Existence of solutions of firstorder differential equations via a fixed point theorem for discontinuous operators
 Rubén Figueroa Sestelo^{1} and
 Rodrigo López Pouso^{1}Email author
https://doi.org/10.1186/s1366301504725
© Sestelo and Pouso 2015
Received: 2 September 2015
Accepted: 17 November 2015
Published: 1 December 2015
Abstract
We use a recent Schaudertype result for discontinuous operators in order to look for solutions for firstorder differential equations subject to initial functional conditions. We show how this abstract fixedpoint result allows us to consider a nonlinearity which can be strongly discontinuous. Some examples of applications and comparison with recent literature are included.
1 Introduction and preliminaries
Our goal is to show that the following general version of Schauder’s theorem proven in [1] can be employed to prove the existence of solutions of (1.1) under very general conditions.
Theorem 1.1
([1], Theorem 3.1)
Let K be a nonempty, convex, and compact subset of a normed space X.
Basically, the use of Theorem 1.1 instead of the classical Schauder’s theorem allows f to be discontinuous over the graphs of countably many functions in the conditions of the following definition. Readers are referred to [2–4] for similar definitions.
Definition 1.1

either \(\gamma'(t)=f(t,\gamma(t))\) for a.a. \(t \in[a,b]\) (and we then say that γ is viable for the differential equation),

eitheror$$ \gamma'(t)+\psi(t)< f(t,y) \quad \mbox{for a.a. }t \in I\mbox{ and all }y \in \bigl[\gamma(t)\varepsilon,\gamma(t)+\varepsilon\bigr], $$(1.3)$$ \gamma'(t)\psi(t)>f(t,y) \quad \mbox{for a.a. }t \in I\mbox{ and all }y \in \bigl[\gamma(t)\varepsilon,\gamma(t)+\varepsilon\bigr]. $$(1.4)
Proposition 1.2
In the conditions of Theorem 1.1, let \(x, y\in K\) be fixed.
 1.
\(y \in{\mathbb {T}}x\) as defined in (1.5).
 2.For every \(\varepsilon>0\) and every \(\rho>0\) there exists a finite family of vectors \(x_{i} \in B_{\varepsilon}(x) \cap K\) and coefficients \(\lambda_{i} \in[0,1]\) (\(i=1,2,\dots,m\)) such that \(\sum \lambda_{i}=1\) and$$\Biggl\Vert y\sum_{i=1}^{m} \lambda_{i} Tx_{i}\Biggr\Vert _{X}< \rho. $$
2 Main result
This section is devoted to the proof of the following existence principle. Notice that f need not be continuous with respect to any of its arguments.
Theorem 2.1
 (H1)
There exist \(N \ge0\) and \(M \in L^{1}(I,[0,\infty))\) such that \(N + \M\_{1} \le R\), \(F(x) \le N\) if \(\x\_{\infty} \le R\), and for a.a. \(t \in I\) and all \(x \in[R,R]\) we have \(f(t,x) \le M(t)\).
 (H2)
The compositions \(t \in I \mapsto f(t,x(t))\) are measurable if \(x \in\mathcal{C}(I)\) and \(\x\_{\infty} \le R\).
 (H3)
There exist admissible discontinuity curves \(\gamma _{n}:I_{n}=[a_{n},b_{n}]\longrightarrow \mathbb {R}\) (\(n \in \mathbb {N}\)) such that for a.a. \(t \in I\) the function \(x \mapsto f(t,x)\) is continuous at every \(x \in [R,R] \setminus\bigcup_{\{n : t \in I_{n}\}} \{\gamma_{n}(t)\}\).
Proof
Case 1  \(m(\{t \in I_{n} : x(t)=\gamma_{n}(t)\})=0\) for all \(n \in \mathbb {N}\). Let us prove that then T is continuous at x, which implies that \({\mathbb {T}}x=\{Tx\}\), and then (1.2) is satisfied.
Case 2  \(m(\{t \in I_{n} : x(t)=\gamma_{n}(t)\})>0\) for some \(n \in \mathbb {N}\) such that \(\gamma_{n}\) is inviable. In this case we can prove that \(x \notin{\mathbb {T}}x\), and so (1.2) obtains.
First, we fix some notation. Let us assume that for some \(n \in \mathbb {N}\) we have \(m(\{t \in I_{n} : x(t)=\gamma_{n}(t)\})>0\) and there exist \(\varepsilon>0\) and \(\psi\in L^{1}(I_{n})\), \(\psi(t)>0\) for a.a. \(t \in I_{n}\), such that (1.4) holds with γ replaced by \(\gamma_{n}\) (The proof is similar if we assume (1.3) instead of (1.4), so we omit it.).
