Karlstad University. Division for Engineering Science, Physics and Mathematics. Yury V. Shestopalov and Yury G. Smirnov. Integral Equations


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1 Krlstd University Division for Engineering Science, Physics nd Mthemtics Yury V. Shestoplov nd Yury G. Smirnov Integrl Equtions A compendium Krlstd
2 Contents 1 Prefce 4 Notion nd exmples of integrl equtions (IEs). Fredholm IEs of the first nd second kind 5.1 Primitive function Exmples of solution to integrl equtions nd ordinry differentil equtions 6 4 Reduction of ODEs to the Volterr IE nd proof of the unique solvbility using the contrction mpping Reduction of ODEs to the Volterr IE Principle of contrction mppings Unique solvbility of the Fredholm IE of the nd kind using the contrction mpping. Neumnn series Liner Fredholm IE of the nd kind Nonliner Fredholm IE of the nd kind Liner Volterr IE of the nd kind Neumnn series Resolvent IEs s liner opertor equtions. Fundmentl properties of completely continuous opertors Bnch spces IEs nd completely continuous liner opertors Properties of completely continuous opertors Elements of the spectrl theory of liner opertors. Spectrum of completely continuous opertor 1 8 Liner opertor equtions: the Fredholm theory The Fredholm theory in Bnch spce Fredholm theorems for liner integrl opertors nd the Fredholm resolvent IEs with degenerte nd seprble kernels Solution of IEs with degenerte nd seprble kernels IEs with degenerte kernels nd pproximte solution of IEs Fredholm s resolvent Hilbert spces. Selfdjoint opertors. Liner opertor equtions with completely continuous opertors in Hilbert spces Hilbert spce. Selfdjoint opertors Completely continuous integrl opertors in Hilbert spce Selfdjoint opertors in Hilbert spce The Fredholm theory in Hilbert spce
3 1.5 The Hilbert Schmidt theorem IEs with symmetric kernels. Hilbert Schmidt theory Selfdjoint integrl opertors Hilbert Schmidt theorem for integrl opertors Hrmonic functions nd Green s formuls Green s formuls Properties of hrmonic functions Boundry vlue problems 5 14 Potentils with logrithmic kernels Properties of potentils Generlized potentils Reduction of boundry vlue problems to integrl equtions Functionl spces nd Chebyshev polynomils Chebyshev polynomils of the first nd second kind Fourier Chebyshev series Solution to integrl equtions with logrithmic singultity of the kernel Integrl equtions with logrithmic singultity of the kernel Solution vi Fourier Chebyshev series Solution to singulr integrl equtions Singulr integrl equtions Solution vi Fourier Chebyshev series Mtrix representtion Mtrix representtion of n opertor in the Hilbert spce Summtion opertors in the spces of sequences Mtrix representtion of logrithmic integrl opertors 74.1 Solution to integrl equtions with logrithmic singulrity of the kernel Glerkin methods nd bsis of Chebyshev polynomils Numericl solution of logrithmic integrl eqution by the Glerkin method
4 1 Prefce This compendium focuses on fundmentl notions nd sttements of the theory of liner opertors nd integrl opertors. The Fredholm theory is set forth both in the Bnch nd Hilbert spces. The elements of the theory of hrmonic functions re presented, including the reduction of boundry vlue problems to integrl equtions nd potentil theory, s well s some methods of solution to integrl equtions. Some methods of numericl solution re lso considered. The compendium contins mny exmples tht illustrte most importnt notions nd the solution techniques. The mteril is lrgely bsed on the books Elements of the Theory of Functions nd Functionl Anlysis by A. Kolmogorov nd S. Fomin (Dover, New York, 1999), Liner Integrl Equtions by R. Kress (nd edition, Springer, Berlin, 1999), nd Logrithmic Integrl Equtions in Electromgnetics by Y. Shestoplov, Y. Smirnov, nd E. Chernokozhin (VSP, Utrecht, ). 4
5 Notion nd exmples of integrl equtions (IEs). Fredholm IEs of the first nd second kind Consider n integrl eqution (IE) f(x) = φ(x) + λ K(x, y)φ(y)dy. (1) This is n IE of the nd kind. Here K(x, y) is given function of two vribles (the kernel) defined in the squre Π = {(x, y) : x b, y b}, f(x) is given function, φ(x) is soughtfor (unknown) function, nd λ is prmeter. We will often ssume tht f(x) is continuous function. The IE f(x) = K(x, y)φ(y)dy () constitutes n exmple of n IE of the 1st kind. We will sy tht eqution (1) (resp. ()) is the Fredholm IE of the nd kind (resp. 1st kind) if the the kernel K(x, y) stisfies one of the following conditions: (i) K(x, y) is continuous s function of vribles (x, y) in the squre Π; (ii) K(x, y) is mybe discontinuous for some (x, y) but the double integrl K (x, y) dxdy <, (3) i.e., tkes finite vlue (converges in the squre Π in the sense of the definition of convergent Riemnn integrls). An IE x f(x) = φ(x) + λ K(x, y)φ(y)dy, (4) is clled the Volterr IE of the nd kind. Here the kernel K(x, y) my be continuous in the squre Π = {(x, y) : x b, y b} for certin b > or discontinuous nd stisfying condition (3); f(x) is given function; nd λ is prmeter, Exmple 1 Consider Fredholm IE of the nd kind x = φ(x) (x + y )φ(y)dy, (5) where the kernel K(x, y) = (x + y ) is defined nd continuous in the squre Π 1 = {(x, y) : x 1, y 1}, f(x) = x is given function, nd the prmeter λ =. Exmple Consider n IE of the nd kind f(x) = φ(x) ln x y φ(y)dy, (6) 5
6 were the kernel K(x, y) = ln x y is discontinuous (unbounded) in the squre Π 1 long the line x = y, f(x) is given function, nd the prmeter λ =. Eqution (6) Fredholm IE of the nd kind becuse the double integrl i.e., tkes finite vlue converges in the squre Π 1. ln x y dxdy <, (7).1 Primitive function Below we will use some properties of the primitive function (or (generl) ntiderivtive) F (x) defined, for function f(x) defined on n intervl I, by F (x) = f(x). (8) The primitive function F (x) for function f(x) (its ntiderivtive) is denoted using the integrl sign to give the notion of the indefinite integrl of f(x) on I, F (x) = f(x)dx. (9) Theorem 1 If f(x) is continuous on [, b] then the function F (x) = x f(x)dx. (1) (the ntiderivtive) is differentible on [, b] nd F (x) = f(x) for ech x [, b]. Exmple 3 F (x) = rctn x is primitive function for the function becuse f(x) = 1 x(1 + x) F (x) = df dx = 1 x(1 + x) = f(x). (11) 3 Exmples of solution to integrl equtions nd ordinry differentil equtions Exmple 4 Consider Volterr IE of the nd kind f(x) = φ(x) λ x e x y φ(y)dy, (1) where the kernel K(x, y) = e x y, f(x) is given continuously differentible function, nd λ is prmeter. Differentiting (1) we obtin f(x) = φ(x) λ x f (x) = φ (x) λ[φ(x) + 6 e x y φ(y)dy x e x y φ(y)dy].
