Liner Differentil Equtions Physics 19 Solutions to Problems 051018 Frnk Porter Revision 11106 F. Porter 1 Exercises 1. Consider the generl liner second order homogeneous differentil eqution in one dimemsion: (x) d d u(x) + b(x) u(x) + c(x)u(x) = 0. (1) Determine the conditions under which this my be written in the form of differentil eqution involving self-djoint (with pproprite boundry conditions) Sturm-Liouville opertor: where Lu = 0, () L = d p(x) d q(x). (3) Note tht prt of the problem is to investigte self-djointness. Solution: Probbly the simplest wy to pproch this is to consider n integrting fctor (e.g., Mthews nd Wlker, chpter 1; Riley, Hobson, nd Bence, chpters 14 nd 15): Assuming (x) 0, divide the eqution by (x): d b(x) u(x) + (x) d c(x) u(x) + u(x) = 0. (4) (x) Then multiply the eqution by integrting fctor exp [ b(x)/(x)]: e b(x)/(x) d u(x)+e b(x)/(x) b(x) d (x) u(x)+e b(x)/(x) c(x) u(x) = 0. (x) (5) This gives Sturm-Liouville form with p(x) = e b(x)/(x) (6) q(x) = e b(x)/(x) c(x) (x). (7) 1
To investigte self-djointness of the Sturm-Liouville opertor, we will ssume tht p nd q re rel. Consider {[ b d Lv u v Lu = p(x) d ] v(x)} q(x) u(x) (8) {[ b d v (x) p(x) d ] } q(x) u(x) (9) {[ b d = p(x) d ] } {[ b d v (x) u(x) v (x) p(x) d ] } u(x) = p(x) dv (x) u(x) b b b v (x)p(x) du(x) = p(x) dv (x) u(x) b + b p(x) dv (x) dv (x) v (x)p(x) du(x) The boundry conditions must be such tht the lst line is zero for L to be self-djoint.. Show tht the opertor du(x) (10) p(x)du(x) (11) b (1) L = d + 1, x [0, π], (13) with homogeneous boundry conditions u(0) = u(π) = 0, is selfdjoint. Solution: Consider {[ π d Lv u v Lu = v(x)} 0 + 1 u(x) v (x) 0 π d v π = 0 (x)u(x) v (x) d u 0 (x) = dv (x)u(x) π v (x) du (x) π 0 ] π [ d + 1 The first term vnishes if u stisfies the boundry conditions, nd the second term vnishes if v stisfies the boundry conditions. Thus, L is self-djoint. 3. Let us consider somewht further the momentum opertor, p = 1 i discussed briefly in the differentil eqution note. We let this opertor 0 d, ] u(x) (14)
be n opertor on the Hilbert spce of squre-integrble (normlizble) functions, with x [, b]. () Find the most generl boundry condition such tht p is Hermitin. (b) Wht is the domin, D P, of p such tht p is self-djoint? (c) Wht is the sitution when [, b] [, ]? Is p bounded or unbounded? 4. Prove tht the different systems of orthogonl polynomils re distinguished by the weight function nd the intervl. Tht is, the system of polynomils in [, b] is uniquely determined by w(x) up to constnt for ech polynomil. 5. We sid tht the recurrence reltion for the orthogonl polynomils my be expressed in the form: see Eqn. 16. Try to verify. f n+1 (x) = ( n + b n x) f n (x) c n f n 1 (x), (15) Solution: The complete sttement is tht the recurrence reltion for the orthogonl polynomils my be expressed in the form: where f n+1 (x) = ( n + b n x) f n (x) c n f n 1 (x), (16) b n = +1 (17) ) n = b n ( k n+1 +1 k n (18) c n = h n +1 1, (19) h n 1 nd c 0 = 0. The nottion here is defined by: k n f n (x) = x n + k nx n 1 + k nx n +... + k (n), (0) where n = 0, 1,,... We lso hve the othogonlity nd normliztion condition: f n f m = h n δ nm, (1) where h n is determined by the k (l) n, l = 0, 1,..., n constnts. 3
