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 Open Access
Solutions and Green’s function of the first order linear equation with reflection and initial conditions
 Alberto Cabada^{1}Email author and
 Fernando Adrián Fernández Tojo^{1}
https://doi.org/10.1186/16872770201499
© Cabada and Tojo; licensee Springer. 2014
 Received: 23 December 2013
 Accepted: 8 April 2014
 Published: 7 May 2014
Abstract
This work is devoted to the study of the existence and sign of Green’s functions for first order linear problems with constant coefficients and initial (one point) conditions. We first prove a result on the existence of solutions of n th order linear equations with involutions via some auxiliary functions to later prove a uniqueness result in the first order case. We study then different situations for which a Green’s function can be obtained explicitly and derive several results in order to obtain information as regards the sign of the Green’s function. Once the sign is known, optimal maximum and antimaximum principles follow.
Keywords
 equations with involutions
 equations with reflection
 Green’s functions
 maximum principles
 comparison principles
 periodic conditions
1 Introduction
The study of functional differential equations with involutions (DEI) can be traced back to the solution of the equation ${x}^{\prime}(t)=x(1/t)$ by Silberstein (see [1]) in 1940. Briefly speaking, an involution is just a function f that satisfies $f(f(x))=x$ for every x in its domain of definition. For most applications in analysis, the involution is defined on an interval of ℝ and in the majority of the cases, it is continuous, which implies it is decreasing and has a unique fixed point. Ever since that foundational paper of Silberstein, the study of problems with DEI has been mainly focused on those cases with initial conditions, with an extensive research in the case of the reflection $f(x)=x$.
Wiener and Watkins study in [2] the solution of the equation ${x}^{\prime}(t)ax(t)=0$ with initial conditions. Equation ${x}^{\prime}(t)+ax(t)+bx(t)=g(t)$ has been treated by Piao in [3, 4]. In [2, 5–8] some results are introduced to transform this kind of problems with involutions and initial conditions into second order ordinary differential equations with initial conditions or first order two dimensional systems, granting that the solution of the last will be a solution to the first. Furthermore, asymptotic properties and boundedness of the solutions of initial first order problems are studied in [9] and [10], respectively. Second order boundary value problems have been considered in [8, 11–13] for Dirichlet and SturmLiouville boundary value conditions, higher order equations has been studied in [14]. Other techniques applied to problems with reflection of the argument can be found in [15–17].
More recently, the papers of Cabada et al. [18, 19] have further studied the case of the second order equation with twopoint boundary conditions, adding a new element to the previous studies: the existence of a Green’s function. Once the study of the sign of the aforementioned function is done, maximum and antimaximum principles follow. Other works in which Green’s functions are obtained for functional differential equations (but with a fairly different setting, like delay or normal equations) are, for instance, [20–25].
In this paper we try to answer to the following question: How is it possible find a solution of an initial problem with a differential equation with reflection? What is more, in which cases can a Green’s function be constructed and how can it be found?
Section 2 will have two parts. In the first one we construct the solutions of the n th order DEI with reflection, constant coefficients and initial conditions. In the second one we find the Green’s function for the order one case. In Section 3 we apply these findings in order to describe exhaustively the range of values for which suitable comparison results are fulfilled and we illustrate them with some examples.
2 Solutions of the problem
In order to prove an existence result for the n th order DEI with reflection, we consider the even and odd parts of a function f, that is, ${f}_{e}(x):=[f(x)+f(x)]/2$ and ${f}_{o}(x):=[f(x)f(x)]/2$ as done in [18].
2.1 The n th order problem
where $h\in {L}_{\mathrm{loc}}^{1}(\mathbb{R})$, ${t}_{0},c,{a}_{k},{b}_{k}\in \mathbb{R}$ for $k=0,\dots ,n1$; ${a}_{n}=0$; ${b}_{n}=1$. A solution to this problem will be a function $u\in {W}_{\mathrm{loc}}^{n,1}(\mathbb{R})$, that is, u is k times differentiable in the sense of distributions and each of the derivatives satisfies ${u}^{(k)}{}_{K}\in {L}^{1}(K)$ for every compact set $K\subset \mathbb{R}$.
