# Math Help - Solution of Diffusion (Heat) equation

1. ## Solution of Diffusion (Heat) equation

Hi, another diff. eq problem:

In my notes, it suggests to use a similarity transformation to solve the heat equation in one dimension $\frac{\partial u}{\partial t} = D\frac{\partial^2 u}{\partial x^2}$.

So if $\xi = \frac{x}{2\sqrt{Dt}}$ and $u(x,t) = v(\xi)$, then:

$\frac{\partial u}{\partial t} = \frac{v}{\xi}\frac{\partial \xi}{\partial t} =. . .$

...What?

any help in explaining this would be much appreciated.

2. You have $\frac{\partial u}{\partial t} = D\frac{\partial^2 u}{\partial x^2}$. Letting $\xi = \frac{x}{2\sqrt{Dt}}$ then $u(x,t)\equiv v(\xi)=v(x,t)$ then:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial \xi} \frac{\partial \xi}{\partial x}=\frac{\partial v}{\partial \xi}\left(\frac{t^{-1/2}}{2\sqrt{D}}\right)$

and:

$\frac{\partial^2 u}{\partial x^2}=\left(\frac{t^{-1/2}}{2\sqrt{D}}\right)\frac{\partial^2 v}{\partial \xi^2}\left(\frac{t^{-1/2}}{2\sqrt{D}}\right)=\frac{\partial^2 v}{\partial \xi^2}\left(\frac{1}{4D t}\right)$

ok, you do $\frac{\partial u}{\partial t}$ in terms of partial of v with respect to $\xi$. When I do that and simplify, I get:

$\frac{d^2 v}{d\xi^2}+2\xi \frac{dv}{d\xi}=0$

Solving that and converting back to x and t, I get an answer in terms of the exponential integral which, when I back-substitute, satisfies the PDE.

3. Thanks for the reply. I'll work through it and see what I get

4. if $\xi = \frac{x}{2\sqrt{Dt}}$ and $u(x,t) = v(\xi)$, then:

$\frac{\partial u}{\partial t} = \frac{v}{\xi}\frac{\partial \xi}{\partial t} =. . .$.
Hi, can somebody explain the logic behind this statement please? Thanks!

5. Originally Posted by harbottle
Hi, can somebody explain the logic behind this statement please? Thanks!
I believe what you want is

$
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \xi} \cdot \frac{\partial \xi}{\partial t} = \frac{d v}{d \xi} \cdot \frac{\partial \xi}{\partial t}
$