heat equation in polar coordinates Consider the heat equation in polar coordinates,
$$\frac{\partial u}{\partial t}=h^2\left(\frac{\partial^2 u}{\partial r^2}+\frac{2}{r}\frac{\partial u}{\partial r}\right), t>0, 0<r<R.$$
A sphere of radius $R$ is initially at constant temperature $u_0$ throughout, then has surface temperature $u_1$ for $t>0$. Find the temperature throughout the sphere for $t>0$ and in particular in the center $u_c$.
I can solve the heat equation when in cartesian form, however polar coordinates has always been my weakness and i have been stumped for a while, i have let $U(r,t)=F(r)G(t)$ then substituted into our original equation to get 
$$G'(t)+(\lambda^2)G(t)=0 \text{ where } \lambda=kh$$
$$F"(r) + F'(r)/r + k^2F(r)=0$$
i have the initial condition that $U(r,0)=u_0$ and the boundary conditions that $u(R,t)=u_1$
however i do not know where to go from here, any help is appreciated.
 A: Hint: Let $u(r,t) = \frac{U(r,t)}{r}$. Then $$\frac{\partial u}{\partial t}  =\frac{1}{r}\frac{\partial U}{\partial t} $$ $$\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial U}{\partial r} - \frac{U}{r^2}$$ and $$\frac{\partial^2 u}{\partial r^2} = -\frac{1}{r^2}\frac{\partial U}{\partial r} + \frac{1}{r} \frac{\partial^2 U}{\partial r^2} + \frac{2 U}{r^3} -\frac{1}{r^2} \frac{\partial U}{\partial r} =  \frac{1}{r} \frac{\partial^2 U}{\partial r^2} - \frac{2}{r^2} \frac{\partial U}{\partial r} + \frac{2 U}{r^3}$$ So the heat equation becomes 
$$\begin{align}\frac{1}{r}\frac{\partial U}{\partial t} &= h^2\left(\frac{1}{r} \frac{\partial^2 U}{\partial r^2} - \frac{2}{r^2} \frac{\partial U}{\partial r} + \frac{2 U}{r^3} + \frac{2}{r}\left(\frac{1}{r}\frac{\partial U}{\partial r} - \frac{U}{r^2}\right)\right)\\& =h^2\left(\frac{1}{r} \frac{\partial^2 U}{\partial r^2} - \frac{2}{r^2} \frac{\partial U}{\partial r} + \frac{2 U}{r^3} + \frac{2}{r^2}\frac{\partial U}{\partial r} - \frac{2U}{r^3}\right)\\& =  \frac{h^2}{r} \frac{\partial^2 U}{\partial r^2}\end{align}$$ which implies $$\frac{\partial U}{\partial t} = h^2 \frac{\partial U}{\partial r^2}$$ Do you think you take  can it from here? 
