Chapter 16 Integral Equations 16.1. Linear Integral Equations of the First Kind with Variable Integration Limit 16.1.1. Volterra Equations of the First Kind 16.1.1-1. Some definitions. Function and kernel classes. A Volterra linear integral equation of the first kind has the general form x a K(x, t)y(t) dt = f(x), (16.1.1.1) where y(x) is the unknown function (a ≤ x ≤ b), K(x, t) is the kernel of the integral equation, and f(x) is a given function, the right-hand side of equation (16.1.1.1). The functions y(x)andf(x) are usually assumed to be continuous or square integrable on [a, b]. The kernel K(x, t) is usually assumed either to be continuous on the square S = {a ≤ x ≤ b, a ≤ t ≤ b} or to satisfy the condition b a b a K 2 (x, t) dx dt = B 2 < ∞,(16.1.1.2) where B is a constant, that is, to be square integrable on this square. It is assumed in (16.1.1.2) that K(x, t) ≡ 0 for t > x. The kernel K(x, t)issaidtobedegenerate if it can be represented in the form K(x, t)= g 1 (x)h 1 (t)+···+ g n (x)h n (t). The kernel K(x, t) of an integral equation is called difference kernel if it depends only on the difference of the arguments, K(x, t)=K(x – t). Polar kernels K(x, t)=L(x, t)(x – t) –β + M (x, t), 0 < β < 1,(16.1.1.3) and logarithmic kernels (kernels with logarithmic singularity) K(x, t)=L(x, t)ln(x – t)+M(x, t), (16.1.1.4) where the functions L(x, t)andM(x, t) are continuous on S and L(x, x) 0,areoften considered as well. Polar and logarithmic kernels form a class of kernels with weak singularity. Equations containing such kernels are called equations with weak singularity. In case the functions K(x, t)andf(x) are continuous, the right-hand side of equa- tion (16.1.1.1) must satisfy the following conditions: 1 ◦ .IfK(a, a) ≠ 0,thenf(x) must be constrained by f (a)=0. 801 802 INTEGRAL EQUATIONS 2 ◦ .IfK(a, a)=K x (a, a)=···= K (n–1) x (a, a)=0, 0 < K (n) x (a, a) < ∞, then the right-hand side of the equation must satisfy the conditions f (a)=f x (a)=···= f (n) x (a)=0. 3 ◦ .IfK(a, a)=K x (a, a)=··· = K (n–1) x (a, a)=0, K (n) x (a, a)=∞, then the right-hand side of the equation must satisfy the conditions f (a)=f x (a)=···= f (n–1) x (a)=0. For polar kernels of the form (16.1.1.4) and continuous f(x), no additional conditions are imposed on the right-hand side of the integral equation. 16.1.1-2. Existence and uniqueness of a solution. Assume that in equation (16.1.1.1) the functions f (x)andK(x, t) are continuous together with their first derivatives on [a, b] and on S, respectively. If K(x, x) ≠ 0 (x [a, b]) and f(a)=0, then there exists a unique continuous solution y(x) of equation (16.1.1.1). Remark. A Volterra equation of the first kind can be treated as a Fredholm equation of the first kind whose kernel K(x, t) vanishes for t > x (see Section 16.3). 16.1.2. Equations with Degenerate Kernel: K(x, t) = g 1 (x)h 1 (t) + ···+ g n (x)h n (t) 16.1.2-1. Equations with kernel of the form K(x, t)=g 1 (x)h 1 (t)+g 2 (x)h 2 (t). Any equation of this type can be rewritten in the form g 1 (x) x a h 1 (t)y(t) dt + g 2 (x) x a h 2 (t)y(t) dt = f(x). (16.1.2.1) It is assumed that g 1 (x)/g 2 (x) ≠ const , h 1 (t)/h 2 (t) ≠ const , 0 < g 2 1 (a)+g 2 2 (a)<∞,and f(a)=0. The change of variables u(x)= x a h 1 (t)y(t) dt,(16.1.2.2) followed by the integration by parts in the second integral in (16.1.2.1) with regard to the relation u(a)=0, yields the following Volterra equation of the second kind: [g 1 (x)h 1 (x)+g 2 (x)h 2 (x)]u(x)–g 2 (x)h 1 (x) x a h 2 (t) h 1 (t) t u(t) dt = h 1 (x)f(x). (16.1.2.3) The substitution w(x)= x a h 2 (t) h 1 (t) t u(t) dt (16.1.2.4) reduces equation (16.1.2.3) to the first-order linear ordinary differential equation [g 1 (x)h 1 (x)+g 2 (x)h 2 (x)]w x – g 2 (x)h 1 (x) h 2 (x) h 1 (x) x w = f(x)h 1 (x) h 2 (x) h 1 (x) x .