Let us write p(x) = ( n + b n x) f n (x) c n f n 1 (x), () where p stnds for proposed, since we hven t demonstrted tht it is relly equl to f n+1 yet. Since b n 0, the recurrence reltion gives polynomil of degree n + 1 for p. The sclr product of p with polynomil f m is: p f m = n f n f m + b n xf n f m c n f n 1 f m. (3) Since p is polynomil of degree n + 1, it is orthogonl to ny f m with m > n + 1, since such polynomils re orthogonl to ny polynomil of lower degree by construction. Consider m = n + 1: Note tht: Thus, p f n+1 = b n xf n f n+1. (4) f n+1 = +1 x n+1 + k n+1x n + k n+1x n 1 +... (5) xf n = x n+1 + x n + x n 1 +... (6) = k ( ) n k f n+1 + n k n 1 n+1 f n + α i f i, (7) +1 +1 where the α i re constnts tht we won t need to determine explicitly, since these terms re ll orthogonl to f n+1 nd f n. Hence, j=0 p f n+1 = b n f n+1 f n+1, (8) +1 nd thus p f n+1 = f n+1 f n+1 if b n = +1 /. Now consider m = n: p f n = n f n f n + b n xf n f n (9) ) = n f n f n + b n ( k n k n+1 +1 f n f n. (30) This is zero if n = b n ( k n+1 +1 k n ). (31) 4
Next, consider m = n 1: Noting tht p f n 1 = b n xf n f n 1 c n f n 1 f n 1 (3) = b n f n xf n 1 c n h n 1. (33) xf n 1 = 1 n 1 f n + β j f j, (34) where gin we needn t evlute the β j coefficients, we find This is zero if j=0 p f n 1 = b n 1 f n f n c n h n 1 (35) = b n 1 h n c n h n 1. (36) c n = h n It finlly remins to consider m < n 1 +1 1. (37) h n 1 k n p f m = b n xf n f m, m < n 1, (38) = b n f n xf m (39) m+1 = b n f n γ j f j (40) j=0 = 0. (41) Hence, p stisfies, for the specified n, b n, c n ll of the orthogonlity nd normliztion conditions, nd hence is equl to f n+1 becuse f n+1 is uniquely specified by these conditions. We note tht the possibly specil cse n = 0 is redily checked. 6. We discussed some theorems for the qulittive behvior of clssicl orthogonl polynomils, nd illustrted this with the one-electron tom rdil wve functions. Now consider the simple hrmonic oscilltor (in one dimension) wve functions. The potentil is Thus, the Schrödinger eqution is V (x) = 1 kx. (4) 1 d m ψ(x) + 1 kx ψ(x) = Eψ(x). (43) 5
Mke sketch showing the qulittive fetures you expect for the wve functions corresponding to the five lowest energy levels. Try to do this with some cre: There is relly lot tht you cn sy in qulittive terms without ever solving the Schrödinger eqution. Include curve of the potentil on your grph. Try to illustrte wht hppens t the clssicl turning points (tht is, the points where E = V (x)). Solution: 8 7 6 5 4 3 1 0 4 3 1 1 0 1 3 4 3 4 5 6 7 V(x) psi_0(x) psi_1(x) psi_(x) psi_3(x) psi_4(x) Figure 1: One dimension simple hrmonic oscilltor wve functions. The potentil energy is V (x) = x /, nd the mss is m = 1. Thus the energy levels re t n + 1/, nd the clssicl turning points my be found s where these vlues intersect the curve for V (x). The wve functions re multiplied by ten for disply. Fetures tht good qulittive sketch should include: () The functions nd their first derivtives should be continuous everywhere. (b) The lowest level should hve one zero, ech dditionl level dds n dditionl zero. (c) The functions should hve negtive second derivtive outside the clssicl turning points, pproching zero. The second derivtive should be positive inside the clssicl turning points. (d) The zeros of successive wve functions should lternte. 6
