On the Question of Oscillation of Solutions of Nonlinear Differential Equations

I. T. Kiguradze

Consider the equation

\[ \frac{d^m u}{dt^m}+f(t,u)=0, \tag{1} \]

where the function \(f(t,u)\) is defined in the domain \(0\le t<\infty\), \(|u|<\infty\), is continuous in \(u\) for every fixed \(t\), summable in \(t\) on every finite interval of the positive half-axis for every fixed \(u\), and, in addition, satisfies the following conditions:

\[ f(t,0)\equiv 0,\qquad f(t,u)\ne 0\quad \text{for }0\le t<\infty,\ \text{if }u\ne 0, \tag{2} \]

\[ f(t,u)u\ \text{does not change sign}, \tag{3} \]

\[ |f(t,u_1)|\le |f(t,u_2)|\quad \text{for } |u_1|<|u_2|,\ u_1u_2>0. \tag{4} \]

In the note [2] we considered the problem of oscillation of solutions of differential equations of the indicated type in the case when \(f(t,u)/u\) does not decrease with respect to \(u\). In the present article we consider the same problem under the following assumption on the growth of \(f(t,u)\) with respect to \(u\):

\[ |f(t,\lambda u)|\le M\lambda^n |f(t,u)|,\quad 0\le n<1,\quad \text{for }0\le t<\infty,\ |u|\ge \delta,\ \lambda\ge \lambda_0>0, \tag{5} \]

where \(\delta\) is an arbitrary positive number, and \(M\), \(n\), and \(\lambda_0\) are constants which in general depend on \(\delta\).

If for \(\delta=1\) we choose the numbers \(M\), \(n\), and \(\lambda_0\) according to condition (5), then by virtue of (4) we find

\[ |f(t,u)|\le |f(t,\lambda_0)|+|f(t,-\lambda_0)| \le M\lambda_0^n\bigl(|f(t,1)|+|f(t,-1)|\bigr) \quad \text{for } |u|\le \lambda_0 \]

and

\[ |f(t,u)|\le M|u|^n\bigl(|f(t,1)|+|f(t,-1)|\bigr) \quad \text{for } |u|\ge \lambda_0. \]

Hence it is clear that for all \(t\) and \(u\) the inequality

\[ |f(t,u)|\le M\bigl(1+\lambda_0^n+|u|^n\bigr) \bigl(|f(t,1)|+|f(t,-1)|\bigr). \tag{6} \]

holds. Consequently, \(f(t,u)\) satisfies all the conditions of Carathéodory’s existence theorem [4].

Let \(u(t)\) be some solution of equation (1). Put

\[ r(t)=1+\lambda_0^n+\sum_{k=0}^{m-1}|u^{(k)}(t)|. \]

Assuming \(M\ge 1\), in accordance with (6), we find

\[ |r'(t)|\le M\bigl(1+|f(t,1)|+|f(t,-1)|\bigr)r(t). \]

It follows from this that

\[ r(t) \leq r(t_0)e^{\left|M\int_{t_0}^{t}(1+|f(\tau,1)|+|f(\tau,-1)|)\,d\tau\right|}. \]

The inequality obtained shows that every solution \(u(t)\) of equation (1) can be continued to the whole interval \([0,\infty)\).

Below we shall use the following definitions.

A solution of equation (1) is called oscillatory if it changes sign an infinite number of times in the interval \((0,\infty)\), and otherwise it is called nonoscillatory.

A nontrivial solution of equation (1) is called singular*, if it is identically equal to zero starting from some value of the argument, and otherwise it is called ordinary.

The minimal interval outside which a singular solution is identically equal to zero is called the support of the singular solution.

Lemma 1. If \(u(t)\) is an absolutely continuous function of constant sign together with its derivatives up to order \((m-1)\), inclusive, in the interval \((t_0,\infty)\), and

\[ u^{(m)}(t)u(t)\leq 0, \tag{7} \]

then there exists a number \(l\), \(0\leq l\leq m-1\), of parity opposite to that of the number \(m\), such that for \(t\geq t_0\) we have

\[ u^{(k)}(t)u(t)\geq 0 \quad (k=0,1,\ldots,l),\quad (-1)^{m+k-1}u^{(k)}(t)u(t)\geq 0 \]

\[ (k=l+1,\ldots,m) \tag{8} \]

and

\[ |u(t)|\geq \frac{(t-t_0)^{m-1}}{(m-1)\cdots(m-l)} \left|u^{(m-1)}(2^{m-l-1}t)\right|. \tag{9} \]

Proof. The validity of inequality (8) is verified quite easily (see [3], Lemma 2). Let us prove that (9) holds. Without loss of generality we may assume that

\[ u(t)\geq 0. \tag{10} \]

