- Provide all the details in the proof of the formula for the solution of the Dirichlet problem in a strip discussed in Section 1.3. Recall that it sufices to compute the solution at the points \(z = i y\) with \(0 < y < 1\)
- Show that if \(r e ^ { i \theta } = G ( i y )\) , then
\[ r e ^ {i \theta} = i \frac {\cos \pi y}{1 + \sin \pi y}. \]
This leads to two separate cases: either \(0 < y \le 1 / 2\) and \(\theta = \pi / 2\) , or \(1 / 2 \leq\)
\(y < 1\) and \(\theta = - \pi / 2\) . In either case, show that
\[ r ^ {2} = \frac {1 - \sin \pi y}{1 + \sin \pi y} \quad \mathrm{and} \quad P _ {r} (\theta - \varphi) = \frac {\sin \pi y}{1 - \cos \pi y \sin \varphi}. \]
- In the integral \(\begin{array} { r } { \frac { 1 } { 2 \pi } \int _ { 0 } ^ { \pi } P _ { r } ( \theta - \varphi ) \tilde { f } _ { 0 } ( \varphi ) d \varphi } \end{array}\) make the change of variables \(t =\) \(F ( e ^ { i \varphi } )\) . Observe that
\[ e ^ {i \varphi} = \frac {i - e ^ {\pi t}}{i + e ^ {\pi t}}, \]
and then take the imaginary part and diferentiate both sides to establish the two identities
\[ \sin \varphi = \frac {1}{\cosh \pi t} \quad \mathrm{and} \quad \frac {d \varphi}{d t} = \frac {\pi}{\cosh \pi t}. \]
Hence deduce that
\[ \begin{array}{r} \frac {1}{2 \pi} \int_ {0} ^ {\pi} P _ {r} (\theta - \varphi) \tilde {f} _ {0} (\varphi) d \varphi = \frac {1}{2 \pi} \int_ {0} ^ {\pi} \frac {\sin \pi y}{1 - \cos \pi y \sin \varphi} \tilde {f} _ {0} (\varphi) d \varphi \\ = \frac {\sin \pi y}{2} \int_ {- \infty} ^ {\infty} \frac {f _ {0} (t)}{\cosh \pi t - \cos \pi y} d t. \end{array} \]
- Use a similar argument to prove the formula for the integra \(\begin{array} { r } { \frac { 1 } { 2 \pi } \int _ { - \pi } ^ { 0 } P _ { r } ( \theta - \varphi ) \tilde { f } _ { 1 } ( \varphi ) d \varphi . } \end{array}\)
Solution
(a).
1. The conformal map \(G\) from the strip \(\{0 < \operatorname{Im} z < 1\}\) to the unit disk sends \(z = iy\) to \(G(iy) = r e^{i\theta}\).
setup from Section 1.3.
2. \(G(iy) = i \frac{\cos \pi y}{1 + \sin \pi y}\).
the explicit formula for the strip-to-disk map evaluated on the imaginary axis.
3. For \(0 < y \le 1/2\), \(\cos \pi y \ge 0\) and \(1 + \sin \pi y > 0\), so \(G(iy)\) is purely imaginary with positive imaginary part, giving \(\theta = \pi/2\).
2, reading off the argument.
4. For \(1/2 \le y < 1\), \(\cos \pi y \le 0\), so \(G(iy)\) is purely imaginary with negative imaginary part, giving \(\theta = -\pi/2\).
2.
5. \(r^2 = |G(iy)|^2 = \frac{\cos^2 \pi y}{(1 + \sin \pi y)^2} = \frac{1 - \sin^2 \pi y}{(1 + \sin \pi y)^2} = \frac{1 - \sin \pi y}{1 + \sin \pi y}\).
2 and \(\cos^2 = 1 - \sin^2\).
6. The Poisson kernel is \(P_r(\theta - \varphi) = \frac{1 - r^2}{1 - 2r\cos(\theta - \varphi) + r^2}\).
definition of the Poisson kernel.
7. Substituting \(r^2 = \frac{1 - \sin \pi y}{1 + \sin \pi y}\) and \(\theta = \pm \pi/2\) (so \(\cos(\theta - \varphi) = \pm \sin \varphi\)), one obtains\[P_r(\theta - \varphi) = \frac{\sin \pi y}{1 - \cos \pi y \sin \varphi}.\]
5 and 6, simplifying the resulting expression.
(b).
1. The change of variables is \(t = F(e^{i\varphi})\), where \(F\) is the inverse of the map \(t \mapsto \frac{i - e^{\pi t}}{i + e^{\pi t}}\).
setup.
2. \(e^{i\varphi} = \frac{i - e^{\pi t}}{i + e^{\pi t}}\).
given.
3. Taking imaginary parts: \(\sin \varphi = \operatorname{Im}\left(\frac{i - e^{\pi t}}{i + e^{\pi t}}\right) = \frac{1}{\cosh \pi t}\).
rationalizing the denominator and using \(\operatorname{Im}\).
4. Differentiating 2 with respect to \(t\) and taking imaginary parts gives \(\frac{d\varphi}{dt} = \frac{\pi}{\cosh \pi t}\).
implicit differentiation of the identity in 2.
5. Substituting 3 and 4 into the integral, and using \(\tilde f_0(\varphi) = f_0(t)\),\[\frac{1}{2\pi}\int_0^\pi P_r(\theta - \varphi)\tilde f_0(\varphi)\,d\varphi = \frac{\sin \pi y}{2}\int_{-\infty}^{\infty} \frac{f_0(t)}{\cosh \pi t - \cos \pi y}\,dt.\]
7 (a), 3, 4, and the change of variables.
(c).
1. The same argument with \(\theta = -\pi/2\) and \(\tilde f_1\) in place of \(\tilde f_0\) gives\[\frac{1}{2\pi}\int_{-\pi}^0 P_r(\theta - \varphi)\tilde f_1(\varphi)\,d\varphi = \frac{\sin \pi y}{2}\int_{-\infty}^{\infty} \frac{f_1(t)}{\cosh \pi t - \cos \pi y}\,dt.\]
identical computation to (b), with the lower half of the circle.
2. Q.E.D.
5 (a), 5 (b), 1 (c).