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16.5. NONLINEAR INTEGRAL EQUATIONS 857 Example 1. Consider the integral equation x 0 y(x – t)y(t) dt = Ax m , m >–1. Applying the Laplace transform to the equation under consideration with regard to the relation L{x m } = Γ(m + 1)p –m–1 , we obtain y 2 (p)=AΓ(m + 1)p –m–1 , where Γ(m) is the gamma function. On extracting the square root of both sides of the equation, we obtain y(p)= AΓ(m + 1)p – m+1 2 . Applying the Laplace inversion formula, we obtain two solutions to the original integral equation: y 1 (x)=– √ AΓ(m + 1) Γ m + 1 2 x m–1 2 , y 2 (x)= √ AΓ(m + 1) Γ m + 1 2 x m–1 2 . 16.5.2-2. Method of differentiation for integral equations. Sometimes, differentiation (possibly multiple) of a nonlinear integral equation with subse- quent elimination of the integral terms by means of the original equation makes it possible to reduce this equation to a nonlinear ordinary differential equation. Below we briefly list some equations of this type. 1 ◦ . The equation y(x)+ x a f t, y(t) dt = g(x)(16.5.2.3) can be reduced by differentiation to the nonlinear first-order equation y x + f(x, y)–g x (x)=0 with the initial condition y(a)=g(a). 2 ◦ . The equation y(x)+ x a (x – t)f t, y(t) dt = g(x)(16.5.2.4) can be reduced by double differentiation (with the subsequent elimination of the integral term by using the original equation) to the nonlinear second-order equation: y xx + f(x, y)–g xx (x)=0.(16.5.2.5) The initial conditions for the function y = y(x) have the form y(a)=g(a), y x (a)=g x (a). (16.5.2.6) 3 ◦ . The equation y(x)+ x a e λ(x–t) f t, y(t) dt = g(x)(16.5.2.7) can be reduced by differentiation to the nonlinear first-order equation y x + f(x, y)–λy + λg(x)–g x (x)=0.(16.5.2.8) The desired function y = y(x) must satisfy the initial condition y(a)=g(a). 858 INTEGRAL EQUATIONS 4 ◦ . Equations of the form y(x)+ x a cosh λ(x – t) f t, y(t) dt = g(x), (16.5.2.9) y(x)+ x a sinh λ(x – t) f t, y(t) dt = g(x), (16.5.2.10) y(x)+ x a cos λ(x – t) f t, y(t) dt = g(x), (16.5.2.11) y(x)+ x a sin λ(x – t) f t, y(t) dt = g(x)(16.5.2.12) can also be reduced to second-order ordinary differential equations by double differentiation. For these equations, see the book by Polyanin and Manzhirov (1998). 16.5.2-3. Successive approximation method. In many cases, the successive approximation method can be applied successfully to solve various types of integral equations. The principles of constructing the iteration process are the same as in the case of linear equations. For Volterra equations of the form y(x)– x a K x, t, y(t) dt = f(x), a ≤ x ≤ b,(16.5.2.13) the corresponding recurrent expression has the form y k+1 (x)=f(x)+ x a K x, t, y k (t) dt, k = 0, 1, 2, (16.5.2.14) It is customary to take the initial approximation either in the form y 0 (x) ≡ 0 or in the form y 0 (x)=f(x). In contrast to the case of linear equations, the successive approximation method has a smaller domain of convergence. Let us present the convergence conditions for the iteration process (16.5.2.14) that are simultaneously the existence conditions for a solution of equation (16.5.2.13). To be definite, we assume that y 0 (x)=f(x). If for any z 1 and z 2 we have the relation |K(x, t, z 1 )–K(x, t, z 2 )| ≤ ϕ(x, t)|z 1 – z 2 | and the relation x a K x, t, f(t) dt ≤ ψ(x) holds, where x a ψ 2 (t) dt ≤ N 2 , b a x a ϕ 2 (x, t) dt dx ≤ M 2 , for some constants N and M, then the successive approximations converge to a unique solution of equation (16.5.2.13) almost everywhere absolutely and uniformly. 