According to (8) and (10), we have

\[ -u^{(m-2)}(t) = -u^{(m-2)}(\infty) + \int_t^\infty u^{(m-1)}(\tau)\,d\tau \geq \int_t^{2t} u^{(m-1)}(\tau)\,d\tau \geq t\,u^{(m-1)}(2t), \]

\[ u^{(m-3)}(t) \geq -\int_t^\infty u^{(m-2)}(\tau)\,d\tau \geq \int_t^{2t}\tau\,u^{(m-1)}(2\tau)\,d\tau \geq t^2u^{(m-1)}(4t), \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ u^{(l)}(t)\geq t^{m-l-1}u^{(m-1)}(2^{m-l-1}t). \]

Since \(u^{(l-1)}(t)\geq 0\), from the last inequality we find

\[ u^{(l-1)}(t) = u^{(l-1)}(t_0) + \int_{t_0}^{t}u^{(l)}(\tau)\,d\tau \geq \int_{t_0}^{t}(\tau-t_0)^{m-l-1}u^{(m-1)}(2^{m-l-1}\tau)\,d\tau \geq \]

\[ \geq \frac{(t-t_0)^{m-l}}{m-l} u^{(m-1)}(2^{m-l-1}t). \]

Hence, after \((l-1)\)-fold integration, the required inequality (9) is obtained. The lemma is proved.

* Clearly, equation (1) has singular solutions only in the case when the conditions for uniqueness of the solution of the Cauchy problem are violated.

Lemma 2. If \(u(t)\) is an absolutely continuous, constant-sign function together with its derivatives up to order \((m-1)\) inclusive on the interval \((t_0,\infty)\), and

\[ u^{(m)}(t)u(t)\geqslant 0, \tag{11} \]

then either

\[ u^{(k)}(t)u(t)\geqslant 0\quad (k=0,1,\ldots,m), \tag{12} \]

or there exists a number \(l\) of the same parity as \(m\), \(0\leq l\leq m-2\), such that

\[ u^{(k)}(t)u(t)\geqslant 0\quad (k=0,1,\ldots,l),\quad (-1)^{m+k}u^{(k)}(t)u(t)\geqslant 0 \]

\[ (k=l+1,\ldots,m) \tag{13} \]

and inequality (9) is satisfied.

Proof. Two cases are possible: either \(u^{(m-1)}(t)u(t)\geqslant 0\), or \(u^{(m-1)}(t)u(t)\leqslant 0\). If the first of these holds, then by virtue of (11) it is easy to conclude that (12) holds; while if \(u^{(m-1)}(t)u(t)\leqslant 0\), then by Lemma 1 there exists a number \(l\), \(0\leq l\leq m-2\), of parity opposite to that of \(m-1\), for which (13) is satisfied and

\[ |u(t)|\geq \frac{(t-t_0)^{m-2}}{(m-2)\ldots(m-l-1)} \left|u^{(m-2)}\left(2^{m-l-2}t\right)\right|. \]

Hence, taking into account that

\[ \left|u^{(m-2)}(t)\right|\geq t\left|u^{(m-1)}(2t)\right|, \]

inequality (9) is obtained immediately.

Lemma 3. Let \(u(t)\) be a solution of the differential equation

\[ \frac{d^m u}{dt^m}+f_1(t,u)=0, \tag{14} \]

defined on the interval \((0,t_0)\), and let \(v(t)\) be a nonnegative solution on the same interval of the differential equation

\[ \frac{d^m v}{dt^m}+f_2(t,v)=0. \tag{15} \]

If

\[ (-1)^{m-1}\bigl[f_1(t,x_1)-f_2(t,x_2)\bigr]>0 \quad\text{for }0\leq t\leq t_0,\ x_1\geq x_2\geq 0 \tag{16} \]

and

\[ (-1)^k\bigl[u^{(k)}(t_0)-v^{(k)}(t_0)\bigr]>0 \quad (k=0,1,\ldots,m-1), \tag{17} \]

then for \(0\leq t\leq t_0\) we shall have

\[ (-1)^k\bigl[u^{(k)}(t)-v^{(k)}(t)\bigr]>0 \quad (k=0,1,\ldots,m-1). \tag{18} \]

Proof. By virtue of (14) and (15) we have

\[ (-1)^k\bigl[u^{(k)}(t)-v^{(k)}(t)\bigr] = \sum_{i=k}^{m-1}(-1)^i \bigl[u^{(i)}(t_0)-v^{(i)}(t_0)\bigr] \frac{(t_0-t)^{i-k}}{(i-k)!} + \]

\[ + \frac{(-1)^{m-1}}{(m-k-1)!} \int_t^{t_0} (\tau-t)^{m-k-1} \bigl[f_1(\tau,u(\tau))-f_2(\tau,v(\tau))\bigr]\,d\tau \]

\[ (k=0,1,\ldots,m-1). \tag{19} \]

