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In very low Reynolds number flows, the Navier-Stokes equations can be reduced to the Stokes equation which is given by \begin{gather} \mu \nabla^2 \textbf{u} = \nabla p \end{gather}

I am looking at analytic solutions to the Stokes equation and I find that even though the governing equation has neglected inertial effects, I can substitute the velocity field into terms like $(\textbf{u} \cdot \nabla)\textbf{u}$ and find that they are non-zero.

How should one interpret $(\textbf{u} \cdot \nabla)\textbf{u} \neq 0$ in a flow that is analytically determined by solving a governing equation that has neglected inertial effects?

A specific example I am considering is the corner flow problem discussed by Batchelor in section 4.8 pp. 224-226. The two-dimensional Stokes flow in a corner is given by \begin{align} \psi &= rf(\theta) = r[A\sin(\theta)+B\cos(\theta)+C\theta\sin(\theta)+D\theta\cos(\theta)]\\ u_r &= \dfrac{1}{r}\dfrac{\partial \psi}{\partial \theta} = f'(\theta)\\ u_\theta &= -\dfrac{\partial \psi}{\partial r} = f(\theta) \end{align}

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The full (dimensionless) equation for steady flow is: $$Re(\mathbf{u}\cdot\nabla\mathbf{u})=-\nabla p+\nabla^2\mathbf{u}$$ where $Re$ is Reynolds number. Stokes flow is obtained in the limit $Re\to 0$. If the physical quantities appearing in the equation are properly non-dimensionalized then $\mathbf{u}\cdot\nabla\mathbf{u}$ remains an order one quantity as $Re\to 0$ (in other words $\mathbf{u}\cdot\nabla\mathbf{u}$ remains finite as $Re\to 0$), so in that limit LHS of the above equation becomes zero. To put it simply, even though $\mathbf{u}\cdot\nabla\mathbf{u}\neq 0$, LHS is set to zero because by definition $Re\to 0$ for Stokes flow.

Physically $\mathbf{u}\cdot\nabla\mathbf{u}$ is the acceleration of a fluid element as it moves (along the flow) from one point in space to another. Stokes flow equation says that inertial forces on the fluid element due to this acceleration is negligible in comparison to viscous forces on it.

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