ON A BOUNDARY VALUE PROBLEM

A. I. PEROV and A. P. MAKMUDOV

UDC 517.911

In the present article the following problem is considered: it is required to find a solution of the vector differential equation of second order

\[ \ddot x=f(t,x,\dot x), \tag{1} \]

satisfying the conditions

\[ x(0)=0,\qquad \|\dot x(0)\|=v>0, \tag{2} \]

which, for some \(t^*\in[0,T]\), assumes a prescribed value \(x^*\)

\[ x(t^*)=x^* \tag{3} \]

(\(v\) is a given number).

An analogous problem for two second-order equations was first posed and studied by V. Nikliborc in [1]. In the present article the proofs of V. Nikliborc are simplified and, at the same time, a whole series of his results is strengthened.

The facts from functional analysis used by us can be found, for example, in [2].

We shall use the following notation. \(E\) is a Banach space; \(C_{[0,T]}\) is the set of all vector functions \(x(t)\), continuous on \([0,T]\), with values in \(E\).

1. In what follows, the following three lemmas play the main role.

Lemma 1. Let \(\varphi(t)\in C_{[0,T]}\) satisfy the Lipschitz condition

\[ \|\varphi(t)-\varphi(\tau)\|\le l|t-\tau|\qquad (0\le t,\tau\le T). \tag{4} \]

Then, under the conditions

\[ \delta+\max_{0\le t\le T}\|\varphi(t)\|\le Tv, \tag{5} \]

\[ l<v, \tag{6} \]

the equation

\[ t=\frac{1}{v}\,\|x^*-\varphi(t)\| \tag{7} \]

has a unique solution \(t^*\in[0,T]\), and this solution can be found by the method of successive approximations \((\delta=\|x^*\|)\).

Proof. Denote the right-hand side of equation (7) by \(F(t)\). Then

\[ 0\le F(t)\le \frac{1}{v}\left(\delta+\max_{0\le t\le T}\|\varphi(t)\|\right) \tag{8} \]

and

\[ |F(t)-F(\tau)|=\frac{1}{v}\left|\|x^*-\varphi(t)\|-\|x^*-\varphi(\tau)\|\right|\le \]

\[ \leq \frac{1}{v}\|\varphi(t)-\varphi(\tau)\|\leq \frac{l}{v}\,|t-\tau|. \tag{9} \]

We see that, when conditions (5) and (6) are satisfied, the operator \(F\) maps the interval \([0,T]\) into itself and satisfies the contraction condition. The assertion of the lemma now follows from the principle of contraction mappings. The lemma is proved.

Let us note that, for the solution \(t^*\) of equation (7), the estimate \(t^*\geq t_{\min}\) is valid, where \(t_{\min}\) is the smallest positive root of the equation

\[ t=\frac{1}{v}\{\delta-\|\varphi(t)\|\}. \tag{10} \]

Here and below we additionally assume that \(\varphi(0)=0\).

Lemma 2. Let the functions \(\varphi(t)\), \(\psi(t)\in C_{[0,T]}\) satisfy the conditions of Lemma 1. Then equation (7) has a unique solution \(t^*\) (for \(\varphi\)) and \(\tau^*\) (for \(\psi\)). Denote by \(\xi^*\) and \(\eta^*\) the following vectors:

\[ \xi^*=\frac{1}{t^*}\bigl(x^*-\varphi(t^*)\bigr),\qquad \eta^*=\frac{1}{\tau^*}\bigl(x^*-\psi(\tau^*)\bigr). \tag{11} \]

Then the estimates

\[ |t^*-\tau^*|\leq \frac{1}{v-l}\max_{0\leq t\leq T}\|\varphi(t)-\psi(t)\|, \tag{12} \]

\[ \|\xi^*-\eta^*\|\leq 2\,\frac{v}{v-l}\,\frac{l+v}{\delta}\, \max_{0\leq t\leq T}\|\varphi(t)-\psi(t)\|. \tag{13} \]

are valid.

