# Existence of positive solutions of the Cauchy problem for a second-order differential equation

## Abstract

In this paper we consider the equation ${u}^{″}\left(t\right)=f\left(t,u\left(t\right),{u}^{\prime }\left(t\right)\right)$ and prove the unique solvability of the Cauchy problem $u\left(0\right)=0$, ${u}^{\prime }\left(0\right)=\lambda$ with $\lambda >0$.

## 1 Introduction

In , Knežević-Miljanović considered the Cauchy problem

$\left\{\begin{array}{l}{u}^{″}\left(t\right)=P\left(t\right){t}^{a}u{\left(t\right)}^{\sigma },\phantom{\rule{1em}{0ex}}t\in \left(0,1\right],\\ u\left(0\right)=0,\phantom{\rule{2em}{0ex}}{u}^{\prime }\left(0\right)=\lambda ,\end{array}$
(1)

where P is continuous, $a,\sigma ,\lambda \in \mathbb{R}$ with $\sigma <0$ and $\lambda >0$, and ${\int }_{0}^{1}|P\left(t\right)|{t}^{a+\sigma }\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty }$. Moreover, in , Kawasaki and Toyoda considered the Cauchy problem

(2)

where f is a mapping from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)$ into and $\lambda \in \mathbb{R}$ with $\lambda >0$. They proved the unique solvability of Cauchy problem (2) using the Banach fixed point theorem. The theorem in  is as follows.

Theorem Suppose that a mapping f from $\left[0,1\right]×\left[0,\mathrm{\infty }\right)$ into satisfies the following.

1. (a)

The mapping $t↦f\left(t,u\right)$ is measurable for any $u\in \left(0,\mathrm{\infty }\right)$, and the mapping $u↦f\left(t,u\right)$ is continuous for almost every $t\in \left[0,1\right]$.

2. (b)

$|f\left(t,{u}_{1}\right)|\ge |f\left(t,{u}_{2}\right)|$ for almost every $t\in \left[0,1\right]$ and for any ${u}_{1},{u}_{2}\in \left[0,\mathrm{\infty }\right)$ with ${u}_{1}\le {u}_{2}$.

3. (c)

There exists $\alpha \in \mathbb{R}$ with $0<\alpha <\lambda$ such that

${\int }_{0}^{1}|f\left(t,\alpha t\right)|\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty }.$
4. (d)

There exists $\beta \in \mathbb{R}$ with $\beta >0$ such that

$|\frac{\partial f}{\partial u}\left(t,u\right)|\le \frac{\beta |f\left(t,u\right)|}{u}$

for almost every $t\in \left[0,1\right]$ and for any $u\in \left(0,\mathrm{\infty }\right)$.

Then there exists $h\in \mathbb{R}$ with $0 such that Cauchy problem (2) has a unique solution in X, where X is a subset

of $C\left[0,h\right]$, which is the class of continuous mappings from $\left[0,h\right]$ into .

The case that $f\left(t,u\left(t\right)\right)=P\left(t\right){t}^{a}u{\left(t\right)}^{\sigma }$ in the above theorem is the theorem of Knežević-Miljanović .

In this paper, we consider the Cauchy problem

where f is a mapping from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)×\mathbb{R}$ into and $\lambda \in \mathbb{R}$ with $\lambda >0$. We prove the unique solvability of this Cauchy problem using the Banach fixed point theorem.

In Section 2, we consider the following four cases for u and v.

1. (I)

Decreasing for u in $f\left(t,u,v\right)$ (b1) and decreasing for v in $f\left(t,u,v\right)$ (b3).

2. (II)

Decreasing for u in $f\left(t,u,v\right)$ (b1) and increasing for v in $f\left(t,u,v\right)$ (b4).

3. (III)

Increasing for u in $f\left(t,u,v\right)$ (b2) and decreasing for v in $f\left(t,u,v\right)$ (b3).

4. (IV)

Increasing for u in $f\left(t,u,v\right)$ (b2) and increasing for v in $f\left(t,u,v\right)$ (b4).

Theorems 2.1, 2.2, 2.3 and 2.4 are the cases of (I), (II), (III) and (IV), respectively.

## 2 Main results

In this section, we consider the Cauchy problem

(3)

where f is a mapping from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)×\mathbb{R}$ into and $\lambda \in \mathbb{R}$ with $\lambda >0$.