Claim  Let \(\varepsilon>0\) be given by our assumptions over \(\gamma_{n}\) and let ρ be as in ( 2.8 ). For every finite family \(x_{i} \in B_{\varepsilon}(x) \cap K\) and \(\lambda_{i} \in[0,1]\) (\(i=1,2,\dots,m\)), with \(\sum\lambda_{i}=1\) , we have \(\x\sum\lambda_{i} Tx_{i}\_{\infty} \ge\rho\) .
Similar computations with \(t_{+}\) instead of \(t_{}\) show that if \(y(\tau _{0}) \le x(\tau_{0})\) then we also have \(\xy\_{\infty} \ge\rho\).
Case 3  \(m(\{t \in I_{n} : x(t)=\gamma_{n}(t)\})>0\) only for some of those \(n \in \mathbb {N}\) such that \(\gamma_{n}\) is viable. Let us prove that in this case the relation \(x \in{\mathbb {T}}x\) implies \(x=Tx\).
Now we assume that \(x \in{\mathbb {T}}x\) and we prove that it implies that \(x'(t)=f(t,x(t))\) a.e. in \(I \setminus A\), and that \(x(t_{0})=F(x)\), thus showing that \(x=Tx\).
Let us denote \(y_{k}=\sum_{i=1}^{m(k)}\lambda_{k,i}Tx_{k,i}\), and notice that \(y_{k} \to x\) uniformly in I and \(\x_{k,i}x\\le1/k\) for all \(k \in \mathbb {N}\) and all \(i \in\{1,2,\dots,m(k)\}\).
3 Examples and nonexistence of extremal solutions
Proposition 3.1
Problem (3.2), with \(f(t,x)\) defined in (3.1), has at least one absolutely continuous solution x such that \(\x\_{\infty} \le5\).
Proof
We have \(\gamma'_{k}(t) <0\) and \(\hat{\gamma}'_{k}(t) <0\) for a.a. \(t \in I\) and all \(k \in K\); however, \(f(t,u) \ge\frac{1}{2}\) for a.a. \(t \in I\) and all \(u \in \mathbb {R}\), and so the discontinuity curves are inviable for the differential equation.
We can conclude by application of Theorem 2.1 that problem (3.2) has at least one absolutely continuous solution x, which, moreover, satisfies \(\x\_{\infty} \le5\). □
In the next example we show that, in general, there is no hope to have extremal solutions of (1.1) in the conditions of Theorem 2.1. By extremal solutions we mean a pair of solutions \(x_{*}\) and \(x^{*}\) (possibly identical) such that any other solution x of (1.1) satisfies \(x_{*}(t) \le x(t) \le x^{*}(t)\) for all \(t \in I\).
Example 3.1
4 Comparison with recent literature
 (C1)
there exists \(M >0\) such that \(0< f(x)<+\infty\) for a.a. \(x\in[0,M]\) and \(\int_{0}^{M} \frac{dx}{f(x)}<+\infty\);
 (C2)
\(h:[0,+\infty)\to[0,+\infty]\) is locally integrable.
In this section we show that Theorem 2.1 guarantees existence of solutions to some cases of (4.1) not covered by the existence results in [6, 7].
Let us consider the discontinuous even function \(\phi: \mathbb {R} \longrightarrow\mathbb {R}\) defined by \(\phi(0)=0\), \(\phi(x)=1/(n+1)\) if \(1/(n+1) \le x < 1/n\) for some \(n \in\mathbb {N}\), and \(\phi(x)=1\) for \(x \ge1\).
However, Theorem 2.1 ensures the existence of solutions of (4.2).
Proposition 4.1
If \(\delta\in(0,1/4)\), then (4.2) has at least one solution defined on the interval \([\delta,\delta]\).
Proof
We note that (4.2) is the particular case of (1.1) corresponding to \(f(t,x)=x+\phi(x)+h(t)\), \(t_{0}=0\), \(L=\delta\), and \(F(x)=0\) for all \(x \in\mathcal{C}(I)\), \(I=[\delta,\delta]\).
For every \(x \in\mathcal{C}(I)\) the sets \(\{t \in I : 1/(n+1) \le x(t) < 1/n\}\), \(n \in\mathbb {N}\), are measurable and \(\phi(x(t))\) is constant on those sets. Hence \(f(\cdot,x(\cdot))\) is measurable.
Since all the conditions in Theorem 2.1 are satisfied, we conclude that (4.2) has at least one solution \(x:[\delta,\delta] \longrightarrow\mathbb {R}\). □
Declarations
Acknowledgements
The first author is partially supported by Xunta de Galicia, Consellería de Cultura, Educación e Ordenación Universitaria, through the project EM2014/032 ‘Ecuacións diferenciais non lineares. The second author is partially supported by Ministerio de Economía y Competitividad, Spain, and FEDER, Projects MTM201015314 and MTM201343014P.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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