Proof. Let
\[ \Delta=\max_{0\leq t\leq T}\|\varphi(t)-\psi(t)\|. \]
We have

\[ t^*=\frac{1}{v}\|x^*-\varphi(t^*)\|,\qquad \tau^*=\frac{1}{v}\|x^*-\psi(\tau^*)\|, \]

\[ |t^*-\tau^*| =\frac{1}{v}\left|\|x^*-\varphi(t^*)\|-\|x^*-\psi(\tau^*)\|\right| \leq \]

\[ \leq \frac{1}{v}\|\varphi(t^*)-\psi(\tau^*)\| \leq \frac{1}{v}\{l|t^*-\tau^*|+\Delta\}. \]

Therefore

\[ |t^*-\tau^*|\leq \frac{\Delta}{v-l}, \]

and estimate (12) is proved. Further,

\[ \|\xi^*-\eta^*\| = v\left\| \frac{x^*-\varphi(t^*)}{\|x^*-\varphi(t^*)\|} - \frac{x^*-\psi(\tau^*)}{\|x^*-\psi(\tau^*)\|} \right\|. \]

By virtue of the elementary inequality

\[ \left\|\frac{x}{\|x\|}-\frac{y}{\|y\|}\right\| \leq \frac{2\|x-y\|}{\max\{\|x\|,\|y\|\}} \]

we obtain that

\[ \|\xi^*-\eta^*\| \leq v\,\frac{2\|\varphi(t^*)-\psi(\tau^*)\|} {\max\{\|x^*-\varphi(t^*)\|,\|x^*-\psi(\tau^*)\|\}}. \]

Since

\[ \|\varphi(t^*)-\psi(\tau^*)\|\leq l|t^*-\tau^*|+\Delta \]

and

\[ \|x^*-\varphi(t^*)\|=t^*v,\qquad \|x^*-\psi(\tau^*)\|=\tau^*v, \]

then, by virtue of the inequality

\[ t^*,\tau^*>\frac{\delta}{l+v}, \]

we obtain

\[ \|\xi^*-\eta^*\|\leq 2\,\frac{l+v}{\delta}\,(l|t^*-\tau^*|+\Delta). \]

Using now estimate (12), we similarly find

\[ \|\xi^*-\eta^*\|\leq 2\,\frac{v}{v-l}\,\frac{l+v}{\delta}\,\max_{0\leq t\leq T}\|\varphi(t)-\psi(t)\|. \]

Inequality (13) is also proved.

Lemma 3. Let \(\varphi(t)\in C^1_{[0,T]}\) have the form

\[ \varphi(t)=\int_0^t (t-s)f(s)\,ds \quad\text{and}\quad \|f(\ )\|_p=\left(\int_0^T \|f(t)\|^p\,dt\right)^{1/p} \tag{14} \]

\[ \bigl(f(t)\in C[0,T],\ \|f(\ )\|_\infty=\max_{0\leq t\leq T}\|f(t)\|\bigr). \]

Then the estimates

\[ \max_{0\leq t\leq T}\|\varphi(t)\|\leq \left(\frac{T^{1+q}}{1+q}\right)^{1/q}\|f(\ )\|_p, \tag{15} \]

\[ \max_{0\leq t\leq T}\|\varphi'(t)\|\leq T^{1/q}\|f(\ )\|_p, \tag{16} \]

are valid, where

\[ \frac1p+\frac1q=1\qquad (1\leq p\leq\infty,\ 1\leq q\leq\infty). \]

In proving the lemma, Hölder’s inequality is used. The estimates used in [1] follow from our lemma when \(q=1\).

1. Consider the boundary value problem for the nonhomogeneous equation

\[ \ddot{x}=f(t), \tag{17} \]

\[ x(0)=0,\qquad \|\dot{x}(0)\|=v>0,\qquad x(t^*)=x^*\ (t^*\in[0,T]), \tag{18} \]

where \(v\) and \(T\) are given, and \(\|x^*\|=\delta,\ f(t)\in C[0,T]\).

