# A harmonic function

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.

• Maybe a silly question: is the function you are looking for unique? – S. Maths Nov 19 at 18:25
• Usually in such problems one adds the condition that the function is bounded. Then it is unique. – Alexandre Eremenko Nov 19 at 18:54

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 $$f(x). This will give an explicit constant $$C$$ in the lower estimate $$Cx^{(1/2\beta)}$$.

• You should map upper half of the straight strip onto the upper half-plane (by an elementary function), and then the upper half-plane to the broken strip (by that integral). The boundary correspondence should be $0\to 0,\; T\to T,\;\infty\to\infty$. By reflection this map extends to a map between the whole strips. – Alexandre Eremenko Nov 20 at 1:10
• If $F$ maps the straight strip onto the broken strip then the solution of Dirichlet problem is $\Re F^{-1}(z)/T$, the inverse function! – Alexandre Eremenko Nov 20 at 13:12
• This was a misprint. I corrected. – Alexandre Eremenko Nov 22 at 1:45
• I do not have the book. Probably I know this principle for harmonic functions but did not know that it is "Giraud". – Alexandre Eremenko Nov 24 at 17:53
• If you state exactly what sort of estimate you need, I may think how to obtain it. – Alexandre Eremenko Nov 24 at 17:55

There is also a probability formulation: For $$z\in \mathbb{C}$$ inside the strip let $$B_t^{(z)}$$ the brownian motion starting at $$z$$. Then we have $$f(z)=\mathbb{P}\big(B_t^{(z)}\text{ touch the right half lines before the left half lines} \big)$$ This kind of formula is useful to get quantitative properties of $$f$$. For example $$f(a+ib)=\frac{a}{T}-\frac{\cos(\alpha)}{\sin(\alpha)}\frac{|b|}{T} + \mathcal{O}(e^{-\gamma|b|})$$ for some $$\gamma>0$$ because the probability that $$B^{(z)}_t$$ cross the axis ($$b=0$$) before touching the right or the left lines decays exponentially.

RELATED QUESTION : I just want to react to the remark of Alexandre Eremenko because I didn't know the theorem of Chebyschev he mention. Do you think that one can find an explicite formula for the eigenvetors and eigenvalues of the Laplacian $$\Delta$$ on a parallelogram (with Dirichlet b.c.) if its angular are rational?