1 Boas, p. 606, problem

Transcription

1 Physics 6C Solutios to Homework Set #4 Fall 2 Boas, p. 66, problem Verify that the differetial equatio of problem.3 Homework set # 2, problem 7) is ot Fuchsia. Solve it by separatio of variables to fid the obvious solutio y =cost. ad a secod solutio i the form of a itegral. Show that the secod solutio is ot expadable i a Frobeius series. The differetial equatio is y + y =. ) x2 A differetial equatio is said Fuchsia whe, give i the form y + fx)y + g x)y =, xfx) ad x 2 gx) are ot sigular i the origi. This is clearly ot the case i ) because xfx) = x is sigular for x. By separatio of variables we have dy y = dx x 2 = y = Ce dx/x 2 = Ce /x = y = C dxe /x +cost. 2) We have foud two solutios of the equatio: oe is the costat fuctio, the other oe is give by dx exp /x). The latter caot be expaded as a Frobeius series: if it could, the secod solutio would be y 2 = x s a x = a +s)x +s, so that we could expad the itegrad i the previous equatio as a power series with a fiite umber of egative powers of x. This is ot the case for e /x = )! x, so the solutio caot be writte as a Frobeius series. 2 Boas, p. 62, problem Solve the Hermite differetial equatio by power series y 2xy +2py = 3) We isert the power series y = a x i to the equatio )a x 2 2 a x +2p a x = 4) +2)+)a +2 x 2 a x +2p a x = 5) +2)+)a p)a = 6) which gives the recursio relatio a +2 = 2 2p +2)+) a. We ca write the solutio as a sum of two series, oe depedet o a ad oe depedet o a : yx) = a + 2p 2 x2 + 2p4 2p) a x+ 2 2p) 3 2 x p)6 2p) 5! x 4 + 2p4 2p)8 2p) x ! ) + 7) ). 8) x p)6 2p) 2p) x !

2 Whe p is a o-egative eve iteger, p = 2N, the a series termiates after N + terms ad we have a polyomial of order 2N; whe p is a positive odd iteger, p = 2N +, the odd series termiates after N + terms ad we have a polyomial of order 2N +. These are the Hermite polyomials. It is covetioal to fix a ad a by the ormalizatio coditio i which the highest power of x of H x) is give by 2x). I this covetio, the first three Hermite polyomials are give by: 3 Boas, p. 62, problem H x) =, H x) = 2x, H 2 x) = 4x ) I the geeratig fuctio for the Hermite polyomials Φx,h) = e 2xh h2, expad the expoetial ad obtai the firs few Hermite polyomials. Verify the idetity 2 x 2Φ 2x x Φ+2h h Φ = ) ad that this proves the polyomials H satisfy Hermite equatio; the verify that the highest term i H is 2x). We have Φx,h) = e 2xh h2 = The first powers i h are: H x) h! = k 2xh h 2 ) k k! = k k k! k! j!k j)! 2xh)j h 2 ) k j ) Φx,h) = +2xh+4x 2 2) h ) 2 which we recogize as yieldig the first three Hermite polyomials [cf. 9)]. Now we verify ): 2 x Φ = 2hΦ, j= Φ = 2x 2h)Φ ; 3) h x 2Φ 2x x Φ+2h h Φ = 4h2 Φ 2x 2hΦ+2h2x 2h)Φ =. 4) From this ad rememberig that Φ = H h /! we ca obtai a differetial equatio for H x): 2 x 2Φ 2x x Φ+2h h Φ = H x)h! 2x H x)h! +2 h H x) )! = 5) = H x) 2xH x)+2h x) =, 6) which is precisely Hermite s equatio 3) for iteger p. Next we look at the highest power of x i H x) i the expressio ) of Φ: Φx,h) = H x) h! = k k k! k! j!k j)! 2x)j ) k j h 2k j 7) j= for fixed = 2k j, that is for j = 2k, we have terms proportioal to 2x) 2k. Because the sum over j was betwee ad k, the terms i the sum over k which cotribute are those with 2 < k <. Thus, the highest power of x is 2 = ad the highest power of H x) is 2x) ; oe also checks that the combiatorial factor i the frot of H is j!k j)! = 2k )! k)! =! so that we have factorized h!. Fially, we have proved that Φ geerates the Hermite polyomials as this is the oly polyomial solutio to Hermite s equatio see previous problem 2). 2