Theorem 1. Suppose that, for some \(p\) and \(\alpha\), the inequalities

\[ \left(\delta+\left(\frac{T^{1+q}}{1+q}\right)^{1/q}\|f(\ )\|_p\right) \leq Tv\left(\frac1p+\frac1q=1\right), \tag{19} \]

\[ T^{1/\beta}\|f(\ )\|_\alpha<v\left(\frac1\alpha+\frac1\beta=1\right) \tag{20} \]

hold. Then the boundary value problem (17), (18) has a unique solution, i.e., there exists only one solution \(x(t)\) of equation (17) for which, for some \(t^*\in[0,T]\), the conditions (18) are satisfied.

Proof. Let \(x(t)\) be a solution of equation (17) satisfying the conditions \(x(0)=0,\ \|\dot{x}(0)\|=v\). Then \(x(t)\) has the form

\[ x(t)=\xi t+\int_0^t (t-s)f(s)\,ds, \tag{21} \]

where \(\|\xi\|=v\).

The required time \(t^*\) is a root of the equation

\[ t=\frac{1}{v}\left\|x^*-\int_0^t (t-s)f(s)\,ds\right\|. \tag{22} \]

It is not difficult to see that the converse is true: if \(t^*\) is a root of equation (22), then the solution (21) is a solution of the boundary-value problem, and in this case

\[ \xi=\frac{1}{t^*}\left\{x^*-\int_0^{t^*}(t^*-s)f(s)\,ds\right\}. \tag{23} \]

Applying Lemmas 1 and 3 to equation (22), we arrive at the required result.

Let us note that Lemma 1 of [1] follows from our theorem for \(p=\alpha=\infty\); moreover, even in this case we obtain a stronger result.

Under the assumptions of Theorem 1 the estimate

\[ t^*\geq t_{\min}, \tag{24} \]

is valid, where \(t_{\min}\) is the smallest positive root of the equation

\[ t=\frac{1}{v}\left\{\delta-\left(\frac{t^{1+q}}{1+q}\right)^{\frac{1}{q}}\|f(\ )\|_p\right\}. \tag{25} \]

We now consider our boundary-value problem for the nonlinear equation

\[ \ddot{x}=f(t,x,\dot{x}), \tag{26} \]

\[ x(0)=0,\qquad \|\dot{x}(0)\|=v>0,\qquad x(t^*)=x^*\quad (x^*\in[0,T]), \tag{27} \]

where the numbers \(v,T\) and the vector \(x^*\) are given.

We shall assume that the vector function \(f(t,x,y)\) is defined and continuous for \(0\leq t\leq T\), \(\|x\|\leq a\), \(\|y\|\leq b\) \((x,y\in E)\), and in this domain satisfies the conditions

\[ \|f(t,x,y)\|\leq M(t), \tag{28} \]

\[ \|f(t,x_1,y_1)-f(t,x_2,y_2)\|\leq L_1\|x_1-y_1\|+L_2\|x_2-y_2\|, \tag{29} \]

where \(L_1\) and \(L_2\) are certain constants.

Theorem 2. Suppose that for some \(p\) and \(\alpha\) the inequalities

\[ \delta+\left(\frac{T^{1+q}}{1+q}\right)^{\frac{1}{q}}\|M(\ )\|_p\leq Tv, \tag{30} \]

\[ T^{\frac{1}{\beta}}\|M(\ )\|_\alpha<v \tag{31} \]

and

\[ a\geq Tv+A,\qquad b\geq v+B, \tag{32} \]

hold, where

\[ A=\left(\frac{T^{1+q}}{1+q}\right)^{\frac{1}{q}}\|M(\ )\|_p,\qquad B=T^{\frac{1}{\beta}}\|M(\ )\|_\alpha. \tag{33} \]

Suppose, finally, that the condition

\[ T^2\left(\frac{1}{2}+\frac{v}{v-l}\frac{T(l+v)}{\delta}\right)L_1+ T\left(1+\frac{v}{v-l}\frac{T(l+v)}{\delta}\right)L_2<1, \tag{34} \]

where

\[ l=T^{\frac{1}{\beta}}\|M(\ )\|_\alpha . \tag{35} \]

Then the boundary-value problem (26), (27) has a unique solution.

