MATHEMATICS

N. V. MARCHENKO

EXISTENCE OF SOLUTIONS FOR A CERTAIN CLASS OF NONLINEAR INTEGRAL EQUATIONS

(Presented by Academician I. G. Petrovskii on 27 X 1960)

Consider the equation

\[ \varphi(x)=\int_0^1 K[x,y,\varphi(y)]\,dy, \tag{1} \]

where the function \(K(x,y,z)\) is defined either on the set \(P(A,B)\), or on \(P_{AB}\), where \(P(A,B)\) is defined by the inequalities: \(0\le x\le 1,\ 0\le y\le 1,\ A<z<B\), and \(P_{AB}\) by the inequalities: \(0\le x\le 1,\ 0\le y\le 1,\ -\infty<A\le z\le B<+\infty\).

We shall call equation (1) \(M\)-solvable if it has a solution and there exist constants \(C,D\) \((A<C<D<B)\) such that if \(\varphi(x)\) is any solution of it, then \(C<\varphi(x)<D\) for \(0\le x\le 1\).

In the present note several sufficient conditions for the \(M\)-solvability of equation (1) are given. The proofs of these conditions are based on the ideas of work \((^1)\).

Lemma 1. Let the function \(K(x,y,z)\) be continuous on \(P_{AB}\), the functions \(K_n(x,y,z)\) converge to it uniformly on \(P_{AB}\), and the equation

\[ \varphi(x)=\int_0^1 K_n[x,y,\varphi(y)]\,dy \]

has a solution for every \(n\). Then equation (1) also has a solution.

Let \(I\) be the unit square \(0\le x\le 1,\ 0\le y\le 1\) of the \(x,y\)-plane. Let \(Q(A,B)\) and \(Q_{AB}\) be the sets of points of the \(y,z\)-plane defined respectively by the inequalities \(0\le y\le 1,\ A<z<B\), and \(0\le y\le 1,\ -\infty<A\le z\le B<+\infty\). Let \(\Delta_i^m\) be the set of points of the \(x\)-axis defined by the inequality \(\frac{i-1}{m}\le x<\frac{i}{m}\) for \(i<m\), and by the inequality \(\frac{m-1}{m}\le x\le 1\) for \(i=m\).

We shall call a function \(K(x,y,z)\) stepwise on a set \(P\) of the space \((x,y,z)\) if it is continuous with respect to the variables \(y,z\) on \(P\) and if \(K(x,y,z)=K\left(\frac{i-1}{m},y,z\right)\) for \(x\in\Delta_i,\ i=1,2,\ldots,m,\ \{x,y,z\}\in P\).

Lemma 2. Let the function \(K(x,y,z)\) be stepwise on \(P_{AB}\), the function \(f(y,z)\) continuous on \(Q_{AB}\),

\[ \left[A-\int_0^1 f(y,A)\,dy\right]\cdot \left[B-\int_0^1 f(y,B)\,dy\right]<0, \]

\[ K^t(x,y,z)\equiv(1-t)K(x,y,z)+tf(y,z) \]

for \(\{x,y,z\}\in P_{AB},\ 0\le t\le 1\). Then, if every solution \(\varphi(x)\) of the equation

\[ \varphi(x)=\int_0^1 K^t[x,y,\varphi(y)]\,dy \tag{2} \]

satisfies the inequality \(A<\varphi(x)<B\), then equation (2) has a solution for any \(0\leq t\leq 1\).

Let the function \(K(x,y,z)\) be defined on \(P(-\infty,+\infty)\). Put, for \(z>0\),

\[ K(z)=\sup |K(x,y,t)| \quad \text{for } \{x,y,t\}\in P_{-z,z}. \]

Lemma 3. Let, for \(z>0\), the function \(g(z)\geq 0\) and

\[ \overline{\lim}_{z\to+\infty}\frac{g(z)}{z}<1. \]

Then there exist constants \(C,D\) \((-\infty<C<D<+\infty)\) such that, for any function \(K(x,y,z)\) defined on \(P(-\infty,+\infty)\) and such that, for \(z>0\), \(K(z)\leq g(z)\), every solution \(\varphi(x)\) of equation (1) satisfies the inequality \(C<\varphi(x)<D\).

Proof. Suppose that the lemma is false. Then there exist functions

\[ \varphi_n(x)=\int_0^1 K_n[x,y,\varphi_n(y)]\,dy, \]

where the functions \(K_n(x,y,z)\) satisfy the conditions of the lemma, such that, putting \(M_n=\sup_x|\varphi_n(x)|\), we obtain

\[ \lim_{n\to+\infty} M_n=+\infty. \]

Since \(g(M_n)\geq K_n(M_n)\geq |K_n[x,y,\varphi_n(y)]|\) for \(\{x,y\}\in I\), \(n=1,2,\ldots\), then for a sequence \(\{\xi_n\}\) such that \(|\varphi(\xi_n)|>M_n-M_n/n\), from (1) we can write

\[ 1=\frac{\int_0^1 K_n[\xi_n,y,\varphi_n(y)]\,dy}{\varphi(\xi_n)} \leq \frac{\int_0^1 |K_n[\xi_n,y,\varphi_n(y)]|\,dy}{|\varphi_n(\xi_n)|} \leq \frac{g(M_n)}{M_n(1-1/n)}, \]

whence, by virtue of the condition

\[ \overline{\lim}_{z\to+\infty}\frac{g(z)}{z}<1, \]

we obtain a contradiction.

