- #1

Hoplite

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I've been looking at a problem that seems simple at first, but appears to be deceptively difficult (unless I'm missing something).

1. Homework Statement

1. Homework Statement

I've been looking at a problem that involves the diffusion of a scalar quantity, ##q(x)##, on the semi-infinite domain, ##\leq x < \infty##. If ##q(x)## represents a scalar quantity, such as heat, then it can diffuse throughout the domain ##\leq x < \infty##, but won't diffuse out of the boundary at ##x=0##. We don't however know what the value of ##q## is at ##x=0##.

On top of this, there is a source at ##x=a## (delta function), and a sink term that is proportional to the magnitude of ##q(x)##.

## Homework Equations

The overall equation is therefore $$\frac{\partial^2 q}{\partial x^2} - q = -\delta (x-a),$$ $$0\leq x <\infty .$$

## The Attempt at a Solution

It seems to me that the tricky part here is accounting for the boundary conditions. We know that since the source is at ##x=a## and there is a sink term, ##q(x) \rightarrow 0## as ##x\rightarrow \infty##. But how to account for the boundary condition at ##x=0##? If ##q(x)## were a vector quantity, I would imagine that the method of images could be used. Is there a useful equivalent for scalars?

If we look at the equation, the solution should be of the form, $$q(x)= Ae^{x}+Be^{-x}.$$ By substitution, we can then turn this into $$q(x)= Ce^{x-a}+De^{-(x-a)}.$$ In the ##x>a## region, therefore, ##C=0## (or else ##q## would diverge at infinity). But we can't do anything so simple in the ##0\leq x \leq a## region. So this leaves

$$ q(x) = \begin{cases}Ce^{x-a}+De^{-(x-a)}, & 0\leq x\leq a \\ Ee^{-(x-a)}, & x>a

\end{cases}$$

If I could reduce the number of unknowns from 3 to 1, it could solve it by integrating the governing equation over ##x \in [a^-,a^+]##.

Does anyone have any suggestion for how to proceed?