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

Abstract

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

1 Introduction

In [1], Knežević-Miljanović considered the Cauchy problem

$$\{\begin{array}{l}{u}^{″}(t)=P(t)t^{a}u{(t)}^{\sigma },t\in (0,1],\\ u(0)=0,u^{\prime }(0)=\lambda ,\end{array}$$

(1)

where P is continuous, $a,\sigma ,\lambda \in \mathbb{R}$ with $\sigma <0$ and $\lambda >0$, and ${\int }_0^1|P(t)|t^{a+\sigma }dt<\mathrm{\infty }$. Moreover, in [2], Kawasaki and Toyoda considered the Cauchy problem

$$\{\begin{array}{l}{u}^{″}(t)=f(t,u(t)),\text{for almost every }t\in [0,1],\\ u(0)=0,u^{\prime }(0)=\lambda ,\end{array}$$

(2)

where f is a mapping from $[0,1]\times (0,\mathrm{\infty })$ into ℝ 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 [2] is as follows.

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

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

• (b) $|f(t,u_1)|\ge |f(t,u_2)|$ for almost every $t\in [0,1]$ and for any $u_1,u_2\in [0,\mathrm{\infty })$ with $u_1\le u_2$.

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

  $${\int }_0^1|f(t,\alpha t)|dt<\mathrm{\infty }.$$

• (d) There exists $\beta \in \mathbb{R}$ with $\beta >0$ such that

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

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

Then there exists $h\in \mathbb{R}$ with $0<h\le 1$ such that Cauchy problem (2) has a unique solution in X, where X is a subset

$$X=\left\{u|\begin{array}{l}u\in C[0,h],u(0)=0,u^{\prime }(0)=\lambda \\ \mathit{\text{and }}\alpha t\le u(t)\mathit{\text{ for any }}t\in [0,h]\end{array}\right\}$$

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

The case that $f(t,u(t))=P(t)t^{a}u{(t)}^{\sigma }$ in the above theorem is the theorem of Knežević-Miljanović [1].

In this paper, we consider the Cauchy problem

$$\{\begin{array}{l}{u}^{″}(t)=f(t,u(t),u^{\prime }(t)),\text{for almost every }t\in [0,1],\\ u(0)=0,u^{\prime }(0)=\lambda ,\end{array}$$

where f is a mapping from $[0,1]\times (0,\mathrm{\infty })\times \mathbb{R}$ into ℝ 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.

• (I) Decreasing for u in $f(t,u,v)$ (b1) and decreasing for v in $f(t,u,v)$ (b3).

• (II) Decreasing for u in $f(t,u,v)$ (b1) and increasing for v in $f(t,u,v)$ (b4).

• (III) Increasing for u in $f(t,u,v)$ (b2) and decreasing for v in $f(t,u,v)$ (b3).

• (IV) Increasing for u in $f(t,u,v)$ (b2) and increasing for v in $f(t,u,v)$ (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

$$\{\begin{array}{l}{u}^{″}(t)=f(t,u(t),u^{\prime }(t)),\text{for almost every }t\in [0,1],\\ u(0)=0,u^{\prime }(0)=\lambda ,\end{array}$$

(3)

where f is a mapping from $[0,1]\times (0,\mathrm{\infty })\times \mathbb{R}$ into ℝ 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 $[0,1]\times (0,\mathrm{\infty })\times \mathbb{R}$ into ℝ satisfies the following:

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

• (b1) $|f(t,u_1,v)|\ge |f(t,u_2,v)|$ for almost every $t\in [0,1]$, for any $u_1,u_2\in (0,\mathrm{\infty })$ with $u_1\le u_2$ and for any $v\in \mathbb{R}$;

• (b3) $|f(t,u,v_1)|\ge |f(t,u,v_2)|$ for almost every $t\in [0,1]$, for any $u\in (0,\mathrm{\infty })$ and for any $v_1,v_2\in \mathbb{R}$ with $v_1\le v_2$;

