864 INTEGRAL EQUATIONS For the case in which Φ(t, y) is a polynomial in y, i.e., Φ(t, y)=p 0 (t)+p 1 (t)y + ···+ p n (t)y n ,(16.5.3.7) where p 0 (t), , p n (t) are, for instance, continuous functions of t on the interval [a, b], system (16.5.3.6) becomes a system of nonlinear algebraic equations for A 1 , , A m . The number of solutions of the integral equation (16.5.3.3) is equal to the number of solu- tions of system (16.5.3.6). Each solution of system (16.5.3.6) generates a solution (16.5.3.5) of the integral equation. 2 ◦ . Consider the Urysohn equation with a degenerate kernel of the special form y(x)+ b a n k=1 g k (x)f k t, y(t) dt = h(x). (16.5.3.8) Its solution has the form y(x)=h(x)+ n k=1 λ k g k (x), (16.5.3.9) where the constants λ k can be defined by solving the algebraic (or transcendental) system of equations λ m + b a f m t, h(t)+ n k=1 λ k g k (t) dt = 0, m = 1, , n.(16.5.3.10) To different roots of this system, there are different corresponding solutions of the nonlinear integral equation. It may happen that (real) solutions are absent. A solution of the Urysohn equation with a degenerate kernel in the general form f x, y(x) + b a n k=1 g k x, y(x) h k t, y(t) dt = 0 (16.5.3.11) can be represented in the implicit form f x, y(x) + n k=1 λ k g k x, y(x) = 0,(16.5.3.12) where the parameters λ k are determined from the system of algebraic (or transcendental) equations: λ k – H k ( λ)=0, k = 1, , n, H k ( λ)= b a h k t, y(t) dt, λ = {λ 1 , , λ n }. (16.5.3.13) Into system (16.5.3.13), we must substitute the function y(x)=y(x, λ), which can be obtained by solving equation (16.5.3.12). The number of solutions of the integral equation is defined by the number of solutions obtained from (16.5.3.12) and (16.5.3.13). It can happen that there is no solution. 16.5. NONLINEAR INTEGRAL EQUATIONS 865 Example 1. Let us solve the integral equation y(x)=λ 1 0 xty 3 (t) dt (16.5.3.14) with parameter λ. We write A = 1 0 ty 3 (t) dt. (16.5.3.15) In this case, it follows from (16.5.3.14) that y(x)=λAx. (16.5.3.16) On substituting y(x) in the form (16.5.3.16) into relation (16.5.3.15), we obtain A = 1 0 tλ 3 A 3 t 3 dt. Hence, A = 1 5 λ 3 A 3 . (16.5.3.17) For λ > 0, equation (16.5.3.17) has three solutions: A 1 = 0, A 2 = 5 λ 3 1/2 , A 3 =– 5 λ 3 1/2 . Hence, the integral equation (16.5.3.14) also has three solutions for any λ > 0: y 1 (x) ≡ 0, y 2 (x)= 5 λ 3 1/2 x, y 3 (x)=– 5 λ 3 1/2 x. For λ ≤ 0, equation (16.5.3.17) has only the trivial solution y(x) ≡ 0. 16.5.3-2. Method of integral transforms. 1 ◦ . Consider the following nonlinear integral equation with quadratic nonlinearity on a semiaxis: μy(x)–λ ∞ 0 1 t y x t y(t) dt = f (x). (16.5.3.18) To solve this equation, the Mellin transform can be applied, which, with regard to the convolution theorem (see Section 11.3), leads to a quadratic equation for the transform ˆy(s)=M{y(x)}: μ ˆy(s)–λ ˆy 2 (s)= ˆ f(s). This implies ˆy(s)= μ μ 2 – 4λ ˆ f(s) 2λ .(16.5.3.19) The inverse transform y(x)=M –1 { ˆy(s)} obtained by means of the Mellin inversion formula (if it exists) is a solution of equation (16.5.3.18). To different signs in the formula for the images (16.5.3.19), there are two corresponding solutions of the original equation. 2 ◦ . By applying the Mellin transform, one can solve nonlinear integral equations of the form y(x)–λ ∞ 0 t β y(xt)y(t) dt = f (x). (16.5.3.20) The Mellin transform (see Table 11.3 in Section 11.3) reduces (16.5.3.20) to the following functional equation for the transform ˆy(s)=M{y(x)}: ˆy(s)–λ ˆy(s)ˆy(1 – s + β)= ˆ f(s). (16.5.3.21) 866 INTEGRAL EQUATIONS On replacing s by 1 – s + β in (16.5.3.21), we obtain the relationship ˆy(1 – s + β)–λ ˆy(s)ˆy(1 – s + β)= ˆ f(1 – s + β). (16.5.3.22) On eliminating the quadratic term from (16.5.3.21) and (16.5.3.22), we obtain ˆy(s)– ˆ f(s)= ˆy(1 – s + β)– ˆ f(1 – s + β). (16.5.3.23) We express ˆy(1 – s + β) from this relation and substitute it into (16.5.3.21). We arrive at the quadratic equation λ ˆy 2 (s)– 1 + ˆ f(s)– ˆ f(1 – s + β) ˆy(s)+ ˆ f(s)=0.