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16.2. LINEAR INTEGRAL EQUATIONS OF THE SECOND KIND WITH VARIABLE INTEGRATION LIMIT 815 Let w = w(x) be a solution of the simpler auxiliary equation with f (x) ≡ 1 and a = 0, Aw(x)+B x 0 K(x – t)w(t) dt = 1.(16.2.3.9) In this case, the solution of the original equation (16.2.3.8) with an arbitrary right-hand side can be expressed via the solution of the auxiliary equation (16.2.3.9) by the formula y(x)= d dx x a w(x – t)f(t) dt = f(a)w(x – a)+ x a w(x – t)f t (t) dt. 16.2.4. Construction of Solutions of Integral Equations with Special Right-Hand Side In this section we describe some approaches to the construction of solutions of integral equations with special right-hand sides. These approaches are based on the application of auxiliary solutions that depend on a free parameter. 16.2.4-1. General scheme. Consider a linear equation, which we shall write in the following brief form: L [y]=f g (x, λ), (16.2.4.1) where L is a linear operator (integral, differential, etc.) that acts with respect to the variable x and is independent of the parameter λ,andf g (x, λ) is a given function that depends on the variable x and the parameter λ. Suppose that the solution of equation (16.2.4.1) is known: y = y(x, λ). (16.2.4.2) Let M be a linear operator (integral, differential, etc.) that acts with respect to the parameter λ and is independent of the variable x. Consider the (usual) case in which M commutes with L. We apply the operator M to equation (16.2.4.1) and find that the equation L [w]=f M (x), f M (x)=M f g (x, λ) (16.2.4.3) has the solution w = M y(x, λ) .(16.2.4.4) By choosing the operator M in a different way, we can obtain solutions for other right- hand sides of equation (16.2.4.1). The original function f g (x, λ) is called the generating function for the operator L. 16.2.4-2. Generating function of exponential form. Consider a linear equation with exponential right-hand side L [y]=e λx .(16.2.4.5) Suppose that the solution is known and is given by formula (16.2.4.2). In Table 16.1 we present solutions of the equation L [y]=f (x) with various right-hand sides; these solutions are expressed via the solution of equation (16.2.4.5). 816 INTEGRAL EQUATIONS Remark 1. When applying the formulas indicated in the table, we need not know the left-hand side of the linear equation (16.2.4.5) (the equation can be integral, differential, etc.) provided that a particular solution of this equation for the exponential right-hand side is known. It is only of importance that the left-hand side of the equation is independent of the parameter λ. Remark 2. When applying formulas indicated in the table, the convergence of the integrals occurring in the resulting solution must be verified. Example 1. We seek a solution of the equation with exponential right-hand side y(x)+ ∞ x K(x – t)y(t) dt = e λx (16.2.4.6) in the form y(x, λ)=ke λx by the method of indeterminate coefficients. Then we obtain y(x, λ)= 1 B(λ) e λx , B(λ)=1 + ∞ 0 K(–z)e λz dz.(16.2.4.7) It follows from row 3 of Table 16.1 that the solution of the equation y(x)+ ∞ x K(x – t)y(t) dt = Ax (16.2.4.8) has the form y(x)= A D x – AC D 2 ,whereD = 1 + ∞ 0 K(–z) dz, C = ∞ 0 zK(–z) dz. For such a solution to exist, it is necessary that the improper integrals of the functions K(–z)andzK(–z) exist. This holds if the function K(–z) decreases more rapidly than z –2 as z →∞. Otherwise a solution can be nonexistent. It is of interest that for functions K(–z) with power-law growth as z →∞in the case λ < 0, the solution of equation (16.2.4.6) exists and is given by formula (16.2.4.7), whereas equation (16.2.4.8) does not have a solution. Therefore, we must be careful when using formulas from Table 16.1 and verify the convergence of the integrals occurring in the solution. It follows from row 15 of Table 16.1 that the solution of the equation y(x)+ ∞ x K(x – t)y(t) dt = A sin(λx)(16.2.4.9) is given by the formula y(x)= A B 2 c + B 2 s B c sin(λx)–B s cos(λx) , where B c = 1 + ∞ 0 K(–z)cos(λz) dz, B s = ∞ 0 K(–z)sin(λz) dz. 16.2.4-3. Power-law generating function. Consider the linear equation with power-law right-hand side L [y]=x λ .