3 4 Boas, p. 62, problem Solve the Laguerre differetial equatio by power series: Isertig the series y = a x, we have xy + x)y +py = 8) )a x + x) a x +p a x = 9) ] [+)a + ++)a + a +pa x = 2) +) 2 a + = p)a = a + = p) +) 2a 2) if p is a iteger, the coefficiet a p+ is zero ad the series termiate, that is, the solutio is a polyomial. Settig the ormalizatio to a =, these are called Laguerre polyomials L x): L x) = x+ ) 2!) 2 x 2 + )2 ) 3!) 2 x )2 ) )!) 2 x 22) From this we ca read the first few Laguerre polyomials: L x) =, L x) = x, L 2 x) = 2x+ x2 2, L 3x) = 3x+ 3 2 x2 6 x3. 23) 5 Boas, p. 64, problem Give the differetial equatio λ y + x 4 ll+) ) x 2 y = 24) where l > is a iteger, fid values of λ such that y for x ad fid the correspodig eigefuctios. We write ad fid a related differetial equatio for v: yx) = x l+ e x/2 vx), 25) y = l+)x l e x/2 v 2 xl+ e x/2 v +x l+ e x/2 v ; 26) y = ll +)x l e x/2 v l + 2 xl e x/2 v +l+)x l e x/2 v 27) l+ 2 xl e x/2 v + 4 xl+ e x/2 v 2 xl+ e x/2 v + 28) +l +)x l e x/2 v 2 xl+ e x/2 v +x l+ e x/2 v ; 29) = l+)x l v +[2l +)x l x l+ ]v +x l+ v +λx l v =, 3) = xv +2l +2 x)v +λ l )v = 3) 3

4 This has the same form of the equatio solved by the associated Laguerre polyomials: xy +k + x)y +y =, y = L k x) 32) The, for a iteger λ > l, we there is a polyomial solutio of the form vx) = L 2l+ λ l x). The solutio to the origial equatio 24) is yx) = x l+ e x/2 L 2l+ λ l x) 33) We ca ote that we just solved a eigevalue problem: we foud that for specific values of λ, the equatio 24) admit solutios related to the associated Laguerre polyomials. Motivatio for the chage of variables To uderstad the motivatio for 25), let us deduce the behavior of the solutio yx) to the differetial equatio 24) i the limit where x ad x respectively. First, as x, the term ll +)/x 2 is much larger tha λ/x ad 4. Hece, the latter two terms ca be eglected, ad we examie y ll +)y x 2 =. Multiplyig by x 2, we see that this differetial equatio has the form of a Euler differetial equatio [cf. Case d) o p. 434 of Boas]. The solutio to this equatio is a power law, y = x p. Pluggig this ito the above equatio yields pp ) = ll+), which has two solutios p = l+ ad p = l. We reject p = l which is egative for positive iteger l, as this would correspod to a solutio yx) that is sigular i.e., ubouded) at x =. Hece, the o-sigular behavior of yx) as x is yx) x l+. As x, we ca eglect the λ/x ad ll+)/x 2 as compared to 4 i 24). Hece, we examie y 4 y =. Thesolutiotothisequatioisaliearcombiatioofe x/2 ade x/2. Werejectthelatterasitcorrespods to a solutio yx) that is sigular i.e., ubouded) as x. Hece, the o-sigular behavior of yx) as x is yx) e x/2. Combiig these two results, it is especially useful to defie yx) = x l+ e x/2 vx), which embodies both the small x ad large x behavior of yx), assumig that vx) is well-behaved i these limits. This is precisely the chage of variables proposed i 25). 6 Boas, p. 64, problem I the theory of the hydroge atom the fuctios of iterest are x ) f x) = x l+ e x/2 L l 2l+ 34) where is a iteger ad so is l, l. For l =, show that f 2 x) = x 2 e x/4, f 3 x) = x 2 e x/6 4 x ), f 4 x) = x 2 e x/8 5x ) x2. 35) 32 We will fid the associated Laguerre polyomials startig from the Laguerre polyomials ad usig L k dk x) = )k dx kl +kx). 36) 4