(16.5.3.24) On solving (16.5.3.24) for ˆy(s), by means of the Mellin inversion formula we can find a solution of the original integral equation (16.5.3.20). 16.5.3-3. Method of differentiating for integral equations. 1 ◦ . Consider the equation y(x)+ b a |x – t|f t, y(t) dt = g(x). (16.5.3.25) Let us remove the modulus in the integrand: y(x)+ x a (x – t)f t, y(t) dt + b x (t – x)f t, y(t) dt = g(x). (16.5.3.26) Differentiating (16.5.3.26) with respect to x yields y x (x)+ x a f t, y(t) dt – b x f t, y(t) dt = g x (x). (16.5.3.27) Differentiating (16.5.3.27), we arrive at a second-order ordinary differential equation for y = y(x): y xx + 2f (x, y)=g xx (x). (16.5.3.28) Let us derive the boundary conditions for equation (16.5.3.28). We assume that –∞ < a < b < ∞. By setting x = a and x = b in (16.5.3.26), we obtain the relations y(a)+ b a (t – a)f t, y(t) dt = g(a), y(b)+ b a (b – t)f t, y(t) dt = g(b). (16.5.3.29) Let us solve equation (16.5.3.28) for f (x, y) and substitute the result into (16.5.3.29). Integrating by parts yields the desired boundary conditions for y(x): y(a)+y(b)+(b – a) g x (b)–y x (b) = g(a)+g(b), y(a)+y(b)+(a – b) g x (a)–y x (a) = g(a)+g(b). (16.5.3.30) Let us point out a useful consequence of (16.5.3.30): y x (a)+y x (b)=g x (a)+g x (b), which can be used together with one of conditions (16.5.3.30). Equation (16.5.3.28) under the boundary conditions (16.5.3.30) determines the solution of the original integral equation (there may be several solutions). 16.5. NONLINEAR INTEGRAL EQUATIONS 867 2 ◦ . The equations y(x)+ b a e λ|x–t| f t, y(t) dt = g(x), y(x)+ b a sinh λ|x – t| f t, y(t) dt = g(x), y(x)+ b a sin λ|x – t| f t, y(t) dt = g(x), can also be reduced to second-order ordinary differential equations by means of the differ- entiation. For these equations, see the book by Polyanin and Manzhirov (1998). 16.5.3-4. Successive approximation method. Consider the nonlinear Urysohn integral equation in the canonical form y(x)= b a K x, t, y(t) dt, a ≤ x ≤ b.(16.5.3.31) The iteration process for this equation is constructed by the formula y k (x)= b a K x, t, y k–1 (t) dt, k = 1, 2, (16.5.3.32) If the function K(x, t, y) is jointly continuous together with the derivative K y (x, t, y) (with respect to the variables x, t,andρ, a ≤ x ≤ b, a ≤ t ≤ b,and|y| ≤ ρ)andif b a sup y |K(x, t, y)| dt ≤ ρ, b a sup y |K y (x, t, y)| dt ≤ β < 1,(16.5.3.33) then for any continuous function y 0 (x) of the initial approximation from the domain {|y| ≤ ρ, a ≤ x ≤ b}, the successive approximations (16.5.3.32) converge to a continu- ous solution y ∗ (x), which lies in the same domain and is unique in this domain. The rate of convergence is defined by the inequality |y ∗ (x)–y k (x)| ≤ β k 1 – β sup x |y 1 (x)–y 0 (x)|, a ≤ x ≤ b,(16.5.3.34) which gives an aprioriestimate for the error of the kth approximation. The a posteriori estimate (which is, in general, more precise) has the form |y ∗ (x)–y k (x)| ≤ β 1 – β sup x |y k (x)–y k–1 (x)|, a ≤ x ≤ b.(16.5.3.35) A solution of an equation of the form (16.5.3.31) with an additional term f(x)onthe right-hand side can be constructed in a similar manner. 868 INTEGRAL EQUATIONS Example 2. Let us apply the successive approximation method to solve the equation y(x)= 1 0 xty 2 (t) dt – 5 12 x + 1. The recurrent formula has the form y k (x)= 1 0 xty 2 k–1 (t) dt – 5 12 x + 1, k = 1, 2, For the initial approximation we take y 0 (x)=1. The calculation yields y 1 (x)=1 + 0.083 x, y 8 (x)=1 + 0.27 x, y 16 (x)=1 + 0.318 x, y 2 (x)=1 + 0.14 x, y 9 (x)=1 + 0.26 x, y 17 (x)=1 + 0.321 x, y 3 (x)=1 + 0.18 x, y 10 (x)=1 + 0.29 x, y 18 (x)=1 + 0.323 x, Thus, the approximations tend to the exact solution y(x)=1 + 1 3 x. We see that the rate of convergence of the iteration process is fairly small. Note that in Paragraph 16.5.3-5, the equation in question is solved by a more efficient method. 