So for heat equation $$T_t - k \Delta T = f(t)1_{r <= R}(r)$$ with initial condition $T(r,0) = 0$. where $r = ||(x,y,z)||$. $f(t)$ could be any positive function that it's integral over time is finite. I want to estimate the heat flux at the boundary $r = R^-$. $q(t) = -k \frac{\partial T}{\partial r} \hat{r}$ is the heat flux since the heat source is radial and the initial condition is also radial.
Here is my approach:
To estimate the flux at the interface boundary $r = R$ at some time $t$. $q(t) = - k \nabla T$. But in polar coordinates, $\nabla = \frac{\partial}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial}{\partial \theta} \hat{\theta}$. Since we have the radial symmetry, then $\nabla = \frac{\partial}{\partial r}\hat{r}$. Then $q(t) = - k \frac{\partial T}{\partial r}\hat{r}$. Then $\iint q(t) \cdot dS$ is the heat energy that is transferring out at time $t$ at the boundary $r = R$. The energy out side the ball is $\iiint_{\mathbb{R}^3 \backslash B_R(0)} T dV$. Then $\iint q(t) \cdot dS = \frac{\partial}{\partial t}\iiint_{\mathbb{R}^3 \backslash B_R(0)} T dV = \iiint_{\mathbb{R}^3 \backslash B_R(0)} T_t dV = \frac{\partial}{\partial t}(\int_{0}^{t} f(s) V ds - \iiint_{B_R(0)} T dV) = f(t)V - \iiint_{B_R(0)} T_t dV$ Since the heat equation has the radial symmetry, then $\iint q(t) \cdot dS = -kT_r(R,t)4\pi R^2$ $T_r(R,t) = \frac{1}{4\pi R^2 k}(\iiint_{B_R(0)} T_t dV - f(t)V) = \frac{1}{4\pi R^2 k}\iiint_{B_R(0)} k \Delta T dV = \frac{1}{4\pi R^2}\iiint_{B_R(0)}\Delta T dV \geq \frac{R}{3}\min_{r \in B_R(0)} \Delta T$ To find the minimum of the laplacian of $T$. Let $T_{ave}(\cdot)$ be the average value of $T$ on the set. Look at the taylor expansion of $T_{ave}$ around some point $x$, $T_{ave}(S_R(x)) = T(x,t) + \frac{\Delta T(x,t)}{6}R^2 + o(R)^4$ where $R$ is the radius of the ball and $S_R(0) = \{x: |x - 0| = R\}$. Then $\Delta T(x,t) \approx 6\frac{T_{ave}(x) - T(x,t)}{R^2}$. But noticing that $T_r(r,t) \leq 0$, $T(0,t)$ has the maximum value. Then we could see that minimum of $\Delta T$ is reached at $(0,t)$. But $T$ has the radial symmetry about the origin, then the average value is just $T(R,t)$. And $\Delta T \approx \frac{6(T(R,t) - T(0,t))}{R^2} + o(R^2)\geq - \frac{6T(0,t)}{R^2}$ Then the maximum flux $\max q(t) = \max(-k \frac{\partial T}{\partial r}) \approx \frac{2T(0,t)}{R}$ Then the temperature drop $\Delta_{drop} T(t) = \frac{q(t)}{G} \approx \frac{2T(0,t)}{RG}$. The relative drop at the boundary, is $\frac{\Delta_{drop} T(t)}{T(R,t)} = \frac{2T(0,t)}{T(R,t)RG} \approx \frac{2}{RG}$
The problem is here: when $R$ goes to sufficiently small, the percentage just blow up. This is estimation is too bad. Is there any way to find the upper bound? And I didn't use $f(t)$ at all which is another confusion. Intuitively, the upper bound should be in terms of $f(t)$ and $R$ in my mind.
I also cannot work out the details for example $T_r(r,t) \leq 0$ for arbitrary $t$. But from graph it is.