# Heat equation volume source vs. heat flux boundary condition

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.

\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.
• Thank you! So if I changed the heat flux in volume source to $$q_\text{for volume source} := q_\text{from boundary condition} + \frac{1}{4\pi}\int_{B\setminus\partial B}\rho c_p\partial_t T\mathrm{d}V$$ then it would work? – StefKKK Jan 28 at 9:22
• My motivation was that I wanted to get a second characteristic number $\Pi_2 := \frac{qt}{\rho c_p T L}$ like the Fourier number $Fo := \frac{\lambda t}{\rho c_p L^2}$ by nondimensionalizing the equation. I wanted to use $\Pi_2$ for some kind of timestep control while solving the heat equation numerically. – StefKKK Jan 29 at 19:37