(16.1.2.5) 1 ◦ . In the case g 1 (x)h 1 (x)+g 2 (x)h 2 (x) 0, the solution of equation (16.1.2.5) satisfying the condition w(a)=0 [this condition is a consequence of the substitution (16.1.2.4)] has 16.1. LINEAR INTEGRAL EQUATIONS OF THE FIRST KIND WITH VARIABLE INTEGRATION LIMIT 803 the form w(x)=Φ(x) x a h 2 (t) h 1 (t) t f(t)h 1 (t) dt Φ(t)[g 1 (t)h 1 (t)+g 2 (t)h 2 (t)] ,(16.1.2.6) Φ(x)=exp x a h 2 (t) h 1 (t) t g 2 (t)h 1 (t) dt g 1 (t)h 1 (t)+g 2 (t)h 2 (t) .(16.1.2.7) Let us differentiate relation (16.1.2.4) and substitute the function (16.1.2.6) into the resulting expression. After integrating by parts with regard to the relations f(a)=0 and w(a)=0,forf const g 2 we obtain u(x)= g 2 (x)h 1 (x)Φ(x) g 1 (x)h 1 (x)+g 2 (x)h 2 (x) x a f(t) g 2 (t) t dt Φ(t) . Using formula (16.1.2.2), we find a solution of the original equation in the form y(x)= 1 h 1 (x) d dx g 2 (x)h 1 (x)Φ(x) g 1 (x)h 1 (x)+g 2 (x)h 2 (x) x a f(t) g 2 (t) t dt Φ(t) ,(16.1.2.8) where the function Φ(x) is given by (16.1.2.7). If f(x) ≡ const g 2 (x), the solution is given by formulas (16.1.2.8) and (16.1.2.7), in which the subscript 1 must be changed by 2 and vice versa. 2 ◦ . In the case g 1 (x)h 1 (x)+g 2 (x)h 2 (x) ≡ 0, the solution has the form y(x)= 1 h 1 d dx (f/g 2 ) x (g 1 /g 2 ) x =– 1 h 1 d dx (f/g 2 ) x (h 2 /h 1 ) x . 16.1.2-2. Equations with general degenerate kernel. A Volterra equation of the first kind with general degenerate kernel has the form n m=1 g m (x) x a h m (t)y(t) dt = f (x). (16.1.2.9) Using the notation w m (x)= x a h m (t)y(t) dt, m = 1, , n,(16.1.2.10) we can rewrite equation (16.1.2.9) as follows: n m=1 g m (x)w m (x)=f(x). (16.1.2.11) On differentiating formulas (16.1.2.10) and eliminating y(x) from the resulting equations, we arrive at the following linear differential equations for the functions w m = w m (x): h 1 (x)w m = h m (x)w 1 , m = 2, , n (16.1.2.12) 804 INTEGRAL EQUATIONS (the prime stands for the derivative with respect to x) with the initial conditions w m (a)=0, m = 1, , n. Any solution of system (16.1.2.11), (16.1.2.12) determines a solution of the original integral equation (16.1.2.9) by each of the expressions y(x)= w m (x) h m (x) , m = 1, , n, which can be obtained by differentiating formula (16.1.2.10). System (16.1.2.11), (16.1.2.12) can be reduced to a linear (n – 1)st-order differen- tial equation for any function w m (x)(m = 1, , n) by multiple differentiation of equa- tion (16.1.2.11) with regard to (16.1.2.12). 16.1.3. Equations with Difference Kernel: K(x, t) = K(x – t) 16.1.3-1. Solution method based on the Laplace transform. Volterra equations of the fi rst kind with kernel depending on the difference of the arguments have the form x 0 K(x – t)y(t) dt = f(x). (16.1.3.1) To solve these equations, the Laplace transform can be used (see Section 11.2). In what follows we need the transforms of the kernel and the right-hand side; they are given by the formulas K(p)= ∞ 0 K(x)e –px dx, f(p)= ∞ 0 f(x)e –px dx.