A square slab of width $L$ and with uniform heat-input $Q$ is insulated perfectly at top and bottom and found at steady state.
Fourier's equation for this case:
$$k(u_{xx}+u_{yy})+Q=0$$ where $k$ is the thermal diffusivity.
The boundary conditions are:
$$u(0,y)=u(L,y)=u(x,0)=u(x,L)=0$$
For the $\text{1D}$ case:
$$ku''(x)+Q=0$$
BC: $u(0)=u(L)=0$
Which solves simply to:
$$u(x)=-\frac{Q}{2k}x^2+\frac{QL}{2k}x=\frac{Q}{2k}(Lx-x^2)$$
But for the $\text{2D}$ case things are harder.
Separation of variables doesn't work here because of the $Q$ term.
And solving the homogeneous case first only yields:
$$u(x,y)=0$$
Someone pointing me in a workable condition would be much appreciated.
Suggested by @jon:
Substitution $u \to w$:
$$u(x,y)=w(x,y)-\frac{Q}{4k}(x^2+y^2)$$
So that:
$$k\Big(w_{xx}-\frac{Q}{2k}+w_{yy}-\frac{Q}{2k}\Big)+Q=0$$
$$kw_{xx}-\frac{Q}{2}+kw_{yy}-\frac{Q}{2}+Q=0$$
$$\Rightarrow w_{xx}+w_{yy}=0$$
But the boundary conditions in $w$ aren't homogeneous...
The new boundary conditions for $w(x,y)$ become:
$w(0,y)=\frac{Q}{4k}y^2$
$w(L,y)=\frac{Q}{4k}y^2$
$w(x,0)=\frac{Q}{4k}x^2$
$w(x,L)=\frac{Q}{4k}x^2$
According to [Paul's online notes][1] this can be solved by starting from:
$$w(x,y)=w_1(x,y)+w_2(x,y)+w_3(x,y)+w_4(x,y)$$
To completely solve Laplace’s equation we’re in fact going to have to solve it four times. Each time we solve it only one of the four boundary conditions can be nonhomogeneous while the remaining three will be homogeneous.
In essence this is applying the Superposition Principle 'in reverse'.
For example for $w_4(x,y)$ we'd have:
$$w_{4,xx}+w_{4,yy}=0$$
$w_4(0,y)=\frac{Q}{4k}y^2$
$w_4(L,y)=0$
$w_4(x,0)=0$
$w_4(x,L)=0$
Ansatz: $w_4=X(x)Y(y)$
Separation of variables:
$$\frac{Y''}{Y}=-\frac{X''}{X}=-m^2$$
Or:
$$Y''+m^2Y=0 \Rightarrow Y(y)=A\sin my+B\cos my$$ $$X''-m^2X=0 \Rightarrow X(x)=C\cosh mx+D\sinh mx$$
BC:
$Y(0)=Y(L)=0 \Rightarrow B=0$
and:
$\sin mL=0 \Rightarrow mL=\pi n \Rightarrow m=\frac{\pi n}{L} \text{ for } n=1,2,3...$
$$Y(y)=\sin\Big(\frac{\pi n y}{L}\Big)$$
Rewrite $X(x)=C\cosh\Big(\frac{\pi n x}{L}\Big) +D\sinh \Big(\frac{\pi n x}{L}\Big)$
to: $X(x)=C\cosh\Big(\frac{\pi n (x-L)}{L}\Big) +D\sinh \Big(\frac{\pi n (x-L)}{L}\Big)$
$X(L)=0 \Rightarrow X(x)=D\sinh \Big(\frac{\pi n (x-L)}{L}\Big)$
$$w_{4,n}=G_n\sinh \Big(\frac{\pi n (x-L)}{L}\Big)\sin\Big(\frac{\pi n y}{L}\Big)\text{ for }n=1,2,3...$$
Using $w_4(0,y)=\frac{Q}{4k}y^2$ and the Fourier sine series, the amplitudes $G_n$ can be determined.
The above process is repeated another $3$ times.
[1] https://tutorial.math.lamar.edu/classes/de/LaplacesEqn.aspx
