Poisson’s Integral Formula

Section 10.3 Poisson’s Integral Formula

The Dirichlet problem for the upper half-plane \(\mathrm{Im}(z)>0\) is to find a function \(\phi(x,y)\) that is harmonic in the upper half-plane and has the boundary values \(\phi(x,0)=U(x)\text{,}\) where \(U(x)\) is a real-valued function of the real variable \(x\text{.}\)

Theorem 10.3.1. Poisson’s integral formula.

Let \(U(t)\) be a real-valued function that is piecewise continuous and bounded for all real \(t\text{.}\) The function

\begin{equation} \phi(x,y) = \frac{y}{\pi}\int\nolimits_{-\infty}^{\infty}\frac{U(t)}{(x-t)^2+y^2}\,dt\tag{10.3.1} \end{equation}

is harmonic in the upper half-plane\(\mathrm{Im}(z) >0\) and has the boundary values

\begin{equation*} \phi(x,0) =U(x) \end{equation*}

wherever \(U\) is continuous.

Proof.

Equation (10.3.1) is easy to determine from the results of Theorem 10.2.5 regarding the Dirichlet problem. Let \(t_1\lt t_2\lt \cdots \lt t_N\) denote \(N\) points that lie along the \(x\) axis. Let \(t_0^*\lt t_1^*\lt \cdots \lt t_N^*\) be \(N+1\) points chosen so that \(t_{k-1}^{\ast }\lt t_k\lt t_k^*\text{,}\) for \(k=1,\,2,\,.\,.\,.\,,\,N\text{,}\) and that \(U( t)\) is continuous at each value \(t_k^*\text{.}\) Then according to Theorem 10.2.5, the function

\begin{equation} \Phi(x,y) = U(t_N^*) + \frac{1}{\pi}\sum\limits_{k=1}^{N}\left[ U(t_{k-1}^*)-U(t_k^*) \right]\mathrm{Arg}(z-t_k)\tag{10.3.2} \end{equation}

is harmonic in the upper half-plane and takes on the boundary values

\begin{align*} {llll} \Phi(x,0) \amp = U(t_0^*), \text{ for } x \lt t_1,\\ \Phi(x,0) \amp = U(t_k^*), \text{ for } t_k \lt x \lt t_{k+1}, \text{ and }\\ \Phi(x,0) \amp = U(t_N^*), \text{ for } x > t_N\text{,} \end{align*}

as shown in Figure 10.3.2.

We use properties of the argument of a complex number (see Section 1.4) to write Equation (10.3.2) in the form

\begin{align*} \Phi(x,y) \amp = \frac{1}{\pi}U(t_0^*)\mathrm{Arg}(z-t_1) + \frac{1}{\pi}\sum_{k=1}^{N-1}U(t_k^*)\mathrm{Arg}\left(\frac{z-t_{k+1}}{z-t_k}\right)\\ \amp + \frac{1}{\pi}U(t_N^*)[\pi-\mathrm{Arg}(z-t_N)]\text{.} \end{align*}

Hence the value \(\Phi\) is given by the weighted mean

\begin{equation} \Phi(x,y)=\frac{1}{\pi}\sum\limits_{k=0}^{N}U(t_k^*)\Delta \theta_k\text{,}\tag{10.3.3} \end{equation}

where the angles \(\Delta \theta_k\text{,}\) for \(k=0, \, 1, \ldots ,N\text{,}\) sum to \(\pi\text{,}\) and are also shown in Figure 10.3.2.

Figure 10.3.2. Boundary values for \(\Phi\)

Using the substitutions

\begin{equation} \theta =\mathrm{Arg}(z-t) =\mathrm{Arctan}\left(\frac{y}{x-t}\right) \text{ and } d\theta=\frac{y}{(x-t)^2+y^2}\,dt\text{,}\tag{10.3.4} \end{equation}

we write Equation (10.3.3) as

\begin{equation*} \Phi(x,y) = \frac{y}{\pi}\sum\limits_{k=0}^{N}\frac{U(t_k^*) \,\Delta t_k}{(x-t_k^*)^2+y^2} \end{equation*}

The limit of this Riemann sum becomes the improper integral

\begin{equation*} \phi(x,y) = \frac{y}{\pi}\int\nolimits_{-\infty}^{\infty}\frac{U(t)}{(x-t)^2+y^2}\,dt \end{equation*}

Example 10.3.3.

Find the function \(\phi(x,y)\) that is harmonic in the upper half-plane \(\mathrm{Im}(z) >0\) and has the boundary values

\begin{equation*} \phi(x,0) = 1, \text{ for } -1 \lt x \lt 1, \text{ and } \phi(x,0) = 0, \text{ for } |x| > 1\text{.} \end{equation*}

Solution.

