Section 5.3 Exercises, 3, 5, 7, 8 and 9 were assigned, and 7,8 were to turn in. You might also note that 5.3.1 was done in class (it is poorly worded, since it is the ODE that is Sturm-iouville, not a PDE). 5.3. Consider the wave equation: ρu tt = T 0 u xx + αu + βu t (a) What signs must α and β be to be physical? SOUTION: There are many possible ways to describe the solution- Here is one: The term αu describes a force that is proportional to displacement. For this force to be restorative, α < 0. Similarly, βu t describes a damping term (damping is typically assumed to be proportional to velocity), so this should also be a negative number. (b) To stay with the book notation for spatial functions, we ll define u = φ(x)t (t) for the separation of variables. Then (prime on φ is the derivative in x, prime on T is the derivative in t: ρ(x)φt = T 0 φ T + αφt + βφt ρ(x)φt βφt = T 0 φ T + αφt In order to remove the dependence on x from the right side of the equation, we would have to be able to factor it out- That means we need to have something of the form: ρ(x)φ(x)(t ct ) = (T 0 φ + α(x)φ(x))t Therefore, if cρ = β(x) (with c constant), then this works and following through with the separation gives: T ct T = T 0φ + αφ ρφ = λ (c) For the last part, show that the spatial equation is in S- form, and solve the time equation: For the spatial equation, we have: Comparing to S- form: T 0 φ + αφ = λρφ ( d p(x) dφ ) + q(x)φ(x) = λσ(x)φ(x) dx dx Therefore, we have S- form with p(x) = T 0, q(x) = α(x), σ(x) = ρ(x). 1
Finally, solving the time equation: T ct = λt T ct + λt = 0 is the usual second order homogeneous DE with constant coefficients. The solutions to the characteristic equation are: r = c ± c 4λ Our solution depends on the sign of the discriminant, c 4λ: If c = 4λ, then we have only one real solution, r = c/, and the general solution in time ODE is: e ct/ (C 1 + tc ) If c 4λ > 0, we have two real solutions, r = r 1, r, and the general solution is C 1 e r 1t + C e r t If c 4λ < 0, we have complex solutions r = γ ± ωi, where γ = c and the solution to the time ODE is: ω = 4λ c e γt (C 1 cos(ωt) + C sin(ωt)) 5.3.3 In this exercise, we want to change the operator into the Sturm-iouville operator using the integrating factor from class. That is, we are given: φ + αφ + (λβ + γ)φ = 0 and we want to transform it into S- form. SOUTION: We already know how to do this- et p(x) = e α(x) dx, then multiply both sides by p. The square brackets denotes where the simplification will occur: [pφ + αpφ ] + (λβp + γp)φ = 0 The expression in the brackets is the derivative of pφ (that is, p = αp), so that ( d p(x) dφ ) + γ(x)p(x)φ(x) = λβ(x)p(x)φ(x) dx dx This is now in S- form, with q(x) = γ(x)p(x) and σ(x) = β(x)p(x).