From (17) it follows that inequalities (18) hold on some interval \((t^*,t_0]\). We shall show that \(t^*=0\). Indeed, otherwise,

in that case we would have \(u^{(m-1)}(t^{**})=v^{(m-1)}(t^{**})\). But this is impossible, since by virtue of (16) and (17), from (19) we have

\[ (-1)^{m-1}\bigl[u^{(m-1)}(t^{**})-v^{(m-1)}(t^{**})\bigr]> (-1)^{m-1}\bigl[u^{(m-1)}(t_0)-v^{(m-1)}(t_0)\bigr]>0. \]

Corollary 1. Let \((-1)^{m-1}f(t,u)u\ge 0\), and let the function \(u(t)\) be an ordinary, bounded, nonoscillatory solution of equation (1). Then the inequalities

\[ (-1)^k u^{(k)}(t)u(t)>0\qquad (k=0,1,\ldots,m-1) \tag{20} \]

hold on the whole interval \((0,\infty)\).

Proof. From conditions (2), (3), and (4) it is clear that \(u(t)\), together with its derivatives up to order \(m-1\) inclusive, is different from zero on some interval \([t_0,\infty)\). In view of the boundedness of \(u(t)\), it follows from Lemmas 1 and 2 that the inequalities (20) hold on \([t_0,\infty)\). Since

\[ \frac{d^m |u|}{dt^m}+f_1(t,|u|)=0, \]

where

\[ (-1)^{m-1}f_1(t,|u|)=(-1)^{m-1}f(t,u)\operatorname{sign}u\ge 0 \]

and

\[ (-1)^k u^{(k)}(t_0)\operatorname{sign}u(t_0)>0\qquad (k=0,1,\ldots,m-1), \]

it follows from Lemma 3 that the inequalities (20) also hold on the interval \([0,t_0]\).

Corollary 2. Let \((-1)^{m-1}f(t,u)u\ge 0\), and let the function \(u(t)\) be a nonoscillatory singular solution of equation (1), whose support is the interval \([0,t_0)\). Then on \([0,t_0)\) the inequalities (20) hold.

Proof. Since \(u(t)\) is nonoscillatory, there exists a number \(t^{**}\), \(0\le t^{**}<t_0\), such that

\[ u(t)\ne 0\quad \text{for } t^{**}\le t<t_0. \]

Therefore, from the equalities

\[ (-1)^k u^{(k)}(t)= \frac{(-1)^{m-1}}{(m-k-1)!} \int_t^{t_0}(\tau-t)^{m-k-1}f(\tau,u(\tau))\,d\tau \]

\[ (k=0,1,\ldots,m-1) \]

it follows that the inequalities (20) hold on \([t^{**},t_0)\). Applying now Lemma 3, we see that these inequalities are preserved also on the interval \([0,t^{**}]\).

Lemma 4. If \((-1)^{m-1}f(t,u)u\le 0\), then equation (1) has no nonoscillatory singular solutions.

Proof. Suppose the contrary, that equation (1) has a nonoscillatory singular solution \(u(t)\), and let \((0,t_0)\) be the support of this solution. From the nonoscillation of \(u(t)\) it follows that \(u(t)\ne 0\) on some interval \([t^{**},t_0)\), and from the equality

\[ u(t)= \frac{(-1)^{m-1}}{(m-1)!} \int_t^{t_0}(\tau-t)^{m-1}f(\tau,u(\tau))\,d\tau \]

we obtain the contradictory result:

\[ |u(t)|= \frac{(-1)^{m-1}}{(m-1)!} \int_t^{t_0}(\tau-t)^{m-1}f(\tau,u(\tau))\operatorname{sign}u(\tau)\,d\tau<0 \quad \text{for } t^{**}\le t<t_0. \]

Lemma 5. If

\[ \int_0^\infty |f(t,t^{m-1})|\,dt \]

or

\[ \int_0^\infty |f(t,-t^{m-1})|\,dt \]

converges, then there exists a solution \(u(t)\) of equation (1) such that

\[ \lim_{t\to\infty} u^{(m-1)}(t)=c_0\ne 0 . \tag{21} \]

Proof. For definiteness, assume that

\[ \int_0^\infty |f(t,t^{m-1})|\,dt<\infty . \]

First consider the case when

\[ f(t,u)u\ge 0 . \tag{22} \]

Choose \(t_0\) so large that

\[ \int_{t_0}^{\infty} |f(t,t^{m-1})|\,dt<\frac12 . \tag{23} \]

Let \(u(t)\) be a solution of equation (1) satisfying the initial conditions

\[ u(t_0)=u'(t_0)=\ldots=u^{(m-2)}(t_0)=0,\quad u^{(m-1)}(t_0)=1 . \tag{24} \]

We shall show that \(u(t)\) is the desired solution.

