Our main result deals with the solvability of the nonlocal BVP

$$ \textstyle\begin{cases} (t^{2\alpha} \vert x^{\prime} \vert ^{\alpha} \operatorname {sgn}x^{\prime})^{\prime }+b(t)\vert x\vert ^{\beta} \operatorname {sgn}x=0,\quad t\geq 1, \alpha< \beta, \\ x(t)>0,\qquad x^{\prime}(t)< 0,\\ \lim_{t\rightarrow\infty}x(t)=0,\qquad \lim_{t\rightarrow\infty }t x(t)=\infty, \end{cases} $$

(BVP)

under the additional assumption

$$ b \in C^{1}[1,\infty). $$

(16)

### Remark 1

In the superlinear case \(\alpha<\beta\), in virtue of (16), any local solution of (1) is a solution, *i.e.* it is continuable to infinity and is proper; see, *e.g.*, [16], Theorem 3.2, or [15], Appendix A. Notice also that, under the weaker assumption \(b(t)\geq0\), \(\sup \{ b(t):t\geq T \} >0\) for any \(T\geq1\), there may exist equations of type (1) with uncontinuable solutions; see, *e.g.*, [16], p.343.

The following holds.

### Theorem 1

*Assume*
\(Z=\infty\). *If* (16) *is satisfied and the function*

$$ G(t)=\frac{1}{t^{2}b(t)}B^{\gamma}(t)\quad \textit{is nonincreasing for }t\geq1, $$

(17)

*where*

$$ \gamma=\frac{1+\alpha\beta+2\alpha}{\alpha(\beta+1)}, $$

(18)

*then* (BVP) *has infinitely many solutions*.

To prove this result, several auxiliary results are needed. Define for any solution *x* of (1) the energy-type function

$$ E_{x}(t)=B(t)\bigl\vert x(t)\bigr\vert ^{\beta+1}+x(t)x^{[1]}(t)+k \frac{t^{2\alpha}}{b(t)}B(t)\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}, $$

(19)

where the quasiderivate \(x^{[1]}\) is defined in (14) and

$$ k=\frac{\alpha(\beta+1)}{\alpha+1}. $$

(20)

### Lemma 3

*For any solution*
*x*
*of* (1) *we have*

$$ \bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime\prime}(t)= \frac{1}{\alpha}\frac{\vert x^{\prime }(t)\vert }{t^{2\alpha}}\frac{d}{dt}x^{[1]}(t)-2t^{-1} \bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha +1}\operatorname {sgn}x^{\prime}(t). $$

(21)

### Proof

Since \(x^{[1]}\) is continuously differentiable on \([t_{x},\infty)\), \(t_{x}\geq1\), and \(t^{-2\alpha}x^{[1]}(t)=\vert x^{\prime }(t)\vert ^{\alpha} \operatorname {sgn}x^{\prime}(t)\), the function \(x^{\prime}\) is continuously differentiable on \([t_{x},\infty)\) as well. If \(x^{\prime}(t)=0\), then the identity (21) is valid. Now, assume \(x^{\prime}(t)\neq0\). We have

$$\begin{aligned} \bigl\vert x^{\prime}(t)\bigr\vert \frac{d}{dt}x^{[1]}(t) & =\bigl\vert x^{\prime}(t)\bigr\vert \bigl( 2\alpha t^{2\alpha-1} \bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha} \operatorname {sgn}x^{\prime}(t)+\alpha t^{2\alpha}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha-1}x^{\prime\prime}(t) \bigr) \\ & =2\alpha t^{2\alpha-1}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}\operatorname {sgn}x^{\prime }(t)+\alpha t^{2\alpha}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime\prime}(t), \end{aligned}$$

from which the assertion follows. □

### Lemma 4

*Assume* (16) *and* (17). *Then for any solution*
*x*
*of* (1) *we have for*
\(t\geq1\)

$$\frac{d}{dt}E_{x}(t)\leq0. $$

### Proof

Let *φ* be a continuously differentiable function on \([1,\infty)\). Then for any positive constant *σ* the function \(\vert \varphi(t)\vert ^{\sigma+1}\) is continuously differentiable and

$$\frac{d}{dt}\bigl\vert \varphi(t)\bigr\vert ^{\sigma+1}=(\sigma+1) \bigl\vert \varphi(t)\bigr\vert ^{\sigma}\varphi^{\prime}(t)\operatorname {sgn}\varphi(t). $$

