Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 8 doc

We have the implicit boundary conditions, u(0, z) = u(2π, z), u θ (0, z) = u θ (0, z) and the boundedness condition, u(θ, +∞) bounded. We expand the solution in a Fourier series. (This ensures that the boundary conditions at θ = 0, 2π are satisfied.) u(θ, z) = ∞  n=−∞ u n (z) e ınθ We substitute the series into the partial differential equation to obtain ordinary differential equations for the u n . −n 2 u n (z) + u  n (z) = 0 The general solutions of this equation are u n (z) =  c 1 + c 2 z, for n = 0, c 1 e nz +c 2 e −nz for n = 0. The bounded solutions are u n (z) =      c e −nz , for n > 0, c, for n = 0, c e nz , for n < 0, = c e −|n|z . We substitute the series into the initial condition at z = 0 to determine the multiplicative constants. u(θ, 0) = ∞  n=−∞ u n (0) e ınθ = f(θ) u n (0) = 1 2π  2π 0 f(θ) e −ınθ dθ ≡ f n 1854 Thus the solution is u(θ, z) = ∞  n=−∞ f n e ınθ e −|n|z . Note that u(θ, z) → f 0 = 1 2π  2π 0 f(θ) dθ as z → +∞. Solution 40.8 The decomposition of the problem is shown in Figure 40.1. w =g (x) w=0∆ w=g (x) w=f (y) w=f (y) 2 1 2 y 1 u =g (x) u=0∆ u=g (x) u=0 v=0∆ v=f (y) v=f (y) = + u=0 v=0 v =0 2 y 1 1 2 y Figure 40.1: Decomposition of the problem. First we solve the problem for u. u xx + u yy = 0, 0 < x < a, 0 < y < b, u(0, y) = u(a, y) = 0, u y (x, 0) = g 1 (x), u(x, b) = g 2 (x) We substitute the separation of variables u(x, y) = X(x)Y (y) into Laplace’s equation. X  X = − Y  Y = −λ 2 1855 We have the eigenvalue problem, X  = −λ 2 X, X(0) = X(a) = 0, which has the solutions, λ n = nπ a , X n = sin  nπx a  , n ∈ N. The equation for Y (y) becomes, Y  n =  nπ a  2 Y n , which has the solutions,  e nπy/a , e −nπy/a  or  cosh  nπy a  , sinh  nπy a  . It will be convenient to choose solutions that satisfy the conditions, Y (b) = 0 and Y  (0) = 0, respectively.  sinh  nπ(b −y) a  , cosh  nπy a   The solution for u(x, y) has the form, u(x, y) = ∞  n=1 sin  nπx a   α n sinh  nπ(b −y) a  + β n cosh  nπy a   . We determine the coefficients from the inhomogeneous b oundary conditions. (Here we see how our choice of solutions 1856 for Y (y) is convenient.) u y (x, 0) = ∞  n=1 − nπ a α n sin  nπx a  cosh  nπb a  = g 1 (x) α n = − a nπ sech  nπb a  2 a  a 0 g 1 (x) sin  nπx a  dx u(x, y) = ∞  n=1 β n sin  nπx a  cosh  nπy a  β n = sech  nπb a  2 a  a 0 g 2 (x) sin  nπx a  dx Now we solve the problem for v. v xx + v yy = 0, 0 < x < a, 0 < y < b, v(0, y) = f 1 (y), v(a, y) = f 2 (y), v y (x, 0) = 0, v(x, b) = 0 We substitute the separation of variables u(x, y) = X(x)Y (y) into Laplace’s equation. X  X = − Y  Y = λ 2 We have the eigenvalue problem, Y  = −λ 2 Y, Y  (0) = Y (b) = 0, which has the solutions, λ n = (2n −1)π 2b , Y n = cos  (2n −1)πy 2b  , n ∈ N. The equation for X(y) becomes, X  n =  (2n −1)π 2b  2 X n . 1857 We choose solutions that satisfy the conditions, X(a) = 0 and X(0) = 0, respectively.  