First, we consider the case of (I).

Theorem 2.1 Let λ be a real number with $\lambda >0$. Suppose that a mapping f from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)×\mathbb{R}$ into satisfies the following:

1. (a)

The mapping $t↦f\left(t,u,v\right)$ is measurable for any $\left(u,v\right)\in \left(0,\mathrm{\infty }\right)×\mathbb{R}$, and the mapping $\left(u,v\right)↦f\left(t,u,v\right)$ is continuous for almost every $t\in \left[0,1\right]$;

2. (b1)

$|f\left(t,{u}_{1},v\right)|\ge |f\left(t,{u}_{2},v\right)|$ for almost every $t\in \left[0,1\right]$, for any ${u}_{1},{u}_{2}\in \left(0,\mathrm{\infty }\right)$ with ${u}_{1}\le {u}_{2}$ and for any $v\in \mathbb{R}$;

3. (b3)

$|f\left(t,u,{v}_{1}\right)|\ge |f\left(t,u,{v}_{2}\right)|$ for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any ${v}_{1},{v}_{2}\in \mathbb{R}$ with ${v}_{1}\le {v}_{2}$;

4. (c1)

There exist ${\alpha }_{1}\in \mathbb{R}$ with $0<{\alpha }_{1}<\lambda$ and ${\alpha }_{2}\in \mathbb{R}$ with ${\alpha }_{2}<\lambda$ such that

${\int }_{0}^{1}|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty };$
5. (d1)

There exists ${\beta }_{1}\in \mathbb{R}$ with ${\beta }_{1}>0$ such that

$|\frac{\partial f}{\partial u}\left(t,u,v\right)|\le \frac{{\beta }_{1}|f\left(t,u,v\right)|}{u}$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

6. (d2)

There exists ${\beta }_{2}\in \mathbb{R}$ with ${\beta }_{2}>0$ such that

$|\frac{\partial f}{\partial v}\left(t,u,v\right)|\le {\beta }_{2}|f\left(t,u,v\right)|$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

7. (e)

There exists the limit

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds$

for any continuously differentiable mapping u from $\left[0,1\right]$ into $\left[0,\mathrm{\infty }\right)$;

8. (f1)

For ${\alpha }_{1}$ and ${\alpha }_{2}$,

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds=0.$

Then there exists $h\in \mathbb{R}$ with $0 such that Cauchy problem (3) has a unique solution in X, where X is a subset

of ${C}^{1}\left[0,h\right]$, which is the class of continuously differentiable mappings from $\left[0,h\right]$ into .

Proof It is noted that ${C}^{1}\left[0,h\right]$ is a Banach space by the maximum norm

$\parallel u\parallel =max\left\{max\left\{|u\left(t\right)|\mid t\in \left[0,h\right]\right\},max\left\{|{u}^{\prime }\left(t\right)|\mid t\in \left[0,h\right]\right\}\right\}.$

Instead of Cauchy problem (3), we consider the integral equation

$u\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

By condition (c1), there exists ${h}_{1}\in \mathbb{R}$ with $0<{h}_{1}\le 1$ such that

${\int }_{0}^{{h}_{1}}|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt

By condition (f1), there exists $h\in \mathbb{R}$ with $0 such that

$\underset{t\in \left(0,h\right]}{sup}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{0}^{{h}_{1}}|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt.$

Let A be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

$Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

Since a mapping $t↦\lambda t$ belongs to X, $X\ne \mathrm{\varnothing }$. Moreover, we have $A\left(X\right)\subset X$. Indeed, by condition (a), $Au\in {C}^{1}\left[0,h\right]$, $Au\left(0\right)=0$ and

${\left(Au\right)}^{\prime }\left(0\right)={\left[\lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right]}_{t=0}=\lambda .$

By conditions (b1) and (b3), we obtain that

$\begin{array}{rcl}Au\left(t\right)& =& \lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda t-t{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda t-t{\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\alpha }_{1}t\end{array}$

and

$\begin{array}{rcl}{\left(Au\right)}^{\prime }\left(t\right)& =& \lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda -{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda -{\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\alpha }_{2}\end{array}$

for any $t\in \left[0,h\right]$. Moreover, by condition (e), there exists the limit