• (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

  $${\int }_0^1|f(t,{\alpha }_{1}t,{\alpha }_2)|dt<\mathrm{\infty };$$

• (d1) There exists ${\beta }_1\in \mathbb{R}$ with ${\beta }_1>0$ such that

  $$|\frac{\partial f}{\partial u}(t,u,v)|\le \frac{{\beta }_1|f(t,u,v)|}{u}$$

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

• (d2) There exists ${\beta }_2\in \mathbb{R}$ with ${\beta }_2>0$ such that

  $$|\frac{\partial f}{\partial v}(t,u,v)|\le {\beta }_2|f(t,u,v)|$$

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

• (e) There exists the limit

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds$$

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

• (f1) For ${\alpha }_1$ and ${\alpha }_2$,

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}s|f(s,{\alpha }_{1}s,{\alpha }_2)|ds=0.$$

Then there exists $h\in \mathbb{R}$ with $0<h\le 1$ such that Cauchy problem (3) has a unique solution in X, where X is a subset

$$X=\left\{u|\begin{array}{l}u\in C^1[0,h],u(0)=0,u^{\prime }(0)=\lambda ,\\ {\alpha }_{1}t\le u(t)\mathit{\text{ and }}{\alpha }_2\le u^{\prime }(t)\mathit{\text{ for any }}t\in [0,h]\\ \mathit{\text{and there exists the limit }}{lim}_{t\to 0+}\frac{t{u}^{\prime }(t)-u(t)}{t^2}\end{array}\right\}$$

of $C^1[0,h]$, which is the class of continuously differentiable mappings from $[0,h]$ into ℝ.

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

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

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

$$u(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds.$$

By condition (c1), there exists $h_1\in \mathbb{R}$ with $0<h_1\le 1$ such that

$${\int }_0^{h_1}|f(t,{\alpha }_{1}t,{\alpha }_2)|dt<min\{\lambda -{\alpha }_1,\lambda -{\alpha }_2,{(\frac{{\beta }_1}{{\alpha }_1}+2{\beta }_2)}^{-1}\}.$$

By condition (f1), there exists $h\in \mathbb{R}$ with $0<h\le h_1$ such that

$$\underset{t\in (0,h]}{sup}\frac{1}{t^2}{\int }_0^{t}s|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\le {\int }_0^{h_1}|f(t,{\alpha }_{1}t,{\alpha }_2)|dt.$$

Let A be an operator from X into $C^1[0,h]$ defined by

$$Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds.$$

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

$${(Au)}^{\prime }(0)={[\lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds]}_{t=0}=\lambda .$$

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

$$\begin{array}{rcl}Au(t)& =& \lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds\\ \ge & \lambda t-t{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \ge & \lambda t-t{\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\\ \ge & {\alpha }_{1}t\end{array}$$

and

$$\begin{array}{rcl}{(Au)}^{\prime }(t)& =& \lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds\\ \ge & \lambda -{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \ge & \lambda -{\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\\ \ge & {\alpha }_2\end{array}$$

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

$$\underset{t\to 0+}{lim}\frac{t{(Au)}^{\prime }(t)-Au(t)}{t^2}=\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds.$$

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

$$\phi [u](t)=\{\begin{array}{ll}\frac{u(t)}{t}& \text{if }t\in (0,h],\\ \lambda & \text{if }t=0.\end{array}$$

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

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

Then we have

$$\phi [X]=\left\{v|\begin{array}{l}v\in C^1[0,h],v(0)=\lambda ,\\ {\alpha }_1\le v(t)\text{ and }{\alpha }_2\le v(t)+t{v}^{\prime }(t)\text{ for any }t\in [0,h]\end{array}\right\}$$

