4 Sturm-Liouville Boundary Value Problems

Transcription

1 4 Sturm-Liouville Boundry Vlue Problems We hve seen tht trigonometric functions nd specil functions re the solutions of differentil equtions. These solutions give orthogonl sets of functions which cn be used to represent functions in generlized Fourier series expnsions. At the sme time we would like to generlize the techniques we hd first used to solve the het eqution in order to solve more generl initil-boundry vlue problems. Nmely, we use seprtion of vribles to seprte the given prtil differentil eqution into set of ordinry differentil equtions. A subset of those equtions provide us with set of boundry vlue problems whose eigenfunctions re useful in representing solutions of the prtil differentil eqution. Hopefully, those solutions will form useful bsis in some function spce. A clss of problems to which our previous exmples belong re the Sturm-Liouville eigenvlue problems. These problems involve self-djoint differentil) opertors which ply n importnt role in the spectrl theory of liner opertors nd the existence of the eigenfunctions needed to solve the interesting physics problems described by the bove initil-boundry vlue problems. In this section we will introduce the Sturm-Liouville eigenvlue problem s generl clss of boundry vlue problems contining the Legendre nd Bessel equtions nd supplying the theory needed to solve vriety of problems. 4.1 Sturm-Liouville Opertors In physics mny problems rise in the form of boundry vlue problems involving second order ordinry differentil equtions. For exmple, we will explore the wve eqution nd the het eqution in three dimensions. Seprting out the time dependence leds to three dimensionl boundry vlue problem in both cses. Further seprtion of vribles leds to set of boundry vlue problems involving second order ordinry differentil equtions.

2 108 prtil differentil equtions In generl, we might obtin equtions of the form 2 x)y + 1 x)y + 0 x)y = f x) 4.1) subject to boundry conditions. We cn write such n eqution in opertor form by defining the differentil opertor L = 2 x)d x)d + 0 x), where D = d/. Then, Eqution 4.1) tkes the form Ly = f. Recll tht we hd solved such nonhomogeneous differentil equtions in Chpter 2. In this section we will show tht these equtions cn be solved using eigenfunction expnsions. Nmely, we seek solutions to the eigenvlue problem Lφ = λφ with homogeneous boundry conditions on φ nd then seek solution of the nonhomogeneous problem, Ly = f, s n expnsion over these eigenfunctions. Formlly, we let yx) = c n φ n x). n=1 The Sturm-Liouville opertor. However, we re not gurnteed nice set of eigenfunctions. We need n pproprite set to form bsis in the function spce. Also, it would be nice to hve orthogonlity so tht we cn esily solve for the expnsion coefficients. It turns out tht ny liner second order differentil opertor cn be turned into n opertor tht possesses just the right properties self-djointedness) to crry out this procedure. The resulting opertor is referred to s Sturm- Liouville opertor. We will highlight some of the properties of these opertors nd see how they re used in pplictions. We define the Sturm-Liouville opertor s L = d px) d + qx). 4.2) The Sturm-Liouville eigenvlue problem. The Sturm-Liouville eigenvlue problem is given by the differentil eqution Ly = λσx)y, or d px) dy ) + qx)y + λσx)y = 0, 4.3) for x, b), y = yx), plus boundry conditions. The functions px), p x), qx) nd σx) re ssumed to be continuous on, b) nd px) > 0, σx) > 0 on [, b]. If the intervl is finite nd these ssumptions on the coefficients re true on [, b], then the problem is sid to be regulr Sturm-Liouville problem. Otherwise, it is clled singulr Sturm-Liouville problem.

