I have the following PDE $$\frac{\partial u}{\partial t}(r,t)=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}(r,t)\right),$$ for which I have determined a numerical scheme. The scheme is however not defined at the boundaries $u_{1}(t)$, where $r=0$ and $u_{N}(t)$ where $r=1$. The PDE has boundary conditions, $$\frac{\partial u}{\partial r}(0,t)=0, \;\; u(1,t)+\frac{1}{B}\frac{\partial u}{\partial r}(1,t)=\sin(\omega t), t>0,$$
where $B$ and $\omega$ are constants. To determine an equation for $\frac{\partial u_{N}}{\partial t}(r,t)$. I use the second boundary condition: $$ \begin{split} \frac{\partial u}{\partial r}(1,t)&=B(\sin(\omega t)-u(1,t)), \\ \frac{\partial u_{N}}{\partial t}(1,t)&=\frac{\partial}{\partial r}(B(\sin(\omega t)-u(1,t)), \\ &=-B\frac{\partial}{\partial r}(u(1,t)), \\ &=-B^{2}(\sin(\omega t)-u_{N}(t)). \end{split} $$ Where in the last step I wrote $u(1,t)=u_{N}(t)$ for my numerical scheme. I believe I determined this correctly so now I move onto $u_{1}(t)$. I was given the hint to apply a Taylor expansion of the RHS of the PDE around $r=0$. However I do not see how this is possible without dividing by zero in the first term of the expansion as, $$ \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial u}{\partial r}(r,t)\right)=\frac{1}{0^2}\frac{\partial}{\partial r}\left(0^{2}\frac{\partial u}{\partial r}(0,t)\right)+r\frac{\partial}{\partial r}\left(\frac{1}{0^2}\frac{\partial}{\partial r}\left(0^{2}\frac{\partial u}{\partial r}(0,t)\right)\right)+O(r^{2}). $$ Is there a step I can do to prevent division by zero here? I see that if I somehow do not have that issue anymore I can use that $\frac{\partial u}{\partial r}(0,t)=0$, but I don't believe I can simply conclude that $\frac{\partial u_{1}}{\partial t}=0$ with the division by zero still there.