NUMERICAL SOLUTION TO INITIAL VALUE PROBLEM FOR ONE CLASS OF DIFFERENTIAL EQUATION WITH MAXIMUM

In many applications one has to deal with infinite dimensional systems which are hardly possible to solve analytically, especially in the nonlinear case. Here we consider a special subclass of functional-differential equations, and look for their numerical solutions. Computation of a solution requires studying functionaldifferential equations. Among the vast majority of functional-differential prob-


Introduction
In many applications one has to deal with infinite dimensional systems which are hardly possible to solve analytically, especially in the nonlinear case.Here we consider a special subclass of functional-differential equations, and look for their numerical solutions.Computation of a solution requires studying functionaldifferential equations.Among the vast majority of functional-differential prob-lems one of the most interesting is differential equation with maximum which contains a maximum of the studied function on a prehistory interval.This kind of problems have a great deal of applications in economics, ecology, biology, oscillation theory, automatic regulation etc. ( [10], [11]).Such problems are often nonlinear, which makes them very challenging in view of control, stabilization, optimization and numerical solution.For example, in 1966 E.P.Popov considered a system for voltage regulation of a generator of constant current.The equation describing the work of the regulator involves the maximum of the unknown function and it has the form where T 0 , q are constants characterizing the object, u(t) is the regulated voltage and f (t) is a perturbation.Similar problems were studied in [1], [2], [8].For instance [2] provides a qualitative analysis of solutions to differential equations with maximum, see also [4].
Among recent developments for systems with maximum we recall stability results [12], averaging methods [9], where the dynamics can depend also on the maximum of the input function (see [6]).However many problems like stabilization or optimal control are not studied yet.In many applications such systems can be nonlinear.Due to nonlinearity analytic solution is in many cases not possible and a numerical method is needed.Standard methods cannot be applied directly in case of non-monotone solutions because of the maximum of the unknown function in the equation.
In this paper we consider the following problem where h > 0 is a constant and T > 0 is finite.We are interested in numerical methods for solution calculation.Sufficient conditions of existence and uniqueness of solutions for this problem were obtained in [13].Recall that in [5] an algorithm for constructions of two sequences of successive approximations of the solution to the initial value problem for nonlinear differential equations with maximum was developed and is based on the monotone iterative techniques.Here we propose a numerical method based on a simpler approach and illustrate it by several examples.

Notation and Preliminaries
Let f from (1) satisfy the following in all variables, and (ii) satisfy the Lipschitz condition where K > 0, L > 0 are Lipschitz constants.
For the reader's convenience and to make our paper self-contained we present the following theorem with proof.
Conversely, let v ∈ C([−h, T ]) be a solution to (5).Define Taking the time derivative of x we obtain a continuous function that satisfies (1),(2) due to (5).

It is known that f ∈ C[0, T ] if and only if lim
Our numerical approach to the problem (1),( 2) is based on left rectangle method of integration, which requires only the C 1 property for f to derive the error estimation.Recall that for the trapezoid or Simpson method more regularity is required.
This proves the following ), then for the error of the left rectangle method the estimation (6) is valid.
We also recall the following related result: Lemma 5. ( [14]) Let on the uniform grid T N = t k = kl, k = 0, N , N l = a the grid functions q : T N → R, w : Then |q n | ≤ w n , n = 0, N , and w n = z(t n ), satisfies (7), where z(t) = Se Qt is a solution to integral equation

Numerical Method
Let l be the step of integration, T = N l, t k = kl, k = 0, N .By x k we denote the approximation of solution x(t k ) at t = t k to the problem (1),(2), also by v k we denote the approximation of solution v(t k ) of ( 5).Assume, that x k and v k , k = 0, N are already found.Let us show how x n+1 , v n+1 can be calculated.
Note, that for t ∈ [0, T ], x(t) = t 0 v(s)ds.By the left rectangle formula we obtain For t ∈ [t k , t k+1 ] we can write From ( 8), (9) and Lemma 4 follows that According to (5 x(τ ) .Then x(τ ) .Let τ be such that x(τ ).Let us calculate an approximation If τ = t n+1−k , then using left rectangle formula we get From (10) we obtain Now consider the case t n+1−k < τ < t n+2−k .Then In this case x n+1 = l n+1−k i=0 v i and the following estimation holds If τ = t n+2−k , then analogically follows that x n+1 = l n+1−k i=0 v i and estimation ( 12) is valid.
Proof.Using our previous notation and from (10), (13) we have Therefor, lim and hence lim Remark 7.This method can be applied for problems with nonzero initial condition: Let y be defined by x(t) = y(t) + u(t).It can be shown (see [13]), that y(t) is the solution to the problem and f (0, y(0), Hence there exists a unique solution to (17),(18).