Using Equation (10.3.1), we obtain

\begin{equation*} \phi(x,y) = \frac{y}{\pi}\int_{-1}^1\frac{1}{(x-t)^2+y^2}\,dt = \frac{1}{\pi}\int_{-1}^1\frac{y}{(x-t)^2+y^2}\,dt \end{equation*}

Using the antiderivative in Equation (10.3.4), we write this solution as

\begin{align*} \phi(x,y) \amp = \left. \frac{1}{\pi}\mathrm{Arctan}\left(\frac{y}{x-t}\right)\right|_{t=-1}^{t=1}\\ \amp = \frac{1}{\pi}\mathrm{Arctan}\left(\frac{y}{x-1}\right)-\frac{1}{\pi}\mathrm{Arctan}\left(\frac{y}{x+1}\right)\text{.} \end{align*}

Example 10.3.4.

Find the function \(\phi(x,y)\) that is harmonic in the upper half-plane \(\mathrm{Im}(z) >0\) and has the boundary values

\begin{equation*} \phi(x,0) =x, \text{ for } -1 \lt x \lt 1, \text{ and } \phi(x,0) = 0, \text{ for } |x| > 1\text{.} \end{equation*}

Solution.

Using Equation (10.3.1), we obtain

\begin{align*} \phi(x,y) \amp = \frac{y}{\pi}\int\nolimits_{-1}^{1}\frac{t}{(x-t)^2+y^2}\,dt\\ \amp = \frac{y}{\pi}\int\nolimits_{-1}^{1}\frac{(x-t)(-1)}{(x-t)^2+y^2}\,dt + \frac{x}{\pi} \int_{-1}^1\frac{y}{(x-t)^2+y^2}\,dt\text{.} \end{align*}

Using calculus techniques and Equations (10.3.4), we write the solution as

\begin{equation*} \phi(x,y) = \frac{y}{2\pi}\ln\left[\frac{(x-1)^2+y^2}{(x+1)^2+y^2}\right] + \frac{x}{\pi}\mathrm{Arctan}\left(\frac{y}{x-1}\right) - \frac{x}{\pi}\mathrm{Arctan}\left(\frac{y}{x+1}\right)\text{.} \end{equation*}

The function \(\phi(x,y)\) is continuous in the upper half-plane, and on the boundary \(\phi(x,0)\text{,}\) it has discontinuities at \(x=\pm 1\) on the real axis. The graph in Figure 10.3.5 shows this phenomenon.

Figure 10.3.5. \(\phi(x,y)\) with boundary values \(\phi(x,0) =x\) for \(|x| \lt 1\text{,}\) and \(\phi(x,0)=0\) for \(|x|>1\)

Example 10.3.6.

Find \(\phi(x,y)\) that is harmonic in the upper half-plane \(\mathrm{Im}(z)>0\) and that has the boundary values

\begin{align*} \phi(x,0) \amp = \;\; x \text{ for } |x|\lt 1,\\ \phi(x,0) \amp = -1 \text{ for } x\lt -1, \text{ and }\\ \phi(x,0) \amp = \;\; 1 \text{ for } x>1\text{.} \end{align*}

Solution.

Using techniques from Section 10.2, we find that the function

\begin{equation*} v(x,y) = 1 - \frac{1}{\pi}\mathrm{Arctan}\left(\frac{y}{x+1}\right) - \frac{1}{\pi}\mathrm{Arctan}\left(\frac{y}{x-1}\right) \end{equation*}

is harmonic in the upper half-plane and has the following boundary values:

\begin{equation*} v(x,0) =0 \text{ for } |x|\lt 1, v(x,0) = -1 \text{ for } x\lt -1, \text{ and } v(x,0)=1 \text{ for } x > 1\text{.} \end{equation*}

This function can be added to the one in Example 10.3.4 to obtain the desired result:

\begin{equation*} \phi(x,y) = 1 + \frac{y}{2\pi}\ln\left[\frac{(x-1)^2+y^2}{(x+1)^2+y^2}\right] + \frac{x-1}{\pi}\mathrm{Arctan}\left(\frac{y}{x-1}\right) - \frac{x+1}{\pi}\mathrm{Arctan}\left(\frac{y}{x+1}\right)\text{.} \end{equation*}

Figure 10.3.7 shows the graph of \(\phi(x,y)\text{.}\)

Figure 10.3.7. \(\phi(x,y)\) with boundary values \(\phi(x,0)=x\) for \(|x|\lt 1, \ \phi(x,0)=-1\text{,}\) for \(x\lt -1\text{,}\) and \(\phi(x,0) =1\) for \(x>1\)

Exercises Exercises

1.

