UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 6C Problem Set 4 Bejami Stahl November 6, 4 BOAS, P. 63, PROBLEM.-5 The Laguerre differetial equatio, x y + ( xy + py =, will be solved by power series. The assumed power series y is stated alog with its first ad secod derivatives with respect to x as follows: y = a x y = a x y = a ( x (. = Substitutig these ito the differetial equatio ad the simplifyig yields: = a ( x + a x a x + p a x = (. Re-idexig the first two sums as + yields: Examiig the case of = gives: = a + ( + x + a + ( + x a x + p a x = (.3 Now examiig the cases of gives the followig relatio: a + ( + x + pa x = a = pa (.4 a + ( + x + a + ( + x a x + pa x = a + = = = = (p ( + a (.5 Usig this relatio, a few more a s are calculated util a patter ca be uderstood. After simplifyig these a s are expressed as: p(p = : a = a = : a 3 = p(p (p 3 a (.6
Geeralizig these results: Thus, the power series solutio is: y = a + a = ( p! (p!(! a (.7 ( p! (p!(! a x p(p = a a px + a x p(p (p a 3 +O(x 4 (.8 From the form of a it is apparet that if p is a iteger, the the series will termiate. If p =, the y is a polyomial of degree ; with a = these polyomials are the Laguerre polyomials L (x. Thus the followig Laguerre polyomials ca be geerated: L (x = L (x = x L (x = x + x L 3 (x = 3x + 3 x 6 x3 (.9 BOAS, P. 6, PROBLEM 3.- The wave equatio is give as: u = v u t (.. A It will ow be show that the expressio u = si(x vt satisfies the wave equatio: si(x vt cos(x vt si(x vt Sice both sides are equivalet, u does ideed satisfy the wave equatio. u v t cos(x vt( v] t v v si(x vt ] v (. Next, it will be show that i geeral a expressio u = f (x ± vt satisfies the wave equatio: f (x ± vt f (x ± vt f (x ± vt Sice both sides are equivalet, u does ideed satisfy the wave equatio. v u v t f (x ± vt(±v ] t v v f (x ± vt ] (.3. B Now it will be show that u(r, t = r f (r ± vt satisfies the wave equatio i spherical coordiates. The Laplacia,, i spherical coordiates is: = ( r r + (.4 r r I must admit I am slightly cofused here, o the left side my derivatives are f f i the above proof but I have doubts about doig that. Is there a better way? f but o the the right there are. I recklessly call these both t
Where is referrig to fuctios of θ, φ, ad their partial derivatives which are all irrelevat i this cotext ad thus ot writte explicitly. Applyig the relevat Laplacia allows the desired to be show: ( r r ] u f (r ± vt r r r v ( r r r r f (r ± vt + ] t r f (r ± vt v ± v ] t r f (r ± vt f (r ± vt + r f r (r ± vt ] v ] (.5 r v r f (r ± vt r f (x ± vt r f (x ± vt Sice both sides are equivalet, u does ideed satisfy the wave equatio. 3 LAPLACE PROBLEM This problem deals with a rectagular plate where x a ad y. It is also give that T (, y = T (a, y =, that T (x, = T, ad that T as y. The steady state temperature T (x, y will be determied startig from the Laplace equatio T =. A solutio of the form T = F (xg(y is assumed. Substitutig this ito the Laplace equatio yields: FG = G d F dx +G d G d y = F d dx = G d G d y = k (3. Where the shorthad F = F (x ad G = G(y have bee adopted. Additioally i the fial step it is oted that both sides of the fial equatio are fuctios of oly oe variable (x o the left ad y o the right, thus both sides must be equivalet to some costat, which has bee chose as k for this problem. From this result, the followig two equatios may be costructed: d F dx = k F & d G d y = k G (3. These equatios are rather trivial to solve so the solutios will be give without expedig much effort. They ca be easily verified by direct substitutio. F = A si(kx + B cos(kx & G = Ce k y + De k y (3.3 Thus the desired steady state temperature fuctio will be 3 : T (x, y = A si(kx + B cos(kx] Ce k y k + De y] = Ae k y si(kx + Be k y si(kx +Ce k y cos(kx + De k y cos(kx Where the arbitrary costats i the secod equatio have bee redefied as compared to the prior equatio. Now the give boudary coditios are applied: T (, y = + +Ce k y + De k y D = Ce k y T (x, y = Ae k y si(kx + Be k y si(kx +Ce k y cos(kx Ce k y cos(kx = Ae k y si(kx + Be k y si(kx T (x, y = Ae k y si(kx + = A = T (x, y = Be k y si(kx T (a, y = Be k y si(ka = ka = k = a ( T (x, t = Be a y si a x Agai, I have questios because of the same reaso as the last footote. 3 This result ad all the work that has bee doe leadig up to it will be quoted i subsequet problems but ot repeated. From this poit forward, boudary coditios specific to the problem will shape the outcome ad the solutio will o loger be geeral. (3.4 (3.5 3