5 We eed to fid L 3, L3, L3 2 ; i additio to the first polyomials i 23), we eed the other L s up to = 5. We make use of the defiitio 22): L 4 x) = 4x+3x x x4, L 5 x) = = 5x+5x x x4 2 x5 37) L 3 = d3 dx 3L 3x) =, Replacig x x we have L 3 = d3 dx 3L 4x) = 4 x, L 3 2 = d3 dx 3L 5x) = 5x+ x2 2 38) x f 2 x) = x 2 e x/4 L 3 = x 2) 2 e x/4, 39) x f 3 x) = x 2 e x/6 L 3 = x 3) 2 e x/6 4 x ), 4) 3 x f 4 x) = x 2 e x/8 L 3 2 = x 4) 2 e x/8 5x ) 4 + x2. 4) 32 which is precisely 35). For fixed l, the fuctios f x) are a orthogoal set o, ) as a cosequece of Sturm-Liouville theory). We ca verify this with these three fuctios: dxf 2 x)f 3 x) = dxx 4 e 5x/2 4 x ) = 4 dxx 4 e 5x/2 dxx 5 e 5x/2 = 42) 3 3 ) 2 5 = 4 dyy 4 e y ) 2 6 ) 2 5 [ dyy 5 e y = 4Γ5) ] Γ6) = dxf 2 x)f 4 x) = dxx 4 e 3x/8 5x ) ) 4 + x2 8 5 = Γ5) 5 ) 8 6 Γ6)+ ) 8 7 Γ7) = ) 8 5 = Γ5)[ ] = 43) dxf 3 x)f 4 x) = dxx 4 e 7x/ x x2 ) 96 x3 = 44) ) 24 5 = Γ5)[ 4 25 ) ) ) ] = 45) where the Γ fuctio is defied as Γz) = dte t t z ad Γz +) = zγz). 7 Boas, p. 68, problem Show that R = lx x 2 )D ad L = lx+ x 2 )D where D = d dx are raisig ad lowerig operators for the Legedre polyomials. More precisely, show that RP l = lp l ad LP l = lp l : This is immediate oce we recall the recursio relatios of the Legedre polyomials: [ RP l = lx x 2 ) d dx LP l = [ lx+ x 2 ) d dx ] P l = lxp l x 2 )P l = lxp l lxp l +lp l = lp l 46) ] P l = lxp l + x 2 )P l = lxp llp l lxp l = lp l 47) 5

6 Assumig P l ) =, we ca fid P as the polyomial aihilated by L ad the fid the other Legedre polyomials usig the raisig operators: LP x) = x 2 )P = = P = cost =, 48) P x) = RP = x x 2 ) d dx = x, P 2x) = 2 RP = 2 2x2 x 2 )) = 2 3x2 ). 49) Note that the choice of costat such that P x) = is a covetio. Oce this covetio has bee chose, the ormalizatio of the other Legedre polyomials is fixed ad determied by applyig the raisig operators. 8 Boas, p , problem 3.-2 a) Show that the expressio ux,t) = six vt) satisfies the wave equatio. Show that, i geeral, u = fx vt) ad u = fx+vt) satisfy the wave equatio. The wave equatio is For a oe-dimesioal problem, = x 2 u 2 v 2 t2u = 5) ad this admits the solutio ux,t) = six vt): x u = cosx vt), 2 x 2u = u, 2 t 2 = v2 u = 2 u 2 v 2 t2u = 5) More geerally, oe ca see that ay fuctio that has a secod derivative) ux,t) = fx±vt) satisfies 5): ξ ± = x±vt, x = ξ ± =, x ξ ± ξ ± t = ξ ± = ±v 52) t ξ ± ξ ± 2 u 2 v 2 t 2u = f ξ ± ) v 2±v)2 f ξ ± ) = 53) where the equatio holds separately for ξ + ad ξ. fx vt) represets a excitatio movig i the positive x directio ad fx+vt) a excitatio movig i the opposite directio. b) Show that ur,t) = r fr vt) ad ur,t) = rfr + vt) satisfy the wave equatio i spherical coordiates. The Laplacia operator i spherical coordiates is 2 = r 2 r 2 ) + r r r 2 siϕ ϕ siϕ ) + ϕ 2 r 2 si 2 ϕθ2. 54) If we are lookig for solutios idepedet of φ, θ, oly the first term cotributes; the wave equatio becomes r 2 r 2 )ur,t) v 2 r r 2 t2ur,t) = 55) As before, if we isert the coordiates ξ ± = r ±vt we see that ur,t) = r fξ ±) is a solutio for ay f: r 2 r 2 ) r r r f = f +rf ) r 2 = r r 2f + r 2f + r f, 56) r 2 r 2 )ur,t) v 2 r r 2 t 2ur,t) = r f ξ ± ) v 2±v)2 r f =. 57) These fuctios represet spherical waves radially comig out of or ito) the origi. 6

