Q39MP - Mathematical Methods in Physical Sciences

Question

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

Use $\mathit{w}\mathbf{=}\left(\frac{z\text{'}-1}{z\text{'}+1}\right)$ to map the half plane $\mathit{v}\mathbf{\ge }\mathbf{0}$ on the upper half plane $\mathit{y}\mathbf{\text{'}}\mathbf{>}\mathbf{0}$, with the positive   axis corresponding to the two rays $\mathit{x}\mathbf{\text{'}}\mathbf{>}\mathbf{1}$ and $\mathit{x}\mathbf{\text{'}}\mathbf{<}\mathbf{-}\mathbf{1}$, and the negative y axis corresponding to the interval $\mathbf{-}\mathbf{1}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{1}$ of the x' axis. Use  z'=-cosz to map the half-strip $\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathbf{\pi }\mathbf{,}\mathbf{ }\mathbf{y}\mathbf{>}\mathbf{0}$ on the Z' half plane described in (a). The interval $\mathbf{-}\mathbf{1}\mathbf{\le }\mathit{x}\mathbf{\text{'}}\mathbf{<}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$ corresponds to the base $\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathbf{\pi }\mathbf{,}\mathbf{ }\mathbf{y}\mathbf{=}\mathbf{0}$ 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 $\mathit{T}\mathbf{=}\left(\frac{100}{\pi }\right)\mathit{a}\mathit{r}\mathit{c}\mathbf{ }\mathbf{tan}\mathbf{ }\left(\frac{v}{u}\right)$. In the z plane you will find $\mathit{T}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\frac{\mathbf{100}}{\mathbf{\pi }}\mathit{a}\mathit{r}\mathit{c}\mathbf{ }\mathbf{tan}\frac{\mathbf{2}\mathbf{sin}\mathbf{x}\mathbf{sin}\mathbf{h}\mathbf{y}}{\mathbf{sin}{\mathbf{h}}^{\mathbf{2}}\mathbf{y}\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}}$ 

Put $\mathit{t}\mathit{a}\mathit{n}\mathit{\alpha }\mathbf{=}\frac{\mathbf{sin}\mathbf{x}}{\mathbf{sin}\mathbf{h}\mathbf{y}}$  and use the formula for $\mathit{t}\mathit{a}\mathit{n}\mathbf{2}\mathit{\alpha }$ to get $\mathit{T}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\frac{\mathbf{200}}{\mathbf{\pi }}\mathit{a}\mathit{r}\mathit{c}\mathbf{ }\mathbf{tan}\frac{\mathbf{sin}\mathbf{x}}{\mathbf{sin}\mathbf{h}\mathbf{y}}$ $$" width="9" height="19" style="max-width: none; vertical-align: -4px;" >

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

Step-by-Step Solution

Answer

The expansion of the steady state temperature in the semi-infinite strip of the z-plane is $T(x,y)=\left(\frac{200}{\mathrm{\pi }}\right)arc \tan \frac{\sin \left(x\right)}{\sin h\left(y\right)}$.

Step 1: Concept of Dirichlet Condition

The Dirichlet conditions are very much sufficient, if a function   is real valued and periodic, at each point it is equal to sum of its Fourier series at each point where   is continuous.

Step 2: Use Dirichlet Condition for calculation

Consider the temperature problem in the plane is the same as the plane.

The temperature equation is as follows:

$T=\left(\frac{100}{\mathrm{\pi }}\right)arc \tan \left(\frac{v}{u}\right)$ 

Consider the mapping of the image plane $w=\frac{z\text{'}-1}{z\text{'}+1}$  use this mapping to the half plane $v\ge 0$ on the upper half plane, y>0 with the positive u axis corresponding to the two rays x'>1 and, x'<-1  and the negative u axis corresponding to the interval $-1\le x\le 1$  of the x axis.

Consider the temperature equation $T=\left(\frac{100}{\mathrm{\pi }}\right)arc \tan \left(\frac{v}{u}\right)$ is in the z plane.

The z plane is defined as follows:

$$\begin{gathered} z(u,v)=u+iv \\ w=\frac{z-1}{z+1} \end{gathered}$$ 

Put values to solve further:

z=-cosz 

So, simplify as follows:

 $$\begin{gathered} w=\frac{z\text{'}-1}{z\text{'}+1} \\ =\frac{(-\cos z-1)}{(-\cos z+1)} \\ =\frac{(\cos z+1)}{(\cos z-1)} \\ =\frac{\left(\cos \right(x+iy)+1)}{\left(\cos \right(x+iy)-1)} \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} =\left(\frac{1}{({\sin }^2\left(x\right)\sin h^2\left(y\right)+\cos \left(x\right)\cos h(y)-1)}\left({\cos }^2\right(x\left)\cos h^2\right(y)+{\sin }^2\left(x\right)\sin h^2(y)+2i\sin (x\left)\sin h\right(y)-1)\right) \\ =\left({\cos }^2\right(x\left)\cos h^2\right(y)-{\cos }^2(x\left)\cos h^2\right(y)+2i\sin (x\left)\sin h\right(y)+(\sin h\left(y\right)+\sin \left(x\right)\left)\right(\sin h\left(y\right)-\sin \left(x\right)\left)\right) \\ =(\sin h^{2}y-{\sin }^{2}x)+i\left(2\sin \right(x\left)\sin h\right(y\left)\right) \\ =u+iv \end{gathered}$$

 Simplify further for solution:

 $$\begin{gathered} T(x,y)=\left(\frac{100}{\mathrm{\pi }}\right)arc \tan \left(\frac{u}{v}\right) \\ T(x,y)=\left(\frac{100}{\mathrm{\pi }}\right)arc \tan \left(\frac{2\sin \left(x\right)\sin h\left(y\right)}{\sin h^2\left(y\right)-{\sin }^2\left(x\right)}\right) \\ T(x,y)=\left(\frac{100}{\mathrm{\pi }}\right)arc \tan \left(\frac{2\sin \left(x\right)\sin h\left(y\right)}{1-{\displaystyle \frac{\sin h^2\left(x\right)}{\sin h^2\left(y\right)}}}\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} T(x,y)=2\left(\frac{100}{\mathrm{\pi }}\right)arc \left(\tan \right(\alpha \left)\right) \\ T(x,y)=\left(\frac{200}{\mathrm{\pi }}\right)arc\left(\frac{\sin \left(x\right)}{\sin h\left(y\right)}\right) \end{gathered}$$