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 $H$?

  • $\begingroup$ If neither the differential operator in the interior nor the boundary conditions depend on $\varphi,$ isn't this a two-dimensional problem? $\endgroup$ – Justpassingby Jan 21 '16 at 9:13
  • $\begingroup$ @Justpassingby: No, it is a 3D problem which has a symmetry with respect to $z$ axis. :) $\endgroup$ – H. R. Jan 21 '16 at 9:13
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    $\begingroup$ 1) $\int_X \partial_z H^2 = \int_{\partial X} H^2$ and $z = \pm \ell$ on $\partial X \implies H = 0$ there. 2) If $b \ge 0$, then $\int_\Omega |\nabla H|^2 + b \int_X H^2$ is a sum of two non-negative terms, it can only be zero when both terms are zero. The first one implies $H$ is constant inside $\Omega$ and since we know $H = 0$ when $z = \pm \ell$, $H = 0$ over whole $\Omega$. $\endgroup$ – achille hui Jan 21 '16 at 11:15
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    $\begingroup$ $\int_X \partial_z H^2 = \int_0^{2\pi} \int_{-\ell}^{\ell} \partial_z H^2 dz d\phi = \int_0^{2\pi} [ H^2 ]_{z=-\ell}^\ell d\phi$. $\endgroup$ – achille hui Jan 21 '16 at 11:21
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    $\begingroup$ The real problem is what happens when $b < 0$. If $a = 0$ and for some $(m,n) \in \mathbb{N} \times \mathbb{Z}_{+}$, following condition is satisfied. $$k_n I'_m(k_n R) + b I_m(k_n R) = 0\quad\text{ where }\quad k_n = \frac{n\pi}{2\ell} $$ Your PDE has non-trivial solution of the form $$I_m(k_n r) \sin(k_n(z+\ell)) \times \begin{cases} A,& m = 0\\ B \cos(m\phi) + C\sin(m\phi) & \text{ otherwise } \end{cases} $$ where $I_m(x)$ are modified Bessel function and $A, B, C$ are constants. $\endgroup$ – achille hui Jan 21 '16 at 12:46

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



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