# Method of images

• Apr 27th 2010, 09:03 PM
ulysses123
Method of images
For the heat equation on a semi infinite domain we have been shown the method of images. Which to my understanding is basically using an even or odd extension to match the boundary condition.ie for $U_x(0,t)=0$
we use the even extension which meets this condition.And when the derivative is not specified but instead $U(0,t)=0$ we use the odd extension which has non zero derivative at x=0 but meets this boundary condition.

For example:

$U_t=kU_xx$
$U(x,0)=sinx$
$U(0,t)=0$

Then using the general solution to solve this i get $2(sinx)e^{-kt}$

The solution does not have a two. Which implies that the initial condition has been extended to an infinite domain and there is only one integral. As we were shown in class i get two integrals and since -sin(-x)=sinx i get the factor of two. I am wondering why there isnt a two in the solution.