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

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

By the mean value theorem, for any $u_1,u_2\in X$, there exist mappings ξ, η such that

$$\begin{array}{r}f(t,u_1(t),u_1^{\prime }(t))-f(t,u_2(t),u_2^{\prime }(t))\\ =\frac{\partial f}{\partial u}(t,\xi (t),u_1^{\prime }(t))(u_1(t)-u_2(t))+\frac{\partial f}{\partial v}(t,u_2(t),\eta (t))(u_1^{\prime }(t)-u_2^{\prime }(t))\\ =(t\frac{\partial f}{\partial u}(t,\xi (t),u_1^{\prime }(t))+\frac{\partial f}{\partial v}(t,u_2(t),\eta (t)))(\phi [u_1](t)-\phi [u_2](t))\\ +t\frac{\partial f}{\partial v}(t,u_2(t),\eta (t))(\phi {[u_1]}^{\prime }(t)-\phi {[u_2]}^{\prime }(t)),\\ min\{u_1(t),u_2(t)\}\le \xi (t)\le max\{u_1(t),u_2(t)\}\end{array}$$

and

$$min\{u_1^{\prime }(t),u_2^{\prime }(t)\}\le \eta (t)\le max\{u_1^{\prime }(t),u_2^{\prime }(t)\}$$

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

$$\begin{array}{r}|f(t,u_1(t),u_1^{\prime }(t))-f(t,u_2(t),u_2^{\prime }(t))|\\ =|(t\frac{\partial f}{\partial u}(t,\xi (t),u_1^{\prime }(t))+\frac{\partial f}{\partial v}(t,u_2(t),\eta (t)))(\phi [u_1](t)-\phi [u_2](t))\\ +t\frac{\partial f}{\partial v}(t,u_2(t),\eta (t))(\phi {[u_1]}^{\prime }(t)-\phi {[u_2]}^{\prime }(t))|\\ \le \left(t\right|\frac{\partial f}{\partial u}(t,\xi (t),u_1^{\prime }(t))|+|\frac{\partial f}{\partial v}(t,u_2(t),\eta (t))\left|\right)|\phi [u_1](t)-\phi [u_2](t)|\\ +t\left|\frac{\partial f}{\partial v}(t,u_2(t),\eta (t))\right|\left|(\phi {[u_1]}^{\prime }(t)-\phi {[u_2]}^{\prime }(t))\right|\\ \le (\frac{{\beta }_1}{{\alpha }_1}+{\beta }_2)|f(t,{\alpha }_{1}t,{\alpha }_2)||\phi [u_1](t)-\phi [u_2](t)|\\ +{\beta }_{2}t|f(t,{\alpha }_{1}t,{\alpha }_2)|\left|(\phi {[u_1]}^{\prime }(t)-\phi {[u_2]}^{\prime }(t))\right|\end{array}$$

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

$$\begin{array}{r}|\mathrm{\Phi }\phi [u_1](t)-\mathrm{\Phi }\phi [u_2](t)|\\ =|\frac{1}{t}{\int }_0^t(t-s)(f(s,u_1(s),u_1^{\prime }(s))-f(s,u_2(s),u_2^{\prime }(s)))ds|\\ \le {\int }_0^t|f(s,u_1(s),u_1^{\prime }(s))-f(s,u_2(s),u_2^{\prime }(s))|ds\\ \le {\int }_0^t\left[(\frac{{\beta }_1}{{\alpha }_1}+{\beta }_2)\right|f(s,{\alpha }_{1}s,{\alpha }_2)\left|\right|\phi [u_1](s)-\phi [u_2](s)|\\ +{\beta }_{2}s|f(s,{\alpha }_{1}s,{\alpha }_2)|\left|(\phi {[u_1]}^{\prime }(s)-\phi {[u_2]}^{\prime }(s))\right|\left]\right]ds\\ \le (\frac{{\beta }_1}{{\alpha }_1}+2{\beta }_2){\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\parallel \phi [u_1]-\phi [u_2]\parallel \end{array}$$

