Lecture IX. Definition 1 A non-singular Sturm 1 -Liouville 2 problem consists of a second order linear differential equation of the form.

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1 Lecture IX Abstract When solving PDEs it is often necessary to represent the solution in terms of a series of orthogonal functions. One way to obtain an orthogonal family of functions is by solving a particular type of boundary value problem for a second order linear ordinary differential equation, called Sturm-Liouville boundary value problem. Sturm-Liouville boundary value problems Till now you dealt with initial value problems (IVPs), that is an ordinary differential equation together with an initial condition for the unknown function and its first derivative at a given point. Before we start to solve some PDEs by means of the method of separation of variables we need to introduce the concept of boundary value problem (BVP), that is is a differential equation together with a set of additional constraints specifying for example the unknown function and its first derivative at two different points. A large class of important boundary value problems are the Sturm-Liouville problems. Definition 1 A non-singular Sturm 1 -Liouville problem consists of a second order linear differential equation of the form d ( p(x) ) + q(x)y = λw(x)y (1) dx dx such that 1. p, its first derivative and w are continuous functions everywhere on the interval [a, b],. q is at least piecewise continuous on [a, b], 3. p and w are both positive on [a, b] together with the following homogeneous unmixed boundary conditions c 1 y(a) + c dx = 0, c 3 y(b) + c 4 x=a dx = 0. () x=b 1 Jacques Charles Francois Sturm ( ) was a French mathematician of German heritage. Joseph Liouville ( ) was a French mathematician. 1

2 There exist other types of Sturm-Liouville problems where the equation has one or more singular points, or the boundary conditions are mixed or periodic but the present definition is general enough to encompass most when not all PDEs we will be solving. Notice that in the previous definition λ is a parameter which is not a priori given and one of the major tasks in a Sturm-Liouville problem is to find λ such that (1) and () admit a nontrivial solution, that is a solution that does not vanish everywhere on the interval [a, b]. It can be shown that non-trivial solutions of a Sturm- Liouville problem exist only if the parameter λ belongs to a certain set of real numbers λ 0, λ 1,, λ n, called eigenvalues of the particular Sturm- Liouville problem. In particular, one can also show that there is a smallest eigenvalue λ 1 with λ 0 < λ 1 < < λ n < and lim λ n =. n When λ = λ n the corresponding solution y(x) = Φ n (x) of equation (1) is called an eigenfunction corresponding to the eigenvalue λ n and each Φ x (x) is unique up to a constant multiple of it. The following result presented in the form of a theorem is very important since it ensures that the set of all eigenfunctions of a certain Sturm-Liouville problem is an orthogonal set ans as such it allows to write the solution of equation (1) as a series expansion on the eigenfunctions, that is y(x) = c n Φ n (x). n=0 Theorem 1 The infinite family of eigenfunctions S = {Φ 0 (x), Φ 1 (x),, Φ n (x), } of the Sturm-Liouville problem (1) with boundary conditions () is an orthogonal family of functions on the interval [a, b] with respect to the weight function w(x) in (1), that is b a dx w(x)φ n (x)φ m (x) = 0 if m n. In view of the above result we discover that there exists another possibility to show that the Legendre polynomials are orthogonal. We just need to show that the Legendre equation can be rewritten as a Sturm-Liouville problem on the interval [ 1, 1]. The orthogonality of the Legendre polynomials will follow from the above theorem.

3 Example 1 Show that the differential equation + λy = 0 (3) dx together with the boundary conditions y(0) = 0, dx = 0 (4) x=1 is a Sturm-Liouville problem on [0, 1] and find all eigenvalues and a corresponding set of orthogonal eigenfunctions. Comparing the ODE in (3) with (1) it is not difficult to see that (3) is a special case of (1) with p(x) = 1 = w(x), q(x) = 0. The boundary conditions (4) correspond to those in () if a = 0, b = 1, c 1 = 1 = c 4, c = 0 = c 3. Thus, we conclude that (3) and (4) represent a Sturm-Liouville problem. To find the eigenvalues we have to determine λ such that (3) has no trivial solution. To find all non-trivial solutions one usually follows the steps summarized here below 1. Find the general solution of (3).. Determine for which values of λ the given boundary conditions are satisfied. I f we try a guess solution in the form e rx we obtain the characteristic equation r + λ = 0. There will be three different cases depending on whether the roots of the above equation are 1. real and unequal (λ < 0),. real and equal (λ = 0), 3. complex conjugate (λ > 0). 3

