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:

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

  • 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 $G_{ij}$?

  • $\begingroup$ Are you sure $g$ is supposed to be an arbitrary constant? Or is it an arbitrary constant vector? I ask because the terms $\nabla p$ and $\mu\nabla^2 u$ are both vector valued, so you shouldn't be able to add to a scalar. $\endgroup$ Commented Jul 8, 2011 at 20:08
  • $\begingroup$ @Willie: In the book(p.19), it is "arbitrary constant". But $g$ is in bold in the book. I think it supposed to be an arbitrary constant vector. $\endgroup$
    – user9464
    Commented Jul 8, 2011 at 20:24
  • 1
    $\begingroup$ Saying an object is an arbitrary constant doesn't necessarily specify what kind of object it is. In this case the context is sufficient to know it's a vector. $\endgroup$
    – anon
    Commented Jul 8, 2011 at 21:56
  • $\begingroup$ What is the difference between the g in Stokes equation with the delta function and the g in the solution of u with the green's function? $\endgroup$
    – Das K Sdp
    Commented Dec 11, 2012 at 9:37

1 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).

  • $\begingroup$ Thanks for your helpful answer. I only learned some undergrad level PDE in Strauss's Partial Differential Equations and some basic measure theory. I don't know if it's too early to touch such book. (Since this is a book in applied mathematics, it usually discuss the topic in an informal way.) $\endgroup$
    – user9464
    Commented Jul 8, 2011 at 21:46
  • 2
    $\begingroup$ Not quite. The book is a bit sloppy on that. That $G$ (at least as I wrote above) only recovers for you the $u$ vector field. You need a second Green's function (which I called $H$) to recover $p$. You think think of the concept of a "Green's function" as a way to invert a linear partial differential operator to get the unknown functions from the known ones. In this case, the unknowns are $(p,u)$, a scalar and a vector. The $G$ is the part which allows you to invert and find the $u$ part, but you also need $H$ to find the $p$ part. $\endgroup$ Commented Jul 9, 2011 at 1:40
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    $\begingroup$ Try Lovelock and Rund's Tensors, differential forms, and variational principles, Wrede's Introduction to Vector and Tensor Analysis, or Synge and Schild's Tensor Calculus. They have the nice benefit of being available in Dover editions: none of those three cost more tha $14 on Amazon. $\endgroup$ Commented Jul 9, 2011 at 14:23
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    $\begingroup$ @Jack: the $G$ is internally consistent (within the confines of my answer). The normalisation may, however, be slightly different from the original question. $\endgroup$ Commented Nov 16, 2011 at 9:29
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    $\begingroup$ @Jack: I think morally speaking the two procedures are the same. It is just that I started by solving the Poisson equation and appealing to the known Green's function for that, whereas in your link the Green's function for the Poisson equation is explicitly "guessed" to be $1/r$. Otherwise the methods look identical. $\endgroup$ Commented Mar 27, 2012 at 7:59

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