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

$$\begin{array}{r}|{(\mathrm{\Phi }\phi [u_1])}^{\prime }(t)-{(\mathrm{\Phi }\phi [u_2])}^{\prime }(t)|\\ =\left|\frac{1}{t^2}{\int }_0^{t}s(f(s,u_1(s),u_1^{\prime }(s))-f(s,u_2(s),u_2^{\prime }(s)))ds\right|\\ \le \frac{1}{t^2}{\int }_0^{t}s|f(s,u_1(s),u_1^{\prime }(s))-f(s,u_2(s),u_2^{\prime }(s))|ds\\ \le \frac{1}{t^2}{\int }_0^{t}s\left[(\frac{{\beta }_1}{{\alpha }_1}+{\beta }_2)\right|f(s,{\alpha }_{1}s,{\alpha }_2)\left|\right|\phi [u_1](s)-\phi [u_2](s)|\\ +{\beta }_{2}s|f(s,{\alpha }_{1}s,{\alpha }_2)|\left|(\phi {[u_1]}^{\prime }(s)-\phi {[u_2]}^{\prime }(s))\right|\left]\right]ds\\ \le \left[(\frac{{\beta }_1}{{\alpha }_1}+{\beta }_2){\int }_0^{h_1}\right|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\\ +{\beta }_2{\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds]\parallel \phi [u_1]-\phi [u_2]\parallel \end{array}$$

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

$$\begin{array}{r}\parallel \mathrm{\Phi }\phi [u_1]-\mathrm{\Phi }\phi [u_2]\parallel \\ \le (\frac{{\beta }_1}{{\alpha }_1}+2{\beta }_2){\int }_0^{h_1}|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\parallel \phi [u_1]-\phi [u_2]\parallel .\end{array}$$

By the Banach fixed point theorem, there exists a unique mapping $\phi [u]\in \phi [X]$ such that $\mathrm{\Phi }\phi [u]=\phi [u]$. 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 $[0,1]\times (0,\mathrm{\infty })\times \mathbb{R}$ into ℝ satisfies the following:

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

• (b1) $|f(t,u_1,v)|\ge |f(t,u_2,v)|$ for almost every $t\in [0,1]$, for any $u_1,u_2\in (0,\mathrm{\infty })$ with $u_1\le u_2$ and for any $v\in \mathbb{R}$;

• (b4) $|f(t,u,v_1)|\le |f(t,u,v_2)|$ for almost every $t\in [0,1]$, for any $u\in (0,\mathrm{\infty })$ and for any $v_1,v_2\in \mathbb{R}$ with $v_1\le v_2$;

• (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

  $${\int }_0^1|f(t,{\alpha }_{1}t,{\alpha }_2)|dt<\mathrm{\infty };$$

• (d1) There exists ${\beta }_1\in \mathbb{R}$ with ${\beta }_1>0$ such that

  $$|\frac{\partial f}{\partial u}(t,u,v)|\le \frac{{\beta }_1|f(t,u,v)|}{u}$$

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

• (d2) There exists ${\beta }_2\in \mathbb{R}$ with ${\beta }_2>0$ such that

  $$|\frac{\partial f}{\partial v}(t,u,v)|\le {\beta }_2|f(t,u,v)|$$

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

• (e) There exists the limit

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds$$

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

• (f1) For ${\alpha }_1$ and ${\alpha }_2$,

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}s|f(s,{\alpha }_{1}s,{\alpha }_2)|ds=0.$$

Then there exists $h\in \mathbb{R}$ with $0<h\le 1$ such that Cauchy problem (3) has a unique solution in X, where X is a subset

$$X=\left\{u|\begin{array}{l}u\in C^1[0,h],u(0)=0,u^{\prime }(0)=\lambda ,\\ {\alpha }_{1}t\le u(t)\mathit{\text{ and }}{u}^{\prime }(t)\le {\alpha }_2\mathit{\text{ for any }}t\in [0,h]\\ \mathit{\text{and there exists the limit }}{lim}_{t\to 0+}\frac{t{u}^{\prime }(t)-u(t)}{t^2}\end{array}\right\}$$

of $C^1[0,h]$.

