0
$\begingroup$

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 $$ \int\int z e^{yt} \prod_{k=0}^{\infty} \frac{\Psi(x-ka)}{y-\Phi(x-ka)} dy dz\quad ? $$

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.