I'm stuck on the following problem and could use a hint on what to do next.
Consider $$ \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} + Q(x,t) $$ $$ -\infty < x < \infty \ ,\ u(x,0)=f(x) $$
(a) Show that a particular solution for the Fourier transform $ \bar U $ is $$ \bar U = e^{-kw^2t} \int_0^t \bar Q(\omega,\tau)e^{k\omega ^2 \tau}d\tau $$
(b) Determine $ \bar U $
(c) Solve for $u(x,t)$ in the simplest form possible.
Here's my attempt for part (a).
Let $ \ \ \bar U (\omega,t) = FT(u(x,t)) \ , \ \bar Q (\omega, t) = FT(Q(x,t)) \ \ $ such that $$ FT \big ( \frac{\partial u}{\partial t} \big) = FT\big( k \frac{\partial^2 u}{\partial x^2} + Q(x,t) \big) $$ $$ \Rightarrow \frac{\partial \bar U}{\partial t} = k \ FT \big( \frac{\partial^2 u}{\partial x^2} \big) + FT\big(Q(x,t) \big) $$ $$ = \ -k\omega ^2 \bar U \ + \ \bar Q $$
This is where I'm stuck. I'm not seeing the next step that leads to the particular solution asked for in part (a). I believe I have ideas on how to solve (b) and (c), but I can't do so without solving part (a) first.
Any hints would be appreciated, thank you.
Since you have been given the form of $\bar{U}$ in the problem and have only been asked to verify that it is a particular solution, you can just check that the $\bar{U}$ solves the ODE.
Rewrite $$ \bar{U}(\omega,t) = \int_0^t e^{k \omega^2 (\tau-t)} \bar{Q}(\omega,\tau) d\tau. $$ Then $$ \partial_t \bar{U}(\omega,t) = e^{-k \omega^2 t} e^{k \omega^2 t} \bar{Q}(\omega,t) -k \omega^2 e^{-k \omega^2 t} \int_0^t e^{k \omega^2 \tau} \bar{Q}(\omega,\tau) d\tau \\ = \bar{Q}(\omega,t) - k \omega^2 \bar{U}(\omega,t), $$ and so $\bar{U}$ is a particular solution the ODE in question.
Now, if you want to know where this particular solution comes from, then what you want is "the method of variation of parameters" or "Duhamel's principle." Actually, we can give a quick proof here. If we want to solve $$ f'(t) = a f(t) + g(t) $$ then we multiply by $e^{-at}$ and rewrite as $$ (e^{-at} f(t))' = e^{-at} f'(t) -a e^{-at} f(t) = e^{-at} g(t), $$ and now we simply integrate $$ e^{-at} f(t) - f(0) = \int_0^t e^{-a \tau} g(\tau) d\tau $$ and hence $$ f(t) = f(0) e^{at} + e^{at} \int_0^t e^{-a \tau} g(\tau) d\tau. $$ This is actually the general solution. The particular solution comes from ignoring the $f(0)$ term.