Numerical solution of linear differential equations with discontinuous coefficients and Henstock integral

In this article we consider the problem of approximative solution of linear differential equations $y'+p(x)y=q(x)$ with discontinuous coefficients $p$ and $q$. We assume that coefficients of such equation are Henstock integrable functions. To find the approximative solution we change the original Cauchy problem to another problem with piecewise-constant coefficients. The sharp solution of this new problems is the approximative solution of the original Cauchy problem. We find the degree approximation in terms of modulus of continuity $\omega_\delta (P),\ \omega_\delta (Q)$, where $P$ and $Q$ are $f$-primitive for coefficients $p$ and $q$.


Introduction
In the classical initial value problem for a linear differential equation of the first order y + p(x)y = q(x), y(a) = y 0 , x ∈ [a, b], (1.1) the coefficients p(x) and q(x) are continuous functions. However, some problems of dry friction and electric circuit with relay are given the equations with the discontinuous functions p and q. For example the RL-electric circuit with relay is described by a linear differential equation di dt + R(t) L i = e(t) L with the discontinuous function R(t). In this case, it is assumed that the functions p(x) and p(x) are (L)-integrable and a function y(x) is called a solution to equation (1.1) if y(x) is absolutely continuous and satisfies the equation (1.1) almost everywhere on [a, b].
There are no effective methods for the approximate solution of equations with unbounded coefficients p(x) and q(x). If the coefficients p(x) and q(x) are unbounded in some neighborhood of the point a then Ruge-Kutta method does not work. If the coefficients p(x) and q(x) are unbounded in some neighborhood of the interior point c ∈ (a, b) then Runge-Kutta method has a very large error, usually more than 1.
Some authors use Haar and Walsh functions to solve linear equations [1], [2], [3]. In [4], [5] G. Gat and R. Toledo propose to approach the solution y(x) by the Walsh polynomial In [4] for continue q(x), (x ∈ [0, 1]) and p(x) = const an estimate for the error |y(x) −ỹ(x)| is obtained. In [5] the authors consider the case when q ∈ L(0, 1) is a continuous function on [0, 1[ and prove thatỹ n (x) converges uniformly to the solution y(x) on the interval [0, 1[. In [8], the authors present the derivative y of the solution y as a Haar expansion and obtain an estimate of the approximate solution in terms of the modulus of continuity of the coefficients p(x) and q(x). This method can also be used for equations with unbounded coefficients p(x) and q(x).
In this article we will assume that p(x) and q(x) are Henstock integrable functions on the interval [a, b]. We construct the approximate solutionỹ(x) and obtain the estimate of the error |y(x) −ỹ(x)| in terms of modulus of continuity ω 1 2 n (e P ), ω 1 2 n (e −P ), and ω 1 2 n (Q), where P and Q are f -primitive for p and q respectively.
The paper is organized as follows. In Sec. 2, we recall some facts from Henstock integral. In Sec. 3, we indicate the necessary and sufficient condition, under which the Cauchy problem has a solution. This solution is given in terms of the Hanstock integral. In Sec. 4, we construct the approximative solution and find the error. In Sec. 5, we give two examples.

Henstock integral on the interval
is called a tagged partition and is denote by It is known that for any gauge δ(x) > 0 on [a, b] there exists a δ-fine partition The number I(f ) is called Henstock integral and is denoted by The collection of all functions that are Henstock integrable on [a, b] is denoted by R * (a, b).
There exists Henstock integrable functions that are not absolutely integrable. If the function on [a, b]\E . We will use next properties of Henstock integral.
In this case, A detailed exposition of the Henstock integral theory can be found in [6], [7].

Linear differential equations and Henstock integral
Let p, q : [a, b] → R be two continuous functions that differentiable on the interval [a, b], with the exception of a countable set E. We will consider the classical Cauchy initial value problem Theorem 2.1 follows that functions p (x) and q (x) are Henstock integrable. This is a weaker condition than p , q ∈ L(a, b). Example 1. Define the function q on [a, b] in the following way. Let Necessity. Let y(x) be a solution of (3.1), that is It means, that the function q (x)e p(x) has c-primitive y(x)e p(x) . Sufficiently. Let q (x)e p(x) has c-primitibe F (x), that is Corollary. A solution of Cauchy initial value problem (3.1)-(3.2) is given by the formula where the integral is an Henstock integral. Proof. Equality holds. Example 2. It is possible to construct the continuous functions p and q so that the function q (x)e p(x) has a c-primitive, but q (x)e p(x) / ∈ L(a, b). For simplicity, we consider the case [a, b] = [0, 1] and select the function q(x) as in Example 1. In this case x n = 2 −n , q(x n ) = q(0) = 0, q(x) is lineal on [x n+1 , ξ n ] and [ξ n , x n ], where ξ n = 1 2 (x x + x n+1 ). Now we define the function p(x) from the conditions: converges. It follows from the Hake theorem that f (x) = q (x)e p(x) ∈ R * (0, 1). Therefore the function F (x) = x 0 f (t)dt is continuous. Since the function f (x) = q (x)e p(x) is continuous on any interval (2 −n−1 , 2 −n ), it follows that F (x) = q (x)e p(x) on any interval (2 −n−1 , 2 −n ). It means that F (x) is c-primitive for q (x)e p(x) . It is not difficult to check that f (x) = q (x)e p(x) / ∈ L(0, 1).

Approximate solution of Cauchy problem (3.1)-(3.1) on interval [0,1]
Now we will find an approximate solution of Cauchy initial value problem We assume that the functions p and q are continuous and have derivatives with the exception of some countable set E. We also assume, that e p(x) q (x) has a c-primitive differentiable on [a, b] \ E.
Choose an arbitrary n ∈ N, define the functionsp(x) andq(x) by equalities and consider the Cauchy initial value problem y +p ỹ =q , (4.3) It is evident that the function ep (x)q has a f -primitive. By theorem 3.1 the functions x 0 q (t)e p(t) d t, are solutions of Cauchy problems (4.1)-(4.2) and (4.3) -(4.4) respectively. The functionỹ(x) is the approximate solution of Cauchy problem (4.1)-(4.2). In the following theorem, we indicate an estimate for the distance y(x) −ỹ(x).   Proof 1)First we estimate the difference y(x) −ỹ(x) for x = k 2 n , k = 0, 1, ..., 2 n . We have

Theorem 4.1 The following inequality
To estimate integrals in I 1 and I 2 we will assume that p and q -are Henstock absolutely integrable. Assume I 1 . Integrating by parts and using the equality q ( j 2 n ) =q ( j 2 n ) we obtain Since the function ep (t) is monotonic on any interval [ j 2 n , j+1 2 n ], it follow that |(e p(t) − ep (t) )| ≤ ω 1 2 n (e p(·) ). Therefore 2) Now we consider the case x ∈ k 2 n , k+1 2 n . Let us write the difference y(x) −ỹ(x) in the form We will estimate A l (l = 1, 2, 3, 4). 1) Since the function e −p(x) is monotonic on any interval [ j 2 n , j+1 2 n ], it follow that 2) Using again (4.6), we get 3) An estimate for A 3 was obtained earlier

4)Let us write A 4 in the form
Since the function ep (t) is monotonic on the interval [ k 2 n , k+1 2 n ], both integrals exist. For the first integral, we have the obvious inequality Integrating by parts in the second integral in (4.9) we have ≤ ω 1 2 n (q)(1 + e p C[0,1] ω 1 2 n (p)).