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406    PARTIAL DIFFERENTIAL EQUATIONS
           9.2  PARABOLIC PDE


           An example of a parabolic PDE is a one-dimensional heat equation describing
           the temperature distribution u(x, t) (x is position, t is time) as

                     2
                    ∂ u(x, t)  ∂u(x, t)
                  A         =             for 0 ≤ x ≤ x f ,  0 ≤ t ≤ T   (9.2.1)
                      ∂x 2       ∂t
              In order for this equation to be solvable, the boundary conditions u(0,t) =
           b 0 (t) & u(x f ,t) = b xf (t) as well as the initial condition u(x, 0) = i 0 (x) should
           be provided.


           9.2.1  The Explicit Forward Euler Method
           To apply the finite difference method, we divide the spatial domain [0,x f ]into
           M sections, each of length  x = x f /M, and divide the time domain [0,T ]into
           N segments, each of duration  t = T/N, and then replace the second partial
           derivative on the left-hand side and the first partial derivative on the right-hand
           side of the above equation (9.2.1) by the central difference approximation (5.3.1)
           and the forward difference approximation (5.1.4), respectively, so that we have

                                         k
                                u k i+1  − 2u + u k i−1  u k+1  − u k
                                         i
                               A                =   i     i              (9.2.2)
                                       x 2             t
           This can be cast into the following algorithm, called the explicit forward Euler
           method, which is to be solved iteratively:

                                                                   t
                              k
                                    k
                                                  k
                      k+1
                     u   = r(u   + u   ) + (1 − 2r)u    with r = A       (9.2.3)
                      i       i+1   i−1           i                 2
                                                                  x
                     for i = 1, 2,.. . , M − 1
              To find the stability condition of this algorithm, we substitute a trial solution
                                  k jiπ/P
                             k
                            u = λ e     (P is any nonzero integer)       (9.2.4)
                             i
           into Eq. (9.2.3) to get

                 λ = r(e jπ/P  + e −jπ/P  ) + (1 − 2r) = 1 − 2r(1 − cos(π/P ))  (9.2.5)

           Sincewemusthave |λ|≤ 1 for nondivergence, the stability condition turns out
           to be
                                             t    1
                                      r = A     ≤                        (9.2.6)
                                             x 2  2
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