[Math] Green’s functions of Stokes flow

ap.analysis-of-pdesreference-request

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:

  1. Is $P({\bf x})$ supposed to be the unknown
    in the Stokes equations:

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

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

  3. How can one get $\frac{1}{8\pi\mu}$?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:
    $${\bf G}({\bf x})=\begin{bmatrix}
    G_1({\bf x})\\ G_2({\bf x})\\ G_3({\bf x})\end{bmatrix}$$
    where $G_i:{\mathbb R}^3\to{\mathbb R}$. What is $G_{ij}$?

  4. What's the Green's function in the most general case?

Best Answer

I moved this question to math.SE a month ago. This is indeed the problem I got from the research, though it may not very appropriate here.

@Willie Wong gave a very nice answer to the question. Instead of closing or deleting the question, I think it's worth putting the link here.