Find solution of particular integral transform

Question

Here's a (not so) fun problem I've encountered trying to complete a proof. I have some constants $b > 0$, $k < 0$, and $c_1, c_2 \in \mathbb R$, and I need to find a function $f:[0, b] \to \mathbb R$ and constant $\lambda \in \mathbb R$ such that $$ c_1 \int_0^x e^{k(x - y)} f(y) ~dy + c_2 \int_x^{b} e^{-k(x - y)} f(y) ~dy = \lambda e^{-kx}, $$ for all $x \in [0, b]$. Does anyone have an approach to find $f$?

Answer 1

Define the following functions:

$$I_1(x):=\int_0^x e^{k(x-y)}f(y) dy~~,~~I_2(x):=\int_{x}^be^{-k(x-y)}f(y)dy\\g(x)=\lambda e^{-kx}$$

and notice, that, very conveniently:

$$\partial_x^2I_1(x)=k^2I_1(x)+kf(x)+f'(x)\\\partial_x^2I_2(x)=k^2I_2(x)+kf(x)-f'(x)$$

which means that, by virtue of the equation $c_1I_1+c_2I_2=g$, $f$ obeys the first order ODE:

$$(c_1-c_2)f'(x)+k(c_1+c_2)f=g''(x)-k^2g(x)$$

which admits the solution:

$$f(x)=f(0)e^{-k\frac{c_1+c_2}{c_1-c_2}x}+\frac{(\lambda-1)k}{2c_2}e^{-kx}$$

However, this is not all that needs to be done, since the function $f$ obeys two linear integral constraints:

$$c_2\int_0^b e^{ky}f(y)dy=\lambda\\c_1\int_0^be^{-ky}f(y)dy=\lambda e^{-2kb}$$

which, as long as $c_2\neq 0$, constitute a linear system in $(f(0),\lambda)$ with a unique solution. The final expressions, albeit easy to obtain, are very lengthy.