Using this equality, we have for \(t\geq1\)

$$\begin{aligned} \frac{d}{dt} \biggl( \frac{t^{2\alpha}}{b(t)}B(t)\bigl\vert x^{\prime}(t) \bigr\vert ^{\alpha +1} \biggr) ={}&\frac{d}{dt} \biggl( \frac{1}{t^{2}b(t)}B(t)t^{2\alpha +2}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1} \biggr) \\ ={}&B(t)t^{2\alpha+2}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1} \frac{d}{dt} \biggl(\frac {1}{t^{2}b(t)} \biggr) +\frac{1}{t^{2}}t^{2\alpha+2} \bigl\vert x^{\prime }(t)\bigr\vert ^{\alpha+1} \\ & {}+(\alpha+1)\frac{B(t)}{t^{2}b(t)} \bigl( 2t^{2\alpha+1}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}+t^{2\alpha+2}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime\prime }(t) \operatorname {sgn}x^{\prime}(t) \bigr) . \end{aligned}$$

Hence we get

$$\begin{aligned} \frac{d}{dt}E_{x}(t) ={}&(\beta+1)\bigl\vert x(t)\bigr\vert ^{\beta}x^{\prime}(t)B(t) \operatorname {sgn}x(t)+t^{2\alpha}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime}(t) \operatorname {sgn}x^{\prime }(t) \\ &{} +kB(t)t^{2\alpha+2}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}\frac{d}{dt} \biggl( \frac {1}{t^{2}b(t)} \biggr) +k \frac{1}{t^{2}}t^{2\alpha+2}\bigl\vert x^{\prime }(t)\bigr\vert ^{\alpha+1} \\ &{} +k(\alpha+1)\frac{B(t)}{t^{2}b(t)} \bigl( 2t^{2\alpha+1}\bigl\vert x^{\prime }(t)\bigr\vert ^{\alpha+1}+t^{2\alpha+2}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime\prime }(t) \operatorname {sgn}x^{\prime}(t) \bigr) \end{aligned}$$

or

$$\frac{d}{dt}E_{x}(t)=t^{2\alpha+2}g(t)+h(t), $$

where

$$g(t)=(1+k)\frac{1}{t^{2}}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}+kB(t)\bigl\vert x^{\prime }(t)\bigr\vert ^{\alpha+1} \frac{d}{dt} \biggl( \frac{1}{t^{2}b(t)} \biggr) $$

and

$$\begin{aligned} h(t)={}&(\beta+1)\bigl\vert x(t)\bigr\vert ^{\beta}x^{\prime}(t)B(t) \operatorname {sgn}x(t) \\ &{} +k(\alpha+1)\frac{B(t)}{t^{2}b(t)} \bigl( 2t^{2\alpha+1}\bigl\vert x^{\prime }(t)\bigr\vert ^{\alpha+1}+t^{2\alpha+2}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime\prime }(t) \operatorname {sgn}x^{\prime}(t) \bigr) . \end{aligned}$$

From (20) we obtain

$$1+k=\gamma k,\qquad k(\alpha+1)=\alpha(\beta+1). $$

Thus, we get

$$ g(t)=k \biggl( \gamma\frac{1}{t^{2}}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}+B(t)\bigl\vert x^{\prime }(t)\bigr\vert ^{\alpha+1}\frac{d}{dt} \biggl( \frac{1}{t^{2}b(t)} \biggr) \biggr) $$

(22)

and

$$\begin{aligned} \frac{h(t)}{(\beta+1)B(t)} ={}&\bigl\vert x(t)\bigr\vert ^{\beta}x^{\prime}(t) \operatorname {sgn}x(t)+\frac{2\alpha}{b(t)}t^{2\alpha-1}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1} \\ &{} +\frac{\alpha}{b(t)}t^{2\alpha}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha}x^{\prime\prime }(t)\operatorname {sgn}x^{\prime}(t). \end{aligned}$$

In view of (17), we have

$$\begin{aligned} &B^{-\gamma+1} (t)\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1} \frac{d}{dt} \biggl( \frac {1}{t^{2}b(t)}B^{\gamma}(t) \biggr) \\ &\quad =B(t)\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1} \frac{d}{dt} \biggl( \frac{1}{t^{2}b(t)} \biggr) +\gamma\frac{1}{t^{2}} \bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}\leq0, \end{aligned}$$

and so, from (22) we obtain

In order to complete the proof, it is sufficient to show that \(h(t)=0\). Using Lemma 3, we have

$$\begin{aligned} \frac{h(t)}{(\beta+1)B(t)} ={}&\bigl\vert x(t)\bigr\vert ^{\beta}x^{\prime}(t) \operatorname {sgn}x(t)+\frac{2\alpha}{b(t)}t^{2\alpha-1}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1} \\ &{} -x^{\prime}(t)\bigl\vert x(t)\bigr\vert ^{\beta} \operatorname {sgn}x(t)- \frac{2\alpha}{b(t)}t^{2\alpha-1}\bigl\vert x^{\prime}(t)\bigr\vert ^{\alpha+1}=0, \end{aligned}$$

thus the assertion follows. □

### Lemma 5

*Assume* (16). *Then* (1) *has proper solutions*
*x*
*for which*
\(E_{x}(1)<0\).

