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Solve the following non-homogenous heat equation:$$ \frac{\partial u}{\partial t} = a^2 \frac{\partial ^2 u}{\partial x^2} + \epsilon \sin(k_0x)$$ on the domain $-\infty \lt x \lt \infty $, with starting conditions $u(x,t=0) = A\delta(x-x_0)$. Assume the solution is bounded everywhere.

I've tried solving this using separation of variables, but I can't seem to figure out how to use the BV's in order to get eigenvalues. Using Fourier transform seems like it would work here, since $|x|<\infty$. And if Fourier transform is the only solution, how would I go about doing it?

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    An idea: Try a solution of the form $u(x,t) = w(x,t) + v(x)$ such that $v(x)$ satisfies the above equation (so it will be of the form $\alpha \sin(k_0x)$. Then use this to solve the homogeneous problem for $w(x,t)$ (be careful of the BV) and add the two solutions. – 2017-02-10

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A substitution is called for, after noticing that \begin{align} \frac{\partial u}{\partial t } & = a^2\frac{\partial^2 u}{\partial x^2}+\epsilon\sin(k_0 x) \\ & = a^2\frac{\partial^2}{\partial x^2}\left(u-\frac{\epsilon}{a^2k_0^2}\sin(k_0 x)\right) \end{align} Let $u = v+\frac{\epsilon}{a^2k_0^2}\sin(k_0 x)$. Then the equation for $v$ is the standard heat equation $$ \frac{\partial v}{\partial t}=a^2\frac{\partial^2v}{\partial x^2}, \;\; v(x,t=0)= A\delta(x-x_0)-\frac{\epsilon}{a^2k_0^2}\sin(k_0x) $$ The solution for the heat equation with initial condition is a standard convolution with a Gaussian kernel against the initial function of $x$.

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    Why is the sine multiplying the time derivative in the first equation? I would guess that that's just a typo, since it's gone in the second equation. – 2017-02-10
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    @Ian : Because my fingers are faster than my brain. – 2017-02-10