The form of this result will satisfy the boudary coditios, ad thus the sum of all solutios that satisfy these boudary coditios is: ( T (x, t = b e a y si a x (3.6 Now the iitial coditios will be examied to reveal the fial solutio: T (x, = b si a x = T (3.7 Now the coefficiets b will be foud usig a Fourier series: b = a a T (x,si a x dx = T a a = T ( ] { eve = odd 4T si a x dx = T ( a cos a x = T cos( ] Comparig the two boxed results to Equatios.9 ad. of Boas Chapter 3 shows agreemet. Thus the specified problem has bee solved. (3.8 4 BOAS, P. 67, PROBLEM 3.- I this problem a rectagular plate of cm by 3 cm is give where two adjacet sides are held at o ad the other two are at o. Startig from the geeral solutio foud i the previous problem ad applyig the boudary coditios that T (, y = T (x, = T (, y = ad T (x,3 = : T (, y = + +Ce k y + De k y = D = Ce k y T (x, y = Ae k y si(kx + Be k y si(kx +Ce k y cos(kx Ce k y = Ae k y si(kx + Be k y si(kx T (x, = A si(kx + B si(kx = B = A T (x, y = Ae k y si(kx Ae k y si(kx = A si(kxsih(k y T (, y = A si(ksih(k y = k = k = T (x, y = A si x sih y T (x,3 = A si x sih(3 T (x, y = B si x sih y A where: B = sih(3 The form of this result will satisfy the boudary coditios, ad thus the sum of all solutios that satisfy these boudary coditios is: T (x, t = b si x sih y (4.3 Now the iitial coditios will be examied to reveal the fial solutio: T (x, = b si x = (4.4 Now the coefficiets b will be foud usig a Fourier series: b = T (x,si dx x = 5 = ( ] { eve = odd 4 si x dx = cos x = cos( ] (4. (4. (4.5 4
Thus the fial solutio for these boudary coditios will be: T (x, y = 4,odd sih(3 si x sih y (4.6 Now a similar situatio with the ew boudary coditios T (, y = T (x, = T (x,3 = ad T (, y = is cosidered. Comparig the old boudary coditios to the ew oes reveals that x y, thus the ew solutio simply becomes: T (x, y = 4,odd sih ( si 3 3 y sih 3 x Takig a superpositio of the solutios to the two differet sets of boudary coditios will yield a solutio for the origially specified problem: T (x, y = 4,odd sih(3 si x sih y + ] sih ( si 3 3 y sih 3 x (4.7 (4.8 5 BOAS, P. 67, PROBLEM 3.-4 I this problem a rectagular plate of width cm ad ifiitely log sides which are isulated is cosidered. At the bottom edge the temperature is give by T = f (x = x 5 ad at a ifiite distace away the temperature is. Give that the sides are isulated we have T = ad T =. Applyig these boudary coditios to the x= x= geeral solutio foud i Problem 3 yields: T (x, y = Ae k y si(kx + +Ce k y cos(kx + = C = A si(kx cos(kx T (x, y = Ae k y si(kx + Be k y si(kx A si(kx cos(kx ek y cos(kx + De k y cos(kx T (x, y = Be k y si(kx + De k y cos(kx T = Bke k y cos( Dke k y si( = Bke k y = B = x= T (x, y = De k y cos(kx T = Dke k y si(k k = k = x= T (x, y = De k y cos x The form of this result will satisfy the boudary coditios, ad thus the sum of all solutios that satisfy these boudary coditios is: ( T (x, y = b e y cos x (5. Now the iitial coditios will be examied to reveal the fial solutio: T (x, = Now the coefficiets b will be foud usig a Fourier series: b = T (x,cos dx x = 5 (5. b cos x = x 5 (5.3 (x 5cos x dx (5.4 5
