SturmLiouville Eigenvalue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (a, b). Here we assume b > a. Let X C 2 1


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1 Ch.4. INTEGRAL EQUATIONS AND GREEN S FUNCTIONS Ronld B Guenther nd John W Lee, Prtil Differentil Equtions of Mthemticl Physics nd Integrl Equtions. Hildebrnd, Methods of Applied Mthemtics, second edition In the study of the prtil differentil equtions of hyperbolic nd prbolic types we solved severl initil nd boundry vlue problems. While solving these equtions we used the method seprtion of vribles which reduces the problem to one of the following types of SturmLiouville problems SturmLiouville Eigenvlue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (, b). Here we ssume b >. Let X C 2 1 (I) C (Ī), then the set of equtions given below d dx dx [p(x) ] + q(x)x = λ r(x)x, dx (1) X() h 0 X (0) = 0, (2) X(b) + h 1 X (b) = 0. (3) is clled the The SturmLiouville Eigenvlue Problem (SLEVP). Here h 0 0 nd h 1 0 re constnts. Let the SturmLiouville opertor L be defined by L = d dx p(x) d + q(x), (4) dx then the SturmLiouville Eigenvlue eqution becomes L X = λ r(x) X, X() h 0 X () = 0, X(b) + h 1 X (b) = 0. (5) Note tht L is selfdjoint, i.e., L = L The SturmLiouville Boundry Vlue Problem: With the sme p, q, r defined bove we hve the SturmLiouville Boundry Vlue Problem (SLBVP) 1
2 L X = f(x), in I, (6) X() h 0 X () = 0, (7) X(b) + h 1 X (b) = 0. (8) Here f(x) is continuous function in I. SturmLiouville Problem: Combintion of SLEVP nd SLBVP my be given s L X = λr(x)x + f(x), in I, (9) X() h 0 x () = 0, (10) X(b) + h 1 x (b) = 0. (11) 4.1. The Sturm Liouville Boundry Vlue Problem nd Green s Functions Let us formulte the SLBVP s L u(x) = f(x), x I, (12) u() h 0 u () = 0, (13) u(b) + h 1 u (b) = 0. (14) We shll solve this problem by the use of Green s function technique. Let u, v C 2 1 (I) C (Ī) then it is esy to prove the Lgrnge identity vlu ulv = d dx [p(x)(vu uv )], (15) Now letting v = G(x, y) nd 2
3 then we hve L x G(x, y) = δ(x y), x, y I, (16) dg(x, y) G(, y) h 0 x= = 0, dx (17) dg(x, y) G(b, y) + h 1 x=b = 0. dx (18) u(x) = G(x, y) f(y)dy, (19) Hence the SLBVP reduces to the construction of the Green s function G(x, y) stisfying (16)(18). In ddition to these properties G(x, y) stisfy lso the following properties 1. G(x, y) is continuous in x, y I (prove it) in prticulr we hve s ε 0 + G(x, y) x=y ε = G(x, y) x=y+ε. (20) 2. dg(x,y) dx prove it is discontinuous t x = y. As ε 0 + we hve the jump condition dg(x, y) dg(x, y) x=y ε x=y+ε = p(y) (21) dx dx Since G(x, y) stisfies the homogeneous eqution for x < y nd x > y we then hve G(x, y) = { 1 (y) v 0 (x) + 2 (y) v 1 (x), b 1 (y) v 0 (x) + b 2 (y) v 1 (x), x < y b y < x b (22) where v 0 (x) nd v 1 (x) re the solutions of the homogeneous SLeqution nd 1, 2, b 1 nd b 2 re functions of y to be determined through the conditions (17), (18), (20), nd (21). If we choose v 0 s solution of the homogeneous SL eqution stisfying the left boundry condition 3
