Boundary-Value Problems in Other Coordinates CHAPTER 14

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

Ch14_ Problems in Polar Coordinates  Laplacian in the Polar Coordinates We already know that

Ch14_4 Thus (1) (2)

Ch14_5 Adding (1) and (2) we have

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,  ).

Ch14_7 Example 1 (2) Introducing the separation constant, we have We are seeking a solution of the form (6)

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.

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

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

Ch14_11 Example 1 (6) Applying the boundary condition at r = c, we get

Ch14_12 Example 2 Find the steady-state temperature u(r,  ) shown in Fig 14.3.

Ch14_13 Example 2 (2) Solution The boundary-value problem is

Ch14_14 Example 2 (3) and (16) (17) The boundary conditions translate into  (0) = 0 and  (  ) = 0.

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 

Ch14_16 Example 2 (5) Thus we have

Ch14_ 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 .

Ch14_18 In this case, (1) take the forms, in turn, (2) where u = u(r, t).

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.

Ch14_20 Example 1 (2) Solution The boundary-value problem is

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.

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).

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)

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).

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):

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.

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.

Ch14_28 Fig 14.8

Ch14_29 Laplacian in Cylindrical Coordinates  See Fig We have x = r cos , y = r sin , z = z and

Ch14_30 Fig 14.10

Ch14_31 Example 2  Find the steady-state temperature shown in Fig

Ch14_32 Example 2 (2) Solution The boundary conditions suggest that the temperature u has radial symmetry. Thus

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

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,

Ch14_35 Example 2 (5)

Ch14_36 Example 2 (6) For the last integral, using t =  n r and d[tJ 1 (t)]/dt = tJ 0 (t), then

Ch14_ Problems in Spherical Coordinates: Legendre Polynomials  Laplacian in Spherical Coordinates See Fig We knew that (1) and (2) We shall consider only a few of the simpler problems that are independent of the azimuthal angle .

Ch14_38 Fig 14.15

Ch14_39 Example 1  Find the steady-state temperature u(r,  ) shown in Fig

Ch14_40 Example 1 (2) Solution The problem is defined as

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, ….

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,

Ch14_43 Example 1 (5) Therefore A n c n are the coefficients of the Fourier- Legendre series (23) of Sec 12.5: