Green's function for 2-D Laplacian within square/rectangular boundary

[Swamp Thing]

From the table of Green functions on Wikipedia we can get the generic 2-D Green's function for the Laplacian operator. But how would one apply boundary conditions like u = 0 along a rectangular boundary? Would we visualize a sort of rectangle-based, tilted pyramid, with logarithmically changing height, with its apex located at our Dirac driving point?

 

[pasmith]

Generally in that case one would look for an eigenfunction expansion, with the eigenfunctions satisfying [tex]
\nabla^2 \phi_{nm} = \lambda_{nm}\phi_{nm}[/tex] subject to [itex]\phi_{nm} = 0[/itex] on the boundary. These turn out to be products of trigonometric functions which are orthogonal with respect to the inner product [tex]
\langle f, g \rangle = \int \int f(x,y)g(x,y)\,dx\,dy[/tex] and the solution of [itex]\nabla^2 u = f[/itex] is [itex]u = \sum_{n,m} a_{nm} \phi_{nm}[/itex] where [tex]
\lambda_{nm}a_{nm}\|\phi_{nm}\|^2 = \langle f, \phi_{nm} \rangle.[/tex]

 

[pasmith]

By setting [itex]f(x,y) = \delta(x - s)\delta(y - t)[/itex] we can recover the Green's function [tex]
G(x,y;s,t) = \sum_{n,m} \frac{\phi_{nm}(s,t)\phi_{nm}(x,y)}{\lambda_{nm}\|\phi_{nm}\|^2}.[/tex] and indeed [tex]
u(x,y) = \iint_A G(x,y;s,t)f(s,t)\,ds\,dt = \sum_{n,m} \frac{\langle f, \phi_{nm}\rangle}{\lambda_{nm}\|\phi_{nm}\|^2}[/tex] by swapping summation and integration.

 

[Swamp Thing]

Thanks for the help!

I am new to this, so maybe this is a silly question...

I am trying this in Mathematica:

X = 10; Y = 10;
gf = GreenFunction[{-Laplacian[u[x, y], {x, y}],
   DirichletCondition[u[x, y] == 0, True]},
  u[x, y], {x, y} \[Element] Rectangle[{0, 0}, {X, Y}], {m, n}]
Plot3D[gf /. {m -> 0.7 X, n -> 0.3 Y} /. {\[Infinity] -> 50} //
   Activate // Evaluate, {x, 0, X}, {y, 0, Y}, PlotRange -> All]

Which gives this expression for the Green Function:
$$
\frac{1}{25} \underset{K[1]=1}{\overset{\infty }{\sum }}\underset{K[2]=1}{\overset{\infty }{\sum }}\frac{\sin \left(\frac{1}{10} m \pi K[1]\right) \sin \left(\frac{1}{10} \pi x K[1]\right) \sin \left(\frac{1}{10} n \pi K[2]\right) \sin \left(\frac{1}{10} \pi y K[2]\right)}{\frac{1}{100} \pi ^2 K[1]^2+\frac{1}{100} \pi ^2 K[2]^2}
$$

I am not sure about which symbols in Mathematica's formula correspond to which ones in your post. I am guessing, for example, that your m's and n's correspond to the K[1] and K[2] in the above formula? I have figured out that the m and n in the above correspond to the coordinates of the point excitation (Dirac function) but I am not sure of some of the other symbols in your post.

 

[pasmith]

It's best not to use [itex]m[/itex] and [itex]n[/itex] as real-valued quanties; by convention they are used as integer variables and doing otherwise will confuse people.

For your particular boundary conditions, the eigenfunctions are [tex]
\phi_{nm}(x,y) = \sin\left(\frac{n\pi x}{10}\right)\sin\left(\frac{m\pi y}{10}\right)[/tex] which has [itex]\|\phi_{nm}\| = 5[/itex] and the corresponding eigenvalues are [tex]
\lambda_{nm} = -\frac{\pi^2(n^2 + m^2)}{100}.[/tex] The result given by Mathematica differs by a sign compared to mine, but that is explained by you asking it to solve [itex]-\nabla^2 u = \delta(x - x_0)[/itex] rather than [itex]\nabla^2 \phi = \delta(x - x_0)[/itex].

 

[Swamp Thing]

Yes, I think I better change the m and n to something else, like ##(x_d,y_d)## as a mnemonic that it's where the Dirac delta is located. The code I posted is from the examples in the Wolfram help page for GreenFunction.

Quick question about the usual workflow when doing this kind of problem from first principles: does the Fourier analysis help to determine the Green's function, OR, is the G.F. found by other means, after which you Fourier-analyze it in order to then get the time evolution?

Thanks again.