16.5. NONLINEAR INTEGRAL EQUATIONS 859 Example 2. Let us apply the successive approximation method to solve the equation y(x)= x 0 1 + y 2 (t) 1 + t 2 dt. If y 0 (x) ≡ 0,then y 1 (x)= x 0 dt 1 + t 2 =arctanx, y 2 (x)= x 0 1 +arctan 2 t 1 + t 2 dt =arctanx + 1 3 arctan 3 x, y 3 (x)= x 0 1 +arctant + 1 3 arctan 3 t 1 + t 2 dt =arctanx + 1 3 arctan 3 x + 2 3⋅5 arctan 5 x + 1 7⋅9 arctan 7 x. On continuing this process, we can observe that y k (x) → tan(arctan x)=x as k →∞, i.e., y(x)=x.The substitution of this result into the original equation shows the validity of the result. 16.5.2-4. Newton–Kantorovich method. A merit of the iteration methods when applied to Volterra linear equations of the second kind is their unconditional convergence under weak restrictions on the kernel and the right-hand side. When solving nonlinear equations, the applicability domain of the method of simple iterations is smaller, and if the process is still convergent, then, in many cases, the rate of convergence can be very low. An effective method that makes it possible to overcome the indicated complications is the Newton–Kantorovich method. Let us apply the Newton–Kantorovich method to solve a nonlinear Volterra equation of the form y(x)=f(x)+ x a K x, t, y(t) dt.(16.5.2.15) We obtain the following iteration process: y k (x)=y k–1 (x)+ϕ k–1 (x), k = 1, 2, ,(16.5.2.16) ϕ k–1 (x)=ε k–1 (x)+ x a K y x, t, y k–1 (t) ϕ k–1 (t) dt,(16.5.2.17) ε k–1 (x)=f(x)+ x a K x, t, y k–1 (t) dt – y k–1 (x). (16.5.2.18) The algorithm is based on the solution of the linear integral equation (16.5.2.17) for the correction ϕ k–1 (x) with the kernel and right-hand side that vary from step to step. This process has a high rate of convergence, but it is rather complicated because we must solve a new equation at each step of iteration. To simplify the problem, we can replace equation (16.5.2.17) with the equation ϕ k–1 (x)=ε k–1 (x)+ x a K y x, t, y 0 (t) ϕ k–1 (t) dt (16.5.2.19) or with the equation ϕ k–1 (x)=ε k–1 (x)+ x a K y x, t, y m (t) ϕ k–1 (t) dt,(16.5.2.20) whose kernels do not vary. In equation (16.5.2.20), m is fixed and satisfies the condition m < k – 1. 860 INTEGRAL EQUATIONS It is reasonable to apply equation (16.5.2.19) with an appropriately chosen initial ap- proximation. Otherwise we can stop at some mth approximation and, beginning with this approximation, apply the simplified equation (16.5.2.20). The iteration process thus ob- tained is the modified Newton–Kantorovich method. In principle, it converges somewhat slower than the original process (16.5.2.16)–(16.5.2.18); however, it is not so cumbersome in the calculations. Example 3. Let us apply the Newton–Kantorovich method to solve the equation y(x)= x 0 [ty 2 (t)–1] dt. The derivative of the integrand with respect to y has the form K y t, y(t) = 2ty(t). For the zero approximation we take y 0 (x) ≡ 0. According to (16.5.2.17) and (16.5.2.18) we obtain ϕ 0 (x)=–x and y 1 (x)=–x.Furthermore,y 2 (x)=y 1 (x)+ϕ 1 (x). By (16.5.2.18) we have ε 1 (x)= x 0 [t(–t) 2 – 1] dt + x = 1 4 x 4 . The equation for the correction has the form ϕ 1 (x)=–2 x 0 t 2 ϕ 1 (t) dt + 1 4 x 4 and can be solved by any of the known methods for Volterra linear equations of the second kind. In the case under consideration, we apply the successive approximation method, which leads to the following results (the number of the step is indicated in the superscript): ϕ (0) 1 = 1 4 x 4 , ϕ (1) 1 = 1 4 x 4 – 2 x 0 1 4 t 6 dt = 1 4 x 4 – 1 14 x 7 , ϕ (2) 1 = 1 4 x 4 – 2 x 0 t 2 1 4 t 4 – 1 14 x 7 dt = 1 4 x 4 – 1 14 x 7 + 1 70 x 10 . We restrict ourselves to the second approximation and obtain y 2 (x)=–x + 1 4 x 4 – 1 14 x 7 + 1 70 x 10 and then pass to the third iteration step of the Newton–Kantorovich method: