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18.3 Dirichlet Problem for a Disk  645


                            6. Apply separation of variables to the problem  8. Solve for the steady-state temperature distribution in
                                       2
                                     ∇ u = 0for0 < x < a,0 < y < b,          a homogeneous, thin, flat plate covering the rect-
                                                                             angle 0 ≤ x ≤ a,0 ≤ y ≤ b if the temperature
                                           ∂u                                on the vertical and lower sides are kept at zero
                                   u(x,0) =  (x,b) = 0for 0 ≤ x ≤ a,
                                           ∂y                                and the temperature along the top side is f (x) =
                                                                                   2
                                                                             x(x − a) .
                                   u(0, y) = 0,u(a, y) = g(y) for 0 ≤ y ≤ b.
                            7. Use separation of variables to solve        9. Solve for the steady-state temperature distribution in
                                                                             a thin, flat plate covering the rectangle 0 ≤ x ≤
                                       2
                                     ∇ u = 0for0 < x < a,0 < y < b,
                                                                             4, 0 ≤ y ≤ 1 if the temperature on the horizontal
                                   u(x,0) = 0,u(x,b) = f (x) for 0 ≤ x ≤ a,  sides is zero while the temperature on the left side
                                                                             is f (y) = sin(πy) and on the right side, g(y) =
                                          ∂u
                                   u(0, y) =  (a, y) = 0for 0 ≤ y ≤ b.       y(1 − y).
                                          ∂x
                            18.3        Dirichlet Problem for a Disk

                                        We will solve the Dirichlet problem for a disk of radius R centered at the origin. The boundary
                                        of this disk is the circle x + y = R . Using polar coordinates, the problem for u(r,θ) is
                                                                2
                                                                    2
                                                            2
                                                                           2
                                                             2
                                                            ∂ u  1 ∂u   1 ∂ u
                                                        2
                                                      ∇ u =    +      +      = 0for0 ≤r < R,−π ≤ θ ≤ π
                                                            ∂r  2  r ∂r  r ∂θ  2
                                                                         2
                                        and
                                                                 u(R,θ) = f (θ) for − π ≤ θ ≤ π.
                                                                                        n
                                                                           n
                                           It is easy to check that the functions 1,r cos(nθ), and r sin(nθ) are harmonic on the entire
                                        plane. Thinking ahead to the possibility of a Fourier series to satisfy the boundary condition,
                                        attempt a solution in a series of these functions:
                                                                        ∞
                                                                  1          n            n
                                                          u(r,θ) = a 0 +  (a n r cos(nθ) + b n r sin(nθ)).
                                                                  2
                                                                        n=1
                                        This would require that
                                                                           ∞
                                                                     1
                                                                                 n
                                                                                              n
                                                      u(R,θ) = f (θ) = a 0 +  (a n R cos(nθ) + b n R sin(nθ)).
                                                                     2
                                                                          n=1
                                                                                                                n
                                        This is a Fourier expansion of f (θ) on [−π,π] if we choose the entire coefficients, a n R and
                                           n
                                        b n R to be the Fourier coefficients of f on [−π,π]. This means that
                                                                           1     π
                                                                      a 0 =     f (ξ)dξ
                                                                          π  −π
                                        and, for n = 1,2,···,
                                                          1     π                    1     π
                                                     a n =      f (ξ)cos(nξ)dξ and b n =   f (ξ)sin(nξ)dξ.
                                                         R n  −π                    R n  −π
                                        Another form of this solution is
                                                   1     π
                                           u(r,θ) =     f (ξ)dξ
                                                   π  −π
                                                     1     r       π                        π
                                                       ∞     n
                                                  +                 f (ξ)cos(nξ)dξ cos(nθ) +  f (ξ)sin(nξ)dξ sin(nθ) .
                                                     π     R     −π                       −π
                                                       n=1
                                                                                                                (18.1)

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