$\underset{t\to 0+}{lim}\frac{t{\left(Au\right)}^{\prime }\left(t\right)-Au\left(t\right)}{{t}^{2}}=\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

We will find a fixed point of A. Let φ be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

Let $\phi \left[X\right]$ be a subset defined by

$\phi \left[X\right]=\left\{\phi \left[u\right]\mid u\in X\right\}.$

Then we have

and $\phi \left[X\right]$ is a closed subset of ${C}^{1}\left[0,h\right]$. Hence it is a complete metric space. Let Φ be an operator from $\phi \left[X\right]$ into $\phi \left[X\right]$ defined by

$\mathrm{\Phi }\phi \left[u\right]=\phi \left[Au\right].$

By the mean value theorem, for any ${u}_{1},{u}_{2}\in X$, there exist mappings ξ, η such that

$\begin{array}{r}f\left(t,{u}_{1}\left(t\right),{u}_{1}^{\prime }\left(t\right)\right)-f\left(t,{u}_{2}\left(t\right),{u}_{2}^{\prime }\left(t\right)\right)\\ \phantom{\rule{1em}{0ex}}=\frac{\partial f}{\partial u}\left(t,\xi \left(t\right),{u}_{1}^{\prime }\left(t\right)\right)\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right)+\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)\left({u}_{1}^{\prime }\left(t\right)-{u}_{2}^{\prime }\left(t\right)\right)\\ \phantom{\rule{1em}{0ex}}=\left(t\frac{\partial f}{\partial u}\left(t,\xi \left(t\right),{u}_{1}^{\prime }\left(t\right)\right)+\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)\right)\left(\phi \left[{u}_{1}\right]\left(t\right)-\phi \left[{u}_{2}\right]\left(t\right)\right)\\ \phantom{\rule{2em}{0ex}}+t\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)\left(\phi {\left[{u}_{1}\right]}^{\prime }\left(t\right)-\phi {\left[{u}_{2}\right]}^{\prime }\left(t\right)\right),\\ min\left\{{u}_{1}\left(t\right),{u}_{2}\left(t\right)\right\}\le \xi \left(t\right)\le max\left\{{u}_{1}\left(t\right),{u}_{2}\left(t\right)\right\}\end{array}$

and

$min\left\{{u}_{1}^{\prime }\left(t\right),{u}_{2}^{\prime }\left(t\right)\right\}\le \eta \left(t\right)\le max\left\{{u}_{1}^{\prime }\left(t\right),{u}_{2}^{\prime }\left(t\right)\right\}$

for almost every $t\in \left[0,h\right]$. Therefore, by conditions (b1), (b3), (d1) and (d2), we obtain that

$\begin{array}{r}|f\left(t,{u}_{1}\left(t\right),{u}_{1}^{\prime }\left(t\right)\right)-f\left(t,{u}_{2}\left(t\right),{u}_{2}^{\prime }\left(t\right)\right)|\\ \phantom{\rule{1em}{0ex}}=|\left(t\frac{\partial f}{\partial u}\left(t,\xi \left(t\right),{u}_{1}^{\prime }\left(t\right)\right)+\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)\right)\left(\phi \left[{u}_{1}\right]\left(t\right)-\phi \left[{u}_{2}\right]\left(t\right)\right)\\ \phantom{\rule{2em}{0ex}}+t\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)\left(\phi {\left[{u}_{1}\right]}^{\prime }\left(t\right)-\phi {\left[{u}_{2}\right]}^{\prime }\left(t\right)\right)|\\ \phantom{\rule{1em}{0ex}}\le \left(t|\frac{\partial f}{\partial u}\left(t,\xi \left(t\right),{u}_{1}^{\prime }\left(t\right)\right)|+|\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)|\right)|\phi \left[{u}_{1}\right]\left(t\right)-\phi \left[{u}_{2}\right]\left(t\right)|\\ \phantom{\rule{2em}{0ex}}+t|\frac{\partial f}{\partial v}\left(t,{u}_{2}\left(t\right),\eta \left(t\right)\right)||\left(\phi {\left[{u}_{1}\right]}^{\prime }\left(t\right)-\phi {\left[{u}_{2}\right]}^{\prime }\left(t\right)\right)|\\ \phantom{\rule{1em}{0ex}}\le \left(\frac{{\beta }_{1}}{{\alpha }_{1}}+{\beta }_{2}\right)|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)||\phi \left[{u}_{1}\right]\left(t\right)-\phi \left[{u}_{2}\right]\left(t\right)|\\ \phantom{\rule{2em}{0ex}}+{\beta }_{2}t|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)||\left(\phi {\left[{u}_{1}\right]}^{\prime }\left(t\right)-\phi {\left[{u}_{2}\right]}^{\prime }\left(t\right)\right)|\end{array}$