By virtue of (22) and (24), it follows from (1) that on some interval \((t_0,t_1)\) the inequalities

\[ u^{(k)}(t)>0\quad (k=0,1,\ldots,m-1),\quad u^{(m)}(t)\le 0, \]

\[ 0<u(t)\le \frac{u^{(m-1)}(t_0)}{(m-1)!}(t-t_0)^{m-1}<t^{m-1} \tag{25} \]

hold.

We shall show that \(t_1=\infty\). Indeed, otherwise we would have \(u^{(m-1)}(t_1)=0\), which is impossible, since by virtue of (23) and (25), from (1) we obtain

\[ u^{(m-1)}(t)=1-\int_{t_0}^{t} f(\tau,u(\tau))\,d\tau >1-\int_{t_0}^{\infty} f(\tau,\tau^{m-1})\,d\tau>\frac12, \tag{26} \]

where \(t_0\le t\le t_1\).

Thus, the inequalities (25) and (26) hold on the entire interval \((t_0,\infty)\). Hence it follows that \(u^{(m-1)}(t)\) tends to a finite limit different from zero as \(t\to\infty\).

Now consider the case when

\[ f(t,u)u\le 0 . \tag{27} \]

Let \(u(t)\) be a solution of equation (1) satisfying the following initial conditions:

\[ u(t_0)=u'(t_0)=\ldots=u^{(m-2)}(t_0)=0,\quad u^{(m-1)}(t_0)=\lambda_0>0 . \tag{28} \]

According to (27) and (28), from (1) we have

\[ u^{(k)}(t)\ge 0\quad (k=0,1,\ldots,m),\quad 0<u(t)<t^{m-1}u^{(m-1)}(t)\quad \text{for } t\ge t_0 . \tag{29} \]

Since \(u^{(m-1)}(t)\) is positive and nondecreasing, to complete the proof of the lemma it is enough to show that \(u^{(m-1)}(t)\) is a bounded function. Indeed, if \(t_0>0\) and \(\lambda_0\) is a sufficiently large number, then by virtue of (5) and (29), from (1) we find

\[ u^{(m-1)}(t) \leq \lambda_0+\int_{t_0}^{t} f(\tau,\tau^{m-1}u^{(m-1)}(\tau))\,d\tau \leq \lambda_0+M|u^{(m-1)}(t)|^n\int_{t_0}^{t} f(\tau,\tau^{m-1})\,d\tau . \]

Consequently,

\[ [u^{(m-1)}(t)]^{1-n}\leq \lambda_0^{1-n}+M\int_{t_0}^{\infty} f(\tau,\tau^{m-1})\,d\tau \quad \text{for } t>t_0 . \]

The lemma is proved.

Theorem 1. If \(f(t,u)u\geq 0\), then for even \(m\) the condition

\[ \int_{0}^{\infty}|f(t,t^{m-1})|\,dt = \int_{0}^{\infty}|f(t,-t^{m-1})|\,dt = \infty \tag{30} \]

is necessary and sufficient for the oscillation of all nontrivial solutions of equation (1), while for odd \(m\) this condition is necessary and sufficient in order that every nontrivial solution of equation (1) either oscillate or tend monotonically to zero as \(t\to\infty\), together with its derivatives up to order \((m-1)\), inclusive.

Proof. The necessity follows from Lemma 5. Let us prove sufficiency. We begin with the case where \(m\) is even. Suppose the contrary: that under condition (30), equation (1) has a nontrivial nonoscillatory solution \(u(t)\). By Lemma 4, \(u(t)\) will be an ordinary solution, and from Lemma 1 it follows that \(|u(t)|\) is nondecreasing on some interval \([t_0,\infty)\). Therefore, from (9) it is clear that for \(t\geq 2^m t_0\) the inequality

\[ |u(t)|\geq |u(2^{1-m}t)|\geq A|u^{(m-1)}(t)|t^{m-1}, \tag{31} \]

holds, where

\[ A=\frac{2^{-m^2}}{(m-1)!}. \]

Taking into account that \(u^{(m-1)}(t)u(t)\geq 0\), from (1) we find

\[ |u^{(m-1)}(t)|\geq \int_t^\infty |f(\tau,u(\tau))|\,d\tau . \]

Therefore, in view of (31), we have

\[ |u(t)|\geq At^{m-1}\int_t^\infty |f(\tau,u(\tau))|\,d\tau . \tag{32} \]

Let, for \(\delta=|u(t_0)|\), the numbers \(M\), \(n\), and \(\lambda_0\) be chosen according to condition (5). Choose \(t_1\geq 2^m t_0\) so large that for \(t\geq t_1\) the inequality

\[ A\int_t^\infty |f(\tau,u(\tau))|\,d\tau \leq \frac{1}{\lambda_0} \tag{33} \]

holds.