3 sturm-liouville boundry vlue problems 109 ns. We lso need to impose the set of homogeneous boundry conditions α 1 y) + β 1 y ) = 0, α 2 yb) + β 2 y b) = ) hve Neumnn boundry conditions. In this cse, y ) = 0 nd y b) = 0. In terms of the het eqution exmple, Dirichlet conditions correspond to mintining fixed temperture t the ends of the rod. The Neumnn boundry conditions would correspond to no het flow cross the ends, or insulting conditions, s there would be no temperture grdient t those points. The more generl boundry conditions llow for prtilly insulted boundries. Another type of boundry condition tht is often encountered is the periodic boundry condition. Consider the heted rod tht hs been bent to form circle. Then the two end points re physiclly the sme. So, we would expect tht the temperture nd the temperture grdient should gree t those points. For this cse we write y) = yb) nd y ) = y b). Boundry vlue problems using these conditions hve to be hndled differently thn the bove homogeneous conditions. These conditions leds to different types of eigenfunctions nd eigenvlues. As previously mentioned, equtions of the form 4.1) occur often. We now show tht ny second order liner opertor cn be put into the form of the Sturm-Liouville opertor. In prticulr, eqution 4.1) cn be put into the form d px) dy ) + qx)y = Fx). 4.5) Another wy to phrse this is provided in the theorem: The proof of this is stright forwrd s we soon show. Let s first consider the eqution 4.1) for the cse tht 1 x) = 2 x). Then, we cn write the eqution in form in which the first two terms combine, f x) = 2 x)y + 1 x)y + 0 x)y = 2 x)y ) + 0 x)y. 4.6) The resulting eqution is now in Sturm-Liouville form. We just identify px) = 2 x) nd qx) = 0 x). Not ll second order differentil equtions re s simple to convert. Consider the differentil eqution x 2 y + xy + 2y = 0. In this cse 2 x) = x 2 nd 2 x) = 2x = 1x). So, this does not fll into this cse. However, we cn chnge the opertor in this eqution, x 2 D + xd, to Sturm-Liouville opertor, Dpx)D for px) tht depends on the coefficients x 2 nd x.. The α s nd β s re constnts. For different vlues, one hs specil types of boundry conditions. For β i = 0, we hve wht re clled Dirichlet boundry conditions. Nmely, y) = 0 nd yb) = 0. For α i = 0, we Dirichlet boundry conditions - the solution tkes fixed vlues on the boundry. These re nmed fter Gustv Lejeune Dirichlet ). Neumnn boundry conditions - the derivtive of the solution tkes fixed vlues on the boundry. These re nmed fter Crl Neumnn ). Differentil equtions of Sturm-Liouville form.

4 110 prtil differentil equtions In the Sturm Liouville opertor the derivtive terms re gthered together into one perfect derivtive, Dpx)D. This is similr to wht we sw in the Chpter 2 when we solved liner first order equtions. In tht cse we sought n integrting fctor. We cn do the sme thing here. We seek multiplictive function µx) tht we cn multiply through 4.1) so tht it cn be written in Sturm-Liouville form. We first divide out the 2 x), giving y + 1x) 2 x) y + 0x) 2 x) y = f x) 2 x). Next, we multiply this differentil eqution by µ, µx)y + µx) 1x) 2 x) y + µx) 0x) f x) y = µx) 2 x) 2 x). The first two terms cn now be combined into n exct derivtive µy ) if the second coefficient is µ x). Therefore, µx) stisfies first order, seprble differentil eqution: dµ = µx) 1x) 2 x). This is formlly solved to give the sought integrting fctor 1 x) µx) = e 2 x). Thus, the originl eqution cn be multiplied by fctor to turn it into Sturm-Liouville form. In summry, Eqution 4.1), µx) 2 x) = 1 2 x) e 1x) 2 x) 2 x)y + 1 x)y + 0 x)y = f x), 4.7) cn be put into the Sturm-Liouville form d px) dy ) + qx)y = Fx), 4.8) where px) = 1 x) e 2 x), qx) = px) 0x) 2 x), Fx) = px) f x) 2 x). 4.9)

5 sturm-liouville boundry vlue problems 111 Exmple 4.1. Convert x 2 y + xy + 2y = 0 into Sturm-Liouville form. We cn multiply this eqution by µx) 2 x) = 1 x 2 e x = 1 x, to put the eqution in Sturm-Liouville form: Conversion of liner second order differentil eqution to Sturm Liouville form. 0 = xy + y + 2 x y = xy ) + 2 y. 4.10) x 4.2 Properties of Sturm-Liouville Eigenvlue Problems There re severl properties tht cn be proven for the regulr) Sturm-Liouville eigenvlue problem in 4.3). However, we will not prove them ll here. We will merely list some of the importnt fcts nd focus on few of the properties. 1. The eigenvlues re rel, countble, ordered nd there is smllest eigenvlue. Thus, we cn write them s λ 1 < λ 2 <.... However, there is no lrgest eigenvlue nd n, λ n. 2. For ech eigenvlue λ n there exists n eigenfunction φ n with n 1 zeros on, b). 3. Eigenfunctions corresponding to different eigenvlues re orthogonl with respect to the weight function, σx). Defining the inner product of f x) nd gx) s Rel, countble eigenvlues. Oscilltory eigenfunctions. f, g = f x)gx)σx), 4.11) then the orthogonlity of the eigenfunctions cn be written in the Orthogonlity of eigenfunctions. form φ n, φ m = φ n, φ n δ nm, n, m = 1, 2, ) 4. The set of eigenfunctions is complete; i.e., ny piecewise smooth function cn be represented by generlized Fourier series expnsion of the eigenfunctions, where f x) c n φ n x), n=1 c n = f, φ n φ n, φ n. Actully, one needs f x) L 2 σ, b), the set of squre integrble functions over [, b] with weight function σx). By squre integrble, we men tht f, f <. One cn show tht such spce is isomorphic to Hilbert spce, complete inner product spce. Hilbert spces ply specil role in quntum mechnics. Complete bsis of eigenfunctions.

