Numerical Solution Methods for a Nonlinear Operator Equation Arising in an Inverse Coefficient Problem

We consider the inverse problem of determining two unknown coefficients in a linear system of partial differential equations using additional information about one of the solution components. The problem is reduced to a nonlinear operator equation for one of the unknown coefficients. The successive approximation method and the Newton method are used to solve this operator equation numerically. Results of calculations illustrating the convergence of numerical methods for solving the inverse problem are presented.


INTRODUCTION
Consider the problem of determining functions u(x, t) and a(x, t) such that This problem is a mathematical model of the sorption dynamic process [1, p. 174; 2, p. 6] under the assumption that the absorbent properties change over time.
The existence and uniqueness of the solution of the following inverse problem were studied in [3]. Assume that the functions µ(t) and ψ(x) are given and the functions γ(t) and ϕ(t) are unknown. It is required to determine γ(t), ϕ(t), u(x, t), and a(x, t) from the following additional information about one of the components of the solution to problem (1.1)-(1.4): We will assume that the given functions µ(t), ψ(x), g(t), and p(t) satisfy the following conditions.
Here are some results in [3] to be used in what follows.
Given a function γ(t), consider the integral equation for the function u(x, t), where Under Conditions A, there exists a unique solution of the integral equation (1.7). To emphasize the dependence of this solution on the function γ(t), we denote it by u(x, t; γ). Let us define the operator It was shown in [3] that solving the inverse problem is reduced to solving the nonlinear operator equation The present paper deals with numerical methods for solving Eq. (1.9) and the above-stated inverse problem. We use two iterative methods, the successive approximation method and the Newton method, to solve the nonlinear operator equation (1.9). There are quite a few papers dealing with iterative methods for solving operator equations to which inverse problems can be reduced (see, e.g., [4][5][6][7][8][9][10][11]). Many of them are determined by the peculiarities and specific features of each specific inverse problem.
Consider the successive approximation method for the operator equation (1.9). Assume that the inequality is satisfied. Define the positive constant Let us introduce the set of functions Consider the sequence of functions γ n (t), n = 0, 1, 2, . . ., recursively defined by the successive approximation method for solving Eq. (1.9), The results in [3] imply an assertion about the convergence of the successive approximation method.
Theorem 1. Let the functions µ(t), ψ(x), g(t), and p(t) satisfy conditions A and inequality (1.10). Then there exists a t 0 ∈ (0, T ] such that, for each function γ 0 (t) ∈ Γ 0 , the sequence of functions γ n (t) belongs to the set Γ 0 and uniformly converges as n → ∞ to a continuous functionγ(t) that is a solution of Eq. (1.9).

NEWTON METHOD
Consider the application of the Newton method to the numerical solution of the inverse problem under study. As was already noted, it suffices to apply the Newton method to solve the nonlinear operator equation (1.9).
To construct the Newton method, one needs to know the derivative of the operator defined by formula (1.8). First, let us study the differentiability of the solution of the integral equation (1.7) with respect to the parameter.
Let us study the Gateaux differentiability of the operator A defined in (1.8). It follows from the results in [3] that there exists a t 0 ∈ (0, T ] such that A maps the set Γ 0 into itself. In what follows, we assume that the number t 0 satisfies this condition.
We introduce the set Theorem 2. If conditions A and inequality (1.10) are satisfied, then, for each function γ ∈ Γ 00 , the operator A is Gateaux differentiable on Γ 00 .
Proof. Let γ(t) be an arbitrary function in Γ 00 , and let γ ∆ (t) be a function and ξ a number such that the function γ(t) + ξγ ∆ (t) belongs to the set Γ 00 as well. Let us show that the functions ((A(γ + ξγ ∆ ))(t) − (Aγ)(t))/ξ converge uniformly on the interval [0, t 0 ] as ξ → 0. They uniformly converge on the interval [0, t 0 ] to the function where the function w(x, t; γ, γ ∆ ) is determined from Eq. (2.4). Thus, the operator A is Gateaux differentiable on the set Γ 00 , and its derivative A [γ]γ ∆ is determined by the right-hand side of Eq. (2.7). The proof of the theorem is complete. The iterative process corresponding to the Newton method [12, p. 669] is defined as follows. One specifies the function γ 0 (t). The subsequent functions γ n+1 (t), n = 0, 1, . . ., are determined by the formula γ n+1 (t) = γ n (t) + γ ∆n (t), where γ ∆n (t) is the solution of the linear integral equation (2.8)

NUMERICAL EXPERIMENTS
Let us present the results of some numerical experiments in which the successive approximation method (1.11) and the Newton method (2.8) were used to solve the inverse problem under study. The general scheme of computational experiments was as follows. The functions µ(t), γ(t), and ϕ(t) were specified on the interval [0, T ], and the function ψ(x) was defined on the interval [0, l]. With these functions, problem (1.1)-(1.4) was solved, and the functions g(t) = u(l, t) and p(t) = u x (l, t) were determined.
Then the operator equation (1.9) with the functions µ(t), g(t), p(t), and ψ(x) was solved by the iterative methods (1.11) and (2.8), and the approximate functionγ(t) was found. To determine the approximate functionφ(t), we used the formula [3] (3.1) One and the same initial approximation γ 0 (t) = γ 0 and the same stopping criterion were used in the approximate solution of the operator equation (1.9) by both iterative methods.
The parameter δ was chosen to be 0.001. By analogy with Fig. 1, Fig. 3 shows the values of the exact function γ(t) as well as the functions obtained on the first two iterations for both methods. The convergence criterion was satisfied at the 7th step of the successive approximation method and at the 9th step of the Newton method. The approximate solutionsγ I (t) = γ I 7 (t) andγ N (t) = γ N 9 (t) found at the final iteration of both methods visually match the exact solution in the scale of Fig. 3.    Figure 4 shows the function ϕ(t) = 2 + 0.5 sin(2πt) as well as the functions ϕ I 1 (t), ϕ I 2 (t), ϕ N 1 (t), and ϕ N 2 (t) corresponding to the first two iterations. The functionsφ I (t) = ϕ I 7 (t) andφ N (t) = ϕ N 9 (t) found by formula (3.1) coincide in the figure with ϕ(t) = 2 + 0.5 sin(2πt).
The above examples, as well as a number of other numerical calculations, allow us to conclude that the convergence of both methods is quite fast and that there are no significant advantages in the rate of convergence of one of the methods in comparison with the other.

OPEN ACCESS
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