# partial differential initial value problem

• Nov 17th 2008, 08:19 AM
gammafunction
partial differential initial value problem
Hi! I am currently trying to find a formula which solves the problem:
$\displaystyle u_t-\triangle u+cu=f$ on $\displaystyle \mathbb{R}^n$ x $\displaystyle (0,\infty)$ and
$\displaystyle u(x,0)=g(x)$ for $\displaystyle x \in \mathbb{R}^n$, where $\displaystyle c \in \mathbb{R}$

I do not have a clue how to do this. How can i handle the $\displaystyle cu$? Does anybody have ideas for a solution?

EDIT: $\displaystyle \triangle u$ meaning the laplacian of u!
• Nov 17th 2008, 01:42 PM
gammafunction
Not regarding the cu, this is called the heat equation. Maybe that helps...?
• Nov 17th 2008, 04:53 PM
ThePerfectHacker
Quote:

Originally Posted by gammafunction
Not regarding the cu, this is called the heat equation. Maybe that helps...?

If $\displaystyle n=2$ I might know of a way using seperation of variables.
The problem is that this is method does not generalize to $\displaystyle n\geq 2$.

By the way your equation is a heat equation in a rod that has a leak at a constant rate. (Wink)
• Nov 20th 2008, 10:05 AM
gammafunction
Thank you for your answer. But i really would like to know how to solve it... I mean there is a formula for the case without cu. It hast to somehow work out to find a formula for this case. How can we somehow detour the cu? Anybody got ideas? i would be very thankful.
• Nov 20th 2008, 02:46 PM
shawsend
I got an idea but I've never tried it before and don't even know if it's applicable but desperate times call for desperate measures. We can solve:

$\displaystyle u_t=ku_{xx}+q(x,t)\quad -\infty<x<\infty, t>0, x\in\mathbb{R}$ by taking the single Fourier Transform of both sides with respect to x. This results in an ODE in t:

$\displaystyle \hat{u}_t(\xi,t)+k\xi^2\hat{u}(\xi,t)=\hat{q}(\xi, t)$

That's solvable via an integrating factor. We solve for $\displaystyle \hat{u}$ and then invert the transform to obtain the solution. Well, how about multiple Fourier Transforms? There are such transforms you know. First solve your problem via Fourier Transforms for a single dimension, if that works, then ramp it up to two dimensions and take the Double Fourier Transform and see if that's tractable, if so then maybe you can ramp it up to n dimensions.