2
$\begingroup$

I have encountered the following problem.

Let $\chi:=\chi_{B(0,1/2)}$ be characteristics function i.e it take $1$ if $x\in B(0,1/2)$ otherwise $0$.

$\nabla\cdot ((1+\chi_{B(0,1/2)})\nabla u )=0 $ in $B(0,1)$ with $u(1,\theta)=f(\theta)$.

I wanted to find $(\nabla u\cdot \nu)|_{B(0,1)} $ from above data where $\nu$ is outer unit normal on disc. I know the given problem is elliptic. But I do not know any technique to tackle that problem. Please suggest some technique or tool to tackle that problem.

Any help or hint will be greatly appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ I assume that you are in 2D and the boundary condition is meant in polar coordinates? In that case the rotational symmetry kind of suggests writing the angular depencency as a Fourier series. This would even allow you to solve for u explicitly by an ODE. $\endgroup$
    – mlk
    Mar 29, 2021 at 9:59
  • $\begingroup$ Perhaps just a couple keywords to get started: look up "Poincaré-Steklov" and more precisely "Dirichlet-to-Neumann map" $\endgroup$ Mar 31, 2021 at 13:36

1 Answer 1

1
$\begingroup$

It is a separable problem. First you notice that the constant part of $f$ will play no role, since it would lead to a constant potential, so you may assume $$ \int_0^{2\pi} f(\theta)d\theta =0. $$ If you write $$ f(\theta) = \sum_{n=1} a_n \cos n\theta + b_n\sin n\theta, $$ then you notice that the solution $u$ can be written $$ u(r,\theta) = \sum_{n=1} u_n(r)(a_n \cos n\theta + b_n\sin n\theta), $$ where $u_n$ is a solution of $$ \frac{1}{r} \left( r (1+\chi_[0,1/2]) u_n^\prime\right)^\prime - (1+\chi_[0,1/2])\frac{n^2}{r^2} u_n =0 $$ with the requirement that $u_n(0)$ is bounded and $u_n(1)=1$. This means in practice $$ u_n = \begin{cases}\alpha_n r^n &\mbox{ for } 0\leq r\leq \frac12\\ \beta_n r^n + \gamma_n r^{-n} &\mbox{ for } \frac12 <r <1 \end{cases} $$ together with continuity, $u_n(\frac 12 -) = u_n(\frac 12 +) $ and continuity of the flux $2 u_n^\prime (\frac 12 -) = u_n^\prime(\frac 12 +)$, and the final condition $u_n(1)=1$. Three unknowns, three equations, you obtain a solution. $$ a_n=\frac{2}{-3+4^{-n}},\quad b_n=\frac{3}{-3+4^{-n}},\quad c_n=\frac{1}{1-3\times4^{n}}, $$ in particular $$ u_n^\prime(1)= n \frac{3+4^{-n}}{3-4^{-n}} $$ And the normal flux at the boundary is $$ \nabla u\cdot \nu (r=1) = \sum_{n=1}^\infty n \frac{3+4^{-n}}{3-4^{-n}} (a_n \cos n\theta + b_n\sin n\theta). $$ This series will not always converge in the usual sense. For the problem to have a solution $u$ in $H^1$, $f$ must be such that $$ \sum_{n=1}^\infty n (a_n^2 + b_n^2) <\infty $$ which makes it $H^{1/2}(S_1)$. This is not enough for the flux to be well defined by the series above : each term is multiplied by $n$, so it is in $H^{-1/2}(S_1)$ : it only converges after an integration by parts against a series which is in $H^{1/2}(S_1)$. For the series to be convergent, you could impose for example $$ \sum_{n=1}^\infty n (|a_n| + |b_n|) <\infty $$ This is one of many approaches to this problem. Another popular one is to use layer potentials to solve this problem. In all cases, what you will notice is

  • There is no silver bullet, giving you all you ever wanted to know about $\nabla u\cdot \nu $ in one simple formula.
  • Different ways of representing the solution all have their uses. Just writing it as the flux on the boundary of the solution of that equation is the most convenient way to represent it for integrations by parts.
  • The rate of convergence of the series depends completely on the regularity of $f$, which in turn decides of the rate of decay of the coefficients $a_n,b_n$.
$\endgroup$

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.