Lemma 4. Let the function \(K(x,y,z)\) be a step function on \(P(-\infty,+\infty)\), and

\[ \overline{\lim}_{z\to+\infty}\frac{K(z)}{z}<1. \]

Then equation (1) is \(M\)-solvable.

Proof. Let

\[ K^t(x,y,z)\equiv (1-t)K(x,y,z)+tK(0,y,z) \]

for \(\{x,y,z\}\in P(-\infty,+\infty)\), \(0\leq t\leq 1\). Then \(K^t(z)\leq K(z)\) for \(z>0\), \(0\leq t\leq 1\). By Lemma 3 there exist constants \(C,D\) \((-\infty<C<D<+\infty)\) such that every solution \(\varphi(x)\) of equation (2) satisfies the inequality \(C<\varphi(x)<D\) for \(0\leq x\leq 1\), \(0\leq t\leq 1\). Using the condition

\[ \overline{\lim}_{z\to+\infty}\frac{K(z)}{z}<1, \]

we obtain

\[ \left[C-\int_0^1 K(0,y,C)\,dy\right]<0,\qquad \left[D-\int_0^1 K(0,y,D)\,dy\right]>0. \]

Applying Lemma 2 to (2), we obtain the required result, since for \(t=0\) (2) becomes (1).

Theorem 1. Let the function \(K(x,y,z)\) be continuous on \(P(-\infty,+\infty)\) and

\[ \overline{\lim}_{z\to+\infty}\frac{K(z)}{z}<1. \]

Then equation (1) is \(M\)-solvable.

Proof. Let

\[ K^m(x,y,z)\equiv K\left(\frac{i-1}{m},y,z\right) \]

for \(x\in\Delta_i\), \(i=1,2,\ldots,m\), \(\{x,y,z\}\in P(-\infty,+\infty)\). The functions \(K^m(x,y,z)\) converge to \(K(x,y,z)\) uniformly on every \(P_{CD}\), and \(K^m(z)\leq K(z)\) for \(z>0\), \(m=1,2,\ldots\).

Using successively Lemmas 4, 3, 1, we obtain the \(M\)-solvability of equation (1).

Let the function \(f(y,z)\geq 0\) be continuous on \(Q(0,+\infty)\). Put

\[ \overline{f}(z)=\sup \frac{f(y,z)}{z}\quad \text{for } 0\leq y\leq 1,\qquad \underline{f}(z)=\inf \frac{f(y,z)}{z}\quad \text{for } 0\leq y\leq 1. \]

We shall say,

that the function \(f(y,z)\) is \(\alpha_1,\alpha_2\)-bounded, if either

\[ \overline{\lim}_{z\to 0} f(z)<\frac{1}{\alpha_2} \]

and

\[ \lim_{z\to+\infty} f(z)>\frac{1}{\alpha_1}, \]

or

\[ \lim_{z\to 0} f(z)>\frac{1}{\alpha_1} \quad\text{and}\quad \overline{\lim}_{z\to+\infty} f(z)<\frac{1}{\alpha_2}, \]

where \(\alpha_2>\alpha_1>0\).

Lemma 5. Let the function \(f(y,z)\) be \(\alpha_1,\alpha_2\)-bounded. Then there exist constants \(C,D\) \((0<C<D<+\infty)\) such that, for every function \(K(x,y,z)\) defined on \(P(0,+\infty)\) and such that
\[ \alpha_1 f(y,z)\leq K(x,y,z)\leq \alpha_2 f(y,z) \]
for \(\{x,y,z\}\in P(0,+\infty)\), every solution \(\varphi(x)\) of equation (1) satisfies the inequality
\[ C<\varphi(x)<D. \]

Proof. Let

\[ \varphi(x)=\int_0^1 K[x,y,\varphi(y)]\,dy, \]

where the function \(K(x,y,z)\) satisfies the conditions of the lemma. Then

\[ \alpha_1\int_0^1 f[y,\varphi(y)]\,dy \leq \int_0^1 K[x,y,\varphi(y)]\,dy \leq \alpha_2\int_0^1 f[y,\varphi(y)]\,dy. \tag{3} \]

