Chapter 14.11, Problem 39MP

Use the following sequence of mappings to find the steady state temperature $T(x,y)$ in the semi-infinite strip $y\ge 0,0\le x\le \pi$ if $T(x,0)={100}^{°},T(0,y)=T(\pi ,y)=0$ and $T(x,y)\to 0$ as $y\to \infty .$ (See Chapter 13, Section 2 and Problem 2.6 .)

(a) Use $w=\left(z^{\prime }-1\right)/\left(z^{\prime }+1\right)$ to map the half plane $v\ge 0$ on the upper half plane $y^{\prime }>0,$ with the positive $u$ axis corresponding to the two rays $x^{\prime }>1$ and $x^{\prime }<-1,$ and the negative $u$ axis corresponding to the interval $-1\le x\le 1$ of the $x^{\prime }$ axis.

(b) Use $z^{\prime }=-\cos z$ to map the half-strip $0<x<\pi ,y>0$ on the $z^{\prime }$ half plane described in (a). The interval $-1\le x^{\prime }<1,y^{\prime }=0$ corresponds to the base $0<x<\pi ,y=0$ of the strip.

Comments: The temperature problem in the $(u,v)$ plane is like the problems shown in the $z$ plane of Figures 10.1 and $10.2,$ and so is given by $T=(100/\pi )\arctan (v/u)$ In the $z$ plane you will find $T(x,y)=\frac{100}{\pi }\arctan \frac{2\sin x\sinh y}{{\sinh }^{2}y-{\sin }^{2}x}$

Put $\tan \alpha =\frac{\sin x}{\sinh y}$ and use the formula for tan $2\alpha$ to get $T(x,y)=\frac{200}{\pi }\arctan \frac{\sin x}{\sinh y}$

Note that this is the same answer as in Chapter $13,$ Problem $2.6,$ if we replace 10 by $\pi$.