 Solution

Conformality and analyticity. Use the orientation-preserving convention for a conformal map, with a continuously differentiable, nonzero differential. Its Jacobian matrix must be a positive scale times a rotation,Thus $u_x=v_y$, $u_y=-v_x$, the Cauchy-Riemann equations; continuous partial derivatives then imply that $u+iv$ is a holomorphic function. If angle preservation is allowed to reverse orientation, the statement needs this qualification: $z\mapsto \bar{z}$ preserves unsigned angles but is an anticonformal map and is not a holomorphic function.

The mapping. Put $a=\pi /L$. The displayed components combine to giveIts derivative does not vanish in the open half-strip $x>0$, $0<y<L$, so it is a conformal map there. Also $\mathrm{Im}w=\sinh (ax)\sin (ay)>0$. The lower edge maps to $[1,\infty )$, the upper edge to $(-\infty ,-1]$, and the vertical edge to $[-1,1]$. The open half-strip maps bijectively onto the upper half-plane. Indeed, $\cosh \zeta =\cosh \eta$ implies $\zeta =\pm \eta +2\pi im$; the half-strip allows only equality. Surjectivity follows by solving $t+t^{-1}=2w$ and choosing the root $t=e^{az}$ with $|t|>1$ and $0<\arg t<\pi$. The two corners are boundary critical points, not interior failures of conformality.

Transformed boundary values. A conformal map preserves Laplace's equation up to a positive scale. Thus the transformed Dirichlet problem is to find a real harmonic function $U(u,v)$ on $v>0$ withThe conflicting values at the original corners are interpreted as one-sided edge limits.

Logarithmic solution. Choose the complex logarithms of $w\pm 1$ with arguments between zero and $\pi$ in the upper half-plane. A linear combination $F(w)=\alpha \log (w+1)+\beta \log (w-1)+\gamma$ is a complex-valued harmonic function, since it is a holomorphic function; the real physical potential must be a real or imaginary part. In particular, setThenOn the three real boundary intervals, the argument difference is respectively $0,\pi ,0$, which gives the required values. This is the bounded harmonic solution, with $U\to V_0$ at infinity; uniqueness holds in this bounded class by the maximum principle. Without a boundedness or far-field condition, the edge data alone do not give uniqueness: one may add $C\sinh (\pi x/L)\sin (\pi y/L)$.

Back substitution. Sincethe required continuous argument difference is $2\arctan [\sin (ay)/\sinh (ax)]$ in the open half-strip. ConsequentlyAs $y\to 0,L$ with $x>0$, the arctangent tends to zero; as $x\to 0^{+}$ with $0<y<L$, it tends to $\pi /2$. These limits verify every open edge of the Dirichlet problem.