A solid, homogeneous object occupies the region $D\subseteq \mathbb{R}^3$, and is completely insulated. Its initial temperature is given by $u(x,0) = \phi(x)$, for some function $\phi$. So $u$ satisfies the heat equation $u_t = k\Delta u$ with boundary condition $\frac{\partial u}{\partial n} = 0$ on $\partial D$. After a long time, the object reaches a steady, uniform temperature. In terms of $f$, what is this temperature?
I normally don't post a question without at least some attempt but I have no idea how to start this. Any suggestions are greatly appreciated.
Best Answer
From the boundary condition, we have $$ \oint_{\partial D} \boldsymbol\nabla u \cdot d\mathbf{A} = \oint_{\partial D}\frac{\partial u}{\partial n} dA = 0. $$ From the divergence theorem, we have $$ \oint_{\partial D} \boldsymbol\nabla u \cdot d\mathbf{A} = \int_D\nabla^2u \,dV = \int_D\frac{1}{k}\frac{\partial u}{\partial t} dV = \frac{d}{dt}\int_D\frac{u}{k}\,dV $$ assuming $k$ is constant in time (if it is also constant in space, you can pull it out of the integral entirely). In other words $$ \frac{d}{dt}\int_D \frac{u}{k}\, dV = 0 $$ This should not be surprising, as heat conduction conserves energy and the boundary condition says there's no energy outflow from $D$. From here I think you can get the answer.
EDIT: So continuing from here, we have that the integral in the final state must equal the integral in the initial state. Thus, $$ \int\frac{u(\mathbf{x},0)}{k}dV = \int_D\frac{\phi(\mathbf{x})}{k}dV = \int_D\frac{u(\mathbf{x},\infty)}{k}dV = \int_D\frac{T}{k}dV = T\int_D\frac{dV}{k}. $$ So $$ T = \frac{\int_D \frac{\phi(\mathbf{x})}{k}dV}{\int_D\frac{dV}{k}} $$ If $k$ is constant in space, we can multiply the top and bottom by $k$ to get $$ T = \frac{\int_D \phi(\mathbf{x}) dV}{\int_D dV} = \frac{1}{V}\int_D\phi(\mathbf{x})dV $$ where $V$ is the volume of $D$.
This is as far as you can go without knowing the actual function $\phi(\mathbf{x})$.