With $u$ as heat, explain intuitively, not mathematically why $\nabla^2 u = 0 \implies \oint \nabla u \cdot \tilde{n} \, ds = 0$
Mathematically, this is a direct application of the divergence theorem. That says that for any vector field $F$:
\begin{align*} \iiint_R \nabla F \, dV &= \oint (F \cdot \tilde{n}) \, dS \\ \end{align*}
Where $F = \nabla u$, we have:
\begin{align*} \iiint_R \nabla^2 u \, dV &= \oint (\nabla u \cdot \tilde{n}) \, dS \\ \end{align*}
Combining with the given of $\nabla^2 u = 0$ yields:
\begin{align*} \iiint_R \nabla^2 u \, dV &= \oint (\nabla u \cdot \tilde{n}) \, dS = 0 \\ \end{align*}
The equation $\nabla^2 u = 0$ means constant heat flow at every position.
The equation $\oint (\nabla u \cdot \tilde{n}) \, dS = 0$ means that all heat coming in equals all heat going out.
I don't see logically why the first implies the second. Can't temperature be rising or falling?
Actually for the heat flow the continuity equation (ce) is $\frac{ \partial u}{\partial t} + \nabla \cdot F = 0$
And the temperature will stop raising or failing when there is a steady state (ss) $\frac{ \partial u}{\partial t}=0$
Only then you can have what you want $\oint (\nabla u \cdot \tilde{n}) \ dS =\iiint_R \nabla^2 u\ dV= \iiint_R \nabla\cdot F\ dV =_{ce}\iiint_R -\frac{ \partial u}{\partial t}\ dV=_{ss}0$
Hope it helps