6 112 prtil differentil equtions The Ryleigh quotient is nmed fter Lord Ryleigh, John Willim Strutt, 3rd Bron Rleigh ). 5. The eigenvlues stisfy the Ryleigh quotient [ p λ n = pφ n dφ n b + φ n, φ n ) ] dφn 2 qφ 2 n. This is verified by multiplying the eigenvlue problem Lφ n = λ n σx)φ n by φ n nd integrting. Solving this result for λ n, we obtin the Ryleigh quotient. The Ryleigh quotient is useful for getting estimtes of eigenvlues nd proving some of the other properties. y x Figure 4.1: Plot of the eigenfunctions φ n x) = sin nx for n = 1, 2, 3, 4. Exmple 4.2. Verify some of these properties for the eigenvlue problem y = λy, y0) = yπ) = 0. This is problem we hd seen mny times. The eigenfunctions for this eigenvlue problem re φ n x) = sin nx, with eigenvlues λ n = n 2 for n = 1, 2,.... These stisfy the properties listed bove. First of ll, the eigenvlues re rel, countble nd ordered, 1 < 4 < 9 <.... There is no lrgest eigenvlue nd there is first one. The eigenfunctions corresponding to ech eigenvlue hve n 1 zeros 0n 0, π). This is demonstrted for severl eigenfunctions in Figure 4.1. We lso know tht the set {sin nx} n=1 is n orthogonl set of bsis functions of length φ n = π 2. Thus, the Ryleigh quotient cn be computed using px) = 1, qx) = 0, nd the eigenfunctions. It is given by φ n φ n π + π 0 0 φ n) 2 R = = 2 π π 0 π 2 n 2 cos nx) 2 = n ) Therefore, knowing the eigenfunction, the Ryleigh quotient returns the eigenvlues s expected. Exmple 4.3. We seek the eigenfunctions of the opertor found in Exmple 4.1. Nmely, we wnt to solve the eigenvlue problem Ly = xy ) + 2 y = λσy 4.14) x subject to set of homogeneous boundry conditions. Let s use the boundry conditions y 1) = 0, y 2) = 0. [Note tht we do not know σx) yet, but will choose n pproprite function to obtin solutions.]

7 sturm-liouville boundry vlue problems 113 Expnding the derivtive, we hve Multiply through by x to obtin xy + y + 2 x y = λσy. x 2 y + xy λxσ) y = 0. Notice tht if we choose σx) = x 1, then this eqution cn be mde Cuchy- Euler type eqution. Thus, we hve The chrcteristic eqution is x 2 y + xy + λ + 2) y = 0. r 2 + λ + 2 = 0. For oscilltory solutions, we need λ + 2 > 0. Thus, the generl solution is yx) = c 1 cos λ + 2 ln x ) + c 2 sin λ + 2 ln x ). 4.15) Next we pply the boundry conditions. y 1) = 0 forces c 2 = 0. This leves The second condition, y 2) = 0, yields This will give nontrivil solutions when yx) = c 1 cos λ + 2 ln x). sin λ + 2 ln 2) = 0. λ + 2 ln 2 = nπ, n = 0, 1, 2, In summry, the eigenfunctions for this eigenvlue problem re nπ ) y n x) = cos ln 2 ln x, 1 x 2 nd the eigenvlues re λ n = nπ ln 2 ) 2 2 for n = 0, 1, 2,.... Note: We include the n = 0 cse becuse yx) = constnt is solution of the λ = 2 cse. More specificlly, in this cse the chrcteristic eqution reduces to r 2 = 0. Thus, the generl solution of this Cuchy-Euler eqution is yx) = c 1 + c 2 ln x. Setting y 1) = 0, forces c 2 = 0. y 2) utomticlly vnishes, leving the solution in this cse s yx) = c 1. We note tht some of the properties listed in the beginning of the section hold for this exmple. The eigenvlues re seen to be rel, countble nd ordered. There is lest one, λ 0 = 2. Next, one cn find the zeros of ech eigenfunction on [1,2]. nπ Then the rgument of the cosine, ln 2 ln x, tkes vlues 0 to nπ for x [1, 2]. The cosine function hs n 1 roots on this intervl.