for almost every $t\in \left[0,h\right]$. Therefore we have

$\begin{array}{r}|\mathrm{\Phi }\phi \left[{u}_{1}\right]\left(t\right)-\mathrm{\Phi }\phi \left[{u}_{2}\right]\left(t\right)|\\ \phantom{\rule{1em}{0ex}}=|\frac{1}{t}{\int }_{0}^{t}\left(t-s\right)\left(f\left(s,{u}_{1}\left(s\right),{u}_{1}^{\prime }\left(s\right)\right)-f\left(s,{u}_{2}\left(s\right),{u}_{2}^{\prime }\left(s\right)\right)\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{t}|f\left(s,{u}_{1}\left(s\right),{u}_{1}^{\prime }\left(s\right)\right)-f\left(s,{u}_{2}\left(s\right),{u}_{2}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{t}\left[\left(\frac{{\beta }_{1}}{{\alpha }_{1}}+{\beta }_{2}\right)|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)||\phi \left[{u}_{1}\right]\left(s\right)-\phi \left[{u}_{2}\right]\left(s\right)|\\ \phantom{\rule{2em}{0ex}}+{\beta }_{2}s|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)||\left(\phi {\left[{u}_{1}\right]}^{\prime }\left(s\right)-\phi {\left[{u}_{2}\right]}^{\prime }\left(s\right)\right)|\right]\right]\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \left(\frac{{\beta }_{1}}{{\alpha }_{1}}+2{\beta }_{2}\right){\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\parallel \phi \left[{u}_{1}\right]-\phi \left[{u}_{2}\right]\parallel \end{array}$

for any $t\in \left[0,h\right]$. Moreover, we have

$\begin{array}{r}|{\left(\mathrm{\Phi }\phi \left[{u}_{1}\right]\right)}^{\prime }\left(t\right)-{\left(\mathrm{\Phi }\phi \left[{u}_{2}\right]\right)}^{\prime }\left(t\right)|\\ \phantom{\rule{1em}{0ex}}=|\frac{1}{{t}^{2}}{\int }_{0}^{t}s\left(f\left(s,{u}_{1}\left(s\right),{u}_{1}^{\prime }\left(s\right)\right)-f\left(s,{u}_{2}\left(s\right),{u}_{2}^{\prime }\left(s\right)\right)\right)\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,{u}_{1}\left(s\right),{u}_{1}^{\prime }\left(s\right)\right)-f\left(s,{u}_{2}\left(s\right),{u}_{2}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{t}^{2}}{\int }_{0}^{t}s\left[\left(\frac{{\beta }_{1}}{{\alpha }_{1}}+{\beta }_{2}\right)|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)||\phi \left[{u}_{1}\right]\left(s\right)-\phi \left[{u}_{2}\right]\left(s\right)|\\ \phantom{\rule{2em}{0ex}}+{\beta }_{2}s|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)||\left(\phi {\left[{u}_{1}\right]}^{\prime }\left(s\right)-\phi {\left[{u}_{2}\right]}^{\prime }\left(s\right)\right)|\right]\right]\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le \left[\left(\frac{{\beta }_{1}}{{\alpha }_{1}}+{\beta }_{2}\right){\int }_{0}^{{h}_{1}}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}+{\beta }_{2}{\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\right]\parallel \phi \left[{u}_{1}\right]-\phi \left[{u}_{2}\right]\parallel \end{array}$

for any $t\in \left[0,h\right]$. Hence we obtain that

$\begin{array}{r}\parallel \mathrm{\Phi }\phi \left[{u}_{1}\right]-\mathrm{\Phi }\phi \left[{u}_{2}\right]\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(\frac{{\beta }_{1}}{{\alpha }_{1}}+2{\beta }_{2}\right){\int }_{0}^{{h}_{1}}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\parallel \phi \left[{u}_{1}\right]-\phi \left[{u}_{2}\right]\parallel .\end{array}$

By the Banach fixed point theorem, there exists a unique mapping $\phi \left[u\right]\in \phi \left[X\right]$ such that $\mathrm{\Phi }\phi \left[u\right]=\phi \left[u\right]$. Then $Au=u$. u is a solution of (3). □

Next, we consider the case of (II).