7 Subtrcting termwise we obtin n ordinry differentil equtin (ODE) of the first order φ (x) (λ + 1)φ(x) = f (x) f(x) (13) with respect to the (sme) unknown function φ(x). Setting x = in (1) we obtin the initil condition for the unknown function φ(x) Thus we hve obtined the initil vlue problem for φ(x) φ() = f(). (14) φ (x) (λ + 1)φ(x) = F (x), F (x) = f (x) f(x), (15) φ() = f(). (16) Integrting (15) subject to the initil condition (16) we obtin the Volterr IE (1), which is therefore equivlent to the initil vlue problem (15) nd (16). Solve (1) by the method of successive pproximtions. Set Write two first successive pproximtions: K(x, y) = e x y y x 1, K(x, y) = x y 1. φ (x) = f(x), φ 1 (x) = f(x) + λ K(x, y)φ (x)dy, (17) Set f(x) + λ K (x, y) = φ (x) = f(x) + λ K(x, y)f(y)dy + λ K(x, t)k(t, y)dt = to obtin (by chnging the order of integrtion) The third pproximtion φ (x) = f(x) + λ K(x, y)φ 1 (y)dy = (18) x y K(x, t)dt K(t, y)f(y)dy. (19) e x t e t y dt = (x y)e x y () K(x, y)f(y)dy + λ K (x, y)f(y)dy. (1) where φ 3 (x) = f(x) + λ K(x, y)f(y)dy + + λ K (x, y)f(y)dy + λ 3 K 3 (x, y) = K(x, t)k (t, y)dt = K 3 (x, y)f(y)dy, () (x y) e x y. (3)! 7
8 The generl reltionship hs the form where n φ n (x) = f(x) + λ m K m (x, y)f(y)dy, (4) m=1 K 1 (x, y) = K(x, y), K m (x, y) = K(x, t)k m (t, y)dt = (x y)m e x y. (5) (m 1)! Assuming tht successive pproximtions (4) converge we obtin the (unique) solution to IE (1) by pssing to the limit in (4) φ(x) = f(x) + λ m K m (x, y)f(y)dy. (6) m=1 Introducing the nottion for the resolvent Γ(x, y, λ) = λ m K m (x, y). (7) m=1 nd chnging the order of summtion nd integrtion in (6) we obtin the solution in the compct form φ(x) = f(x) + λ Γ(x, y, λ)f(y)dy. (8) Clculte the resolvent explicitly using (5): Γ(x, y, λ) = e x y so tht (8) gives the solution Note tht ccording to (17) m=1 m (x y)m λ (m 1)! φ(x) = f(x) + λ nd the series (7) converges for every λ. = e x y e λ(x y) = e (λ+1)(x y), (9) e (λ+1)(x y) f(y)dy. (3) Γ(x, y, λ) = e (λ+1)(x y), y x 1, Γ(x, y, λ) =, x y 1. 4 Reduction of ODEs to the Volterr IE nd proof of the unique solvbility using the contrction mpping 4.1 Reduction of ODEs to the Volterr IE Consider n ordinry differentil equtin (ODE) of the first order dy dx = f(x, y) (31) 8
9 with the initil condition y(x ) = y, (3) where f(x, y) is defined nd continuous in twodimensionl domin G which contins the point (x, y ). Integrting (31) subject to (3) we obtin φ(x) = y + x x f(t, φ(t))dt, (33) which is clled the Volterr integrl eqution of the second kind with respect to the unknown function φ(t). This eqution is equivlent to the initil vlue problem (31) nd (3). Note tht this is generlly nonliner integrl eqution with respect to φ(t). Consider now liner ODE of the second order with vrible coefficients with the initil condition y + A(x)y + B(x)y = g(x) (34) y(x ) = y, y (x ) = y 1, (35) where A(x) nd B(x) re given functions continuous in n intervl G which contins the point x. Integrting y in (34) we obtin y (x) = x x A(t)y (x)dx x x B(x)y(x)dx + Integrting the first integrl on the righthnd side in (36) by prts yields y (x) = A(x)y(x) x Integrting second time we obtin y(x) = x x A(x)y(x)dx Using the reltionship we trnsform (38) to obtin y(x) = x x x x x (B(x) A (x))y(x)dx + x (B(t) A (t))y(t)dtdx+ x x x x f(t)dtdx = x x {A(t)+(x t)[(b(t) A (t))]}y(t)dt+ x x x x x x g(x)dx + y 1. (36) x g(x)dx + A(x )y + y 1. (37) x g(t)dtdx+[a(x )y +y 1 ](x x )+y. (38) x (x t)f(t)dt, (39) x x (x t)g(t)dt+[a(x )y +y 1 ](x x )+y. Seprte the known functions in (4) nd introduce the nottion for the kernel function Then (4) becomes (4) K(x, t) = A(t) + (t x)[(b(t) A (t))], (41) f(x) = x x (x t)g(t)dt + [A(x )y + y 1 ](x x ) + y. (4) x y(x) = f(x) + K(x, t)y(t))dt, (43) x which is the Volterr IE of the second kind with respect to the unknown function φ(t). This eqution is equivlent to the initil vlue problem (34) nd (35). Note tht here we obtin liner integrl eqution with respect to y(x). 9
10 Exmple 5 Consider homogeneous liner ODE of the second order with constnt coefficients nd the initil conditions (t x = ) We see tht here y + ω y = (44) y() =, y () = 1. (45) A(x), B(x) ω, g(x), y =, y 1 = 1, if we use the sme nottion s in (34) nd (35). Substituting into (4) nd clculting (41) nd (4), K(x, t) = ω (t x), f(x) = x, we find tht the IE (43), equivlent to the initil vlue problem (44) nd (45), tkes the form y(x) = x + x 4. Principle of contrction mppings (t x)y(t)dt. (46) A metric spce is pir of set X consisting of elements (points) nd distnce which is nonnegtive function stisfying ρ(x, y) = if nd only if x = y, ρ(x, y) = ρ(y, x) (Axiom of symmetry), (47) ρ(x, y) + ρ(y, z) ρ(x, z) (Tringle xiom) A metric spce will be denoted by R = (X, ρ). A sequence of points x n in metric spce R is clled fundmentl sequence if it stisfies the Cuchy criterion: for rbitrry ɛ > there exists number N such tht ρ(x n, x m ) < ɛ for ll n, m N. (48) Definition 1 If every fundmentl sequence in the metric spce R converges to n element in R, this metric spce is sid to be complete. Exmple 6 The (Euclidin) ndimensionl spce R n of vectors ā = ( 1,..., n ) with the Euclidin metric ρ(ā, b) = ā b = n ( i b i ) is complete. i=1 1
11 Exmple 7 The spce C[, b] of continuous functions defined in closed intervl x b with the metric ρ(φ 1, φ ) = mx x b φ 1(x) φ (x) is complete. Indeed, every fundmentl (convergent) functionl sequence {φ n (x)} of continuous functions converges to continuous function in the C[, b]metric. Exmple 8 The spce C (n) [, b] of componentwise continuous ndimensionl vectorfunctions f = (f 1,..., f n ) defined in closed intervl x b with the metric ρ(f 1, f ) = mx x b f 1 i (x) f i (x) is complete. Indeed, every componentwisefundmentl (componentwiseconvergent) functionl sequence { φ n (x)} of continuous vectorfunctions converges to continuous function in the bovedefined metric. Definition Let R be n rbitrry metric spce. A mpping A of the spce B into itself is sid to be contrction if there exists number α < 1 such tht ρ(ax, Ay) αρ(x, y) for ny two points x, y R. (49) Every contrction mpping is continuous: x n x yields Ax n Ax (5) Theorem Every contrction mpping defined in complete metric spce R hs one nd only one fixed point; tht is the eqution Ax = x hs only one solution. We will use the principle of contrction mppings to prove Picrd s theorem. Theorem 3 Assume tht function f(x, y) stisfies the Lipschitz condition f(x, y 1 ) f(x, y ) M y 1 y in G. (51) Then on some intervl x d < x < x + d there exists unique solution y = φ(x) to the initil vlue problem dy dx = f(x, y), y(x ) = y. (5) Proof. Since function f(x, y) is continuous we hve f(x, y) k in region G G which contins (x, y ). Now choose number d such tht (x, y) G if x x d, y y kd Md < 1. Consider the mpping ψ = Aφ defined by ψ(x) = y + x x f(t, φ(t))dt, x x d. (53) 11