Since \(|u(t)|\geq \delta\) for \(t\geq t_1\), and (32) and (33) are satisfied, it follows that, according to (4) and (5), for \(t\geq t_1\) we shall have

\[ |f(t,t^{m-1}\operatorname{sign}u(t))| \leq \left| f\left[ t,u(t)\left(A\int_t^\infty |f(\tau,u(\tau))|\,d\tau\right)^{-1} \right] \right| \leq \]

\[ \leq MA^{-n}|f(t,u(t))| \left( \int_t^\infty |f(\tau,u(\tau))|\,d\tau \right)^{-n}. \]

Integrating both sides of this inequality from \(t_1\) to \(\infty\), we find

\[ \int_{t_1}^{\infty} |f(\tau,\tau^{m-1}\operatorname{sign} u(\tau))|\,d\tau \le \frac{M A^{-n}}{1-n} \left( \int_{t_1}^{\infty} |f(\tau,u(\tau))|\,d\tau \right)^{1-n} <\infty, \]

which contradicts condition (30). Thus, the assumption that for even \(m\) equation (1) has nonoscillatory solutions is false.

Suppose now that \(m\) is an odd number, and \(u(t)\) is a nonoscillatory solution of equation (1). If \(u(t)\) is a special solution, then, by Corollary 2 of Lemma 3, it will be monotone together with its derivatives up to order \((m-1)\), inclusive, on \([0,\infty)\). Let \(u(t)\) be an ordinary solution. We shall show that \(\lim_{t\to\infty} u(t)=0\). Indeed, otherwise, by (9), we conclude that for large \(t\) the function \(u(t)\) satisfies inequality (31), where
\[ A=\frac{2^{-m^2}}{(m-1)!}\inf_{t>t_0}\frac{u(t)}{u(2^{1-m}t)}. \]
But this inequality, as was already noted above, leads to a contradiction. Thus, \(\lim_{t\to\infty}u(t)=0\). Therefore, by Corollary 1 of Lemma 3, we easily conclude that \(u(t)\) is monotone together with its derivatives up to order \((m-1)\), inclusive, on the whole interval \((0,\infty)\), and \(\lim_{t\to\infty}u^{(k)}(t)=0\) \((k=1,2,\ldots,m-1)\). The theorem is proved.

For
\[ f(t,u)=\sum_{i=1}^{k} a_i(t)u^{n_i}, \]
where \(a_i(t)\ge 0\), \(0\le n_i<1\) \((i=1,2,\ldots,k)\), Theorem 1 yields the theorems of Sh. Belogorets [1] and M. Shvets [5]*.

Theorem 2. If \(f(t,u)u\le 0\), then for even \(m\) condition (30) is necessary and sufficient in order that every nontrivial solution of equation (1) either oscillate or tend monotonically to zero or to infinity as \(t\to\infty\), together with its derivatives up to order \((m-1)\), inclusive; and if \(m\) is an odd number, then this condition is necessary and sufficient in order that every nontrivial solution of equation (1) either oscillate or tend monotonically to infinity, together with its derivatives up to order \((m-1)\), inclusive.

Proof. Necessity again follows from Lemma 5. We prove sufficiency. Let \(u(t)\) be a nontrivial nonoscillatory solution of equation (1). If \(m\) is an odd number, then by Lemma 4 \(u(t)\) will be an ordinary solution, while if \(m\) is an even number, then it will be either ordinary or special. If \(u(t)\) is a special solution, then by Corollary 2 of Lemma 3 it will be monotone together with its derivatives up to order \((m-1)\), inclusive, on the whole interval \([0,\infty)\). Therefore below we shall assume that \(u(t)\) is a nonoscillatory, ordinary solution of equation (1). By Lemma 2, there are two possibilities: either (12) holds, or (9) and (13) hold.

We shall show that if (12) holds, then \(\lim_{t\to\infty}|u^{(k)}(t)|=\infty\) \((k=0,1,\ldots,m-1)\). It is clear that for this it suffices to show
\[ \lim_{t\to\infty}|u^{(m-1)}(t)|=\infty. \tag{34} \]
From (12) it follows that
\[ |u(t)|\ge c_0 t^{m-1}\quad\text{for } t\ge t_0, \]

* We note that in [5] a theorem on the oscillation of solutions of the equation
\[ \frac{d^m u}{dt^m}+a(t)u^n=0 \]
for \(a(t)\ge 0\), \(n>1\), established earlier in [2] for a more general equation (see [2], Theorem 5), is also proved.

where \(c_0>0\), and \(t_0\) is a sufficiently large number. On the other hand, if \(\lambda\) is a sufficiently large positive number, then, by virtue of (4) and (5), for \(t\geq t_0\) we shall have

\[ |f(t,\pm c_0 t^{m-1})| \geq \frac{1}{M\lambda^n}\,|f(t,\pm \lambda c_0 t^{m-1})| \geq \frac{1}{M\lambda^n}\,|f(t,\pm t^{m-1})|. \]

Therefore from (1) we find

\[ |u^{(m-1)}(t)| \geq |u^{(m-1)}(t_0)| + \frac{1}{M\lambda^n} \int_{t_0}^{t} |f(\tau,\tau^{m-1}\operatorname{sign}u(\tau))|\,d\tau, \]

whence, according to (30), the validity of condition (34) follows.

