Separation of variables for PDE

Question

Consider the PDE $$ \partial_t f(t,x) = \Phi(x) f(t,x)+ \Psi(x)f(t,x-a)$$ $$f(0,x)=1$$ $$f(t,0)=1$$ where $a$ is a constant and $\Phi$ and $\Psi$ are some differentiable functions. To solve this I used separation of variables $f(t,x)=g(x)h(t)$ and I get $$ \frac{\partial_t h(t)} {h(t)}=\Phi(x)+\Psi(x)\frac{g(x-a)}{g(x)}=c_i$$

Where $c_i$ is a constant so we get $h(t)=d_i e^{c_i t}$ and we obtain $$ g(x)=\prod_{k=0}^{\infty} \frac{\Psi(x-ka)}{c_i-\Phi(x-ka)}$$ so we have two free parameter $c_i$ and $d_i$ so the question is what is the general solution of the PDE based on these two free parameters? Is that correct to write the general solution as $$ \int\int z e^{yt} \prod_{k=0}^{\infty} \frac{\Psi(x-ka)}{y-\Phi(x-ka)} dy dz\quad ? $$

Answer 1

Assuming everything converges I think the correct candidate for a general solution for your PDE (leaving boundary conditions aside) is rather $$ \int q(y) e^{yt} \prod_{k=0}^{\infty} \frac{\Psi(x-ka)}{y-\Phi(x-ka)} dy $$ where $q(y)$ is an arbitrary function of $y$.

Formally, it satisfies your PDE (again assuming everything converges and you can differentiate under the integral and product signs) and contains an arbitrary function of one variable, and so should be a general solution.