Examples
Example 1.Consider the following differential equation with h = 0.5, T = 1.For this example we can compare the numerical solution which we get by the proposed method with the analytic solution.Using theorem about differential inequalities it easy to prove that the solution x is a monotonically increasing function, that is why max The solution to the problem (20) is x(t) = e t − t − 1. Figure 1 illustrates the approximations of the solution to the problem with maximum (19) by the proposed numerical method and the analytic solution to the problem (20).Table 1 shows the values of x k -numerical solution by the proposed method and x(t k ) -the analytic solution calculated at points t k .In the same table the estimation between analytic and numerical solutions for steps l = 0.1, l = 0.05, l = 0.0025 are presented.In particular, for the step l = 0.1 we get maximum value of |x(t k ) − x k | on an interval [0, 1] which is 0.1245, for the step l = 0.05 the maximum value is 0.0650, and for l = 0.0025 is 0.0034.Thus for step l we can point numbers ε l such that |x(t k ) − x k | ≤ ε l , and ε 0.1 = 0.1245, ε 0.05 = 0.0650, ε 0.0025 = 0.0034.
Example 2. Now we consider differential equation with maximum and with constant initial function where the solution is not monotone, so that the "max" sign cannot be omitted, but such that an analytic solution can be derived with some more effort: where h = 0.5, T = 5.The analytic solution to the problem (21) can be written (after simplifications of numbers given by large expressions with fractions with roots) as follows According to Remark 7 in order to apply our numerical method to the problem (21) we first have to solve the following problem  Example 3. In the last example we consider a system with maximum described by the following differential equation where h = 1, T = 4.In this case one can verify that ẋ(t) < 0, t ∈ [0, T ] so that the problem (23) can be reduce to the following problem with time delay because of the monotonicity of the solution ẋ(t) = e −t sin t − tx(t − h), t ∈ [0, T ] On the following Figure 3 we present the graphs of numerical solutions calculated by the proposed method to the problem (23) (note that we apply Remark 7) with the steps l = 0.1, l = 0.05, l = 0.0025.There exist ε l such that |x(t k ) − x k | ≤ ε l and for particular steps we can specify ε 0.1 = 0.0983, ε 0.05 = 0.0407, ε 0.0025 = 0.0018.

Conclusion
In this work we have proposed a simple but efficient numerical method to calculate solutions for initial value problems of one type of differential equations with maximum and proved its convergence.From examples it can be seen that the estimation in (10) is confirmed.As a continuation of this research our numerical method can be extended to a problem of a more general type such as ẋ(t) = f (t, x(t), max τ ∈[g(t),γ(t)] x(τ )), t ∈ [t 0 , T ], x(t) = ϕ(t), t ∈ [inf t≥0 g(t), t 0 ].
where g(t) ≤ γ(t) ≤ t, t 0 ≥ inf t≥0 g(t).Also a numerical method for computation of solutions to initial value problem for differential equations with maximum in case when the right hand side is a measurable function will be a matter of our future investigation.

Figure 1 :
Figure 1: Solutions to (19) and (20) for different values of l where h = 0.5, T = 5.Then we compute the solution to (21) as x(t) = y(t) + 1, where y(t) is the solution to the problem (22) by the proposed numerical method with step l = 0.05.Note that the graph of the true solution coincides with the graph of the numerical solution.Here |x(t k ) − x k | ≤ ε 0.05 = 0.0125.

Figure 2 :
Figure 2: Plot of numerical solution to the problem (21)

Figure 3 :
Figure 3: Plot of solutions to problem (23) with different steps.