So one of my least favorite things that textbooks do is using the words "clearly", "it should be obvious", etc.
In my PDEs class, we've started the Fourier Transform, and I missed the first day of it so I am trying to read through my book. Regarding the heat equation on an infinite domain, it tells me this:
From our previous experience, we note that the expression $\sin{\frac{n\pi x} L } e^{-k(\frac{n\pi} L)^2t}$ solves the heat equation [$u_t=k\cdot u_{xx}$] for integer $n$, as well as $\cos{\frac{n\pi x} L } e^{-k(\frac{n\pi} L)^2t}$. In fact, it is clear that $$u(x,t)=e^{-i\omega x}e^{-k\omega^2t}$$ solves [the heat equation as well], for arbitrary $\omega$ both positive and negative.
It's not "clear" to me why this happens, so I tried 'deriving' this form for a bit by using $\omega=\frac{n\pi}L$ and writing both of the trig functions in their exponential forms
$$\sin x = \frac 1 2(e^{ix}-e^{-ix})$$ $$\cos x = \frac 1 2(e^{ix}+e^{-ix})$$
(and with terms like $e^{i\omega x}$ as well) and added, multiplied, etc, but to no avail.
To be clear (no pun intended), I know that the $e^{-k\omega^2t}$ term is the same as the exponential term in both of the expressions which solve the equation, but I'm failing to see where $e^{-i\omega x}$ came into play.
Any advice as to how I can figure this out? (If possible, please give me some advice on 'deriving' it myself before giving a full answer?)
Let $u(x,t) = X(x)T(t)$, then $$ X(x)T'(t) = k X''(x)T(t)$$ $$ \frac{T'}{T} = k \frac{X''}{X}$$
Since the LHS is only in terms of $t$ and the RHS is only in terms of $x$, they have to both be equal to a constant, so we let $$ \frac{X''}{X} = -\omega^2 $$ $$ \frac{T'}{T} = -k\omega^2 $$
Solving the above ODEs give $X(x) = e^{-i\omega x}$ and $T(t) = e^{-k\omega^2t}$
Note: You can also verify the solution by just taking the derivatives and see that $$ u_t = -k\omega^2 e^{-i\omega x}e^{-k\omega^2t}$$ $$ u_{xx} = -\omega^2 e^{-i\omega x}e^{-k\omega^2t} $$