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

Passo 1

O enunciado dá o campo de velocidade
\boldsymbol{u}(x_{1},x_{2},x_{3},t) = \bigl(u_{1},u_{2},u_{3}\bigr) = \bigl(x_{1}^{2}x_{2}^{2},\,-\alpha\,t^{3},\,2x_{1}x_{3}\bigr),
com \alpha constante.
Queremos a aceleração Euleriana (também chamada local)
\boldsymbol{a}_{E} =\frac{\partial \boldsymbol{u}}{\partial t},
e a aceleração Lagrangeana (ou material)
\boldsymbol{a}_{L} =\frac{D\boldsymbol{u}}{Dt} =\frac{\partial \boldsymbol{u}}{\partial t} +(\boldsymbol{u}\cdot\nabla)\boldsymbol{u}.
Para isso precisamos da derivada parcial de \boldsymbol{u} em relação a t e do termo convectivo (\boldsymbol{u}\cdot\nabla)\boldsymbol{u}.

Passo 2

Calcular a aceleração Euleriana
\frac{\partial u_{1}}{\partial t}=0, \quad \frac{\partial u_{2}}{\partial t} =-\alpha\cdot 3\,t^{2} =-3\alpha\,t^{2}, \quad \frac{\partial u_{3}}{\partial t}=0.
Logo
\boldsymbol{a}_{E} =\bigl(0,\,-3\alpha\,t^{2},\,0\bigr).

Passo 3

Calcular o termo convectivo (\boldsymbol{u}\cdot\nabla)\boldsymbol{u}. Primeiro as derivadas parciais em relação a x_{i}:
\begin{aligned} \frac{\partial u_{1}}{\partial x_{1}} &=2\,x_{1}\,x_{2}^{2}, & \frac{\partial u_{1}}{\partial x_{2}} &=2\,x_{1}^{2}\,x_{2}, & \frac{\partial u_{1}}{\partial x_{3}} &=0,\\\ \frac{\partial u_{2}}{\partial x_{i}} &=0\quad(i=1,2,3),\\\ \frac{\partial u_{3}}{\partial x_{1}} &=2\,x_{3}, & \frac{\partial u_{3}}{\partial x_{2}} &=0, & \frac{\partial u_{3}}{\partial x_{3}} &=2\,x_{1}. \end{aligned}
Agora
(\boldsymbol{u}\cdot\nabla)u_{1} = u_{1}\frac{\partial u_{1}}{\partial x_{1}} +u_{2}\frac{\partial u_{1}}{\partial x_{2}} +u_{3}\frac{\partial u_{1}}{\partial x_{3}} = x_{1}^{2}x_{2}^{2}\cdot(2x_{1}x_{2}^{2}) +(-\alpha\,t^{3})\cdot(2x_{1}^{2}x_{2}) +2x_{1}x_{3}\cdot0
\quad =2\,x_{1}^{3}x_{2}^{4} -2\alpha\,t^{3}x_{1}^{2}x_{2},
(\boldsymbol{u}\cdot\nabla)u_{2} =0\quad(\text{pois }u_{2}\text{ independe de }x_{i}),
(\boldsymbol{u}\cdot\nabla)u_{3} = x_{1}^{2}x_{2}^{2}\cdot(2x_{3}) +(-\alpha\,t^{3})\cdot0 +2x_{1}x_{3}\cdot(2x_{1}) =2\,x_{1}^{2}x_{2}^{2}x_{3} +4\,x_{1}^{2}x_{3}^{2}.
Portanto
(\boldsymbol{u}\cdot\nabla)\boldsymbol{u} = \bigl( 2x_{1}^{3}x_{2}^{4}-2\alpha\,t^{3}x_{1}^{2}x_{2},\, 0,\, 2x_{1}^{2}x_{2}^{2}x_{3}+4x_{1}^{2}x_{3}^{2} \bigr).

Passo 4

Somando local e convectivo para obter a aceleração Lagrangeana
\boldsymbol{a}_{L} =\boldsymbol{a}_{E}+(\boldsymbol{u}\cdot\nabla)\boldsymbol{u} = \bigl(0,\,-3\alpha\,t^{2},\,0\bigr) + \bigl( 2x_{1}^{3}x_{2}^{4}-2\alpha\,t^{3}x_{1}^{2}x_{2},\, 0,\, 2x_{1}^{2}x_{2}^{2}x_{3}+4x_{1}^{2}x_{3}^{2} \bigr)
\quad = \Bigl( 2x_{1}^{3}x_{2}^{4}-2\alpha\,t^{3}x_{1}^{2}x_{2},\, -3\alpha\,t^{2},\, 2x_{1}^{2}x_{2}^{2}x_{3}+4x_{1}^{2}x_{3}^{2} \Bigr).

Resposta

Aceleração Euleriana:
\boldsymbol{a}_{E} =\bigl(0,\,-3\alpha\,t^{2},\,0\bigr).
Aceleração Lagrangeana (material):
\boldsymbol{a}_{L} =\Bigl( 2x_{1}^{3}x_{2}^{4}-2\alpha\,t^{3}x_{1}^{2}x_{2},\, -3\alpha\,t^{2},\, 2x_{1}^{2}x_{2}^{2}x_{3}+4x_{1}^{2}x_{3}^{2} \Bigr).

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