, nd so Hence Im[(1+z/(1 z] Re[(1+z/(1 z] = 2t 1 s 2 t 2 = 2sin(πx/Le πy/l 1 e 2πy/L = sin(πx/l sinh(πy/l. T(x,y = 2T ( sin(πx/l π tn 1. (22 sinh(πy/l Note: For x, sin(πx/l nd so T(,y =. For x L, sin(πx/l nd so T(L,y =. For y, sinh(πy/l nd so lim T(x,y =. y For y, sinh(πy/l, so tn 1 [sin(πx/l/sinh(πy/l] tn 1 = π/2 nd so T(x, = T. Hence the closed form expression in Eq. (22 stisfies the boundry conditions s required. 8