Proof By condition (c2), there exists $h_1\in \mathbb{R}$ with $0<h_1\le 1$ such that

$${\int }_0^{h_1}|f(t,{\alpha }_{1}t,{\alpha }_2)|dt<min\{\lambda -{\alpha }_1,{\alpha }_2-\lambda ,{(\frac{{\beta }_1}{{\alpha }_1}+2{\beta }_2)}^{-1}\}.$$

By condition (f1), there exists $h\in \mathbb{R}$ with $0<h\le h_1$ such that

$$\underset{t\in (0,h]}{sup}\frac{1}{t^2}{\int }_0^{t}s|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\le {\int }_0^{h_1}|f(t,{\alpha }_{1}t,{\alpha }_2)|dt.$$

Let A be an operator from X into $C^1[0,h]$ defined by

$$Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds.$$

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

$${(Au)}^{\prime }(0)={[\lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds]}_{t=0}=\lambda .$$

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

$$\begin{array}{rcl}Au(t)& =& \lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds\\ \ge & \lambda t-t{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \ge & \lambda t-t{\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\\ \ge & {\alpha }_{1}t\end{array}$$

and

$$\begin{array}{rcl}{(Au)}^{\prime }(t)& =& \lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds\\ \le & \lambda +{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \le & \lambda +{\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\\ \le & {\alpha }_2\end{array}$$

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

$$\underset{t\to 0+}{lim}\frac{t{(Au)}^{\prime }(t)-Au(t)}{t^2}=\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds.$$

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

$$\phi [u](t)=\{\begin{array}{ll}\frac{u(t)}{t}& \text{if }t\in (0,h],\\ \lambda & \text{if }t=0,\end{array}$$

and

$$\begin{array}{rcl}\phi [X]& =& \{\phi [u]\mid u\in X\}\\ =& \left\{v|\begin{array}{l}v\in C^1[0,h],v(0)=\lambda ,\\ {\alpha }_1\le v(t)\text{ and }v(t)+t{v}^{\prime }(t)\le {\alpha }_2\text{ for any }t\in [0,h]\end{array}\right\}.\end{array}$$

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

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

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping $\phi [u]\in \phi [X]$ such that $\mathrm{\Phi }\phi [u]=\phi [u]$ 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 $[0,1]\times (0,\mathrm{\infty })\times \mathbb{R}$ into ℝ satisfies the following:

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

• (b2) $|f(t,u_1,v)|\le |f(t,u_2,v)|$ for almost every $t\in [0,1]$, for any $u_1,u_2\in (0,\mathrm{\infty })$ with $u_1\le u_2$ and for any $v\in \mathbb{R}$;

• (b3) $|f(t,u,v_1)|\ge |f(t,u,v_2)|$ for almost every $t\in [0,1]$, for any $u\in (0,\mathrm{\infty })$ and for any $v_1,v_2\in \mathbb{R}$ with $v_1\le v_2$;

• (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

  $${\int }_0^1\left|f(t,(2\lambda -{\alpha }_1)t,{\alpha }_2)\right|dt<\mathrm{\infty };$$

• (d1) There exists ${\beta }_1\in \mathbb{R}$ with ${\beta }_1>0$ such that

  $$|\frac{\partial f}{\partial u}(t,u,v)|\le \frac{{\beta }_1|f(t,u,v)|}{u}$$

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

• (d2) There exists ${\beta }_2\in \mathbb{R}$ with ${\beta }_2>0$ such that

  $$|\frac{\partial f}{\partial v}(t,u,v)|\le {\beta }_2|f(t,u,v)|$$

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

• (e) There exists the limit

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds$$

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

• (f2) For ${\alpha }_1$ and ${\alpha }_2$,

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}s\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds=0.$$

Then there exists $h\in \mathbb{R}$ with $0<h\le 1$ such that Cauchy problem (3) has a unique solution in X, where X is a subset

$$X=\left\{u|\begin{array}{l}u\in C^1[0,h],u(0)=0,u^{\prime }(0)=\lambda ,\\ {\alpha }_{1}t\le u(t)\le (2\lambda -{\alpha }_1)t\mathit{\text{ and }}{\alpha }_2\le u^{\prime }(t)\mathit{\text{ for any }}t\in [0,h]\\ \mathit{\text{and there exists the limit }}{lim}_{t\to 0+}\frac{t{u}^{\prime }(t)-u(t)}{t^2}\end{array}\right\}$$

of $C^1[0,h]$.

