Tricky Partial Differential problem

Hi all, I wish to verify

$\displaystyle u_t=ku_x_x$ with the shorthand notations being $\displaystyle u_t=\frac{\partial u}{\partial t}$ and $\displaystyle u_x_x=\frac{\partial^2 u}{\partial x^2}$ for the following expression

$\displaystyle u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi$ where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives $\displaystyle u_t=ku_x_x$? If so, how does one tackle the integrand with the infinity signs etc?

Thanks, tips will be appreciated.

Looking forward to hearing from some one.

Bugatti79

Tricky Partial Differential problem

Loooking at that Leibniz expression...i dont know how one would tackle the problem because I believe the derivative of infinity is undefined ( I am referring to $\displaystyle \frac{\partial\beta}{\partial x}$ ). A function must be continous to be differentiable....

bugatti79

Tricky Partial Differential problem

I dont know where I am gone wrong. $\displaystyle u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi$

Using the chain rule I calculate for $\displaystyle u_x, u_x_x$

$\displaystyle u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$ and

$\displaystyle u_x_x=\frac{\frac{4(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$

and the product rule for $\displaystyle u_t$ since t is on both numerator and denominator

$\displaystyle \frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}{\pi k t}^{-1/2})}{4\pi kt}$

Can anyone confirm im going correct so far? (Happy)

Tricky Partial Differential problem

Still cant get $\displaystyle u_t=ku_x_x$

I dont know wheter im getting my algebra in a twist or my derivatives are still wrong.

$\displaystyle u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi$

I recalculate

$\displaystyle u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$

and for $\displaystyle u_x_x$ I use product rule in the numerator

$\displaystyle u_x_x=\frac{\frac{(-2(x-\xi))}{4kt}\frac{(-2(x-\xi))}{4kt}(e^\frac{-(x-\xi)^2}{4kt})+e^\frac{-(x-\xi)^2}{4kt}(\frac{-2}{4kt})}{2(\pi kt)^\frac{1}{2}}$

and qoutient rule for $\displaystyle u_t$

$\displaystyle \frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}({\pi k t})^{-1/2})}{4\pi kt}$

If these derivatives can be confirmed correct then I know the rest is basic algebra....

Thanks

bugatti79