Suppose that the lemma is false. Then there exist functions
\[ \varphi_n(x)=\int_0^1 K_n[x,y,\varphi_n(y)]\,dy, \]
where the functions \(K_n(x,y,z)\) satisfy the conditions of the lemma, such that, if
\[ M_n=\sup_x \varphi_n(x),\qquad m_n=\inf_x \varphi_n(x), \]
then either
\[ \lim_{n\to+\infty} M_n=+\infty, \]
or
\[ \lim_{n\to+\infty} m_n=0. \]
We restrict ourselves to the case
\[ \lim_{n\to+\infty} M_n=+\infty. \]

From (3) we have
\[ \lim_{n\to+\infty} m_n=+\infty. \]
Let
\[ \lim_{z\to+\infty} f(z)>\frac{1}{\alpha_1}. \]
Then, for a sequence \(\{\xi_n\}\) such that
\[ \varphi(\xi_n)<m_n+\frac{m_n}{n}, \]
using (1), (3), and the continuity of \(f(y,z)\), we obtain

\[ 1= \frac{\displaystyle\int_0^1 K_n[\xi_n,y,\varphi_n(y)]\,dy}{\varphi(\xi_n)} \geq \alpha_1 \frac{\displaystyle\int_0^1 f[y,\varphi_n(y)]\,dy}{m_n+m_n/n} \geq \alpha_1 \frac{f(z_n)z_n}{(1+1/n)m_n}, \]

where \(0\leq y_n\leq 1\), \(m_n\leq z_n\leq M_n\); whence, taking into account
\[ \lim_{n\to+\infty} z_n=+\infty \]
and
\[ \lim_{z\to+\infty} f(z)>\frac{1}{\alpha_1}, \]
we obtain a contradiction. In the case
\[ \overline{\lim}_{z\to+\infty} f(z)<\frac{1}{\alpha_2}, \]
we take as \(\{\xi_n\}\) a sequence such that
\[ \varphi(\xi_n)>M_n-\frac{M_n}{n}; \]
the subsequent arguments are analogous.

Lemma 6. Let the function \(K(x,y,z)\) be stepwise on \(P(0,+\infty)\), and
\[ \alpha_1 f(y,z)\leq K(x,y,z)\leq \alpha_2 f(y,z) \]
for \(\{x,y,z\}\in P(0,+\infty)\), where the function \(f(y,z)\) is \(\alpha_1,\alpha_2\)-bounded. Then equation (1) is \(M\)-solvable.

Theorem 2. Let the function \(K(x,y,z)\) be continuous on \(P(0,+\infty)\), and
\[ \alpha_1 f(y,z)\leq K(x,y,z)\leq \alpha_2 f(y,z) \]
for \(\{x,y,z\}\in P(0,+\infty)\), where the function \(f(y,z)\) is \(\alpha_1,\alpha_2\)-bounded. Then equation (1) is \(M\)-solvable.

Lemma 6 and Theorem 2 are proved analogously to Lemma 4 and Theorem 1, except that Lemma 5 is used instead of Lemma 3.

If, in equation (1),
\[ K(x,y,z)=K(x,y)f(y,z), \]
then we obtain the equation

\[ \varphi(x)=\int_0^1 K(x,y)f[y,\varphi(y)]\,dy, \tag{4} \]

and from Theorem 2 there follows Theorem 3.

Theorem 3. Let the function \(K(x,y)\) be continuous on \(I\), the function \(f(y,z)\) continuous on \(Q(0,+\infty)\), and
\[ \overline{\lim}_{z\to+\infty} K\frac{g(z)}{z}<1, \]
where
\[ K=\sup_{\{x,y\}\in I}|K(x,y)|, \qquad g(z)=\sup_{\{y,t\}\in Q_{-z,z}} |f(y,t)|. \]
Then equation (4) is \(M\)-solvable.

Further, from Theorem 3, for \(f(y,z)=z^\alpha\), it follows:

Theorem 4. Let the function \(K(x,y)>0\) and be continuous on \(I\). Then the equation

\[ \varphi(x)=\int_0^1 K(x,y)\varphi^\alpha(y)\,dy \tag{5} \]

for \(\alpha\ne 1\) is \(M\)-solvable.

Let us note that for \(\alpha=1\), (5) is a Fredholm equation of the second kind and may either have or fail to have solutions under the conditions of Theorem 4. For \(-1\le \alpha<1\), the equation has a unique solution \(\varphi(x)\), with \(\varphi(x)=\lim_{n\to+\infty}\varphi_n(x)\), where \(\varphi_0(x)\) is any positive function continuous for \(0\le x\le 1\), and

\[ \varphi_{n+1}(x)=\varphi_n^{\frac{\alpha}{\alpha-1}}(x) \left[\int_0^1 K(x,y)\varphi_n^\alpha(y)\,dy\right]^{\frac{1}{1-\alpha}}. \]

This fact was proved by A. S. Kronrod for \(-1\le \alpha\le 0\). For \(|\alpha|>1\), for each \(\alpha\) one can find an infinitely differentiable function \(K(x,y)\) such that equation (5) has at least 3 solutions.

Received
6 X 1960

References

1. J. Leray, J. Schauder, UMN, 1, no. 3–4 (13–14), 23 (1943).