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Boundary-Value Problems in Other Coordinates CHAPTER 14
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Ch14_2 Contents 14.1 Problems in Polar Coordinates 14.1 Problems in Polar Coordinates 14.2 Problems in Cylindrical Coordinates 14.2 Problems in Cylindrical Coordinates 14.3 Problems in Spherical Coordinates 14.3 Problems in Spherical Coordinates
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Ch14_3 14.1 Problems in Polar Coordinates Laplacian in the Polar Coordinates We already know that
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Ch14_4 Thus (1) (2)
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Ch14_5 Adding (1) and (2) we have
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Ch14_6 Example 1 Solve Laplace’s Equation (3) subject to u(c, ) = f( ), 0 < < 2 . Solution Since (r, + 2 ) is equivalent to (r, ), we must have u(r, ) = u(r, + 2 ). If we seek a product function u = R(r) ( ), then (r, + 2 ) = (r, ).
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Ch14_7 Example 1 (2) Introducing the separation constant, we have We are seeking a solution of the form (6)
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Ch14_8 Example 1 (3) Of the three possible general solutions of (5): (7) (8) (9) we can dismiss (8) as an inherently non-periodic unless c 1 = c 2 = 0. Similarly (7) is non-periodic unless c 2 = 0. The solution = c 1 0 can be assigned any period and so = 0 is an eigenvalue.
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Ch14_9 Example 1 (4) When we take = n, n = 1, 2, …, (9) is 2 -periodic. The eigenvalues of (6) are then 0 = 0 and n = n 2, n = 1, 2, …. If we correspond 0 = 0 with n = 0, the eigenfunctions are When n = n 2, n = 0, 1, 2, … the solutions of (4) are
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Ch14_10 Example 1 (5) Note we should define c 4 = 0 to guarantee that the solution is bounded at he center of the plate (r = 0). Finally we have
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Ch14_11 Example 1 (6) Applying the boundary condition at r = c, we get
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Ch14_12 Example 2 Find the steady-state temperature u(r, ) shown in Fig 14.3.
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Ch14_13 Example 2 (2) Solution The boundary-value problem is
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Ch14_14 Example 2 (3) and (16) (17) The boundary conditions translate into (0) = 0 and ( ) = 0.
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Ch14_15 Example 2 (4) Together with (17) we have (18) The familiar problem possesses n = n 2 and eigenfunctions ( ) = c 2 sin n , n = 1, 2, … Similarly, R(r) = c 3 r n and u n = R(r) ( ) = A n r n sin n
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Ch14_16 Example 2 (5) Thus we have
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Ch14_17 14.2 Problems in Polar Coordinates and Cylindrical Coordinates: Bessel Functions Radial Symmetry The two-dimensional heat and wave equations expressed in polar coordinated are, in turn (1) where u = u(r, , t). The product solution is defined as u = R(r) ( )T(t). Here we consider a simpler problems that possesses radial symmetry, that is, u is independent of .
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Ch14_18 In this case, (1) take the forms, in turn, (2) where u = u(r, t).
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Ch14_19 Example 1 Find the displacement u(r, t) of a circular membrane of radius c clamped along its circumference if its initial displacement is f(r) and its initial velocity is g(r). See Fig 14.7.
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Ch14_20 Example 1 (2) Solution The boundary-value problem is
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Ch14_21 Example 1 (3) Substituting u = R(r)T(t) into the PDE, then (3) The two equations obtained from (3) are (4) (5) This problem suggests that we use only = 2 > 0, > 0.
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Ch14_22 Example 1 (4) Now (4) is the parametric Bessel differential equation of order v = 0, that is, rR” + R’ + 2 rR = 0. The general solution is (6) The general solution of (5) is T = c 3 cos a t + c 4 sin a t Recall that Y 0 ( r) − as r 0 + and so the implicit assumption that the displacement u(r, t) should be bounded at r = 0 forces c 2 = 0 in (6).
