Please help evaluate this Green's problem.

  • Thread starter yungman
  • Start date
In summary, the conversation discusses how to evaluate the integral of x times the gradient of a Green function with respect to the normal direction on a unit disk region with a positive oriented boundary. The homework equations are provided and it is mentioned that the function u is harmonic in the region. The conversation then goes on to discuss how to find the Green function and the difficulties the speaker is facing. It is mentioned that the problem is not explained in the book and the speaker asks for help.
  • #1
yungman
5,755
293

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]

Homework Equations



[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]

[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]

In this case [itex] u= x \Rightarrow \nabla^2 u = 0 [/itex] which means u is harmonic in [itex]\Omega[/tex]

[tex]G=v + h \hbox { where }\; v=\frac{1}{2}ln[(x-x_0)^2 + (y-y_0)^2] = ln|r| [/tex]

[tex] h = -v \hbox { on }\; \Gamma \hbox { and h is harmonic in } \Omega [/tex]

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.

The Attempt at a Solution



I have no idea how to approach this and no idea how to find G. Please help.

Thanks

Alan
 
Physics news on Phys.org
  • #2
Anyone please?
 
  • #3
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
 

FAQ: Please help evaluate this Green's problem.

What is a Green's problem?

A Green's problem is a boundary value problem in mathematics that involves solving a partial differential equation subject to specified boundary conditions.

What is the purpose of evaluating a Green's problem?

The purpose of evaluating a Green's problem is to find a solution to the given partial differential equation that satisfies the specified boundary conditions.

How is a Green's problem solved?

A Green's problem is typically solved using a method called the Green's function, which involves breaking down the problem into simpler parts and using known solutions to solve for the desired solution.

What are some real-world applications of Green's problems?

Green's problems have many applications in physics, engineering, and other fields where partial differential equations are used to model physical phenomena. For example, they can be used to solve problems related to heat transfer, fluid dynamics, and electromagnetism.

Are there any limitations to using Green's problems?

While Green's problems are a powerful tool for solving certain types of partial differential equations, they do have some limitations. They may not be applicable to all types of boundary value problems, and they may be difficult to solve for more complex equations or boundary conditions.

Back
Top