(16.2.4.10) Suppose that the solution is known and is given by formula (16.2.4.2). In Table 16.2, solutions of the equation L[y]=f(x) with various right-hand sides are presented, which can be expressed via the solution of equation (16.2.4.10). 16.2. LINEAR INTEGRAL EQUATIONS OF THE SECOND KIND WITH VARIABLE INTEGRATION LIMIT 817 TABLE 16.1 Solutions of the equation L [y]=f(x) with generating function of the exponential form No. Right-hand side f (x) Solution y Solution method 1 e λx y(x, λ) Original equation 2 A 1 e λ 1 x + ···+ A n e λ n x A 1 y(x, λ 1 )+···+ A n y(x, λ n ) Follows from linearity 3 Ax + B A ∂ ∂λ y(x, λ) λ=0 + By(x,0) Follows from linearity and the results of row 4 4 Ax n , n = 0, 1, 2, A ∂ n ∂λ n y(x, λ) λ=0 Follows from the results of row 6 for λ = 0 5 A x + a , a > 0 A ∞ 0 e –aλ y(x,–λ) dλ Integration with respect to the parameter λ 6 Ax n e λx , n = 0, 1, 2, A ∂ n ∂λ n y(x, λ) Differentiation with respect to the parameter λ 7 a x y(x,lna) Follows from row 1 8 A cosh(λx) 1 2 A[y(x, λ)+y(x,–λ) Linearity and relations to the exponential 9 A sinh(λx) 1 2 A[y(x, λ)–y(x,–λ) Linearity and relations to the exponential 10 Ax m cosh(λx), m = 1, 3, 5, 1 2 A ∂ m ∂λ m [y(x, λ)–y(x,–λ) Differentiation with respect to λ and relation to the exponential 11 Ax m cosh(λx), m = 2, 4, 6, 1 2 A ∂ m ∂λ m [y(x, λ)+y(x,–λ) Differentiation with respect to λ and relation to the exponential 12 Ax m sinh(λx), m = 1, 3, 5, 1 2 A ∂ m ∂λ m [y(x, λ)+y(x,–λ) Differentiation with respect to λ and relation to the exponential 13 Ax m sinh(λx), m = 2, 4, 6, 1 2 A ∂ m ∂λ m [y(x, λ)–y(x,–λ) Differentiation with respect to λ and relation to the exponential 14 A cos(βx) A Re y(x, iβ) Selection of the real part for λ = iβ 15 A sin(βx) A Im y(x, iβ) Selection of the imaginary part for λ = iβ 16 Ax n cos(βx), n = 1, 2, 3, A Re ∂ n ∂λ n y(x, λ) λ=iβ Differentiation with respect to λ and selection of the real part for λ = iβ 17 Ax n sin(βx), n = 1, 2, 3, A Im ∂ n ∂λ n y(x, λ) λ=iβ Differentiation with respect to λ and selection of the imaginary part for λ = iβ 18 Ae μx cos(βx) A Re y(x, μ + iβ) Selection of the real part for λ = μ + iβ 19 Ae μx sin(βx) A Im y(x, μ + iβ) Selection of the imaginary part for λ = μ + iβ 20 Ax n e μx cos(βx), n = 1, 2, 3, A Re ∂ n ∂λ n y(x, λ) λ=μ+iβ Differentiation with respect to λ and selection of the real part for λ = μ + iβ 21 Ax n e μx sin(βx), n = 1, 2, 3, A Im ∂ n ∂λ n y(x, λ) λ=μ+iβ Differentiation with respect to λ and selection of the imaginary part for λ = μ + iβ 818 INTEGRAL EQUATIONS TABLE 16.2 Solutions of the equation L [y]=f(x) with generating function of power-law form No. Right-hand side f (x) Solution y Solution method 1 x λ y(x, λ) Original equation 2 n k=0 A k x k n k=0 A k y(x, k) Follows from linearity 3 A ln x + B A ∂ ∂λ y(x, λ) λ=0 + By(x,0) Follows from linearity and from the results of row 4 4 A ln n x, n = 0, 1, 2, A ∂ n ∂λ n y(x, λ) λ=0 Follows from the results of row 5 for λ = 0 5 Ax λ ln n x, n = 0, 1, 2, A ∂ n ∂λ n y(x, λ) Differentiation with respect to the parameter λ 6 A cos(β lnx) A Re y(x, iβ) Selection of the real part for λ = iβ 7 A sin(β lnx) A Im y(x, iβ) Selection of the imaginary part for λ = iβ 8 Ax μ cos(β lnx) A Re y(x, μ + iβ) Selection of the real part for λ = μ + iβ 9 Ax μ sin(β lnx) A Im y(x, μ + iβ) Selection of the imaginary part for λ = μ + iβ Example 2. We seek a solution of the equation with power-law right-hand side y(x)+ x 0 1 x K t x y(t) dt = x λ in the form y(x, λ)=kx λ by the method of indeterminate coefficients. We finally obtain y(x, λ)= 1 1 + B(λ) x λ , B(λ)= 1 0 K(t)t λ dt. It follows from row 3 of Table 16.2 that the solution of the equation with logarithmic right-hand side y(x)+ x 0 1 x K t x y(t) dt = A ln x has the form y(x)= A 1 + I 0 ln x – AI 1 (1 + I 0 ) 2 ,whereI 0 = 1 0 K(t) dt, I 1 = 1 0 K(t)lntdt. Remark. The cases where the generating function is defined sine or cosine are treated likewise. 