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Consider the homogeneous functional equations

\begin{equation*} au(x+s)+bu(x+r)=0,\text{ }x\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion . \end{equation*}

It is known that for $a=1,b=-1$ and $u$ is periodic that $u$ is constant if and only if $\frac{s}{r}\in %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion .$ For example, The sum of two continuous periodic functions is periodic if and only if the ratio of their periods is rational? or A real continuous periodic function with two incommensurate periods is constant.

My question: What can happens if $a,b$ are arbitrary coefficients with $\frac{s}{r}\notin %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion $.? Can we prove the non-existence of such a function $u$ under a suitable conditions on $a$ and $b$?.

If we look for a solution of the exponential form $u(x)=Ce^{\lambda x},$ then we find that under a suitable assumption on $a$ and $b$ the solution exists if and only if $\frac{s}{r}\in %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion .$ However, we don't know if this form of solution is the unique one.

Is there any reference that treats this case or any ideas?.

Thank you.

Bill Dubuque
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Gustave
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1 Answers1

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If $a=b=0$ then $u$ can be anything. Otherwise, without loss of generality let $b \ne 0$ and define

$$v(x) \equiv u(x+s),$$ $$c \equiv -a/b,$$ $$ q \equiv r - s.$$

Then, the original equation becomes

$$v(x+q) = c v(x).$$

Then, we could specify the value of $v(x)$ for $0 \le x < q$ and otherwise obtain the solution

$$v(x) = v(x- nq)c^n,$$

where $n$ is an integer chosen so that $0 \le x - n q < q$. Then, we just undo my redefinitions to obtain $u(x)$.

David
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  • thank you sir for the answer; what if the equations is pose on (0,1) instead of the real line?. – Gustave Jul 24 '21 at 00:56
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    @Gustave thanks. If everything else is the same but $x$ is restricted to $(0,1)$ the answer stays the same except the result only applies on that restricted domain, right? Are you trying to enforce some sort of periodicity, like that $u(x+1)=u(x)$? Based on the analysis already presented, I think the solutions can only be periodic if $b = \pm a$. – David Jul 24 '21 at 13:25
  • Thank you @David, I want to find a necessary and sufficient conditions on the coefficients for the existence of such solution. I'm not looking for periodicity properties. – Gustave Jul 24 '21 at 15:39
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    @Gustave well if I'm understanding the setup properly, getting a nontrivial periodicity requires some positive integer power of $b/a$ to be unity. Since $b/a$ is a real number, that means you need $b = \pm a$. – David Jul 24 '21 at 20:23
  • Thank you @David again. I'm not looking for the existence of periodic solutions. My question is the following: Under what assumption on the coefficients so that the unique solution is the null one ( r/s is irrational). Do you know any référence which deals with this kind of problems?. Thank you in advance. – Gustave Jul 25 '21 at 10:58
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    @Gustave Per my original answer, I don't think there is a set parameters for which the null solution is the only one. As I said, the answer only seems to depend (nontrivially) on $r,s$ through the difference, $r-s$. I don't see why it would matter at all whether $r/s$ or anything else is rational or irrational. – David Jul 25 '21 at 22:21
  • Thank you @David for helping me. – Gustave Jul 25 '21 at 23:35
  • @Gustave No problem, best of luck with your study. – David Jul 26 '21 at 00:28