Proof By condition (c3), there exists $h_1\in \mathbb{R}$ with $0<h_1\le 1$ such that

$${\int }_0^{h_1}\left|f(t,(2\lambda -{\alpha }_1)t,{\alpha }_2)\right|dt<min\{\lambda -{\alpha }_1,\lambda -{\alpha }_2,{(\frac{{\beta }_1}{{\alpha }_1}+2{\beta }_2)}^{-1}\}.$$

By condition (f2), there exists $h\in \mathbb{R}$ with $0<h\le h_1$ such that

$$\underset{t\in (0,h]}{sup}\frac{1}{t^2}{\int }_0^{t}s\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\le {\int }_0^{h_1}\left|f(t,(2\lambda -{\alpha }_1)t,{\alpha }_2)\right|dt.$$

Let A be an operator from X into $C^1[0,h]$ defined by

$$Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds.$$

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

$${(Au)}^{\prime }(0)={[\lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds]}_{t=0}=\lambda ,$$

by conditions (b2) and (b3),

$$\begin{array}{r}Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds\\ \phantom{Au(t)}\ge \lambda t-t{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \phantom{Au(t)}\ge \lambda t-t{\int }_0^h\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\\ \phantom{Au(t)}\ge {\alpha }_{1}t,\\ Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds\\ \phantom{Au(t)}\le \lambda t+t{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \phantom{Au(t)}\le \lambda t+t{\int }_0^h\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\\ \phantom{Au(t)}\le (2\lambda -{\alpha }_1)t,\\ {(Au)}^{\prime }(t)=\lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds\\ \phantom{{(Au)}^{\prime }(t)}\ge \lambda -{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \phantom{{(Au)}^{\prime }(t)}\ge \lambda -{\int }_0^h\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\\ \phantom{{(Au)}^{\prime }(t)}\ge {\alpha }_2\end{array}$$

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

$$\underset{t\to 0+}{lim}\frac{t{(Au)}^{\prime }(t)-Au(t)}{t^2}=\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds.$$

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

$$\phi [u](t)=\{\begin{array}{ll}\frac{u(t)}{t}& \text{if }t\in (0,h],\\ \lambda & \text{if }t=0,\end{array}$$

and

$$\begin{array}{rcl}\phi [X]& =& \{\phi [u]\mid u\in X\}\\ =& \left\{v|\begin{array}{l}v\in C^1[0,h],v(0)=\lambda ,\\ {\alpha }_1\le v(t)\le 2\lambda -{\alpha }_1\text{ and }{\alpha }_2\le v(t)+t{v}^{\prime }(t)\text{ for any }t\in [0,h]\end{array}\right\}.\end{array}$$

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

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

Then we can show, just like Theorem 2.1, that by the Banach fixed point theorem there exists a unique mapping $\phi [u]\in \phi [X]$ such that $\mathrm{\Phi }\phi [u]=\phi [u]$ 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 $[0,1]\times (0,\mathrm{\infty })\times \mathbb{R}$ into ℝ satisfies the following:

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

• (b2) $|f(t,u_1,v)|\le |f(t,u_2,v)|$ for almost every $t\in [0,1]$, for any $u_1,u_2\in (0,\mathrm{\infty })$ with $u_1\le u_2$ and for any $v\in \mathbb{R}$;

• (b4) $|f(t,u,v_1)|\le |f(t,u,v_2)|$ for almost every $t\in [0,1]$, for any $u\in (0,\mathrm{\infty })$ and for any $v_1,v_2\in \mathbb{R}$ with $v_1\le v_2$;

• (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

  $${\int }_0^1\left|f(t,(2\lambda -{\alpha }_1)t,{\alpha }_2)\right|dt<\mathrm{\infty };$$

• (d1) There exists ${\beta }_1\in \mathbb{R}$ with ${\beta }_1>0$ such that

  $$|\frac{\partial f}{\partial u}(t,u,v)|\le \frac{{\beta }_1|f(t,u,v)|}{u}$$

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

• (d2) There exists ${\beta }_2\in \mathbb{R}$ with ${\beta }_2>0$ such that

  $$|\frac{\partial f}{\partial v}(t,u,v)|\le {\beta }_2|f(t,u,v)|$$

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

• (e) There exists the limit

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds$$

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

• (f2) For ${\alpha }_1$ and ${\alpha }_2$,

  $$\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}s\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds=0.$$