sinh  (2n −1)π(a −x) 2b  , sinh  (2n −1)πx 2b  The solution for v(x, y) has the form, v(x, y) = ∞  n=1 cos  (2n −1)πy 2b  γ n sinh  (2n −1)π(a −x) 2b  + δ n sinh  (2n −1)πx 2b  . We determine the coefficients from the inhomogeneous boundary conditions. v(0, y) = ∞  n=1 γ n cos  (2n −1)πy 2b  sinh  (2n −1)πa 2b  = f 1 (y) γ n = csch  (2n −1)πa 2b  2 b  b 0 f 1 (y) cos  (2n −1)πy 2b  dy v(a, y) = ∞  n=1 δ n cos  (2n −1)πy 2b  sinh  (2n −1)πa 2b  = f 2 (y) δ n = csch  (2n −1)πa 2b  2 b  b 0 f 2 (y) cos  (2n −1)πy 2b  dy With u and v determined, the solution of the original problem is w = u + v. 1858 Chapter 41 Waves 1859 41.1 Exercises Exercise 41.1 Consider the 1-D wave equation u tt − u xx = 0 on the domain 0 < x < 4 with initial displacement u(x, 0) =  1, 1 < x < 2 0, otherwise, initial velocity u t (x, 0) = 0, and subject to the following boundary conditions 1. u(0, t) = u(4, t) = 0 2. u x (0, t) = u x (4, t) = 0 In each case plot u(x, t) for t = 1 2 , 1, 3 2 , 2 and combine onto a general plot in the x, t plane (up to a sufficiently large time) so the behavior of u is clear for arbitrary x, t. Exercise 41.2 Sketch the solution to the wave equation: u(x, t) = 1 2 (u(x + ct, 0) + u(x −ct, 0)) + 1 2c  x+ct x−ct u t (τ, 0) dτ, for various values of t corresponding to the initial conditions: 1. u(x, 0) = 0, u t (x, 0) = sin ωx where ω is a constant, 1860 2. u(x, 0) = 0, u t (x, 0) =      1 for 0 < x < 1 −1 for − 1 < x < 0 0 for |x| > 1. Exercise 41.3 1. Consider the solution of the wave equation for u(x, t): u tt = c 2 u xx on the infinite interval −∞ < x < ∞ with initial displacement of the form u(x, 0) =  h(x) for x > 0, −h(−x) for x < 0, and with initial velocity u t (x, 0) = 0. Show that the solution of the wave equation satisfying these initial conditions also solves the following semi-infinite problem: Find u(x, t) satisfying the wave equation u tt = c 2 u xx in 0 < x < ∞, t > 0, with initial conditions u(x, 0) = h(x), u t (x, 0) = 0, and with the fixed end condition u(0, t) = 0. Here h(x) is any given function with h(0) = 0. 2. Use a similar idea to explain how you could use the general solution of the wave equation to solve the finite interval problem (0 < x < l) in which u(0, t) = u(l, t) = 0 for all t, with u(x, 0) = h(x) and u t (x, 0) = 0. Take h(0) = h(l) = 0. Exercise 41.4 The deflection u(x, T ) = φ(x) and velocity u t (x, T ) = ψ(x) for an infinite string (governed by u tt = c 2 u xx ) are measured at time T , and we are asked to determine what the ini tial displacement and velocity profiles u(x, 0) and u t (x, 0) must have been. An alert student suggests that this problem is equivalent to that of determining the solution of the wave equation at time T when initial conditions u(x, 0) = φ(x), u t (x, 0) = −ψ(x) are prescribed. Is sh e correct? If not, can you rescue her idea? 1861 Exercise 41.5 In obtaining the general solution of the wave equation the interval was chosen to be infinite in order to simplify the evaluation of the functions α(ξ) and β(ξ) in the general solution u(x, t) = α(x + ct) + β(x − ct). But this general solution is in fact valid for