Use Poisson’s integral formula to find the harmonic function \(\phi(x,y)\) in the upper half-plane that takes on the boundary values

\begin{align*} \phi(t,0) \amp = U(t)=0, \text{ for } t \lt 0;\\ \phi(t,0) \amp = U(t)=t, \text{ for } 0 \lt t \lt 1;\\ \phi(t,0) \amp = U(t)=0, \text{ for } 1 \lt t\text{.} \end{align*}

Solution.

\(\phi (x,y)=\frac{y}{2\pi}\ln\Big[\frac{(x-1)^2+y^2}{(x+1)^2+y^2}\Big] + \frac{x-1}{\pi}\mathrm{Arctan}(\frac{y}{x-1})-\frac{x+1}{\pi}\mathrm{Arctan}(\frac{y}{x+1})+1\)

2.

Use Poisson’s integral formula to find the harmonic function \(\phi(x,y)\) in the upper half-plane that takes on the boundary values

\begin{align*} \phi(t,0) \amp = U(t) =0, \text{ for } t \lt 0;\\ \phi(t,0) \amp = U(t) =t, \text{ for } 0 \lt t \lt 1;\\ \phi(t,0) \amp = U(t) =1, \text{ for } 1 \lt t\text{.} \end{align*}

3.

Use Poisson’s integral formula for the upper half-plane to conclude

\begin{equation*} \phi(x,y) =e^{-y}\cos x=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\cos t}{(x-t)^2+y^2}\,dt \end{equation*}

Solution.

Both \(e^y\cos x\) and \(e^{-y}\cos x\) are harmonic in the upper half plane and satisfy the boundary conditions. Also, \(\lim\limits_{y \to \infty }e^{-y}\cos x=0\text{.}\) It can be show that the Poisson integral formula defines a bounded function in the upper half plane, therefore the desired solution is \(\phi(x,y)=e^{-y}\cos x\text{.}\)

4.

Use Poisson’s integral formula for the upper half-plane to conclude

\begin{equation*} \phi(x,y) = e^{-y}\sin x=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\sin t}{(x-t)^2+y^2}\,dt \end{equation*}

5.

Show that the function \(\phi(x,y)\) given by Poisson’s integral formula is harmonic by applying Leibniz’s rule, which permits you to write

\begin{equation*} \left(\frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2}\right) \phi(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty} U(t)\left[\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) \frac{y}{(x-t)^2+y^2}\right]dt\text{.} \end{equation*}

Solution.

Apply Leibniz’s rule \(\phi _{xx}\)\(+\phi _{yy}=\frac{1}{\pi} \int_{-\infty }^{\infty}U(t) \left[\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) \frac{y}{(x-t)^2+y^2}\right]dt\text{.}\) The term in brackets in the integrand is \(\frac{\partial^2}{\partial x^2}\frac{y}{(x-t)^2+y^2} + \frac{\partial^2}{\partial y^2}\frac{y}{(x-t)^2+y^2} = \frac{2(3t^2y-6txy+3x^2y-y^{3})}{((x-t)^2+y^2)^3}+\frac{2(-3t^2y+6txy-3x^2y + y^3)}{((x-t)^2+y^2)^3} = 0\text{.}\) Hence the integrand vanishes and \(\phi_{xx}(x,y) + \phi_{yy}(x,y) = 0\text{,}\) which implies that \(\phi(x,y)\) is harmonic.

6.

Let \(U(t)\) be a real-valued function that satisfies the conditions for Poisson’s integral formula for the upper half-plane. If \(U(t)\) is an even function so that \(U(-t) =U(t)\text{,}\) then show that the harmonic function \(\phi(x,y)\) has the property \(\phi(-x,y)=\phi(x,y)\text{.}\)

7.

Let \(U(t)\) be a real-valued function that satisfies the conditions for Poisson’s integral formula for the upper half-plane. If \(U(t)\) is an odd function so that for all \(t\) \(U(-t) =-U(t)\text{,}\) then show that the harmonic function \(\phi(x,y)\) has the property \(\phi(-x,y) = -\phi(x,y)\text{.}\)

Solution.

\(\phi(-x,y) = \frac{y}{\pi}\int_{-\infty }^{\infty}\frac{U(t)\,dt}{(-x-t)^2+y^2} = \frac{y}{\pi}\int_{-\infty }^{\infty}\frac{U(t)\,dt}{(x+t)^2+y^2} = \frac{y}{\pi}\int_{\infty }^{-\infty}\frac{U(-t)(-1)\,dt}{(x-t)^2+y^2}\) \(=\frac{y}{\pi}\int_{-\infty }^{\infty}\frac{U(-t)\,dt}{(x-t)^2+y^2} = -\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{U(t)\,dt}{(x-t)^2+y^2} = -\phi(x,y)\text{.}\)

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