7 9 Boas, p. 626, problem Solve the semi-ifiite plate problem if the bottom edge of width 3 is held at T = { x, < x < 5, 3 x, 5 < x < 3, 58) ad the other sides are at o C. The temperature T iside the plate satisfies Laplace s equatio with the boudary coditios give by 58). Solvig by separatio of variables, we have { }{ } 2 e ky sikx Tx,y) =, Tx,y) = Xx)Yy) = e ky 59) coskx Note that the boudary heat distributio 58) does ot ifluece the form of the solutio iside the plate; it will oly select a differet solutio of that form. We ow apply the boudary coditio: sice T as y, o solutio of the form e ky is acceptable; similarly, as T,y) =, o solutio with coskx ca exist; T3,y) = so that si3k =, or k = π 3 ; The solutio must have the form Tx,y) = Fially, we must reproduce the temperature distributio 58) for y = : c e πy/3 si πx 3 ; 6) Tx,) = c si πx 3 = { x, < x < 5, 3 x, 5 < x < 3 ; 6) we ca fid the coefficiet c s because this is a Fourier series: c = 2 3 = 6 3 /2 Tx,)si πx 3 = 2 3 ydysiπy+6 [ = 6 π ycosπy = 6 π 2 π22si 2 /2 + π 5 /2 /2 xsi πx 3 dx x)si πx 3 dx = 62) y)dysiπy = 63) cosπydy π y)cosπy /2 ] cos πydy π /2 For eve = 2k that tells us c 2 k =, while for odd = 2k+ we have c = 2 2 π 2 ) k. We ca ow write the solutio to Laplace s equatio with the boudary coditio 58): 64) 65) Tx,y) = k 2 ) k 2k+)πy 2k +)πx 2k +) 2 π 2e 3 si 3 66) The plots for the temperature distributio are i figure. 7

8 Figure : Temperature distributio 66), with the sum trucated at k = 2 8

9 Boas, p. 627, problem Show that the series ca be summed to get T = 4 π k 2k+)πy 2k +)πx 2k + e si 67) T = 2 ) siπx/) π arcta. 68) sihπy/) We have six = 2i eix e ix ) = Ime ix, the T = 4 π odd e πy/ Ime iπx/ = 4 π Im odd eiπx+iy)/ = 4 π Im odd z, 69) where z e iπx+iy)/ ad odd meas that we sum over =,3,5,... To evaluate this sum, recall that ) + L+z) = z, for z, z, 7) = where z is a complex umber ad L is the pricipal value of the complex logarithm. We would like to write: ) +z ) + L = L+z) L z) = z ) + z) z = = = 2 z, for z, z ±. 7) odd However, recall that Lz /z 2 ) = L z Lz is valid oly whe π < Arg z Arg z 2 π, where Argz is the pricipal value of the argumet of the complex umber z as discussed at great legth i the Physics 6A hadout etitled, The complex logarithm, expoetial ad power fuctios). Nevertheless, it is straightforward to check that for z = +z ad z 2 = z, this coditio is satisfied whe z ad z ±. Hece, it follows that: T = 2 π ImL +e iπx+iy)/ e iπx+iy)/ ) = 2 π Arg +e iπx+iy)/ e iπx+iy)/ where we have used Lz = L z +iargz for the pricipal value of the complex logarithm. To evaluate the argumet of the expressio above, it is coveiet to rewrite the complex umber i a+ib form, +e iπx+iy)/ e iπx+iy)/ = +eiπx+iy)/ ) e iπx iy)/ ) e iπx+iy)/) e iπx iy)/ ) = e πy/5 +2ie πy/ siπx/) +e πy/5 2e πy/ cosπx/). I the Physics 6A hadout etitled, The argumet of a complex umber, I show that if a > the Arga + ib) = Arctab/a), where Arcta is the pricipal value of the arctaget fuctio. I the preset applicatio, we have: a = e πy/5 +e πy/5 2e πy/ cosπx/), b = 2e πy/ siπx/) +e πy/5 2e πy/ cosπx/). 72) Note that sice the right had side of 7) is a sigle-valued fuctio, the left had side must be sigle-valued as well. Choosig z = yields L = as expected for the pricipal value. ), 9

