Green's functions of Stokes flow

Question

In mathematics, a Green's function is a type of function used to solve inhomogeneous differential equations subject to specific initial conditions or boundary conditions. A fundamental solution for a linear partial differential operator L is a formulation in the language of distribution theory of the older idea of a Green's function.

In C. POZRIKIDIS's Boundary integral and singularity methods for linearized viscous flow,

The Green's functions of Stokes flow represent solutions of the continuity equation $\nabla\cdot {\bf u}=0$ and the singularly forced Stokes equation $$-\nabla P+\mu \nabla^2{\bf u}+{\bf g}\delta({\bf x-x_0})=0 $$

where ${\bf g}$ is an arbitrary constant, ${\bf x_0}$ is an arbitrary point, and $\delta$ is the three-dimensional delta function. Introducing the Green's function ${\bf G}$, we write the solution of (2.1.1) in the form $$u_i({\bf x})=\frac{1}{8\pi\mu}G_{ij}({\bf x,x_0})g_j$$

I am confused with the Green's function in this text.

Here are my questions:

• Is $p(x)$ supposed to be the unknown in the following Stokes equations?

$$ \begin{align} -\nabla p+\mu \nabla^2 u+\rho b=0\\ \nabla \cdot u =0 \end{align} $$

• What does the Green's function mean here? (Is it "with respect to" $u$?) Why is the solution of this kind of form?

• What's the Green's function in the most general case?

• What is the relation between ${\bf G}$ and $G_{ij}$? As I understand, $G_{ij}$ are the components and ${\bf G}:{\mathbb R}^3\to{\mathbb R}^3$. Then one should write: $${\bf G}({\bf x})=\left[ \begin{array}{cc} G_1({\bf x})\\G_2({\bf x})\\G_3({\bf x})\end{array}\right]$$ where $G_i:{\mathbb R}^3\to{\mathbb R}$. What is $G_{ij}$?

Answer 1

I'm not sure if the following will answer your question exactly. But here goes:

1. Yes, $p$ is the pressure of the fluid, as in the usual Stokes equations
2. The Green's function means precisely what it usually means. The equation is linear. So there exists some function $G_{ij}(x,x_0)$ and some other function $H_i(x,x_0)$ such that to solve the arbitrary Stokes equation $$ - \nabla p + \mu \nabla^2 u + f = 0 $$ with $u$ divergence free, you have (up to some normalizing constants) $$u_i(x) = \int G_{ij}(x,y)f_j(y) dy$$ and $$p(x) = \int H_i(x,y)f_i(y) dy$$ The reason the solution looks kinda funny is because you have a rotational symmetry in your problem: if you rotate the vector $g$, you should get a solution for the problem with $u$ and $p$ both also rotated. If you scale $g$, you should also get appropriately scaled versions of $u$ and $p$ as solutions. So the particular form of the listed Green's function is to capture this invariance of the solution under certain linear transformations.
3. If you are looking for the actual Green's function for the Stokes equation, they are given on the Wikipedia page you linked to. A sketch of how you obtain it starts by taking the divergence of your equation. Using that $u$ is divergence free, you get that $\nabla^2 p = g\cdot \nabla\delta(x)$ (we can assume $x_0 = 0$ by translation invariance). You can check that $$ H_i(x,x_0) = H_i(x-x_0) \qquad H_i(x,0) = \frac{1}{4\pi} \frac{x_i}{|x|^3} $$ by plugging it in to the above equation for $p$. Once you have that, you then plug $p$ back into Stokes equation, and you now have a Poisson equation for $u$, which you can solve by components and verify it to be $G_{ij}(x,0) = \mathbb{J}(x)$ where $\mathbb{J}$ is the Oseen tensor given in that Wikipedia page you linked to.

Response to Edited Question

$\mathbf{G}$ is a rank 2 tensor, NOT a vector. (Which is why it has two indices.) Like I said above, $\mathbf{G}$ is the Oseen tensor given in the Wikipedia entry that you linked to. What I wrote in (2) above explains why $\mathbf{G}$ must be a matrix/tensor-valued function, and why instead of a function on $\mathbb{R}^3\times\mathbb{R}^3$, it can be a function on $\mathbb{R}^3$ (that $\mathbf{G}(x,x_0)$ depends only on the difference $x-x_0$; this is translation invariance of the problem).

So, $\mathbf{G}$ is a function that, at every point in $\mathbb{R}^3$, evaluates to a rank-2 tensor, and so we can represent it at every point as a matrix. hence

$$ \mathbf{G} = \left( G_{ij} \right) $$

where each of $G_{ij}$ (9 components total) are functions from $\mathbb{R}^3$ to $\mathbb{R}$. Now, what are the component functions of $G_{ij}$? we have that

$$ G_{ij}(x) = \frac{1}{8\pi \mu}\left( \frac{\delta_{ij}}{|x|} + \frac{x_i x_j}{|x|^3}\right) $$

where $\delta_{ij}$ is the Kronecker delta ($=1$ if $i=j$ and $0$ elsewise).