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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – StefKKK Jan 28 at 9:22
  • $\begingroup$ That change would fix the one integral that I solved, but that is only a necessary (not sufficient) condition for making the boundary condition correct at each point. In the bigger picture though, you will still need some boundary condition to solve your new system, so I am not sure that transforming things as you suggest actually makes things easier to work with. What is the motivation for the transformation? $\endgroup$ – user1476176 Jan 28 at 17:35
  • $\begingroup$ 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. $\endgroup$ – StefKKK Jan 29 at 19:37

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