12 Let us prove tht this is contrction mpping of the complete set C = C[x d, x + d] of continuous functions defined in closed intervl x d x x + d (see Exmple 7). We hve x x x ψ(x) y = f(t, φ(t))dt f(t, φ(t)) dt k dt = k(x x ) kd. (54) x x x x ψ 1 (x) ψ (x) f(t, φ 1 (t)) f(t, φ (t)) dt Md mx φ 1(x) φ (x). (55) x x d x x +d Since Md < 1, the mpping ψ = Aφ is contrction. From this it follows tht the opertor eqution φ = Aφ nd consequently the integrl eqution (33) hs one nd only one solution. This result cn be generlized to systems of ODEs of the first order. To this end, denote by f = (f 1,..., f n ) n ndimensionl vectorfunction nd write the initil vlue problem for system of ODEs of the first order using this vector nottion dȳ dx = f(x, ȳ), ȳ(x ) = ȳ. (56) where the vectorfunction f is defined nd continuous in region G of the n + 1dimensionl spce R n+1 such tht G contins the n+1dimensionl point x, ȳ. We ssume tht f stisfies the the Lipschitz condition (51) termwise in vribles ȳ. the initil vlue problem (56) is equivlent to system of IEs φ(x) = ȳ + x x f(t, φ(t))dt. (57) Since function f(t, ȳ) is continuous we hve f(x, y) K in region G G which contins (x, ȳ ). Now choose number d such tht (x, ȳ) G if x x d, ȳ ȳ Kd Md < 1. Then we consider the mpping ψ = A φ defined by ψ(x) = ȳ + x x f(t, φ(t))dt, x x d. (58) nd prove tht this is contrction mpping of the complete spce C = C[x d, x + d] of continuous (componentwise) vectorfunctions into itself. The proof is termwise repetitiion of the resoning in the sclr cse bove. 5 Unique solvbility of the Fredholm IE of the nd kind using the contrction mpping. Neumnn series 5.1 Liner Fredholm IE of the nd kind Consider Fredholm IE of the nd kind f(x) = φ(x) + λ 1 K(x, y)φ(y)dy, (59)
13 where the kernel K(x, y) is continuous function in the squre Π = {(x, y) : x b, y b}, so tht K(x, y) M for (x, y) Π. Consider the mpping g = Af defined by of the complete spce C[, b] into itself. We hve g(x) = φ(x) + λ K(x, y)φ(y)dy, (6) g 1 (x) = λ g (x) = λ ρ(g 1, g ) = mx g 1(x) g (x) x b K(x, y)f 1 (y)dy + φ(x), K(x, y)f (y)dy + φ(x), λ K(x, y) f 1 (y) f (y) dy λ M(b ) mx f 1(y) f (y) (61) y b 1 < ρ(f 1, f ) if λ < (6) M(b ) Consequently, the mpping A is contrction if λ < 1 M(b ), (63) nd (59) hs unique solution for sufficiently smll λ stisfying (63). The successive pproximtions to this solution hve the form f n (x) = φ(x) + λ K(x, y)f n (y)dy. (64) 5. Nonliner Fredholm IE of the nd kind The method is pplicble to nonliner IEs where the kernel K nd φ re continuous functions, nd f(x) = φ(x) + λ K(x, y, f(y))dy, (65) K(x, y, z 1 ) K(x, y, z ) M z 1 z. (66) If λ stisfies (63) we cn prove in the sme mnner tht the mpping g = Af defined by g(x) = φ(x) + λ K(x, y, f(y))dy, (67) of the complete spce C[, b] into itself is contrction becuse for this mpping, the inequlity (61) holds. 13
14 5.3 Liner Volterr IE of the nd kind Consider Volterr IE of the nd kind f(x) = φ(x) + λ x K(x, y)φ(y)dy, (68) where the kernel K(x, y) is continuous function in the squre Π for some b >, so tht K(x, y) M for (x, y) Π. Formulte generliztion of the principle of contrction mppings. Theorem 4 If A is continuous mpping of complete metric spce R into itself such tht the mpping A n is contrction for some n, then the eqution Ax = x hs one nd only one solution. In fct if we tke n rbitrry point x R nd consider the sequence A kn x, k =, 1,,..., the repetition of the proof of the clssicl principle of contrction mppings yields the convergence of this sequence. Let x = lim k A kn x, then Ax = x. In fct Ax = lim k A kn Ax. Since the mpping A n is contrction, we hve ρ(a kn Ax, A kn x) αρ(a (k)n Ax, A (k)n x)... α k ρ(ax, x). Consequently, lim k ρ(akn Ax, A kn x) =, i.e., Ax = x. Consider the mpping g = Af defined by g(x) = φ(x) + λ of the complete spce C[, b] into itself. 5.4 Neumnn series x K(x, y)φ(y)dy, (69) Consider n IE f(x) = φ(x) λ K(x, y)φ(y)dy, (7) In order to determine the solution by the method of successive pproximtions nd obtin the Neumnn series, rewrite (7) s φ(x) = f(x) + λ K(x, y)φ(y)dy, (71) nd tke the righthnd side f(x) s the zero pproximtion, setting φ (x) = f(x). (7) Substitute the zero pproximtion into the righthnd side of (73) to obtin the first pproximtion φ 1 (x) = f(x) + λ K(x, y)φ (y)dy, (73) 14
15 nd so on, obtining for the (n + 1)st pproximtion φ n+1 (x) = f(x) + λ K(x, y)φ n (y)dy. (74) Assume tht the kernel K(x, y) is bounded function in the squre Π = {(x, y) : x b, y b}, so tht K(x, y) M, (x, y) Π. or even tht there exists constnt C 1 such tht The the following sttement is vlid. K(x, y) dy C 1 x [, b], (75) Theorem 5 Assume tht condition (75) holds Then the sequence φ n of successive pproximtions (74) converges uniformly for ll λ stisfying λ 1 B, B = The limit of the sequence φ n is the unique solution to IE (7). K(x, y) dxdy. (76) Proof. Write two first successive pproximtions (74): Set f(x) + λ φ 1 (x) = f(x) + λ K(x, y)f(y)dy, (77) φ (x) = f(x) + λ K(x, y)f(y)dy + λ K (x, y) = to obtin (by chnging the order of integrtion) φ (x) = f(x) + λ In the sme mnner, we obtin the third pproximtion K(x, y)φ 1 (y)dy = (78) K(x, t)dt K(t, y)f(y)dy. (79) K(x, t)k(t, y)dt (8) K(x, y)f(y)dy + λ K (x, y)f(y)dy. (81) where φ 3 (x) = f(x) + λ K(x, y)f(y)dy + + λ K (x, y)f(y)dy + λ 3 K 3 (x, y)f(y)dy, (8) K 3 (x, y) = K(x, t)k (t, y)dt. (83) 15