### Proof

Consider on \([0,\infty)\) the scalar function

$$ \phi(u)=c_{1}u^{\alpha+1}-mu^{\alpha}+cm^{\beta+1}, $$

(23)

where

$$ c_{1}=\frac{\alpha(\beta+1)}{\alpha+1}\frac{c}{b(1)} $$

(24)

and *m* is a positive parameter. A standard calculation shows that when *m* is sufficiently small, then *ϕ* attains negative values in a neighborhood of the point

$$ \overline{u}=\frac{\alpha m}{(\alpha+1)c_{1}}. $$

(25)

Indeed, consider the local solution *x* of (1) with the initial condition

$$ x(1)=m,\qquad x^{\prime}(1)=-\overline{u}. $$

(26)

In view of Remark 1, *x* is continuable to infinity and proper. From (19) we obtain

$$E_{x}(1)=cm^{\beta+1}-m \biggl( \frac{\alpha m}{(\alpha+1)c_{1}} \biggr) ^{\alpha}+k\frac{c}{b(1)} \biggl( \frac{\alpha m}{(\alpha+1)c_{1}} \biggr) ^{\alpha+1}, $$

or, in view of (20),

$$E_{x}(1)=\phi(\overline{u}). $$

Using (25), we get

$$\begin{aligned} E_{x}(1)&=\phi(\overline{u})=cm^{\beta+1}- \biggl( \frac{\alpha}{(\alpha +1)c_{1}} \biggr) ^{\alpha}m^{\alpha+1}+c_{1} \biggl( \frac{\alpha}{(\alpha+1)c_{1}}m \biggr) ^{\alpha+1} \\ &=m^{\alpha+1} \biggl[ cm^{\beta-\alpha}+ \biggl( \frac{1}{(\alpha+1)c_{1}} \biggr) ^{\alpha}\biggl(-1+\frac{\alpha}{\alpha+1}\biggr) \biggr] \\ &=m^{\alpha+1} \biggl[ cm^{\beta-\alpha}-\frac{1}{\alpha+1} \biggl( \frac {1}{(\alpha+1)c_{1}} \biggr) ^{\alpha} \biggr] . \end{aligned}$$

From this and \(\beta>\alpha\), choosing *m* sufficiently small such that

$$ 0< m^{\beta-\alpha}< \frac{1}{c}\frac{1}{\alpha+1} \biggl( \frac{1}{(\alpha+1)c_{1}} \biggr) ^{\alpha}, $$

(27)

we get \(\phi(\overline{u})<0\), which is the assertion. □

### Proof of Theorem 1

From Lemma 4, the function \(E_{x}\) is nonincreasing on \([1,\infty)\) for any solution *x* of (1).

Fixed *m* satisfying (27), consider the local solution *x* of (1) with the initial condition (26), where \(\overline{u}\) is given by (25). In view of Remark 1, this solution is also continuable to infinity and proper. Moreover, it is uniquely determined, because in the superlinear case the uniqueness of solutions with respect to the initial conditions holds; see, *e.g.*, [7]. Moreover, in virtue of the proof of Lemma 5, we have \(E_{x}(1)<0\), and so, from Lemma 4, we obtain

$$ E_{x}(t)\leq E_{x}(1)< 0\quad \mbox{on }[1,\infty). $$

(28)

Let us show that *x* and \(x^{\prime}\) cannot have zeros for \(t\geq1\). By contradiction, if there exists \(t_{1}>1\) such that \(x(t_{1})=0\), then, in virtue of the uniqueness with respect to the initial data, we have \(x^{\prime }(t_{1})\neq0\). Hence \(E(t_{1})>0\), which contradicts (28). Similarly, if there exists \(t_{2}\geq1\) such that \(x^{\prime}(t_{2})=0\), we obtain \(E(t_{2})>0\), which is again a contradiction.

