Hi all,

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

Q:

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

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

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

2nd proof:

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

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

ending with

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

3rd proof:

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

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

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

reducing the orders of the differential terms? I suspect this is not possible since one of the terms contains a $\displaystyle \phi' \xi$..... if we cannot directly integrate to reduce the power, is it possible to use a power reduction? I suspect not since $\displaystyle \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)