10 Sice y, we shall treat y = ad y > separately. Whe y >, it follows that a >, sice the umerator of a is positive ad the deomiator of a is +e πy/5 2e πy/ cosπx/) e πy/ ) 2, after otig that cosπx/). Hece, ) +e iπx+iy)/ ) b Arg e iπx+iy)/ = Arcta = Arcta a where i the last step, we used the fact that: Hece, we coclude that: ) 2e πy/ ) siπx/) siπx/) e πy/5 = Arcta, sihπy/) 2e πy/ e πy/5 = 2 e πy/ e πy/5 ) = 2 e πy/ e πy/ = sihπy/). I the case of y = ad < x <, we have 2 T = 2 ) siπx/) π Arcta. 73) sihπy/) a =, b = siπx/) = cotπx/2) >. cosπx/) If a = ad b >, the it follows that Arga +bi) = 2π. Hece, i 73), if y = ad < x <, the arctaget is equal to 2π ad we fid T =, which is the boudary coditio for the bottom of the rectagular plate. Fially, we ca use 73) to calculate T5,5) = o C. Boas, p. 627, problem Fid the steady state temperature distributio i a rectagular plate coverig the area < x <, < y < 2 if the two adjacet sides alog the axes are held at temperatures T = x ad T = y ad the other two sides at o C. The solutio to Laplace s equatio is always the same, but this time we have differet boudary coditios. { }{ } 2 sihky sikx Tx,y) =, Tx,y) = Xx)Yy) =, 74) cosh ky coskx Tx,) = x, T,y) = y, Tx,2) = T,y) =. 75) Here we have substituted the expoetials i y with the hyperbolic sie ad cosie. We ca do it because these are liear combiatios of the expoetials ad still solutios to Laplace s equatio. Now, because Laplace s equatio is a liear differetial equatio, the sum of two solutios is still a solutio; the we will fid a solutio T satisfyig the boudary coditio T x,) = x, T,y) = ad oe T 2 satisfyig T 2,y) =, T 2,y) = y ad add them together; the sum will satisfy Laplace s equatio ad the boudary coditios 75). We first look at the boudary coditio T x,2) =, T,y) =,T,y) = : T,y) = T,y) = = X = si πx, Tx,2) = = Y = sihk2 y) 76) 2 We do ot cosider the poits x = y = or x =, y = sice the temperature is ot well defied at these two poits o the boudary of the rectagular plate.

11 so that the solutio takes the form T = A sih π πx 2 y)si. 77) By applyig the last coditio T x,) = x we fid the coefficiets A : A sih2π = 2 [ dxxsi πx = 2 πx xcos π = 2 [ ] π ) + π ] cos πx dx = 78) = 2 π )+ 79) TofidtheothersolutiowesolvethesameequatiowithboudarycoditiosT 2,y) =, T 2,y) = y, T 2 x,2) = T 2 x,) = ; this goes i the same way we just did so we ca give the aswer exchagig the roles of x ad y ad beig careful about the differet sides legths): T 2 = B sih π πy x)si ) B sih π 2 = 2 [ 2 ] 2 π 2 ) = 4 π )+ 8) The solutio satisfyig the origial boudary coditios 75) is the Tx,y) = 2 π + 4 π ) + sih2π ) + sihπ/2 2 Boas, p , problem π πx sih 2 y)si + 82) π πy sih x)si 2 2 The heat flow across a edge is proportioal to the derivative alog the directio ormal to that edge, T/. For a plate with a isulated edge, the boudary coditio is that the heat flow alog that edge is zero. Fid the steady-state temperature of a semi-ifiite plate of width cm where the two log edges are isulated, the far ed is at o C ad the bottom edge is at Tx,) = x 5. We solve Laplace s equatio with this boudary coditio: { 2 e ky Tx,y) =, Tx,y) = Xx)Yy) = Tx,) = x 5, T lim Tx,y) =, y e ky 83) }{ } sikx, 84) coskx T,y) =,y) = ; 85) x x Because of the secod coditio, we elimiate the solutio e ky. The others give: { } T coskx x,y) = = o sikx terms i T 86) sikx T π,y) = = sik = = k = 87) x So the solutio takes the form Tx,y) = = b e πy/ cos πx 88)

12 The coefficiets are give by b = 2 dxx 5)cos πx = 2 2 π π 2 x π [ 2 )cosxdx = 2 π 2 x π 2 )six π π ] six = 2 2 π 2 ) ) = 4 2 π 2 for odd 89) Fially, the solutio is T = 4 π 2 odd 2e πy/ cos πx. 9) We elimiated the solutio e ky because we wated T for y. If we require T to stay fiite ot ecessarily zero) for y we ca admit the solutio e ky k=, or equivaletly admit = i the sum 88). We ca solve the same equatio with this ew boudary coditio ad the source fx) = x. With respect to the previous boudary coditio, we have fx) = f old x) + 5; the we ca use the solutio to the previous case ad add a solutio that respects the boudary coditio Tx,) = 5. A costat respects all the required boudary coditio, the the solutio to this ew problem is T = 5 4 π 2 odd 2e πy/ cos πx. 9) 2