1 Boas, p. 643, problem (b)

Transcription

1 Physics 6C Solutions to Homework Set #6 Fall Boas, p. 643, problem 3.5-3b Fin the steay-state temperature istribution in a soli cyliner of height H an raius a if the top an curve surfaces are hel at an the base at. We have to solve the equation subject to the bounary conitions ur,θ,z = ur,θ,h = ua,θ,z =, ur,θ, = After the usual separation of variables in cylinrical coorinates ur, θ, z = RrΘθZz we are left with rr rr + r Θ Θ = Z Z = K 3 The R Θ equation can also be separate an gives r R rr +K r = Θ Θ = n = { Θθ = e ±inθ, Rr = J n Kr 4 n has to be an integer so that Θθ+π = Θθ. The Z equation gives Zz = AsinhKz +BcoshKz = A sinhkh z, were we have parametrize the solution so that ZH=. The full solution is ur,θ,z = J n KrsinhKH ze ±inθ 5 We now have to impose the bounary conitions, but first we notice that they have a rotational symmetry aroun the z axis that is, there is no θ-epenence: then there will be no θ-epenence in the answer, that is, we have n =. For the other bounary conitions, ua,θ,z = = J Ka = = Ka = x m is a zero of the Bessel function J x 6 u = m c m J x m r/asinh[x m H z/a] 7 ur,θ, = = m c m J x m r/asinhx m H/a 8 Because the set {J x m r/a,m =,,...} is an orthogonal set in,a, we can fin the coefficient c n by multiplying expression 8 by J x n r/a an integrating between an a. Then we get following Boas, p.64 sinhx n H/ac n = 9 x n J x n or ur,θ,z = m= x m J x m sinhx m H/a J x m r/asinh[x m H z/a].

2 Boas, p. 643, problem A flat circular plate of raius a has a temperature istribution ur,θ = rsinθ. At t =, the circumference of the plate is hel at. Fin the time epenent temperature istribution. We have to solve the problem u = α t u, We solve the ifferential equation in the usual way: { ur,θ, = rsinθ, ua,θ,t =, t > u = Fx,yTt = F F = α T The solution for T is Tt = e k α t, while for F we have T = k Fx,y = RrΘθ = r R rr +k r = Θ Θ = n = { Θθ = e ±inθ, Rr = J n kr 3 The bounary conition Ra = tells us that ka is a zero of the n-th Bessel function; if we let x mn be the m-th zero of the Bessel function J n, then ka = x mn. Unlike in the last problem, we now have a θ-epenence in the initial conition: ur,θ, = rsinθ = m,nj n x mn r/aa mn cosnθ+b mn sinnθ 4 By inspection, we alreay see there will be no cosnθ epenence on the right han sie A mn =. If we multiply the previous equation by sinlθ an integrate over θ,π, we have π r sinlθsinθθ = J rδ l π = n x mn r/ab ml πδ ln, 5 n,mj = B ml = for l, r = m J x m r/ab m. 6 If we now multiply by rj x k r/a an integrate over r,a, we get Hence, = m a So that the final solution is u = k= a r ar J x k r/ar = x k r rj x k r/ar = a3 J x k = 7 x k a a B m rj x m r/aj x k r/ar = B k J x k 8 B k = a x k J x k J x k r/ae x k α /a t a x k J x k. 9 with x k =zeros of J x.

3 3 Boas, p. 643, problem Fin the steay state temperature istribution in a soli cyliner of height an raius 3 if the flat ens are hel at an the curve surface is at. We have to solve the problem ur,θ,z =, { ur,θ, = ur,θ, =, ua,θ,z =. Following the erivation on pp of Boas, we write an obtain Rr r ur, θ, z = RrΘθZz, r R r + Θ Θr θ + Z Z =. 3 z Choosing the opposite sign for the separation constant as compare to eq. 5.4 on p. 639 of Boas, we have { Z Z z = sinkz, k, Z = 4 coskz. Inserting this result back into 3, Rr r r R r + Θ Θr θ k =. 5 We can separate the variables by multiplying through by r, r r R + Θ Rr r Θ θ k r =. 6 Thus we have, { Θ Θ θ = sinnθ, n, Θ = cosnθ. Inserting this back into 6, we obtain the raial equation, r r R n k r =, 8 Rr r or r R r +rr r k r +n R =. 9 Comparing this ifferential equation with eq. 7. on p. 595 of Boas, we ientify the possible solutions as: { I n kr, Rr = 3 K n kr. However, since the present problem inclues the origin, we eman that the raial solution shoul be finite as r. In light of the small r behavior of I n kr an K n kr as specifie on p. 64 of Boas, we must exclue K n kr as a possible solution. Hence, 7 Rr = I n kr. 3 3