4 v 0 () h 0 v 0() = 0, (23) nd v 1 (x) s the solution of the SL eqution stisfying the right boundry condition v 1 (b) + h 1 v 1(b) = 0, (24) then 2 = b 1 = 0 nd due to the symmetry G(x, y) = G(y, x) nd remining conditions we show tht G(x, y) = { v 0 (x) v 1 (y), v 0 (y) v 1 (x), x < y b y < x b (25) The jump condition (21) becomes p(x) W (v 0, v 1 ) = 1 (26) where W (u, v) = uv vu is the Wronskin of u nd v. We then hve the following theorem: Theorem 1. The SLBVP (12) hs Green s function if nd only if the corresponding homogeneous SLEVP [with f(x) = 0] hs only the trivil solution, in which cse the Green s function is given in (25) with the boundry conditions in (23) nd (24). proof: () If we ssume the existence of the Green s function we hve the solution (19) for ech function f(x) in I. The corresponding homogeneous solution (with f = 0 in (19) goes to the trivil solution u = 0. (b) Let us ssume tht the only solution of homogeneous SLBVP is only the trivil one. Hence we should be ble to find nontrivil solutions so tht pw = C. Let C = 0. This mens v 0 nd v 1 re proportionl hence, for exmple v 0 stisfies the homogeneous SLBVP which is nontrivil, hence we obtin contrdiction. This mens C 0. By simple scling C = 1 (by redefining 4
5 v 0 for instnce). Hence pw = 1. hence we hve v 0 nd v 0 solving the homogeneous SL eqution, v 0 stisfying the left boundry condition (23) nd v 1 the right boundry condition (24). Then there exist unique Green s function given in (25). To summrize ll properties of the Green s function we formulte the following theorem Theorem 2. Assume tht the homogeneous SLBVP (f = 0 problem hs only the trivil solution. Let G(x, y) be the function given in (25) nd with the conditions (23) nd (24). Then i G(x,y) is continuous on the squre x, y b nd hs continuous second derivtives on ech of the tringles x < y b nd y < x b. On ech tringle L x G(x, y) = 0 (ii) G(x,y) stisfies the boundry conditions t x = nd x = b, (iii) For ech y (, b), G(x,y) stisfies the jump condition (21) Conversely, properties (i)(iii) uniquely determine G such tht (19) solves the SLBVP in (19). We hve remrked tht, since L is self djoint, the Green s function is symmetric. If p(x) > 0, q 0 nd h 0, h 1 0 the we hve the following result Theorem 3. The Green s function of the SLBVP exists if p(x) > 0, q(x) 0, nd h 0, h 1 0. proof. Th.1 sys tht the Green s function exists if nd only if the homogeneous SLBVP hs only the trivil solution. The homogeneous SLBVP is given s follows. Let z C 2 1 (I) C (Ī) stisfy [p(x)z ] + q(x)z = 0, x I, (27) z() h z () = 0, z(b) + h 1 z (b) = 0. (28) 5
6 Multiplying the differentil eqution by Z nd integrting over I nd using integrtion prts we obtin p(b)h 1 z (b) 2 + p()h 0 z () 2 + [p(x)z 2 + q(x)z 2 ]dx = 0 (29) Here, since p > 0 in I then z (x) = 0 in I. Hence z = constnt everywhere in I, but this constnt should vnish t the boundry points nd b, then z(x) must vnish t ll pints in I. This proves tht under the given circumstnces the only solution of the SLBVP is the trivil one,z = 0. Exmple 1. Solve u = f(x), x (, b) nd u() = u(b) = 0. solution. Let G(x, y) = { 1 x + 2, b 1 x + b 2, x < y b y < x b (30) where the coefficient re to be determined through the conditions (17), (18), (20) nd (21). We find G(x, y) = Hence the solution is given by { (x )(y b) b, x < y b (y )(x b) b, y < x b (31) x (y )(x b) f(y)dy + b = x b b x u(x) = x (y ) f(y)dy + x b G(x, y)f(y)dy = (32) (y b)(x ) f(y)dy, (33) b x (y b) f(y)dy (34) Exmple 2. Solve the sme problem by chnging the boundry conditions to u() = 0, u (b) = 0. 6
