Finite volume method for 1D heat equation in 1D

Question

I wish to solve the following using the finite volume method: $$\frac{\partial u}{\partial t}=\frac{D}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+Q(t,r)$$ with the following boundary conditions: $$\frac{\partial u}{\partial r}\Bigg|_{r=0}=0,\quad -D\frac{\partial u}{\partial r}\Bigg|_{r=R}=hT(t,R)$$ Integrating the PDE (multiplied with $r$) from $r_{j-\frac{1}{2}}$ to $r_{j+\frac{1}{2}}$ to get: $$\frac{d}{dt}\int_{r_{j}-\frac{1}{2}}^{r_{j}+\frac{1}{2}}rudr=D\left[ r\frac{\partial u}{\partial r}\right]_{r_{j-\frac{1}{2}}}^{r_{j+\frac{1}{2}}}+\int_{r_{j}-\frac{1}{2}}^{r_{j}+\frac{1}{2}}rQ(t,r)dr$$ I denote for general $f(t, r) $: $$\tilde{f}_{j}(t)=\frac{1}{\delta r}\int_{r_{j}-\frac{1}{2}}^{r_{j}+\frac{1}{2}}rf(t,r)dr$$ which makes my equation: $$\frac{d\tilde{u}_{j}}{dt}=\frac{D}{\delta r}\left[ r\frac{\partial u}{\partial r}\right]_{r_{j-\frac{1}{2}}}^{r_{j+\frac{1}{2}}}+\tilde{Q}_{j}$$ Now is this a valid approximation? $$r\frac{\partial u}{\partial r}\Bigg|_{r=r_{j+\frac{1}{2}}}=r_{j+\frac{1}{2}}\cdot\frac{\tilde{u}_{j+1}-\tilde{u}_{j}}{\delta r}$$ and likewise for the other flux. I'm also having issues with implementing the inner boundary condition at $r=0$ which is confusing me a lot.

Answer 1

The simple answer is, as I now realise, that you are free to compute the fluxes however you please, and the fluxes are evaluated on the cell boundaries, so I can just insert the actual values for the flux and both the inner and outer boundaries.