Jim Lambers MAT 415/515 Fall Semester 2013-14 Lecture 10 Notes These notes correspond to Sections 8.1 and 8.2 in the text. Introduction to Sturm-Liouville Theory We have learned various techniques for solving certain ODEs, and in a first course in differential equations, such ODEs are generally accompanied by one or more initial conditions. However, many ODEs that arise naturally in physics are accompanied with boundary conditions, in which the value of the solution, or some derivative of the solution, is specified on the boundary of the domain on which the ODE is defined. The ODE, together with the boundary conditions, is called a boundary value problem. Furthermore, the ODEs in such boundary value problems often have a parameter that is only allowed to assume certain values in order to obtain solutions that satisfy the boundary conditions. Such an ODE generally has the form L(x)y(x) = λy(x); where L is a differential operator, and λ is a parameter. If a solution to this ODE exists that also satisfies the given boundary conditions, then λ is called an eigenvalue of the operator L(x), and the accompanying solution y(x) is called an eigenfunction. The ODE itself is called an eigenvalue problem. Example Consider the eigenvalue problem d2y + λ2y = 0; 0 < x < `; dx2 with boundary conditions y(0) = 0; y(`) = 0: This problem arises from modeling the standing waves of a vibrating string of length ` that is clamped at its ends. The solution y(x) represents the amplitude of the string at the point x. This ODE can be written as Ly = λ2y; where L = −d2=dx2. The values of λ2 for which solutions exist that also satisfy the boundary conditions are eigenvalues of L. Using standard techniques for solving second-order linear ODEs with constant coefficients, we can obtain the general solution of the ODE, y(x) = A cos λx + B sin λx, where A and B are arbitrary constants. Substituting x = 0 and x = ` into y(x) and imposing the boundary conditions, we obtain the equations y(0) = 0 = A; y(`) = 0 = A cos λ` + B sin λ`, 1 which yields a solution of the form y(x) = B sin λx, where λ` = kπ, for some integer k, to ensure that the boundary condition y(`) = 0 is satisfied. It follows that for each positive integer k, the function kπx y (x) = B sin k ` is a solution of the eigenvalue problem. We do not consider negative values of k, as they yield solutions that are scalar multiples of those obtained from positive values, and are therefore not linearly independent. The constant B is arbitrary, because the ODE and boundary conditions are homogeneous. However, we can impose the normalization condition kyk(x)k = 1; where the norm is induced by the scalar product Z ` hfjgi = f ∗(x)g(x) dx: 0 This yields the one-parameter family of solutions r2 kπx y (x) = sin ; k = 1; 2;:::: k ` ` These are the eigenfunctions of L, with corresponding eigenvalues kπ 2 λ2 = ; k = 1; 2;:::: k ` It is worth noting that the eigenvalues are real and positive. We will see that in general, a Hermitian operator such as L always has real eigenvalues. 2 Hermitian Operators We consider a general linear second-order differential operator of the form d2 d L(x) = p (x) + p (x) + p (x): 0 dx2 1 dx 2 We are interested in obtaining eigenvalues and eigenfunctions of such an operator. The study of these operators is known as Sturm-Liouville theory. Suppose L is Hermitian with respect to an appropriate scalar product, and that y(x) is an eigenfunction of L with corresponding eigenvalue λ. Then hyjLyi = hyjλyi = λkyk2; but we also have hyjLyi = hLyjyi = hyjLyi∗: It follows that hyjLyi is real, and therefore λ itself is real. 2 Self-Adjoint ODEs When is L Hermitian? We assume boundary conditions are imposed at x = a and x = b, and use the scalar product Z b hfjgi = f ∗(x)g(x) dx: a Then, using integration by parts, we obtain 2 d u du hvjLui = v p0 + p1 + p2u dx2 dx 2 d u du = v p0 + v p1 + hv jp2ui dx2 dx 2 d u du = p0v + p1v + hp2v jui dx2 dx 2 ∗ 0 ∗ 0 b d = p0(x)v (x)u (x) − (p0(x)v (x)) u(x) + (p0v) u + a dx2 ∗ b d [p1(x)v (x)u(x)]j − (p1v) u + hp2v jui a dx ∗ 0 ∗ 0 0 ∗ b = p0(x)v (x)u (x) − p0(x)(v ) (x)u(x) + (p1(x) − p0(x))v (x)u(x) a + 2 2 d v dp0 dv d p0 dp1 p0 + 2 − p1 + − + p2 v u : dx2 dx dx dx2 dx We assume that each function u(x) in the underlying Hilbert space H satisfies boundary conditions of the form u(a) = 0; u(b) = 0; known as Dirichlet boundary conditions, or of the form u0(a) = 0; u0(b) = 0; known as Neumann boundary conditions. It follows that all of the boundary terms vanish if 0 p0(x) = p1(x): If this condition is satisfied, then we obtain 2 d v dv hvjLui = p0 + p1 + p2v u = hLvjui; dx2 dx and since hvjLui = hLyjui as well, we conclude that L is Hermitian, or self-adjoint. It is essential to note that whether an operator is self-adjoint depends on both the scalar product being used, and the boundary conditions. It is also worth noting that if u and v are eigenfunctions of a self-adjoint operator L with eigenvalues λu and λv, respectively, then hvjLui = hvjλuui = λuhvjui; but because L is self-adjoint, and therefore has real eigenvalues, we also have hvjLui = hLvjui = hλvvjui = λvhvjui: 3 It follows that (λv − λu)hujvi = 0; and therefore we can conclude that eigenfunctions of a self-adjoint operator corresponding to distinct eigenvalues are orthogonal. Making an ODE Self-Adjoint Suppose that an operator L is not self-adjoint. We consider whether we can scale both sides of the eigenvalue problem Ly = λy by a function w(x) so that it becomes self-adjoint. Then we consider the modified operator d2 d w(x)L(x) = w(x)p (x) + w(x)p (x) + w(x)p (x): 0 dx2 1 dx 2 For it to be self-adjoint, we must have 0 (w(x)p0(x)) = w(x)p1(x); or 0 0 w (x)p0(x) = w(x)[p1(x) − p0(x)]: This is a first-order, linear, separable ODE that has the solution Z p (x) p0 (x) Z p (x) 1 Z p (x) w(x) = exp 1 − 0 dx = e− ln p0(x) exp 1 dx = exp 1 dx : p0(x) p0(x) p0(x) p0(x) p0(x) It follows from the fact that wL is self-adjoint that Z b Z b hvjwLui = v∗(x)w(x)Lu(x) dx = u(x)w(x)Lv∗(x) dx = hwLvjui; a a and therefore L is self-adjoint with respect to the scalar product Z b hfjgi = f ∗(x)g(x)w(x) dx: a Using this scalar product, it can be shown that as before, the eigenvalues of L are real, and eigen- functions corresponding to distinct eigenvalues are orthogonal, with respect to this scalar product. Example Consider the eigenvalue problem Ly = λy, where d2 d L = x + (1 − x) : dx2 dx 0 This operator is not self-adjoint, as p0(x) = x and p1(x) = 1 − x, so p0(x) 6= p1(x). We can make it self-adjoint using the weight function R p (x) 1 1 dx 1 R 1−x dx 1 R 1 −1 dx 1 ln x−x −x w(x) = e p0(x) = e x = e x = e = e : p0(x) x x x If we work with the domain 0 ≤ x < 1, with boundary condition limx!1 u(x) = 0, then the eigenfunctions are orthogonal with respect to the scalar product Z 1 hfjgi = e−xf ∗(x)g(x) dx; 0 and they are known as the Laguerre polynomials. 2 4.