# Heat Equation with Weird Boundary Conditions

• September 29th 2011, 09:24 AM
Creebe
Heat Equation with Weird Boundary Conditions
After a few trivial steps I reached:

$X'' + \lambda^2 X = 0$
$X(x) = Asin(\lambda x) + Bcos(\lambda x)$

But my boundary conditions are:

$X(-2) = 0, X(2) = 0$

How am I supposed to solve for A and B and $\lambda$?
• September 29th 2011, 02:26 PM
Jester
Re: Heat Equation with Weird Boundary Conditions
If you sub. in your BC's you get

$-A \sin 2 \lambda + B \cos 2\lambda = 0,$

$A \sin 2 \lambda + B \cos 2\lambda = 0,$

or

$A \sin 2\lambda = 0$ and $B \cos 2 \lambda = 0$.

This gives rise to two cases

$(1)\;\;A = 0,\;\; \cos 2 \lambda = 0,$

$(2)\;\; B = 0, \;\; \sin 2 \lambda = 0.$