We want to prove that the unique solution to the following difference equation is the null one: $$ au(x)+b\mathbf{1}_{(0,\frac{1}{2})}(x)u(x+\frac{1}{2})+c\mathbf{1}_{(\frac{1% }{2},1)}(x)u(x-\frac{1}{2})=0,\text{ }x\in (0,1). $$ Extending $u$ by zero outside $(0,1)$ and taking the Laplace transform yields $$ a\int_{0}^{1}e^{-px}u(x)dx+be^{\frac{p}{2}}\int_{\frac{1}{2}% }^{1}e^{-px}u(x)dx+ce^{-\frac{p}{2}}\int_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,% \text{ }p\in %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion , $$ that is $$ \left( a+be^{\frac{p}{2}}\right) \int_{\frac{1}{2}}^{1}e^{-px}u(x)dx+\left( a+ce^{-\frac{p}{2}}\right) \int_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,\text{ }% p\in %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion . $$ If we let for instance $p=\gamma +4n\pi i,n\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ with $a+ce^{-\frac{\gamma }{2}}=0$ we get $$ \left( a+be^{\frac{\gamma }{2}}\right) e^{-\gamma }\int_{\frac{1}{2}% }^{1}e^{-4n\pi ix}u(x)dx=0, $$ which yields that $u=0$ on $(\frac{1}{2},1)$ if $a+be^{\frac{\gamma }{2}% }\neq 0$ which is equivalent to $a^2-bc \neq 0$. With the same manner, by choosing this time $p=\delta +4n\pi i,n\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ with $a+be^{\frac{\delta }{2}}=0$ we get $$ \left( a+ce^{-\frac{\delta }{2}}\right) e^{-\delta }\int_{0}^{\frac{1}{2}% }e^{-4n\pi ix}u(x)dx=0, $$ so if $a+ce^{-\frac{\delta }{2}}\neq 0$ we get $u=0$ on $(0,\frac{1}{2})$, which is equivalent to $a^2-bc \neq 0$.

I don't know if this kind of reasoning is correct since the Laplace transform of $u$ is zero on subintervals with different choices of $p.$


1 Answer 1


Your condition can be rewritten as the system of three equations:
$$au(x)+bu(x+1/2)=0\ \forall x\in(0,1/2), \tag{1}$$ $$au(x)+cu(x-1/2)=0\ \forall x\in(1/2,1), \tag{2} $$ $$au(1/2)=0. \tag{3} $$ In turn, (2) can be rewritten as $$cu(x)+au(x+1/2)=0\ \forall x\in(0,1/2). \tag{2a} $$ So, if the determinant $a^2-bc$ of the system (1)--(2a) of linear equations for $u(x),u(x+1/2)$ is nonzero, then $u(x)=u(x+1/2)=0$ $\forall x\in(0,1/2)$, that is, $u(x)=0$ $\forall x\in(0,1)\setminus\{1/2\}$. Together with (3), this yields $u(x)=0$ $\forall x\in(0,1)$ if $a\ne0$.

The cases when $a^2-bc=0$ or $a=0$ are considered similarly.

In particular, if $a^2-bc=0$ but $a\ne0$, then $a\ne0$ and $b\ne0$, then the system (1)--(3) reduces to the conditions $$u(x)=-(b/a)u(x+1/2)\ \forall x\in(0,1/2)\tag{1a}$$ and $u(1/2)=0$. So, here one may assign any values to $u$ on $(1/2,1)$, and then use (1a) to determine the values of $u$ on $(1/2,1)$.

The OP requested in a comment that the solution be given in terms of the Laplace transform, say $L$. This can be done as follows.

Of course, to use the Laplace transform, we have to assume that $u$ is integrable on $(0,1)$. Let $$U(x):=\begin{cases}u(x)&\text{ if }0<x<1/2,\\ 0&\text{ if }x>1/2,\end{cases}$$ $$V(x):=\begin{cases}u(x+1/2)&\text{ if }0<x<1/2,\\ 0&\text{ if }x>1/2.\end{cases}$$ Then (1) and (2) imply $$aL(U)+bL(V)=0,$$ $$cL(U)+aL(V)=0.$$ So, if $a^2-bc\ne0$, then $L(U)=L(V)=0$ and hence $U=V=0$ almost everywhere (a.e.), so that $u=0$ a.e., so that $u=0$ if $u$ is continuous.

As shown above, there is no uniqueness if $a^2-bc=0$.

  • $\begingroup$ Thank you sir, I have already solved the equation by this approach. The goal us to use Laplace transform to solve the system. The unique approach that I found is what I have written above. I don't know if there is more general approaches to deal with difference equations on bounded domains. $\endgroup$
    – Gustave
    Oct 13, 2021 at 14:16
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    $\begingroup$ @Gustave : Why did you not say you had already solved it by this method? Why did you not state your goal, to necessarily use the Laplace transform? And why is that a goal? $\endgroup$ Oct 13, 2021 at 14:25
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    $\begingroup$ @Gustave : Now you have it for all $p$. $\endgroup$ Oct 14, 2021 at 11:45
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    $\begingroup$ @Gustave : Will you now mark my answers to this question and to the other one (at mathoverflow.net/a/406054/36721) accordingly? $\endgroup$ Oct 15, 2021 at 17:45
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    $\begingroup$ @Gustave : You may want to look at these guidelines: mathoverflow.net/help/someone-answers and mathoverflow.net/help/accepted-answer $\endgroup$ Oct 17, 2021 at 23:16

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