16.5.3-5. Newton–Kantorovich method. We consider the Newton–Kantorovich method in connection with the Urysohn equation in the canonical form (16.5.3.31). The iteration process is constructed as follows: y k (x)=y k–1 (x)+ϕ k–1 (x), k = 1, 2, ,(16.5.3.36) ϕ k–1 (x)=ε k–1 (x)+ b a K y x, t, y k–1 (t) ϕ k–1 (t) dt,(16.5.3.37) ε k–1 (x)= b a K x, t, y k–1 (t) dt – y k–1 (x). (16.5.3.38) At each step of the algorithm, a linear integral equation for the correction ϕ k–1 (x) is solved. Under some conditions, the process (16.5.3.36) has a high rate of convergence; however, it is rather complicated because at each iteration we must obtain the new kernel K y x, t, y k–1 (t) for equations (16.5.3.37). The algorithm can be simplified by using the equation ϕ k–1 (x)=ε k–1 (x)+ b a K y x, t, y 0 (t) ϕ k–1 (t) dt (16.5.3.39) instead of (16.5.3.37). If the initial approximation is chosen successfully, then the differ- ence between the integral operators in (16.5.3.37) and (16.5.3.39) is small, and the kernel in (16.5.3.39) remains the same in the course of the solution. The successive approximation method that consists of the application of formulas (16.5.3.36), (16.5.3.38), and (16.5.3.39) is called the modified Newton–Kantorovich method. In principle, its rate of convergence is less than that of the original (unmodified) method; however, this version of the method is less complicated in calculations, and therefore it is frequently preferable. Let the function K(x, t, y) be jointly continuous together with the derivatives K y (x, t, y) and K yy (x, t, y) with respect to the variables x, t, y,wherea ≤ x ≤ b and a ≤ t ≤ b,andlet the following conditions hold: 16.5. NONLINEAR INTEGRAL EQUATIONS 869 1 ◦ . For the initial approximation y 0 (x), the resolvent R(x, t) of the linear integral equa- tion (16.5.3.37) with the kernel K y x, t, y 0 (t) satisfies the condition b a |R(x, t)| dt ≤ A < ∞, a ≤ x ≤ b. 2 ◦ . The residual ε 0 (x) of equation (16.5.3.38) for the approximation y 0 (x) satisfies the inequality |ε 0 (x)| = b a K x, t, y 0 (t) dt – y 0 (x) ≤ B < ∞. 3 ◦ . In the domain |y(x)–y 0 (x)| ≤ 2(1 + A)B, the following relation holds: b a sup y K yy (x, t, y) dt ≤ D < ∞. 4 ◦ . The constants A, B,andD satisfy the condition H =(1 + A) 2 BD ≤ 1 2 . In this case, under assumptions 1 ◦ – 4 ◦ , the process (16.5.3.36) converges to a solution y ∗ (x) of equation (16.5.3.31) in the domain |y(x)–y 0 (x)| ≤ (1 – √ 1 – 2H)H –1 (1 – A)B, a ≤ x ≤ b. This solution is unique in the domain |y(x)–y 0 (x)| ≤ 2(1 + A)B, a ≤ x ≤ b. The rate of convergence is determined by the estimate |y ∗ (x)–y k (x)| ≤ 2 1–k (2H) 2 k –1 (1 – A)B, a ≤ x ≤ b. Thus, the above conditions establish the convergence of the algorithm and the existence, the position, and the uniqueness domain of a solution of the nonlinear equation (16.5.3.31). These conditions impose certain restrictions on the initial approximation y 0 (x) whose choice is an important independent problem that has no unified approach. As usual, the initial approximation is determined either by more detailed apriorianalysis of the equation under consideration or by physical reasoning implied by the essence of the problem described by this equation. Under a successful choice of the initial approximation, the Newton– Kantorovich method provides a high rate of convergence of the iteration process to obtain an approximate solution with given accuracy. Remark. Let the right-hand side of