16 The generl reltionship hs the form where n φ n (x) = f(x) + λ m K m (x, y)f(y)dy, (84) m=1 K 1 (x, y) = K(x, y), K m (x, y) = K(x, t)k m (t, y)dt. (85) nd K m (x, y) is clled the mth iterted kernel. One cn esliy prove tht the iterted kernels stisfy more generl reltionship K m (x, y) = K r (x, t)k m r (t, y)dt, r = 1,,..., m 1 (m =, 3,...). (86) Assuming tht successive pproximtions (84) converge we obtin the (unique) solution to IE (7) by pssing to the limit in (84) φ(x) = f(x) + λ m K m (x, y)f(y)dy. (87) m=1 In order to prove the convergence of this series, write nd estimte C m. Setting r = m 1 in (86) we hve K m (x, y) = K m (x, y) dy C m x [, b], (88) Applying to (89) the Schwrtz inequlity we obtin K m (x, y) = Integrting (9) with respect to y yields which gives K m (x, y) dy B K m (x, t) B C m (B = nd finlly the required estimte K m (x, t)k(t, y)dt, (m =, 3,...). (89) K m (x, t) K(t, y) dt. (9) K(x, y) dxdy) (91) C m B C m, (9) C m B m C 1. (93) Denoting D = f(y) dy (94) nd pplying the Schwrtz inequlity we obtin K m (x, y)f(y)dy K m (x, y) dy f(y) dy D C 1 B m. (95) 16
17 Thus the common term of the series (87) is estimted by the quntity D C 1 λ m B m, (96) so tht the series converges fster thn the progression with the denomintor λ B, which proves the theorem. If we tke the first n terms in the series (87) then the resulting error will be not greter thn λ D C n+1 B n 1 1 λ B. (97) 5.5 Resolvent Introduce the nottion for the resolvent Γ(x, y, λ) = λ m K m (x, y) (98) m=1 Chnging the order of summtion nd integrtion in (87) we obtin the solution in the compct form φ(x) = f(x) + λ Γ(x, y, λ)f(y)dy. (99) One cn show tht the resolvent stisfies the IE Γ(x, y, λ) = K(x, y) + λ K(x, t)γ(t, y, λ)dt. (1) 6 IEs s liner opertor equtions. Fundmentl properties of completely continuous opertors 6.1 Bnch spces Definition 3 A complete normed spce is sid to be Bnch spce, B. Exmple 9 The spce C[, b] of continuous functions defined in closed intervl x b with the (uniform) norm f C = mx x b f(x) [nd the metric ρ(φ 1, φ ) = φ 1 φ C = mx x b φ 1(x) φ (x) ] is normed nd complete spce. Indeed, every fundmentl (convergent) functionl sequence {φ n (x)} of continuous functions converges to continuous function in the C[, b]metric. Definition 4 A set M in the metric spce R is sid to be compct if every sequence of elements in M contins subsequence tht converges to some x R. 17
18 Definition 5 A fmily of functions {φ} defined on the closed intervl [, b] is sid to be uniformly bounded if there exists number M > such tht for ll x nd for ll φ in the fmily. φ(x) < M Definition 6 A fmily of functions {φ} defined on the closed intervl [, b] is sid to be equicontinuous if for every ɛ > there is δ > such tht φ(x 1 ) φ(x ) < ɛ for ll x 1, x such tht x 1 ) x < δ nd for ll φ in the given fmily. Formulte the most common criterion of compctness for fmilies of functions. Theorem 6 (Arzel s theorem). A necessry nd sufficient condition tht fmily of continuous functions defined on the closed intervl [, b] be compct is tht this fmily be uniformly bounded nd equicontinuous. 6. IEs nd completely continuous liner opertors Definition 7 An opertor A which mps Bnch spce E into itself is sid to be completely continuous if it mps nd rbitrry bounded set into compct set. Consider n IE of the nd kind f(x) = φ(x) + λ which cn be written in the opertor form s Here the liner (integrl) opertor A defined by f(x) = K(x, y)φ(y)dy, (11) f = φ + λaφ. (1) K(x, y)φ(y)dy (13) nd considered in the (Bnch) spce C[, b] of continuous functions in closed intervl x b represents n extensive clss of completely continuous liner opertors. Theorem 7 Formul (13) defines completely continuous liner opertor in the (Bnch) spce C[, b] if the function K(x, y) is to be bounded in the squre Π = {(x, y) : x b, y b}, nd ll points of discontinuity of the function K(x, y) lie on the finite number of curves where ψ k (x) re continuous functions. y = ψ k (x), k = 1,,... n, (14) 18
19 Proof. To prove the theorem, it is sufficient to show (ccording to Arzel s theorem) tht the functions (13) defined on the closed intervl [, b] (i) re continuous nd the fmily of these functions is (ii) uniformly bounded nd (iii) equicontinuous. Below we will prove sttements (i), (ii), nd (iii). Under the conditions of the theorem the integrl in (13) exists for rbitrry x [, b]. Set M = sup K(x, y), Π nd denote by G the set of points (x, y) for which the condition y ψ k (x) < ɛ, k = 1,,... n, (15) 1Mn holds. Denote by F the complement of G with respect to Π. Since F is closed set nd K(x, y) is continuous on F, there exists δ > such tht for points (x, y) nd (x, y) in F for which x x < δ the inequlity K(x, y) K(x, y) < (b ) 1 3 ɛ (16) holds. Now let x nd x be such tht x x < δ holds. Then f(x ) f(x ) cn be evluted by integrting over the sum A of the intervls y ψ k (x ) < y ψ k (x ) < K(x, y) K(x, y) φ(y) dy (17) ɛ, k = 1,,... n, 1Mn ɛ, k = 1,,... n, 1Mn nd over the complement B of A with respect to the closed intervl [, b]. The length of A does not exceed ɛ. Therefore 3M K(x, y) K(x, y) φ(y) dy ɛ φ. (18) A 3 The integrl over the complement B cn be obviously estimted by K(x, y) K(x, y) φ(y) dy ɛ φ. (19) B 3 Therefore f(x ) f(x ) ɛ φ. (11) Thus we hve proved the continuity of f(x) nd the equicontinuity of the functions f(x) (ccording to Definition (6)) corresponding to the functions φ(x) with bounded norm φ. The uniform boundedness of f(x) (ccording to Definition (5)) corresponding to the functions φ(x) with bounded norms follows from the inequlity φ(x) K(x, y) φ(y) dy M(b ) φ. (111) B 19
