A unique continuation problem

Question

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?.

Thank you in advance.

Answer 1

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 $0<k<n$ and $(k,n)=1$. Then the $n$ intervals $I_j:=\tau^j([0,\frac1n))$, $0\le j<n$ 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 $f(x)=\Big[\prod_{0\le j<n}\alpha(\tau^j(x))\Big]f(x).$ Note that for any $x\in[0,1)$ the $n$ iterates $\{\tau^j(x)\}_{0\le j<n}$ 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.