In this paper we study the following problem

The problem (1.1)-(1.2) has been studied in the papers R. Driver [4], I.A. Rus and C. Iancu [9] and V.A. Darzu [2]. This problem is known in literature as WheelerFeynman problem. For this problem the above authors studied the existence and uniqueness of the solution using the step method.

The purpose of this paper is to elaborate an algorithm based on the step method and the successive approximation method and to apply it on the problem (1.1)-(1.2).

Algorithm. At each step we have a problem like this

where $f\in C\left([a+h,a+2h]\times\mathbb{R}^{3},\mathbb{R}\right),\theta\in C([a,a+2h],\mathbb{R})$ and $x:[a,a+3h]\rightarrow\mathbb{R}$.

It follows that

We denote $\xi:=t+h,\xi\in[a+2h,a+3h]$.
Then

We denote $F(\xi,x(\xi)):=f(\xi-h,\theta(\xi-h),\theta(\xi-2h),x(\xi))-\theta^{\prime}(\xi-h)$. So

The purpose is to impose conditions on $f$ such that equation (1.4) has a unique solution who can be approximated by Newton’s method.

In order to study the problem (1.1)-(1.2) we need the following well known results.
Implicit function theorem. ([1]) We suppose that $F:[a,b]\times\mathbb{R}\rightarrow\mathbb{R}$ satisfy the following conditions
(i) $F\in C^{1}([a,b]\times\mathbb{R})$;
(ii) there exists $\frac{\partial F(t,u)}{\partial u}\in\mathbb{R}^{*}$ and $\left|\frac{\partial F(t,u)}{\partial u}\right|\leq M_{1},\forall t\in[a,b],u\in\mathbb{R}$;
(iii) for each $t_{0}\in[a,b]$, there exists $u_{0}\in\mathbb{R}$ such that $F\left(t_{0},u_{0}\right)=0$.

Then, there exists a unique solution $x\in C^{1}[a,b]$ such that $F(t,x(t))=0$, $\forall(t,x(t))\in[a,b]\times\mathbb{R}$, solution that can be obtained using the successive approximations method.

In terms of $f$, for the problem (1.1)-(1.2), the conditions from the above theorem are:
$\left(\mathrm{C}_{1}\right)f\in C^{\infty}\left([-T,T]\times\mathbb{R}^{3},\mathbb{R}\right),\varphi\in C([-h,h],\mathbb{R});$
(C2) $\frac{\partial f(t,u,v,w)}{\partial w}\in\mathbb{R}^{*},\forall t\in[-T,T],\forall u,v,w\in\mathbb{R}$;
$\left(\mathrm{C}_{3}\right)\left|\frac{\partial f(t,u,v,w)}{\partial w}\right|\leq M_{1},\forall t\in[-T,T],\forall u,v,w\in\mathbb{R};$
( $\mathrm{C}_{4}$ ) $\forall t\in[-T,T],u,v,\eta\in\mathbb{R}$, the equation $f(t,u,v,w)-\eta=0$ has a unique solution.
We shall use the notations, the terminology and some results given by I.A. Rus in the paper [6], and the following result, that is a generalization of the fibre contraction theorem (see [6]).
Fibre contraction theorem. (Theorem 9.1., [7]) Let $\left(X_{i},d_{i}\right),i=\overline{0,m},m\geq 1$, be some metric spaces. Let

be some operators. We suppose that:
(i) $\left(X_{i},d_{i}\right),i=\overline{1,m}$, are complete metric spaces;
(ii) the operator $A_{0}$ is WPO;
(iii) there exists $\alpha_{i}\in(0;1)$ such that:

are $\alpha_{i}$-contractions;
(iv) the operators $A_{i},i=\overline{1,m}$, are continuous.

The operator $A$ : $X_{0}\times\cdots\times X_{m}\rightarrow X_{0}\times\cdots\times X_{m}$,

is $WPO$. If $A_{0}$ is PO, then $A$ is PO.

In this section we apply the algorithm from section 1 for the problem (1.1)-(1.2).
Let $n\in\mathbb{N}^{*}$ be such that $nh\leq T,(n+1)h>T$. In the conditions $\left(C_{1}\right)-\left(C_{3}\right)$, the step method consists in the following:

For $t\in[h,2h]$ we have

where $x(t)=\varphi(t)=\left\{\begin{array}[]{l}x_{-1}(t),t\in[-h,0],\\ x_{0}(t),t\in[0,h].\end{array}\right.$
We denote $x(t):=x_{1}(t),t\in[h,2h]$.
Let

From the implicit function theorem there exists a solution $x_{1}^{*}\in C^{1}[h,2h]$ such that

The key of each step is to approximate the solution $x_{1}^{*}\in[h,2h]$ with the method of Newton:

where $G\left(t,x_{1}^{*}(t)\right)\neq 0$ and $x_{1,m-1}(t)-G\left(t,x_{1}^{*}(t)\right)F\left(t,x_{1,m-1}(t)\right)$ is a contraction.
We choose the function $G:[h,2h]\times\mathbb{R}\rightarrow\mathbb{R}$ with $G\left(t,x_{1}^{*}(t)\right):=M\left(\frac{\partial F\left(h,x_{1}(h)\right)}{\partial x_{1}}\right)^{-1}$, where $M\in(0,1)$ is a constant. It is obvious that $G\left(t,x_{1}^{*}(t)\right)\neq 0$.