20 6.3 Properties of completely continuous opertors Theorem 8 If {A n } is sequence of completely continuous opertors on Bnch spce E which converges in (opertor) norm to n opertor A then the opertor A is completely continuous. Proof. Consider n rbitrry bounded sequence {x n } of elements in Bnch spce E such tht x n < c. It is necessry to prove tht sequence {Ax n } contins convergent subsequence. The opertor A 1 is completely continuous nd therefore we cn select convergent subsequence from the elements {A 1 x n }. Let x (1) 1, x (1),..., x (1) n,..., (11) be the inverse imges of the members of this convergent subsequence. We pply opertor A to ech members of subsequence (11)). The opertor A is completely continuous; therefore we cn gin select convergent subsequence from the elements {A x (1) n }, denoting by x () 1, x (),..., x () n,..., (113) the inverse imges of the members of this convergent subsequence. The opertor A 3 is lso completely continuous; therefore we cn gin select convergent subsequence from the elements {A 3 x () n }, denoting by x (3) 1, x (3),..., x (3) n,..., (114) the inverse imges of the members of this convergent subsequence; nd so on. Now we cn form digonl subsequence x (1) 1, x (),..., x (n) n,.... (115) Ech of the opertors A n trnsforms this digonl subsequence into convergent sequence. If we show tht opertor A lso trnsforms (115)) into convergent sequence, then this will estblish its complete continuity. To this end, pply the Cuchy criterion (48) nd evlute the norm We hve the chin of inequlities Ax (n) n Choose k so tht A A k < ɛ c Ax (n) n Ax (m) m Ax n (n) A k x (n) + A k x (n) n Ax (m) m. (116) n + A k x (m) m A A k ( x (m) m nd N such tht A k x (n) n + A k x (m) m + x (m) m A k x (m) m < ɛ Ax (m) m ) + A k x (n) n A k x (m) m. holds for rbitrry n > N nd m > N (which is possible becuse the sequence A k x (n) n converges). As result, Ax (n) n i.e., the sequence Ax (n) n is fundmentl. Ax (m) m < ɛ, (117)
21 Theorem 9 (Superposition principle). If A is completely continuous opertor nd B is bounded opertor then the opertors AB nd BA re completely continuous. Proof. If M is bounded set in Bnch spce E, then the set BM is lso bounded. Consequently, the set ABM is compct nd this mens tht the opertor AB is completely continuous. Further, if M is bounded set in E, then AM is compct. Then, becuse B is continuous opertor, the set BAM is lso compct, nd this completes the proof. Corollry A completely continuous opertor A cnnot hve bounded inverse in finitedimensionl spce E. Proof. In the contrry cse, the identity opertor I = AA would be completely continuous in E which is impossible. Theorem 1 The djoint of completely opertor is completely continuous. 7 Elements of the spectrl theory of liner opertors. Spectrum of completely continuous opertor In this section we will consider bounded liner opertors which mp Bnch spce E into itself. Definition 8 The number λ is sid to be chrcteristic vlue of the opertor A if there exists nonzero vector x (chrcteristic vector) such tht Ax = λx. In n ndimensionl spce, this definition is equivlent to the following: the number λ is sid to be chrcteristic vlue of the opertor A if it is root of the chrcteristic eqution Det A λi =. The totlity of those vlues of λ for which the inverse of the opertor A λi does not exist is clled the spectrum of the opertor A. The vlues of λ for which the opertor A λi hs the inverse re clled regulr vlues of the opertor A. Thus the spectrum of the opertor A consists of ll nonregulr points. The opertor R λ = (A λi) is clled the resolvent of the opertor A. In n infinitedimensionl spce, the spectrum of the opertor A necessrily contins ll the chrcteristic vlues nd mybe some other numbers. In this connection, the set of chrcteristic vlues is clled the point spectrum of the opertor A. The remining prt of the spectrum is clled the continuous spectrum of the opertor A. The ltter lso consists, in its turn, of two prts: (1) those λ for which A λi hs n unbounded inverse with domin dense in E (nd this prt is clled the continuous spectrum); nd () the 1
22 reminder of the spectrum: those λ for which A λi hs (bounded) inverse whose domin is not dense in E, this prt is clled the residul spectrum. If the point λ is regulr, i.e., if the inverse of the opertor A λi exists, then for sufficiently smll δ the opertor A (λ + δ)i lso hs the inverse, i.e., the point δ + λ is regulr lso. Thus the regulr points form n open set. Consequently, the spectrum, which is complement, is closed set. Theorem 11 The inverse of the opertor A λi exists for rbitrry λ for which λ > A. Proof. Since A λi = λ (I 1 ) λ A, the resolvent R λ = (A λi) = 1 λ ( I A ) = 1 λ λ This opertor series converges for A/λ < 1, i.e., the opertor A λi hs the inverse. Corollry. The spectrum of the opertor A is contined in circle of rdius A with center t zero. Theorem 1 Every completely continuous opertor A in the Bnch spce E hs for rbitrry ρ > only finite number of linerly independent chrcteristic vectors which correspond to the chrcteristic vlues whose bsolute vlues re greter thn ρ. k= A k λ k. Proof. Assume, d bs, tht there exist n infinite number of linerly independent chrcteristic vectors x n, n = 1,,..., stisfying the conditions Ax n = λ n x n, λ n > ρ > (n = 1,,...). (118) Consider in E the subspce E 1 generted by the vectors x n (n = 1,,...). In this subspce the opertor A hs bounded inverse. In fct, ccording to the definition, E 1 is the totlity of elements in E which re of the form vectors nd these vectors re everywhere dense in E 1. For these vectors x = α 1 x α k x k, (119) Ax = α 1 λ 1 x α k λ k x k, (1) nd, consequently, nd A x = (α 1 /λ 1 )x (α k /λ k )x k (11) A x < 1 ρ x. Therefore, the opertor A defined for vectors of the form (119) by mens of (1) cn be extended by continuity (nd, consequently, with preservtion of norm) to ll of E 1. The opertor
23 A, being completely continuous in the Bnch spce E, is completely continuous in E 1. According to the corollry to Theorem (9), completely continuous opertor cnnot hve bounded inverse in infinitedimensionl spce. The contrdiction thus obtined proves the theorem. Corollry. Every nonzero chrcteristic vlue of completely continuous opertor A in the Bnch spce E hs only finite multiplicity nd these chrcteristic vlues form bounded set which cnnot hve single limit point distinct from the origin of coordintes. We hve thus obtined chrcteriztion of the point spectrum of completely continuous opertor A in the Bnch spce E. One cn lso show tht completely continuous opertor A in the Bnch spce E cnnot hve continuous spectrum. 8 Liner opertor equtions: the Fredholm theory 8.1 The Fredholm theory in Bnch spce Theorem 13 Let A be completely continuous opertor which mps Bnch spce E into itself. If the eqution y = x Ax is solvble for rbitrry y, then the eqution x Ax = hs no solutions distinct from zero solution. Proof. Assume the contrry: there exists n element x 1 such tht x 1 Ax 1 = T x 1 = Those elements x for which T x = form liner subspce E 1 of the bnch spce E. Denote by E n the subspce of E consisting of elements x for which the powers T n x =. It is cler tht subspces E n form nondecresing subsequence E 1 E... E n.... We shll show tht the equlity sign cnnot hold t ny point of this chin of inclusions. In fct, since x 1 nd eqution y = T x is solvble for rbitrry y, we cn find sequence of elements x 1, x,..., x n,... distinct from zero such tht T x = x 1, T x 3 = x,..., T x n = x n,.... The element x n belongs to subspce E n for ech n but does not belong to subspce E n. In fct, but T n x n = T n x n =... = T x 1 =, T n x n = T n x n =... = T x = x 1. 3