Thus, *x* is nonoscillatory. Moreover, in view of Lemma 2, we have

$$\lim_{t\rightarrow\infty}x(t)=0. $$

Hence, *x* is positive decreasing in the half-line \([1,\infty)\), that is,

$$x(t)>0,\qquad x^{\prime}(t)< 0\quad \mbox{for }t\geq1. $$

From (1), the quasiderivative \(x^{[1]}\) is negative decreasing, *i.e.*

$$\lim_{t\rightarrow\infty}-x^{[1]}(t)=\ell_{x},\quad 0< \ell_{x}\leq\infty. $$

If \(\ell_{x}<\infty\), we get \(\lim_{t\rightarrow\infty}x(t)x^{[1]}(t)=0\), and from (19) we obtain \(\liminf_{t\rightarrow\infty}E_{x}(t) \geq 0\), that is, a contradiction with (28). Hence \(\ell_{x}=\infty\), *i.e.*

$$\lim_{t\rightarrow\infty}t^{2}x^{\prime}(t)=-\infty. $$

Using the l’Hospital rule we get

$$\lim_{t\rightarrow\infty}tx(t)=\infty. $$

Hence, *x* is a solution of (BVP). Since there are infinitely many solutions which satisfy (26) with the choice of *m* taken with (27), the proof is now complete. □

From Theorem 1 and its proof, we get the following.

### Corollary 1

*Under assumptions of Theorem *
1, (1) *has infinitely many slowly decaying solutions*, *which are positive decreasing on the whole interval*
\([1,\infty)\). *Moreover*, (1) *has also infinitely many strongly decaying solutions and every nonoscillatory solution of* (1) *tends to zero as*
\(t\rightarrow\infty\).

### Proof

In virtue of Theorem 1 and its proof, the boundary value problem (BVP) is solvable by every solution *x* which satisfies (26) and (27). Clearly, these solutions are slowly decaying solutions.

Consequently, (1) has nonoscillatory solutions and, in view of Lemma 2(iii) we get \(Y<\infty\). Then the existence of infinitely many strongly decaying solutions follows from Lemma 2(ii). □

When the monotonicity condition (17) is valid only for large *t*, reasoning as in the proof of Theorem 1, we obtain the following.

### Corollary 2

*Assume*
\(Z=\infty\). *If* (16) *is satisfied and the function*
*G*, *given in* (17) *is nonincreasing for any large*
*t*, *then* (1) *has infinitely many slowly decaying solutions*, *which are eventually positive decreasing*. *Moreover*, (1) *has also infinitely many strongly decaying solutions and every nonoscillatory solution of* (1) *tends to zero as*
\(t\rightarrow\infty\).

Finally, when also the initial starting point is fixed, we have the following.

### Corollary 3

*Assume*
\(Z=\infty\). *If* (16) *and* (17) *are satisfied*, *then the BVP*

$$ \textstyle\begin{cases} (t^{2\alpha} \vert x^{\prime} \vert ^{\alpha} \operatorname {sgn}x^{\prime})^{\prime }+b(t)\vert x\vert ^{\beta} \operatorname {sgn}x=0, \quad t\geq 1, \alpha< \beta, \\ x(1)=x_{1},\qquad x(t)>0,\qquad x^{\prime}(t)< 0,\\ \lim_{t\rightarrow\infty}x(t)=0,\qquad \lim_{t\rightarrow\infty }t x(t)=\infty, \end{cases} $$

*has infinitely many solutions for every initial data*
\(x_{1}\)
*such that*

$$ 0< (x_{1})^{\beta-\alpha}< \biggl( \frac{1}{c} \biggr) ^{\alpha+1}\frac {1}{\alpha+1} \biggl( \frac{b(1)}{\alpha(\beta+1)} \biggr) ^{\alpha}. $$

(29)

### Proof

The assertion follows by a reasoning as in the proof of Theorem 1 and choosing \(m=x_{1}\) in (26). Taking into account that *m* satisfies (27) and \(c_{1}\) is given by (24), we get (29). The details are left to the reader. □

### Remark 2

It is worth to note that the condition (17) may depend on the choice of the constant *c* in (15), *i.e.* on the choice of a primitive to *b*.

If (17) is satisfied for a fixed *B*, then (17) remains to hold for \(B(t)+\overline{c}\), where \(\overline{c}>0\). Indeed, setting \(\Psi(t)=t^{-2}b^{-1}(t)\), from \(G^{\prime}(t)\leq0\) we get

$$0\geq\Psi^{\prime}(t)+\gamma\Psi(t)b(t)\frac{1}{B(t)}= \Psi^{\prime}(t)+\gamma\Psi(t)b(t)\frac{1}{B(t)+\overline{c}} \frac{B(t)+\overline{c}}{B(t)}$$

or

$$0\geq\Psi^{\prime}(t)+\gamma\Psi(t)b(t)\frac{1}{B(t)+\overline{c}}$$

*i.e.* (17) is satisfied also for \(B(t)+\overline{c}\) with \(\overline {c}>0\). However, if (17) is valid for *B* given by (15), then it is possible that (17) is not valid for \(\widetilde{B}(t)=B(t)+\tilde {c}\), where \(0<\tilde{c}<c\). This fact is illustrated below in Example 2.