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| − | <math> \left\{\begin{array}{l}
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| − | \vec{u}_t + (\vec{u} \cdot \nabla) \vec{u} = -\nabla p +\displaystyle\frac{1}{Re}\nabla^2 \vec{u} \\
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| − | \nabla \cdot u = 0\\
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| − | \end{array}
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| − | \right. </math>
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| − | Vorticidade: <math>\zeta = \nabla \times \vec{u} \cdot \vec{k}</math>
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| − | Função corrente <math> \psi </math>: <math>\vec{u} = (\nabla\psi)\times k</math>
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| − | Assim
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| − | <math>\nabla\cdot\vec{u} = \nabla \cdot(\nabla\psi\times k) = \frac{\partial}{\partial x}\left( \frac{\partial \psi}{\partial y} \right) + \frac{\partial}{\partial y}\left( -\frac{\partial \psi}{\partial x} \right) =0 </math>
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| − | Identidade 1) <math> (\vec{u}\cdot\nabla)\vec{u} = (\nabla \times \vec{u})\times \vec{u} + \nabla\left( \frac{1}{2}||\vec{u}||^2 \right) = \zeta k \times \vec{u} + \nabla\left( \frac{1}{2}||\vec{u}||^2 \right) </math>
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| − | Identidade 2) <math> \nabla \times (\vec{a} \times \vec{b}) = \vec{a} \nabla \cdot \vec{b} - \vec{b} \nabla \cdot \vec{a} + (\vec{b} \cdot \nabla)\vec{a} - (\vec{a} \cdot \nabla)\vec{b} </math>
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| − | Aplicando <math> \nabla \times </math> na equação
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| − | <math> (\nabla \times \vec{u})_t + \nabla \times \{ (\vec{u} \cdot \nabla)\vec{u} \} = \frac{1}{Re}\nabla^2(\nabla \times \vec{u}) </math>
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| − | Usando a identidade (1)
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| − | <math> (\nabla \times \vec{u})_t + \nabla \times (\zeta \vec{k} \times \vec{u}) = \frac{1}{Re}\nabla^2(\nabla \times \vec{u}) </math>
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| − | <math> (\zeta\vec{k})_t + \nabla\times(\zeta\vec{k}\times\vec{u}) = \frac{1}{Re}\nabla^2(\zeta\vec{k}) </math>
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| − | Da identidade (2) com <math> \vec{a}=\nabla\times\vec{u} </math> e <math> \vec{b}=\vec{u} </math> e usando que <math> \nabla\cdot\vec{u}=0 </math>, temos
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| − | <math> \nabla\times(\zeta\vec{k}\times\vec{u}) = \nabla \times \{(\nabla\times\vec{u})\times\vec{u}\} = (\vec{u}\cdot\nabla)\nabla\times\vec{u} - (\nabla\times\vec{u}\cdot\nabla)\vec{u} </math>
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| − | <math> \vec{u} = \vec{u}(x,y,t) \Longrightarrow (\nabla\times\vec{u}\cdot\nabla)\vec{u}=0 </math>
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| − | pois <math> \vec{u} </math> não depende de <math> z </math> .
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| − | Segue-se que
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| − | <math> \nabla\times(\zeta\vec{k}\times\vec{u}) = (\vec{u}\cdot\nabla)\nabla\times\vec{u} = (\vec{u}\cdot\nabla)\zeta\vec{u} </math>
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| − | Portanto
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| − | <math> (\zeta\vec{k})_t + (\vec{u}\cdot\nabla)\zeta\vec{k} = \frac{1}{Re}\nabla^2(\zeta\vec{k}) </math>
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| − | <math> \vec{u} = \nabla\psi\times\vec{k} \Longrightarrow (\vec{u}\cdot\nabla)\zeta\vec{k} = \frac{\partial(\zeta,\psi)}{\partial(x,y)}\vec{k} = J(\zeta,\psi)\vec{k} </math>
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| − | Então
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| − | <math>(\zeta\vec{k})_t + J(\zeta,\psi)\vec{k} = \frac{1}{Re}\nabla^2(\zeta\vec{k}) </math>
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| − | Além disso, como <math> \vec{u} = \nabla\psi\times\vec{k} </math> e <math> \zeta = \nabla\times\vec{u}\cdot\vec{k} </math> temos
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| − | <math> \nabla\times(\nabla\psi\times\vec{k})=\nabla\times(\psi_y,-\psi_x) = -\psi_{xx}-\psi_{yy} = -\nabla^2\psi </math>
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| − | Ou seja
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| − | <math>-\nabla^2\psi = \zeta </math>
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| − | E as equações ficam
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| − | <math> \left\{\begin{array}{l}
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| − | \zeta_t + J(\zeta,\psi) = \displaystyle\frac{1}{Re}\nabla^2\zeta \\
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| − | -\nabla^2\psi = \zeta \\
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| − | \end{array}
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| − | \right. </math>
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