24 All the subspces E n re liner nd closed; therefore for rbitrry n there exists n element y n+1 E n+1 such tht y n+1 = 1 nd ρ(y n+1, E n ) 1, where ρ(y n+1, E n ) denotes the distnce from y n+1 to the spce E n : ρ(y n+1, E n ) = inf { y n+1 x, x E n }. Consider the sequence {Ay k }. We hve (ssuming p > q) Ay p Ay q = y p (y q + T y p T y q ) 1, since y q + T y p T y q E p. It is cler from this tht the sequence {Ay k } cnnot contin ny convergent subsequence which contrdicts the complete continuity of the opertor A. This contrdiction proves the theorem Corollry 1. If the eqution y = x Ax is solvble for rbitrry y, then this eqution hs unique solution for ech y, i.e., the opertor I A hs n inverse in this cse. We will consider, together with the eqution y = x Ax the eqution h = f A f which is djoint to it, where A is the djoint to A nd h, f re elements of the Bnch spce Ē, the conjugte of E. Corollry. If the eqution h = f A f is solvble for ll h, then the eqution f A f = hs only the zero solution. It is esy to verify this sttement by reclling tht the opertor djoint to completely continuous opertor is lso completely continuous, nd the spce Ē conjugte to Bnch spce is itself Bnch spce. Theorem 14 A necessry nd sufficient condition tht the eqution y = x Ax be solvble is tht f(y) = for ll f for which f A f =. Proof. 1. If we ssume thst the eqution y = x Ax is solvble, then f(y) = f(x) f(ax) = f(y) = f(x) A f(x) i.e., f(y) = for ll f for which f A f =.. Now ssume tht f(y) = for ll f for which f A f =. For ech of these functionls we consider the set L f of elements for which f tkes on the vlue zero. Then our ssertion is equivlent to the fct tht the set L f consists only of the elements of the form x Ax. Thus, it is necessry to prove tht n element y 1 which cnnot be represented in the form x Ax cnnot be contined in L f. To do this we shll show tht for such n element y 1 we cn construct functionl f 1 stisfying f 1 (y 1 ), f 1 A f 1 =. 4
25 These conditions re equivlent to the following: In fct, f 1 (y 1 ), f 1 (x Ax) = x. (f 1 A f 1 )(x) = f 1 (x) A f 1 (x) = f 1 (x) f 1 (Ax) = f 1 (x Ax). Let G be subspce consisting of ll elements of the form x Ax. Consider the subspce {G, y 1 } of the elements of the form z + αy 1, z G nd define the liner functionl by setting f 1 (z + αy 1 ) = α. Extending the liner functionl to the whole spce E (which is possible by virtue of the Hhn Bnch theorem) we do obtin liner functionl stisfying the required conditions. This completes the proof of the theorem. Corollry If the eqution f A f = does not hve nonzero solution, then the eqution x Ax = y is solvble for ll y. Theorem 15 A necessry nd sufficient condition tht the eqution f A f = h be solvble is tht h(x) = for ll x for which x Ax =. Proof. 1. If we ssume tht the eqution f A f = h is solvble, then h(x) = f(x) A f(x) = f(x Ax) i.e., h(x) = if x Ax =.. Now ssume tht h(x) = for ll x such tht x Ax =. We shll show tht the eqution f A f = h is solvble. Consider the set F of ll elements of the form y = x Ax. Construct functionl f on F by setting f(t x) = h(x). This eqution defines in fct liner functionl. Its vlue is defined uniquely for ech y becuse if T x 1 = T x then h(x 1 ) = h(x ). It is esy to verify the linerity of the functionl nd extend it to the whole spce E. We obtin f(t x) = T f(x) = h(x). i.e., the functionl is solution of eqution f A f = h. This completes the proof of the theorem. Corollry If the eqution x Ax = does not hve nonzero solution, then the eqution f A f = h is solvble for ll h. The following theorem is the converse of Thereom 13. Theorem 16 If the eqution x Ax = hs x = for its only solution, then the eqution x Ax = y hs solution for ll y. 5
26 Proof. If the eqution x Ax = hs only one solution, then by virtue of the corollry to the preceding theorem, the eqution f A f = h is solvble for ll h. Then, by Corollry to Thereom 13, the eqution f A f = hs f = for its only solution. Hence Corollry to Thereom 14 implies tht the eqution x Ax = y hs solution for ll y. Thereoms 13 nd 16 show tht for the eqution x Ax = y (1) only the following two cses re possible: (i) Eqution (1) hs unique solution for ech y, i.e., the opertor I A hs n inverse. (ii) The corresponding homogeneous eqution x Ax = hs nonzero solution, i.e., the number λ = 1 is chrcteristic vlue for the opertor A. All the results obtined bove for the eqution x Ax = y remin vlid for the eqution x λax = y (the djoint eqution is f λa f = h). It follows tht either the opertor I λa hs n inverse or the number 1/λ is chrcteristic vlue for the opertor A. In other words, in the cse of completely continuous opertor, n rbitrry number is either regulr point or chrcteristic vlue. Thus completely continuous opertor hs only point spectrum. Theorem 17 The dimension n of the spce N whose elements re the solutions to the eqution x Ax = is equl to the dimension of the subspce N whose elements re the solutions to the eqution f A f =. This theorem is proved below for opertors in the Hilbert spce. 8. Fredholm theorems for liner integrl opertors nd the Fredholm resolvent One cn formulte the Fredholm theorems for liner integrl opertors Aφ(x) = K(x, y)φ(y)dy (13) in terms of the Fredholm resolvent. The numbers λ for which there exists the Fredholm resolvent of Fredholm IE φ(x) λ K(x, y)φ(y)dy = f(x) (14) will be clled regulr vlues. The numbers λ for which the Fredholm resolvent Γ(x, y, λ) is not defined (they belong to the point spectrum) will be chrcteristic vlues of the IE (14), nd they coincide with the poles of the resolvent. The inverse of chrcteristic vlues re eigenvlues of the IE. Formulte, for exmple, the first Fredholm theorem in terms of regulr vlues of the resolvent. If λ is regulr vlue, then the Fredholm IE (14) hs one nd only one solution for rbitrry f(x). This solution is given by formul (99), φ(x) = f(x) + λ Γ(x, y, λ)f(y)dy. (15) 6
