Коэффициенты Ламе

[Charlz_Klug]

Проверьте, пожалуйста, решение.

Задача:
Представить оператор Лапласа в параболоидальных координатах.

$$\left\{ \begin{array}{rcl} x &= \sqrt{\xi\eta} \cos{\varphi};& \\ y &=\sqrt{\xi\eta} \sin{\varphi};& \\ z &=\dfrac{1}{2} (\xi - \eta) .&\\ \end{array} \right.$$

Решение:
Для того, чтобы удобнее было пользоваться коэффициентами Ламе обозначу переменные таким образом:

$$ q_1 = \xi, q_2 = \eta, q_3 = \varphi. $$

А функции $x$, $y$ и $z$ как $\omega_1$, $\omega_2$ и $\omega_3$ соответственно.
Общая формула оператора Лапласа в криволинейных координатах следующая:

$$ \Delta u = \dfrac {1} {H_1 H_2 H_3} \left[ \dfrac {\partial} {\partial q_1} \left( \dfrac {H_2 H_3} {H_1} \dfrac {\partial u} {\partial q_1} \right) + \dfrac {\partial} {\partial q_2} \left( \dfrac {H_1 H_3} {H_2} \dfrac {\partial u} {\partial q_2} \right) + \dfrac {\partial} {\partial q_3} \left( \dfrac {H_2 H_1} {H_3} \dfrac {\partial u} {\partial q_3} \right)\right]. $$

Здесь

$$H_i^2 = \left( \dfrac {\partial \omega_1} {\partial q_i}\right)^2 + \left( \dfrac {\partial \omega_2} {\partial q_i}\right)^2 + \left( \dfrac {\partial \omega_3} {\partial q_i}\right)^2.$$

Вычисляю частные производные:

$$\dfrac {\partial \omega_1} {\partial q_1} = \dfrac {\partial (\sqrt {\xi \eta} \cos {\varphi}) } {\partial \xi} = \dfrac {1} {2} \dfrac {1} {\sqrt {\xi}} \sqrt {\eta} \cos {\varphi} = \dfrac {\sqrt {\eta} \cos {\varphi}} {2 \sqrt {\xi}},$$

$$\dfrac {\partial \omega_2} {\partial q_1} = \dfrac {\partial (\sqrt {\xi \eta} \sin {\varphi}) } {\partial \xi} = \dfrac {\sqrt {\eta} \sin {\varphi}} {2 \sqrt {\xi}},$$

$$\dfrac {\partial \omega_3} {\partial q_1} = \dfrac {\partial \left(\dfrac {1} {2} (\xi - \eta) \right) } {\partial \xi} = \dfrac {1} {2} (1) = \dfrac {1} {2},$$

$$\dfrac {\partial \omega_1} {\partial q_2} = \dfrac {\partial (\sqrt {\xi \eta} \cos {\varphi}) } {\partial \eta} = \dfrac {\sqrt {\xi} \cos {\varphi}} {2 \sqrt {\eta}},$$

$$\dfrac {\partial \omega_2} {\partial q_2} = \dfrac {\partial (\sqrt {\xi \eta} \sin {\varphi}) } {\partial \eta} = \dfrac {\sqrt {\xi} \sin {\varphi}} {2 \sqrt {\eta}},$$

$$\dfrac {\partial \omega_3} {\partial q_2} = \dfrac {\partial \left( \dfrac {1} {2} (\xi - \eta) \right) } {\partial \eta} = \dfrac {1} {2} (-1) = - \dfrac {1} {2},$$

$$\dfrac {\partial \omega_1} {\partial q_3} = \dfrac {\partial (\sqrt {\xi \eta} \cos {\varphi}) } {\partial \varphi} = \sqrt {\xi \eta} (- \sin {\varphi}) = - \sqrt {\xi \eta} \sin {\varphi},$$

$$\dfrac {\partial \omega_2} {\partial q_3} = \dfrac {\partial (\sqrt {\xi \eta} \sin {\varphi}) } {\partial \varphi} = \sqrt {\xi \eta} \cos {\varphi},$$

$$\dfrac {\partial \omega_3} {\partial q_3} = \dfrac {\partial \left( \dfrac {1} {2} (\xi - \eta) \right) } {\partial \varphi} = 0.$$

