Задача 48. Записать проекции вектора $\Delta \vec{A}$ на оси цилиндрической системы координат.

Решение.

\[\Delta (A_i \vec{e}_i) = \vec{e}_i \Delta A_i + A_i \Delta \vec{e}_i + 2 (\nabla A_i \cdot \nabla) \vec{e}_i\] \[\begin{aligned} & \vec{e}_r = \cos \alpha \vec{e}_x + \sin \alpha \vec{e}_y, \\ & \vec{e}_\alpha = - \sin \alpha \vec{e}_x + \cos \alpha \vec{e}_y , \\ & \vec{e}_z = \vec{e}_z. \end{aligned}\] \[\Delta = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2}{\partial \alpha^2} + \frac{\partial^2}{\partial z^2}\] \[\begin{aligned} & \Delta \vec{e}_r = - \frac{1}{r^2} \vec{e}_r, \\ & \Delta \vec{e}_\alpha = - \frac{1}{r^2} \vec{e}_\alpha \\ & \Delta \vec{e}_z = 0. \end{aligned}\] \[\begin{alignedat}{} & \dfrac{\partial \vec{e}_r}{\partial r} = 0, & \dfrac{\partial \vec{e}_\alpha}{\partial r} = 0, & \dfrac{\partial \vec{e}_z}{\partial r} = 0, \\ & \dfrac{\partial \vec{e}_r}{\partial \alpha} = \vec{e}_\alpha, & \dfrac{\partial \vec{e}_\alpha}{\partial \alpha} = -\vec{e}_r, & \dfrac{\partial \vec{e}_z}{\partial \alpha} = 0, \\ & \dfrac{\partial \vec{e}_r}{\partial z} = 0, & \dfrac{\partial \vec{e}_\alpha}{\partial z} = 0, & \dfrac{\partial \vec{e}_z}{\partial z} = 0, \end{alignedat}\]

Собираем всё вместе:

\[(\Delta \vec{A})_r = \Delta A_r - \frac{1}{r^2} A_r - \frac{2}{r^2} \frac{\partial A_\alpha}{\partial \alpha}\] \[(\Delta \vec{A})_\alpha = \Delta A_\alpha - \frac{1}{r^2} A_\alpha + \frac{2}{r^2} \frac{\partial A_r}{\partial \alpha}\] \[(\Delta \vec{A})_z = \Delta A_z\]