7 Exmple 3. Solve the sme problem by chnging the boundry conditions to u() = 0, u () = 0. Exmple 4. Solve the sme problem by chnging the boundry conditions to u() α 2 u () = 0 nd u(b) + β 2 u (b) = 0. Here α nd β re some constnts. 4.2 The Neumnn Series Let us now consider the generl SL problem L u = λ r(x) u + f(x), x (, b), (35) u() h 0 u () = 0, u(b) + h 1 u (b) = 0. (36) Since the Green s function G(x, y) outlined in the first section belongs to the opertor L we then hve u(x) = or, equivlently we hve G(x, y) [f(y) + λ r(y)u(y)]dy (37) u(x) = g(x) + λ where g(x) = G(x, y) r(y) u(y)dy, (38) G(x, y)f(y)dy Eq(38) is clled the The Fredholm integrl eqution of the second kind which cn be put into more symmetricl one. Let y(x) = r(x) u(x), g(x, s) = r(x) r(s) G(x, s), f 1 (x) = r(x) g(x). then (39) 7
8 y(x) = f 1 (x) + λ g(x, s) y(s)ds, (40) In our future nlysis we let f 1 f. Here g(x, s) is clled the kernel of the integrl eqution, f(x) is given nd λ is in generl complex prmeter. In most cses it is rel. And lso we cn ssume tht g(x, s) is continuous in x, s b nd f(x) is continuous in x b. The integrl eqution given in (40) my be solved by using severl methods. Let us ssume tht y is power series in λ y(x) = λ n y n (x), (41) n=0 where the coefficients y n (x) will be determined by the use of the integrl eqution (40) (we let f 1 = f). We obtin y 1 (x) = y 2 (x) = y n+1 = y 0 (x) = f(x), (42) g(x, s) y 0 (s)ds, (43) g(x, s) y 1 (s)ds, (44) g(x, s) y n (s)ds, (45) for n = 0, 1, Hence ll y n (x) cn be clculted recursively. We hve shown tht the series (41) must be uniformly convergent to justify the term by term integrtion in (40). Proposition 4. Assume tht g(x, s) is continuous in x, s b nd f(x) is continuous in x b. The series in (40) is uniformly convergent if λ M(b ) 1 proof. Since both g nd f re continuous we hve M = mx g(x, s), N = mx f(x) 8
9 in x, s b. Using (42), (43, (44), nd (45) we obtin y 0 (x) N, (46) y 1 (x) MN(b ), (47) y 2 (x) NM 2 (b ) 2, y n (x) NM n (b ) n (48) which leds to λ n y n (x) N n=0 [ λ M(b ) n < n=0 This result provides lso solution of the integrl eqution (40) Proposition 5. Let y(x) = f(x) + λ g(x, s) y(s)ds (49) be the integrl eqution. Let g(x, s) nd f(x) be continuous in x, s b. If λ < 1 M(b ) then y(x) = f(x) + λ γ(x, s) f(s)ds (50) where γ(x, s) is clled the resolvent kernel nd given by γ(x, s) = λ n 1 k n (x, s) (51) n=1 proof: Using (42)(45) we obtin y 0 (x) = f(x), (52) y 1 (x) = g(x, s) f(s)ds, (53) 9
10 y 2 (x) = = g(x, s) y s ds, (54) g 2 (x, s) f(s)ds, (55) (56) where k 2 (x, s) = g(x, t) g(t, s)dt. Defining, in generl g 1 (x, s) = g(x, s), (57) g n (x, s) = then it is strightforwrd to estblish the reltion g(x, t)g n 1 (t, s)dt, n = 2, 3, (58) nd hence y n (x) = g n (x, s) f(s)ds, n = 1, 2, where y(x) = λ n y n (x) (59) n=0 = f(x) + = f(x) + = f(s) + λ λ n y n (x) (60) n=1 λ n y n (s)f(s)ds (61) n=1 γ(x, s)f(s)ds, (62) γ(x, s) = λ n 1 g n (x, s) (63) n=1 10