Вычисляю $H_1$, $H_2$ и $H_3$.

$$H_1^2 = \left( \dfrac {\partial \omega_1} {\partial q_1} \right)^2 + \left( \dfrac {\partial \omega_2} {\partial q_1} \right)^2 + \left( \dfrac {\partial \omega_3} {\partial q_1} \right)^2 = \left( \dfrac {\sqrt {\eta} \cos {\varphi}} {2 \sqrt {\xi}} \right)^2 + \left( \dfrac {\sqrt {\eta} \sin {\varphi}} {2 \sqrt {\xi}} \right)^2 + \left( \dfrac {1} {2} \right)^2 = \dfrac {\eta \cos^2 {\varphi}} {4 \xi} + \dfrac {\eta \sin^2 {\varphi}} {4 \eta} + \dfrac {1} {4} = $$

$$= \dfrac {\eta} {4 \xi} (\cos^2 {\varphi} + \sin^2 {\varphi}) + \dfrac {1} {4} = \dfrac {\eta + \xi} {4 \xi},$$

$$H_2^2 = \left( \dfrac {\partial \omega_1} {\partial q_2} \right)^2 + \left( \dfrac {\partial \omega_2} {\partial q_2} \right)^2 + \left( \dfrac {\partial \omega_3} {\partial q_2} \right)^2 = \left( \dfrac {\sqrt {\xi} \cos {\varphi}} {2 \sqrt {\eta}} \right)^2 + \left( \dfrac {\sqrt {\xi} \sin {\varphi}} {2 \sqrt {\eta}} \right)^2 + \left( -\dfrac {1} {2} \right)^2 = \dfrac {\xi \cos^2 {\varphi}} {4 \eta} + \dfrac {\xi \sin^2 {\varphi}} {4 \eta} + \dfrac {1} {4} = $$

$$= \dfrac {\xi} {4 \eta} (\cos^2 {\varphi} + \sin^2 {\varphi}) + \dfrac {1} {4} = \dfrac {\eta + \xi} {4 \eta},$$

$$H_3^2 = \left( \dfrac {\partial \omega_1} {\partial q_3} \right)^2 + \left( \dfrac {\partial \omega_2} {\partial q_3} \right)^2 + \left( \dfrac {\partial \omega_3} {\partial q_3} \right)^2 = \left( - \sqrt {\xi \eta} \sin {\varphi} \right)^2 + \left( \sqrt {\xi \eta} \cos {\varphi} \right)^2 + 0^2 = \xi \eta \sin^2 {\varphi} + \xi \eta \cos^2 {\varphi} = \xi \eta.$$

Или же

$$H_1 = \dfrac {1} {2} \sqrt { \dfrac {\eta + \xi} {\xi}}, H_2 = \dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}}, H_3 = \sqrt {\xi \eta}.$$

Подставляю в формулу в криволинейных координатах (закрывающая квадратная скобка не отображается по техническим причинам):

$$\Delta u = \dfrac {1} { \dfrac {1} {2} \sqrt {\dfrac {\eta + \xi} {\xi}} \dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}} \sqrt {\xi \eta}} \left[ \dfrac {\partial} {\partial \xi} \left( \dfrac {\dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}} \sqrt {\xi \eta}} {\dfrac {1} {2} \sqrt {\dfrac {\eta + \xi} {\xi}}} \dfrac {\partial u} {\partial \xi} \right) + \dfrac {\partial} {\partial \eta} \left( \dfrac {\dfrac {1} {2} \sqrt {\dfrac {\eta + \xi} {\xi}} \sqrt {\xi \eta}} {\dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}} } \dfrac {\partial u} {\partial \eta} \right) + $$

