Need help with change of variables in very strange heat equation problem in the ring $0< a<r<b$ if $u_{r}(a,t)=1$ and $u_{r}(b,t) + u(b,t) = 2$.
Also, am I interpreting $\nabla^{2}u$ wrong? Since this is a ring, should it be in polar coordinates? That part was not clear to me from reading the question
 A: I suspect that you were given 
$$
u_{t} = \nabla^{2} u, 
0< a<r<b\\
u_{r}(a,t)=1 \\
u_{r}(b,t) + u(b,t) = 2.
$$
Now what exactly $u(a, t)$ means isn't clear to me, but I suspect it's this: 
The problem has circular symmetry, so although $u$ would normally be expressed (in polar coordinates) as $u(r, \theta, t)$, the author's omitted $\theta$ entirely. 
The Laplace operator in polar coordinates (see this wiki article) looks like
$$
\nabla^2 u = \frac{1}{r} \frac{\partial}{\partial r}(r u_r) + \frac{1}{r^2} u_{\theta\theta}
$$
By independence of $\theta$, the last term is trivial, and your equation becomes
$$
u_t = \frac{1}{r} \frac{\partial}{\partial r}(r u_r) 
$$
where $u = u(r, t)$ satisfies $u_r(a, t) = 1$ for all $t$, and $u_r(b, t) + u(b, t) = 2$ for all $t$. 
My guess is that this is amenable to some sort of separation of variables, but it's been 30 years since i've solved a differential equation for real. :)
A: To find equilibrium solution, you need to solve $\nabla^2 u = 0$ with the given domain and boundary condition. Now, in Cartesian coordinate $(x,y)$, $\nabla^2=\partial_{xx} + \partial_{yy}$. In polar coordinate $(r,\theta)$, it is given by
$$ \nabla^2 = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}. \label{eq:1}\tag{1}$$
Now, use the method of separation of variables. More precisely, we guess an ansatz of the form
$$ u(r,\theta) = R(r)\Theta(\theta), $$
but since the boundary condition is independent of $\theta$, we guess 
$$ u(r,\theta) = R(r). $$
Substituting this into $\nabla^2 u=0$ and using \eqref{eq:1} yields
$$ \frac{1}{r}(rR')' = 0 \implies rR' = C\implies R = C\ln(r) + D. $$
To find the constants $C,D$, we apply the boundary conditions (BCs) $R'(a) = 1$ and $R'(b) + R(b) = 2$. The first BC yields
$$ 1 = R'(a) = \frac{C}{a}\implies C=a,$$
while the second BC yields
$$ 2 = \frac{C}{b} + C\ln(b) + D = \frac{a}{b} + a\ln(b) + D\implies D = 2 - \frac{a}{b} - a\ln(b). $$
Hence, the equilibrium solution has the form
$$ u(r,\theta) = u(r) = \frac{a}{r} + 2- \frac{a}{b} - a\ln(b). $$