Theorem 2.2 Let λ be a real number with $\lambda >0$. Suppose that a mapping f from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)×\mathbb{R}$ into satisfies the following:

1. (a)

The mapping $t↦f\left(t,u,v\right)$ is measurable for any $\left(u,v\right)\in \left(0,\mathrm{\infty }\right)×\mathbb{R}$, and the mapping $\left(u,v\right)↦f\left(t,u,v\right)$ is continuous for almost every $t\in \left[0,1\right]$;

2. (b1)

$|f\left(t,{u}_{1},v\right)|\ge |f\left(t,{u}_{2},v\right)|$ for almost every $t\in \left[0,1\right]$, for any ${u}_{1},{u}_{2}\in \left(0,\mathrm{\infty }\right)$ with ${u}_{1}\le {u}_{2}$ and for any $v\in \mathbb{R}$;

3. (b4)

$|f\left(t,u,{v}_{1}\right)|\le |f\left(t,u,{v}_{2}\right)|$ for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any ${v}_{1},{v}_{2}\in \mathbb{R}$ with ${v}_{1}\le {v}_{2}$;

4. (c2)

There exist ${\alpha }_{1}\in \mathbb{R}$ with $0<{\alpha }_{1}<\lambda$ and ${\alpha }_{2}\in \mathbb{R}$ with ${\alpha }_{2}>\lambda$ such that

${\int }_{0}^{1}|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty };$
5. (d1)

There exists ${\beta }_{1}\in \mathbb{R}$ with ${\beta }_{1}>0$ such that

$|\frac{\partial f}{\partial u}\left(t,u,v\right)|\le \frac{{\beta }_{1}|f\left(t,u,v\right)|}{u}$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

6. (d2)

There exists ${\beta }_{2}\in \mathbb{R}$ with ${\beta }_{2}>0$ such that

$|\frac{\partial f}{\partial v}\left(t,u,v\right)|\le {\beta }_{2}|f\left(t,u,v\right)|$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

7. (e)

There exists the limit

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds$

for any continuously differentiable mapping u from $\left[0,1\right]$ into $\left[0,\mathrm{\infty }\right)$;

8. (f1)

For ${\alpha }_{1}$ and ${\alpha }_{2}$,

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds=0.$

Then there exists $h\in \mathbb{R}$ with $0 such that Cauchy problem (3) has a unique solution in X, where X is a subset

of ${C}^{1}\left[0,h\right]$.

Proof By condition (c2), there exists ${h}_{1}\in \mathbb{R}$ with $0<{h}_{1}\le 1$ such that

${\int }_{0}^{{h}_{1}}|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt

By condition (f1), there exists $h\in \mathbb{R}$ with $0 such that

$\underset{t\in \left(0,h\right]}{sup}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{0}^{{h}_{1}}|f\left(t,{\alpha }_{1}t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt.$

Let A be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

$Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

Since a mapping $t↦\lambda t$ belongs to X, $X\ne \mathrm{\varnothing }$. Moreover, we have $A\left(X\right)\subset X$. Indeed, by condition (a), $Au\in {C}^{1}\left[0,h\right]$, $Au\left(0\right)=0$ and

${\left(Au\right)}^{\prime }\left(0\right)={\left[\lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right]}_{t=0}=\lambda .$

By conditions (b1) and (b4), we obtain that

$\begin{array}{rcl}Au\left(t\right)& =& \lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda t-t{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \ge & \lambda t-t{\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \ge & {\alpha }_{1}t\end{array}$

and

$\begin{array}{rcl}{\left(Au\right)}^{\prime }\left(t\right)& =& \lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & \lambda +{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \le & \lambda +{\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\alpha }_{2}\end{array}$

for any $t\in \left[0,h\right]$. Moreover, by condition (e), there exists the limit

