ON THE CAUCHY PROBLEM FOR SINGULAR SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS

I. T. Kiguradze

Consider the problem

\[ \frac{dx}{dt}=f(t,x), \tag{1} \]

\[ x(a)=x_0, \tag{2} \]

where \(x\) is an \(n\)-dimensional vector, and \(f(t,x)\) is an \(n\)-dimensional vector function defined in the domain

\[ R:\ a<t\le b,\quad |x-x_0|<r^*, \]

with \(f(t,x)\) measurable in \(t\) for every fixed \(x\), continuous in \(x\) for every fixed \(t\), and throughout the domain \(R\) satisfying the inequality

\[ |f(t,x)|\le l(t), \tag{3} \]

where \(l(t)\) is a summable function on the interval \(a+\delta\le t\le b\) for every \(\delta>0\).

By a solution of problem (1)—(2) we mean an \(n\)-dimensional vector function \(x(t)\), continuous on some interval \(a\le t\le b_0\), where \(a<b_0\le b\), and absolutely continuous on the interval \(a+\delta\le t\le b_0\) for every \(\delta>0\), satisfying equation (1) for almost all \(t\) and condition (2).

For the case when \(l(t)\) is summable on the interval \(a\le t\le b\), i.e., when \(f(t,x)\) satisfies the Carathéodory condition, problem (1)—(2) has been studied sufficiently well (see, for example, [1], [3], [5]).

In the present article we consider the case when \(f(t,x)\), having a singularity at \(t=a\), is not bounded by a summable function, i.e.,

\[ \int_a^b l(t)\,dt=\infty . \]

This case was previously investigated by V. A. Chechik [6] (see also [4]).

In § 1 conditions \((A)\), \((B)\), \((C)\), and \((D)\) are introduced, and sufficient criteria for the fulfillment of these conditions are indicated.

In § 2 it is proved that the fulfillment of condition \((A)\) is necessary and sufficient for the solvability of problem (1)—(2).

In §§ 3 and 4 it is shown that under conditions \((B)\), \((C)\), and \((D)\) the classical theorems on the structure of the integral funnel and on the correctness of problem (1)—(2) hold.

\[ \text{* The norm of the vector } x=(x_1,\ldots,x_n) \text{ is defined by the equality } |x|=\sum_{i=1}^{n}|x_i|. \]

§ 1. Conditions \((A)\)—\((D)\)

Definition 1.1. We shall say that equation (1) satisfies condition \((A)\) at the point \((a,x_0)\) if there exists a sequence \(\{x_k(t)\}\) of solutions of this equation such that each solution \(x_k(t)\) is defined\(^*\) on the interval \(t_k \leq t \leq b_0\), where \(b_0\) is a number independent of \(k\), \(a<b_0\leq b\), \(t_k\to a\) as \(k\to\infty\), and

\[ \lim_{t\to a}\left(\limsup_{k\to\infty}|x_k(t)-x_0|\right)=0. \tag{1.1} \]

Definition 1.2. Suppose that problem (1)—(2) has a solution. Then we shall say that equation (1) satisfies condition \((B)\) at the point \((a,x_0)\) if there exists a number \(b_0\), \(a<b_0\leq b\), such that all solutions of problem (1)—(2) are defined on the interval \(a\leq t\leq b_0\), and, moreover, whatever the solutions \(x^{(1)}(t)\) and \(x^{(2)}(t)\) of problem (1)—(2) and the sequence \(\{x_k(t)\}\) of solutions of equation (1), for large \(k\) each solution \(x_k(t)\) is defined on the interval \(t_k\leq t\leq b_0\) and condition (1.1) is satisfied, if
\[ x_k(t_k)=\nu_k x^{(1)}(t_k)+(1-\nu_k)x^{(2)}(t_k), \]
where \(0\leq \nu_k\leq 1\), \(a\leq t_k\leq b_0\), and \(t_k\to a\) as \(k\to\infty\).

Definition 1.3. We shall say that equation (1) satisfies condition \((C)\) at the point \((a,x_0)\) if for any sequence of points \(\{(t_k,x_k)\}\) from the domain \(R\), converging to the point \((a,x_0)\), and for any sequence of solutions \(\{x_k(t)\}\) of equation (1) satisfying the condition \(x_k(t_k)=x_k\) \((k=1,2,\ldots)\), each solution \(x_k(t)\) for large \(k\) is defined on the interval \(t_k\leq t\leq b_0\), where \(b_0\) is a number independent of \(k\), \(a<b_0\leq b\), and condition (1.1) is satisfied.

It is easy to verify that if condition \((C)\) is satisfied, then condition \((B)\) is satisfied, and if condition \((B)\) is satisfied, then condition \((A)\) is satisfied.

In addition to equation (1), below we shall consider the equation

\[ \frac{dx}{dt}=f(t,x;\mu), \tag{1.2} \]

where \(x\) is an \(n\)-dimensional vector, \(\mu\) is a \(p\)-dimensional vector, and \(f(t,x;\mu)\) is an \(n\)-dimensional vector-function defined in the domain

\[ R(\mu): a<t\leq b,\quad |x-x_0|<r,\quad |\mu|\leq 1, \]

with \(f(t,x;\mu)\) measurable in \(t\) for fixed \(x\) and \(\mu\), continuous in \(x\) and \(\mu\) for fixed \(t\), and throughout the domain \(R(\mu)\) satisfying the inequality

\[ |f(t,x;\mu)|\leq l(t), \tag{1.3} \]

where \(l(t)\) is a summable function on the interval \(a+\delta\leq t\leq b\) for every \(\delta>0\).

Definition 1.4. We shall say that equation (1.2) satisfies condition \((D)\) at the point \((a,x_0)\) if, for every \(\mu\), problem (1.2)—(2) has a solution, all solutions of this problem are defined on the interval \(a\leq t\leq b_0\), where \(b_0\) is a number independent of \(\mu\), and for any sequence \(\{x_k(t;\mu_k)\}\), where \(x_k(t;\mu_k)\) is a solution of problem (1.2)—(2) for \(\mu=\mu_k\), we have

\[ \lim_{t\to a}\left[\limsup_{k\to\infty}|x_k(t;\mu_k)-x_0|\right]=0. \tag{1.4} \]

\(^*\) In saying that a solution \(x(t)\) is defined on some interval \(t_1\leq t\leq t_2\), here and below we mean that \(|x(t)-x_0|<r\) for \(t_1<t\leq t_2\).

We shall say that a scalar function \(V(t,x)\) satisfies one or another condition along the solutions of equation (1), if the function \(V(t,x(t))\) satisfies this condition for every solution \(x(t)\) of equation (1).

By \(C^k_{[a_0,b_0]}\) we denote the space of \(k\)-dimensional continuous vector-functions on the interval \(a_0 \leqslant t \leqslant b_0\), with norm
\[ \|x(t)\|=\max_{a_0\leqslant t\leqslant b_0}\sum_{i=1}^k |x_i(t)|. \]
By \(S^k_{[a_0,b_0]}(x_0(t);r_0)\) and \(\overline S^k_{[a_0,b_0]}(x_0(t);r_0)\) we denote, respectively, the open and the closed ball of this space with center at \(x_0(t)\) and radius \(r_0\).

Below it will be more convenient for us to write problem (1)—(2) in the following form:
\[ \frac{dy}{dt}=g(t,y,z),\quad \frac{dz}{dt}=h(t,y,z), \tag{1.5} \]
\[ y(a)=y_0,\quad z(a)=z_0, \tag{1.6} \]
where \(y\) and \(g(t,y,z)\) are \(m\)-dimensional vectors, \(z\) and \(h(t,y,z)\) are \((n-m)\)-dimensional vectors, and the functions \(g(t,y,z)\) and \(h(t,y,z)\) satisfy the same conditions in the domain \(R\) as the function \(f(t,x)\).

Theorem 1.1. Let the functions \(V_1(t,y)\) and \(V_2(t,z)\) be continuous in the domain
\[ a<t\leqslant b_0,\quad |y-y_0|\leqslant r/4,\quad |z-z_0|\leqslant r/4, \]
\[ \liminf_{t\to a} V_1(t,y_0)=0,\quad V_2(b_0,z_0)=0,\quad V_1(t,y)>0,\quad V_2(t,z)>0 \]
for
\[ a<t\leqslant b_0,\quad |y-y_0|=r/4,\quad |z-z_0|=r/4 \tag{1.7} \]
and for every \(\varepsilon>0\) the conditions
\[ \liminf_{t\to a}\left[\inf_{|y-y_0|\geqslant \varepsilon} V_1(t,y)\right]>0,\quad \liminf_{t\to a}\left[\inf_{|z-z_0|\geqslant \varepsilon} V_2(t,z)\right]>0. \tag{1.8} \]
are satisfied.

If, for arbitrary
\[ \zeta(t)\in \overline S^{\,n-m}_{[a_0,b_0]}\left(z_0;\frac r4\right),\quad \eta(t)\in \overline S^{\,m}_{[a_0,b_0]}\left(y_0;\frac r4\right), \]
the function \(V_1(t,y)\) does not increase along solutions of the equation
\[ \frac{dy}{dt}=g(t,y,\zeta(t)), \tag{1.9} \]
and the function \(V_2(t,z)\) does not decrease along solutions of the equation
\[ \frac{dz}{dt}=h(t,\eta(t),z), \tag{1.10} \]
then equation (1.5) satisfies condition (A) at the point \((a,y_0,z_0)\).