4 If we now impose the requirement that the temperature is hel at at z = an on the surface of the soli cyliner, then it follows that n = since the bounary value temperature has no θ-epenence. Hence, the allowe solutions for the temperature are given by: ur,θ,z = I krsinkz. 3 Next, we impose the bounary conition that the temperature is hel at at z =. This implies that k = nπ or k = mπ/, for m =,,3,... Hence, the solution must be of the form, ur,θ,z = c m I πmr/sinπmz/. 33 m= Finally, we impose the bounary conition that u3,θ,z =. This yiels = c m I 3πm/sinπmz/, 34 m= which we recognize as a Fourier sine series. Thus, we can solve for the Fourier coefficients, c m I 3πm/, c m I 3πm/ = sinπmz/z = mπ cosπmz/ = mπ [ m ] =, for even m, 35 4 mπ, for o m. This equation etermines the c m, which we can insert back into 33 to obtain the final solution, ur,θ,z = 4 π o m 4 Boas, p. 644, problem 3.5- mi 3mπ/ I mπr sin mπz. 36 Solve the Laplace s equation in imensions in polar coorinates an solve the r an θ equation. Solve the problem of the steay-state temperature in a circular plate if the upper semicircular bounary is hel at an the lower at. The Laplacian in polar coorinates is = r r + r r r θ 37 so that Laplace s equation u = becomes with u = RrΘθ r R rr = Θ Θ = n 38 The angular epenence is Θ = e ±inθ an n must be an integer for the function to be single-value, Θθ+π = Θθ. The raial equation is r R +rr n R = 39 4

5 To solve this we try a solution R = r α an substituting we fin α = ±n. The solution to Laplace s equation is then u = n=[ r n A n sinnθ+b n cosnθ+r n A nsinnθ+b ncosnθ ] 4 To solve the requeste problem, first we put A = B = because the temperature woul iverge in the center of the plate. Then we apply the bounary conition ua,θ = H θ +π where H is the Heavisie step function: ua,θ = H θ +π = n=a n A n sinnθ+b n cosnθ 4 Now we must multiply this equation by sinmθ or cosmθ an integrate between,π in orer to fin respectively A n an B n. Because π cosmθθ = πδ m, only the constant term in the cosine series is non-zero. It is given by π π θ = B θ = π = B = 5 4 a n A n = π We can rewrite the solution as π sinnθθ = nπ u = 5+ π 5 Boas, p. 65, problem o n for o n, an otherwise 43 r a n sinnθ n Fin the steay state temperature istribution insie a sphere of raius with the following surface temperature istribution: { cosθ, < θ < π/, u,θ,φ = 45, π/ < θ < π. In spherical coorinates, the Laplacian is = r r r r + r sinθ sinθ + θ θ r sin θ 44 φ 46 After separating the variables, ur, θ, φ = RrΘθΦφ, we have r R = kr, sinθ Θ+kΘ m sinθθ θ sin θ Θ =, Φ = m Φ 47 The R equation is solve by setting k = ll + an the solutions are r l, r l. The secon solution iverges at the origin so we o not consier it. The Θ equation gives the associate Legenre polynomials in cosθ, while the Φ equation is solve by the sine an the cosine. The full solution to Laplace s equation is ur,θ,φ = r l P m l cosθe ±imφ 48 m must be integer for Φ to be single value, while l must be integer for Pl m to not iverge at θ =,π. As the initial conition 45 has no φ-epenence, the only allowe value will be m =, which gives us the Legenre polynomials P l cosθ. We can write our solution as ur,θ,φ = c l r l P l cosθ. 49 l= 5