equation (16.5.3.31) contain an additional term f(x). Then such an equation can be represented in the form (16.5.3.31), where the integrand is K x, t, y(t) +(b – a) –1 f(x). Example 3. Let us apply the Newton–Kantorovich method to solve the equation y(x)= 1 0 xty 2 (t) dt – 5 12 x + 1. (16.5.3.40) For the initial approximation we take y 0 (x)=1. According to (16.5.3.38), we find the residual ε 0 (x)= 1 0 xty 2 0 (t) dt – 5 12 x + 1 – y 0 (x)=x 1 0 tdt– 5 12 x + 1 – 1 = 1 12 x. 870 INTEGRAL EQUATIONS The y-derivative of the kernel K(x, t, y)=xty 2 (t), which is needed in the calculations, has the form K y (x, t, y)= 2xty(t). According to (16.5.3.37), we form the following equation for ϕ 0 (x): ϕ 0 (x)= 1 12 x + 2x 1 0 ty 0 (t)ϕ 0 (t) dt, where the kernel turns out to be degenerate, which makes it possible to obtain the solution ϕ 0 (x)= 1 4 x directly. Now we define the first approximation to the desired function: y 1 (x)=y 0 (x)+ϕ 0 (x)=1 + 1 4 x. We continue the iteration process and obtain ε 1 (x)= 1 0 xt 1 + 1 4 t dt + 1 – 5 12 x – 1 + 1 4 x = 1 64 x. The equation for ϕ 1 (x)hastheform ϕ 1 (x)= 1 64 x + 2x 1 0 t 1 + 1 4 t dt + 1 – 5 12 x – 1 + 1 4 x , and the solution is ϕ 1 (x)= 3 40 x. Hence, y 2 (x)=1 + 1 4 x + 3 40 x = 1 + 0.325 x. The maximal difference between the exact solution y(x)=1 + 1 3 x and the approximate solution y 2 (x) is observed at x = 1 and is less than 0.5%. This solution is not unique. The other solution can be obtained by taking the function y 0 (x)=1 + 0.8 x for the initial approximation. In this case we can repeat the above sequence of approximations and obtain the following results (the numerical coefficient of x is rounded): y 1 (x)=1 + 0.82 x, y 2 (x)=1 + 1.13 x, y 3 (x)=1 + 0.98 x, , and the subsequent approximations tend to the exact solution y(x)=1 + x. We see that the rate of convergence of the iteration process performed by the Newton–Kantorovich method is significantly higher than that performed by the method of successive approximations (see Example 2 in Paragraph 16.5.3-4). To estimate the rate of convergence of the performed iteration process, we can compare the above results with the realization of the modified Newton–Kantorovich method. In connection with the latter, for the above versions of the approximations we can obtain y n (x)=1 + k n x; k 0 k 1 k 2 k 3 k 4 k 5 k 6 k 7 k 8 0 0.25 0.69 0.60 0.51 0.44 0.38 0.36 0.345 . The iteration process converges to the exact solution y(x)=1 + 1 3 x. We see that the modified Newton–Kantorovich method is less efficient than the Newton–Kantorovich method, but more efficient than the method of successive approximations (see Example 2). 16.5.3-6. Quadrature method. To solve an arbitrary nonlinear equation, we can apply the method based on the application of quadrature formulas. The procedure of composing the approximating system of equations is the same as in the linear case (see Subsection 16.4.11). We consider this procedure for an example of the Urysohn equation y(x)– b a K x, t, y(t) dt = f(x), a ≤ x ≤ b.(16.5.3.41) We set x = x i (i = 1, , n). Then we obtain y(x i )– b a K x i , t, y(t) dt = f(x i ). i = 1, , n.(16.5.3.42) . system of nonlinear algebraic equations for A 1 , , A m . The number of solutions of the integral equation (16.5.3.3) is equal to the number of solu- tions of system (16.5.3.6). Each solution of. 4λ ˆ f(s) 2λ .(16.5.3.19) The inverse transform y(x)=M –1 { ˆy(s)} obtained by means of the Mellin inversion formula (if it exists) is a solution of equation (16.5.3.18). To different signs in the formula for the images. solving (16.5.3.24) for ˆy(s), by means of the Mellin inversion formula we can find a solution of the original integral equation (16.5.3.20). 16.5.3-3. Method of differentiating for integral equations. 1 ◦ .