(e) The wvelength of oscilltions should become shorter s the potentil decreses (i.e., the second derivtive increses there). (f) Becuse the potentil is symmetric, the even wve functions should hve even prity nd the odd wve functions should hve odd prity. Note tht the signs of the wve functions (in fct, their complex phses) re rbitrrily chosen. 7. Find the Green s function for the opertor L = d + k, (44) where k is constnt, nd with boundry conditions u(0) = u(1) = 0. For wht vlues of k does your result brek down? You my ssume x [0, 1]. Solution: We look for G(x, y) such tht L x G(x, y) = δ(x y). This is in the form of Sturm-Liouville opertor with p(x) = 1 nd q(x) = k. Thus, we cn write the Green s function s: ( ) u1 (x)u (y) u (x)u 1 (y) G(x, y) = A(y)u 1 (x)+b(y)u (x)+ + W x y x y. (45) The solutions u 1 nd u to the homogeneous eqution need not stisfy the boundry conditions. Possible choices re u 1 = sin kx nd u = cos kx. The Wronskin is: Thus, W = u 1 (x)u (x) u 1(x)u (x) (46) = k. (47) ( ) + sin kx cos ky cos kx sin ky G(x, y) = A(y) sin kx+b(y) cos kx+ k (48) Now we turn our ttention to the boundry conditions. The problem doesn t explicitly specify the rnge for x. We ll mke the typicl presumption tht the rnge is implied by the limits in the boundry conditions, when not explicitly specified otherwise. Different ssumptions re lso cceptble. x y x y. 7
Thus, u(0) = 0 gives: or B(y) = or 0 = A(y) sin k + 0 = B(y) sin ky. The u(1) = 0 condition gives: k cos k sin ky k A(y) = Finlly, we hve the Green s function: sin ky k, (49) 1 (sin k cos ky cos k sin ky), (50) k sin k cos ky cos k sin ky. (51) k sin k G(x, y) = 1 [ ] sin kx sin ky θ(y x) sin kx cos ky + θ(x y) cos kx sin ky. k tn k (5) Note the symmetry in x y, s expected for this self-djoint opertor. The eqution breks down for k = 0, where the Wronskin vnishes. We hve lredy delt with the Green s function for the k = 0 cse in the notes. 8. An integrl tht is encountered in clculting rditive corrections in e + e collisions is of the form: I(t;, b) = where 0 < b 1, nd t 0. b x t 1, (53) 1 x Show tht this integrl my be expressed in terms of the hypergeometric function F 1. Mke sure to check the t = 0 cse. Solution: We re given the integrl representtion: F 1 (α, β; γ, y) = Γ(γ) 1 x β 1 (1 x) γ β 1 (1 yx) α. (54) Γ(β)Γ(γ β) 0 Thus, since I(t; 0, 1) = 1 t F 1 (α = 1, β = t; γ = 1 + t, y = 1), (55) Γ(1 + t) Γ(t)Γ(1) = t. (56) 8
We wnt the integrl from to b, so write: b x t 1 x t 1 I(t;, b) = 0 1 x (57) 0 1 x 1 = b t x t 1 1 x t 1 t (58) 0 1 bx 0 1 x = bt t F 1 (1, t; 1 + t, b) t t F 1 (1, t; 1 + t, ). (59) Wht bout the t = 0 cse? Note tht F 1 (1, t = 0; γ, y) is not relly defined. However, we my be ble to mke sense out of our expression nywy. Consider the series expnsion: Thus, F 1 (α, β; γ, y) = I(t;, b) = = Γ(γ) Γ(α)Γ(β) n=0 Γ(α + n)γ(β + n) y n Γ(γ + n) n!. (60) ( ) Γ(1 + t) 1 Γ(n + 1)Γ(n + t) b n+t n+t (61) Γ(t)Γ(1) t n=0 Γ(n + t + 1) n! n! b n+t n+t. (6) n + t n=0 Tking the t 0 limit: b n+t n+t lim t 0 n=0 n + t This my be compred with I(0;, b) = b t t = lim + t 0 t b n n n=1 n (63) = e t ln b e t ln lim + ln 1 t 0 t 1 b (64) = 1 + t ln b 1 t ln lim + ln 1 t 0 t 1 b (65) = b(1 ) ln (1 b). (66) = b b = ln (67) x(1 x) ( 1 x + 1 ) (68) 1 x b(1 ) (1 b). (69) As long s > 0 nd b < 1 we obtin the correct nswer even for t = 0. 9
