1
$\begingroup$

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 $H$?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $\int_X \partial_z H^2 = \int_0^{2\pi} \int_{-\ell}^{\ell} \partial_z H^2 dz d\phi = \int_0^{2\pi} [ H^2 ]_{z=-\ell}^\ell d\phi$. $\endgroup$ – achille hui Jan 21 '16 at 11:21
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

$$H=0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.