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2-d problem with Neumann boundary conditions
Let us redo the above calculation, replacing the Dirichlet boundary conditions (149)
with the following simple Neumann boundary conditions:
![\begin{displaymath}
\frac{\partial u(x,y=0)}{\partial y} = \frac{\partial u(x,y=L)}{\partial y} = 0.
\end{displaymath}](img761.png) |
(161) |
In this case, we can express
in the form
![\begin{displaymath}
u(x,y) = \sum_{j=0}^{\infty} U_j(x)\,\cos(j\,\pi\,y/L),
\end{displaymath}](img762.png) |
(162) |
which automatically satisfies the boundary conditions in the
-direction. Likewise,
we can write the source term
as
![\begin{displaymath}
v(x,y) = \sum_{j=0}^{\infty} V_j(x)\,\cos(j\,\pi\,y/L),
\end{displaymath}](img764.png) |
(163) |
where
![\begin{displaymath}
V_j(x) = \frac{2}{L} \int_0^L v(x,y)\,\cos(j\,\pi\,y/L)\,dy,
\end{displaymath}](img765.png) |
(164) |
since
![\begin{displaymath}
\frac{2}{L}\int_0^L \cos(j\,\pi\,y/L)\,\cos(k\,\pi\,y/L)\,dy = \delta_{jk}.
\end{displaymath}](img766.png) |
(165) |
Finally, the boundary conditions in the
-direction become
![\begin{displaymath}
\alpha_l\,U_j(x) + \beta_l\,\frac{dU_j(x)}{dx} = \Gamma_{l\,j},
\end{displaymath}](img742.png) |
(166) |
at
, and
![\begin{displaymath}
\alpha_h\,U_j(x)+ \beta_h\,\frac{dU_j(x)}{dx} = \Gamma_{h\,j},
\end{displaymath}](img743.png) |
(167) |
at
, where
![\begin{displaymath}
\Gamma_{l\,j} = \frac{2}{L} \int_0^L \gamma_l(y)\,\cos(j\,\pi\,y/L)\,dy,
\end{displaymath}](img767.png) |
(168) |
etc. Note, however, that the factor in front of the integrals in Eqs. (164)
and (168) takes the special value
for the
harmonic.
As before, we truncate the Fourier expansion in the
-direction, and discretize in the
-direction, to obtain the set of tridiagonal matrix equations specified in Eqs. (158), (159), and (160). We can solve these equations to obtain the
, and then reconstruct
the
from Eq. (162). Hence, we have solved the problem.
Next: The fast Fourier transform
Up: Poisson's equation
Previous: 2-d problem with Dirichlet
Richard Fitzpatrick
2006-03-29