Then problem (2.1) has a solution.
Observe that φ is odd, ψ is even and $h=\phi \tilde{u}+\psi \tilde{v}$. So, in order to ensure the existence of solution of problem (2.1) it is enough to find y and z such that $Ly=\phi \tilde{u}$ and $Lz=\psi \tilde{v}$ for, in that case, defining $u=y+z$, we can conclude that $Lu=h$. We will deal with the initial condition later on.
Hence, $Ly=\phi \tilde{u}$.
All the same, by taking $z=\tilde{\psi}\tilde{v}$ with $\tilde{\psi}(t):=\frac{1}{(n1)!}{\int}_{0}^{t}{(ts)}^{n1}\psi (s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$, we have $Lz=\psi \tilde{v}$.
Hence, defining $\overline{u}:=y+z=\tilde{\phi}\tilde{u}+\tilde{\psi}\tilde{v}$ we find that $\overline{u}$ satisfies $L\overline{u}=h$ and $\overline{u}(0)=0$.
If we assume (h1), $w=\overline{u}+\frac{c\overline{u}({t}_{0})}{\tilde{u}({t}_{0})}\tilde{u}$ is clearly a solution of problem (2.1).
When (h2) is fulfilled a solution of problem (2.1) is given by $w=\overline{u}+\frac{c\overline{u}({t}_{0})}{\tilde{v}({t}_{0})}\tilde{v}$.
If (h3) holds, using the aforementioned construction we can find ${w}_{1}$ such that $L{w}_{1}=1$ and ${w}_{1}(0)=0$. Now, ${w}_{2}:={w}_{1}1/({a}_{0}+{b}_{0})$ satisfies $L{w}_{2}=0$. Observe that the second part of condition (h2) is precisely ${w}_{2}({t}_{0})\ne 0$, and hence, defining $w=\overline{u}+\frac{c\overline{u}({t}_{0})}{{w}_{2}({t}_{0})}{w}_{2}$ we see that w is a solution of problem (2.1). □
2.2 The first order problem
with $h\in {L}_{\mathrm{loc}}^{1}(\mathbb{R})$ and ${t}_{0},a,b,c\in \mathbb{R}$. A solution of this problem will be $u\in {W}_{\mathrm{loc}}^{1,1}(\mathbb{R})$.
Let $\omega :=\sqrt{{a}^{2}{b}^{2}}$. Equation (2.7) presents three different cases:
of (2.6) with $\alpha \in \mathbb{R}$.
of (2.6) with $\alpha \in \mathbb{R}$.
(C3) ${a}^{2}={b}^{2}$. In this a case, $u(t)=\alpha t+\beta $ is a solution of (2.7) for every $\alpha ,\beta \in \mathbb{R}$. So, (2.6) holds provided that one of the two following cases is fulfilled:
is the general solution of (2.6) with $\alpha \in \mathbb{R}$, and
is the general solution of (2.6) with $\alpha \in \mathbb{R}$.
Observe that $\tilde{u}$ and $\tilde{v}$ can be obtained from the explicit expressions of the cases (C1)(C3) by taking $\alpha =1$.
Remark 2.2 Note that if u is in the case (C3.1), v is in the case (C3.2) and vice versa.
We have now the following properties of functions $\tilde{u}$ and $\tilde{v}$.
 (I)
${\tilde{u}}_{e}\equiv {\tilde{v}}_{e}$, ${\tilde{u}}_{o}\equiv k{\tilde{v}}_{o}$ for some real constant k a.e.,
 (II)
${\tilde{u}}_{e}(s){\tilde{v}}_{e}(t)={\tilde{u}}_{e}(t){\tilde{v}}_{e}(s)$, ${\tilde{u}}_{o}(s){\tilde{v}}_{o}(t)={\tilde{u}}_{o}(t){\tilde{v}}_{o}(s)$,
 (III)
${\tilde{u}}_{e}{\tilde{v}}_{e}{\tilde{u}}_{o}{\tilde{v}}_{o}\equiv 1$,
 (IV)
$\tilde{u}(s)\tilde{v}(s)+\tilde{u}(s)\tilde{v}(s)=2[{\tilde{u}}_{e}(s){\tilde{v}}_{e}(s){\tilde{u}}_{o}(s){\tilde{v}}_{o}(s)]=2$.
Proof (I) and (III) can be checked by inspection of the different cases. (II) is a direct consequence of (I). (IV) is obtained from the definition of even and odd parts and (III). □
Now, Theorem 2.1 has the following corollary.
Corollary 2.3 Problem (2.5) has a unique solution if and only if $\tilde{u}({t}_{0})\ne 0$.
Proof Considering Lemma 2.2(III), $\tilde{u}$ and $\tilde{v}$, defined as in (2.8) and (2.9), respectively, satisfy the hypothesis of Theorem 2.1, (h1), therefore a solution exists.
Now, assume ${w}_{1}$ and ${w}_{2}$ are two solutions of (2.5). Then ${w}_{2}{w}_{1}$ is a solution of (2.6). Hence, ${w}_{2}{w}_{1}$ is of one of the forms covered in the cases (C1)(C3) and, in any case, a multiple of $\tilde{u}$, that is, ${w}_{2}{w}_{1}=\lambda \tilde{u}$ for some $\lambda \in \mathbb{R}$. Also, it is clear that $({w}_{2}{w}_{1})({t}_{0})=0$, but we have $\tilde{u}({t}_{0})\ne 0$ as a hypothesis, therefore $\lambda =0$ and ${w}_{1}={w}_{2}$. This is, problem (2.5) has a unique solution.
Assume now that w is a solution of (2.5) and $\tilde{u}({t}_{0})=0$. Then $w+\lambda \tilde{u}$ is also a solution of (2.5) for every $\lambda \in \mathbb{R}$, which proves the result. □
This last theorem raises an obvious question: In which circumstances $\tilde{u}({t}_{0})\ne 0$? In order to answer this question, it is enough to study the cases (C1)(C3). We summarize this study in the following lemma, which can be checked easily.
Lemma 2.4 $\tilde{u}({t}_{0})=0$ only in the following cases:

