Problem : Find a function which is harmonic in the upper half plane $y>0$ and which on the $x$-axis takes the value $-1$ if $x<0$ and $1$ if $x>0$ ,(using Poisson's integral formulas for a Half plane) my book gives the answer $1-(2/\pi)\arctan(y/x)$ … and although I know what the Possion's formula for the Upper half plane , I cannot see how this answer was obtained … Any help is greatly appreciated….
[Math] function which is harmonic in the upper half plane (Poisson’s formula)
complex-analysis
Related Solutions
If you assume that $f = u+iv$ is bounded, say $|f| \le M$, you can argue like this: Since $|(\zeta - z)(\zeta-\bar z)| > \frac12R^2$ on $C_R$ for $R$ large, we get
$$ \left|\int_{C_R}{\frac{2iy \cdot f(\zeta)}{(\zeta - z)(\zeta - \bar{z})}d\zeta} \right| \le \frac{4yM}{R^2} \cdot \pi R \to 0 $$ as $R\to\infty$ by the "standard estimation lemma" for complex integrals.
It is possible to reduce the problem to this case by approximating $u$ with functions that are harmonic on a neighbourhood of the closed upper half-plane and then the corresponding $f$'s are automatically bounded on the upper half plane. You may find it easier to map the upper half-plane conformally to the unit disc and use that version of Poisson's integral formula and then map back.
I would be happier with assuming that $u$ extends continuously to the real axes. Otherwise you have to worry about what you actually mean by the integral.
$T$ is holomorphic in $B(0,|z_{0}|^{-1})$. Therefore, if $u$ is harmonic in $B(0,1)$, it follows that $u \circ T$ is harmonic in $B(0,1)$.
This is a general principle in complex analysis. If $D$ is an open set, $u : D \to \mathbb{R}$ is harmonic, and $T : D \to \mathbb{C}$ is holomorphic, then $u \circ T$ is harmonic. There are a number of ways to see this; I'll follow the elementary route. Write $T(z) = f(z) + i g(z)$, where $f$ and $g$ are the real and imaginary parts respectively. First, $$\frac{\partial}{\partial x}(u \circ T)(z) = u_{x}(z) f_{x}(z) + u_{y}(z) g_{x}(z), \quad \frac{\partial}{\partial y}(u \circ T)(z) = u_{x}(z) f_{y}(z) + u_{y}(z) g_{y}(z).$$ Differentiating once more, we obtain \begin{align*} &\frac{\partial^{2}}{\partial x^{2}}(u \circ T)(z) = u_{xx}(z) f_{x}(z) + u_{x}(z) f_{xx}(z) + u_{yx}(z) g_{x}(z) + u_{y}(z) g_{xx}(z) \\ &\frac{\partial^{2}}{\partial y^{2}}(u \circ T)(z) = u_{xy}(z) f_{y}(z) + u_{x}(z) f_{yy}(z) + u_{yy}(z) g_{y}(z) + u_{y}(z) g_{yy}(z).\end{align*} Recall that since $T$ is holomorphic, $f$ and $g$ are both harmonic. Adding what we computed above and using the fact that $u_{xy} = u_{yx}$, we find \begin{align*} \Delta(u \circ T)(z) &= \Delta u(z) (f_{x}(z) + g_{y}(z)) + u_{x}(z) \Delta f(z) + u_{y}(z) \Delta g(z) + u_{xy}(z) (g_{x}(z) + f_{y}(z)) \\ &= u_{xy}(z)(g_{x}(z) + f_{y}(z)) \end{align*} Finally, since $T$ is holomorphic, $f$ and $g$ satisfy the Cauchy-Riemann equations $$f_{x} = g_{y}, \quad f_{y} = - g_{x}.$$ Thus, $$\Delta (u \circ T)(z) = 0.$$ Therefore, $u \circ T$ is harmonic.
Best Answer
Recall Poisson's integral formula for the half-plane says that if $U : \mathbb{R} \to \mathbb{R}$ is piecewise continuous and bounded, then $$ f(x,y) = \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{U(t)}{(x-t)^{2} +y^{2}} dt, $$ is harmonic in the upper half-plane and has boundary values $f(x,0) = U(x)$.
So, given your boundary values, we have $$ U(x) = \begin{cases} -1 & x < 0 \\ 1 & x > 0. \end{cases} $$ Note, as we are integrating, we don't care what the value of $U(0)$ is.
Then,
\begin{equation} f(x,y) = \frac{y}{\pi} \int_{-\infty}^{\infty} \frac{U(t) dt}{(x-t)^{2} +y^{2}} = \frac{y}{\pi} \left[ \int_{0}^{\infty} \frac{dt}{(x-t)^{2} + y^{2}} - \int_{-\infty}^{0} \frac{dt}{(x-t)^{2} + y^{2}}\right]. \end{equation}
I'll compute the first integral, and leave the second as practice. Using the change of variables $u = x-t$, we see that $d u = -dt$. So,
\begin{align*} \int_{0}^{\infty} \frac{dt}{(x-t)^{2} + y^{2}} &= \int_{0}^{\infty} \frac{-du}{u^{2} + y^{2}}\\ & = \frac{-1}{y} \arctan\left( \frac{x-t}{y} \right) \bigg|_{t=0}^{\infty}\\ & = \frac{1}{y} \arctan\left(\frac{t-x}{y}\right) \bigg|_{t=0}^{\infty} \\ & = \frac{1}{y} \left( \frac{\pi}{2} - \arctan \frac{x}{y} \right). \end{align*}
Now use similar technique to perform the other integral. Don't forget the $\frac{y}{\pi}$ out in front of Poisson's formula, and be wary of negative signs.