I have the following equation :
\begin{equation} \forall x \in \mathbb{R}, \quad \ u(x) = e^{-|x|} + \beta\int_{\mathbb{R}}e^{-|x-s|}u(s)ds, \end{equation} with $\beta > 0,\ \beta \in \mathbb{R} $ and $u \in L^1(\mathbb{R})$.
I need to prove that the solution of the equation exists and is unique. To prove that it exists, I rewrote the expression as such : \begin{equation} \forall x \in \mathbb{R}, \quad u(x) = f(x) + \beta (f * u)(x), \end{equation} $f(x)$ being equal to $e^{-|x|}$. By applying the Fourier transform to both sides (both $u$ and $f$ being integrable), I obtain: \begin{equation} \forall \xi \in \mathbb{R}, \quad \hat{u}(\xi) = \frac{2}{1+\xi^2} + \frac{2 \beta}{1+\xi^2} \hat{u}(\xi) \end{equation} \begin{equation} \Longleftrightarrow \hat{u}(\xi)\Big(1-\frac{2\beta}{1+\xi^2} \Big) = \frac{2}{1+\xi^2} \end{equation} \begin{equation} \Longleftrightarrow \hat{u}(\xi) = \frac{2}{1-2\beta+\xi^2}. \end{equation} This expression is well defined when $\beta \neq \frac{1+\xi^2}{2}, \ \ \forall \xi \in \mathbb{R}$. $\ \beta$ being strictly greater than $0$, and $\forall \xi \in \mathbb{R}, \ \frac{1+\xi^2}{2} \geq \frac{1}{2}$, I deduce that for $\beta \in \Big(0, \frac{1}{2} \Big)$, the problem has a solution.
How do I prove the uniqueness of the solution? I tried setting $v$ just like the initial equation and then subtracted $v$ from $u$, but I'm stuck from there on. I was also wondering what is the solution? Many thanks.
Note: I used the following result in my calculations: \begin{equation} \mathfrak{F}\Big(e^{-|\xi|} \Big) = \frac{2}{1+\xi^2}. \end{equation}
I will prove the existence and uniqueness using a Banach fixed-point argument.
Let us define a map $\Phi:L^1(\mathbb{R})\rightarrow L^1(\mathbb{R})$ by \begin{align} [\Phi(u)](x) = e^{-|x|} +\beta \int^\infty_{-\infty} e^{-|x-s|}u(s)\ ds \end{align} then we see that \begin{align} \|\Phi(u)-\Phi(v)\|_{L^1} \leq&\ \beta \left\|\int^\infty_{-\infty} e^{-|x-s|}[u(s)-v(s)]\ ds\right\|_{L^1} = \beta\|e^{-|\cdot|}\ast[u-v]\|_{L^1}\\ \leq&\ \beta\|e^{-|\cdot|}\|_{L^1}\|u-v\|_{L^1} = 2\beta\|u-v\|_{L^1}. \end{align} Thus $\Phi$ is a contraction mapping provided $0<\beta<\frac{1}{2}$. Hence by the Banach fixed point theorem there exist a unique $\phi \in L^1(\mathbb{R})$ such that \begin{align} \phi(x) = e^{-|x|}+\int_{\mathbb{R}} e^{-|x-s|}\phi(s)\ ds. \end{align}
Note: I have used the inequality \begin{align} \|f\ast g\|_{L^1} \leq \|f\|_{L^1}\|g\|_{L^1}. \end{align}