27 In prticulr, if λ is regulr vlue, then the homogeneous Fredholm IE (14) hs only the zero solution φ(x) =. φ(x) λ K(x, y)φ(y)dy = (16) 9 IEs with degenerte nd seprble kernels 9.1 Solution of IEs with degenerte nd seprble kernels An IE with degenerte kernel φ(x) λ K(x, y)φ(y)dy = f(x) (17) n K(x, y) = i (x)b i (y) (18) i=1 cn be represented, by chnging the order of summtion nd integrtion, in the form φ(x) λ n i=1 i (x) b i (y)φ(y)dy = f(x). (19) Here one my ssume tht functions i (x) (nd b i (y) ) re linerly independent (otherwise, the numer of terms in (18)) cn be reduced). It is esy to solve such IEs with degenerte nd seprble kernels. Denote c i = which re unknown constnts. Eqution (19)) becomes b i (y)φ(y)dy, (13) n φ(x) = f(x) + λ c i i (x), (131) i=1 nd the problem reduces to the determintion of unknowns c i. To this end, substitute (131) into eqution (19) to obtin, fter simple lgebr, { n [ ] } b n i (x) c i b i (y) f(y) + λ c k k (y) dy =. (13) i=1 k=1 Since functions i (x) re linerly independent, equlity (13) yields [ ] b n c i b i (y) f(y) + λ c k k (y) dy =, i = 1,,... n. (133) k=1 7
28 Denoting f i = b i (y)f(y)dy, ik = nd chnging the order of summtion nd integrtion, we rewrite (133) s b i (y) k (y)dy (134) n c i λ ik c k = f i, i = 1,,... n, (135) k=1 which is liner eqution system of order n with respect to unknowns c i. Note tht the sme system cn be obtined by multiplying (131) by k (x) (k = 1,,... n) nd integrting from to b System (135) is equivlent to IE (19) (nd (17), (18)) in the following sense: if (135) is uniquely solvble, then IE (19) hs one nd only one solution (nd vice vers); nd if (135) hs no solution, then IE (19) is not solvble (nd vice vers). The determinnt of system (135) is D(λ) = 1 λ 11 λ 1... λ 1n λ 1 1 λ... λ n λ n1 λ n... 1 λ nn. (136) D(λ) is polynomil in powers of λ of n order not higher thn n. Note tht D(λ) is not identiclly zero becuse D() = 1. Thus there exist not more thn n different numbers λ = λ k such tht D(λ k) =. When λ = λ k, then system (135) together with IE (19) re either not solvble or hve infinitely mny solutions. If λ λ k, then system (135) nd IE (19) re uniquely solvble. Exmple 1 Consider Fredholm IE of the nd kind Here where φ(x) λ (x + y)φ(y)dy = f(x). (137) K(x, y) = x + y = i (x)b i (y) = 1 (x)b 1 (y) + (x)b (y), i=1 1 (x) = x, b 1 (y) = 1, (x) = 1, b (y) = y, is degenerte kernel, f(x) is given function, nd λ is prmeter. Look for the solution to (137) in the form (131) (note tht here n = ) Denoting φ(x) = f(x) + λ c i i (x) = f(x) + λ(c 1 1 (x) + c (x)) = f(x) + λ(c 1 x + c ). (138) i=1 f i = b i (y)f(y)dy, ik = 8 b i (y) k (y)dy, (139)
29 so tht nd 11 = 1 = 1 = = f 1 = f = b 1 (y) 1 (y)dy = b 1 (y) (y)dy = b (y) 1 (y)dy = b (y) (y)dy = b 1 (y)f(y)dy = b (y)f(y)dy = ydy = 1, dy = 1, (14) y dy = 1 3, ydy = 1, f(y)dy, (141) yf(y)dy, nd chnging the order of summtion nd integrtion, we rewrite the corresponding eqution (133) s or, explicitly, c i λ ik c k = f i, i = 1,, (14) k=1 (1 λ/)c 1 λc = f 1, (λ/3)c 1 + (1 λ/)c = f, which is liner eqution system of order with respect to unknowns c i. The determinnt of system (143) is 1 λ/ λ D(λ) = λ/3 1 λ/ = 1 + λ /4 λ λ /3 = 1 λ λ /1 = 1 1 (λ + 1λ 1)(143) is polynomil of order. (D(λ) is not identiclly, nd D() = 1). There exist not more thn different numbers λ = λ k such tht D(λ k) =. Here there re exctly two such numbers; they re λ 1 = , (144) λ = If λ λ k, then system (143) nd IE (137) re uniquely solvble. To find the solution of system (143) pply Crmer s rule nd clculte the determinnts of system (143) f D 1 = 1 λ f 1 λ/ = f 1(1 λ/) + f λ = = (1 λ/) f(y)dy + λ yf(y)dy = 9 [(1 λ/) + yλ]f(y)dy, (145)
30 f D 1 = 1 λ f 1 λ/ = f 1(1 λ/) + f λ = = (1 λ/) f(y)dy + λ yf(y)dy = [(1 λ/) + yλ]f(y)dy, (146) (1 λ/) f D = 1 λ/3 f = f (1 λ/) + f 1 λ/3 = = (1 λ/) yf(y)dy + (λ/3) f(y)dy = [y(1 λ/) + λ/3]f(y)dy, (147) Then, the solution of system (143) is c 1 = D 1 D = 1 = 1 λ + 1λ 1 λ + 1λ 1 c = D D = 1 = 1 λ + 1λ 1 λ + 1λ 1 [(1 λ/) + yλ]f(y)dy = [(6λ 1) 1yλ]f(y)dy, (148) According to (138), the solution of the IE (137) is φ(x) = f(x) + λ(c 1 x + c ) = = λ f(x) + λ + 1λ 1 [y(1 λ/) + λ/3]f(y)dy = [y(6λ 1) 4λ]f(y)dy (149) [6(λ )(x + y) 1xy 4λ]f(y)dy. (15) When λ = λ k (k = 1, ), then system (143) together with IE (137) re not solvble. Exmple 11 Consider Fredholm IE of the nd kind Here π φ(x) λ sin x cos yφ(y)dy = f(x). (151) K(x, y) = sin x cos y = (x)b(y) is degenerte kernel, f(x) is given function, nd λ is prmeter. Rewrite IE (151) in the form (131) (here n = 1) φ(x) = f(x) + λc sin x, c = π cos yφ(y)dy. (15) Multiplying the first equlity in (15) by b(x) = cos x nd integrting from to π we obtin c = π f(x) cos xdx (153) 3
31 Therefore the solution is φ(x) = f(x) + λc sin x = f(x) + λ π Exmple 1 Consider Fredholm IE of the nd kind x = φ(x) sin x cos yf(y)dy. (154) (x + y )φ(y)dy, (155) where K(x, y) = x +y is degenerte kernel f(x) = x is given function, nd the prmeter λ = IEs with degenerte kernels nd pproximte solution of IEs Assume tht in n IE φ(x) λ K(x, y)φ(y)dy = f(x) (156) the kernel K(x, y nd f(x) re continuous functions in the squre Π = {(x, y) : x b, y b}. Then φ(x) will be lso continuous. Approximte IE (156) with n IE hving degenerte kernel. To this end, replce the integrl in (156) by finite sum using, e.g., the rectngle rule where The resulting pproximtion tkes the form n K(x, y)φ(y)dy h K(x, y k )φ(y k ), (157) k=1 h = b n, y k = + kh. (158) n φ(x) λh K(x, y k )φ(y k ) = f(x) (159) k=1 of n IE hving degenerte kernel. Now replce vrible x in (159) by finite number of vlues x = x i, i = 1,,..., n. We obtin liner eqution system with unknowns φ k = φ(x k ) nd righthnd side contining the known numbers f i = f(x i ) n φ i λh K(x i, y k )φ k = f i, i = 1,,..., n. (16) k=1 We cn find the solution {φ i } of system (16) by Crmer s rule, φ i = D(n) i (λ), i = 1,,..., n, (161) D (n) (λ) 31