$\underset{t\to 0+}{lim}\frac{t{\left(Au\right)}^{\prime }\left(t\right)-Au\left(t\right)}{{t}^{2}}=\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

We will find a fixed point of A. Let φ be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

and

Then $\phi \left[X\right]$ is a closed subset of ${C}^{1}\left[0,h\right]$ and hence it is a complete metric space. Let Φ be an operator from $\phi \left[X\right]$ into $\phi \left[X\right]$ defined by

$\mathrm{\Phi }\phi \left[u\right]=\phi \left[Au\right].$

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping $\phi \left[u\right]\in \phi \left[X\right]$ such that $\mathrm{\Phi }\phi \left[u\right]=\phi \left[u\right]$ and hence $Au=u$. □

Next, we consider the case of (III).

Theorem 2.3 Let λ be a real number with $\lambda >0$. Suppose that a mapping f from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)×\mathbb{R}$ into satisfies the following:

1. (a)

The mapping $t↦f\left(t,u,v\right)$ is measurable for any $\left(u,v\right)\in \left(0,\mathrm{\infty }\right)×\mathbb{R}$, and the mapping $\left(u,v\right)↦f\left(t,u,v\right)$ is continuous for almost every $t\in \left[0,1\right]$;

2. (b2)

$|f\left(t,{u}_{1},v\right)|\le |f\left(t,{u}_{2},v\right)|$ for almost every $t\in \left[0,1\right]$, for any ${u}_{1},{u}_{2}\in \left(0,\mathrm{\infty }\right)$ with ${u}_{1}\le {u}_{2}$ and for any $v\in \mathbb{R}$;

3. (b3)

$|f\left(t,u,{v}_{1}\right)|\ge |f\left(t,u,{v}_{2}\right)|$ for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any ${v}_{1},{v}_{2}\in \mathbb{R}$ with ${v}_{1}\le {v}_{2}$;

4. (c3)

(c3) There exist ${\alpha }_{1}\in \mathbb{R}$ with $0<{\alpha }_{1}<\lambda$ and ${\alpha }_{2}\in \mathbb{R}$ with ${\alpha }_{2}<\lambda$ such that

${\int }_{0}^{1}|f\left(t,\left(2\lambda -{\alpha }_{1}\right)t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty };$
5. (d1)

There exists ${\beta }_{1}\in \mathbb{R}$ with ${\beta }_{1}>0$ such that

$|\frac{\partial f}{\partial u}\left(t,u,v\right)|\le \frac{{\beta }_{1}|f\left(t,u,v\right)|}{u}$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

6. (d2)

There exists ${\beta }_{2}\in \mathbb{R}$ with ${\beta }_{2}>0$ such that

$|\frac{\partial f}{\partial v}\left(t,u,v\right)|\le {\beta }_{2}|f\left(t,u,v\right)|$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

7. (e)

There exists the limit

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds$

for any continuously differentiable mapping u from $\left[0,1\right]$ into $\left[0,\mathrm{\infty }\right)$;

8. (f2)

For ${\alpha }_{1}$ and ${\alpha }_{2}$,

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds=0.$

Then there exists $h\in \mathbb{R}$ with $0 such that Cauchy problem (3) has a unique solution in X, where X is a subset

of ${C}^{1}\left[0,h\right]$.

Proof By condition (c3), there exists ${h}_{1}\in \mathbb{R}$ with $0<{h}_{1}\le 1$ such that

${\int }_{0}^{{h}_{1}}|f\left(t,\left(2\lambda -{\alpha }_{1}\right)t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt

By condition (f2), there exists $h\in \mathbb{R}$ with $0 such that

$\underset{t\in \left(0,h\right]}{sup}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{0}^{{h}_{1}}|f\left(t,\left(2\lambda -{\alpha }_{1}\right)t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt.$

Let A be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

$Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

Since a mapping $t↦\lambda t$ belongs to X, $X\ne \mathrm{\varnothing }$. Moreover, $A\left(X\right)\subset X$. Indeed, by condition (a), $Au\in {C}^{1}\left[0,h\right]$, $Au\left(0\right)=0$,

