Introduction to Sturm-Liouville Theory and the Theory of Generalized Fourier Series

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1 CHAPTER 5 Introduction to Sturm-Liouville Theory and the Theory of Generalized Fourier Series We start with some introductory examples. 5.. Cauchy s equation The homogeneous Euler-Cauchy equation (Leonhard Euler and Augustin-Louis Cauchy is a linear homogeneous ODE which can be written as (* x 2 y + axy + by =. Example 5.. Solve the equation (*. Solution: Set y(x = x r, then y (x = rx r and y (x = r(r x r 2. If we insert this into (* we get which gives us the equation r(r x r + arx r + bx r =, (** r(r + ar + b =. This is the so-called Characteristic equation corresponding to (*. Assume that the solutions of (** are r and r 2. We have three different cases:. If r and r 2 are real and different, r r 2 then y(x = Ax r + Bx r If r and r 2 are real and equal, r = r 2 = r then y(x = Ax r + Bx r lnx. 3. If r and r 2 are complex conjugates, r = α + iβ, r 2 = α iβ then REMARK 6. Observe that and y(x = Ax α+iβ + Bx α iβ. x α+iβ = x α e iβlnx = x α (cos(βlnx + isin(βlnx x α iβ = x α (cos(βlnx isin(βlnx, hence we can write the solution of case 3 in the example above as y(x = x α ((A + Bcos(βlnx + i(a Bsin(βlnx. 27

2 28 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES If we only consider constants A and B such that C = A + B and D = i(a B are real numbers then we see that y(x = x α (C cos(βlnx + Dsin(βlnx is a real-valued solution to (*. Example 5.2. i.e. Solve the differential equation Solution: The characteristic equation is which has the solutions x 2 y + 2xy 6y =. r(r + 2r 6 =, r 2 + r 6 = r = 2, r 2 = 3. Since we have two different real solutions we are in case above, and the general solution to the differential equation is given by y(x = Ax 2 + Bx 3. Example 5.3. Solve the equation x 2 y + 2xy + λy =, λ > 4. Solution: The characteristic equation is r 2 + r + λ =, with solutions r = λ 2 ± 4 λ = 2 ± i 4. Since we now have two complex conjugate solutions r = α + iβ and r = α iβ we are in case 3 above, and the general solution to the differential equation is given by ( y(x = Ax 2 sin ( λ 4 lnx + Bcos ( λ 4 lnx Examples of Sturm-Liouville Problems In the next section we will describe in more details what is meant by a Sturm-Liouville problem (Charles- Fran cois Sturm and Joseph Liouville, but first we will look at some examples. Example 5.4. Solve y + λy =, y( = y(l =.

3 5.2. EXAMPLES OF STURM-LIOUVILLE PROBLEMS 29 Solution: Previously (cf. section 4.8, p. 22 we saw that this problem can be solved if and only if ( nπ 2 λ = λ n =, n =,2,3,...(eigenvalues l with the corresponding solutions ( nπ y n (x = a n sin l x (eigenfunctions. Example 5.5. Solve X (x λx(x =, x, X( =, X ( = 3X(. Solution: We have three different cases: λ = λ > λ < X(x = Ax + B, X( = B =, and X ( = 3X( A = 3A A =. Hence, we only get the trivial solution X(x. With λ = p 2 the solutions are given by X(x = Ae px + Be px. The boundary conditions X( = and X ( = 3X( gives us the system X( = A + B = X ( + 3X( = A ( pe p + pe p + 3A ( e p e p =, i.e. B = A and A = or e p (p+3+e p (p 3 =, but this expression is never for p (show this! and hence we must have A = B =, and also in this case we get only the trivial solution X. With λ = p 2 we get the solution X(x = Acos px + Bsin px, and the boundary conditions are X( = A =, and X ( = 3X( which gives pbcos px = 3Bsin px B(pcospx + 3sin px =, hence either B =, (and we get the trivial solution, or (pcospx + 3sin px =, i.e. p must satisfy the equation tan p = p 3. Thus, we see that we only have non-trivial solutions when λ is an eigenvalue λ = λ n = p 2 n, n =,2,..., where p n is a solution of tan p = p (see Fig. 5.2., and then we have the corresponding eigenfunctions 3 Example 5.6. Solve Solution: The characteristic equation is X n (x = a n sin p n x. x 2 X (x + 2xX (x + λx =, X( =, X(e =. r(r + 2r + λ = r 2 + r + λ =

4 3 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES FIGURE Solutions to tan p = p 3 y y = tan p p p2 p3 p y = p 3 which has the solutions r = 2 ± 4 λ = 2 ± i λ 4, hence the cases we must investigate are λ < 4, λ = 4 and λ > (cf. Example λ < 4 λ = 4 λ > 4 With r,2 = 2 ± 4 λ (different real we get the solutions X(x = Axr + Bx r 2 and the boundary conditions gives X( =, X(e =, A + B =, Ae r + Be r 2 =, A = B, A(e r e r 2 =, i.e. since e r e r 2 we must have A = and we only get the trivial solution X. Now we get a double root r = 2 and the solutions are X(x = Ax 2 + Bx 2 lnx. The boundary conditions give X( = A = and X(e = Be 2 =, i.e. A = B = and we only get the trivial solution X. The two complex roots r = λ 2 ± i give the solutions 4 ( λ ( λ 4 lnx 4 lnx X(x = A x sin + B x cos, ( and we get X( = B =, and X(e = A sin e λ 4 = hence λ must satisfy λ 4 = nπ, for some positive integer n. We get the eigenvalues λ n = 4 + (nπ2, n Z +,