11 Convergence of the series in (62) is gurnteed by Prop.4, nd the solution given in Prop.5 is unique. Theorem 6. Let g(x,s) nd f(x) be continuous on x, s b. Let M = mx g(x, s) for ll (x, s) [, b]. If λ < 1 M(b ) is unique nd continuous. y(x) = f(x) + λ γ(x, s) f(s)ds the the solution proof: Let us ssume tht there re two different solutions z 1 (x) nd z 2 (x) of the integrl eqution (49). Then their difference w(x) = z 2 (x) z 1 (x) stisfies w(x) = λ g(x, s)w(s)ds Hence we hve for ll x [, b] we hve where W = mx w(x). Then w(x) λ MW (b ) [1 λ M(b )]W 0 This implies tht W = 0. Hence z 1 = z 2 everywhere in [, b]. Exmple 5. Find the solution of y(x) = f(x) + λ e x s y(s)ds Solution: Try to find the following solution. Given continuous f(x) for x [, b] we hve y(x) = f(x) + λ e x 1 λ (b ) e s f(s)ds 11
12 where λ 1 1. (i) Discuss the cse λ = (ii) nd s n exmple let b b f(x) = x 2. Discuss lso the ppliction of (iii) Prop.4, (iv) Prop.5 nd (v) Thm.6 to this exmple. (vi) find lso the Neumnn series corresponding to this exmple Fredholm Eqution with Seprble Kernels. A seprble kernel is given s g(x, s) = N p n (x) q n (s) (64) n=1 Here we ssume tht the N functions p n re linerly independent. Then the integrl eqution (49) tkes the form where n y(x) = f(x) + λ = f(x) + λ = f(x) + λ = f(x) + λ g(x, s) y(s)ds, (65) [ N ] p n (x) q n (s) y(s)ds, (66) n=1 N [ p n (x) n=1 ] q n (s)y(s)ds, (67) N c n p n (x), (68) n=1 c n = q n (s)y(s)ds, n = 1, 2, (69) Using (68) in (69) we obtin the following lgebric liner equtions for the constnts c n s N c n = β n + λ α nm c m, (70) 12 m=1
13 where for ll n, m = 1, 2, β n = α mn = q n (s) f(s)ds, (71) p n (s) q m (s)ds (72) let A denote the N N mtrix corresponding to α mn B be the column N vector corresponding to β n, nd C be the column unknown Nvector to be determined then (70) simply becomes [I λa]c = B, ) (73) where I is the N N unit mtrix. The bove liner eqution for C is esily solved, but we hve to consider ll possible cses. In the bove eqution (70) we need to determine the unknown coefficients c n s in terms of the known coefficients β n nd α mn for ll m, n = 1, 2, cse (). f(x) = 0 or B = 0 the eqution (73) becomes homogeneous. For nontrivil solutions det[i λa] must vnish. Otherwise there is only the trivil solution c n = 0, for ll n = 1, 2,. If det[i λa] = 0 t lest (depending upon the rnk of mtrix A one of the c n s is left rbitrry. In such cse there re infinitely mny solutions. To remind you the terminology: Those vlues of λ where det[i λa] = 0 re clled chrcteristic or eigenvlues nd ny nontrivil solution of the homogeneous integrl eqution is clled the corresponding chrcteristic or eigenfunction. If there re k number of constnts c n s n = 1, 2,, k for given eigenvlue λ, then k number of linerly independent eigenfunctions re obtined. cse (b). f(x) 0 but β n = 0, n = 1, 2, this mens tht f(x) is orthogonl to ll functions q n (x), n = 1, 2,. Hence B = 0. The cse () pplies lso here except for the fct tht here the solution (68) contins the function f(x). Hence the trivil solution C = 0 correspond to the solution 13