For the proof of this theorem we shall need the following

Lemma 1.1. Let \(a<a_1<b_1\leqslant b\), \(|y_1-y_0|<r/4\), and \(|z_1-z_0|<r/4\).

If, for arbitrary
\[ \zeta(t)\in \overline S^{\,(n-m)}_{[a_1,b_1]}\left(z_0;\frac r4\right) \]
and
\[ \eta(t)\in \overline S^{\,m}_{[a_1,b_1]}\left(y_0;\frac r4\right), \]
for arbitrary solutions \(y(t)\) and \(z(t)\) of the problems
\[ \frac{dy}{dt}=g(t,y,\zeta(t)),\quad y(a_1)=y_1 \tag{1.11} \]
and
\[ \frac{dz}{dt}=h(t,\eta(t),z),\quad z(b_1)=z_1 \tag{1.12} \]

we have

\[ y(t) \in S^{m}_{[a_1,b_1]}\left(z_0;\frac{r}{4}\right),\quad z(t) \in S^{n-m}_{[a_1,b_1]}\left(z_0;\frac{r}{4}\right), \tag{1.13} \]

then there exists at least one solution of equation (1.5) satisfying the boundary conditions

\[ y(a_1)=y_1,\quad z(b_1)=z_1. \tag{1.14} \]

Proof. By the assumption adopted above, for \((t,y,z)\in R\) we have

\[ |g(t,y,z)|+|h(t,y,z)|\leq l(t), \tag{1.15} \]

where \(l(t)\) is a summable function on the interval \(a_1\leq t\leq b_1\).

Let \(M\) be the set of continuous \(n\)-dimensional vector-functions \(x(t)=(y(t),z(t))\) on the interval \(a_1\leq t\leq b_1\) satisfying the conditions

\[ \max_{a_0\leq t\leq b_0}\bigl(|y(t)-y_1|+|z(t)-z_1|\bigr)\leq r_1 \quad\text{and}\quad |y(t_2)-y(t_1)|+|z(t_2)-z(t_1)|\leq \]

\[ \leq \int_{t_1}^{t_2} l(s)\,ds \quad\text{for } a_1\leq t_1<t_2\leq b_1, \tag{1.16} \]

where

\[ r_1=\int_{a_1}^{b_1} l(s)\,ds. \]

It is clear that \(M\) is a closed, convex, and compact set of elements of the space \(C^n_{[a_1,b_1]}\).

Consider the operator \(Lx(t)=(L_1y(t),\,L_2z(t))\), defined on \(C^n_{[a_1,b_1]}\) as follows:

\[ L_1y(t)=y_1+\int_{a_1}^{t} g\bigl(\tau,\sigma_1[y(\tau)],\sigma_2[z(\tau)]\bigr)\,d\tau,\quad L_2z(t)=z_1+ \]

\[ +\int_{b_1}^{t} h\bigl(\tau,\sigma_1[y(\tau)],\sigma_2[z(\tau)]\bigr)\,d\tau, \tag{1.17} \]

where

\[ \sigma_1(y)= \begin{cases} y, & \text{if } |y-y_0|<\dfrac{r}{4},\\[6pt] y_0+\dfrac{r(y-y_0)}{4|y-y_0|}, & \text{if } |y-y_0|\geq\dfrac{r}{4}, \end{cases} \]

\[ \sigma_2(z)= \begin{cases} z, & \text{if } |z-z_0|<\dfrac{r}{4},\\[6pt] z_0+\dfrac{r(z-z_0)}{4|z-z_0|}, & \text{if } |z-z_0|\geq\dfrac{r}{4}. \end{cases} \tag{1.18} \]

Taking into account (1.15), (1.16), (1.17), and (1.18), it is easy to verify that the operator \(Lx(t)\) is continuous and maps the set \(M\) into itself. Therefore, by the well-known Schauder theorem [2], there exists a function \(x(t)=(y(t),z(t))\in M\) such that

\[ y(t)=y_1+\int_{a_1}^{t} g\bigl(\tau,\sigma_1[y(\tau)],\sigma_2[z,(\tau)]\bigr)\,d\tau, \]

\[ z(t)=z_1+\int_{b_1}^{t} h(\tau,\sigma_1[y(\tau)],\,\sigma_2[z(\tau)])\,d\tau . \tag{1.19} \]

We shall show that \(y(t)\in S_{[a_1,b_1]}^{m}\left(y_0;\dfrac r4\right)\). Indeed, otherwise, as is clear from (1.19), there would be a number \(t_0\), \(a_1<t_0\leq b_1\), such that

\[ |y(t)-y_0|<r/4 \quad \text{for } a_1\leq t<t_0 \quad \text{and} \quad |y(t_0)-y_0|=r/4 . \tag{1.20} \]

But this contradicts condition (1.13), since, as follows from (1.18) and (1.19), \(y(t)\) is a solution of problem (1.11) with \(\zeta(t)=\sigma_2[z(t)]\). Similarly we show that
\[ z(t)\in S_{[a_1,b_1]}^{\,n-m}\left(z_0;\dfrac r4\right). \]

Thus, \(x(t)=(y(t),z(t))\) satisfies condition (1.13) and the integral equation (1.19). Therefore, by (1.18), we conclude that it is a solution of problem (1.5)—(1.14). The lemma is proved.

Proof of Theorem 1.1. We shall first show that if

\[ V_1(a_1,y_1)<V_1 \quad \text{and} \quad V_2(b_1,z_1)<V_2, \]

where \(a<a_1<b_1\leq b_0\), \(V_1=\inf V_1(t,y)\) and \(V_2=\inf V_2(t,z)\), \(a<t\leq b_1\), \(|z-z_0|=r/4\), then problem (1.5)—(1.14) has a solution. For this it is enough to verify that the conditions of Lemma 1.1 are satisfied.

Let \(y(t)\) be some solution of problem (1.11). We shall show that
\[ y(t)\in S_{[a_1,b_1]}^{m}\left(y_0;\dfrac r4\right). \]
Indeed, otherwise, for some \(t_0\), \(a_1<t_0\leq b_1\), condition (1.20) holds. On the other hand, we have

\[ V_1(t_0,y(t_0))\leq V_1(a_1,y_1)<V_1, \]

since the function \(V_1(t,y(t))\) does not increase on the interval \(a_1\leq t\leq t_0\). But this is impossible, since \(V_1(t,y)\geq V_1\) when \(|y-y_0|=r/4\).

In exactly the same way we show that if \(z(t)\) is a solution of problem (1.12), then
\[ z(t)\in S_{[a_1,b_1]}^{\,n-m}\left(z_0;\dfrac r4\right). \]

Thus, all the conditions of Lemma 1.1 are satisfied and, consequently, problem (1.5)—(1.14) has a solution.

By virtue of (1.7) and (1.8), there exists a sequence \(\{t_k\}\) such that

\[ a<t_k<b_0,\quad V_1(t_k,y_0)<V_1 \quad \text{and} \]

\[ \lim_{k\to\infty} V_1(t_k,y_0)=0 . \tag{1.21} \]

According to the preceding argument, it is clear that if \(V_2(b_0,z_1)\leq 0\), then for every \(k\) equation (1.5) has a solution \((y_k(t),z_k(t))\) satisfying the boundary conditions

\[ y_k(t_k)=y_0,\qquad z_k(b_0)=z_1 . \]

Since

\[ V_1(t,y_k(t))\leq V_1(t_k,y_0),\qquad V_2(t,z_k(t))\leq 0 \quad \text{for } t_k\leq t\leq b_0 \]

and condition (1.21) holds, it follows, by condition (1.8), that for any \(\varepsilon>0\) there is a \(\delta>0\) such that, for \(t_k\leq t\leq a+\delta\), we have

\[ |y_k(t)-y_0|<\varepsilon \quad \text{and} \quad |z_k(t)-z_0|<\varepsilon, \]

which proves that equation (1.5) satisfies condition \((A)\) at the point \((a,x_0)\).

Corollary. If throughout the domain \(R\) the inequalities
\[ g(t,y,z)\cdot \operatorname{grad}|y-y_0|\leqslant \varphi(t), \]
\[ h(t,y,z)\cdot \operatorname{grad}|z-z_0|\geqslant \psi(t)|z-z_0|-\varphi(t), \tag{1.22} \]
hold, where the function \(\varphi(t)\) is nonnegative and summable on the interval \(a\leqslant t\leqslant b\), while the function \(\psi(t)\) is nonnegative, summable for every \(\delta>0\) on the interval \(a+\delta\leqslant t\leqslant b\), and
\[ \int_a^b \psi(t)\,dt=\infty, \tag{1.23} \]
then equation (1.4) satisfies condition \((A)\) at the point \((a,y_0,z_0)\).