if${a}^{2}>{b}^{2}$and${t}_{0}=\frac{1}{\omega}arctan\frac{\omega}{a+b}+k\pi $for some$k\in \mathbb{Z}$,

if${a}^{2}<{b}^{2}$, $ab>0$and${t}_{0}=\frac{1}{\omega}arctanh\frac{\omega}{a+b}$,

if$a=b$and${t}_{0}=\frac{1}{2a}$.
Also, ${\chi}_{{t}_{1}}^{{t}_{2}}={\chi}_{{t}_{2}}^{{t}_{1}}$.
The following corollary gives the expression of the Green’s function for problem (2.5).
Thus, adding (2.11) and (2.12), it is clear that ${u}^{\prime}(t)+au(t)+bu(t)=h(t)$.
which proves the result. □
Denote now by ${G}_{a,b}$ the Green’s function for problem (2.5) with coefficients a and b. The following lemma is analogous to [[18], Lemma 4.1].
Lemma 2.6 ${G}_{a,b}(t,s)={G}_{a,b}(t,s)$, for all $t,s\in I$.
therefore we can conclude that ${G}_{a,b}(t,s)={G}_{a,b}(t,s)$ for all $t,s\in I$. □
As a consequence of the previous result, we arrive at the following immediate conclusion.
Corollary 2.7 ${G}_{a,b}$ is positive if and only if ${G}_{a,b}$ is negative on ${I}^{2}$.
3 Sign of the Green’s function
In this section we use the above obtained expressions to obtain the explicit expression of the Green’s function, depending on the values of the constants a and b. Moreover, we study the sign of the function and deduce suitable comparison results.
We separate the study in three cases, taking into consideration the expression of the general solution of (2.6).
3.1 The case (C1)
Studying the expression of G we can obtain maximum and antimaximum principles. In order to do this, we will be interested in those maximal strips (in the sense of inclusion) of the kind $[\alpha ,\beta ]\times \mathbb{R}$ where G does not change sign depending on the parameters.
So, we are in a position to study the sign of the Green’s function in the different triangles of definition. The result is the following.
Then the Green’s function of problem (2.5) is

positive on$\{(t,s),0<s<t\}$if and only if$t\in (0,\eta (a,b))$,

negative on$\{(t,s),t<s<0\}$if and only if$t\in (\eta (a,b),0)$.
If $a>0$, the Green’s function of problem (2.5) is

positive on$\{(t,s),t<s<0\}$if and only if$t\in (0,\pi /\sqrt{{a}^{2}{b}^{2}})$,

positive on$\{(t,s),0<s<t\}$if and only if$t\in (\pi /\sqrt{{a}^{2}{b}^{2}},0)$,
and, if $a<0$, the Green’s function of problem (2.5) is