y 3 (x)=y 2 (x)+ϕ 2 (x), ε 2 (x)= 1 160 x 10 – 1 1820 x 13 – 1 7840 x 16 + 1 9340 x 19 + 1 107800 x 22 , ϕ 2 (x)=ε 2 (x)+2 x 0 t –t + 1 4 t 4 – 1 14 t 7 + 1 70 t 10 ϕ 2 (t) dt. When solving the last equation, we restrict ourselves to the zero approximation and obtain y 3 (x)=–x + 1 4 x 4 – 1 14 x 7 + 23 112 x 10 – 1 1820 x 13 – 1 7840 x 16 + 1 9340 x 19 + 1 107800 x 22 . The application of the successive approximation method to the original equation leads to the same result at the fourth step. As usual, in the numerical solution the integral is replaced by a quadrature formula. The main difficulty of the implementation of the method in this case is in evaluating the derivative of the kernel. The problem can be simplified if the kernel is given as an analytic expression that can be differentiated in the analytic form. However, if the kernel is given by a table, then the evaluation must be performed numerically. 16.5. NONLINEAR INTEGRAL EQUATIONS 861 16.5.2-5. Collocation method. When applied to the solution of a nonlinear Volterra equation x a K x, t, y(t) dt = f (x), a ≤ x ≤ b,(16.5.2.21) the collocation method is as follows. The interval [a, b] is divided into N parts on each of which the desired solution can be presented by a function of a certain form y(x)=Φ(x, A 1 , , A m ), (16.5.2.22) involving free parameters A i , i = 1, , m. On the (k +1)st part x k ≤ x ≤ x k+1 ,wherek = 0, 1, , N – 1, the solution can be written in the form x x k K x, t, y(t) dt = f (x)–Ψ k (x), (16.5.2.23) where the integral Ψ k (x)= x k a K x, t, y(t) dt (16.5.2.24) can always be calculated for the approximate solution y(x), which is known on the interval a ≤ x ≤ x k and was previously obtained for k – 1 parts. The initial value y(a) of the desired solution can be found by an auxiliary method or is assumed to be given. To solve equation (16.5.2.23), representation (16.5.2.22) is applied, and the free param- eters A i (i = 1, , m) can be defined from the condition that the residuals vanish: ε(A i , x k,j )= x k,j x k K x k,j , t, Φ(t, A 1 , , A m ) dt – f(x k,j )–Ψ k (x k,j ), (16.5.2.25) where x k,j (j = 1, , m) are the nodes that correspond to the partition of the interval [x k , x k+1 ]intom parts (subintervals). System (16.5.2.25) is a system of m equations for A 1 , , A m . For convenience of the calculations, it is reasonable to present the desired solution on any part as a polynomial y(x)= m i=1 A i ϕ i (x), (16.5.2.26) where ϕ i (x) are linearly independent coordinate functions. For the functions ϕ i (x), power and trigonometric polynomials are frequently used; for instance, ϕ i (x)=x i–1 . In applications, the concrete form of the functions ϕ i (x) in formula (16.5.2.26), as well as the form of the functions Φ in (16.5.2.20), can sometimes be given on the basis of physical reasoning or defined by the structure of the solution of a simpler model equation. 16.5.2-6. Quadrature method. To solve a nonlinear Volterra equation, we can apply the method based on the use of quadrature formulas. The procedure of constructing the approximate system of equations is the same as in the linear case (see Subsection 16.2.7). 