Proof. Put
\[ V_1(t,y)=|y-y_0|-\int_a^t \varphi(s)\,ds \quad\text{and}\quad V_2(t,z)= \]
\[ = e^{\int_t^{b_0}\psi(s)\,ds}|z-z_0| -\int_t^{b_0} e^{\int_t^{b_0}\psi(\tau)\,d\tau}\varphi(s)\,ds, \]
where \(a<b_0\leqslant b\) and
\[ \int_a^{b_0}\varphi(t)\,dt<\frac r4, \]
then, taking (1.22) and (1.23) into account, it is easy to verify that all the conditions of Theorem 1.1 are satisfied.

Theorem 1.2. Let the function \(V(t,x)\) be continuous in the domain \(a<t\leqslant b_0\), \(|x-x_0|<r\), where \(a<b_0\leqslant b\),
\[ \liminf_{t\to a} V(t,x_0)=0,\qquad V(t,x)>0\quad\text{for }a<t\leqslant b_0,\ |x-x_0|=r, \tag{1.24} \]
and for every \(\varepsilon>0\) we have
\[ \liminf_{t\to a}\left[\inf_{|x-x_0|\geqslant \varepsilon} V(t,x)\right]>0. \tag{1.25} \]
If, moreover, the function \(V(t,x)\) does not increase along solutions of equation (1), and for any solutions \(x^{(1)}(t)\) and \(x^{(2)}(t)\) of problem (1)—(2) the condition
\[ \lim_{t\to a}\left[\sup_{0\leqslant \nu\leqslant 1} V\bigl(t,\nu x^{(1)}(t)+(1-\nu)x^{(2)}(t)\bigr)\right]=0, \tag{1.26} \]
holds, then equation (1) satisfies condition \((B)\) at the point \((a,x_0)\).

Proof. In analogy with the preceding, we shall show that equation (1) satisfies condition \((A)\) at the point \((a,x_0)\) and, consequently, as will be shown in § 2, problem (1)—(2) has a solution.

Let \(x(t)\) be some solution of problem (1)—(2). Since \(V(t,x(t))\) does not increase and
\[ \lim_{t\to a}V(t,x(t))=0, \]
therefore, according to (1.24), it is easy to conclude that
\[ x(t)\in S^n_{[a,b_0]}(x_0;r). \]

* Here and in what follows, \(x_1\cdot x_2\) denotes the scalar product of the two vectors \(x_1\) and \(x_2\).

From (1.24) and (1.25) it follows that

\[ V(t,x)\geqslant V_0>0 \quad \text{for } a<t\leqslant b_0,\quad |x-x_0|=r. \tag{1.27} \]

Suppose now that \(x^{(1)}(t)\) and \(x^{(2)}(t)\) are arbitrary solutions of problem (1)—(2), and \(\{x_k(t)\}\) is a sequence of solutions of equation (1) satisfying the conditions

\[ x_k(t_k)=\nu_k x^{(1)}(t_k)+(1-\nu_k)x^{(2)}(t_k)\quad (k=1,2,\ldots), \]

where \(0\leqslant \nu_k\leqslant 1\) and \(t_k\to a\) as \(k\to\infty\). From (1.26) it follows that, for large \(k\), the inequality

\[ V(t_k,x_k(t_k))<V_0 \]

holds. Hence \(x_k(t)\in S^n_{[t_k,b_0]}(x_0;r)\) and

\[ V(t,x_k(t))\leqslant V(t_k,x_k(t_k))\quad \text{for } t_k\leqslant t\leqslant b_0, \]

since \(V(t,x_k(t))\) is nonincreasing and (1.27) holds.

In view of the fact that \(\lim\limits_{k\to\infty} V(t_k,x_k(t_k))=0\) and condition (1.25) is satisfied, it follows from the last inequality that the sequence \(\{x_k(t)\}\) satisfies condition (1.1). The theorem is proved.

Theorem 1.3. If the function \(V(t,x)\) is continuous in the domain \(a\leqslant t\leqslant b_0,\ |x-x_0|<r\), where \(a<b_0\leqslant b\), is nonincreasing along the solutions of equation (1), and

\[ V(t,x)>0 \quad \text{for } a\leqslant t\leqslant b_0,\ |x-x_0|=r, \tag{1.28} \]

\[ V(a,x_0)=0 \quad \text{and} \quad V(a,x)>0 \quad \text{for } x\ne x_0, \tag{1.29} \]

then equation (1) satisfies condition (C) at the point \((a,x_0)\).

Proof. According to (1.28), we have

\[ V_0=\min_{\substack{a\leqslant t\leqslant b_0\\ |x-x_0|=r}} V(t,x)>0. \tag{1.30} \]

Let \(\{(t_k,x_k)\}\) be any sequence of points of the domain \(R\) converging to the point \((a,x_0)\), and let \(\{x_k(t)\}\) be a sequence of solutions of equation (1) satisfying the conditions \(x_k(t_k)=x_k\) \((k=1,2,\ldots)\). Since \(V(t,x_k(t))\) is nonincreasing and, for large \(k\), the inequality \(V(t_k,x_k)<V_0\) holds, therefore, according to (1.30), we conclude that \(x_k(t)\in S^n_{[t_k,b_0]}(x_0;r)\) and

\[ V(t,x_k(t))\leqslant V(t_k,x_k)\quad \text{for } t_k\leqslant t\leqslant b_0. \]

Hence, by virtue of (1.29), it easily follows that \(\{x_k(t)\}\) satisfies condition (1.1). The theorem is proved.

Corollary. If throughout the domain \(R\) the inequality

\[ f(t,x)\cdot \operatorname{grad}|x-x_0|\leqslant \varphi(t), \]

holds, where \(\varphi(t)\) is a summable function on the interval \(a\leqslant t\leqslant b\), then equation (1) satisfies condition (C) at the point \((a,x_0)\).

To verify the validity of this proposition, it is enough to check that the function

\[ V(t,x)=|x-x_0|-\int_a^t \varphi(s)\,ds \]

satisfies all the conditions of Theorem 1.3 in the domain \(a \leq t \leq b_0,\ |x-x_0|<r\), where \(a<b_0\leq b\) and \(\displaystyle \int_a^{b_0}\varphi(s)\,ds<r\).

Theorem 1.4. If the function \(V(t,x)\) is continuous in the domain \(a<t\leq b_0,\ |x-x_0|<r\), where \(a<b_0\leq b\), satisfies conditions (1.24) and (1.25), does not increase along the solutions of equation (1.2) for any \(\mu\), and for any solution \(x(t;\mu)\) of problem (1.2)—(2) we have

\[ \lim_{t\to a} V(t,x(t;\mu))=0, \tag{1.31} \]

then equation (1.2) satisfies condition \((D)\) at the point \((a,x_0)\).

Proof. Indeed, \(V(t,x)\) does not increase along the solutions of equation (1.2) and condition (1.31) is satisfied. Therefore, according to (1.24), it is clear that every solution \(x(t;\mu)\in S^n_{[a,b_0]}(x_0;r)\) and

\[ V(t,x(t;\mu))\leq 0\quad \text{for } a\leq t\leq b_0 . \]

Hence, by virtue of (1.25), it follows directly that equation (1.2) satisfies condition \((D)\) at the point \((a,x_0)\).

From the theorem just proved the following follows easily.

Corollary. If throughout the whole domain \(R(\mu)\) the inequality

\[ f(t,x;\mu)\cdot \operatorname{grad}|x-x_0|\leq \varphi(t) \]

holds, where \(\varphi(t)\) is a summable function on the interval \(a\leq t\leq b\), then equation (1.2) satisfies condition \((D)\) at the point \((a,x_0)\).

Remark. In Theorems 1.2—1.4 it was required that the function \(V(t,x)\) not increase along the solutions of equation (1) or (1.2). This restriction can be somewhat weakened by requiring instead that, for any \(t_0,\ a<t_0'<b\), along the solutions of equation (1) or (1.2) for \(t\geq t_0\) the inequality

\[ V(t,x(t))\leq \sigma[V(t_0,x(t_0))]+\int_{t_0}^{t}\omega[s,V(s,x(s))]\,ds, \tag{1.32} \]

hold, where \(\sigma(\lambda)\) is a continuous function equal to zero for \(\lambda=0\), and \(\omega(t,\lambda)\) is nondecreasing in \(\lambda\) and satisfies the Carathéodory conditions in every finite rectangle of the domain \(a\leq t\leq b,\ -\infty<\lambda<\infty\). Indeed, if \(V(t,x(t))\) does not increase, then we have

\[ V(t,x(t))\leq V(t_0,x(t_0))\quad \text{for } t\geq t_0, \tag{1.33} \]

and if (1.32) is satisfied, then, according to Lemma 5.1, we obtain

\[ V(t,x(t))\leq \bar{\lambda}(t;t_0,\sigma_0)\quad \text{for } t\geq t_0, \tag{1.34} \]

where \(\sigma_0=\sigma[V(t_0,x(t_0))]\), and \(\bar{\lambda}(t;t_0,\sigma_0)\) is the upper solution of the problem

\[ \frac{d\lambda}{dt}=\omega(t,\lambda),\qquad \lambda(t_0)=\sigma_0 . \]

Thus the entire course of the proof of Theorems 1.2—1.4 is preserved; only instead of inequality (1.33) we must apply inequality (1.34).