Finally, let us show that condition (13) can be satisfied only for even \(m\), and if it is satisfied, then \(l=0\) and

\[ \lim_{t\to\infty} u^{(k)}(t)=0 \qquad (k=0,1,\ldots,m-1). \]

Indeed, in the opposite case, according to (9), it is easy to conclude that, for large \(t\), \(u(t)\) satisfies inequality (31), where \(A\) is a positive constant. But in exactly the same way as above, in the proof of Theorem 1, one can show that this inequality contradicts condition (30). The theorem is proved.

From Theorems 1 and 2 it follows that if \(m=2\) and condition (30) is satisfied, then, when \(f(t,u)u\geq 0\), all solutions of equation (1) are oscillatory, while when \(f(t,u)u\leq 0\), all solutions of this equation tend monotonically to zero as \(t\to\infty\). The question of exactly what kinds of solutions equation (1) has for \(m\geq 3\) is answered by the following

Theorem 3. Let \(m\geq 3\), let condition (30) be satisfied, and suppose that for any numbers \(t_0\) and \(u_0^{(k)}\) \((k=0,1,\ldots,m-1)\), where \(u_0^{(0)}\ne 0\), there exists a unique solution \(u(t)\) of equation (1) satisfying the conditions \(u^{(k)}(t_0)=u_0^{(k)}\) \((k=0,1,\ldots,m-1)\). Then equation (1) always has oscillatory solutions; but, besides such solutions, it has nontrivial solutions tending monotonically to zero as \(t\to\infty\) when \((-1)^{m-1}f(t,u)u\geq 0\), and solutions tending monotonically to infinity as \(t\to\infty\) when \(f(t,u)u\leq 0\).

Proof. First we shall prove the existence of the nonoscillatory solutions mentioned in the theorem. We begin with the case when the condition

\[ (-1)^{m-1}f(t,u)u\geq 0 \]

is satisfied, and show that in this case equation (1) possesses nontrivial solutions tending monotonically to zero. Since the special solution belongs to the named type, we may therefore, without loss of generality, assume that equation (1) has no such solutions.

Let \(u(t)\) be a solution of equation (1) such that

\[ (-1)^i u^{(i)}(k)=B \qquad (i=0,1,\ldots,m-1). \]

From Lemma 3 we easily conclude that

\[ (-1)^i u^{(i)}(t)\geq B \qquad (i=0,1,\ldots,m-1) \quad \text{for } 0\leq t\leq k. \]

Consequently, if \(B>\dfrac{A}{m}\), where \(A\) is an arbitrary positive number, then

\[ \sum_{i=0}^{m-1}|u^{(i)}(0)|>A. \]

On the other hand, for small \(B\), according to

by the theorem on continuous dependence of the solution of the Cauchy problem on the initial data, we shall have

\[ \sum_{i=0}^{m-1} |u^{(i)}(0)| < A. \]

Therefore, by the theorem just mentioned, for any \(k\) there will be a positive number \(B_k\) such that, for the solution \(u_k(t)\) satisfying the condition \((-1)^i u_k^{(i)}(k)=B_k\) \((i=0,1,\ldots,m-1)\), we shall have

\[ \sum_{i=0}^{m-1} |u_k^{(i)}(0)| = A \qquad (k=1,2,\ldots). \]

Accordingly, without loss of generality we may assume that the sequences \(\{u_k^{(i)}(0)\}\) \((i=0,\alpha,\ldots,m-1)\) are convergent. Denote their limits respectively by \(u_0^{(i)}\) \((i=0,1,\ldots,m-1)\). It is clear that

\[ \sum_{i=0}^{m-1} |u_0^{(i)}|=A. \]

Let \(u_0(t)\) be the solution of equation (1) satisfying the initial condition

\[ u_0^{(i)}(0)=u_0^{(i)} \qquad (i=0,1,\ldots,m-1). \]

It is clear that \(u_0(t)\not\equiv 0\) and that \(\{u_k^{(i)}(t)\}\) \((i=0,1,\ldots,m-1)\) converge uniformly to \(u_0^{(i)}(t)\) \((i=0,1,\ldots,m-1)\) on every finite interval. Therefore we have \((-1)^i u_0^{(i)}(t)\ge 0\) \((i=0,1,\ldots,m-1)\) for \(t\ge 0\). Hence, by Theorems 1 and 2, it follows that

\[ \lim_{t\to\infty} u_0(t)=0. \]

