- #1

steve1763

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- TL;DR Summary
- How do I reach the expression for pressure?

So I've just started learning about Greens functions and I think there is some confusion. We start with the Stokes equations in Cartesian coords for a point force.

$$-\nabla \textbf{P} + \nu \nabla^2 \textbf{u} + \textbf{F}\delta(\textbf{x})=0$$

$$\nabla \cdot \textbf{u}=0$$

We can apply the second relation to the first to get

$$- \nabla^2 \textbf{P} + \textbf{F} \cdot \nabla \delta(\textbf{x})=0 $$

$$\nabla^2 \textbf{P} = \textbf{F} \cdot \nabla \delta(\textbf{x}) $$

The Greens function for the 3D laplacian, according to Wikipedia, is $$-1 \over{4 \pi r}$$ where $$r=(x^2+y^2+z^2)^{1 \over 2}$$ Generally, with Greens functions

$$u(x)= \int G(x,s) f(s) ds$$

So would we get something like

$$\textbf{P}= {-1\over {4 \pi}} \int {1 \over r} \textbf{F} \cdot \nabla \delta(\textbf{x}) ds$$

I believe you might have to convert $$\delta(\textbf{x})={1\over{4 \pi r^2}}\delta(r)$$ I'm a little confused as to where to go next and how to deal with an integral that might have delta functions in it.

Thank you

$$-\nabla \textbf{P} + \nu \nabla^2 \textbf{u} + \textbf{F}\delta(\textbf{x})=0$$

$$\nabla \cdot \textbf{u}=0$$

We can apply the second relation to the first to get

$$- \nabla^2 \textbf{P} + \textbf{F} \cdot \nabla \delta(\textbf{x})=0 $$

$$\nabla^2 \textbf{P} = \textbf{F} \cdot \nabla \delta(\textbf{x}) $$

The Greens function for the 3D laplacian, according to Wikipedia, is $$-1 \over{4 \pi r}$$ where $$r=(x^2+y^2+z^2)^{1 \over 2}$$ Generally, with Greens functions

$$u(x)= \int G(x,s) f(s) ds$$

So would we get something like

$$\textbf{P}= {-1\over {4 \pi}} \int {1 \over r} \textbf{F} \cdot \nabla \delta(\textbf{x}) ds$$

I believe you might have to convert $$\delta(\textbf{x})={1\over{4 \pi r^2}}\delta(r)$$ I'm a little confused as to where to go next and how to deal with an integral that might have delta functions in it.

Thank you