Solving pde on a strip 
Let $\Omega = \lbrace x_1,x_2 \in \mathbb{R}^2 | x_1>0, 0<x_2<1\rbrace$. Solve $$
\begin{aligned}
u_t(\boldsymbol{x},t) &= \nabla^2 u(\boldsymbol{x},t) , \qquad \boldsymbol{x}\in \Omega, t> 0,\\
u(\boldsymbol{x},0) &= g(\boldsymbol{x}) , \\
u_{x_1}(0,x_2,t) &= 0 ,\\
u(x_1,0,t) &= u(x_1,1,t) = 0,
\end{aligned} $$

Im trying to solve this equation but Im stuck. I know to use fourier transform but this only works for infinite domain. Here we have a infinite strip in the plane. How do we handle this heat equation with the boundary condition ?
 A: Ignoring the initial condition, the generic separated solution has the form
$$
         A(n,s)\sin(n\pi x_2)\cos(s x_1)e^{-(s^2+n^2\pi^2)t} \\
             n=1,2,3,\cdots,\;\;\; s \ge 0.
$$
These are required because $\sin(n\pi x_2)$ vanishes at $x_2=0,1$, and because $\cos'(sx_1)=0$ at $x_1=0$. $A(n,s)$ is an unknown coefficient function that is determined by the initial condition $u(x_1,x_2,0)=g(x_1,x_2)$.
The general solution is an integral over $s$ and a sum over $n$ of the separated solutions:
$$
       u(x_1,x_2,t)= \\ \int_0^\infty\left(\sum_{n=1}^{\infty}A(n,s)\sin(n\pi x_2)e^{-n^2\pi^2 t}\right)\cos(sx_1)e^{-s^2 t}ds
$$
The coefficients $A(n,s)$ are determined by the initial condition $u(x_1,x_2,0)=g(x_1,x_2)$, where $g$ is given:
$$
        g(x_1,x_2) = \int_0^\infty\left(\sum_{n=1}^{\infty}A(n,s)\sin(n\pi x_2)\right)\cos(sx_1) ds
$$
The expression in parentheses on the right is the inverse cosine transform of $g$ in the variable $x_1$. That is,
$$
      \frac{2}{\pi}\int_0^{\infty}g(x_1,x_2)\cos(sx_1)dx_1
 = \sum_{n=1}^{\infty}A(n,s)\sin(n\pi x_2)
$$
The coefficients $A(n,s)$ are completely determined by the orthogonality of the $\sin(n\pi x_2)$ functions on $0 < x_2 < 1$ and the "orthogonality" of the $\cos(sx_1)$ functions on $[0,\infty)$ through the Fourier cosine transform:
$$
   A(n,s) =  
\frac{\int_0^1\left(\frac{2}{\pi} \int_0^\infty g(x_1,x_2)\cos(sx_1)dx_1 \right)\sin(n\pi x_2)dx_2}{\int_0^1\sin^2(n\pi x_2)dx_2}
$$