(16.1.3.2) Applying the Laplace transform L to equation (16.1.3.1) and taking into account the fact that an integral with kernel depending on the difference of the arguments is transformed to the product by the rule (see Paragraph 11.2.2-1) L x 0 K(x – t)y(t) dt = K(p)y(p), we obtain the following equation for the transform y(p): K(p)y(p)= f(p). (16.1.3.3) The solution of equation (16.1.3.3) is given by the formula y(p)= f(p) K(p) .(16.1.3.4) On applying the Laplace inversion formula (if it is applicable) to (16.1.3.4), we obtain a solution of equation (16.1.3.1) in the form y(x)= 1 2πi c+i∞ c–i∞ f(p) K(p) e px dp.(16.1.3.5) To evaluate the corresponding integrals, tables of direct and inverse Laplace transforms can be applied(see Sections T3.1 andT3.2), and, in many cases, to find theinverse transform, 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). 16.1. LINEAR INTEGRAL EQUATIONS OF THE FIRST KIND WITH VARIABLE INTEGRATION LIMIT 805 16.1.3-2. Case in which the transform of the solution is a rational function. Consider the important special case in which the transform (16.1.3.4) of the solution is a rational function of the form y(p)= f(p) K(p) ≡ R(p) Q(p) , where Q(p)andR(p) are polynomials in the variable p and the degree of Q(p) exceeds that of R(p). If the zeros of the denominator Q(p) are simple, i.e., Q(p) ≡ const (p – λ 1 )(p – λ 2 ) (p – λ n ), and λ i ≠ λ j for i ≠ j, then the solution has the form y(x)= n k=1 R(λ k ) Q (λ k ) exp(λ k x), where the prime stands for the derivatives. Example 1. Consider the Volterra integral equation of the first kind x 0 e –a(x–t) y(t) dt = A sinh(bx). We apply the Laplace transform to this equation and obtain (see Subsections T3.1.1 and T3.1.4) 1 p + a y(p)= Ab p 2 – b 2 . This implies y(p)= Ab(p + a) p 2 – b 2 = Ab(p + a) (p – b)(p + b) . We have Q(p)=(p – b)(p + b), R(p)=Ab(p + a), λ 1 = b,andλ 2 =–b. Therefore, the solution of the integral equation has the form y(x)= 1 2 A(b + a)e bx + 1 2 A(b – a)e –bx = Aa sinh(bx)+Ab cosh(bx). 16.1.3-3. Convolution representation of a solution. In solving Volterra integral equations of the first kind with difference kernel K(x – t)by means of the Laplace transform, it is sometimes useful to apply the following approach. Let us represent the transform (16.1.3.4) of a solution in the form y(p)= N(p) M(p) f(p), N(p) ≡ 1 K(p) M(p) .(16.1.3.6) If we can find a function M(p) for which the inverse transforms L –1 M(p) = M (x), L –1 N(p) = N(x)(16.1.3.7) exist and can be found in a closed form, then the solution can be written as the convolution y(x)= x 0 N(x – t)F (t) dt, F(t)= t 0 M(t – s)f(s) ds.(16.1.3.8) 806 INTEGRAL EQUATIONS Example 2. Consider the equation x 0 sin k √ x – t y(t) dt = f(x), f(0)=0.(16.1.3.9) Applying the Laplace transform, we obtain (see Subsections T3.1.1 and T3.1.6) y(p)= 2 √ πk p 3/2 exp(α/p) f(p), α = 1 4 k 2 . (16.1.3.10) Let us rewrite the right-hand side of (16.1.3.10) in the equivalent form y(p)= 2 √ πk p 2 p –1/2 exp(α/p) f(p), α = 1 4 k 2 , (16.1.3.11) where the factor in the square brackets corresponds to M(p) in formula (16.1.3.6) and N(p) = const p 2 . By applying the Laplace inversion formula according to the above scheme to formula (16.1.3.11) with regard to the relation (see Subsections T3.2.1 and T3.2.5) L –1 p 2 ϕ(p) = d 2 dx 2 ϕ(x), L –1 p –1/2 exp(α/p) = 1 √ πx cosh k √ x , we find the solution y(x)= 2 πk d 2 dx 2 x 0 cosh k √ x – t √ x – t f(t) dt. 16.1.3-4. Application of an auxiliary equation. Consider the equation x a K(x – t)y(t) dt = f(x), (16.1.3.12) where the kernel K(x) has an integrable singularity at x = 0. Let w = w(x) be the solution of the simpler auxiliary equation with f(x) ≡ 1 and a = 0, x 0 K(x – t)w(t) dt = 1.