14 y(x) = f(x). Solutions corresponding to eigenvlues of λ should be expresses s the sum of f(x) nd liner sum of the corresponding eigenfunctions. cse (c). B 0. We ssume tht t lest for some n, β n 0.if det[i λa] 0 unique nontrivil solution of (73) exists, leding to unique nontrivil solution y(x) of the integrl eqution (68). if det[i λa] = 0 either there is no solution or the solution is not unique mening tht there re infinitely mny solutions. Exmple 6. Let g(x, s) = 1 3xs. Solve the corresponding integrl eqution by considering ll three cses bove 4.4. HilbertSchmidt Theory When the kernel g(x, s) is not of type (64) there re, in generl, infinitely mny eigenvlues nd eigenfunctions of the homogeneous Fredholm eqution. In ddition, there my lso be more thn one eigenfunctions corresponding to one eigenvlue. This is clled the degenercy nd the number of eigenfunctions corresponding to single eigenvlue is clled the multiplicity. In this section we ssume tht the kernel g(x, s) is symmetric with respect to the vribles x nd s. We lso ssume tht multiplicity is one. Remrk: The homogeneous Fredholm eqution y(x) = λ g(x, s) y(s) ds (74) cn not hve zero eigenvlue, becuse the corresponding eigenfunction is lso zero. We the hve the following result. Proposition 7. Eigenfunctions of the homogeneous Fredholm integrl eqution with symmetric kernel corresponding to different eigenvlues re orthogonl. proof: By its definition we hve tht 14
15 y m = λ m g(x, s) y m ds, m = 1, 2, (75) Multiplying by y n (x) nd integrting over (, b) we get n which leds to y m () y n (x) dx = λ m y n (x)dx = λ m y m (s)ds = λ m λ n g(x, s) y m (s)ds, (76) y n (x) g(x, s)dx, (77) y m (s) y n (s)ds (78) y m (x) y n (x) dx = 0 (79) when λ m λ n Remrk 1: If there re more thn one eigenfunctions corresponding to n eigenvlue λ n then orthogonliztion of such set is performed by the stndrd GrmSchmidt procedure. In the sequel we ssume tht such sets re orthogonlized. Remrk 2: In the cse of complex functions we generlize the Prop. 7 in the following wy Proposition 7. Eigenfunctions of the homogeneous Fredholm integrl eqution with hermitin kernel, g(x, s) = ḡ(s, x) corresponding to different eigenvlues re orthogonl. Note tht the inner product in this cse is defined by < f, g >= f(x)ḡ(x)dx 15
16 where br over letter denotes complex conjugtion. This proposition implies tht y m (x) ȳ n (x)dx = 0 for m n. A Corollry of this proposition is tht λ n for such homogeneous Fredholm integrl eqution with hermitin kernel nd hence lso for rel symmetric kernels the eigenvlues re rel (Prove these sttements). We hve now theorem which will be used very often in our future nlysis. Theorem 8. Let g(x, s) be rel nd symmetric continuous kernel over (, b). Let H(x) be ny continuous function over (, b) the ny function defined s h(x) = g(x, s)h(s)ds (80) cn be represented s liner superposition of the eigenfunctions y n (x), n = 1, 2,, of the homogeneous Fredholm equtions with sme kernel over (, b). Hence we hve h(x) = n=1 n y n (x) nd with ρ n = y n(x) 2 dx. Remrk 3: n = 1 h(x)y n (x)ds ρ n If there re finite number of eigenfunctions then the functions generted by the opertion g(x, s) H(s)ds form restricted clss of functions, irrespective the form of the function H(x). 16