Let us now consider the problem

\[ u^{(n)}=f_0(t,u,u',\ldots,u^{(n-1)};\mu), \tag{1.35} \]

\[ u^{(i)}(a)=x_{0i+1}\quad (i=0,1,\ldots,n-1), \tag{1.36} \]

which is a particular case of problem (1.2)—(2) for \(u^{(i)}=x_{i+1}\), \(x=(x_1,x_2,\ldots,x_n)\), \(x_0=(x_{01},x_{02},\ldots,x_{0n})\), and

\[ f(t,x;\mu)=\bigl(x_2,\ldots,x_{n-1}, f_0(t,x;\mu)\bigr). \tag{1.37} \]

From the corollaries of Theorems 1.3 and 1.4 it follows directly that if throughout the entire domain \(R(\mu)\) the inequality

\[ f_0(t,x_1,\ldots,x_n)\operatorname{sign}(x_n-x_{0n})\leq \varphi(t), \]

holds, where \(\varphi(t)\) is a summable function on the interval \(a\leq t\leq b\), then equation (1.35) satisfies condition \((D)\), and, for any fixed \(\mu\), condition \((C)\) at the point \((a,x_{01},\ldots,x_{0n})\).

The following also holds.

Theorem 1.5. If throughout the entire domain \(R(\mu)\) the inequality

\[ f_0(t,x_1,\ldots,x_n;\mu)\operatorname{sign}(x_n-x_{0n}) \leq \omega\left(t,\sum_{j=1}^{n}(t-a)^{j-n}\left|x_j-\sigma^{(j-1)}(t)\right|\right), \tag{1.38} \]

holds, where

\[ \sigma(t)=\sum_{i=1}^{n}\frac{x_{0i}}{(i-1)!}(t-a)^{i-1}, \]

and the function \(\omega(t,\lambda)\) is nondecreasing in \(\lambda\) and satisfies the Carathéodory conditions in every finite rectangle of the domain \(a\leq t\leq b,\ -\infty<\lambda<\infty\), then equation (1.35) satisfies condition \((D)\), and, for any fixed \(\mu\), condition \((B)\) at the point \((a,x_{01},\ldots,x_{0n})\).

Proof. Put

\[ V(t,x)=\sum_{j=1}^{n}(t-a)^{j-n}\left|x_j-\sigma^{(j-1)}(t)\right|. \]

According to (1.37) and (1.38), for any solution
\(x(t;\mu)=(x_1(t;\mu),\ldots,x_n(t;\mu))\) of equation (1.2) with \(t\geq t_0>a\) we have

\[ \begin{aligned} \left|x_{n-i+1}(t;\mu)-\sigma^{(n-i)}(t)\right| &\leq \sum_{j=1}^{i} \left|x_{n-j+1}(t_0;\mu)-\sigma^{(n-j)}(t_0)\right| \frac{(t-t_0)^{i-j}}{(i-j)!} \\ &\quad+ \frac{1}{(i-1)!}\int_{t_0}^{t} (t-\tau)^{i-1}\omega[s,V(s,x(s;\mu))]\,ds \quad (i=1,2,\ldots,n). \end{aligned} \]

Dividing both sides of each of these inequalities by \((t-a)^{i-1}\) and summing, we obtain

\[ V(t,x(t;\mu))\leq 3V(t_0,x(t_0;\mu)) +3\int_{t_0}^{t}\omega[s,V(s,x(s;\mu))]\,ds. \]

Since, moreover, for any solution \(x(t;\mu)\) of problem (1.2)—(2) we have
\(\lim_{t\to a}V(t,x(t;\mu))=0\), according to the remark to Theorem 1.4, the validity of Theorem 1.5 becomes evident.

We note that under the conditions of Theorem 1.5, equation (1.35) does not always satisfy condition (C) at the point \((a, x_{01}, \ldots, x_{0n})\) for every fixed \(\mu\).

Indeed, consider the equation

\[ u^{(n)}=\sum_{i=1}^{n} t^{1-i} u^{(n-i)}, \tag{1.39} \]

where \(n \geqslant 2\). According to Theorem 1.5, it satisfies condition (B) at the point \(t=0,\ x_i=0\ (i=1,2,\ldots,n)\). Denote by \(u_k(t)\) the solution of equation (1.39) satisfying the initial conditions

\[ u_k^{(i)}\left(\frac{1}{k}\right)=\frac{1}{\ln k}\quad (i=0,1,\ldots,n-1). \]

It is clear that \(u_k^{(i)}(t)>\dfrac{1}{\ln k}\) for \(\dfrac{1}{k}<t<\infty\). Therefore, from (1.39) we find

\[ u_k^{(n-1)}(t)> \int_{1/k}^{t}\frac{u_k^{(n-2)}(s)}{s}\,ds > \frac{1}{\ln k}(\ln t+\ln k). \]

Hence it follows that

\[ u_k^{(n-1)}\left(\frac{1}{\sqrt{k}}\right)>\frac{1}{2}\quad (k=2,3,\ldots). \]

The inequality obtained shows that equation (1.39) does not satisfy condition (C) at the point \(t=0,\ x_i=0\ (i=1,2,\ldots,n)\).

§ 2. Existence Theorem

Lemma 2.1. Let a sequence \(\{x_k(t;\mu_k)\}\) be given, where \(x_k(t;\mu_k)\) is a solution of equation (1.2) for \(\mu=\mu_k\), defined on the interval \(t_k\leqslant t\leqslant b_0\), and \(b_0\) is a number independent of \(k\) and \(\mu_k\). If \(t_k\to a\) and \(\mu_k\to\mu_0\) as \(k\to\infty\), and condition (1.4) is satisfied, then from \(\{x_k(t;\mu_k)\}\) one can extract a subsequence converging uniformly to some solution \(x(t;\mu_0)\) of problem (1.2)—(2) for \(\mu=\mu_0\).

Proof. Put

\[ \tilde{x}_k(t;\mu_k)= \begin{cases} x_k(t_k;\mu_k), & \text{for } a\leqslant t<t_k,\\ x_k(t;\mu_k), & \text{for } t_k\leqslant t\leqslant b_0, \end{cases} \]

and prove that the sequence \(\{\tilde{x}_k(t;\mu_k)\}\) is equicontinuous on the interval \(a\leqslant t\leqslant b_0\).

According to condition (1.4), for any \(\varepsilon>0\) there exist a number \(\delta_1>0\) and a natural number \(N\) such that

\[ |\tilde{x}_k(t;\mu_k)-x_0|<\frac{\varepsilon}{2} \quad \text{for } a\leqslant t\leqslant a+2\delta_1,\ k\geqslant N. \tag{2.1} \]

Let \(\delta_2>0\) be so small a number that

\[ |\tilde{x}_k(t;\mu_k)-\tilde{x}_k(\tau;\mu_k)|<\varepsilon \tag{2.2} \]

for \(1 \leq k \leq N,\ a \leq t,\tau \leq b_0\) and \(|t-\tau| \leq \delta_2\), and

\[ \left|\int_{\tau}^{t} l(s)\,ds\right|<\varepsilon \tag{2.3} \]

for \(a+\delta_1 \leq t,\tau \leq b_0\) and \(|t-\tau| \leq \delta_2\), where \(l(t)\) is the function occurring in inequality (1.3).

Put \(\delta=\min\{\delta_1,\delta_2\}\). It is clear that if \(a \leq t,\tau \leq b_0\) and \(|t-\tau|\leq\delta\), then for \(1\leq k\leq N\) inequality (2.2) holds. We shall show that it also holds for \(k>N\). Indeed, if \(|t-\tau|\leq\delta\), then either \(a\leq t,\tau \leq a+2\delta_1\), or \(a+\delta_1 \leq t,\tau \leq b_0\). In the first case, by (2.1), we have

\[ |\widetilde{x}_k(t;\mu_k)-\widetilde{x}_k(\tau;\mu_k)| \leq |\widetilde{x}_k(t;\mu_k)-x_0| + |\widetilde{x}_k(\tau;\mu_k)-x_0| <\varepsilon, \]

while in the second case, by virtue of (1.3) and (2.3), we obtain

\[ |\widetilde{x}_k(t;\mu_k)-\widetilde{x}_k(\tau;\mu_k)| = \left|\int_{\tau}^{t} f(s,\widetilde{x}_k(s;\mu_k);\mu_k)\,ds\right| \leq \left|\int_{\tau}^{t} l(s)\,ds\right| <\varepsilon . \]

Thus the equicontinuity of the sequence \(\{\widetilde{x}_k(t;\mu_k)\}\) is proved. Since it is also uniformly bounded, by the well-known Arzelà–Ascoli theorem one can select from it a uniformly convergent subsequence \(\{\widetilde{x}_{k_i}(t;\mu_{k_i})\}\). Denote its limit by \(x(t;\mu_0)\).

