# 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 \$\displaystyle U_x(0,t)=0\$
we use the even extension which meets this condition.And when the derivative is not specified but instead \$\displaystyle U(0,t)=0\$ we use the odd extension which has non zero derivative at x=0 but meets this boundary condition.

For example:

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

Then using the general solution to solve this i get \$\displaystyle 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.