Deriving the Solution for a PDE with Constants: What Method is Used?

[ShayanJ]

In a physics textbook I'm reading, the PDE ## \frac{\partial p}{\partial t}=-\mu E \frac{\partial p}{\partial x}+D \frac{\partial^2 p}{\partial x^2}-\frac{p-p_o}{\tau} ## is given where ## \mu, \ E, \ D, \ p_o ## and ## \tau ## are constants. It is then stated(yeah, just stated!) that the solution for E=0 is ## p(x,t)=\frac{N}{\sqrt{4 \pi D t}} \exp{\left( -\frac{x^2}{4 D t}-\frac{t}{\tau} \right)}+p_o## and that for nonzero E, the only change to the solution is ## x \rightarrow x-\mu E t ##. But I'm really wondering how did the author get this solution. Its obvious that he didn't use separation of variables. But I know no other method for solving it. What method is used?
Thanks

 

[the_wolfman]

The equation is similar to the heat equation, and the solution is similar to the fundamental solution to the heat equation. This solution to the heat equation is a well known results, and you can find multiple derivations on-line. The general idea is to note that the heat equation is invariant under the transform [itex]x'=ax [/itex] and [itex]t' =a^2 t [/itex]. This implies a solution of the form [itex]u(x,t) = u\left(x^2/t \right) [/itex].

It's not clear how to go from the heat equation to your equation. I suspect that there is a clever substitution.

 

[ShayanJ]

The equation is similar to the heat equation, and the solution is similar to the fundamental solution to the heat equation. This solution to the heat equation is a well known results, and you can find multiple derivations on-line. The general idea is to note that the heat equation is invariant under the transform [itex]x'=ax [/itex] and [itex]t' =a^2 t [/itex]. This implies a solution of the form [itex]u(x,t) = u\left(x^2/t \right) [/itex].

It's not clear how to go from the heat equation to your equation. I suspect that there is a clever substitution.

But the solution I stated in the OP contains ## \frac{1}{\sqrt t} ## and ## t ##, not only ## \frac{x^2}{t} ##.

 

[Chestermiller]

Do you understand the problem that was being solved for the case where E = 0, say with τ infinite?

Chet

 

[HomogenousCow]

Shift p to p-p0 (p nought) to get rid of the inhomogeneous term. (Doesn't affect the derivatives)
Then Fourier transform the equation, the derivation is similar to that of the heat kernel. Throw p0 to the other side and you get that expression.
Sorry for the informal language.

EDIT:I'm bad with LaTex so here's a photo of the first few steps. Just do the same trick with the heat kernel here and you're good.

 

[ShayanJ]

Do you understand the problem that was being solved for the case where E = 0, say with τ infinite?

Chet

Its Haynes-Shockley experiment in semiconductor physics and the equation is basically the continuity equation for holes. The first two terms in the R.H.S. are the divergence of drift and diffusion current and the third term is the sink term corresponding to the recombination process. With ## \tau \to \infty ##, no recombination happens and with E=0, the peak of the hole distribution doesn't move.

Shift p to p-p0 (p nought) to get rid of the inhomogeneous term. (Doesn't affect the derivatives)
Then Fourier transform the equation, the derivation is similar to that of the heat kernel. Throw p0 to the other side and you get that expression.
Sorry for the informal language.

EDIT:I'm bad with LaTex so here's a photo of the first few steps. Just do the same trick with the heat kernel here and you're good.

Thanks man, that worked out nicely.

 

[HomogenousCow]

Its Haynes-Shockley experiment in semiconductor physics and the equation is basically the continuity equation for holes. The first two terms in the R.H.S. are the divergence of drift and diffusion current and the third term is the sink term corresponding to the recombination process. With ## \tau \to \infty ##, no recombination happens and with E=0, the peak of the hole distribution doesn't move.Thanks man, that worked out nicely.

No prob. Did you have difficulty with the inverse Fourier transform

 

[ShayanJ]

No prob. Did you have difficulty with the inverse Fourier transform

I'd have if I've done it myself, but I used wolframalpha.com!