# Unique zero solution to a difference equation via Laplace transform

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 $$p.$$

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 $$U(x):=\begin{cases}u(x)&\text{ if }01/2,\end{cases}$$ $$V(x):=\begin{cases}u(x+1/2)&\text{ if }01/2.\end{cases}$$ 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 $$a^2-bc=0$$.

• 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. Oct 13, 2021 at 14:16
• @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? Oct 13, 2021 at 14:25
• @Gustave : Now you have it for all $p$. Oct 14, 2021 at 11:45
• @Gustave : Will you now mark my answers to this question and to the other one (at mathoverflow.net/a/406054/36721) accordingly? Oct 15, 2021 at 17:45
• @Gustave : You may want to look at these guidelines: mathoverflow.net/help/someone-answers and mathoverflow.net/help/accepted-answer Oct 17, 2021 at 23:16