any interval be it infinite or finite. We need only choose appropriate functions α(ξ), β(ξ) to satisfy the appropriate initial and boundary conditions. This is not always convenient but there are other situations besides the solution for u(x, t) in an infinite domain in which the general solution is of use. Consider the “whip-cracking” problem, u tt = c 2 u xx , (with c a constant) in the domain x > 0, t > 0 with initial conditions u(x, 0) = u t (x, 0) = 0 x > 0, and boundary conditions u(0, t) = γ(t) prescribed for all t > 0. Here γ(0) = 0. Find α and β so as to determine u for x > 0, t > 0. Hint: (From physical considerations conclude that you can take α(ξ) = 0. Your solution will corroborate this.) Use the initial conditions to determine α(ξ) and β(ξ) for ξ > 0. Then use the initial condition to determine β(ξ) for ξ < 0. Exercise 41.6 Let u(x, t) satisfy the equation u tt = c 2 u xx ; (with c a constant) in some region of the (x, t) plane. 1. Show that the quantity (u t − cu x ) is constant along each straight line defined by x − ct = constant, and that (u t + cu x ) is constant along each straight line of the form x + ct = constant. These straight lines are called characteristics; we will refer to typical members of the two families as C + and C − characteristics, respectively. Thus the line x −ct = constant is a C + characteristic. 1862 [...]... ut (x, 0) = −1 for − 1 < x < 0   0 for |x| > 1 ut (x, 0) = −H(x + 1) + 2H(x) − H(x − 1) 187 1 1 0 .5 u 0 -0 .5 -1 6 4 t -5 2 0 x 5 0 Figure 41.4: Solution of the wave equation We integrate the Heaviside function  0 for b < c  b H(x − c) dx = b − a for a > c  a  b − c otherwise If a < b, we can express this as b H(x − c) dx = min(b − a, max(b − c, 0)) a 187 2 Now we find an expression for the solution... ∞) 13 1 − x− − 8n 2 2 , ut (x, 0) = 0 We use D’Alembert’s solution to solve this problem ∞ 1 H u(x, t) = 2 n=−∞ 1 3 − x − − 8n − t 2 2 −H +H 1 3 − x − − 8n + t 2 2 1 13 − x− − 8n − t 2 2 −H 1 13 − x− − 8n + t 2 2 The solution for several times is plotted in Figure 41.1 Note that the solution is periodic in time with period 8 Figure 41.3 shows the solution in the phase plane for 0 < t < 8 Note the odd... x ∈ (−∞ ∞), 3 1 − x − − 8n 2 2 +H t ∈ (0 ∞) 1 13 − x− − 8n 2 2 , ut (x, 0) = 0 We use D’Alemberts solution to solve this problem ∞ 1 u(x, t) = H 2 n=−∞ 1 3 − x − − 8n − t 2 2 +H +H 1 3 − x − − 8n + t 2 2 1 13 − x− − 8n − t 2 2 186 9 +H 1 13 − x− − 8n + t 2 2 The solution for several times is plotted in Figure 41.2 Note that the solution is periodic in time with period 8 Figure 41.3 shows the solution... speed? 