Now, the itegral will be solved usig the separatio of variables techique: b = 5 (x 5si x ] si x dx = + + ] 5 cos x = ( ] { 4 odd = eve Thus the fial solutio for these boudary coditios will be: T (x, y = 4,odd e y cos x (5.5 (5.6 If f (x = x istead, the oly chage would be to add 5 to the result: I both cases, b =. T (x, y = 4,odd e y cos x + 5 (5.7 6 BOAS, P. 637, PROBLEM 3.3-4 It is give that at t =, two flat slabs each 5 cm thick, oe at o ad oe at o, are stacked together, ad the the surfaces are kept at o. The temperature T (x, t as a fuctio of x ad t for t > will be foud. Startig from the heat flow equatio ad the assumig a solutio of the form T = F (xg(t yields: T = α T t G d F dx = F dg α dt F d F dx = dg α G dt = k (6. Where the shorthad F = F (x ad G = G(t have bee adopted. Additioally i the fial step it is oted that both sides of the fial equatio are fuctios of oly oe variable (x o the left ad t o the right, thus both sides must be equivalet to some costat, which has bee chose as k for this problem. From this result, the followig two equatios may be costructed: d F dx = k F & dg dt = α k G (6. These equatios are rather trivial to solve so the solutios will be give without expedig much effort. They ca be easily verified by direct substitutio. F = A si(kx + B cos(kx & G = Ce α k t (6.3 Thus the desired temperature distributio will be 4 : T (x, y = FG = A si(kx + B cos(kx]e α k t (6.4 Where the costat C has bee absorbed by A ad B. Now the boudary coditios give i the iitial wordig of the problem will be applied: T (, t = + B]e α k t = B = T (x, t = A si(kxe α k t T ( 5, t = A si(ke α k t = si(k = k = k = T (x, t = A si x e α t 4 This result ad all the work that has bee doe leadig up to it will be quoted i subsequet problems but ot repeated. From this poit forward, boudary coditios specific to the problem will shape the outcome ad the solutio will o loger be geeral. (6.5 6
The form of this result will satisfies the boudary coditios, ad thus the sum of all solutios that satisfy these boudary coditios is: T (x, t = b si x e ( α t (6.6 Now the iitial coditios will be examied to reveal the fial solutio: { < x < 5 T (x, = b si x = 5 < x < Now the coefficiets b will be foud usig a Fourier series: (6.7 b = { 5 T (x,si x dx = 5 si x dx < x < 5 5 5 si x dx 5 < x < Notig that for < x < 5 the itegral is, the remaiig itegral will be foud: 4 5 si x dx = 4 ( cos x 5 = 4 cos( cos ] = 4 ] ( cos (6.8 (6.9 Thus the solutio ca ow be expressed completely: T (x, t = 4 cos ( ] si x e ( α t (6. 7 BOAS, P. 63, PROBLEM 3.3-6 It is give that the eds of a bar of legth l are iitially at o ad 5 o, ad that at t = the 5 o ed is chaged to 5 o. The followig kow solutio is take from Equatio 3.5 of Boas Chapter 3: T (x, t = ( e ( α t l si( l x (7. Now this result ca be made to apply to this situatio by addig a fial steady state T f. This solutio will be assumed as liear because the heat should flow liearly from oe ed of the bar to aother. The arbitrary costat of this liear relatio will be foud by applyig the boudary coditios: T f (x, t = ax + b T f (, t = a( + b = b = T f (l, t = al + = 5 a = 3 l (7. T f (x, t = 3 l x + Therefore, the problem is easily solved usig the assumed T (x, t: T ( x, t = 3 l x + + ( e ( α t l si( l x (7.3 8 BOAS, P. 633, PROBLEM 3.3-8 It is give that a bar of legth is iitially at o. From t = o, the x = ed is held at o ad the x = ed is held at o. The depedet temperature distributio T (x, t will be foud. Startig from the geeral solutio T (x, t 7
that was foud i Problem 6, the boudary coditios are applied: T (, t = + B]e α k t = B = T (x, t = A si(kxe α k t T (, t = A si(ke α k t = si(k = k = k = T (x, t = A si x e α t Where T (x, t has bee set equal to eve though it should be. This was doe i the iterest of makig the problem reasoable to solve, the flaw i the problem caused by this will be rectified ear the ed. The sum of all solutios that satisfy these boudary coditios is: T (x, t = (8. b si x e ( α t (8. Where b = A. Now the iitial coditios will be examied to reveal the fial solutio: T (x, = Now T f (x, t is determied from the specified coditios: b si x = T f (x, t (8.3 T f = ax + b T f (, t = a( + b b = T f (, t = a( + = a = 5 T f (x, t = 5x (8.4 Thus, T (x, is ow re-expressed: T (x, = b si x = 5x (8.5 Now the coefficiets b will be foud usig a Fourier series: b = T (x,si x dx = 5 Now, the itegral will be foud usig the itegratio by parts techique: b = 5 x cos x + Thus the solutio ca ow be expressed completely: x si x dx (8.6 ] cos x dx = 5 4 ( 4 ] si x = ( (8.7 T (x, t = 5x ( si x e ( α t (8.8 8