Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
Differential Equations
Green's function for Stokes equation
Reply to thread
Message
[QUOTE="pasmith, post: 6610351, member: 415692"] Pressure is a scalar quantity; it doesn't need a bold symbol. The Green's function is used to decompose a non-homogenous linear problem [itex]L(u) = f[/itex] into a superposition of problems [itex]L(G_{x_0}) = \delta(x - x_0)[/itex] from which it follows that [tex]u(x) = \int f(x_0) G_{x_0}(x)\,dx_0.[/tex] If we want the solution for a point source of strength [itex]k[/itex] at [itex]x_1[/itex], then we get [tex] u(x) = \int k\delta(x_0 - x_1)G_{x_0}(x)\,dx_0 = kG_{x_1}(x)[/tex] and we are looking for a constant multiple of the Green's function [itex]G_{x_1}[/itex]. To solve [itex]L(G_{x_0}) = \delta(x - x_0)[/itex], we treat it as a homogenous problem [itex]L(G_{x_0}) = 0[/itex] subject to appropriate conditions on [itex]G_{x_0}[/itex] and its derivatives. In this case, we want [itex]P[/itex] and [itex]\mathbf{u}[/itex] to be linear in [itex]\mathbf{F}[/itex] and we want [itex]-\nabla P + \mu\nabla^2 \mathbf{u}[/itex] to be proportional to [itex]\nabla^2(1/r)[/itex] with [tex] \int_{r \leq a} \nabla P - \mu \nabla^2 \mathbf{u}\,dV = \mathbf{F}.[/tex] That leads us to [tex]\begin{split} P &= \frac{1}{4\pi} \frac{\mathbf{F} \cdot \mathbf{r}}{r^3} = -\frac{1}{4\pi}\mathbf{F} \cdot \nabla\left(\frac1r\right),\\ \mathbf{u} &= \frac{1}{8\pi\mu}\left(\frac{\mathbf{F}}{r} + \frac{(\mathbf{F} \cdot \mathbf{r})\mathbf{r}}{r^3}\right) = \frac{1}{8\pi \mu}\left(\frac{\mathbf{F}}{r} - (\mathbf{F} \cdot \mathbf{r}) \nabla\left(\frac1r\right)\right).\end{split}[/tex] [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
Differential Equations
Green's function for Stokes equation
Back
Top