8 114 prtil differentil equtions Orthogonlity cn be checked s well. We set up the integrl nd use the substitution y = π ln x/ ln 2. This gives 2 nπ ) mπ ) y n, y m = cos 1 ln 2 ln x cos ln 2 ln x x = ln 2 π cos ny cos my dy π 0 = ln 2 2 δ n,m. 4.16) Adjoint Opertors In the study of the spectrl theory of mtrices, one lerns bout the djoint of the mtrix, A, nd the role tht self-djoint, or Hermitin, mtrices ply in digonliztion. Also, one needs the concept of djoint to discuss the existence of solutions to the mtrix problem y = Ax. In the sme spirit, one is interested in the existence of solutions of the opertor eqution Lu = f nd solutions of the corresponding eigenvlue problem. The study of liner opertors on Hilbert spce is generliztion of wht the reder hd seen in liner lgebr course. Just s one cn find bsis of eigenvectors nd digonlize Hermitin, or self-djoint, mtrices or, rel symmetric in the cse of rel mtrices), we will see tht the Sturm-Liouville opertor is self-djoint. In this section we will define the domin of n opertor nd introduce the notion of djoint opertors. In the lst section we discuss the role the djoint plys in the existence of solutions to the opertor eqution Lu = f. We begin by defining the djoint of n opertor. The djoint, L, of opertor L stisfies u, Lv = L u, v for ll v in the domin of L nd u in the domin of L. Here the domin of differentil opertor L is the set of ll u L 2 σ, b) stisfying given set of homogeneous boundry conditions. This is best understood through exmple. Exmple 4.4. Find the djoint of L = 2 x)d x)d + 0 x) for D = d/. In order to find the djoint, we plce the opertor inside n integrl. Consider the inner product u, Lv = u 2 v + 1 v + 0 v). We hve to move the opertor L from v nd determine wht opertor is cting on u in order to formlly preserve the inner product. For simple opertor like L = d, this is esily done using integrtion by prts. For the given opertor, we will need to pply severl integrtions by prts to the individul terms. We consider ech derivtive term in the integrnd seprtely. For the 1 v term, we integrte by prts to find ux) 1 x)v x) = 1 x)ux)vx) b ux) 1 x)) vx). 4.17)

9 sturm-liouville boundry vlue problems 115 Now, we consider the 2 v term. In this cse it will tke two integrtions by prts: ux) 2 x)v x) = 2 x)ux)v x) b ux) 2 x)) vx) Combining these results, we obtin u, Lv = = [ 2 x)ux)v x) 2 x)ux)) vx) ] b + ux) 2 x)) vx). 4.18) u 2 v + 1 v + 0 v) = [ 1 x)ux)vx) + 2 x)ux)v x) 2 x)ux)) vx) ] b + [ 2 u) 1 u) + 0 u ] v. 4.19) Inserting the boundry conditions for v, one hs to determine boundry conditions for u such tht This leves [ 1 x)ux)vx) + 2 x)ux)v x) 2 x)ux)) vx) ] b u, Lv = [ 2 u) 1 u) + 0 u ] v L u, v. = 0. Therefore, L = d2 2 2x) d 1x) + 0 x). 4.20) When L = L, the opertor is clled formlly self-djoint. When the domin of L is the sme s the domin of L, the term self-djoint is used. As the domin is importnt in estblishing self-djointness, we need to do complete exmple in which the domin of the djoint is found. Self-djoint opertors. Exmple 4.5. Determine L nd its domin for opertor Lu = du where u stisfies the boundry conditions u0) = 2u1) on [0, 1]. We need to find the djoint opertor stisfying v, Lu = L v, u. Therefore, we rewrite the integrl v, Lu >= 1 0 v du = uv u dv = L v, u. From this we hve the djoint problem consisting of n djoint opertor nd the ssocited boundry condition or, domin of L.): 1. L = d. 2. uv = 0 0 = u1)[v1) 2v0)] v1) = 2v0). 1 0

