Hi all,

I have been looking at the following problem and have hit a wall.

Q:

$\displaystyle \frac {\partial^2u} {\partial x^2}= \frac {\partial u} {\partial t}$ where $\displaystyle u(x,0) = 0$ , $\displaystyle \frac {\partial u}{\partial x} (0,t) = -1$ , $\displaystyle u(x,t) -> 0$ as $\displaystyle x -> \infty$

let $\displaystyle v(x,p) = \int_{0}^{\infty} e^{-pt}u(x,t)dt$

Prove $\displaystyle \frac {\partial^2 v}{\partial x^2} = pv$ and $\displaystyle \frac {\partial v}{\partial x} (0,p)= - \frac {1}{p} $ and $\displaystyle v(x,p) -> 0 $ as $\displaystyle x -> \infty$

I can find the first proof, but am having problems figuring out the second part

$\displaystyle \frac {\partial v}{\partial x} (0,p)= - \frac {1}{p}$

.... I am not sure where to start- any one have any ideas on the method for this one?

Thanks for reading!

(original post subsequently corrected)