User:Egm6321.f09.Team06.sada/HW5

Problem 2
p.29-4 Find equation (4)for i=3.

Solution

We know that, from p.29-3

$$\displaystyle \Delta \psi = \frac{1}{h_1 h_2 h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left[ \frac{h_1 h_2 h_3}{h_i^2} \frac{\partial \psi}{\partial \xi_i} \right] $$

Using $$\displaystyle h_1 = 1, \quad h_2 = r \cos \theta, \quad h_3 = r $$

and

$$\displaystyle \xi_1 = r, \quad \xi_2 = \varphi, \quad \xi_3 = \theta $$

For i=1 by using the above relations, we get equ. 2 in p.29-4

$$\displaystyle \frac{1}{r^2 \cos \theta} \frac{\partial}{\partial r} \left[ \frac{r^2 \cos \theta}{1} \frac{\partial \varphi}{\partial r} \right] $$

For i=2

$$\displaystyle \Delta \psi_2 = \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial \xi_2} \left[ \frac{h_1 h_2 h_3}{h_2^2} \frac{\partial \psi}{\partial \xi_2} \right] = \frac{1}{r^2 \cos \theta} \frac{\partial}{\partial \varphi} \left[ \frac{r^2 \cos \theta}{r^2 \cos^2 \theta} \frac{\partial \psi}{\partial \varphi} \right] $$

$$\mathcal {4}\psi_{2}={1\over r^2cos^2\theta}\Bigg[{\partial^2 \psi \over \partial \phi^2}\Bigg]$$

For i=3

$$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Bigg[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Bigg]$$ becomes

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[{r^2cos\theta\over (r)^2}{\partial \psi \over \partial \theta}\Bigg]$$

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[{r^2cos\theta\over r^2}{\partial \psi \over \partial \phi}\Bigg]$$

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Bigg[cos\theta{\partial \phi\over \partial\theta}\Bigg]$$

$$\mathcal {4}\psi_{i=3}=\frac{1}{r^2 \cos \theta} \left[ \sin \theta \frac{\partial \psi}{\partial \theta} + \cos \theta \frac{\partial^2 \psi}{\partial \theta^2} \right] $$