10 116 prtil differentil equtions Lgrnge s nd Green s Identities Before turning to the proofs tht the eigenvlues of Sturm-Liouville problem re rel nd the ssocited eigenfunctions orthogonl, we will first need to introduce two importnt identities. For the Sturm-Liouville opertor, we hve the two identities: L = d p d ) + q, Lgrnge s Identity: ulv vlu = [puv vu )]. Green s Identity: ulv vlu) = [puv vu )] b. The proof of Lgrnge s identity follows by simple mnipultions of the opertor: [ d ulv vlu = u p dv ) = u d p dv ) = u d p dv = d [ pu dv ] + qv v d ) + p du pv du [ d v ) p du p du ) ] + qu p du ) p du dv dv v d ]. 4.21) Green s identity is simply proven by integrting Lgrnge s identity Orthogonlity nd Relity We re now redy to prove tht the eigenvlues of Sturm-Liouville problem re rel nd the corresponding eigenfunctions re orthogonl. These re esily estblished using Green s identity, which in turn is sttement bout the Sturm-Liouville opertor being self-djoint. Exmple 4.6. The eigenvlues of the Sturm-Liouville problem 4.3) re rel. Let φ n x) be solution of the eigenvlue problem ssocited with λ n : Lφ n = λ n σφ n. We wnt to show tht Nmely, we show tht λ n = λ n, where the br mens complex conjugte. So, we lso consider the complex conjugte of this eqution, Lφ n = λ n σφ n. Now, multiply the first eqution by φ n, the second eqution by φ n, nd then subtrct the results. We obtin φ n Lφ n φ n Lφ n = λ n λ n )σφ n φ n.

11 sturm-liouville boundry vlue problems 117 Integrting both sides of this eqution, we hve ) φn Lφ n φ n Lφ n = λn λ n ) We pply Green s identity to the left hnd side to find σφ n φ n. [pφ n φ n φ n φ b n )] b = λ n λ n ) σφ n φ n. Using the homogeneous boundry conditions 4.4) for self-djoint opertor, the left side vnishes. This leves 0 = λ n λ n ) σ φ n 2. The integrl is nonnegtive, so we must hve λ n = λ n. Therefore, the eigenvlues re rel. Exmple 4.7. The eigenfunctions corresponding to different eigenvlues of the Sturm-Liouville problem 4.3) re orthogonl. This is proven similr to the lst exmple. Let φ n x) be solution of the eigenvlue problem ssocited with λ n, Lφ n = λ n σφ n, nd let φ m x) be solution of the eigenvlue problem ssocited with λ m = λ n, Lφ m = λ m σφ m, Now, multiply the first eqution by φ m nd the second eqution by φ n. Subtrcting these results, we obtin φ m Lφ n φ n Lφ m = λ m λ n )σφ n φ m Integrting both sides of the eqution, using Green s identity, nd using the homogeneous boundry conditions, we obtin 0 = λ m λ n ) σφ n φ m. Since the eigenvlues re distinct, we cn divide by λ m λ n, leving the desired result, σφ n φ m = 0. Therefore, the eigenfunctions re orthogonl with respect to the weight function σx) The Ryleigh Quotient The Ryleigh quotient is useful for getting estimtes of eigenvlues nd proving some of the other properties ssocited with Sturm-Liouville

12 118 prtil differentil equtions eigenvlue problems. The Ryleigh quotient is generl nd finds pplictions for both mtrix eigenvlue problems s well s self-djoint opertors. For Hermitin mtrix M the Ryleigh quotient is given by Rv) = v, Mv v, v. One cn show tht the criticl vlues of the Ryleigh quotient, s function of v, re the eigenvectors of M nd the vlues of R t these criticl vlues re the corresponding eigenvectors. In prticulr, minimizing Rv over the vector spce will give the lowest eigenvlue. This leds to the Ryleigh-Ritz method for computing the lowest eigenvlues when the eigenvectors re not known. This definition cn esily be extended to Sturm-Liouville opertors, Rφ n ) = φ nlφ n φ n, φ n. We begin by multiplying the eigenvlue problem Lφ n = λ n σx)φ n by φ n nd integrting. This gives [ d φ n p dφ ) ] n + qφn 2 = λ n φ 2 nσ. One cn solve the lst eqution for λ to find λ n = [ ) ] b d φ n p dφ n + qφn 2 = Rφ n ). φ2 nσ It ppers tht we hve solved for the eigenvlues nd hve not needed the mchinery we hd developed in Chpter 4 for studying boundry vlue problems. However, we relly cnnot evlute this expression when we do not know the eigenfunctions, φ n x) yet. Nevertheless, we will see wht we cn determine from the Ryleigh quotient. One cn rewrite this result by performing n integrtion by prts on the first term in the numertor. Nmely, pick u = φ n nd dv = d for the stndrd integrtion by prts formul. Then, we hve d φ n p dφ ) n = pφ n dφ n [ b p dφn ) 2 qφ 2 n ] p dφ n. ) Inserting the new formul into the expression for λ, leds to the Ryleigh Quotient dφ pφ n n b + [ ) ] b dφn 2 p qφ 2 n λ n =. 4.22) φ2 nσ In mny pplictions the sign of the eigenvlue is importnt. As we hd seen in the solution of the het eqution, T + kλt = 0. Since we expect

