Green's Formula for Laplace Eqaution : For any BC's ?

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The discussion centers on the application of Green's Function to solve the Laplace equation on a rectangle with various boundary conditions. The user A.S. inquires whether the specified boundary conditions, including a combination of Neumann and Dirichlet types, can lead to an integral equation. The response clarifies that homogeneous boundary conditions yield a trivial solution, while inhomogeneous conditions produce a non-trivial solution involving Green's functions, specifically u(x,y) = u1(x,y) + u2(x,y).

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avinashsahoo
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Hi all,
I don't know anything about Green's Function approach to solve pdes.
But I want to know if It can be used to solve a laplace eqn ,on a rectangle having all the boundary conditions of the type :


d[Psi]/dx@some x = k[Psi@some x +f(y)] .

Will it result in an Integral equation ?


Thanks,
A.S
 
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If you impose the following on the a rectangle:

[tex] u(0,y) = u(x,L_2) = u(L_1,y) = u(x,0) = 0[/tex]
You get a very uneventful solution of [tex]u = (x,y) = 0[/tex]
Once you start having inhomogeneous boundary conditions with linear combinations of Neuman and Dirichlet conditions you get interesting stuff:
[tex] \alpha_1 u(0,y) + \alpha_2 u_x(0,y)= f_1(x) \\[/tex]
[tex] \beta_1 u(L_1,y) + \beta_2 u_x(L_1,y)= f_2(x) \\[/tex]
[tex] \gamma_1 u(x,0) + \gamma_2 u_y(x,0)= g_1(x) \\[/tex]
[tex] \delta_1 u(x,L_2) + \gamma_2 u_y(x,L_2)= g_2(x) \\[/tex]
You should get u1 and u2. By linearity u(x,y)=u1(x,y)+u2(x,y)

In the end you will get a Green's function
 

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