16.2.5. Method of Model Solutions 16.2.5-1. Preliminary remarks. ∗ Consider a linear equation that we briefly write out in the form L [y(x)] = f(x), (16.2.5.1) where L is a linear (integral) operator, y(x) is an unknown function, and f(x) is a known function. * Before reading this section, it is useful to look over Subsection 16.2.4. 16.2. LINEAR INTEGRAL EQUATIONS OF THE SECOND KIND WITH VARIABLE INTEGRATION LIMIT 819 We first define arbitrarily a test solution y 0 = y 0 (x, λ), (16.2.5.2) which depends on an auxiliary parameter λ (it is assumed that the operator L is independent of λ and y 0 const). By means of equation (16.2.5.1) we define the right-hand side that corresponds to the test solution (16.2.5.2): f 0 (x, λ)=L [y 0 (x, λ)]. Let us multiply equation (16.2.5.1), for y = y 0 and f = f 0 , by some function ϕ(λ)and integrate the resulting relation with respect to λ over an interval [a, b]. We finally obtain L [y ϕ (x)] = f ϕ (x), (16.2.5.3) where y ϕ (x)= b a y 0 (x, λ)ϕ(λ) dλ, f ϕ (x)= b a f 0 (x, λ)ϕ(λ) dλ.(16.2.5.4) It follows from formulas (16.2.5.3) and (16.2.5.4) that, for the right-hand side f = f ϕ (x), the function y = y ϕ (x) is a solution of the original equation (16.2.5.1). Since the choice of the function ϕ(λ) (as well as of the integration interval) is arbitrary, the function f ϕ (x) can be arbitrary in principle. Here the main problem is how to choose a function ϕ(λ)to obtain a given function f ϕ (x). This problem can be solved if we can find a test solution such that the right-hand side of equation (16.2.5.1) is the kernel of a known inverse integral transform (we denote such a test solution by Y (x, λ) and call it a model solution). 16.2.5-2. Description of the method. Indeed, let P be an invertible integral transform that takes each function f(x)tothe corresponding transform F (λ)bytherule F (λ)=P{f(x)}.(16.2.5.5) Assume that the inverse transform P –1 has the kernel ψ(x, λ) and acts as follows: P –1 {F (λ)} = f(x), P –1 {F (λ)} ≡ b a F (λ)ψ(x, λ) dλ.(16.2.5.6) The limits of integration a and b and the integration path in (16.2.5.6) may well lie in the complex plane. Suppose that we succeeded in finding a model solution Y (x, λ) of the auxiliary problem for equation (16.2.5.1) whose right-hand side is the kernel of the inverse transform P –1 : L [Y (x, λ)] = ψ(x, λ). (16.2.5.7) Let us multiply equation (16.2.5.7) by F (λ) and integrate with respect to λ within the same limits that stand in the inverse transform (16.2.5.6). Taking into account the fact that the operator L is independent of λ and applying the relation P –1 {F (λ)} = f (x), we obtain L b a Y (x, λ)F (λ) dλ = f (x). 820 INTEGRAL EQUATIONS Therefore, the solution of equation (16.2.5.1) for an arbitrary function f(x) on the right- hand side is expressed via a solution of the simpler auxiliary equation (16.2.5.7) by the formula y(x)= b a Y (x, λ)F (λ) dλ,(16.2.5.8) where F (λ) is the transform (16.2.5.5) of the function f(x). For the right-hand side of the auxiliary equation (16.2.5.7) we can take, for instance, exponential, power-law, and trigonometric functions, which are the kernels of the Laplace, Mellin, and sine and cosine Fourier transforms (up to a constant factor). Sometimes it is rather easy to find a model solution by means of the method of indeterminate coefficients (by prescribing its structure). Afterward, to construct a solution of the equation with an arbitrary right-hand side, we can apply formulas written out below in Paragraphs 16.2.5-3–16.2.5-6. 16.2.5-3. Model solution in the case of an exponential right-hand side. Assume that we have found a model solution Y = Y (x, λ) that corresponds to the exponential right-hand side: L [Y (x, λ)] = e λx .(16.2.5.9) Consider two cases: 1 ◦ . Equations on the semiaxis, 0 ≤ x < ∞. Let f(p) be the Laplace transform of the function f(x): f(p)=L{f (x)}, L{f(x)} ≡ ∞ 0 f(x)e –px dx.