9. We consider the Helmholtz eqution in three dimensions: u + k u = 0 (70) inside sphere of rdius, subject to the boundry condition u(r = ) = 0. Such sitution my rise, for exmple, if we re interested in the electric field inside conducting sphere. Our gol is to find G(x, y) such tht ( x + k )G(x, y) = δ(x y), (71) with G(r =, y) = 0. We ll do this vi one pproch in this problem, nd try nother pproch in the next problem. Find G(x, y) by obtining solutions to the homogeneous eqution ( + k )G = 0, (7) on either side of r = y ; stisfying the boundry conditions t r =, nd the pproprite mtching conditions t r = y. Solution: The symmetry of the problem is such tht it is convenient to work in sphericl polr coordintes. Let r = x nd r = y. Let r < be the smller of (r, r ) nd let r > be the lrger of (r, r ). Let (θ, φ) be the polr nd zimuthl ngles of x, nd likewise let (θ, φ ) be the polr nd zimuthl ngles of y. Let ψ be the ngle between x nd y, tht is, The finl nswer is: G(x, y) = l=0 cos ψ = cos θ cos θ + sin θ sin θ cos(φ φ ). (73) k(l + 1)P l (cos ψ) 4π { } jl (kr < ) j l (k) [j l(k)n l (kr > ) n l (k)j l (kr > )]. (74) 10. We return to the preceding problem. This is the problem of the Helmholtz eqution: u + k u = 0 (75) inside sphere of rdius, subject to the boundry condition u(r = ) = 0. Such sitution my rise, for exmple, if we re interested in the electric field inside conducting sphere. Our gol is to find G(x, y) such tht ( x + k )G(x, y) = δ(x y), (76) 10
with G(r =, y) = 0. In problem 9, you found G(x, y) by obtining solutions to the homogeneous eqution ( + k )G = 0, (77) on either side of r = y ; stisfying the boundry conditions t r =, nd the pproprite mtching conditions t r = y. Now we tke different pproch: Find G by directly solving ( x + k )G(x, y) = δ(x y). You should ignore the boundry conditions t first nd obtin solution by integrting the eqution over smll volume contining y. Then stisfy the boundry conditions by dding suitble function g(x, y) tht stisfies ( x +k )g(x, y) = 0 everywhere. 11. Let s continue our discussion of the preceding two problems. This is the problem of the Helmholtz eqution: u + k u = 0 (78) inside sphere of rdius, subject to the boundry condition u(r = ) = 0. Our gol is to find G(x, y) such tht with G(r =, y) = 0. ( x + k )G(x, y) = δ(x y), (79) In problem10, you found G(x, y) by directly solving ( x+k )G(x, y) = δ(x y), ignoring the boundry conditions t first. This is clled the fundmentl solution becuse it contins the desired singulrity structure, nd hence hs to do with the source. Now find the fundmentl solution by nother technique: Put the origin t y nd solve the eqution ( x + k )f(x) = δ(x), (80) by using Fourier trnsforms. Do you get the sme nswer s lst week? 1. Referring still to the Helmholz problem (problems 10 11), discuss the reltive merits of the solutions found in problems 9 nd 10. In prticulr, nlyze, by mking suitble expnsion, cse where the problem 10 solution is likely to be preferred, stting the necessry ssumptions clerly. 13. We noted tht the Green s function method is pplicble beyond the Sturm-Liouville problem. For exmple, consider the differentil opertor: L = d4 + d 4. (81) 11