32 clculting the determinnts D (n) (λ) = 1 hλk(x 1, y 1 ) hλk(x 1, y )... hλk(x 1, y n ) hλk(x, y 1 ) 1 hλk(x, y )... hλk(x, y n ) hλk(x n, y 1 ) hλk(x n, y )... 1 hλk(x n, y n ). (16) nd so on. To reconstruct the pproximte solution φ (n) using the determined {φ i } one cn use the IE (159) itself: φ(x) φ n (n) = λh K(x, y k )φ k + f(x). (163) k=1 One cn prove tht φ (n) φ(x) s n where φ(x) is the exct solution to IE (159). 9.3 Fredholm s resolvent As we hve lredy mentioned the resolvent Γ(x, y, λ) of IE (159) stisfies the IE Γ(x, y, λ) = K(x, y) + λ K(x, t)γ(t, y, λ)dt. (164) Solving this IE (164) by the described pproximte method nd pssing to the limit n we obtin the resolvent s rtio of two power series Γ(x, y, λ) = D(x, y, λ), (165) D(λ) where D(x, y, λ) = D(λ) = () n B n (x, y)λ n, n= n! (166) () n c n λ n, n= n! (167) B n (x, y) =... B (x, y) = K(x, y), K(x, y) K(x, y 1 )... K(x, y n ) K(y 1, y) K(y 1, y 1 )... K(y 1, y n ) K(y n, y) K(y n, y 1 )... K(y n, y n ) dy 1... dy n ; (168) c = 1, 3
33 c n =... K(y 1, y 1 ) K(y 1, y )... K(y 1, y n ) K(y, y 1 ) K(y, y )... K(y, y n ) K(y n, y 1 ) K(y n, y )... K(y n, y n ) dy 1... dy n. (169) D(λ) is clled the Fredholm determinnt, nd D(x, y, λ), the first Fredholm minor. If the integrl K(x, y) dxdy <, (17) then seires (167) for the Fredholm determinnt converges for ll complex λ, so tht D(λ) is n entire function of λ. Thus (164) (167) define the resolvent on the whole complex plne λ, with n exception of zeros of D(λ) which re poles of the resolvent. We my conclude tht for ll complex λ tht do not coincide with the poles of the resolvent, the IE φ(x) λ K(x, y)φ(y)dy = f(x) (171) is uniquely solvble, nd its solution is given by formul (99) φ(x) = f(x) + λ Γ(x, y, λ)f(y)dy. (17) For prcticl computtion of the resolvent, one cn use the following recurrent reltionships: B (x, y) = K(x, y), c = 1, c n = B n (x, y) = c n K(x, y) n B n (y, y)dy, n = 1,,..., (173) K(x, t)b n (t, y)dt, n = 1,,.... (174) Exmple 13 Find the resolvent of the Fredholm IE (137) of the nd kind φ(x) λ We will use recurrent reltionships (173) nd (174). Here (x + y)φ(y)dy = f(x). (175) K(x, y) = x + y is degenerte kernel. We hve c = 1, B i = B i (x, y) (i =, 1, ), nd B = K(x, y) = x + y, c 1 = B (y, y)dy = B 1 = c 1 K(x, y) ydy = 1, K(x, t)b (t, y)dt = = x + y (x + t)(t + y)dt = 1 (x + y) xy 1 3, ( 1 c = B 1 (y, y)dy = (y + y) y 1 dy = 3) 1 6, B = c K(x, y) K(x, t)b 1 (t, y)dt = = 1 ( 1 6 (x + y) (x + t) (t + y) ty 1 dt =. 3) 33
34 Since B (x, y), ll subsequent c 3, c 4,..., nd B 3, B 4,..., vnish (see (173) nd (174)), nd we obtin nd D(x, y, λ) = 1 () n = B n (x, y)λ n = B (x, y) λb 1 (x, y) = n= n! ( 1 = x + y λ (x + y) xy 1, (176) 3) D(λ) = = () n c n λ n = c c 1 λ + 1 n= n! c λ = = 1 λ 1 1 λ = 1 1 (λ + 1λ 1) Γ(x, y, λ) = D(x, y, λ) D(λ) = x + y λ ( 1 (x + y) xy ) 1 3 ). (177) λ + 1λ 1 The solution to (175) is given by formul (99) so tht we obtin, using (176), φ(x) = f(x) + φ(x) = f(x) + λ Γ(x, y, λ)f(y)dy, λ 1 [6(λ )(x + y) 1xy 4λ]f(y)dy, (178) λ + 1λ 1 which coincides with formul (15). The numbers λ = λ k (k = 1, ) determined in (144), λ 1 = nd λ = 6 4, re (simple, i.e., of multiplicity one) poles of the resolvent becuse thy re (simple) zeros of the Fredholm determinnt D(λ). When λ = λ k (k = 1, ), then IE (175) is not solvble. 1 Hilbert spces. Selfdjoint opertors. Liner opertor equtions with completely continuous opertors in Hilbert spces 1.1 Hilbert spce. Selfdjoint opertors A rel liner spce R with the inner product stisfying (x, y) = (y, x) x, y R, (x 1 + x, y) = (x 1, y) + (x, y), x 1, x, y R, (λx, y) = λ(x, y), x, y R, (x, x), x R, (x, x) = if nd only if x =. 34
35 is clled the Euclidin spce. In complex Euclidin spce the inner product stisfies (x, y) = (y, x), (x 1 + x, y) = (x 1, y) + (x, y), (λx, y) = λ(x, y), Note tht in complex liner spce (x, x), (x, x) = if nd only if x =. (x, λy) = λ(x, y). The norm in the Euclidin spce is introduced by x = x, x. The Hilbert spce H is complete (in the metric ρ(x, y) = x y ) infinitedimensionl Euclidin spce. In the Hilbert spce H, the opertor A djoint to n opertor A is defined by (Ax, y) = (x, A y), x, y H. The selfdjoint opertor A = A is defined from (Ax, y) = (x, Ay), x, y H. 1. Completely continuous integrl opertors in Hilbert spce Consider n IE of the second kind φ(x) = f(x) + K(x, y)φ(y)dy. (179) Assume tht K(x, y) is Hilbert Schmidt kernel, i.e., squreintegrble function in the squre Π = {(x, y) : x b, y b}, so tht nd f(x) L [, b], i.e., K(x, y) dxdy, (18) f(x) dx. Define liner Fredholm (integrl) opertor corresponding to IE (179) Aφ(x) = K(x, y)φ(y)dy. (181) If K(x, y) is Hilbert Schmidt kernel, then opertor (181) will be clled Hilbert Schmidt opertor. Rewrite IE (179) s liner opertor eqution φ = Aφ(x) + f, f, φ L [, b]. (18) 35
36 Theorem 18 Equlity (18) nd condition (18) define completely continuous liner opertor in the spce L [, b]. The norm of this opertor is estimted s A K(x, y) dxdy. (183) Proof. Note first of ll tht we hve lredy mentioned (see (75)) tht if condition (18) holds, then there exists constnt C 1 such tht K(x, y) dy C 1 lmost everywhere in [, b], (184) i.e., K(x, y) L [, b] s function of y lmost for ll x [, b]. Therefore the function ψ(x) = K(x, y)φ(y)dy is defined lmost everywhere in [, b] (185) We will now show tht ψ(x) L [, b]. Applying the Schwrtz inequlity we obtin ψ(x) = K(x, y)φ(y)dy = φ K(x, y) dy. K(x, y) dy φ(y) dy = (186) Integrting with respect to x nd replcing n iterted integrl of K(x, y) by double integrl, we obtin Aφ = ψ(y) dy φ K(x, y) dxdy, (187) which proves (1) tht ψ(x) is squreintegrble function, i.e., ψ(x) dx, nd estimtes the norm of the opertor A : L [, b] L [, b]. Let us show tht the opertor A : L [, b] L [, b] is completely continuous. To this end, we will use Theorem 8 nd construct sequence {A n } of completely continuous opertors on the spce L (Π) which converges in (opertor) norm to the opertor A. Let {ψ n } be n orthonorml system in L [, b]. Then {ψ m (x)ψ n (y)} form n orthonorml system in L (Π). Consequently, By setting K(x, y) = K N (x, y) = m, N m, mn ψ m (x)ψ n (y). (188) mn ψ m (x)ψ n (y). (189) define n opertor A N which is finitedimensionl (becuse it hs degenerte kernel nd mps therefore L [, b] into finitedimensionl spce generted by finite system {ψ m } N m=1) nd therefore completely continuous. 36
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