(16.1.3.13) Then the solution of the original equation (16.1.3.12) with arbitrary right-hand side can be expressed as follows via the solution of the auxiliary equation (16.1.3.13): 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.(16.1.3.14) Example 3. Consider the generalized Abel equation x a y(t) dt (x – t) μ = f(x), 0 < μ < 1. (16.1.3.15) We seek a solution of the corresponding auxiliary equation x 0 w(t) dt (x – t) μ = 1, 0 < μ < 1, (16.1.3.16) by the method of indeterminate coefficients in the form w(x)=Ax β . (16.1.3.17) Let us substitute (16.1.3.17) into (16.1.3.16) and then perform the change of variable t = xξ in the integral. Taking into account the relationship B(p, q)= 1 0 ξ p–1 (1 – ξ) 1–q dξ = Γ(p)Γ(q) Γ(p + q) 16.1. LINEAR INTEGRAL EQUATIONS OF THE FIRST KIND WITH VARIABLE INTEGRATION LIMIT 807 between the beta and gamma functions, we obtain A Γ(β + 1)Γ(1 – μ) Γ(2 + β – μ) x β+1–μ = 1. From this relation we find the coefficients A and β: β = μ – 1, A = 1 Γ(μ)Γ(1 – μ) = sin(πμ) π . (16.1.3.18) Formulas (16.1.3.17) and (16.1.3.18) define the solution of the auxiliary equation (16.1.3.16) and make it possible to find the solution of the generalized Abel equation (16.1.3.15) by means of formula (16.1.3.14) as follows: y(x)= sin(πμ) π d dx x a f(t) dt (x – t) 1–μ = sin(πμ) π f(a) (x – a) 1–μ + x a f t (t) dt (x – t) 1–μ . 16.1.4. Reduction of Volterra Equations of the First Kind to Volterra Equations of the Second Kind 16.1.4-1. First method. Suppose that the kernel and the right-hand side of the equation x a K(x, t)y(t) dt = f (x)(16.1.4.1) have continuous derivatives with respect to x and that the condition K(x, x) 0 holds. In this case, after differentiating relation (16.1.4.1) and dividing the resulting expression by K(x, x), we arrive at the following Volterra equation of the second kind: y(x)+ x a K x (x, t) K(x, x) y(t) dt = f x (x) K(x, x) .(16.1.4.2) Equations of this type are considered in Section 16.2. If K(x, x) ≡ 0, then, on differentiating equation (16.1.4.1) with respect to x twice and assuming that K x (x, t)| t=x 0, we obtain the Volterra equation of the second kind y(x)+ x a K xx (x, t) K x (x, t)| t=x y(t) dt = f xx (x) K x (x, t)| t=x . If K x (x, x) ≡ 0, we can again apply differentiation, and so on. 16.1.4-2. Second method. Let us introduce the new variable Y (x)= x a y(t) dt and integrate the right-hand side of equation (16.1.4.1) by parts taking into account the relation f(a)=0. After dividing the resulting expression by K(x, x), we arrive at the Volterra equation of the second kind Y (x)– x a K t (x, t) K(x, x) Y (t) dt = f(x) K(x, x) , for which the condition K(x, x) 0 must hold. . T3.1 andT3.2), and, in many cases, to find theinverse transform, methods of the theory of functions of a complex variable are applied, including formulas for the calculation of residues and the. f (n–1) x (a)=0. For polar kernels of the form (16.1.1.4) and continuous f(x), no additional conditions are imposed on the right-hand side of the integral equation. 16.1.1-2. Existence and uniqueness of a. (16.1.3.4) of the solution is a rational function of the form y(p)= f(p) K(p) ≡ R(p) Q(p) , where Q(p)andR(p) are polynomials in the variable p and the degree of Q(p) exceeds that of R(p). If