17 Exmple 7. Let g(x, s) = sin(x + s) with (, b) = (0, 2π) then it is esy to show tht λ 1 = 1 π, y 1 = sin x + cos x, (81) λ 2 = 1 π, y 2 = sin x cos x. (82) Hence we hve finite number of eigenfunctions. Then ny function h(x) given s h(x) = tkes the form 2π 0 g(x, s) H(s)ds g(x) = C 1 sin x + C 2 cos x whtever the function H(x) is. It is obvious tht h(x) cn lso be written s h(x) = 1 y 1 (x) + 2 y 2 (x) Remrk 4. In some cses the eigenfunctions of the homogeneous Fredholm integrl eqution my not form complete set (see DK for the definition). This mens tht ny continuous function f(x) defined in (, b) my not be represented over the sme intervl by series of y n s. In the sequel we ssume tht y n s form complete set over the intervl (, b). The essence of the eigenfunctions of the homogeneous Fredholm integrl eqution with rel nd symmetric kernel shows up when we wish to solve the inhomogeneous Fredholm integrl eqution y(x) = f(x) + λ g(x, s)y(s)ds) (83) where f(x) is given continuous function over the intervl (, b). First we shll use the orthonormlized set of eigenfunctions φ n (x) which re defined by φ n = C n y n (x) 17
18 where C n = 1 ρn nd φ m (x) φ n (x)dx = 0, m n Then from (83) by letting λ g(x, s) y(s)ds = n=1 n φ n (x) Hence we get y(x) = f(x) + n=1 n φ n (x), x b where By defining then c n = n = [y(s) f(s)]φ n (s)ds y(s) φ n ds, β n = f(s) φ n (s)ds n = c n β n Multiplying (83) by φ n (x) over (, b) nd using the symmetry of the kernel we obtin [1 λ λ n ] c n = β n, n = 1, 2, (84) cse. If λ = λ k where λ k is of the eigenvlues then c k becomes rbitrry nd the solution nd β k = 0 or f(s) φ k(s)ds = 0 nd the solution becomes y(x) = f(x) + c k φ k (x) + λ k n k β n λ n λ φ n(x) (85) cse b. If λ λ k, k = 1, 2,, ny one of the eigenvlues. Then the solution is unique. 18
19 y(x) = f(x) + λ n=1 β n λ n λ φ n(x) 4.4 Singulr Integrl Equtions So fr we hve ssumed tht (i) the intervl (, b) of the integrl equtions re finite nd (ii) the kernel g(x, s) ws continuous. If n integrl eqution hs either n infinite intervl or hs discontinuous kernel then such n integrl eqution is clled singulr integrl eqution. The first two of the following re singulr Fredholm equtions of the first kind nd the lst one is the singulr Volterr integrl eqution of the second kind (known lso s the Abel s eqution). Here y(x) re unknown functions to be determined in ech cses. ) F (x) = b) F (x) = c) F (x) = 0 0 x 0 e xs y(s)ds, (86) sin(xs) y(s)ds, (87) y(s) x s ds (88) Here in ech cse F (x) is given function. The solutions of the homogeneous integrl equtions with y(s) = λ g(x, s)y(s)ds, (89) (i): (, b) = (0, ) or (ii): g(x, s) is not continuous in (, b) do not shre the sme properties s solutions of the homogeneous equtions in the previous sections. For exmple the first singulr eqution bove hs 19
20 continuous eigenvlues, the second one hs two eigenvlues with infinite multiplicity nd the lst one cn be solved exctly. Here we shll only present the first cse. Recll the definition of the Gmm function 0 e xs s α 1 ds = Γ(α) x α, α > 0 (90) Chnging α to 1 α nd rewriting the bove eqution once more we get 0 e xs s α ds = Γ(1 α) x α 1, α < 1 (91) Dividing first one by Γ(α) the second one by Γ(1 α) nd dding them we obtin 0 1 [ s α s α ]ds (92) Γ(α) Γ(1 α) = Γ(α) x α + Γ(1 α) x α 1, 0 < α < 1 (93) This lst eqution gives us the eigenvlues nd the eigenfunctions of the homogeneous singulr Fredholm integrl equtuion (89) λ α = 1 Γ(α) Γ(1 α), (94) By using the identity y α = Γ(1 α) x α 1 + Γ(α) x α (95) Γ(α) Γ(1 α) = then the eigenvlues become more simpler sin απ λ α = π π sin α π, 0 < α < 1 As we observe tht α tkes ny vlues in the intervl (0, 1), hence eigenvlues tke lso continuous vlues. 20
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