(16.2.5.10) The solution of equation (16.2.5.1) for an arbitrary right-hand side f (x) can be expressed via the solution of the simpler auxiliary equation with exponential right-hand side (16.2.5.9) for λ = p by the formula y(x)= 1 2πi c+i∞ c–i∞ Y (x, p) f(p) dp.(16.2.5.11) 2 ◦ . Equations on the entire axis, –∞ < x < ∞. Let f(u) the Fourier transform of the function f(x): f(u)=F{f(x)}, F{f(x)} ≡ 1 √ 2π ∞ –∞ f(x)e –iux dx.(16.2.5.12) The solution of equation (16.2.5.1) for an arbitrary right-hand side f (x) can be expressed via the solution of the simpler auxiliary equation with exponential right-hand side (16.2.5.9) for λ = iu by the formula y(x)= 1 √ 2π ∞ –∞ Y (x, iu) f(u) du.(16.2.5.13) In the calculation of the integrals on the right-hand sides in (16.2.5.11) and (16.2.5.13), methods of the theory of functions of a complex variable are applied, including formulas for the calculation of residues and the Jordan lemma (see Subsection 11.1.2). 16.2. LINEAR INTEGRAL EQUATIONS OF THE SECOND KIND WITH VARIABLE INTEGRATION LIMIT 821 Remark 1. The structure of a model solution Y (x, λ) can differ from that of the kernel of the Laplace or Fourier inversion formula. Remark 2. When applying the method under consideration, the left-hand side of equation (16.2.5.1) need not be known (the equation can be integral, differential, functional, etc.) if a particular solution of this equation is known for the exponential right-hand side. Here only the most general information is important, namely, that the equation is linear, and its left-hand side is independent of the parameter λ. Example 1. Consider the following Volterra equation of the second kind with difference kernel: y(x)+ ∞ x K(x – t)y(t) dt = f(x). (16.2.5.14) This equation cannot be solved by direct application of the Laplace transform because the convolution theorem cannot be used here. In accordance with the method of model solutions, we consider the auxiliary equation with exponential right-hand side y(x)+ ∞ x K(x – t)y(t) dt = e px . (16.2.5.15) Its solution has the form (see Example 1 of Section 16.2.4) Y (x, p)= 1 1 + K(–p) e px , K(–p)= ∞ 0 K(–z)e pz dz. (16.2.5.16) This, by means of formula (16.2.5.11), yields a solution of equation (16.2.5.14) for an arbitrary right-hand side, y(x)= 1 2πi c+i∞ c–i∞ f(p) 1 + K(–p) e px dp, (16.2.5.17) where f(p) is the Laplace transform (16.2.5.10) of the function f (x). 16.2.5-4. Model solution in the case of a power-law right-hand side. Suppose that we have succeeded in finding a model solution Y = Y (x, s) that corresponds to a power-law right-hand side of the equation: L [Y (x, s)] = x –s , λ =–s.(16.2.5.18) Let ˆ f(s) be the Mellin transform of the function f(x): ˆ f(s)=M{f(x)}, M{f(x)} ≡ ∞ 0 f(x)x s–1 dx.(16.2.5.19) The solution of equation (16.2.5.1) for an arbitrary right-hand side f (x) can be expressed via the solution of the simpler auxiliary equation with power-law right-hand side (16.2.5.18) by the formula y(x)= 1 2πi c+i∞ c–i∞ Y (x, s) ˆ f(s) ds.(16.2.5.20) In the calculation of the corresponding integrals on the right-hand side of formula (16.2.5.20), one can use tables of inverse Mellin transforms (e.g., see Section T3.6, as well as methods of the theory of functions of a complex variable, including formulas for the calculation of residues and the Jordan lemma (see Subsection 11.1.2). . iβ) Selection of the real part for λ = iβ 7 A sin(β lnx) A Im y(x, iβ) Selection of the imaginary part for λ = iβ 8 Ax μ cos(β lnx) A Re y(x, μ + iβ) Selection of the real part for λ = μ. right-hand side of formula (16.2.5.20), one can use tables of inverse Mellin transforms (e.g., see Section T3.6, as well as methods of the theory of functions of a complex variable, including formulas. λ and selection of the real part for λ = μ + iβ 21 Ax n e μx sin(βx), n = 1, 2, 3, A Im ∂ n ∂λ n y(x, λ) λ=μ+iβ Differentiation with respect to λ and selection of the imaginary part for
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