If

\[ f(t,u)u \le 0, \tag{35} \]

then, as follows easily from (1), for any solution \(u(t)\) of equation (1) satisfying the conditions

\[ u^{(i)}(0)>0 \qquad (i=0,1,\ldots,m-1), \]

we shall have

\[ u^{(i)}(t)\ge 0 \qquad (i=0,1,\ldots,m-1) \quad \text{for } t\ge 0 \tag{36} \]

and, consequently,

\[ \lim_{t\to\infty} u(t)=\infty. \]

We proceed to the proof of the existence of oscillatory solutions. Let

\[ f(t,u)u\ge 0. \]

Then, for even \(m\), all solutions are oscillatory, and for odd \(m\), as is clear from Corollary 1 of Lemma 3, for the nonoscillation of a solution \(u(t)\) it is necessary that \((-1)^i u^{(i)}(0)u(0)>0\) \((i=0,1,\ldots,m-1)\). Consequently, if at least one of these conditions is violated, then the solution \(u(t)\) will be oscillatory.

Consider now the case when condition (35) is satisfied. Let \(u(t)\) be a solution of equation (1) satisfying the initial condition

\[ u(0)=A>0,\quad u'(0)=\ldots=u^{(m-2)}(0)=0,\quad u^{(m-1)}(0)=B. \]

If \(B\ge 0\), then (36) is satisfied. We show that if

\[ B<-(m-1)!\,A+\int_0^1 f(\tau,A)\,d\tau, \tag{37} \]

then

\[ u^{(i)}(t)\le 0 \qquad (i=0,1,\ldots,m-1) \quad \text{for } t\ge t_0, \tag{38} \]

where \(0<t_0<1\). For this it is enough to show that

\[ u(t_0)=0,\quad u^{(i)}(t)<0\quad (i=1,2,\ldots,m-1)\quad \text{for }0<t\leqslant 1. \]

Suppose the contrary. Then either

\[ u(t)>0,\quad u^{(i)}(t)<0\quad (i=1,2,\ldots,m-1)\quad \text{for }0<t\leqslant 1, \]

or there exists a number \(t_1\), \(0<t_1\leqslant 1\), such that \(u(t)>0,\ u^{(i)}(t)<0\) \((i=1,2,\ldots,m-1)\) for \(0<t<t_1\) and \(u^{(m-1)}(t_1)=0\). In the first case, according to (37), we have

\[ u(1)=A+\frac{B}{(m-1)!}-\frac{1}{(m-1)!}\int_0^1 (1-\tau)^{m-1} f(\tau,u(\tau))\,d\tau\leqslant \]

\[ \leqslant A+\frac{B}{(m-1)!}-\frac{1}{(m-1)!}\int_0^1 f(\tau,A)\,d\tau<0, \]

and in the second case

\[ u^{(m-1)}(t_1)=B-\int_0^{t_1} f(\tau,u(\tau))\,d\tau \leqslant B-\int_0^1 f(\tau,A)\,d\tau<0. \]

The contradictions obtained show that, under condition (37), condition (38) is satisfied.

Thus, for any natural number \(k\) we shall have \(u(k)>0\), if \(B\geqslant 0\), and \(u(k)<0\), if \(B\) satisfies inequality (37). Hence it is clear that there exists a sequence \(\{B_k\}\) such that

\[ -(m-1)!A+\int_0^1 f(\tau,A)\,d\tau<B_k<0\quad (k=1,2,\ldots), \]

and, for the solution \(u_k(t)\) satisfying the initial conditions

\[ u_k(0)=A>0,\quad u_k'(0)=\cdots=u_k^{(m-2)}(0)=0,\quad u_k^{(m-1)}(0)=B_k, \]

we shall have

\[ u_k(k)=0\quad (k=1,2,\ldots). \tag{39} \]

Without loss of generality, the sequence \(\{B_k\}\) may be assumed convergent. Let \(\lim_{k\to\infty}B_k=B_0\), and let \(u_0(t)\) be a solution of equation (1) satisfying the initial conditions

\[ u_0(0)=A,\quad u_0'(0)=\cdots=u_0^{(m-2)}(0)=0,\quad u_0^{(m-1)}(0)=B_0. \]

It is clear that \(\{u_k^{(i)}(t)\}\) \((i=0,1,\ldots,m-1)\) converge uniformly to \(\{u_0^{(i)}(t)\}\) \((i=0,1,\ldots,m-1)\) on every finite interval of the positive half-axis \(t>0\).

We shall prove that \(u_0(t)\) is an oscillatory solution. According to Theorem 2, for this it is enough to show that \(u_0(t)\) cannot tend monotonically to zero or to infinity as \(t\to\infty\).

Since \(u_0'(0)=0\), it is clear from Corollary 1 of Lemma 3 that \(u_0(t)\) does not tend monotonically to zero as \(t\to\infty\).

