Identidad muestra lo siguiente, divergencia.

Question

Calcula $\nabla\cdot F$ y $\nabla\wedge$$F$ para la función vectorial $F(x,y,z)=(a\wedge r)\wedge r$ donde $r=xi+yj+zk$ y $a$ es un vector constante.

Estoy un poco atascado aquí, he usado el hecho de que $F(x,y,z)=(a\wedge r)\wedge r$ $= (a\cdot r)r-(r\cdot r)a$ luego descompongo todos los términos usando $a_{1}, a_{2}, a_{3}$ etc. sin embargo, esto se vuelve muy tedioso y debe haber una mejor manera, debo estar perdiendo alguna identidad,

cualquier ayuda sería muy apreciada, muchas gracias.

Answer 1

\begin{align*} \boldsymbol \nabla \boldsymbol \cdot \mathbf F &= \boldsymbol \nabla \boldsymbol \cdot \Big((\mathbf a \boldsymbol \times \mathbf r)\boldsymbol \times \mathbf r)\Big) \\ &= \boldsymbol \nabla \boldsymbol \cdot \Big((\mathbf a \boldsymbol \cdot \mathbf r)\mathbf r \boldsymbol - (\mathbf r \boldsymbol \cdot \mathbf r)\mathbf a\Big) \\ &= (\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \cdot \mathbf r + \boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \cdot \mathbf r - (\mathbf r \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \cdot \mathbf a - \boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) \boldsymbol \cdot \mathbf a \tag{$1$}\\ &= 3 \mathbf a \boldsymbol \cdot \mathbf r + \mathbf a \boldsymbol \cdot \mathbf r - 0 - 2 \mathbf r \boldsymbol \cdot \mathbf a \tag{$2$} \\ &= 2 \mathbf a \boldsymbol \cdot \mathbf r\\\\ \boldsymbol \nabla \boldsymbol \times \mathbf F &= \boldsymbol \nabla \boldsymbol \times\Big((\mathbf a \boldsymbol \times \mathbf r)\boldsymbol \times \mathbf r)\Big) \\ &= \boldsymbol \nabla \boldsymbol \times \Big((\mathbf a \boldsymbol \cdot \mathbf r)\mathbf r \boldsymbol - (\mathbf r \boldsymbol \cdot \mathbf r)\mathbf a\Big) \\ &= (\mathbf a \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \times \mathbf r \boldsymbol + \boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r) \boldsymbol \times \mathbf r \boldsymbol- (\mathbf r \boldsymbol \cdot \mathbf r)\boldsymbol \nabla \boldsymbol \times \mathbf a \boldsymbol - \boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) \boldsymbol \times \mathbf a \tag{$3$}\\ &= \mathbf 0 \boldsymbol+ \mathbf a \boldsymbol \times\mathbf r \boldsymbol- \mathbf 0 \boldsymbol- 2\mathbf r \boldsymbol \times \mathbf a \tag{$4$}\\ &= 3\mathbf a \boldsymbol \times \mathbf r, \end{align*} donde

• $(1)$ sigue ya que $\boldsymbol \nabla \boldsymbol \cdot (\mathbf F + \mathbf G) = \boldsymbol \nabla \boldsymbol \cdot \mathbf F + \boldsymbol \nabla \boldsymbol \cdot \mathbf G$ y $\boldsymbol \nabla \boldsymbol \cdot (\phi \mathbf F) = \phi \boldsymbol \nabla \boldsymbol \cdot \mathbf F + \boldsymbol \nabla \phi \boldsymbol \cdot \mathbf F$;
• $(2)$ sigue ya que $\begin{aligned}[t]&\boldsymbol \nabla \boldsymbol \cdot \mathbf r = \boldsymbol \nabla \boldsymbol \cdot (x,y,z)=3, \\ &\boldsymbol \nabla(\mathbf a \boldsymbol \cdot \mathbf r) = \boldsymbol \nabla(a_1x + a_2y+a_3y)=(a_1,a_2,a_3)=\mathbf a, \text{ y}\\ &\boldsymbol \nabla(\mathbf r \boldsymbol \cdot \mathbf r) = \boldsymbol \nabla\left(x^2+y^2+z^2\right) = (2x,2y,2z)=2\mathbf r;\end{aligned}$
• $(3)$ sigue ya que $\boldsymbol \nabla \boldsymbol \times (\mathbf F + \mathbf G) = \boldsymbol \nabla \boldsymbol \times \mathbf F + \boldsymbol \nabla \boldsymbol \times \mathbf G$ y $\boldsymbol \nabla \boldsymbol \times (\phi \mathbf F) = \phi \boldsymbol \nabla \boldsymbol \times \mathbf F + \boldsymbol \nabla \phi \boldsymbol \times \mathbf F$; y
• $(4)$ sigue ya que $\boldsymbol \nabla \boldsymbol \times \mathbf r = \boldsymbol \nabla \boldsymbol \times(x,y,z)=\mathbf 0$.