8- Dado o seguinte campo de velocidade:
\boldsymbol{u}=\left(x_{1}^{2} x_{2}^{2},-\alpha t x_{3}, 2 x_{1} x_{3}\right)
Onde \alpha é um valor constante, calcule as aceleraçōes Euleriana e Lagrangeana deste escoamento.

Passo 1

O enunciado dá o campo de velocidade
\boldsymbol{u}(x_{1},x_{2},x_{3},t)=\bigl(x_{1}^{2}x_{2}^{2},\,-\alpha\,t\,x_{3},\,2\,x_{1}\,x_{3}\bigr),
com \alpha constante. Precisamos calcular:
  1. A aceleração Euleriana (ou local) \displaystyle \frac{\partial \boldsymbol{u}}{\partial t}.
  2. A aceleração Lagrangeana (ou material) \displaystyle \frac{D\boldsymbol{u}}{Dt}=\frac{\partial \boldsymbol{u}}{\partial t}+(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}.

Passo 2

Calcular a derivada parcial em relação ao tempo de cada componente de \boldsymbol{u}:
\frac{\partial \boldsymbol{u}}{\partial t} =\Bigl( \frac{\partial}{\partial t}(x_{1}^{2}x_{2}^{2}),\, \frac{\partial}{\partial t}(-\alpha\,t\,x_{3}),\, \frac{\partial}{\partial t}(2\,x_{1}\,x_{3}) \Bigr) =(0,\,-\alpha\,x_{3},\,0).

Passo 3

Calcular o termo convectivo (\boldsymbol{u}\cdot\nabla)\boldsymbol{u}. Vale lembrar
(\boldsymbol{u}\cdot\nabla)\boldsymbol{u} =\bigl(u_{1}\partial_{x_{1}}+u_{2}\partial_{x_{2}}+u_{3}\partial_{x_{3}}\bigr)\,(u_{1},u_{2},u_{3}).
Fazendo componente a componente:
  1. Componente 1:
u_{1}=x_{1}^{2}x_{2}^{2},\quad \partial_{x_{1}}u_{1}=2x_{1}x_{2}^{2},\quad \partial_{x_{2}}u_{1}=2x_{1}^{2}x_{2},\quad \partial_{x_{3}}u_{1}=0.
Logo
(u\cdot\nabla)u_{1} =x_{1}^{2}x_{2}^{2}\,2x_{1}x_{2}^{2} +(-\alpha t x_{3})\,2x_{1}^{2}x_{2} +2x_{1}x_{3}\cdot0 =2x_{1}^{3}x_{2}^{4}-2\alpha t\,x_{1}^{2}x_{2}x_{3}.
  1. Componente 2:
u_{2}=-\alpha t x_{3},\quad \partial_{x_{1}}u_{2}=0,\quad \partial_{x_{2}}u_{2}=0,\quad \partial_{x_{3}}u_{2}=-\alpha t.
Logo
(u\cdot\nabla)u_{2} =2x_{1}x_{3}\,(-\alpha t) =-2\alpha t\,x_{1}x_{3}.
  1. Componente 3:
u_{3}=2x_{1}x_{3},\quad \partial_{x_{1}}u_{3}=2x_{3},\quad \partial_{x_{2}}u_{3}=0,\quad \partial_{x_{3}}u_{3}=2x_{1}.
Logo
(u\cdot\nabla)u_{3} =x_{1}^{2}x_{2}^{2}\,2x_{3} +2x_{1}x_{3}\,2x_{1} =2x_{1}^{2}x_{2}^{2}x_{3}+4x_{1}^{2}x_{3} =2x_{1}^{2}x_{3}\bigl(x_{2}^{2}+2\bigr).

Passo 4

Agora somamos as parcelas local e convectiva para obter a aceleração material:
\frac{D\boldsymbol{u}}{Dt} =\frac{\partial \boldsymbol{u}}{\partial t} +(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}
=\Bigl( 0,\,-\alpha x_{3},\,0 \Bigr) + \Bigl( 2x_{1}^{3}x_{2}^{4}-2\alpha t\,x_{1}^{2}x_{2}x_{3},\, -2\alpha t\,x_{1}x_{3},\, 2x_{1}^{2}x_{3}(x_{2}^{2}+2) \Bigr).

Resposta

A aceleração Euleriana (local) é
\frac{\partial \boldsymbol{u}}{\partial t} =(0,\,-\alpha\,x_{3},\,0),
e a aceleração Lagrangeana (material) é
\frac{D\boldsymbol{u}}{Dt} =\bigl( 2x_{1}^{3}x_{2}^{4}-2\alpha t\,x_{1}^{2}x_{2}x_{3},\, -\alpha x_{3}-2\alpha t\,x_{1}x_{3},\, 2x_{1}^{2}x_{3}(x_{2}^{2}+2) \bigr).

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