Please help evaluate this Green's problem.

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The discussion centers on evaluating the integral of the Green's function for a unit disk region, specifically the expression \int_{\Gamma} x\frac{\partial}{\partial n} G(x,y,\frac{1}{2}, \frac{1}{3}) ds. The user, Alan, struggles with understanding how to derive the Green's function G and its implications for the harmonic function u(x,y) = x. The discussion highlights that while u is harmonic in the region \Omega, the Green's function G is not due to the singularity at (x_0, y_0).

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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]
 
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Anyone please?
 
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
 

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