temperature distribution using Fourier/Laplace

Question

Consider room that occupies quarter space That is formed by two walls. One wall is fully insulated and has a constant temperature $0$. Another wall has a window of length L that with one edge at the origin. Window satisfies Neumann condition $\frac{dT}{dn}=T_{out}$

Set up mathematical problem and solve for temperature distribution after long time using Laplace or Fourier transform.

Attempt to solve: Setting up mathematical model

Quarter space wlog can consider stationary diffusion problem in 1st quadrant $T_{xx}+T_{yy}=0$

With BC

$x=0:\quad \quad \quad \quad \space \space\space\space T(x,y)=0$

$y=0;\space 0\leq x \leq L: \space \space T_y(x,y)=T_{out}$

$\quad\quad\space\space\space x > L: \quad\quad T_y(x,y)=0$

Take Laplace transform wrt to $y$

$\bar T_{xx}(x,s)+s^2\bar T (x,s) - s T(x,0) - T_y(x,0)=0$

Leaves with $\hat T_{xx} + s^2\hat T = T_y(x,0)$

Method of variation states that

1. $C_1’ (x,s)\cos(sx) + C_2’(x,s) \sin(sx)=0$
2. $-C_1’ (x,s)s\sin(sx) + C_2’(x,s) s\cos(sx)=T_y (x,0)$

From 2 when $s=0$ we have $u_y(x,0)=0$ because LHS has s factors. Or is it wrong and $T_y (x,0)= T_y (x,y=0)$ meaning can’t use $s=0$ value?

Any hints how to proceed further? I feel completely lost due to given boundary conditions. Because whatever direction I use needs to make use of two boundary conditions on one wall. One is given one is not. It feels like Laplace over Fourier due to semi infinite in both directions.

Answer 1

Using separation of variable, you will see that the only solution that satisfying the boundary condtion $T(0,y)=0$ is $e^{\omega x}\sin(\omega y)$. Thus you can write your general solution as $T(x,y)=\int_0^\infty\, C(\omega)e^{\omega x}\sin(\omega y) \,d\omega$. From now on, you should be able to solve what $C(\omega)$ is based on the second boundary condition $T_y(x,0)$.