862 INTEGRAL EQUATIONS 1 ◦ . We consider the nonlinear Volterra equation of the form y(x)– x a K x, t, y(t) dt = f (x)(16.5.2.27) on an interval a ≤ x ≤ b. Assume that K x, t, y(t) and f (x) are continuous functions. From equation (16.5.2.27) we find that y(a)=f(a). Let us choose a constant integration step h and consider the discrete set of points x i = a + h(i – 1), where i = 1, , n.For x = x i , equation (16.5.2.27) becomes y(x i )– x i a K x i , t, y(t) dt = f (x i ). (16.5.2.28) Applying the quadrature formula (see Subsection 16.1.5) to the integral in (16.5.2.28), choosing x j (j = 1, , i) to be the nodes in t, and neglecting the truncation error, we arrive at the following system of nonlinear algebraic (or transcendental) equations: y 1 = f 1 , y i – i j=1 A ij K ij (y j )=f i , i = 2, , n,(16.5.2.29) where A ij are the coefficients of the quadrature formula on the interval [a, x i ]; y i are the approximate values of the solution y(x) at the nodes x i ; f i = f(x i ); and K ij (y j )= K(x i , t j , y j ). Relations (16.5.2.29) can be rewritten as a sequence of recurrent nonlinear equations, y 1 = f 1 , y i – A ii K ii (y i )=f i + i–1 j=1 A ij K ij (y j ), i = 2, , n,(16.5.2.30) for the approximate values of the desired solution at the nodes. 2 ◦ . When applied to the Volterra equation of the second kind in the Hammerstein form y(x)– x a Q(x, t)Φ t, y(t) dt = f (x), (16.5.2.31) the main relations of the quadrature method have the form (x 1 = a) y 1 = f 1 , y i – i j=1 A ij Q ij Φ j (y j )=f i , i = 2, , n,(16.5.2.32) where Q ij = Q(x i , t j )andΦ j (y j )=Φ(t j , y j ). These relations lead to the sequence of nonlinear recurrent equations y 1 = f 1 , y i – A ii Q ii Φ i (y i )=f i + i–1 j=1 A ij Q ij Φ j (y j ), i = 2, , n,(16.5.2.33) whose solutions give approximate values of the desired function. 16.5. NONLINEAR INTEGRAL EQUATIONS 863 Example 4. In the solution of the equation y(x)– x 0 e –(x–t) y 2 (t) dt = e –x , 0 ≤ x ≤ 0.1, where Q(x, t)=e –(x–t) , Φ t, y(t) = y 2 (t), and f(x)=e –x , the approximate expression has the form y(x i )– x i 0 e –(x i –t) y 2 (t) dt = e –x i . On applying the trapezoidal rule to evaluate the integral (with step h = 0.02)andfinding the solution at the nodes x i = 0, 0.02, 0.04, 0.06, 0.08, 0.1, we obtain, according to (16.5.2.33), the following system of computational relations: y 1 = f 1 , y i – 0.01 Q ii y 2 i = f i + i–1 j=1 0.02 Q ij y 2 j , i = 2, , 6. Thus, to find an approximate solution, we must solve a quadratic equation for each value y i ,whichmakesit possible to write out the answer y i = 50 50 1 – 0.04 f i + i–1 j=1 0.02 Q ij y 2 j 1/2 , i = 2, , 6. 16.5.3. Equations with Constant Integration Limits 16.5.3-1. Nonlinear equations with degenerate kernels. 1 ◦ . Consider a Urysohn equation of the form y(x)= b a Q(x, t)Φ t, y(t) dt,(16.5.3.1) where Q(x, t)andΦ(t, y) are given functions and y(x) is the unknown function. Let the kernel Q(x, t) be degenerate, i.e., Q(x, t)= m k=1 g k (x)h k (t). (16.5.3.2) In this case equation (16.5.3.1) becomes y(x)= m k=1 g k (x) b a h k (t)Φ t, y(t) dt.(16.5.3.3) We write A k = b a h k (t)Φ t, y(t) dt, k = 1, , m,(16.5.3.4) where the constants A k are yet unknown. Then it follows from (16.5.3.3) that y(x)= m k=1 A k g k (x). (16.5.3.5) On substituting the expression (16.5.3.5) for y(x) into relations (16.5.3.4), we obtain (in the general case) m transcendental equations of the form A k = Ψ k (A 1 , , A m ), k = 1, , m,(16.5.3.6) which contain m unknown numbers A 1 , , A m . . that correspond to the partition of the interval [x k , x k+1 ]intom parts (subintervals). System (16.5.2.25) is a system of m equations for A 1 , , A m . For convenience of the calculations, it. derivative of the integrand with respect to y has the form K y t, y(t) = 2ty(t). For the zero approximation we take y 0 (x) ≡ 0. According to (16.5.2.17) and (16.5.2.18) we obtain ϕ 0 (x)=–x and. = 1 4 x 4 . The equation for the correction has the form ϕ 1 (x)=–2 x 0 t 2 ϕ 1 (t) dt + 1 4 x 4 and can be solved by any of the known methods for Volterra linear equations of the second kind. In
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