Heat Equation Problem

**padawan** · September 2nd 2010, 01:19 PM

Hi all,

Working though this problem, I have reached a proof I am not sure of the method, any opinions much appreciated!

Q:
$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ where $u(x,t) = t^{\alpha} \phi (\xi)$ ; $\xi = xt^{-1/2}$ then $\phi(\xi)$ satisfies :

$\alpha \phi - \frac {1}{2} \xi \phi' = \phi ''$ where $' = \frac {\partial}{\partial \xi}$

------ This part is fine with some partial differentiation, chain rule and substitution, I can get the proof.

2nd proof:

show that $\int_{-\infty}^{\infty} u(x,t) dx = \int_{-\infty}^{\infty} t^{\alpha} \phi(\xi) dx$ is independant of $t$ if $\alpha = -1/2$

------ This proof is also fine with some partial differentiation and change of variable

ending with

$\int_{-\infty}^{\infty} u(x,t) dx = \int_{-\infty}^{\infty} \phi(\xi) d \xi$

3rd proof:

further show $C - \frac {1}{2} \xi \phi = \phi '$ when $\alpha = -1/2$ where $C$ is an arbitary constant....

Now given the second proof, is it possible to integrate the entire expression

$\alpha \phi - \frac {1}{2} \xi \phi' = \phi ''$ where $' = \frac {\partial}{\partial \xi}$ w r t : $\d \xi$ ?

reducing the orders of the differential terms? I suspect this is not possible since one of the terms contains a $\phi' \xi$ ..... if we cannot directly integrate to reduce the power, is it possible to use a power reduction? I suspect not since $\phi$ is an unknown function- not an explicit solution that allows this method..........

Thanks for reading!
(I realise the post is rather long winded, but I didnt want to miss info that the proof relied upon in previous sub part)

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