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Existence of positive solutions of the Cauchy problem for a secondorder differential equation
Journal of Inequalities and Applications volume 2013, Article number: 465 (2013)
Abstract
In this paper we consider the equation {u}^{\u2033}(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], KneževićMiljanović considered the Cauchy problem
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}\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty}. Moreover, in [2], Kawasaki and Toyoda considered the Cauchy problem
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)\phantom{\rule{0.2em}{0ex}}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
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 KneževićMiljanović [1].
In this paper, we consider the Cauchy problem
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
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})\phantom{\rule{0.2em}{0ex}}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))\phantom{\rule{0.2em}{0ex}}dsfor 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}sf(s,{\alpha}_{1}s,{\alpha}_{2})\phantom{\rule{0.2em}{0ex}}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
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
Instead of Cauchy problem (3), we consider the integral equation
By condition (c1), there exists {h}_{1}\in \mathbb{R} with 0<{h}_{1}\le 1 such that
By condition (f1), there exists h\in \mathbb{R} with 0<h\le {h}_{1} such that
Let A be an operator from X into {C}^{1}[0,h] defined by
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
By conditions (b1) and (b3), we obtain that
and
for any t\in [0,h]. Moreover, by condition (e), there exists the limit
We will find a fixed point of A. Let φ be an operator from X into {C}^{1}[0,h] defined by
Let \phi [X] be a subset defined by
Then we have
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
By the mean value theorem, for any {u}_{1},{u}_{2}\in X, there exist mappings ξ, η such that
and
for almost every t\in [0,h]. Therefore, by conditions (b1), (b3), (d1) and (d2), we obtain that
for almost every t\in [0,h]. Therefore we have
for any t\in [0,h]. Moreover, we have
for any t\in [0,h]. Hence we obtain 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]. 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})\phantom{\rule{0.2em}{0ex}}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))\phantom{\rule{0.2em}{0ex}}dsfor 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}sf(s,{\alpha}_{1}s,{\alpha}_{2})\phantom{\rule{0.2em}{0ex}}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
of {C}^{1}[0,h].
Proof By condition (c2), there exists {h}_{1}\in \mathbb{R} with 0<{h}_{1}\le 1 such that
By condition (f1), there exists h\in \mathbb{R} with 0<h\le {h}_{1} such that
Let A be an operator from X into {C}^{1}[0,h] defined by
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
By conditions (b1) and (b4), we obtain that
and
for any t\in [0,h]. Moreover, by condition (e), there exists the limit
We will find a fixed point of A. Let φ be an operator from X into {C}^{1}[0,h] defined by
and
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
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}\leftf(t,(2\lambda {\alpha}_{1})t,{\alpha}_{2})\right\phantom{\rule{0.2em}{0ex}}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))\phantom{\rule{0.2em}{0ex}}dsfor 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\leftf(s,(2\lambda {\alpha}_{1})s,{\alpha}_{2})\right\phantom{\rule{0.2em}{0ex}}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
of {C}^{1}[0,h].
Proof By condition (c3), there exists {h}_{1}\in \mathbb{R} with 0<{h}_{1}\le 1 such that
By condition (f2), there exists h\in \mathbb{R} with 0<h\le {h}_{1} such that
Let A be an operator from X into {C}^{1}[0,h] defined by
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,
by conditions (b2) and (b3),
for any t\in [0,h], and by condition (e), there exists the limit
We will find a fixed point of A. Let φ be an operator from X into {C}^{1}[0,h] defined by
and
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
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}\leftf(t,(2\lambda {\alpha}_{1})t,{\alpha}_{2})\right\phantom{\rule{0.2em}{0ex}}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))\phantom{\rule{0.2em}{0ex}}dsfor 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\leftf(s,(2\lambda {\alpha}_{1})s,{\alpha}_{2})\right\phantom{\rule{0.2em}{0ex}}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
of {C}^{1}[0,h].
Proof By condition (c4), there exists {h}_{1}\in \mathbb{R} with 0<{h}_{1}\le 1 such that
By condition (f2), there exists h\in \mathbb{R} with 0<h\le {h}_{1} such that
Let A be an operator from X into {C}^{1}[0,h] defined by
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,
by conditions (b2) and (b4),
for any t\in [0,h], and by condition (e), there exists the limit
We will find a fixed point of A. Let φ be an operator from X into {C}^{1}[0,h] defined by
and
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
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. □
References
KneževićMiljanović J: On the Cauchy problem for an EmdenFowler equation. Differ. Equ. 2009, 45(2):267–270. 10.1134/S0012266109020141
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.
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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 secondorder differential equation. J Inequal Appl 2013, 465 (2013). https://doi.org/10.1186/1029242X2013465
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DOI: https://doi.org/10.1186/1029242X2013465