13 sturm-liouville boundry vlue problems 119 the het energy to diffuse, the solutions should decy in time. Thus, we would expect λ > 0. In studying the wve eqution, one expects vibrtions nd these re only possible with the correct sign of the eigenvlue positive gin). Thus, in order to hve nonnegtive eigenvlues, we see from 4.22) tht. qx) 0, nd dφ b. pφ n n b 0. Furthermore, if λ is zero eigenvlue, then qx) 0 nd α 1 = α 2 = 0 in the homogeneous boundry conditions. This cn be seen by setting the numertor equl to zero. Then, qx) = 0 nd φ nx) = 0. The second of these conditions inserted into the boundry conditions forces the restriction on the type of boundry conditions. One of the properties of Sturm-Liouville eigenvlue problems with homogeneous boundry conditions is tht the eigenvlues re ordered, λ 1 < λ 2 <.... Thus, there is smllest eigenvlue. It turns out tht for ny continuous function, yx), λ 1 = min yx) py dy b + [ p y2 σ ) dy 2 qy 2] 4.23) nd this minimum is obtined when yx) = φ 1 x). This result cn be used to get estimtes of the minimum eigenvlue by using tril functions which re continuous nd stisfy the boundry conditions, but do not necessrily stisfy the differentil eqution. Exmple 4.8. We hve lredy solved the eigenvlue problem φ + λφ = 0, φ0) = 0, φ1) = 0. In this cse, the lowest eigenvlue is λ 1 = π 2. We cn pick nice function stisfying the boundry conditions, sy yx) = x x 2. Inserting this into Eqution 4.23), we find Indeed, 10 π 2. λ x)2 1 0 x x2 ) 2 = The Eigenfunction Expnsion Method In this section we solve the nonhomogeneous problem Ly = f using expnsions over the bsis of Sturm-Liouville eigenfunctions. We hve seen tht Sturm-Liouville eigenvlue problems hve the requisite set of orthogonl eigenfunctions. In this section we will pply the eigenfunction expnsion method to solve prticulr nonhomogeneous boundry vlue problem.

14 120 prtil differentil equtions Recll tht one strts with nonhomogeneous differentil eqution Ly = f, where yx) is to stisfy given homogeneous boundry conditions. The method mkes use of the eigenfunctions stisfying the eigenvlue problem Lφ n = λ n σφ n subject to the given boundry conditions. Then, one ssumes tht yx) cn be written s n expnsion in the eigenfunctions, yx) = c n φ n x), n=1 nd inserts the expnsion into the nonhomogeneous eqution. This gives f x) = L c n φ n x) n=1 ) = n=1 c n λ n σx)φ n x). The expnsion coefficients re then found by mking use of the orthogonlity of the eigenfunctions. Nmely, we multiply the lst eqution by φ m x) nd integrte. We obtin Orthogonlity yields f x)φ m x) = Solving for c m, we hve n=1 c n λ n φ n x)φ m x)σx). f x)φ m x) = c m λ m φmx)σx) 2. c m = f x)φ mx) λ m φ2 mx)σx). Exmple 4.9. As n exmple, we consider the solution of the boundry vlue problem xy ) + y x = 1, x [1, e], 4.24) x y1) = 0 = ye). 4.25) This eqution is lredy in self-djoint form. So, we know tht the ssocited Sturm-Liouville eigenvlue problem hs n orthogonl set of eigenfunctions. We first determine this set. Nmely, we need to solve xφ ) + φ x = λσφ, φ1) = 0 = φe). 4.26) Rerrnging the terms nd multiplying by x, we hve tht x 2 φ + xφ λσx)φ = 0.