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Ch14_23 Example 1 (5) Thus R = J 0 ( r). Since the boundary condition u(c, t) = 0 implies R(c) = 0,we must have c 1 J 0 ( c) = 0. We rule out c 1 = 0, so J 0 ( c) = 0(7) If x n = n c are the positive roots of (7) then n = x n /c and so the eigenvalues are n = n 2 = x n 2 /c 2 and the eigenfunctions are c 1 J 0 ( n r). The product solutions are (8)
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Ch14_24 Example 1 (6) where we have done the useful relabeling of the constants. The superposition principle gives (9) Setting t = 0 in (9) and using u(r, 0) = f(r) give (10) This is recognized as the Fourier-Bessel expansion of f on the interval (0, c).
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Ch14_25 Example 1 (7) Now we have (11) Next differentiating (9) with respect to t, set t = 0, and use u t (r, 0) = g(r):
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Ch14_26 Standing Waves The solution (8) are called standing waves. For n = 1, 2, 3, …, they are basically the graph of J 0 ( n r) with the time-varying amplitude A n cos n t + B n sin n t The zeros of each standing wave in the interval (0, c) are the roots of J 0 ( n r) = 0 and correspond to the set of points of a standing wave where there is no motion. This set is called a nodal line.
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Ch14_27 As in Example 1, the zeros of standing waves are determined from J 0 ( n r) = J 0 (x n r/c) = 0 Now from Table 5.2 and for n = 1, the first positive root of J 0 (x 1 r/c) = 0 is 2.4r/c = 2.4 or r = c Since the desired interval is (0, c), the last result has no nodal line. For n = 2, the roots of J 0 (x 2 r/c) = 0 are 5.5r/c = 2.4 and 5.5r/c = 5.5 We have r = 2.4c/5.5 that has one nodal line. See Fig 14.8.
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Ch14_28 Fig 14.8
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Ch14_29 Laplacian in Cylindrical Coordinates See Fig 14.10. We have x = r cos , y = r sin , z = z and
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Ch14_30 Fig 14.10
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Ch14_31 Example 2 Find the steady-state temperature shown in Fig 14.11.
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Ch14_32 Example 2 (2) Solution The boundary conditions suggest that the temperature u has radial symmetry. Thus
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Ch14_33 Example 2 (3) Using u = R(r)Z(z) and separation constant, (13) (14) (15) For the choice = 2 > 0, > 0, the solution of (14) is R(r) = c 1 J 0 ( r) + c 2 Y 0 ( r) Since the solution of (15) is defined on [0, 2], we have Z(z) = c 3 cosh z + c 4 sinh z
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Ch14_34 Example 2 (4) As in Example 1, the assumption that u is bounded at r = 0 demands c 2 = 0. The condition u(2, z) = 0 implies R(2) = 0. Then J 0 (2 ) = 0 (16) defines the eigenvalues n = n 2. Last, Z(0) = 0 implies c 3 = 0. Hence we have R(r) = c 1 J 0 ( r), Z(z) = c 4 sinh z,
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Ch14_35 Example 2 (5)
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Ch14_36 Example 2 (6) For the last integral, using t = n r and d[tJ 1 (t)]/dt = tJ 0 (t), then
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Ch14_37 14.3 Problems in Spherical Coordinates: Legendre Polynomials Laplacian in Spherical Coordinates See Fig 14.15. We knew that (1) and (2) We shall consider only a few of the simpler problems that are independent of the azimuthal angle .
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Ch14_38 Fig 14.15
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Ch14_39 Example 1 Find the steady-state temperature u(r, ) shown in Fig 14.16.
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Ch14_40 Example 1 (2) Solution The problem is defined as
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Ch14_41 Example 1 (3) and so (2) (3) After letting x = cos , 0 , (3) becomes (4) This is a form of Legendre’s equation. Now the only solutions of (4) that are continuous and have continuous derivatives on [-1, 1] are the Legendre polynomials P n (x) corresponding to 2 = n(n+1), n = 0, 1, 2, ….
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Ch14_42 Example 1 (4) Thus we take the solutions of (3) to be = P n (cos ) When = n(n + 1), the solution of (2) is R = c 1 r n + c 2 r –(n+1) Since we again expect u to be bounded at r = 0, we define c 2 = 0. Hence,
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Ch14_43 Example 1 (5) Therefore A n c n are the coefficients of the Fourier- Legendre series (23) of Sec 12.5:
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