${\left(Au\right)}^{\prime }\left(0\right)={\left[\lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right]}_{t=0}=\lambda ,$

by conditions (b2) and (b3),

$\begin{array}{r}Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\ge \lambda t-t{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\ge \lambda t-t{\int }_{0}^{h}|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\ge {\alpha }_{1}t,\\ Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\le \lambda t+t{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\le \lambda t+t{\int }_{0}^{h}|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\le \left(2\lambda -{\alpha }_{1}\right)t,\\ {\left(Au\right)}^{\prime }\left(t\right)=\lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{\left(Au\right)}^{\prime }\left(t\right)}\ge \lambda -{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{\left(Au\right)}^{\prime }\left(t\right)}\ge \lambda -{\int }_{0}^{h}|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{\left(Au\right)}^{\prime }\left(t\right)}\ge {\alpha }_{2}\end{array}$

for any $t\in \left[0,h\right]$, and by condition (e), there exists the limit

$\underset{t\to 0+}{lim}\frac{t{\left(Au\right)}^{\prime }\left(t\right)-Au\left(t\right)}{{t}^{2}}=\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

We will find a fixed point of A. Let φ be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

and

Then $\phi \left[X\right]$ is a closed subset of ${C}^{1}\left[0,h\right]$, and hence it is a complete metric space. Let Φ be an operator from $\phi \left[X\right]$ into $\phi \left[X\right]$ defined by

$\mathrm{\Phi }\phi \left[u\right]=\phi \left[Au\right].$

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping $\phi \left[u\right]\in \phi \left[X\right]$ such that $\mathrm{\Phi }\phi \left[u\right]=\phi \left[u\right]$ and hence $Au=u$. □

Finally, we consider the case of (IV).

Theorem 2.4 Let λ be a real number with $\lambda >0$. Suppose that a mapping f from $\left[0,1\right]×\left(0,\mathrm{\infty }\right)×\mathbb{R}$ into satisfies the following:

1. (a)

The mapping $t↦f\left(t,u,v\right)$ is measurable for any $\left(u,v\right)\in \left(0,\mathrm{\infty }\right)×\mathbb{R}$, and the mapping $\left(u,v\right)↦f\left(t,u,v\right)$ is continuous for almost every $t\in \left[0,1\right]$;

2. (b2)

$|f\left(t,{u}_{1},v\right)|\le |f\left(t,{u}_{2},v\right)|$ for almost every $t\in \left[0,1\right]$, for any ${u}_{1},{u}_{2}\in \left(0,\mathrm{\infty }\right)$ with ${u}_{1}\le {u}_{2}$ and for any $v\in \mathbb{R}$;

3. (b4)

$|f\left(t,u,{v}_{1}\right)|\le |f\left(t,u,{v}_{2}\right)|$ for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any ${v}_{1},{v}_{2}\in \mathbb{R}$ with ${v}_{1}\le {v}_{2}$;

4. (c4)

There exist ${\alpha }_{1}\in \mathbb{R}$ with $0<{\alpha }_{1}<\lambda$ and ${\alpha }_{2}\in \mathbb{R}$ with ${\alpha }_{2}>\lambda$ such that

${\int }_{0}^{1}|f\left(t,\left(2\lambda -{\alpha }_{1}\right)t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty };$
5. (d1)

There exists ${\beta }_{1}\in \mathbb{R}$ with ${\beta }_{1}>0$ such that

$|\frac{\partial f}{\partial u}\left(t,u,v\right)|\le \frac{{\beta }_{1}|f\left(t,u,v\right)|}{u}$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

6. (d2)

There exists ${\beta }_{2}\in \mathbb{R}$ with ${\beta }_{2}>0$ such that

$|\frac{\partial f}{\partial v}\left(t,u,v\right)|\le {\beta }_{2}|f\left(t,u,v\right)|$

for almost every $t\in \left[0,1\right]$, for any $u\in \left(0,\mathrm{\infty }\right)$ and for any $v\in \mathbb{R}$;

7. (e)

There exists the limit

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds$

for any continuously differentiable mapping u from $\left[0,1\right]$ into $\left[0,\mathrm{\infty }\right)$;

8. (f2)

For ${\alpha }_{1}$ and ${\alpha }_{2}$,

$\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds=0.$

Then there exists $h\in \mathbb{R}$ with $0 such that Cauchy problem (3) has a unique solution in X, where X is a subset

of ${C}^{1}\left[0,h\right]$.

