# A unique continuation problem

Let $$f\in L^{2}(0,1).$$ Consider the following unique continuation problem: $$\left\{ \begin{array}{ccc} af(x-r)+bf(x)=0, & \mathrm{if} & x\in (r,1) \\ & & \\ cf(x+1-r)+df(x)=0 & \mathrm{if} & x\in (0,r)% \end{array}% \right. \Longrightarrow f=0?.\text{ }r\in (0,1).$$

This is what I have done:

I have started by writing Fourier expansion of $$f:$$ $$f(x)=\sum_{n\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion }c_{n}e^{2in\pi x}\text{ with }c_{n}=\int_{0}^{1}f(t)e^{-2in\pi t}dt.$$

The goal is to prove that $$c_{n}=0,$$ $$\forall n\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion .$$ $$\begin{eqnarray*} c_{n} &=&\int_{0}^{1}f(t)e^{-2in\pi t}dt=\int_{0}^{r}f(t)e^{-2in\pi t}dt+\int_{r}^{1}f(t)e^{-2in\pi t}dt \\ &&\overset{\text{by definition}}{=}-\frac{c}{d}\int_{0}^{r}f(t+1-r)e^{-2in% \pi t}dt-\frac{a}{b}\int_{r}^{1}f(t-r)e^{-2in\pi t}dt \\ &=&-\frac{c}{d}e^{-2in\pi r}\int_{1-r}^{1}f(t)e^{-2in\pi t}dt-\frac{a}{b}% e^{-2in\pi r}\int_{0}^{1-r}f(t)e^{-2in\pi t}dt. \end{eqnarray*}$$

So, $$\left( 1+\frac{c}{d}e^{-2in\pi r}\right) \int_{1-r}^{1}f(t)e^{-2in\pi t}dt+\left( 1+\frac{a}{b}e^{-2in\pi r}\right) \int_{0}^{1-r}f(t)e^{-2in\pi t}dt=0.$$

For example if $$1+\frac{c}{d}e^{-2in\pi r}=1+\frac{a}{b}e^{-2in\pi r},\text{ }\forall n\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion ,$$

the we get $$c_{n}=\int_{0}^{1}f(t)e^{-2in\pi t}dt=0,\text{ }\forall n\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion ,$$

but this condition is clearly not necessary.

Any ideas or other approaches to handle this problem?.

A partial answer: some cases of uniqueness and some cases of non-uniqueness.

Since $$f$$ is only defined a.e., we may identify $$0$$ and $$1$$ and pose the problem in $$\mathbb{R}/\mathbb{Z}$$, which we may identify with $$[0,1)$$ as a measure space. If we denote $$\tau:[0,1)\to[0,1)$$ the translation $$x\mapsto x-r\mod 1$$, and define $$\alpha:=-\frac c d\chi_{[0,r)}-\frac ab\chi_{[r,1)}$$ the conditions can be rewritten as a fixed point equation: $$f(x)=\alpha(x)f(\tau(x))$$ (a.e.).

Situation 1. Assume $$|c|<|d|$$ and $$|a|<|b|$$ (or also, $$|c|>|d|$$ and $$|a|>|b|$$). Then $$\|\alpha\|_\infty<1$$ (respectively, $$\|\frac1 \alpha\|_\infty<1$$ ) and taking the $$L_2$$ norms $$\|f\|_2\le\|\alpha\|_\infty\|f\circ\tau\|_2=\|\alpha\|_\infty\|f\|_2$$ whence $$f=0$$ (the other case is analogous).

Situation 2. Now assume $$r$$ is rational, thus $$r=\frac kn$$ with $$0 and $$(k,n)=1$$. Then the $$n$$ intervals $$I_j:=\tau^j([0,\frac1n))$$, $$0\le j are a partition of $$[0,1)$$, $$k$$ of which are included in $$[0,r)$$, the other $$n-k$$ being included in $$[r,1)$$.

If we define freely $$f$$in the interval $$[0,\frac1n)$$ the equation determines it uniquely on $$[0,1)$$, with a compatibility condition, namely $$f(x)=\Big[\prod_{0\le j Note that for any $$x\in[0,1)$$ the $$n$$ iterates $$\{\tau^j(x)\}_{0\le j are a choice of representatives for the mentioned partition, so the compatibility condition writes $$\Big(-\frac cd\Big)^k\Big(-\frac ab\Big)^{n-k}=1,$$ and we have a closed infinite dimensional linear space of solutions, or no nonzero solutions, according whether this condition holds or does not.

• Thank you Mr. Majer for this great answer. This helped me so much.
– Goga
Aug 8, 2021 at 19:58
• You're welcome. An interesting subcase of 2 is when both $-a/b$ and $-c/d$ are positive real numbers, because the condition for existence of non-zero solutions writes equivalently $(-c/d)^r(-a/b)^{1-r}=1$, that also make sense for irrational $r$, and suggests approximation arguments with rational $r_m\to r$ to prove existence of nontrivial solutions. Aug 8, 2021 at 22:08
• Thank you a lot sir for the remark.
– Goga
Aug 9, 2021 at 18:32
• Actually i think there is more. I'll try to add something in a few days Aug 9, 2021 at 20:26
• Thank you Mr. Majer in advance.
– Goga
Aug 10, 2021 at 19:51