User:Egm6321.f11.team4.allen/HW7

= Problem R*7.4 Laplacian in Elliptic Coordinates = From [[media:Pea1.f11.mtg41.djvu|Mtg 41-1]]

Given
The Laplacian operator for general coordinates is given by $$\nabla^{2} u = \Delta u = \frac{1}{h_1h_2h_3} \sum^3_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2h_3}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The elliptic coordinates are given by $$\displaystyle \xi_1 = \mu $$ $$\displaystyle \xi_2 = \nu $$. The scale factors for the elliptic coordinates $$\displaystyle (\mu, \nu)$$ are given by $$\displaystyle h_{\mu} = h_{\nu} = a \sqrt{\sinh^2{\mu} + \sin^2{\nu}}$$.

Find
Verify the Laplacian in elliptic coordinates is given by $$\displaystyle \nabla^{2} u = \Delta u= \frac{1}{a^{2} \left( \sinh^{2} \mu + \sin^{2}\nu \right)} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

Solution
Since this is 2-D problem, the Laplacian operator for general coordinates is reduces to $$\Delta u = \frac{1}{h_1h_2} \sum^2_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The above equation expands to $$\Delta u = \frac{1}{h_1 h_2} \left[ \frac{\partial}{\partial\xi_1} \left( \frac{h_1 h_2}{(h_1)^2} \frac{\partial u}{\partial\xi_1} \right) + \frac{\partial}{\partial\xi_2} \left( \frac{h_1h_2}{(h_2)^2} \frac{\partial u}{\partial\xi_2} \right) \right] $$. Since $$\displaystyle h_{\mu} = h_{\nu} $$, the following terms cancel out: $$\displaystyle \cancel{\frac{h_1 h_2}{(h_1)^2}} $$ and $$\displaystyle \cancel{\frac{h_1 h_2}{(h_2)^2}} $$. With $$\displaystyle \xi_1 = \mu $$, $$\displaystyle \xi_2 = \nu $$ and $$\displaystyle h_{\mu} = h_{\nu} = a \sqrt{\sinh^2{2\mu} - \sin^2{2\nu}}$$, the Laplacian reduces to $$\displaystyle \Delta u= \frac{1}{a^{2} \left( \sinh^{2} \mu + \sin^{2}\nu \right)} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

= Problem R7.5 Laplacian in Parabolic Coordinates = From [[media:Pea1.f11.mtg41.djvu|Mtg 41-2]]

Given
The Laplacian operator for general coordinates is given by $$\nabla^{2} u = \Delta u = \frac{1}{h_1h_2h_3} \sum^3_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2h_3}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The parabolic coordinates are given by $$\displaystyle \xi_1 = \mu $$ $$\displaystyle \xi_2 = \nu $$. The scale factors for the parabolic coordinates $$\displaystyle (\mu, \nu)$$ are given by $$\displaystyle h_{\mu} = h_{\nu} = \sqrt{\mu^2 + \nu^2}$$.

Find
Verify the Laplacian in elliptic coordinates is given by $$\displaystyle \nabla^{2} u = \Delta u= \frac{1}{\mu^2 + \nu^2} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.

Solution
Since this is 2-D problem, the Laplacian operator for general coordinates is reduces to $$\Delta u = \frac{1}{h_1h_2} \sum^2_{i=1}\frac{\partial}{\partial\xi_i} \left[ \frac{h_1h_2}{(h_i)^2} \frac{ \partial u}{ \partial\xi_i} \right]$$. The above equation expands to $$\Delta u = \frac{1}{h_1 h_2} \left[ \frac{\partial}{\partial\xi_1} \left( \frac{h_1 h_2}{(h_1)^2} \frac{\partial u}{\partial\xi_1} \right) + \frac{\partial}{\partial\xi_2} \left( \frac{h_1h_2}{(h_2)^2} \frac{\partial u}{\partial\xi_2} \right) \right] $$. Since $$\displaystyle h_{\mu} = h_{\nu} $$, the following terms cancel out: $$\displaystyle \cancel{\frac{h_1 h_2}{(h_1)^2}} $$ and $$\displaystyle \cancel{\frac{h_1 h_2}{(h_2)^2}} $$. With $$\displaystyle \xi_1 = \mu $$, $$\displaystyle \xi_2 = \nu $$ and $$\displaystyle h_{\mu} = h_{\nu} = \sqrt{\mu^2 + \nu^2} $$, the Laplacian reduces to $$\displaystyle \Delta u= \frac{1}{\mu^2 + \nu^2} \left( \frac{\partial^{2} u}{\partial \mu^{2}} + \frac{\partial^{2} u}{\partial \nu^{2}} \right) $$.