Let \(\tau\) be an arbitrary number, \(a<\tau<b_0\). It is clear that for large \(i\) and for \(\tau\leq t\leq b_0\) we have

\[ \widetilde{x}_{k_i}(t;\mu_{k_i}) = \widetilde{x}_{k_i}(\tau,\mu_{k_i}) + \int_{\tau}^{t} f(s,x_{k_i}(s;\mu_{k_i});\mu_{k_i})\,ds . \]

Passing to the limit in both sides of this equality as \(i\to\infty\), we find

\[ x(t;\mu_0) = x(\tau;\mu_0) + \int_{\tau}^{t} f(s,x(s;\mu_0);\mu_0)\,ds \quad \text{for } \tau\leq t\leq b_0 . \]

In view of the arbitrariness of \(\tau\), it follows that \(x(t;\mu_0)\) is a solution of equation (1.2) for \(\mu=\mu_0\) on the interval \(a\leq t\leq b_0\). On the other hand, from (1.4) it is clear that \(x(a;\mu_0)=x_0\). Consequently, \(x(t;\mu_0)\) is a solution of the problem (1.2)—(2) for \(\mu=\mu_0\). The lemma is proved.

For \(f(t,x;\mu)\equiv f(t,x)\), this theorem yields

Lemma 2.1′. Let \(\{x_k(t)\}\) be a sequence of solutions of equation (1), such that each solution \(x_k(t)\) is defined on the interval \(t_k\leq t\leq b_0\), where \(t_k\to a\) as \(k\to\infty\), and \(b_0\) is a number independent of \(k\), and suppose that condition (1.1) is satisfied. Then from \(\{x_k(t)\}\) one can select a subsequence uniformly convergent to a solution \(x(t)\) of the problem (1)—(2).

It follows immediately from Lemma 2.1′ that if equation (1) satisfies condition (A) at the point \((a,x_0)\), then the problem (1)—(2) has a solution. On the other hand, it is clear (see Definition 1.1) that if the problem (1)—(2) has a solution, then equation (1) satisfies condition (A) at the point \((a,x_0)\). Consequently, the following theorem holds.

Theorem 2.1. In order that the problem (1)—(2) have a solution, it is necessary and sufficient that equation (1) satisfy condition (A) at the point \((a,x_0)\).

Remark. Suppose equation (1) can be represented in the form (1.5), where \(g(t,y,z)\) and \(h(t,y,z)\) are respectively \(m\)- and \((n-m)\)-dimensional vector functions \((m\ne n)\) satisfying the conditions of the corollary to Theorem 1.1. Analyzing the proofs of Theorem 1.1 and Lemma 2.1, we easily conclude that if

\[ \int_a^{b_0}\varphi(s)\,ds<\frac r4, \]

then for any \(z_1,\ |z_1-z_0|<\dfrac r4-\int_a^{b_0}\varphi(s)\,ds\), there exists a solution of problem (1.5)—(1.6) satisfying the condition \(z(b_0)=z_1\), i.e., problem (1.5)—(1.6) has a set of solutions depending on \(n-m\) arbitrary parameters.

§ 3. Structure of the integral funnel

Definition 3.1. Let \((\tau,\xi)\in R\) and let all solutions of equation (1) satisfying the initial condition

\[ x(\tau)=\xi, \tag{3.1} \]

be defined on the interval \(\tau\le t\le b_0\). Then the set of all points of the form \((t,x(t))\), where \(\tau\le t\le b_0\), while \(x(t)\) ranges over the set of all solutions of problem (1)—(3.1), will be called the right integral funnel of problem (1)—(3.1), defined on the interval \(\tau\le t\le b_0\), and will be denoted by \(X(\tau,\xi;b_0)\).

The left integral funnel is defined analogously.

By an integral funnel, unless the contrary is specified, we shall henceforth always mean the right integral funnel.

The integral funnel of problem (1)—(2) will be denoted simply by \(X\).

Theorem 3.1. If equation (1) satisfies condition (B) at the point \((a,x_0)\), then the set \(X_\tau\), obtained by intersecting the funnel \(X\) with the hyperplane \(t=\tau,\ a\le \tau\le b_0\), is closed and connected.

Proof. First we prove the closedness of the set \(X_\tau\), i.e., we show that if a sequence of points \(P_k=(\tau,x_k(\tau))\), where \(x_k(t)\) is a solution of problem (1)—(2), converges to a point \(P\), then \(P\in X_\tau\). Indeed, by condition (B) and Lemma 2.1, the sequence \(\{x_k(t)\}\) may be assumed to converge to some solution \(x(t)\) of problem (1)—(2), i.e.

\[ P=\lim_{k\to\infty}(\tau,x_k(\tau))=(\tau,x(\tau))\in X_\tau. \]

We now prove that \(X_\tau\) is a connected set. Suppose, on the contrary, that

\[ X_\tau=X_\tau^{(1)}\cup X_\tau^{(2)} \quad\text{and}\quad \rho\bigl(X_\tau^{(1)},X_\tau^{(2)}\bigr)=\rho_0>0^*, \tag{3.2} \]

where \(X_\tau^{(1)}\) and \(X_\tau^{(2)}\) are closed sets. Then there exist solutions \(x^{(1)}(t)\) and \(x^{(2)}(t)\) of problem (1)—(2) such that

\[ Q_i=(\tau,x^{(i)}(\tau))\in X_\tau^{(i)}. \tag{3.3} \]

Let \(a<t_k<b_0\) and \(t_k\to a\) as \(k\to\infty\). Consider the problem

\[ \frac{dx}{dt}=f(t,x),\qquad x(t_k^*)=\nu x^{(1)}(t_k)+(1-\nu)x^{(2)}(t_k), \tag{3.4} \]

where \(0\le \nu\le 1\). It is equivalent to the integral equation

\(*\) Here and in what follows, \(\rho(X,Y)\) denotes the distance between the sets \(X\) and \(Y\).

\[ x(t)=v x^{(1)}(t_k)+(1-v)x^{(2)}(t_k)+\int_{t_k}^{t} f(s,x(s))\,ds. \tag{3.5} \]

According to condition \((B)\), one may assume that all solutions of problem (3.4) are defined on the interval \(t_k \leq t \leq b_0\).

Let \(X_{k\tau}(v)\) be the section of the integral funnel of problem (3.4) by the hyperplane \(t=\tau\). Put

\[ X_{k\tau}=\bigcup_{0\leq v\leq 1} X_{k\tau}(v). \]

By virtue of (3) from (3.5) we easily conclude that the set \(X_{k\tau}\) is closed. Let us show that it is also connected. Suppose that

\[ X_{k\tau}=X_{k\tau}^{(1)}\cup X_{k\tau}^{(2)} \quad\text{and}\quad \rho\bigl(X_{k\tau}^{(1)},X_{k\tau}^{(2)}\bigr)>0, \tag{3.6} \]

where \(X_{k\tau}^{(i)}\) \((i=1,2)\) are closed sets. By the Kneser—Bokshtěin theorem [1], for any \(v\) the set \(X_{k\tau}(v)\) is connected. Therefore, according to (3.6), there is a sequence \(\{v_i\}\) such that \(v_i\to v_0\) as

\[ i\to\infty,\quad X_{k\tau}(v_i)\subset X_{k\tau}^{(1)}\ (i=1,2,\ldots) \quad\text{and}\quad X_{k\tau}(v_0)\subset X_{k\tau}^{(2)}. \]

Let \(x_k(t;v_i)\) be some solution of problem (3.4) for \(v=v_i\). By virtue of (3) from (3.5) it follows at once that the sequence \(\{x_k(t;v_i)\}_{i=1}^{\infty}\) is uniformly bounded and equicontinuous on the interval \(t_k\leq t\leq b_0\). Therefore it may be assumed to converge to some solution \(x_k(t;v_0)\) of problem (3.4) for \(v=v_0\). Consequently,

\[ (\tau,x_k(\tau;v_i))\to(\tau,x_k(\tau;v_0))\quad\text{as } i\to\infty, \]

which contradicts condition (3.6), since

\[ (\tau,x_k(\tau;v_i))\in X_{k\tau}^{(1)}\ (i=1,2,\ldots) \quad\text{and}\quad (\tau,x_k(\tau;v_0))\in X_{k\tau}^{(2)}. \]

The contradiction obtained proves the connectedness of the set \(X_{k\tau}\).

Since \(X_{k\tau}\) is a connected set and \(Q_i\in X_{k\tau}\) \((i=1,2)\), there exists a number \(v_k\) and a corresponding solution \(x_k(t)\) of problem (3.4) such that

\[ \rho\bigl(X_{\tau}^{(1)},P_k\bigr)=\frac{\rho_0}{2}, \tag{3.7} \]

where \(P_k=(\tau,x_k(\tau))\).

The sequence \(\{x_k(t)\}\) satisfies all the conditions of Lemma \(2.1'\). Therefore it may be assumed to converge to a solution \(x(t)\) of problem (1)—(2). Passing to the limit in equality (3.7), we find

\[ \rho\bigl(X_{\tau}^{(1)},P\bigr)=\frac{\rho_0}{2}, \tag{3.8} \]

where \(P=(\tau,x(\tau))\). According to (3.2), it follows from (3.8) that \(P\in \overline{X_{\tau}^{(i)}}\) \((i=1,2)\), which is impossible, since \(P\in X_\tau\). The contradiction obtained proves the theorem.

