Please help evaluate this Green's problem.

  • Thread starter Thread starter yungman
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
yungman
Messages
5,741
Reaction score
291

Homework Statement



Evaluate

[tex]\int_{\Gamma} x\frac{\partial}{\partial n} G(x,y,\frac{1}{2}, \frac{1}{3}) ds[/tex]

On a unit disk region [itex]\Omega[/tex] with positive oriented boundary [itex]\Gamma[/itex]<br /> <br /> <h2>Homework Equations</h2><br /> <br /> [tex]u(x_0, y_0) = \frac{1}{2\pi}\int_{\Gamma} ( u\frac{\partial v}{\partial n} - v\frac{\partial u}{\partial n}) ds[/tex]<br /> <br /> [tex]u(x_0, y_0) = \frac{1}{2\pi}\int_{\Gamma} [ u\frac{\partial }{\partial n} G(x,y,x_0,y_0)] ds[/tex]<br /> <br /> In this case [itex]u= x \Rightarrow \nabla^2 u = 0[/itex] which means u is harmonic in [itex]\Omega[/tex]<br /> <br /> [tex]G=v + h \hbox { where }\; v=\frac{1}{2}ln[(x-x_0)^2 + (y-y_0)^2] = ln|r|[/tex]<br /> <br /> [tex]h = -v \hbox { on }\; \Gamma \hbox { and h is harmonic in } \Omega[/tex]<br /> <br /> Since v is not harmonic in [itex]\Omega [/tex] because [itex]v\rightarrow -\infty \hbox { as } (x,y) \rightarrow (x_0,y_0)[/itex]. This mean G is not harmonic.<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> I have no idea how to approach this and no idea how to find G. Please help.<br /> <br /> Thanks<br /> <br /> Alan[/itex][/itex][/itex]
 
Physics news on Phys.org
I have been reading the books over and over, the problem is the book asked this question without ever showing methods on how to solve the problem.

The two example in the book basically solve the poissons equation with normal ways of separation of variable and then put into the formula of

[tex]u(x_0,y_0) = \frac{1}{2\pi}\int_{\Omega} f(x,y) G(x,y,x_0,y_0) ds[/tex]

Then just equate Green function G.

But in this case, [itex]u(x,y)\;=\; x\;=\; r\; cos(\theta)[/itex] where u is harmonic function with boundary condition of [itex]u(1,\theta)= cos \theta[/itex]. Solving this Laplace equation with boundary condition will quickly give [itex]u(r,\theta) = r cos\theta[/itex]! Which is going nowhere.

Please help.

Alan