Proof By condition (c4), there exists ${h}_{1}\in \mathbb{R}$ with $0<{h}_{1}\le 1$ such that

${\int }_{0}^{{h}_{1}}|f\left(t,\left(2\lambda -{\alpha }_{1}\right)t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt

By condition (f2), there exists $h\in \mathbb{R}$ with $0 such that

$\underset{t\in \left(0,h\right]}{sup}\frac{1}{{t}^{2}}{\int }_{0}^{t}s|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{0}^{{h}_{1}}|f\left(t,\left(2\lambda -{\alpha }_{1}\right)t,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}dt.$

Let A be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

$Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

Since a mapping $t↦\lambda t$ belongs to X, $X\ne \mathrm{\varnothing }$. Moreover, $A\left(X\right)\subset X$. Indeed, by condition (a), $Au\in {C}^{1}\left[0,h\right]$, $Au\left(0\right)=0$,

${\left(Au\right)}^{\prime }\left(0\right)={\left[\lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right]}_{t=0}=\lambda ,$

by conditions (b2) and (b4),

$\begin{array}{r}Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\ge \lambda t-t{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\ge \lambda t-t{\int }_{0}^{h}|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\ge {\alpha }_{1}t,\\ Au\left(t\right)=\lambda t+{\int }_{0}^{t}\left(t-s\right)f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\le \lambda t+t{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\le \lambda t+t{\int }_{0}^{h}|f\left(s,\left(2\lambda -{\alpha }_{1}\right)s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{Au\left(t\right)}\le \left(2\lambda -{\alpha }_{1}\right)t,\\ {\left(Au\right)}^{\prime }\left(t\right)=\lambda +{\int }_{0}^{t}f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{\left(Au\right)}^{\prime }\left(t\right)}\le \lambda +{\int }_{0}^{h}|f\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{\left(Au\right)}^{\prime }\left(t\right)}\le \lambda +{\int }_{0}^{h}|f\left(s,{\alpha }_{1}s,{\alpha }_{2}\right)|\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{\left(Au\right)}^{\prime }\left(t\right)}\le {\alpha }_{2}\end{array}$

for any $t\in \left[0,h\right]$, and by condition (e), there exists the limit

$\underset{t\to 0+}{lim}\frac{t{\left(Au\right)}^{\prime }\left(t\right)-Au\left(t\right)}{{t}^{2}}=\underset{t\to 0+}{lim}\frac{1}{{t}^{2}}{\int }_{0}^{t}sf\left(s,u\left(s\right),{u}^{\prime }\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.$

We will find a fixed point of A. Let φ be an operator from X into ${C}^{1}\left[0,h\right]$ defined by

and

Then $\phi \left[X\right]$ is a closed subset of ${C}^{1}\left[0,h\right]$ and hence it is a complete metric space. Let Φ be an operator from $\phi \left[X\right]$ into $\phi \left[X\right]$ defined by

$\mathrm{\Phi }\phi \left[u\right]=\phi \left[Au\right].$

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping $\phi \left[u\right]\in \phi \left[X\right]$ such that $\mathrm{\Phi }\phi \left[u\right]=\phi \left[u\right]$ and hence $Au=u$. □

## References

1. Knežević-Miljanović J: On the Cauchy problem for an Emden-Fowler equation. Differ. Equ. 2009, 45(2):267–270. 10.1134/S0012266109020141

2. Kawasaki T, Toyoda M: Existence of positive solution for the Cauchy problem for an ordinary differential equation. Advances in Intelligent and Soft Computing 100. In Nonlinear Mathematics for Uncertainly and Its Applications. Edited by: Li S, Wang X, Okazaki Y, Kawabe J, Murofushi T, Guan L. Springer, Berlin; 2011:435–441.

## Author information

Authors

### Corresponding author

Correspondence to Toshiharu Kawasaki.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

TK wrote first draft. MT wrote final manuscript. All authors read and approved the final manuscript.

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Kawasaki, T., Toyoda, M. Existence of positive solutions of the Cauchy problem for a second-order differential equation. J Inequal Appl 2013, 465 (2013). https://doi.org/10.1186/1029-242X-2013-465

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• DOI: https://doi.org/10.1186/1029-242X-2013-465

### Keywords

• Differential Equation
• Banach Space
• Continuous Mapping
• Real Number
• Integral Equation 