Remark. If condition \((B)\) is satisfied, then the integral funnel \(X\) of problem (1)—(2), just as in the classical cases, has a “funnel-shaped” form, i.e. the diameter of the set \(X_\tau\) tends to zero as \(\tau\to a\). It is natural to pose the following question: will it remain true—

the conclusion of Theorem 3.1 is also valid for such cases when \(X\) does not have the classical configuration. The answer to this question is affirmative, since the integral funnel of problem (1)—(2) coincides with the cylinder
\(a < t \leq b,\ |x-x_0| \leq r\), if

\[ f(t,x_0)\equiv 0,\qquad f(t,x)\equiv 0 \quad \text{for } |x-x_0|=r \]

and

\[ f(t,x)\cdot \operatorname{grad}|x-x_0|\geq \psi(t)\varphi(|x-x_0|)\quad \text{for } (t,x)\in R, \tag{3.9} \]

where \(\psi(t)\) is positive on the interval \(a<t\leq b\), summable on the interval \(a+\delta\leq t\leq b\) for every \(\delta>0\), and

\[ \int_a^b \psi(t)\,dt=\infty, \tag{3.10} \]

while the function \(\varphi(t)\) is continuous and positive for \(0<t<r\) and

\[ \int_0^r \frac{dt}{\varphi(t)}<\infty \tag{3.11} \]

for every \(\delta>0\). Indeed, according to (3.9), along any integral curve passing through a point \((t_0,x)\in R\), we have

\[ \int_{|x(t)-x_0|}^{|x-x_0|} \frac{ds}{\varphi(s)} \geq \int_t^{t_0}\psi(s)\,ds \quad \text{for } t\leq t_0. \]

Hence, by virtue of (3.10), we easily conclude that \(|x(t)-x_0|\leq |x-x_0|\) for \(a\leq t\leq t_0\) and \(x(a)=x_0\). It remains to show that through every point \((t_0,\bar{x})\), where \(a<t_0\leq b\) and \(|\bar{x}-x_0|=r\), there passes at least one integral curve of problem (1)—(2). Consider a sequence of points \((t_k,x_k)\in R\) converging to the point \((t_0,\bar{x})\), and a sequence of solutions \(\{x_k(t)\}\) of equation (1) satisfying the condition \(x_k(t_0)=x_k\). Since the sequence \(\{x_k(t)\}\) is uniformly bounded and equicontinuous on every interval \(t_1\leq t\leq t_0\), where \(a<t_1<t_0\), it may be regarded as uniformly convergent to a solution \(\bar{x}(t)\) of equation (1). If \(x(t_2)=x_0\) for some \(t_2,\ t_1<t_2<t_0\), then, as easily follows from (3.9), \(x(t)\equiv x_0\) for \(a\leq t\leq t_2\). If, however, \(\bar{x}(t)\ne x_0\) for \(t_1\leq t\leq t_0\), then, according to (3.11), we may pass to the limit in the inequality

\[ \int_{|x_k(t_1)-x_0|}^{|x_k-x_0|} \frac{ds}{\varphi(s)} \geq \int_{t_1}^{t_0}\psi(s)\,ds \]

and obtain

\[ \int_{|\bar{x}(t_1)-x_0|}^{|\bar{x}-x_0|} \frac{ds}{\varphi(s)} \geq \int_{t_1}^{t_0}\psi(s)\,ds. \]

Consequently, \((t_1,\bar{x}(t_1))\in R\). Therefore, by what was proved above, \(\bar{x}(t)\) exists also on the interval \(a\leq t\leq t_1\), and \(\bar{x}(a)=x_0\).

Theorem 3.2. If equation (1) satisfies condition (B) at the point \((a,x_0)\), then through every point \((\tau,\xi)\), \(a<\tau<b_0\), on the boundary of the integral funnel \(X\) there passes at least one integral curve of problem (1)—(2) which, between the points \((a,x_0)\) and \((\tau,\xi)\), belongs entirely to the boundary \(X\).

Proof. We first prove that for any number \(\tau_1\), \(a<\tau_1<\tau\), there exists an integral curve of problem (1)—(2) whose point with abscissa \(\tau_1\) belongs to the boundary \(X\).

According to condition (B), there exists a number \(r_0\), \(0<r_0<r\), such that every solution of problem (1)—(2) satisfies the condition

\[ |x(t)-x_0|<r_0 \quad \text{for } a\leq t\leq b_0 . \tag{3.12} \]

Consider the equation

\[ \frac{dx}{dt}=\tilde f(t,x), \tag{3.13} \]

where

\[ \tilde f(t,x)= \begin{cases} f(t,x), & \text{for } |x-x_0|\leq r_0,\\[6pt] f\!\left(t,x_0+\dfrac{r(x-x_0)}{|x-x_0|}\right), & \text{for } |x-x_0|>r_0 . \end{cases} \tag{3.14} \]

Every solution of problem (1)—(2) satisfies condition (3.12). Therefore, according to (3.14), it is easy to conclude that problems (1)—(2) and (3.13)—(2) have the same solutions on the interval \(a\leq t\leq b_0\).

According to (3) and (3.14), it is clear that all solutions of equation (3.13) satisfying the initial condition

\[ x(\tau)=\xi, \tag{3.15} \]

are defined on the interval \(a<t\leq \tau\). Let \(\widetilde X\) be the left integral funnel of problem (3.13)—(3.15), and let \(X_{\tau_1}\) and \(\widetilde X_{\tau_1}\) be the sections of the funnels \(X\) and \(\widetilde X\) by the hyperplane \(t=\tau_1\).

Consider a sequence of points \(\{(\tau,\xi_k)\}\), not belonging to \(X\), tending to \((\tau,\xi)\) as \(k\to\infty\). Let \(\{\tilde x_k(t)\}\) be a sequence of solutions of equation (3.13) satisfying the condition \(\tilde x_k(\tau)=\xi_k\). We show that

\[ (t,\tilde x_k(t))\notin X \quad \text{for } a<t\leq \tau \quad (k=1,2,\ldots). \tag{3.16} \]

Assume the contrary, that for some \(k\) and \(\tau_2\), \(a<\tau_2<\tau\), we have \((\tau_2,\tilde x_k(\tau_2))\in X\). It is clear that the function

\[ \tilde x(t)= \begin{cases} x(t), & \text{for } a\leq t\leq \tau_2,\\ \tilde x_k(t), & \text{for } \tau_2\leq t\leq \tau, \end{cases} \]

where \(x(t)\) is a solution of problem (1)—(2) satisfying the condition
\[ x(\tau_2)=\tilde x_k(\tau_2), \]
will be a solution of problem (3.13)—(2) and, consequently, of problem (1)—(2). But, on the other hand, \((\tau,\tilde x(\tau))=(\tau,\xi_k)\notin X\). The contradiction obtained proves that (3.16) holds.

From (3) and (3.14) it follows easily that the sequence \(\{\tilde x_k(t)\}\) is uniformly bounded and equicontinuous on the interval \(t_1\leq t\leq \tau\), where \(a<t_1\). Therefore it may be assumed to converge uniformly.

Let

\[ \lim_{k\to\infty}\tilde x_k(t)=\tilde x(t). \]

By virtue of (3.16) it is clear that the arc \((t,\tilde{x}(t))\), \(t_1 \leq t \leq \tau\), contains no interior point of the funnel \(X\). Thus the point \((t_1,\tilde{x}(t_1))\) belongs to the set \(\widetilde{X}_{t_1}\) and is an exterior or boundary point of the set \(X_{t_1}\). But, on the other hand, \(X_{t_1}\) and \(\widetilde{X}_{t_1}\) intersect, since there exists an integral curve of equation (1) connecting the points \((a,x_0)\) and \((\tau,\xi)\). According to the aforementioned Kneser–Boksztejn theorem and Theorem 3.1, the sets \(X_{t_1}\) and \(\widetilde{X}_{t_1}\) are closed and connected. Consequently, \(\widetilde{X}_{t_1}\) contains at least one boundary point \((t_1,\bar{\xi})\) of the set \(X_{t_1}\). Since the problems (1)—(2) and (3.13)—(2) have the same solutions on the interval \(a \leq t \leq b_0\), the integral curve of equation (3.13) joining the point \((\tau,\bar{\xi})\) to the points \((a,x_0)\) and \((\tau,\xi)\) is the required integral curve of problem (1)—(2).

Let us now consider a sequence \(\{t_k\}\), everywhere dense on the interval \(a < t \leq \tau\). According to the preceding argument, for every \(k\) it is easy to construct a solution \(x_k(t)\) of problem (1)—(2) such that \((t_i,x_k(t_i))\) \((i=1,2,\ldots,k)\) are boundary points of the set \(X\). By condition \((B)\), the sequence \(\{x_k(t)\}\) satisfies all the conditions of Lemma \(2.1'\), and therefore it may be regarded as converging to a solution \(x(t)\) of problem (1)—(2). Since the points \((t_k,x(t_k))\), for all \(k\), belong to the boundary of \(X\), it is clear that the arc \((t,x(t))\), \(a \leq t \leq \tau\), lies entirely on the boundary of the set \(X\). The theorem is proved.

