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.