Laplace Equation in a Cylinder with Some Uncommon Boundary Conditions While I was working on some theorems in PDEs, I encountered the following axisymmetric boundary value problem
$$\matrix{
   {{\nabla ^2}H = 0} \hfill & {{\rm{in}}} \hfill & \Omega  \hfill  \cr 
   {\partial_r H+a \partial_z H+bH=0} \hfill & {{\rm{on}}} \hfill & {r = R} \hfill  \cr 
   {H = 0} \hfill & {{\rm{on}}} \hfill & {z =  - \ell } \hfill  \cr 
   {H = 0} \hfill & {{\rm{on}}} \hfill & {z = \ell } \hfill  \cr 
 } $$
such that $\nabla^2=\partial_{r}^{2}+\frac 1r \partial_r+\partial_{z}^{2}$ is the axisymmetric Laplace operator and $H:\mathbb{R}^3\to\mathbb{R}$ is an infinitely differentiable scalar field $C^{\infty}(\mathbb{R}^3)$. Also, $a$ and $b$ are some real constants. The domain $\Omega$ is a cylinder defined as
$$\Omega  = \left\{ {(r,\phi ,z)|0 \le r \lt R,0 \lt \phi  < 2\pi , - \ell  \lt z \lt \ell } \right\}$$
where $(r,\phi,z)$ is the usual cylindrical coordinates.
What can we say about the uniqueness of $H$?
 A: I am just rewriting the discussions, above in the comments, with  achille hui as an answer.
First, we note the general identity
$$\psi \nabla^2 \phi = \nabla \cdot (\psi \nabla \phi) - \nabla \phi \cdot \nabla \psi$$
and by choosing $\psi=\phi=H$ we get
$$H \nabla^2 H = \nabla \cdot (H \nabla H) -  |\nabla H|^2 \tag{2}$$
Let $X$ be defined as the lateral surface of the cylinder
$$X=\left\{ {(r,\varphi ,z)|r=R,0 \lt \varphi  < 2\pi , - \ell  \lt z \lt \ell } \right\}$$
Next, integrate $(2)$ over $\Omega$, use the divergence theorem and the boundary conditions on $X$ to get
$$\begin{align}
\int_{\Omega} \nabla \cdot (H \nabla H) dV - \int_{\Omega} |\nabla H|^2 dV &= \int_{\Omega} H \nabla^2 H\\
\int_{\Omega} \nabla \cdot (H \nabla H) dV - \int_{\Omega} |\nabla H|^2 dV  &= 0\\
\int_{\partial \Omega} H \nabla H \cdot {\bf{n}} dS - \int_{\Omega} |\nabla H|^2 dV &= 0\\
\int_{X} H \nabla H \cdot {\bf{n}} dS - \int_{\Omega} |\nabla H|^2 dV &= 0\\
\int_{X} H \partial_{r}H dS - \int_{\Omega} |\nabla H|^2 dV &= 0\\
-a\int_{X} H \partial_{z}H dS - b \int_{X} H^2 dS - \int_{\Omega} |\nabla H|^2 dV &= 0 \\
\frac{1}{2}a\int_{X} \partial_{z}H^2 dS + b \int_{X} H^2 dS + \int_{\Omega} |\nabla H|^2 dV &= 0 \\
\frac{1}{2}aR \int_{0}^{2\pi}\int_{-\ell}^{\ell} \partial_{z}H^2 dzd\phi + b \int_{X} H^2 dS + \int_{\Omega} |\nabla H|^2 dV &= 0 \\
\frac{1}{2}aR \int_{0}^{2\pi} [H^2(R,\phi,\ell)-H^2(R,\phi,-\ell)] d\phi + b \int_{X} H^2 dS + \int_{\Omega} |\nabla H|^2 dV &= 0 \\
b \int_{X} H^2 dS + \int_{\Omega} |\nabla H|^2 dV &= 0
\end{align}$$
Next, if we assume that $b \gt 0$ we can conclude 
$$\begin{align}
\int_{X} H^2 dS &= 0 \\
\int_{\Omega} |\nabla H|^2 dV &=0
\end{align}$$
and hence
$$\begin{align}
H^2 &=0 & \text{on} & \quad X \\
|\nabla H|^2 &=0 & \text{in} & \quad \Omega
\end{align}$$
which finally leads to
$$H=0$$