Proof. Consider the set \(R\) of all continuously differentiable vector functions and define the generalized norm [3] by the equality

\[ |x(\ )|= \left\{ \begin{array}{c} \max\limits_{0\le t\le T}\|x(t)\|\\[2mm] \max\limits_{0\le t\le T}\|\dot x(t)\| \end{array} \right\}. \tag{36} \]

In the space \(R\), distinguish the set \(F\) by the inequalities \(\|x(t)\|\le a\), \(\|\dot x(t)\|\le b\). Define on \(F\) the operator \(D\) in the following way. Let \(x(t)\in F\).

Consider the equation

\[ t=\frac{1}{v}\left\|x^*-\int_0^t (t-s)f\bigl(s,x(s),\dot x(s)\bigr)\,ds\right\|. \tag{37} \]

By virtue of conditions (30) and (31) and Theorem 1, this equation has a unique solution \(t(x)\).

Put now

\[ Dx(t)=\xi(x)t+\int_0^t (t-s)f\bigl(s,x(s),\dot x(s)\bigr)\,ds, \tag{38} \]

where

\[ \xi(x)=\frac{1}{t(x)} \left\{x^*-\int_0^{t(x)}(t(x)-s)f\bigl(s,x(s),\dot x(s)\bigr)\,ds\right\}. \tag{39} \]

It is not difficult to see that \(Dx(t)\in R\). Since

\[ \|Dx(t)\|\le Tv+A,\qquad \|\dot D x(t)\|\le v+B, \tag{40} \]

the operator \(D\) maps the set \(F\) into itself. We show that the operator \(D\) satisfies the conditions of Theorem 3 of [3]. Let \(x_1(t)\), \(x_2(t)\in F\). Then, by Lemma 2, we have

\[ \|Dx_1(t)-Dx_2(t)\|\le T^2\left(\frac{1}{2}+\frac{v}{v-l}\frac{T(l+v)}{\delta}\right) \left( L_1\max_{0\le t\le T}\|x_1(t)-x_2(t)\| \right. \]

\[ \left. +\,L_2\max_{0\le t\le T}\|\dot x_1(t)-\dot x_2(t)\| \right), \]

\[ \|\dot D x_1(t)-\dot D x_2(t)\|\le T\left(1+\frac{v}{v-l}\frac{T(l+v)}{\delta}\right)\times \]

\[ \times\left( L_1\max_{0\le t\le T}\|x_1(t)-x_2(t)\| + L_2\max_{0\le t\le T}\|\dot x_1(t)-\dot x_2(t)\| \right) \tag{41} \]

or

\[ |Dx_1-Dx_2|\le S|x_1-x_2|, \tag{42} \]

where the matrix \(S\) has the form

\[ \begin{pmatrix} \sigma L_1 & \sigma L_2\\ \rho L_1 & \rho L_2 \end{pmatrix}, \tag{43} \]

where

\[ \sigma=T^2\left(\frac{1}{2}+\frac{v}{v-l}\frac{T(l+v)}{\delta}\right); \quad \rho=T\left(1+\frac{v}{v-l}\frac{T(l+v)}{\delta}\right). \]

It is not difficult to see that the matrix \(S\) is an \(a\)-matrix if and only if condition (34) is satisfied.

Therefore, by the generalized principle of contraction mappings, the operator \(D\) has a unique fixed point. This fixed point \(x^*(t)\) is the solution of the boundary value problem posed. The theorem is proved.

Let us note that the existence of a solution can be proved using only conditions (30)—(33) of Theorem 2, provided, however, that \(f(t,x,y)\) is compact, since under these conditions the operator \(D\) constructed above, which maps the closed convex bounded set \(F\) into itself, will be continuous and compact. The existence of a solution then follows from the well-known Schauder principle.

References

1. Nicliborc W. Ber. Verhandl, Sächsich. Akad. der Wiss. zu Leipzig, Math-phys. Kb., 82, 227—242, 1930.

2. Lyusternik L. A., Sobolev V. I. Elements of Functional Analysis. Gostekhizdat, 1951.

3. Perov A. I. Approximate methods for solving differential equations, issue 2. Kiev, 1964, pp. 115—134.

Received by the editors
April 24, 1965.

Azerbaijan State University
named after S. M. Kirov,
Voronezh State University