Then there exists $h\in \mathbb{R}$ with $0<h\le 1$ such that Cauchy problem (3) has a unique solution in X, where X is a subset

$$X=\left\{u|\begin{array}{l}u\in C^1[0,h],u(0)=0,u^{\prime }(0)=\lambda ,\\ {\alpha }_{1}t\le u(t)\le (2\lambda -{\alpha }_1)t\mathit{\text{ and }}{u}^{\prime }(t)\le {\alpha }_2\mathit{\text{ for any }}t\in [0,h]\\ \mathit{\text{and there exists the limit }}{lim}_{t\to 0+}\frac{t{u}^{\prime }(t)-u(t)}{t^2}\end{array}\right\}$$

of $C^1[0,h]$.

Proof By condition (c4), there exists $h_1\in \mathbb{R}$ with $0<h_1\le 1$ such that

$${\int }_0^{h_1}\left|f(t,(2\lambda -{\alpha }_1)t,{\alpha }_2)\right|dt<min\{\lambda -{\alpha }_1,{\alpha }_2-\lambda ,{(\frac{{\beta }_1}{{\alpha }_1}+2{\beta }_2)}^{-1}\}.$$

By condition (f2), there exists $h\in \mathbb{R}$ with $0<h\le h_1$ such that

$$\underset{t\in (0,h]}{sup}\frac{1}{t^2}{\int }_0^{t}s\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\le {\int }_0^{h_1}\left|f(t,(2\lambda -{\alpha }_1)t,{\alpha }_2)\right|dt.$$

Let A be an operator from X into $C^1[0,h]$ defined by

$$Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds.$$

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

$${(Au)}^{\prime }(0)={[\lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds]}_{t=0}=\lambda ,$$

by conditions (b2) and (b4),

$$\begin{array}{r}Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds\\ \phantom{Au(t)}\ge \lambda t-t{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \phantom{Au(t)}\ge \lambda t-t{\int }_0^h\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\\ \phantom{Au(t)}\ge {\alpha }_{1}t,\\ Au(t)=\lambda t+{\int }_0^t(t-s)f(s,u(s),u^{\prime }(s))ds\\ \phantom{Au(t)}\le \lambda t+t{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \phantom{Au(t)}\le \lambda t+t{\int }_0^h\left|f(s,(2\lambda -{\alpha }_1)s,{\alpha }_2)\right|ds\\ \phantom{Au(t)}\le (2\lambda -{\alpha }_1)t,\\ {(Au)}^{\prime }(t)=\lambda +{\int }_0^{t}f(s,u(s),u^{\prime }(s))ds\\ \phantom{{(Au)}^{\prime }(t)}\le \lambda +{\int }_0^h\left|f(s,u(s),u^{\prime }(s))\right|ds\\ \phantom{{(Au)}^{\prime }(t)}\le \lambda +{\int }_0^h|f(s,{\alpha }_{1}s,{\alpha }_2)|ds\\ \phantom{{(Au)}^{\prime }(t)}\le {\alpha }_2\end{array}$$

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

$$\underset{t\to 0+}{lim}\frac{t{(Au)}^{\prime }(t)-Au(t)}{t^2}=\underset{t\to 0+}{lim}\frac{1}{t^2}{\int }_0^{t}sf(s,u(s),u^{\prime }(s))ds.$$

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

$$\phi [u](t)=\{\begin{array}{ll}\frac{u(t)}{t}& \text{if }t\in (0,h],\\ \lambda & \text{if }t=0,\end{array}$$

and

$$\begin{array}{rcl}\phi [X]& =& \{\phi [u]\mid u\in X\}\\ =& \left\{v|\begin{array}{l}v\in C^1[0,h],v(0)=\lambda ,\\ {\alpha }_1\le v(t)\le 2\lambda -{\alpha }_1\text{ and }v(t)+t{v}^{\prime }(t)\le {\alpha }_2\text{ for any }t\in [0,h]\end{array}\right\}.\end{array}$$

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

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

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