§ 4. Dependence of solutions on the initial data and on a parameter

Theorem 4.1. Let \(X\) be the integral funnel of problem (1)—(2), defined on the interval \(a \leq t \leq b_1\). Then, in order that the integral funnel of problem (1)—(3.1), for sufficiently small \(|\tau-a|+|\xi-x_0|\), \((\tau,\xi)\in R\), be defined on the interval \(\tau \leq t \leq b_1\) and

\[ \max_{M\in X(\tau;\xi;b_0)} \rho(M,X) \to 0 \quad \text{as } \quad (\tau,\xi)\to(a,x_0), \tag{4.1} \]

it is necessary and sufficient that equation (1) satisfy condition \((C)\) at the point \((a,x_0)\).

Proof. Necessity is obvious. Let us prove sufficiency. Since \(X\) is defined on the interval \(a \leq t \leq b_1\), there is a number \(r_0\), \(0<r_0<r\), such that for any solution \(x(t)\) of problem (1)—(2) we have

\[ |x(t)-x_0|<r_0 \quad \text{for } \quad a \leq t \leq b_1 . \tag{4.2} \]

We shall show that if \(\delta>0\) is sufficiently small and \(|\tau-a|+|\xi-x_0|\leq \delta\), then for any solution \(\tilde{x}(t)\) of problem (3.13)—(3.1) we have

\[ |\tilde{x}(t)-x_0|<r_0 \quad \text{for } \quad \tau \leq t \leq b_1 . \tag{4.3} \]

Suppose the contrary. Then there is a sequence \(\{\tilde{x}_k(t)\}\) of solutions of equation (3.13) such that

\[ \max_{t_k \leq t \leq b_1} |\tilde{x}_k(t)-x_0| \geq r_0, \tag{4.4} \]

\[ t_k \to a \quad \text{and} \quad |\tilde{x}_k(t_k)-x_0| \to 0 \quad \text{as } \quad k\to\infty . \tag{4.5} \]

According to (4.5), from conditions (3.14) and \((C)\) it follows that each function \(\tilde{x}_k(t)\), for large \(k\), is a solution of equation (1) on the interval \(t_k \leq\)

\(\leq t \leq b_0\), where \(b_0\) is a number independent of \(k\). Therefore, from the equality

\[ \tilde{x}_k(t)=\tilde{x}_k(b_0)+\int_{b_0}^{t}\tilde{f}(s,\tilde{x}_k(s))\,ds \]

it follows that

\[ |\tilde{x}_k(t)-x_0|<r+\int_{b_0}^{b_1} l(s)\,ds \quad \text{for } t_k \leq t \leq b_1, \]

i.e., the sequence \(\{\tilde{x}_k(t)\}\) is uniformly bounded on the interval \(t_k \leq t \leq b_1\). Thus, the sequence \(\{\tilde{x}_k(t)\}\) satisfies all the conditions of Lemma \(2.1'\). Therefore it may be regarded as converging to a solution \(\tilde{x}(t)\) of problem (3.13)—(2).

Since any solution of problem (1)—(2) satisfies condition (4.2), it is clear from (3.14) that problems (1)—(2) and (3.13)—(2) have the same solutions on the interval \(a \leq t \leq b_1\). Consequently, \(\tilde{x}(t)\) is a solution of problem (1)—(2). On the other hand, from (4.4) we obtain

\[ \max_{a\leq t\leq b_1} |\tilde{x}(t)-x_0| \geq r_0, \]

which contradicts condition (4.2). The contradiction obtained shows that inequality (4.3) holds. Hence, by virtue of (3.14), it follows that, for \(|\tau-a|+|\xi-x_0|\leq \delta\), problem (1)—(3) has an integral funnel defined on the interval \(\tau \leq t \leq b_1\).

Finally, let us prove that condition (4.1) holds. Indeed, if we assume the contrary, then there exists a sequence \(\{x_k(t)\}\) of solutions of equation (1) such that each solution \(x_k(t)\) is defined on the interval \(t_k \leq t \leq b_1\), \(t_k^* \to a\), \(|x_k(t_k)-x_0| \to 0\) as \(k\to\infty\), and

\[ \limsup_{k\to\infty}\left[\max_{t_k\leq t\leq b_1}\rho(P_k(t),X)\right]\geq \varepsilon>0, \]

where \(P_k(t)=(t,x_k(t))\), which is impossible, since, according to condition \((C)\), the sequence \(\{x_k(t)\}\) satisfies all the requirements of Lemma \(2.1'\), and therefore it may be regarded as uniformly convergent to a solution \(x(t)\) of problem (1)—(2). The contradiction obtained proves the theorem.

Corollary. Let problem (1)—(2) have a unique solution \(x(t)\), defined on the interval \(a \leq t \leq b_1\), and let, for any point \((\tau,\xi)\in R\), problem (1)—(3.1) have a unique solution \(x(t;\tau,\xi)\). Then, in order that \(x(t;\tau,\xi)\), for sufficiently small \(|\tau-a|+|\xi-x_0|\), be defined on the interval \(\tau \leq t \leq b_1\) and

\[ \max_{\tau\leq t\leq b_1}|x(t;\tau,\xi)-x(t)|\to 0 \quad \text{as } (\tau,\xi)\to(a,x_0), \]

it is necessary and sufficient that equation (1) satisfy condition \((C)\) at the point \((a,x_0)\).

Theorem 4.2. Let problem (1.2)—(2) for \(\mu=\mu_0\) have an integral funnel \(X(\mu_0)\), defined on the interval \(a \leq t \leq b_1\). Then, in order that the funnel \(X(\mu)\) of problem (1.2)—(2) for sufficiently small \(|\mu-\mu_0|\) be defined on the interval \(a \leq t \leq b_1\) and

\[ \max_{M\in X(\mu)} \rho(M,X(\mu_0))\to 0 \quad \text{as } \mu\to\mu_0, \tag{4.6} \]

it is necessary and sufficient that equation (1.2) satisfy condition \((D)\) at the point \((a,x_0)\).

Proof. We shall prove only sufficiency, since necessity is obvious. According to the conditions of the theorem, for any solution \(x(t;\mu_0)\) of problem (1.2)—(2) with \(\mu=\mu_0\) we have

\[ |x(t;\mu_0)-x_0|<r_0 \quad \text{for } a\leq t\leq b_1, \tag{4.7} \]

where \(0<r_0<r\). We shall show that, for sufficiently small \(\delta\), every solution \(x(t;\mu)\) of problem (1.2)—(2) also satisfies the condition

\[ |x(t;\mu)-x_0|<r_0 \quad \text{for } a\leq t\leq b_1, \tag{4.8} \]

if \(|\mu-\mu_0|\leq\delta\).

Consider the equation

\[ \frac{dx}{dt}=\tilde f(t,x;\mu), \tag{4.9} \]

where

\[ \tilde f(t,x;\mu)= \begin{cases} f(t,x;\mu), & \text{if } |x-x_0|<r_0,\\[6pt] f\!\left(t,x_0+\dfrac{r(x-x_0)}{|x-x_0|};\mu\right), & \text{if } |x-x_0|\geq r_0. \end{cases} \tag{4.10} \]

According to (4.9), the integral curves of problems (1.2)—(2) and (4.9)—(2) in the domain \(a\leq t\leq b_1,\ |x-x_0|<r_0\) coincide. Therefore, in order to prove (4.8), it is enough to show that any solution \(\tilde x(t;\mu)\) of problem (4.9)—(2) satisfies the inequality

\[ |\tilde x(t;\mu)-x_0|<r_0 \quad \text{for } a\leq t\leq b_1, \tag{4.11} \]

if \(|\mu-\mu_0|\leq\delta\). Suppose the contrary. Then there exists a sequence \(\{\tilde x(t;\mu_k)\}\), where \(\tilde x(t;\mu_k)\) is a solution of problem (4.9)—(2) with \(\mu=\mu_k\), such that \(\mu_k\to\mu_0\) as \(k\to\infty\) and

\[ \max_{a\leq t\leq b_1}|\tilde x(t;\mu_k)-x_0|\geq r_0. \tag{4.12} \]

According to condition \((D)\), it is easy to verify that \(\{\tilde x(t;\mu_k)\}\) satisfies all the requirements of Lemma 2.1. Therefore it may be regarded as converging to a solution \(\tilde x(t;\mu_0)\) of problem (4.9)—(2) for \(\mu=\mu_0\). Every solution of problem (1.2)—(2) satisfies inequality (4.7). Therefore, according to (4.10), it is clear that \(\tilde x(t;\mu_0)\) is a solution of problem (1.2)—(2), which is impossible, since from (4.12) it follows that

\[ \max_{a\leq t\leq b_1}|\tilde x(t;\mu_0)-x_0|\geq r_0. \]

This proves that condition (4.11), and consequently condition (4.8), holds.

According to condition \((D)\) and (4.8), we now easily conclude that condition (4.6) holds.

