Heat loss of a cube insulated on four sides.

Question

I'm looking for a way to make sense of this. A cube, perfectly insulated on four sides is heated to an initial temperature of 100C. The two non-insulated sides will be the top and the bottom of the cube. We place the cube on a plate which maintains a temperature of 0C. The loss of heat through the top is given, and we are left to solve a pretty standard PDE.

Here's the issue. I'm saying the insulated sides correspond to the BCs of the $X(x)$ and $Y(y)$ functions after separation. When I go to compute the Fourier coefficients I end up with 0! The entire problem collapses. I've never encountered a problem where this happens... Am I just supposed to say the eigenvalues must be 0? I'm not sure if this physically makes sense, but would this just be equivalent to an infinite number of thin rods, implying that solution is independent of $x$ and $y$ values?

Answer 1

Looks like you are right. From the physical standpoint, there is no horizontal heat transfer. Consider a thought experiment: if instead of the cube you had an entire thick plane, due to symmetry there will be no horizontal heat transfer. It means that you may cut the plane into cubes and insulate them. Nothing will change.

It is indeed a one-dimensional problem, and your computations confirmed that.

Answer 2

Assume a unit cube $-1/2 < x < 1/2$, $-1/2 < y < -1/2$ and $0 \le z \le 1$. The heat equation is $$ u_t = k^2\nabla^2u $$ The bottom is at held at $T=0$: $$ u(t,x,y,0)=0. $$ The heat dissipation on the top surface is specified: $$ u_{z}(t,x,y,1)=C(x,y). $$ The four sides are insulated: $$ u_x(t,-1/2,y,z)=u_{x}(t,1/2,y,z)=0 \\ u_y(t,x,-1/2,z)=u_{y}(t,x,1/2,z)=0 $$ I don't believe this has a $0$ solution.