Green's functions of Stokes flow
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 $G_{ij}$?
I'm not sure if the following will answer your question exactly. But here goes:
- Yes, $p$ is the pressure of the fluid, as in the usual Stokes equations
- 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.
- 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).