Derive Poisson’s formulas for the half plane [see page 146].

Question

Accepted Answer

Let C be the boundary of a semicircle of radius R [see Fig. 5-10] containing [latex]\zeta[/latex] as an interior point. Since C encloses [latex]\zeta[/latex] but does not enclose [latex]\bar{\zeta}[/latex], we have by Cauchy's integral formula,

[latex]f(\zeta)=\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-\zeta} d z, \quad 0=\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-\bar{\zeta}} d z[/latex]

Then, by subtraction,

[latex]f(\zeta)=\frac{1}{2 \pi i} \oint_C f(z)\left\{\frac{1}{z-\zeta}-\frac{1}{z-\bar{\zeta}} ight\} d z=\frac{1}{2 \pi i} \oint_C \frac{(\zeta-\bar{\zeta}) f(z) d z}{(z-\zeta)(z-\bar{\zeta})}[/latex]

[latex] ext { Letting } \zeta=\xi+i \eta, \bar{\zeta}=\xi-i \eta ext {, this can be written }[/latex]

[latex]f(\zeta)=\frac{1}{\pi} \int_{-R}^R \frac{\eta f(x) d x}{(x-\xi)^2+\eta^2}+\frac{1}{\pi} \int_{\Gamma} \frac{\eta f(z) d z}{(z-\zeta)(z-\bar{\zeta})}[/latex]

where Γ is the semicircular arc of C. As R →∞, this last integral approaches zero [see Problem 5.76] and we have

[latex]f(\zeta)=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\eta f(x) d x}{(x-\xi)^2+\eta^2}[/latex]

[latex] ext { Writing } f(\zeta)=f(\xi+i \eta)=u(\xi, \eta)+i v(\xi, \eta), \quad f(x)=u(x, 0)+i v(x, 0) ext {, we obtain as required, }[/latex]

[latex]u(\xi, \eta)=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\eta u(x, 0) d x}{(x-\xi)^2+\eta^2}, \quad v(\xi, \eta)=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\eta v(x, 0) d x}{(x-\xi)^2+\eta^2}[/latex]

Question 5.22: Derive Poisson’s formulas for the half plane [see page 146].

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