negative on$\{(t,s),t<s<0\}$if and only if$t\in (0,\pi /\sqrt{{a}^{2}{b}^{2}})$,

negative on$\{(t,s),0<s<t\}$if and only if$t\in (\pi /\sqrt{{a}^{2}{b}^{2}},0)$.
Proof For $0<b<a$, the argument of the sin in (3.1c) is positive, so (3.1c) is positive for $t<\pi /\omega $. On the other hand, it is easy to check that (3.1a) is positive as long as $t<\eta (a,b)$.
The rest of the proof continues similarly. □
As a corollary of the previous result we obtain the following one.
Lemma 3.2 Assume ${a}^{2}>{b}^{2}$. Then we have the following:

if$a>0$, the Green’s function of problem (2.5) is nonnegative on$[0,\eta (a,b)]\times \mathbb{R}$,

if$a<0$, the Green’s function of problem (2.5) is nonpositive on$[\eta (a,b),0]\times \mathbb{R}$,

the Green’s function of problem (2.5) changes sign in any other strip not a subset of the aforementioned.
□
Remark 3.1 Realize that the rectangles defined in the previous lemma are optimal in the sense that G changes sign in a bigger rectangle. The same observation applies to similar results we will prove for the other cases. This fact implies that we cannot have maximum or antimaximum principles on bigger intervals for the solution, something that is widely known and which the following results, together with Example 3.4, illustrate.
Since $G(t,0)$ changes sign at $t=\eta (a,b)$, it is immediate to verify that, by defining function ${h}_{\u03f5}(s)=1$ for all $s\in (\u03f5,\u03f5)$ and ${h}_{\u03f5}(s)=0$ otherwise, we have a solution of problem (2.5) that crosses the 0 line as close to the right of $\eta (a,b)$ as necessary. So the estimates are optimal for this case.
However, one can study problems with particular nonhomogeneous part h for which the solution crosses 0 for a bigger interval. This is showed in the following example.
Example 3.1 Consider the problem ${x}^{\prime}(t)5x(t)+4x(t)={cos}^{2}3t$, $x(0)=0$.
$\overline{u}(0)=0$, so $\overline{u}$ is the solution of our problem.
If we use Lemma 3.2, we find that, a priori, $\overline{u}$ is nonpositive on $[4/15,0]$ which we know is true by the study we have done of $\overline{u}$, but this estimate is, as expected, far from the interval $[\gamma 1,0]$ in which $\overline{u}$ is nonpositive. This does not contradict the optimality of the a priori estimate, as we have showed before, some other examples could be found for which the interval where the solution has constant is arbitrarily close to the one given by the a priori estimate.
3.2 The case (C2)
Studying the expression of G we can obtain maximum and antimaximum principles. With this information, we can state the following lemma.
Then we have the following:

if$a>0$, the Green’s function of problem (2.5) is positive on$\{(t,s),t<s<0\}$and$\{(t,s),0<s<t\}$,

if$a<0$, the Green’s function of problem (2.5) is negative on$\{(t,s),t<s<0\}$and$\{(t,s),0<s<t\}$,

if$b>0$, the Green’s function of problem (2.5) is negative on$\{(t,s),t<s<0\}$,

if$b>0$, the Green’s function of problem (2.5) is positive on$\{(t,s),0<s<t\}$if and only if$t\in (0,\sigma (a,b))$,

if$b<0$, the Green’s function of problem (2.5) is positive on$\{(t,s),0<s<t\}$,

if$b<0$, the Green’s function of problem (2.5) is negative on$\{(t,s),t<s<0\}$if and only if$t\in (\sigma (a,b),0)$.
Proof For $0<a<b$, he argument of the sinh in (3.1d) is negative, so (3.2d) is positive. The argument of the sinh in (3.1c) is positive, so (3.2c) is positive. It is easy to check that (3.2a) is positive as long as $t<\sigma (a,b)$.
On the other hand, (3.2b) is always negative.
The rest of the proof continues similarly. □
As a corollary of the previous result we obtain the following one.
Lemma 3.4 Assume ${a}^{2}<{b}^{2}$. Then we have the following.

