Delay equations In an effort to solve a delay partial differential equation
$$\partial_t f(t,x)= \Phi(x) f(t,x)+\Psi(x) f(t,x-\alpha),$$
with
$$f(0,x)=1,\hspace{0.3cm} f(t,0)=1$$
Where $\alpha$ is the delay ( a real number) and in particular I have $\Phi(x)=e^{-x}+e^{x}-2$ and $\Psi(x)=1-e^{-x}$, 
I used Laplace transform $$ L(f)=\int_{0}^{t} e^{-st} f(x,t) dt: =U(s,t) $$
and I got a delay equation with no derivatives involved :
$$s\hspace{0.1cm}U(s,x)-1 = \Phi(x) U(s,x)+\Psi(x) U(s,x-\alpha)$$
This is where I stopped. Does anyone know how I can find an explicit formula for $U(s,x)$ from the last equation? Or is there any other way to solve the delay PDE expressed on top? Obviously the separation of variables method doesn't work here because it doesn't satisfy initial and boundary conditions.
 A: Firstly, we remove the $\Phi$ contribution by writing
$$
f(t,x)=e^{t\Phi(x)}h(t,x).
$$
This gives
$$
\partial h(t,x)=e^{-t\Phi(x)}\Psi(x)e^{-t\Phi(x-\alpha)}h(t,x-\alpha).
$$
that grants that $h(t,0)=1$ and $h(0,x)=1$. Then, we notice that
$$
h(t,x-a)=\sum_{n=0}^\infty\frac{(-\alpha)^n}{n!}h^{(n)}(t,x)
$$
that means that the equation is non-local. This can be rewritten formally as
$$
h(t,x-\alpha)=e^{-\alpha\partial_x}h(t,x).
$$
If we take
$$
h(t,x)=\int_{-\infty}^\infty\frac{dk}{2\pi}e^{ikx}{\tilde h}(t,k)
$$
it is easy to see that
$$
\int_{-\infty}^\infty\frac{dk}{2\pi}e^{ikx}\partial_t{\tilde h}(t,k)=
A(t,x)\int_{-\infty}^\infty\frac{dk}{2\pi}e^{ik(x-a)}{\tilde h}(t,k)
$$
where we have set $A(t,x)=e^{-t\Phi(x)}\Psi(x)e^{-t\Phi(x-\alpha)}$. Therefore,
$$
\int_{-\infty}^\infty dxe^{-ik'x}\int_{-\infty}^\infty\frac{dk}{2\pi}e^{ikx}\partial_t{\tilde h}(t,k)=
\int_{-\infty}^\infty dxe^{-ik'x}A(t,x)\int_{-\infty}^\infty\frac{dk}{2\pi}e^{ik(x-\alpha)}{\tilde h}(t,k)
$$
i.e.
$$
\partial_t{\tilde h}(t,k')=\int_{-\infty}^\infty\frac{dk}{2\pi}{\tilde A}(t,k'-k)e^{-ik\alpha}{\tilde h}(t,k)
$$
being
$$
{\tilde A}(t,k'-k)=\int_{-\infty}^\infty dxe^{-i(k'-k)x}A(t,x).
$$
This result says that this equation can be amenable to numerical methods but is not solvable exactly.