It remains to show that \(u_0(t)\) does not tend to infinity as \(t\to\infty\). Indeed, in the contrary case, by Theorem 2, we would have

\[ u_0^{(i)}(t)u_0(t)>0\quad (i=0,1,\ldots,m-1)\quad \text{for }t\geqslant \bar t, \]

where \(\bar t\) is a sufficiently large number. Hence it is clear that, for large \(k\), the inequality

\[ u_k^{(i)}(\bar t)u_k(\bar t)>0\quad (i=0,1,\ldots,m-1), \]

is satisfied, from which, according to (35), it follows that

\[ u_k^{(i)}(t)u_k(t)>0\quad (i=0,1,\ldots,m-1)\quad \text{for } t\geq \bar t, \]

which contradicts condition (39). The theorem is proved.

If the Cauchy problem for equation (1), for arbitrary initial data, has a unique solution, then, of course, all solutions of equation (1) that tend monotonically to zero as \(t\to\infty\) are ordinary; but if the uniqueness condition is violated, it may turn out that equation (1) has no ordinary solutions tending monotonically to zero at all. This is indicated by the following

Theorem 4. Suppose that \((-1)^{m-1}f(t,u)u\geq 0\) and that the conditions of Theorem 3 are satisfied. If, in addition, for arbitrary \(t\) and \(u\) we have

\[ |f(t,u)|\geq \frac{L}{(1+t)^m}|u|^\sigma, \tag{40} \]

where \(L>0\) and \(0<\sigma<1\), then all nonoscillatory nontrivial solutions of equation (1) that tend monotonically to zero as \(t\to\infty\) are singular.

Proof. Suppose the contrary, that under condition (40) equation (1) possesses a nontrivial ordinary nonoscillatory solution \(u(t)\) tending monotonically to zero as \(t\to\infty\). Without loss of generality we may assume that \(u(t)>0\) for \(t\geq 0\). Then, by virtue of Corollary 1 of Lemma 3, we shall have

\[ (-1)^i u^{(i)}(t)>0 \quad \text{for } 0\leq t<\infty \quad (i=0,1,\ldots,m-1). \tag{41} \]

Consider the function

\[ v_k(t)=c\left(\frac{k-t}{1+k}\right)^{\frac{m}{1-\sigma}}, \quad \text{where }\quad c=\left[\frac{(1-\sigma)^m L}{\displaystyle\prod_{i=0}^{m-1}(m-i(1-\sigma))}\right]^{\frac{1}{1-\sigma}}. \]

It is not difficult to verify that \(v_k(t)\) is a solution of the equation

\[ \frac{d^m v}{dt^m}+(-1)^{m-1}\frac{L}{(1+k)^m}v^\sigma=0. \]

According to (40) and (41), it is clear that

\[ (-1)^{m-1}\left[f(t,v)-\frac{(-1)^{m-1}L}{(1+k)^m}v^\sigma\right]>0 \quad \text{for } 0\leq t\leq k,\ v>0 \]

and

\[ (-1)^i\bigl(u^{(i)}(k)-v_k^{(i)}(k)\bigr)=(-1)^i u^{(i)}(k)>0 \quad (i=0,1,\ldots,m-1). \]

Therefore, according to Lemma 3, we have

\[ u(t)>v_k(t)\quad \text{for } 0\leq t<k, \]

by virtue of which, from the inequality

\[ u(0) > (-1)^{m-1}\int_{0}^{k}\tau^{m-1} f(\tau,u(\tau))\,d\tau \ge L \int_{1}^{k/2}\frac{\tau^{m-1}}{(1+\tau)^m}\times \]

\[ {}\times w^\sigma(\tau)\,d\tau > L2^{-m} w^\sigma\!\left(\frac{k}{2}\right)\ln\frac{k}{2} \]

we find

\[ u(0) > L2^{-m} w_k^\sigma\!\left(\frac{k}{2}\right)\ln\frac{k}{2} > Lc^\sigma 2^{-\frac{m+m\sigma}{1-\sigma}}\ln\frac{k}{2} \to \infty \quad \text{as } k\to\infty . \]

The contradiction obtained proves the theorem.

References

1. Belohorec Š. Matematicko-fyzikálny Časopis SAV, 11, 250—255, 1961.
2. Kiguradze I. T. DAN SSSR, 144, No. 1, 33—36, 1962.
3. Kiguradze I. T. Matem. sbornik, 65, (107), No. 2, 172—187, 1964.
4. Coddington E. A., Levinson N. Theory of Ordinary Differential Equations. IL, 1958.
5. Ličko Imrich, Švec Marko. Czechoslovak Mathematical Journal, 13, No. 4, 481—491, 1963.

Received by the editors
March 22, 1965

Tbilisi State
University