6 We now impose the bounary conition at r =. Defining x cosθ, { } x, < x < q, ur = = = c, < x < l P l x. 5 We can now use the orthogonality of the Legenre polynomials, xp lp m = δ lm l+ l= an fin c l = l+ ur = P l xx = l + xp l xx. 5 To evaluate this integral, we first employ the recursion relation given in eq 5.8 on p. 57 of Boas, Shifting l l+, we obtain Hence, lp l x = l xp l x l P l x. 5 xp l x = l + [l+p l+x+lp l x]. 53 xp l xx = [ l+ l+ We first consier the cases of l = an l =. xp xx = P l+ xx+l xx =, xp xx = For l, we make use of the result of problem 3.3 on p. 65 of Boas, ] P l xx. 54 x x = P n xx =, for n >, an P n+ xx = n n!! n+ n+!. 56 Thus 54 an 56 yiel xp n+ xx =, for n =,,3,..., 57 Hence, 5 yiels xp n xx = [ n ] n+!! 4n+ n+ n n 3!! n+! n n! = n n 3!! 4n+ n+ n+! [n+n 4nn+] = n+ n 3!! n+ n+! c = 4, c =, c n = n+ 4n+ n 3!! n+ n+!, for n =,,3,... 58, c n+ =, for n =,,3,

7 We can rea the first few terms as c = 4, c =, c = 5 6, c 3 =, c 4 = Inserting these results into 49, we obtain ur,θ,φ = 4 + rcosθ+ n+ 4n+ n 3!! n+ r n P n cosθ. 6 n+! The first few terms of the series are: n= ur,θ,φ = 4 + rcosθ r P cosθ 3 3 r4 P 4 cosθ Boas, p. 65, problem 3.7- Fin the steay state temperature istribution insie a hemisphere if the spherical surface is hel at an the equatorial plane at. We alreay have the solution for a full sphere with the upper half hel at an the lower half at Boas, p.649, eq. 7.5: u = l c l r l P l cosθ, c l = l + xfxp l x, fx = {, < x <, < x < 63 c l = xfx[p l+ P l ] = [P l+ P l ] = P l P l+ = 64 { c = l+ = l l!! l+ l+!! l l! = l l!! l+ l+! + l+ l l! l+, 65 c l =, c = u = [ P cosθ+ 3 4 rp cosθ 7 6 r3 P 3 cosθ+ 3 r5 P 5 cosθ+... ] 66 Let u be a solution corresponing to the upper half hel at an the lower half at : u = l r l l + {, < x < P l cosθ, l = xgxp l x, gx = l, < x < l = l+ xgxp l x = l+ xp l x = l+ This tells us that =, l =, l+ = c l+ : then u is xp l x = l c l 67 u = [ P cosθ+ 3 4 rp cosθ 7 6 r3 P 3 cosθ+ 3 r5 P 5 cosθ+... ] 68 Finally, u +u will still be a solution of Laplace s equation, an satisfy the bounary conitions with the two hemispheres hel at o an. This also correspons to have the central plane at o, so that the if we consier u = u +u θ<π/ we got the solution to our original problem: u = [ 3 4 rp cosθ 7 6 r3 P 3 cosθ+ 3 r5 P 5 cosθ+... ] 69 7

8 7 Boas, p. 65, problem Separate the wave equation in spherical coorinates an fin that the θ, φ solutions are the spherical harmonics Yl m θ,φ = Pl m cosθe ±imφ an the r solutions are the spherical Bessel functions j l kr an y l kr The wave equation is u = u v t, 7 where v is the spee of the wave. In spherical coorinates, the Laplacian is = r r r r + r sinθ θ sinθ θ + r sin θ After separating the variables, ur,θ,φ,t = Fr,θ,φTt = RrΘθΦφTt, we have which implies that φ 7 F F = T c T = k = Tt = e ±ikvt, F +k F =, 7 R r R +k r = sinθ Θ Φ + Θsinθθ θ Φsin = ll + 73 θ φ sinθ Θ sinθ sinθ Θ θ θ θ sinθ θ = Φ Φ φ = m, r R +k r ll+r, Θ+ll +Θ m sin θ Θ =, Φ = m Φ 74 TheR equation is precisely thespherical Bessel equation inthe variable kr, x y +xy +[x ll+]y = so that its solutions are j l kr an y l kr. The Θ equation gives the associate Legenre polynomials in cosθ, P m l cosθ, while the Φ equation is solve by e ±imφ. 8 Boas, p. 658, problem 3.8- Show that the gravitational potential V = Gm r satisfy Laplace s equation. We have to show that V r =. For that we express the Laplacian in spherical coorinates an note that the only contribution comes from the raial part: r = r r r r r = r =, for r. 75 r With this we have proven that the gravitational potential satisfies Laplace s equation at all points in space away from the origin. 8

