Soultion of a particular functional differential equation

Question

I need to solve the following functional differential equation:

$(4\mu-\lambda r - \lambda x+\lambda)f'(x) = \lambda( f(x) - f(1-x-r))$, where $x\in (0, 1-r)$, $r\in (\frac{1}{2}, 1)$

Or, a more general version of it. Please help.

Thanks

Answer 1

EDIT:

Write $f(x) = g(t) + h(t)$ where $$ \eqalign{t &= x - (1-r)/2\cr g(t) &= (f(x) + f(1-r-x))/2\cr h(t) &= (f(x) - f(1-r-x))/2\cr}$$ Thus $g$ is an even function and $h$ is an odd function. Your equation becomes $$ (4\mu + \lambda(1-r)/2 - \lambda t) (g'(t) + h'(t)) = 2 \lambda h(t) $$ For convenience, let $A = 4 \mu/\lambda + (1-r)/2$, so this says $$ (A - t)(g'(t) + h'(t)) = 2 h(t)$$ Changing $t$ to $-t$ (and remembering that $g'$ is odd and $h'$ is even) we get $$ (A + t)(-g'(t) + h'(t)) = -2 h(t) $$ We can then solve for $g'$ and $h'$, obtaining $$\eqalign{g'(t) &= \dfrac{2 A \; h(t)}{A^2 - t^2}\cr h'(t) &= \dfrac{2 t\; h(t)}{A^2 - t^2}}$$ But since $h$ is odd, we need $h(0) = 0$ and then the solution of this initial value problem for $h$ is $h(t) = 0$. Integrating the other equation, $g(t)$ is constant. So you get the rather uninteresting (but obvious, on hindsight!) solutions $$f(x) = constant$$