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The formula is from the first paragraph in the paper "Second Kind Integral Equation Formulation of Stokes' Flows Past a Particle of Arbitrary Shape" by Power and Miranda:

... the governing equations for the auxiliary perturbed fluid velocity $\vec{v}=(v_1,v_2,v_3)$ and pressure $p$ can be approximated by the creeping motion and continuity equations $$ \begin{align} &\frac{\partial^2 v_i}{\partial x_j\partial x_j}(x)=\frac{\partial p}{\partial x_i}(x)\\ &\frac{\partial v_i}{\partial x_i}(x)=0 \end{align} $$

Here are my questions:

  • How should I understand the first formula? Is the Einstein summation convention applied here, i.e., $$ \sum_j\frac{\partial^2 v_i}{\partial x_j\partial x_j}(x)=\frac{\partial p}{\partial x_i}(x)? $$

  • Is there any reference for the derivation of the first formula? (I don't find any in the paper.)

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en.wikipedia.org/wiki/Creeping_motion Einstein summation convention is indeed implied. In other words, the equation can be written as $$ \nabla ^2\mathbf{v}=\mathbf{\nabla}p. $$ As for the derivation, according to the article, it follows directly from the Navier-Stokes equations, but as I don't know what they mean by ". . . the inertial forces are assumed to be negligible . . .", I'm afraid I can't explain this. –  Jonathan Gleason Nov 20 '11 at 0:25
    
@JonathanGleason, Thanks for your comment. It is the answer I'm looking for. When I read your $\nabla^2 v=\nabla p$, things are clear to me. May I suggest you to write it down as an answer? –  Jack Nov 20 '11 at 1:14
    
The second formula in the question, I think, means $\nabla\cdot v=0$, instead of $\nabla v=0$. –  Jack Nov 20 '11 at 1:18
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1 Answer

up vote 1 down vote accepted

See the Wikipedia article Creeping Motion.

Einstein summation convention is indeed implied. In other words, the equation can be written as $\nabla ^2\mathbf{v}=\mathbf{\nabla}p$.

As for the derivation, according to the article, it follows directly from the Navier-Stokes equations, but as I don't know what they mean by ". . . the inertial forces are assumed to be negligible . . .", so I'm afraid I can't explain this, although perhaps this will help.

Note: The original reason I did not want to post this as an answer was because I don't know how to completely answer your second point. In any case, if you are happy with this answer, then I am certainly happy to post it as one :).

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