Arendrogfkl

2022-11-16

I'm interested in solving a heat equation with a discontinuous source term in one dimension on the real line:
$\frac{\mathrm{\partial }u}{\mathrm{\partial }t}\left(x,t\right)=-x\theta \left(x\right)u\left(x,t\right)-x\theta \left(-x\right)u\left(-x,t\right)+D\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{x}^{2}}\left(x,t\right)$
where $\theta \left(\cdot \right)$ is the Heaviside function. The boundary conditions are that $\underset{|x|\to \mathrm{\infty }}{lim}u\left(x,t\right)=0$ The initial condition is arbitrary, but even for a symmetric Gaussian or a symmetric exponential an analytical solution would be interesting.
Because of the discontinuity in the right-hand side induced by $\theta \left(\cdot \right)$, my intuition would be to solve in the two domains x>0 and x<0 and patch them together at x=0. However, the source terms have the effect of causing exponential decay with rate x at the point x for x>0 and corresponding growth at the point −x. This introduces an asymmetry, which should cause a diffusive flux back to the side x>0. Solving for the solution in the two domains does not seem to capture this feedback, and I'm not sure how to solve globally. What are some avenues to proceed?

Samsonitew7b

As I understand your equation, it can be written as
${u}_{t}=-xu\left(|x|,t\right)+{u}_{xx}.$
Take first the case where x>0. A separation of variables $u\left(x,t\right)={e}^{-ct}X\left(x\right)$ gives
$-c=-x+\frac{{X}^{″}}{X}.$
A bounded solution to this uses the Airy function Ai,
$X\left(x\right)=\mathrm{A}\mathrm{i}\left(x-c\right),$
because ${\mathrm{A}\mathrm{i}}^{″}\left(x\right)=x\phantom{\rule{thinmathspace}{0ex}}\mathrm{A}\mathrm{i}\left(x\right)$ Take c to be minus a negative root of Ai; there are infinitely many choices.
So we have
$u\left(x,t\right)={e}^{-ct}\mathrm{A}\mathrm{i}\left(x-c\right),\phantom{\rule{2em}{0ex}}\left(x>0\right).$
Then for x<0 we need to solve
${u}_{t}=-x{e}^{-ct}\mathrm{A}\mathrm{i}\left(-x-c\right)+{u}_{xx}.$
For this it is sufficient to take
$u\left(x,t\right)=-{e}^{-ct}\mathrm{A}\mathrm{i}\left(-x-c\right),\phantom{\rule{2em}{0ex}}\left(x<0\right).$
Thus u is an odd function of x, and with the choice of c, u is continuous and once differentiable with respect to x at x=0.
To get existence of ${u}_{xx}\left(0,t\right)$ this can be achieved by a linear combination of two such functions with different values of c, arranged so that the second derivative is zero at the origin.

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