# Green's function for Stokes equation

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• steve1763

#### 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

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 $$\begin{split} P &= \frac{1}{4\pi} \frac{\mathbf{F} \cdot \mathbf{r}}{r^3} = -\frac{1}{4\pi}\mathbf{F} \cdot \nabla\left(\frac1r\right),\\ \mathbf{u} &= \frac{1}{8\pi\mu}\left(\frac{\mathbf{F}}{r} + \frac{(\mathbf{F} \cdot \mathbf{r})\mathbf{r}}{r^3}\right) = \frac{1}{8\pi \mu}\left(\frac{\mathbf{F}}{r} - (\mathbf{F} \cdot \mathbf{r}) \nabla\left(\frac1r\right)\right).\end{split}$$
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