18 65 41.2 Hints Hint 41.1 Hint 41.2 Hint 41.3 Hint 41.4 Hint 41 .5 From physical considerations conclude that you can take α(ξ) = 0 Your solution will corroborate this Use the initial conditions to determine α(ξ) and β(ξ) for ξ > 0 Then use the initial condition to determine β(ξ) for ξ < 0 Hint 41.6 Hint 41.7 a) Substitute u(x, t) = (A eıωt−αx ) into the partial differential equation and solve for. .. ωyear be the frequency for annual temperature variation, then ωday = 3 65 year If xyear is the depth that a particular yearly temperature variation reaches and xday is the depth that this same variation in daily temperature reaches, then ωyear ωday exp − xyear = exp − xday , 2κ 2κ 188 0 ωyear ωday xyear = xday , 2κ 2κ xyear √ = 3 65 xday Solution 41 .8 We seek a periodic solution of the form, u(r, θ, t) =... single partial differential equation for I We differentiate the two partial differential equations with respect to x and t, respectively and then eliminate the Vxt terms −Ixx = CVtx + GVx , −Vxt = LItt + RIt −Ixx + LCItt + RCIt = GVx 18 85 We use the initial set of equations to write Vx in terms of I −Ixx + LCItt + RCIt + G(LIt + RI) = 0 Itt + RC + GL GR 1 It + I− Ixx = 0 LC LC LC Now we derive a single partial... ct − 1, 0)) u(x, t) = Figure 41 .5 shows the solution for c = 1 1 0 .5 u 0 -0 .5 -1 -4 3 2 t -2 1 0 x 2 4 0 Figure 41 .5: Solution of the wave equation 187 3 Solution 41.3 1 The solution on the interval (−∞ ∞) is 1 u(x, t) = (h(x + ct) + h(x − ct)) 2 Now we solve the problem on (0 ∞) We define the odd extension of h(x) ˆ h(x) = h(x) for x > 0, = sign(x)h(|x|) −h(−x) for x < 0, Note that d ˆ h (0− )... satisfies the wave equation vτ τ = c2 vxx The initial conditions for v are v(x, 0) = u(x, T ) = φ(x), vτ (x, 0) = −ut (x, T ) = −ψ(x) Thus we see that the student was correct 18 75 Direct Solution D’Alembert’s solution is valid for all x and t We formally substitute t − T for t in this solution to solve the problem with deflection u(x, T ) = φ(x) and velocity ut (x, T ) = ψ(x) u(x, t) = 1 1 (φ(x + c(t − T... u(x, t) = 1 1 (u(x − ct, 0) + u(x + ct, 0)) + 2 2c 187 8 x+ct ut (τ, 0) dτ x−ct Solution 41.7 a) We substitute u(x, t) = A eıωt−αx into the partial differential equation and take the real part as the solution We assume that α has positive real part so the solution vanishes as x → +∞ ıωA eıωt−αx = κα2 A eıωt−αx ıω = κα2 α = (1 + ı) ω 2κ A solution of the partial differential equation is, u(x, t) = A exp ıωt... the solution is periodic in time with period 8 Figure 41.3 shows the solution in the phase plane for 0 < t < 8 Note the even reflections at the boundaries 1 0 .8 0.6 0.4 0.2 1 0 .8 0.6 0.4 0.2 1 2 3 4 1 0 .8 0.6 0.4 0.2 1 2 3 4 1 2 3 4 1 0 .8 0.6 0.4 0.2 1 2 3 4 Figure 41.2: The solution at t = 1/2, 1, 3/2, 2 for the boundary conditions ux (0, t) = ux (4, t) = 0 Solution 41.2 1 u(x, t) = 1 1 (u(x + ct, 0) . =      1 for 0 < x < 1 −1 for − 1 < x < 0 0 for |x| > 1. u t (x, 0) = −H(x + 1) + 2H(x) −H(x −1) 187 1 -5 0 5 x 0 2 4 6 t -1 -0 .5 0 0 .5 1 u -5 0 5 x Figure 41.4: Solution of the. the phase plane for 0 < t < 8. Note the even reflections at the boundaries. 1 2 3 4 0.2 0.4 0.6 0 .8 1 1 2 3 4 0.2 0.4 0.6 0 .8 1 1 2 3 4 0.2 0.4 0.6 0 .8 1 1 2 3 4 0.2 0.4 0.6 0 .8 1 Figure 41.2:. −1)πa 2b  2 b  b 0 f 2 (y) cos  (2n −1)πy 2b  dy With u and v determined, the solution of the original problem is w = u + v. 1 85 8 Chapter 41 Waves 1 85 9 41.1 Exercises Exercise 41.1 Consider the 1-D