Heat equation volume source vs. heat flux boundary condition

Question

I want to solve the heat equation in the 3D unit sphere $B$ with a general heat flux boundary condition, no volume sources and some given constant initial temperature: $$ \rho c_p\partial_t T - \lambda \Delta T = 0,\,\text{inside}\, B$$ $$ -\lambda \operatorname{grad}(T) \cdot \vec{n} = q\,\text{on the boundary}\, \partial B,$$ $$ T\rvert_{t=0} = T_0 = \text{const,}\, \text{inside } B$$

My question is the following: Do I get the same solution for the temperature $T$ inside the sphere $B$ if I instead solve the following equation?

$$ \rho c_p\partial_t T - \lambda \Delta T = q\delta(\lVert \vec{x} \rVert - 1),\,\text{on the whole space}\, \mathbb{R}^3$$ $$ T\rvert_{t=0} = T_0 = \text{const,}\, \text{on the whole space } \mathbb{R}^3.$$

My reasoning is that I can view the heat flux on the boundary $\partial B$ as being generated from a heat source concentrated on the boundary of the sphere. This is why I chose $q\delta(\lVert \vec{x} \rVert - 1)$ with a $\delta$ function as the heat source term.

Answer 1

The solution inside the sphere will be the same if the conditions in and on the sphere are the same. It's easy to show that the second system reduces to the first inside the sphere - what about on the boundary?

Consider integrating over the volume of the sphere:

\begin{align} \int_{B} \rho c_p \partial_t T\ \text{d} V - \int_B \lambda \nabla \cdot \nabla T \ \text{d}V &= \int_B q \delta(||\vec{x}||-1) \ \text{d}V \\ \int_{B} \rho c_p \partial_t T\ \text{d} V - \int_{\partial B} \lambda \nabla T \cdot \hat{n} \ \text{d}A &= 4\pi q \end{align} where the divergence theorem has been applied and the factor of $4\pi$ comes from the fact that the delta is integrating to one at each point on the unit sphere, which has area $4\pi$. Let's assume that the original boundary condition holds and see what the implications are. Since the surface integral above is on the relevant boundary, we can substitute the original boundary condition \begin{align} \int_{B} \rho c_p \partial_t T\ \text{d} V - \int_{\partial B} (-q) \ \text{d}A &= 4\pi q \\ \int_{B} \rho c_p \partial_t T\ \text{d} V + 4 \pi q &= 4\pi q \\ \int_{B} \rho c_p \partial_t T\ \text{d} V &= 0 \\ \partial_t \underbrace{\int_{B} \rho c_p T\ \text{d} V}_{U} &= 0 \end{align}

This equation tells us that the second system only returns the first boundary condition when the total energy $U$ inside the sphere is constant - i.e. at steady state. The systems will have the same steady state solution, but not the same transient behaviour.