4 1. Case λ < 0 Then, from the characteristic equation r = λ > 0 and we have two unequal roots ± λ. The general solution of (3) will be a linear combination of exponential functions Taking into account that y(x) = c 1 e λx + c e λx. dx = ( λ c 1 e ) λx c e λx, it is straightforward to verify that the boundary conditions (4) require that c 1 = c = 0. This means that the only solution of (3) with λ < 0 satisfying the boundary conditions (4) is the trivial solution y(x) = 0 x [0, 1] and therefore, there exist no negative eigenvalues.. Case λ = 0 There is a double root at r = 0 and the general solution is y(x) = c 1 e 0 x + c xe 0 x = c 1 + c x. Taking into account that dx = c, the boundary conditions (4) implies that c 1 = c = 0. Again, the only solution is the zero solution. Therefore, λ = 0 is not an eigenvalue. 3. Case λ > 0 We have two complex conjugate roots ±i λ and the general solution of (3) is y(x) = c 1 cos ( λx) + c sin ( λx). If we apply the first condition in (3) we obtain y(0) = c c 0 = c 1 = 0 since y(0) = 0. Thus, the solution simplifies to y(x) = c sin ( λx). 4

5 Applying the second condition in (4) to dx = c λ cos ( λx) we end up with the equation c λ cos ( λ) = 0. (5) Since in the present case λ cannot vanish we can divide the above equation by λ. Moreover, in order to obtain a non trivial solution of (3) the constant c should be taken different from zero. Thus, (5) becomes cos ( λ) = 0. (6) Recall that the equation cos α = 0 is satisfied whenever α = ± π, ±3π + 1)π,, ±(n, n N. Taking into account that in the present case α = λ > 0, we conclude that (6) is satisfied for that is λ = π, 3π (n + 1)π,,, n N, λ = π 4, 9π 4,, (n + 1) π, n N. 4 We conclude that the eigenvalues λ n are given by λ n = (n + 1) π, n N. 4 The corresponding eigenfunctions are ( ) (n + 1)π Φ n (x) = c n sin x and Theorem 1 ensures that the set ( π ) ( ) ( 3π (n + 1)π S = { sin x, sin x,, sin ) x, } is an orthogonal set. Finally, the solution of (3) satisfying (4) can be written as an expansion on the eigenfunctions Φ n (x) as follows ( ) (n + 1)π y(x) = c n sin x. n=0 Notice that the orthogonal set we derived in the previous example is not the family of functions we used to generate the trigonometric Fourier series! 5

6 Practice problems 1. Find all eigenvalues and corresponding eigenfunctions for the Sturm- Liouville problems below (a) d y + λy = 0, y(0) = 0, y(π) = 0. dx (b) d y dx + λy = 0, dx = 0, y(l) = 0. x=0 (c) d y dx + λy = 0, dx = 0, x=0 dx = 0. x=π. Let n N. Show that 1 0 dx sin ( (n + 1)π ) x = For the Sturm-Liouville problem d y + λy = 0, y(0) = 0, y(1) + dx dx = 0 x=1 (a) Show that there are no negative eigenvalues. (b) Show that λ = 0 is not an eigenvalue. (c) Show that the positive eigenvalues are of the form λ n = K n where K n is the n-th solution of the equation tan(k n ) = K n and that the corresponding eigenfunctions are Φ n (x) = c n sin (K n x). (d) Find numerical values for λ 1, λ and λ 3 and the graph of the corresponding three eigenfunctions Φ 1 (x), Φ (x) and Φ 3 (x). (e) Find the numerical value of the integral 1 What should it equal? Why? 0 dx Φ 1 (x)φ (x). 6