# How to find $\nabla u\cdot \nu|_{B(0,1)}$ where $u$ is solution of given conductivity equation?

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.

• 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.
– mlk
Mar 29, 2021 at 9:59
• Perhaps just a couple keywords to get started: look up "Poincaré-Steklov" and more precisely "Dirichlet-to-Neumann map" Mar 31, 2021 at 13:36

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 $$u_n = \begin{cases}\alpha_n r^n &\mbox{ for } 0\leq r\leq \frac12\\ \beta_n r^n + \gamma_n r^{-n} &\mbox{ for } \frac12 together with continuity, $$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$$ in one simple formula.
• The rate of convergence of the series depends completely on the regularity of $$f$$, which in turn decides of the rate of decay of the coefficients $$a_n,b_n$$.