Задача 37. Воспользовавшись декартовыми, сферическими и цилиндрическими координатами, вычислить $\mathrm{div\,} \vec{r}$, $\mathrm{rot\,} \vec{r}$, $\mathrm{grad\,} (\vec{l}\cdot\vec{r})$, $(\vec{l}\cdot\nabla)\vec{r}$, где $\vec{r}$ - радиус вектор, а $\vec{l}$ - постоянный вектор.

Решение.

В декартовых координатах:

\[\mathrm{div\,} \vec{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3\] \[\mathrm{rot\,} \vec{r} = \begin{vmatrix} \vec{e}_x & \vec{e}_y & \vec{e}_z \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ x & y & z \end{vmatrix} = 0\] \[\mathrm{grad\,} (\vec{l}\cdot\vec{r}) = \left\{ \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\} (l_x x + l_y y + l_z z) = \{l_x, l_y, l_z\} = \vec{l}\] \[(\vec{l}\cdot\nabla)\vec{r} = \left(l_x\frac{\partial}{\partial x} + l_y\frac{\partial}{\partial y} + l_z\frac{\partial}{\partial z}\right) \{x, y, z\} = \{l_x, l_y, l_z\} = \vec{l}\]

В сферических координатах:

\[\vec{r} = r \vec{e}_r\] \[\mathrm{div\,} \vec{r} = \mathrm{div\,} (r \vec{e}_r) = \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial r} \left(r^3 \sin \theta\right) = 3\] \[\mathrm{rot\,} \vec{r} = \begin{vmatrix} \dfrac{1}{r^2 \sin \theta} \vec{e}_r & \dfrac{1}{r \sin \theta} \vec{e}_\theta & \dfrac{1}{r} \vec{e}_\alpha \\ \dfrac{\partial}{\partial r} & \dfrac{\partial}{\partial \theta} & \dfrac{\partial}{\partial \alpha} \\ r & 0 & 0 \end{vmatrix} = 0\]

Сонаправим ось $z$ с $\vec{l}$. Тогда:

\[\vec{l} = l (\vec{e}_r \cos \theta - \vec{e}_\theta \sin \theta)\] \[\mathrm{grad\,} (\vec{l}\cdot\vec{r}) = \left(\vec{e}_r \frac{\partial}{\partial r} + \vec{e}_\theta \frac{1}{r} \frac{\partial}{\partial \theta} + \vec{e}_\alpha \frac{1}{r \sin \theta} \frac{\partial}{\partial \alpha} \right) (l r \cos \theta) = l \left(\vec{e}_r \cos \theta - \vec{e}_\theta \sin \theta \right) =\vec{l}\] \[(\vec{l}\cdot\nabla)\vec{r} = \left(l \cos \theta\frac{\partial}{\partial r} - l \sin \theta \frac{1}{r} \frac{\partial}{\partial \theta} \right) (r\vec{e}_r) = l \left(\cos \theta \vec{e}_r + r \cos \theta \frac{\partial \vec{e}_r}{\partial r}- \sin \theta \frac{\partial \vec{e}_r}{\partial \theta} \right) (r\vec{e}_r) = (1)\]

Но:

\[\begin{aligned} & \vec{e}_r = \vec{e}_x \sin \theta \cos \alpha + \vec{e}_y \sin \theta \sin \alpha + \vec{e}_z \cos \theta \\ & \vec{e}_\theta = \vec{e}_x \cos \theta \cos \alpha + \vec{e}_y \cos \theta \sin \alpha - \vec{e}_z \sin \theta \\ & \vec{e}_\alpha = - \vec{e}_x \sin \alpha + \vec{e}_y \cos\alpha \end{aligned}\]

Поэтому:

\[\frac{\partial \vec{e}_r}{\partial r} = 0\] \[\frac{\partial \vec{e}_r}{\partial \theta} = \vec{e}_\theta\]

и

\[(1) = \vec{l}\]

В цилиндрических координатах:

\[\vec{r} = \rho \vec{e}_\rho + z \vec{e}_z\] \[\mathrm{div\,} \vec{r} = \frac{1}{\rho} \left(\frac{\partial \rho^2}{\partial \rho} + \rho \frac{\partial z}{\partial z} \right) = 3\] \[\mathrm{rot\,} \vec{r} = \begin{vmatrix} \dfrac{1}{\rho}\vec{e}_\rho & \vec{e}_\alpha & \dfrac{1}{\rho}\vec{e}_z \\ \dfrac{\partial}{\partial \rho} & \dfrac{\partial}{\partial \alpha} & \dfrac{\partial}{\partial z} \\ \rho & 0 & z \end{vmatrix} = 0\]

Сонаправим $z$ c $\vec{l}$:

\[\mathrm{grad\,} (\vec{l}\cdot\vec{r}) = \left( \vec{e}_\rho \frac{\partial}{\partial \rho} + \vec{e}_\alpha \frac{1}{\rho}\frac{\partial}{\partial \alpha} + \vec{e}_z \frac{\partial}{\partial z} \right) (l z) = l \vec{e}_z = \vec{l}\] \[(\vec{l}\cdot\nabla)\vec{r} = l\frac{\partial}{\partial z}\left(\rho \vec{e}_\rho + z \vec{e}_z\right) = l \vec{e}_z = \vec{l}\]