15 sturm-liouville boundry vlue problems 121 This is lmost n eqution of Cuchy-Euler type. σx) = 1 x, we hve x 2 φ + xφ λ)φ = 0. Picking the weight function This is esily solved. The chrcteristic eqution is r λ) = 0. One obtins nontrivil solutions of the eigenvlue problem stisfying the boundry conditions when λ > 1. The solutions re φ n x) = A sinnπ ln x), n = 1, 2,.... where λ n = n 2 π 2 1. It is often useful to normlize the eigenfunctions. This mens tht one chooses A so tht the norm of ech eigenfunction is one. Thus, we hve 1 = e 1 = A 2 e φ n x) 2 σx) 1 = A sinnπ ln x) 1 x sinnπy) dy = 1 2 A ) Thus, A = 2. Severl of these eigenfunctions re show in Figure 4.2. We now turn towrds solving the nonhomogeneous problem, Ly = 1 x. We first expnd the unknown solution in terms of the eigenfunctions, 1 e x yx) = c n 2 sinnπ ln x). n=1 Inserting this solution into the differentil eqution, we hve Figure 4.2: Plots of the first five eigenfunctions, yx) = 2 sinnπ ln x). 1 x = Ly = 1 c n λ n 2 sinnπ ln x) x. n=1 Next, we mke use of orthogonlity. Multiplying both sides by the eigenfunction φ m x) = 2 sinmπ ln x) nd integrting, gives λ m c m = e 1 2 sinmπ ln x) 1 x = 2 mπ [ 1)m 1]. Solving for c m, we hve n=1 c m = 2 [ 1) m 1] mπ m 2 π 2 1. Finlly, we insert these coefficients into the expnsion for yx). The solution is then 2 [ 1) yx) = n 1] nπ n 2 π 2 sinnπ lnx)). 1 We plot this solution in Figure Figure 4.3: Plot of the solution in Exmple 4.9.

16 122 prtil differentil equtions 4.4 Appendix: The Fredholm Alterntive Theorem Given tht Ly = f, when cn one expect to find solution? Is it unique? These questions re nswered by the Fredholm Alterntive Theorem. This theorem occurs in mny forms from sttement bout solutions to systems of lgebric equtions to solutions of boundry vlue problems nd integrl equtions. The theorem comes in two prts, thus the term lterntive. Either the eqution hs exctly one solution for ll f, or the eqution hs mny solutions for some f s nd none for the rest. The reder is fmilir with the sttements of the Fredholm Alterntive for the solution of systems of lgebric equtions. One seeks solutions of the system Ax = b for A n n m mtrix. Defining the mtrix djoint, A through < Ax, y >=< x, A y > for ll x, y, C n, then either Theorem 4.1. First Alterntive The eqution Ax = b hs solution if nd only if < b, v >= 0 for ll v stisfying A v = 0. or Theorem 4.2. Second Alterntive A solution of Ax = b, if it exists, is unique if nd only if x = 0 is the only solution of Ax = 0. The second lterntive is more fmilir when given in the form: The solution of nonhomogeneous system of n equtions nd n unknowns is unique if the only solution to the homogeneous problem is the zero solution. Or, equivlently, A is invertible, or hs nonzero determinnt. Proof. We prove the second theorem first. Assume tht Ax = 0 for x = 0 nd Ax 0 = b. Then Ax 0 + αx) = b for ll α. Therefore, the solution is not unique. Conversely, if there re two different solutions, x 1 nd x 2, stisfying Ax 1 = b nd Ax 2 = b, then one hs nonzero solution x = x 1 x 2 such tht Ax = Ax 1 x 2 ) = 0. The proof of the first prt of the first theorem is simple. Let A v = 0 nd Ax 0 = b. Then we hve < b, v >=< Ax 0, v >=< x 0, A v >= 0. For the second prt we ssume tht < b, v >= 0 for ll v such tht A v = 0. Write b s the sum of prt tht is in the rnge of A nd prt tht in the spce orthogonl to the rnge of A, b = b R + b O. Then, 0 =< b O, Ax >=< A b, x > for ll x. Thus, A b O. Since < b, v >= 0 for ll v in the nullspce of A, then < b, b O >= 0. Therefore, < b, v >= 0 implies tht 0 =< b, b O >=< b R + b O, b O >=< b O, b O >. This mens tht b O = 0, giving b = b R is in the rnge of A. So, Ax = b hs solution.

