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?
  • 1
    $\begingroup$ Yes, p is the pressure in the Stokes equation. Yes, the Green's function given here is for representing the velocity. Why is its form strange, and what is "the most general case"? $\endgroup$ Jun 3 '11 at 17:34
  • $\begingroup$ @Michael Renardy: Thanks for the comment. Actually I don't understand how can one get $\frac{1}{8\pi\mu}$ in the solution formula. $\endgroup$
    – Jack
    Jun 3 '11 at 17:46
  • $\begingroup$ I think it is just a matter of how you define $G_{ij}$. $\endgroup$ Jun 3 '11 at 18:57

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.