# I'm having trouble with this one

• Oct 18th 2009, 04:18 AM
MathTragic
I'm having trouble with this one
A semi-infinite cylinder of metal lies along the positive x-axis with its sides and the face at $x = 0$ insulated. The initial temperature distribution is given by $u(x,0) = u_0 e^{\frac{-x}{L}}$ where $u_0$ and $L$ are positive constants. Show that the temperature in the cylinder is given for $t > 0$ and $x > 0$ by

$u(x,t) = \frac{u_0}{2} e^{\frac{c^2 t}{L^2}}$ $\Biggl( \ e^{-\frac{x}{L}} \Biggl[ 1 + \mbox{erf}$ $\ \ \Biggl( \frac{x - \frac{2c^2t}{L}}{\sqrt{(4c^2t)}} \Biggr) \Biggr]$ $+ \ e^{\frac{x}{L}}$ $\Biggl[1 - \mbox{erf}$ $\ \ \Biggl( \frac{x - \frac{2c^2t}{L}}{\sqrt{(4c^2t)}} \Biggr) \Biggr] \Biggr)$
• Jan 18th 2010, 05:00 AM
Rebesques
Quote:

Faces insulated, initial temperature distribution $u(x,0) = u_0 e^{\frac{-x}{L}}$.

So, the solution is given by

$u(x,t)=\int H(x-y,t)u(y,0)dy$, where $H$ is the heat kernel $H(x,t)=\frac{1}{2c\sqrt{\pi t}}\exp\left(\frac{-x^2}{4c^2t}\right)$. Have fun calculating :p