As usul, we wish to find the solution to Lu = φ. Let us consider the cse of boundry conditions u(0) = u (0) = u (0) = u (0) = 0. () Find the Green s function for this opertor. (b) Find the solution for x [0, ] nd φ(x) = e x. You re encourged to notice, t lest in hindsight, tht you could probbly hve solved this problem by elementry mens. 14. Using the Green s function method, we derived in clss the time development trnsformtion for the free-prticle Schrödinger eqution in one dimension: U(x, y; t) = 1 ( 1 i t t ) m π t exp [ ] im(x y). (8) t This should hve the property tht if you do trnsformtion by time t, followed by trnsformtion by time t, you should get bck to where you strted. Check whether this is indeed the cse or not. 15. Using the Christoffel-Drboux formul, find the projection opertor onto the subspce spnned by the first three Chebyshev polynomils. 16. We discussed the rdil solutions to the one-electron Schrödinger eqution. Investigte orthogonlity of the result re our wve functions orthogonl or not? 17. In clss we considered the problem with the Hmiltonin d H = 1 m. (83) Let us modify the problem somewht nd consider the configurtion spce x [, b] ( infinite squre well ). () Construct the Green s function, G(x, y; z) for this problem. (b) From your nswer to prt (), determine the spectrum of H. (c) Notice tht, using G(x, y; z) = k=1 φ k (x)φ k(y), (84) ω k z the normlized eigenstte, φ k (x), cn be obtined by evluting the residue of G t the pole z = ω k. Do this clcultion, nd check tht your result is properly normlized. 1
(d) Consider the limit, b. Show, in this limit tht G(x, y; z) tends to the Green s function we obtined in clss for this Hmiltonin on x (, ): G(x, y; z) = i m z eiρ x y. (85) 18. Let us investigte the Green s function for slightly more complicted sitution. Consider the potentil: V (x) = { V x 0 x > (86) V _ _ Δ Δ x Figure : The finite squre potentil. () Determine the Green s function for prticle of mss m in this potentil. Remrks: You will need to construct your left nd right solutions by considering the three different regions of the potentil, mtching the functions nd their first derivtives t the boundries. Note tht the right solution my be very simply obtined from the left solution by the symmetry of the problem. In your solution, let ρ = m(z V ) (87) ρ 0 = mz. (88) Mke sure tht you describe ny cuts in the complex plne, nd your selected brnch. You my find it convenient to express your nswer to some extent in terms of the force-free Green s function: G 0 (x, y; z) = im ρ eiρ 0 x y. (89) 13
(b) Assume V > 0. Show tht your Green s function G(x, y; z) is nlytic in your cut plne, with brnch point t z = 0. (c) Assume V < 0. Show tht G(x, y; z) is nlytic in your cut plne, except for finite number of simple poles t the bound sttes of the Hmiltonin. 19. In clss, we obtined the free prticle propgtor for the Schrödinger eqution in quntum mechnics: U(x, t; x 0, t 0 ) = 1 ( 1 i t t ) [ 0 m im(x t t 0 π t t 0 exp x0 ) ]. (t t 0 ) (90) Let s ctully use this to evolve wve function. Thus, let the wve function t time t = t 0 = 0 be: ψ(x 0, t 0 = 0) = ( ) 1 1/4 ( ) exp x 0 π + ip 0x 0, (91) where nd p 0 re rel constnts. Since the bsolute squre of the wve function gives the probbility, this wve function corresponds to Gussin probbility distribution (i.e., the probbility density function to find the prticle t x 0 ) t t = t 0 : ψ(x 0, t 0 ) = ( ) 1 1/ e x 0 π. (9) The stndrd devition of this distribution is σ = /. Find the probbility density function, ψ(x, t), to find the prticle t x t some lter (or erlier) time t. You re encourged to think bout the physicl interprettion of your result. 14