9 9 Boas, p. 658, problem Solve Poisson s equation for a charge q insie a groune sphere to obtain the potential V insie the sphere. Sum the series solution an state the image metho of solving the problem. Let the charge q be at,,a insie the sphere, a < R. Poisson s equation is V = πρ, ρ = qδxδyδz a 76 This is solve by ux,y,z = 4π ρx,y,z x y z x x +y y +z z = q x +y +z a 77 We still have to satisfy the bounary conition Vr = R =. For this, we search for another solution satisfying it an a it to our particular solution 77; the solutions to Laplace s equation in spherical coorinates is as we foun earlier solving eq. 47 A lm r l +B lm r l P m l cosθe ±imφ 78 As we are searching for a solution in the insie of the sphere, we take B = so that our result oes not iverge at r =. Because the problem is symmetric aroun the z axis, there will be no φ epenence, that is m =. The sum of our two solutions is V = q r arcosθ +a + l A l r l P l cosθ 79 where we have expresse the particular solution 77 in spherical coorinates. The coefficients are chosen in orer to satisfy bounary conition Vr = R = = q R arcosθ +a + l A l R l P l cosθ = q l a l R l+p lcosθ+ l A l R l P l cosθ, = A l = qal R l+ where in the first line we have expane the potential in terms of Legenre polynomials. The solution satisfying the bounary conition is 8 V = q r arcosθ+a q l a l r l R l+p lcosθ 8 We can sum the secon term using l hl P l x = / xh+h : q l a l r l R l+p lcosθ = q R l ar R lpl cosθ = q R ar/r cosθ +a r /R 4 = qr/a r rr /a cosθ +R 4 /a This has the same form of 77, but for a charge qr a situate at,, R a. Then the problem of the groune sphere with a charge insie it is equivalent to the electrostatic problem of having two charges of ifferent magnitue locate at,,a an,, R a. The secon charge is calle image charge. 9

10 Boas, p. 664, problem 3.-9 A long conucting cyliner is place parallel to the z axis in an originally uniform electric fiel in the negative x irection. The cyliner is hel at zero potential. Fin the potential in the region outsie the cyliner. First, we recall the solution to Laplace s equation in cylinrical coorinates that we foun in Problem 4, eq. 4 V = n= [ r n A n sinnθ+b n cosnθ+r n A n sinnθ+b n cosnθ] = n r n A n sinnθ+b ncosnθ 8 As we will be looking for the potential outsie the cyliner, the coefficients multiplying r n must be zero. The electric fiel E is in the x irection, that is, E = E i = V so that V = E x = E rcosθ. We want a solution to Laplace s equation not Poisson s, as there are no charges such that Vr = a = an V = E rcosθ for large r far away from the cyliner. Then, we pick V = E rcosθ + n r n A n cosnθ+b n sinnθ, satisfying Vr = a = 83 The coefficients are quickly picke by inspection: B n =, A n = except for A, which satisfies = E acosθ +A a cosθ = A = E a 84 an finally our solution is V = E r a cos θ 85 r Boas, p. 664, problem 3.- Use Problem 7 to fin the characteristic vibration frequencies of a soun in a spherical cavity. The vibration moes of a soun in a spherical cavity are solutions to the wave equation 7 subject to the bounary conition ur = a =. The general solution was given by ur,θ,φ,t = { jl kr y l kr } P m l cosθe ±imφ { coskvt sin kvt where v is the spee of soun. Because we are looking at the solution insie the sphere, the y l oes not contribute. The bounary conition will be satisfie if j l ka =, that is, if ka = λ l is a zero of the spherical Bessel function 87 The time epenence of the solution is given by sines an cosines, that is, oscillatory moes with an angular frequency ω = kv so that the frequencies of the normal moes are } 86 ν = ω π = λ lv πa 88 where λ l is a zero of the spherical Bessel j l.