5.3.5 We did something like this in class, but will repeat here: Given φ = λφ with Newmann boundary conditions, the eigenvalues/eigenfunctions were found to be: ( ) nπ ( ) nπ λ = φ n (x) = cos x, n = 0, 1, 3, Now we consider the properties: (a) There is an infinite number of eigenvalues with a smallest but no largest. SOUTION: Check. The smallest is 0, and there is no largest. (b) The n th eigenfunction has n 1 zeros. SOUTION: The first eigenfunction is a constant, and has no zeros. This is going to throw off our count, since by the theorem: φ 1 (x) = 1 λ 1 = 0 That means that we should count: ( ) (n 1)π φ n (x) = cos x n =, 3, 4, With that note, here we go: The second eigenfunction is cos ( π x), and its only zero on [0, ] is at x = 1. The third eigenfunction is cos ( π x = 1 4 x = 3 4 The fourth eigenfunction is cos ( 3π x = 1 6 x = 3 6 = 1 x = 5 6 The fifth eigenfunction is cos ( 4π x = 1 8 x = 3 8, x = 5 8, x = 7 8 Continuing, the (k + 1) st eigenfunction is (with k fixed, x varying): ( ) kπ cos x = 0 kπ x = π, 3π, 5π,, Therefore (There are k zeros) x = 1 k, 3 k, 5 k, 7 k 1,, k k 3
(c) The eigenfunctions are complete and orthonormal: This is a result of our study of the Fourier series. Any piecewise smooth function on [0, ] can be represented by the Fourier cosine series: f(x) a n φ n (x) n=1 (with φ 1 (x) = 1, then φ and the others are cosines), and we also know that for n m from Chapter 3. 0 ( ) ( ) nπ mπ cos x cos x dx = 0 (d) We ve also looked at the Rayleigh quotient: λ n = pφ nφ n 0 + 0 p(φ n) qφ n dx 0 φ σ(x) dx For this problem, p = 1, q = 0, σ = 1, and the given boundary conditions imply the first term in the numerator is zero. Therefore, the Rayleigh quotient simplifies to: 0 λ n = (φ n) dx 0 φ dx which is always positive unless φ n = 0, or if φ n is constant (which occurs in this case for the first eigenvalue). 5.3.7* 5.3.8* 5.3.9 We re given the Cauchy-Euler equation (the text calls this equation equidimensional ): x φ + xφ = λφ φ(1) = 0 φ(b) = 0 (a) The book asks us to verify that multiplying by 1/x puts this is S- form, but we can figure it out ourselves by the technique discussed in class. First put the equation in the right form, then find the integrating factor. Therefore, the equation becomes: φ + 1 x φ = λ x φ p(x) = e 1/x dx = x xφ + φ = λ x φ (xφ ) = λ 1 x φ Therefore, this is in S- form with p(x) = x, q(x) = 0 and σ(x) = 1 x. 4
(b) For part (b), use the Rayleigh quotient (the BCs make the first term 0): λ n = 0 + b 1 x(φ n) dx b 1 φ 1 x dx Now, with 1 x b, the numerator is always positive or zero. Similarly, with the restriction on x, the denominator is always positive. Therefore, λ n 0. (c) We can solve for the eigenvalues by solving the Cauchy-Euler equation. Recall that we use φ = x r as the ansatz, and in that case, the characteristic equation of the original ODE in the problem is: r(r 1) + r + λ = 0 r = λ r = ± λ i since by the previous part, λ 0. Now, can λ = 0? In that case, φ = 0, or φ is constant. But with the boundary conditions, that implies that φ = 0. Therefore, we only have the case that λ > 0. With only positive eigenvalues, we ll continue: The solution is the made up the real and imaginary parts of x r, which is: x r = x i λ = e ln(xi λ ) = e i λ ln(x) = cos( λ ln(x)) + i sin( λ ln(x)) The general solution to the Cauchy-Euler equation is found by taking the real part and the imaginary parts as the fundamental set: φ(x) = C 1 cos( λ ln(x)) + C sin( λ ln(x)) Now incorporating the boundary values (recall that ln(1) = 0): and φ(1) = 0 C 1 = 0 φ(b) = 0 C sin( λ ln(b)) = 0 λ ln(b) = nπ λ n = with φ n (x) = sin ( ) nπ ln(b) ln(x) (d) For part (d), the easiest method is to do substitution. That is: b 1 φ n (x)φ m (x) 1 et w = ln(x), so dw = (1/x) dx x dx and x = 1 w = 0 with x = b w = ln(b) = then the integral becomes our familiar one: ( ) ( ) nπ mπ sin 0 w sin w dw = 0, n m ( ) nπ ln(b) (e) For part (e), we just make the same argument that we did in Exercise 5.3.5(b), remembering that the zeros do NOT count the endpoints x = 1 and x = b. 5