$$+ \dfrac {\partial} {\partial \varphi} \left( \dfrac {\dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\xi}} \dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}}} {\sqrt {\xi \eta}} \dfrac {\partial u} {\partial \varphi} \right) \right]=$$

$$= \dfrac {1} {\dfrac {\xi + \eta} {4}} \left[ \dfrac {\partial} {\partial \xi} \left( \xi \dfrac {\partial u} {\partial \xi} \right) + \dfrac {\partial} {\partial \eta} \left( \eta \dfrac {\partial u} {\partial \eta} \right) + \dfrac {\partial} {\partial \varphi} \left( \dfrac {\xi + \eta} {4 \xi \eta} \dfrac {\partial u} {\partial \varphi} \right)\right] =$$

$$= \dfrac {4} {\xi + \eta} \dfrac {\partial u} {\partial \xi} + \dfrac {4 \xi} {\xi + \eta} \dfrac {\partial^2 u} {\partial \xi^2} + \dfrac {4} {\xi + \eta} \dfrac {\parital u} {\partial \eta} + \dfrac {4 \eta} {\xi + \eta} \dfrac {\parital^2 u} {\partial \eta^2} + \dfrac {1} {\xi \eta } \dfrac {\partial^2 u} {\partial \varphi^2}$$

[ellipse]

Опечатки в коде, \patial вместо \partial.

$$\dfrac {{\color{red}\partial} \omega_1} {\partial q_1} = \dfrac {\partial (\sqrt {\xi \eta} \cos {\varphi}) } {\partial \xi} = \dfrac {1} {2} \dfrac {1} {\sqrt {\xi}} \sqrt {\eta} \cos {\varphi} = \dfrac {\sqrt {\eta} \cos {\varphi}} {2 \sqrt {\xi}},$$

[Charlz_Klug]

ellipse, спасибо, поправил.

[Charlz_Klug]

Ещё поправка:

$$\Delta u = \dfrac {1} { \dfrac {1} {2} \sqrt {\dfrac {\eta + \xi} {\xi}} \dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}} \sqrt {\xi \eta}} \left[ \dfrac {\partial} {\partial \xi} \left( \dfrac {\dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}} \sqrt {\xi \eta}} {\dfrac {1} {2} \sqrt {\dfrac {\eta + \xi} {\xi}}} \dfrac {\partial u} {\partial \xi} \right) + \dfrac {\partial} {\partial \eta} \left( \dfrac {\dfrac {1} {2} \sqrt {\dfrac {\eta + \xi} {\xi}} \sqrt {\xi \eta}} {\dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}} } \dfrac {\partial u} {\partial \eta} \right) + $$

$$+ \dfrac {\partial} {\partial \varphi} \left( \dfrac {\dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\xi}} \dfrac {1} {2} \sqrt {\dfrac {\xi + \eta} {\eta}}} {\sqrt {\xi \eta}} \dfrac {\partial u} {\partial \varphi} \right) \right]=$$

$$= \dfrac {1} {\dfrac {\xi + \eta} {4}} \left[ \dfrac {\partial} {\partial \xi} \left( \xi \dfrac {\partial u} {\partial \xi} \right) + \dfrac {\partial} {\partial \eta} \left( \eta \dfrac {\partial u} {\partial \eta} \right) + \dfrac {\partial} {\partial \varphi} \left( \dfrac {\xi + \eta} {4 \xi \eta} \dfrac {\partial u} {\partial \varphi} \right)\right] =$$

$$= \dfrac {4} {\xi + \eta} \dfrac {\partial u} {\partial \xi} + \dfrac {4 \xi} {\xi + \eta} \dfrac {\partial^2 u} {\partial \xi^2} + \dfrac {4} {\xi + \eta} \dfrac {\partial u} {\partial \eta} + \dfrac {4 \eta} {\xi + \eta} \dfrac {\partial^2 u} {\partial \eta^2} + \dfrac {1} {\xi \eta } \dfrac {\partial^2 u} {\partial \varphi^2}$$