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402    PARTIAL DIFFERENTIAL EQUATIONS
           9.1  ELLIPTIC PDE

           As an example, we will deal with a special type of elliptic equation called
           Helmholtz’s equation, which is written as

                                  2
                                ∇ u(x, y) + g(x, y)u(x, y) =
                                 2
                      2
                     ∂ u(x, y)  ∂ u(x, y)
                              +         + g(x, y)u(x, y) = f (x, y)      (9.1.1)
                       ∂x 2       ∂y 2
           over a domain D ={(x, y)|x 0 ≤ x ≤ x f ,y 0 ≤ y ≤ y f } with some boundary con-
           ditions of
                          u(x 0 ,y) = b x0 (y),  u(x f ,y) = b xf (y),
                                                                         (9.1.2)
                          u(x, y 0 ) = b y0 (x),  and u(x, y f ) = b yf (x)

           (cf) Equation (9.1.1) is called Poisson’s equation if g(x, y) = 0 and it is called Laplace’s
               equation if g(x, y) = 0and f (x, y) = 0.

              To apply the difference method, we divide the domain into M x sections, each
           of length  x = (x f − x 0 )/M x along the x-axis and into M y sections, each of
           length  y = (y f − y 0 )/M y along the y-axis, respectively, and then replace the
           second derivatives by the three-point central difference approximation (5.3.1)
               2
             ∂ u(x, y)      u i,j+1 − 2u i,j + u i,j−1
                          ∼
                          =                     with x j = x 0 + j x, y i = y 0 + i y
                ∂x 2                 x 2
                      x j ,y i
                                                                        (9.1.3a)
             2
                         ∼
            ∂ u(x, y)      u i+1,j − 2u i,j + u i−1,j
                         =                     with u i,j = u(x j ,y i )  (9.1.3b)
               ∂y 2                 y 2
                     x j ,y i
           so that, for every interior point (x j ,y i ) with 1 ≤ i ≤ M y − 1and 1 ≤ j ≤ M x − 1,
           we obtain the finite difference equation
               u i,j+1 − 2u i,j + u i,j−1  u i+1,j − 2u i,j + u i−1,j
                                  +                     + g i,j u i,j = f i,j  (9.1.4)
                        x 2                  y 2
           where
                    u i,j = u(x j ,y i ),  f i,j = f(x j ,y i ),  and g i,j = g(x j ,y i )

              These equations can somehow be arranged into a system of simultaneous
           equations with respect to the (M y − 1)(M x − 1) variables {u 1,1 ,u 1,2 ,...,u 1,M x −1 ,
           u 2,1 ,...,u 2,M x −1 ,..., u M y −1,1 ,u M y −1,2 ,... ,u M y −1,M x −1 }, but it seems to be
           messy to work with and we may be really in trouble as M x and M y become
           large. A simpler way is to use the iterative methods introduced in Section 2.5.
           To do so, we first need to shape the equations and the boundary conditions into
           the following form:
            u i,j = r y (u i,j+1 + u i,j−1 ) + r x (u i+1,j + u i−1,j ) + r xy (g i,j u i,j − f i,j )  (9.1.5a)
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