Green's function for Stokes equation

[steve1763]

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

 

[pasmith]

Pressure is a scalar quantity; it doesn't need a bold symbol.

The Green's function is used to decompose a non-homogenous linear problem $L(u) = f$ into a superposition of problems $L(G_{x_0}) = \delta(x - x_0)$ from which it follows that $$u(x) = \int f(x_0) G_{x_0}(x)\,dx_0.$$ If we want the solution for a point source of strength $k$ at $x_1$, then we get $$u(x) = \int k\delta(x_0 - x_1)G_{x_0}(x)\,dx_0 = kG_{x_1}(x)$$ and we are looking for a constant multiple of the Green's function $G_{x_1}$.

To solve $L(G_{x_0}) = \delta(x - x_0)$, we treat it as a homogenous problem $L(G_{x_0}) = 0$ subject to appropriate conditions on $G_{x_0}$ and its derivatives.

In this case, we want $P$ and $\mathbf{u}$ to be linear in $\mathbf{F}$ and we want $-\nabla P + \mu\nabla^2 \mathbf{u}$ to be proportional to $\nabla^2(1/r)$ with $$\int_{r \leq a} \nabla P - \mu \nabla^2 \mathbf{u}\,dV = \mathbf{F}.$$ That leads us to $$\begin{split} P &= \frac{1}{4\pi} \frac{\mathbf{F} \cdot \mathbf{r}}{r^3} = -\frac{1}{4\pi}\mathbf{F} \cdot \nabla\left(\frac1r\right),\\ \mathbf{u} &= \frac{1}{8\pi\mu}\left(\frac{\mathbf{F}}{r} + \frac{(\mathbf{F} \cdot \mathbf{r})\mathbf{r}}{r^3}\right) = \frac{1}{8\pi \mu}\left(\frac{\mathbf{F}}{r} - (\mathbf{F} \cdot \mathbf{r}) \nabla\left(\frac1r\right)\right).\end{split}$$