Corollary. Suppose that for every \(\mu\) problem (1.2)—(2) has a unique solution \(x(t;\mu)\), and suppose that \(x(t;\mu_0)\) is defined on the interval \(a\leq t\leq b_1\). Then, in order that \(x(t;\mu)\), for sufficiently small \(|\mu-\mu_0|\), be defined on the interval \(a\leq t\leq b_1\) and

\[ \max_{a\leq t\leq b_1} |x(t;\mu)-x(t;\mu_0)|\to 0 \quad \text{as } \mu\to\mu_0, \]

it is necessary and sufficient that equation (1.2) satisfy condition (D) at the point \((a,x_0)\).

§ 5. Uniqueness theorem

Before formulating the theorem, we introduce the following

Definition 5.1. We shall say that the function \(W(t,x)\) is positive definite in the domain \(a<t\leq b,\ |x|\leq r_0\), if \(W(t,x)=0\) for \(|x|=0\) and \(W(t,x)>0\) for \(|x|\ne 0\).

Theorem 5.1. Let the function \(W(t,x)\) be continuous and positive definite in the domain \(a<t\leq b,\ |x|\leq 2r\), and let the function \(\omega(t,\lambda)\) satisfy the Carathéodory conditions in every finite rectangle of the domain \(a\leq t\leq b,\ -\infty<\lambda<\infty\), be nondecreasing in \(\lambda\), \(\omega(t,0)=0\), and let the problem

\[ \frac{d\lambda}{dt}=\omega(t,\lambda),\quad \lambda(a)=0 \tag{5.1} \]

have only the trivial solution. If for any two solutions \(x(t)\) and \(y(t)\) of problem (1)—(2) the conditions

\[ \lim_{t\to a} W(t,x(t)-y(t))=0 \tag{5.2} \]

and

\[ W(t,x(t)-y(t))\leq \int_a^t \omega\{s,W(s,x(s)-y(s))\}\,ds, \tag{5.3} \]

hold, then problem (1)—(2) has no more than one solution.

The proof of this theorem is based on the following simple lemma.

Lemma 5.1. Let the function \(\omega(t,\lambda)\) be nondecreasing in \(\lambda\) and satisfy the Carathéodory conditions in every finite rectangle of the domain \(a\leq t\leq b,\ -\infty<\lambda<\infty\), and let the function \(\bar{\lambda}(t)\) be an upper solution of the equation

\[ \frac{d\lambda}{dt}=\omega(t,\lambda) \tag{5.4} \]

with initial condition \(\lambda(t_0)=\lambda_0\), defined on the interval \(t_0\leq t\leq t_1\). Then for any continuous function \(W(t)\) satisfying on the interval \(t_0\leq t\leq t_1\) the inequality

\[ W(t)\leq \lambda_0+\int_{t_0}^t \omega(s,W(s))\,ds, \tag{5.5} \]

we have

\[ W(t)\leq \bar{\lambda}(t) \quad \text{for } t_0\leq t\leq t_1. \tag{5.6} \]

Proof. Let \(\lambda_k>\lambda_0\) and \(\lambda_k\to\lambda_0\) as \(k\to\infty\), and let \(\{\bar{\lambda}_k(t)\}\) be a sequence of upper solutions of equation (5.4) with initial conditions \(\bar{\lambda}_k(t_0)=\lambda_k\) \((k=1,2,\ldots)\). We shall prove that for every \(k\) the condition

\[ W(t)<\bar{\lambda}_k(t) \quad \text{for } t_0\leq t\leq t_1. \tag{5.7} \]

is satisfied. Suppose the contrary. Then for some \(k\) and \(t_2,\ t_0<t_2\leq t_1\), we shall have

\[ W(t)<\bar{\lambda}_k(t)\quad \text{for } t_0\leq t<t_2 \quad \text{and}\quad W(t_2)=\lambda_k(t_2). \]

On the other hand, taking into account that \(\omega(t,\lambda)\) is nondecreasing in \(\lambda\), by (5.5) we find

\[ W(t_2)-\bar{\lambda}_k(t_2)\leq \lambda_k-\lambda_0+ \int_{t_0}^{t_1}[\omega(s,W(s))-\omega(s,\lambda(s))]\,ds<0. \]

The contradiction obtained proves the validity of inequality (5.7). Passing to the limit in this inequality, as \(k\to\infty\), we obtain the required inequality (5.6). The lemma is proved.

Proof of Theorem 5.1. Suppose that problem (1)—(2) has two distinct solutions \(x(t)\) and \(y(t)\), defined on the interval \(a\leq t\leq b_0\). Since inequalities (5.2) and (5.3) are satisfied and problem (5.1) has only the trivial solution, it follows, by Lemma 5.1, that

\[ W(t,x(t)-y(t))\leq 0\quad \text{for } a\leq t\leq b_0. \tag{5.8} \]

Taking into account that \(W(t,x)\) is positive definite, we obtain from this \(x(t)\equiv y(t)\). This contradiction proves the theorem.

Corollary 1. Let the function \(W(t,x)\) be continuous and positive definite in the domain \(a<t\leq b,\ |x|\leq 2r\). If for any two solutions \(x(t)\) and \(y(t)\) of problem (1)—(2) the function \(W(t,x(t)-y(t))\) is nonincreasing and condition (5.2) is satisfied, then problem (1)—(2) has at most one solution.

Indeed, since \(W(t,x(t)-y(t))\) is nonincreasing and (5.2) is satisfied, inequality (5.8) holds; hence, as noted above, it follows that \(x(t)\equiv y(t)\).

Corollary 2. If, for \((t,x),(t,y)\in R\), the inequality

\[ [f(t,x)-f(t,y)]\cdot \operatorname{grad}|x-y|\leq \varphi(t)\omega(|x-y|) \tag{5.9} \]

is satisfied, where \(\varphi(t)\) is a function summable on the interval \(a\leq t\leq b\), and \(\omega(t)\) is continuous and positive on the interval \(0<t\leq 2r\), and

\[ \int_0^{2r}\frac{dt}{\omega(t)}=\infty, \tag{5.10} \]

then problem (1)—(2) has at most one solution.

Proof. Put

\[ W(t,x)=e^{-\int_a^t\varphi(s)\,ds-\int_{|x|}^{2r}\frac{ds}{\omega(s)}}. \tag{5.11} \]

Let \(x(t)\) and \(y(t)\) be arbitrary solutions of problem (1)—(2). By virtue of (5.9) and (5.10), from (5.11) we conclude that

\[ W'(t,x(t)-y(t))\leq 0 \]

and (5.2) holds. Consequently, all the conditions of the preceding corollary are satisfied, whence the uniqueness of the solution of problem (1)—(2) follows.

Corollary 3. If, for \((t,x_1,\ldots,x_n),(t,y_1,\ldots,y_n)\in R\), the inequality is satisfied

\[ [f_0(t,x_1,\ldots,x_n;\mu)-f_0(t,y_1,\ldots,y_n;\mu)]\operatorname{sign}(x_n-y_n)\le \]

\[ \le \omega\left(t,\sum_{k=1}^{n}(t-a)^{k-n}|x_k-y_k|\right), \tag{5.12} \]

where \(\omega(t,\lambda)\) satisfies all the conditions of Theorem 5.1, then for any \(\mu\) the problem (1.35)—(1.36) has at most one solution.

Proof. Put

\[ W(t,x)=\sum_{k=1}^{n}(n-k)!(t-a)^{k-n}|x_k|. \tag{5.13} \]

According to (1.37), from (5.13) we conclude that for any two solutions
\(x(t)=(x_1(t),\ldots,x_n(t))\), \(y(t)=(y_1(t),\ldots,y_n(t))\) of problem (1.2)—(2) condition (5.2) is satisfied. On the other hand, if \(t_0>a\), then, according to (5.12), we find

\[ k!|x_{n-k}(t)-y_{n-k}(t)|\le k!\sum_{i=0}^{k}\frac{(t-t_0)^i}{i!}|x_{n-i}(t_0)-y_{n-i}(t_0)|+ \]

\[ +\int_{t_0}^{t}(t-s)^k\omega[s,W(s,x(s)-y(s))]\,ds. \]

Dividing both sides of these inequalities by \((t-t_0)^k\), passing to the limit as \(t_0\to a\), and summing over \(k\), we obtain

\[ W(t,x(t)-y(t))\le \int_{a}^{t}\omega[s,W(s,x(s)-y(s))]\,ds. \]

Thus, all the conditions of Theorem 5.1 are satisfied, whence the validity of our assertion follows.

The author expresses deep gratitude to L. G. Magnaradze for valuable advice in the preparation of the present article.

References

1. Bokshtein M. F. Uch. zap. MGU, issue 15, mathematics, 3—72, 1939.
2. Kantorovich L. V., Akilov G. P. Functional Analysis in Normed Spaces. Moscow, 1959.
3. Caratheodory C. Vorlesungen über reelle Funktionen. Leipzig, 1927.
4. Kiguradze I. T. Communications of the Academy of Sciences of the Georgian SSR, 37, No. 1, 19—24, 1965.
5. Coddington E. A., Levinson N. Theory of Ordinary Differential Equations. IL, 1958.
6. Chechik V. A. Investigation of systems of ordinary differential equations with a singularity. Proceedings of the Moscow Mathematical Society, 8, 155—198, 1959.

Received by the editors
March 22, 1965

Tbilisi State
University