Consider the vertical strip of angle $\alpha=\frac{\pi}{2}$


In this case, the harmonic function which is $0$ on the left line and $1$ on the right line is given by $$f(a+ib)=\frac{a}{T}.$$
Now, when the angle $0<\alpha <\frac{\pi}{2}$ the strip becomes bent.


My question is: can we determine explicitly the harmonic function that is $0$ on left half lines and $1$ on the right ones. My attempt gave the following function
$$f(a+ib)=\frac{a}{T}- \frac{\cos(\alpha)}{\sin(\alpha)}\frac{|b|}{T},$$
but this is not a harmonic function. Thank you for any hint.



A harmonic function
Yes, it can be found explicitly, though not in elementary functions but in terms of a combination of elementary and hypergeometric functions.
The problem is (almost) equivalent to finding a conformal map
from your straight strip onto the broken strip. For this it is sufficient to find 
a conformal map from the upper half-plane onto the UPPER half of the broken strip (and then apply the Schwarz Symmetry principle). This map of the upper half-plane on the upper half
of the broken strip is given by the Christoffel-Schwarz formula
$$C\int\_0^z\zeta^{\beta-1}(\zeta-T)^{-\beta} d\zeta,$$
where $\alpha=\pi\beta$. This integral can be explicitly expressed in terms of
hypergeometric function. I hope these hints are enough to make an explicit computation.


In a comment you ask about dependence on $T$. Dependence on $T$ is trivial:
If $F$ is the conformal mapping corresponding to $T=1$, normalized such that
$F(0)=0,F(1)=1,F(\infty\_j)=\infty\_j, j=1,2$ then $TF(z/T)$ is the
map corresponding to $T$.


Remark. According to a theorem of Chebyshev, this integral is not an elementary function when $\beta$ is irrational. Therefore I suppose that the solution of
your problem is also not an elementary function.


Here are some more detail. Let $T=1$. Let $F(z)$ be the integral I wrote above. Then $C$
is determined from the boundary correspondence: $C=1/F(1)$. Now 
$$G(z)=(-\cos(\pi z)+1)/2$$
maps the rectangular half-strip onto the upper half-plane, $(0,1,\infty)\mapsto (0,1,\infty)$. Therefore 
$$H(z)=F(G(z))/F(1)$$
maps the upper half of the rectilinear strip on the upper half of the slanted strip (both with $T=1$). So the solution of the Dirichlet problem (with $T=1$) is 
$$\Re(H^{-1}(z)).$$
Not very explicit, of course, since one has to invert the integral $F$. But OK for numerical computation.


Remark 2. The integral 
$$B\_x(\beta,1-\beta):=\int\_0^x\zeta^{\beta-1}(1-\zeta)^{-\beta}d\zeta$$
is called the incomplete Beta-function, and there is even the standard notation
$$I\_x(\beta,1-\beta):=\frac{B\_x(\beta,-\beta)}{B(\beta,1-\beta)}=F(x)/F(1).$$
To obtain simple estimates, expand it to the power series:
$$B\_x(\beta,1-\beta)=\beta^{-1}x^\beta+\sum\_{n=1}^\infty\frac{\beta(\beta+1)\ldots(\beta+n-1)}{n!(n+\beta)}x^{n+\beta}.$$
This is sufficient to determine your function on $(0,c)$ with any desired accuracy, unless $c$ is close to $1$. Near $1$ use a similar expansion
into powers of $x-1$.


For example, near $0$, $G(x)\sim x^2/4,$ so $$H(x)\sim\frac{1}{\beta 4^{\beta}B(\beta,1-\beta)}x^{2\beta},$$
$H^{-1}(x)\sim Cx^{1/(2\beta)}.$ Also notice that the series for $B\_x$ has
positive coefficients which simplifies the estimates.


To obtain a lower estimate, you need the upper estimate to $B\_x(\beta,1-\beta)$. We can write it as
$$B(x):=B\_x(\beta,1-\beta)=\beta^{-1}x^\beta+f(x),$$
where $f$ is increasing (because it is a power series with positive coefficients), and we also know that $B(1)=B(\beta,1-\beta),$ therefore
$f(x)<f(1)=B(\beta,1-\beta)-1/\beta$. This will give an explicit constant $C$
in the lower estimate $Cx^{(1/2\beta)}$.