17 sturm-liouville boundry vlue problems 123 Exmple Determine the llowed forms of b for solution of Ax = b to exist, where ) 1 2 A =. 3 6 First note tht A = A T. This is seen by looking t < Ax, y > = < x, A y > n n i=1 j=1 ij x j ȳ i = = n n x j ij ȳ i j=1 j=1 n n x j ā T ) ji y i. 4.28) j=1 j=1 For this exmple, A = ). We next solve A v = 0. This mens, v 1 + 3v 2 = 0. So, the nullspce of A is spnned by v = 3, 1) T. For solution of Ax = b to exist, b would hve to be orthogonl to v. Therefore, solution exists when b = α So, wht does the Fredholm Alterntive sy bout solutions of boundry vlue problems? We extend the Fredholm Alterntive for liner opertors. A more generl sttement would be Theorem 4.3. If L is bounded liner opertor on Hilbert spce, then Ly = f hs solution if nd only if < f, v >= 0 for every v such tht L v = 0. The sttement for boundry vlue problems is similr. However, we need to be creful to tret the boundry conditions in our sttement. As we hve seen, fter severl integrtions by prts we hve tht 1 3 ) < Lu, v >= Su, v)+ < u, L v >, where Su, v) involves the boundry conditions on u nd v. Note tht for nonhomogeneous boundry conditions, this term my no longer vnish. Theorem 4.4. The solution of the boundry vlue problem Lu = f with boundry conditions Bu = g exists if nd only if < f, v > Su, v) = 0 for ll v stisfying L v = 0 nd B v = 0. Exmple Consider the problem u + u = f x), u0) u2π) = α, u 0) u 2π) = β..

18 124 prtil differentil equtions Only certin vlues of α nd β will led to solutions. We first note tht Solutions of L = L = d L v = 0, v0) v2π) = 0, v 0) v 2π) = 0 re esily found to be liner combintions of v = sin x nd v = cos x. Next, one computes Su, v) = [ u v uv ] 2π 0 = u 2π)v2π) u2π)v 2π) u 0)v0) + u0)v 0). For vx) = sin x, this yields Su, sin x) = u2π) + u0) = α. 4.29) Similrly, Su, cos x) = β. Using < f, v > Su, v) = 0, this leds to the conditions tht we were seeking, 2π 0 2π 0 f x) sin x = α, f x) cos x = β. Problems 1. Prove the if ux) nd vx) stisfy the generl homogeneous boundry conditions t x = nd x = b, then α 1 u) + β 1 u ) = 0, α 2 ub) + β 2 u b) = ) for the gen- px)[ux)v x) vx)u x)] x=b x= = Prove Green s Identity ulv vlu) = [puv vu )] erl Sturm-Liouville opertor L. b 3. Find the djoint opertor nd its domin for Lu = u + 4u 3u, u 0) + 4u0) = 0, u 1) + 4u1) = Show tht Sturm-Liouville opertor with periodic boundry conditions on [, b] is self-djoint if nd only if p) = pb). [Recll, periodic boundry conditions re given s u) = ub) nd u ) = u b).]

19 sturm-liouville boundry vlue problems The Hermite differentil eqution is given by y 2xy + λy = 0. Rewrite this eqution in self-djoint form. From the Sturm-Liouville form obtined, verify tht the differentil opertor is self djoint on, ). Give the integrl form for the orthogonlity of the eigenfunctions. 6. Find the eigenvlues nd eigenfunctions of the given Sturm-Liouville problems.. y + λy = 0, y 0) = 0 = y π). b. xy ) + λ x y = 0, y1) = ye2 ) = The eigenvlue problem x 2 y λxy + λy = 0 with y1) = y2) = 0 is not Sturm-Liouville eigenvlue problem. Show tht none of the eigenvlues re rel by solving this eigenvlue problem. 8. In Exmple 4.8 we found bound on the lowest eigenvlue for the given eigenvlue problem.. Verify the computtion in the exmple. b. Apply the method using { yx) = Is this n upper bound on λ 1 x, 0 < x < x, 1 2 < x < 1. c. Use the Ryleigh quotient to obtin good upper bound for the lowest eigenvlue of the eigenvlue problem: φ + λ x 2 )φ = 0, φ0) = 0, φ 1) = Use the method of eigenfunction expnsions to solve the problems:. y = x 2, y0) = y1) = 0. b. y + 4y = x 2, y 0) = y 1) = Determine the solvbility conditions for the nonhomogeneous boundry vlue problem: u + 4u = f x), u0) = α, u π/4) = β.

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