Now we consider the operator $A_{1}:C[h,2h]\rightarrow C[h,2h]$, defined by

Proving that $A_{1}$ is a contraction we have the uniqueness of the solution $x_{1m}$ on $[h,2h]$. For all $t\in[h,2h]$ we have the inequality

We have that $x_{1m}\xrightarrow{\text{ unif }}x_{1}^{*}$ on $[h,2h]$, so in the next step we shall use $x_{1m}$ instead of $x_{1}^{*}$.

For $t\in[2h,3h]$ we have

We denote $x(t):=x_{2}(t),t\in[2h,3h]$.
Let

Applying the implicit function theorem, we have that there exists the solution $x_{2}^{*}\in C^{1}[2h,3h]$ such that

Now we approximate the solution $x_{2}^{*}\in[2h,3h]$ with the method of Newton:

where $G\left(t,x_{2}^{*}(t)\right)\neq 0$ and $x_{2,m-1}(t)-G\left(t,x_{2}^{*}(t)\right)F\left(t,x_{2,m-1}(t)\right)$ is a contraction.
We choose $G:[2h,3h]\times\mathbb{R}\rightarrow\mathbb{R}$ with $G\left(t,x_{2}^{*}(t)\right):=M\left(\frac{\partial F\left(2h,x_{2}(2h)\right)}{\partial x_{2}}\right)^{-1}$, where $M\in(0,1)$ is a constant. Then we have $G\left(t,x_{2}^{*}(t)\right)\neq 0$.

Let us consider the operator $A_{2}:C[2h,3h]\rightarrow C[2h,3h]$, defined by

In the same way as in the previous step we prove that $A_{2}$ is a contraction. Follows that $x_{2m}\xrightarrow{\text{ unif }}x_{2}^{*}$ on $[2h,3h]$, so in the next step we shall use $x_{2m}$ instead of $x_{2}^{*}$.

By induction, for $t\in[nh,T]$ we have

We denote $x(t):=x_{n}(t),t\in[nh,T]$.
Let

Applying implicit function theorem, there exists the solution $x_{n}^{*}\in C^{1}[nh,T]$ such that $F\left(t,x_{n}^{*}(t)\right)=0,\forall t\in[nh,T]$. We approximate the solution $x_{n}^{*}\in[nh,T]$ with the method of Newton, by $x_{nm}(t)=x_{n,m-1}(t)-G\left(t,x_{n}^{*}(t)\right)F\left(t,x_{n,m-1}(t)\right)$, where $G\left(t,x_{n}^{*}(t)\right)\neq 0$ and $x_{n,m-1}(t)-G\left(t,x_{n}^{*}(t)\right)F\left(t,x_{n,m-1}(t)\right)$ is a contraction.

The function chosen here is $G:[nh,T]\times\mathbb{R}\rightarrow\mathbb{R},G\left(t,x_{n}^{*}(t)\right):=M\left(\frac{\partial F\left(nh,x_{n}(nh)\right)}{\partial x_{n}}\right)^{-1}$, where $M\in(0,1)$ is a constant. Then $G\left(t,x_{n}^{*}(t)\right)\neq 0$.

Let the operator $A_{n}:C[nh,T]\rightarrow C[nh,T]$ defined by

Notice that $A_{n}$ is a contraction. Then we have that $x_{nm}\xrightarrow{\text{ unif }}x_{n}^{*}$ on $[nh,T]$.
So, the following convergence takes place

In what follows we present the step method for the solution determined with the above algorithm.

Thus we have the following theorem

Theorem 2.1. In the conditions $\left(C_{1}\right)-\left(C_{3}\right)$ we have:
a) the problem (1.1)-(1.2) has in $C[-T,T]$ a unique solution

b) the sequence define by

Theorem 2.1 gives a uniqueness result for the solution of the problem (1.1)-(1.2) by successive approximation method and now we want to improve the convergence of this solution. So, here cames the question: can we put $x_{i-1,m}(t)$ instead of $x_{i-1}^{*}(t),i=\overline{2,n}$ in the conclusion b) of Theorem 2.1? The answer of this question is given by the following theorem.

Theorem 2.2. We suppose that the conditions $\left(C_{1}\right)-\left(C_{3}\right)$ and
$\left(\mathrm{C}_{4}\right)$ there exists $L_{f}>0$ such that

are satisfied. Then the sequence defined by

is convergent and $\lim_{m\rightarrow\infty}x_{km}=x_{k}^{*},k=\overline{1,n}$.

Proof. We consider the Banach spaces

and the operators