if$0<a<b$, the Green’s function of problem (2.5) is nonnegative on$[0,\sigma (a,b)]\times \mathbb{R}$,

if$b<a<0$, the Green’s function of problem (2.5) is nonnegative on$[0,+\mathrm{\infty})\times \mathbb{R}$,

if$b<a<0$, the Green’s function of problem (2.5) is nonpositive on$[\sigma (a,b),0]\times \mathbb{R}$,

if$b>a>0$, the Green’s function of problem (2.5) is nonpositive on$(\mathrm{\infty},0]\times \mathbb{R}$,

the Green’s function of problem (2.5) changes sign in any other strip not a subset of the aforementioned.
with $\lambda >0$.
Observe that for $\lambda =1$, $c=sinh1$, $w(t)=sinht$. Lemma 3.4 guarantees the nonnegativity of w on $[0,1.52069\dots ]$, but it is clear that the solution w is positive on the whole positive real line.
3.3 The case (C3)
Studying the expression of G we can obtain maximum and antimaximum principles. With this information, we can prove the following lemma as we did with the analogous ones for cases (C1) and (C2).
Lemma 3.5 Assume $a=b$. Then, if $a>0$, the Green’s function of problem (2.5) is

positive on$\{(t,s),t<s<0\}$and$\{(t,s),0<s<t\}$,

negative on$\{(t,s),t<s<0\}$,

positive on$\{(t,s),0<s<t\}$if and only if$t\in (0,1/a)$,
and, if $a<0$, the Green’s function of problem (2.5) is

negative on$\{(t,s),t<s<0\}$and$\{(t,s),0<s<t\}$,

positive on$\{(t,s),0<s<t\}$,

negative on$\{(t,s),t<s<0\}$if and only if$t\in (1/a,0)$.
As a corollary of the previous result we obtain the following one.
Lemma 3.6 Assume $a=b$. Then we have the following:

if$0<a$, the Green’s function of problem (2.5) is nonnegative on$[0,1/a]\times \mathbb{R}$,

if$a<0$, the Green’s function of problem (2.5) is nonpositive on$[1/a,0]\times \mathbb{R}$,

the Green’s function of problem (2.5) changes sign in any other strip not a subset of the aforementioned.
For this particular case we have another way of computing the solution to the problem.
The initial condition is also satisfied for, clearly, $u({t}_{0})=c$. □
where $\overline{u}(t)=\frac{1}{p+1}t{t}^{p}$ and $\tilde{u}(t)=12\lambda t$. $\overline{u}$ is positive in $(0,+\mathrm{\infty})$ and negative in $(\mathrm{\infty},0)$ independently of λ, so the solution has better properties than the ones guaranteed by Lemma 3.6.
The next example shows that the estimate is sharp.
Lemma 3.8 Assume $a=b$. Then, if $a>0$, the Green’s function of problem (2.5) is

positive on$\{(t,s),t<s<0\}$, $\{(t,s),0<s<t\}$and$\{(t,s),0<s<t\}$,

negative on$\{(t,s),t<s<0\}$if and only if$t\in (1/a,0)$,
and, if $a>0$, the Green’s function of problem (2.5) is

negative on$\{(t,s),t<s<0\}$, $\{(t,s),t<s<0\}$and$\{(t,s),0<s<t\}$,

positive on$\{(t,s),0<s<t\}$if and only if$t\in (0,1/a)$.
As a corollary of the previous result we obtain the following one.
Lemma 3.9 Assume $a=b$. Then we have the following:

if$a>0$, the Green’s function of problem (2.5) is nonnegative on$[0,+\mathrm{\infty})\times \mathbb{R}$,

if$a<0$the Green’s function of problem (2.5) is nonpositive on$(\mathrm{\infty},0]\times \mathbb{R}$,

the Green’s function of problem (2.5) changes sign in any other strip not a subset of the aforementioned.
Again, for this particular case we have another way of computing the solution to the problem.
The initial condition is also satisfied for, clearly, $u({t}_{0})=c$. □
where $\overline{u}(t)=\frac{1}{1+{t}^{2}}+\lambda (1+2\lambda t)arctant{\lambda}^{2}ln(1+{t}^{2})1$.
is positive on ℝ if $\lambda >1$ and negative on ℝ for all $\lambda <1$. Therefore, Lemma 3.9 guarantees that $\overline{u}$ will be positive on $(0,\mathrm{\infty})$ for $\lambda >1$ and in $(\mathrm{\infty},0)$ when $\lambda <1$.
Declarations
Acknowledgements
The authors are thankful to the anonymous referees for the careful reading of the manuscript and suggestions. This work was supported in part by FEDER and Ministerio de Educación y Ciencia, Spain, project MTM201015314. The second author was supported by FPU scholarship, Ministerio de Educación, Cultura y Deporte, Spain.
Authors’ Affiliations
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