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  fundamental 2D conductivity boundary value problem

+ 4 like - 0 dislike

Consider the following homogeneous boundary value problem for function/potential $u(x,y)$ on the infinite strip $[-\infty,\infty]\times[0,\pi/4]$ w/positive periodic coefficient/conductivity $\gamma(x+1,y)=\gamma(x,y)>0$: $$\begin{cases} \operatorname{div}(\gamma\nabla(u))=0,\\ u(x+1,y)=e^\mu u(x,y),\\ u_y(x,y+\pi/4)=0,\\ \gamma u_y(x,0)=-\lambda u(x,0). \end{cases}$$ The b.v. problem can be moved conformally to an anullus w/a slit.

In the uniform medium $\gamma\equiv const$, $\Delta u_k=0$ and the solution is of the separable form: $$u_k(x,y)=ce^{\mu_k x}(\cos(\lambda_k y)+\sin(\lambda_ky)),c\in R,$$ $\lambda_k=|k|$ for an integer $k\in Z$ and, therefore, $$\lambda_k=|\mu_k|=|k|\beta(|k|)>0,$$ where $\beta\equiv 1.$ Let $$ z=\tau(\mu)=i(\sqrt\mu-1/\sqrt\mu)$$ and $z_k=\tau(\mu_k)$. The analysis of the reflected/levant finite-difference problem suggests for any positive periodic $\gamma(x,y)>0$: $$\lambda_k=z_k\beta(z_k)>0,$$ for an analytic function $\beta(z)>0$ for $z>0$, that maps the right complex halfplane $C^+=\{z:\Re(z)>0\}$ into itself: $$\beta: C^+\rightarrow C^+.$$ In particular $\lambda_k$'s are not negative.

Any intuition for the existence and formula for $\beta(z)$? Orientation preserving deformations and winding #s?

The second part of the question is to replace positive coefficient $\gamma(x,y)>0$ w/separable $\alpha(y)\delta(x)$, such that $\beta_{\alpha\delta}=\beta_\gamma$...

This post imported from StackExchange Physics at 2016-01-12 16:10 (UTC), posted by SE-user DVD
asked Jan 4, 2016 in Theoretical Physics by DVD (20 points) [ no revision ]

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