5 5.2. EXAMPLES OF STURM-LIOUVILLE PROBLEMS 3 y FIGURE The Bessel function J (x J (x α α2 α3 α4 α5 x and the corresponding eigenfunctions X n (x = A n x sin(nπlnx. Example 5.7. (The Bessel equation An important ordinary differential equation in mathematical physics is the Bessel equation (Wilhelm Bessel of order m: r 2 w + rw + (r 2 m 2 w =. The solutions (there are two linearly independent to this equation are called Bessel functions of order m. (For more information see e.g. Besselfunktions at engineering fundamentals. Here we will only consider a special case. Solve the following problem involving the Bessel equation of order : d 2 w dr 2 + dw r dr + k2 w =, w(r =, w (r <. Solution: A general solution is given by w(r = C J (kr +C 2 Y (kr, where J and Y are the Bessel functions of the first and second kind of order. It is known that Y is not bounded and if we impose the condition that w (r must be bounded we get C 2 =. The boundary condition implies that w(r = C J (kr =, and if we want a non-trivial solution (C then k and R must satisfy J (kr =. It is well-known that J has infinitely many zeros α n (α = , α 2 = , α 3 = ,... etc., see Fig Hence we only get non-trivial solutions for the eigenvalues k n = α n R, n Z+, with the corresponding solutions are the eigenfunctions ( αn w n (r = J R r, n Z +.

6 32 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES 5.3. Inner Product and Norm To construct an orthonormal basis in a vector space we must be able to measure lengths and angles. Hence we must introduce an inner product (a scalar product. With the help of an inner product we can easily determine which elements are orthogonal to each other. There are two examples of vector spaces and inner products we will consider here. The plane R 2 together with the usual scalar product, and a vector space consisting of functions on an interval together with an inner product defined by an integral. Vectors in R 2 If we have two vectors x = (x,x 2 and y = (y,y 2, the inner product of x and y is defined by x y = x y + x 2 y 2. The norm of x, x, is defined by x 2 = x x = x 2 + x2, 2 and the distance between x and y, x y, is given by x y 2 = (x y 2 + (x 2 y 2 2. The angle θ between x and y can now be computed using the relation x y = x y cosθ, and we say that two vectors are orthogonal (perpendicular to each other, x y, if θ = π, i.e. if 2 A Function Space x y =. We now consider the vector space consisting of functions f (x defined on the interval [,l] (for some l > together with a positive weight-function r(x. The generalizations of the concepts above are f,g = f 2 = f g 2 = f (xg(xr(xdx, f (x 2 r(xdx, (norm (inner product f (x g(x 2 r(xdx, (distance f,g = f g cosθ, (angle f g f,g = (orthogonality f (xg(xr(xdx = Sturm-Liouville Problems A general Sturm-Liouville problem can be written as ( P (xy + ( q(x + λr(xy =, < x < l, c y( + c 2 y ( =, c 3 y(l + c 4 y (l =. Here r(x, q(x and P (x are given functions, c,...,c 4 given constants and λ a constant which can only take certain values, the eigenvalues corresponding to the problem. r(x is usually called a weight function. It is also customary to assume that r(x >.

7 5.4. STURM-LIOUVILLE PROBLEMS 33 If P (x > and c,...,c 4 we say that the problem is regular, and if P or r is in some endpoint we say that it is singular (note that there are other examples of both regular and singular SL problem, e.g. the following problem is regular. Example 5.8. r(x =, P (x =, q(x =, c = c 3 =, c 4 = c 2 =. y + λy =, y( =, y(l =. (Cf. Example 5.4. In this case we have ( nπ 2 λ n =, n =,2,3,..., (eigen values l ( nπ y n = sin l x, (eigen functions and y n,y m = y n 2 = ( nπ sin sin l x sin ( nπ l x 2 dx = ( mπ x dx =, if n m, l ( ( nπ cos 2 l x dx = l 2. If f is a function on the interval [,l] we can define the Fourier series of f, S(x, by (cf. Section 6.: S(x = c n = ( nπ c n sin l x, where y n 2 f,y n = 2 ( nπ f (xsin l l x dx. REMARK 7. Examples is also a Sturm-Liouville problem. For a regular Sturm-Liouville problem we have: (i (ii The eigenvalues are real and to every eigenvalue the corresponding eigenfunction is unique up to a constant multiple. The eigenvalues form an infinite sequence λ,λ 2,... and they can be ordered as with λ < λ 2 < λ 3 <, lim λ n =. n (iii If y and y 2 are two eigenfunctions corresponding to two different eigenvalues, λ i λ i2, they are orthogonal with the respect to the inner product defined by r(x, i.e. y,y 2 = y (xy 2 (xr(xdx =.

8 34 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES 5.5. Generalized Fourier Series We will now see how we can generalize the concept of Fourier series from the usual trigonometric basis functions to an orthonormal basis consisting of eigenfunctions to a Sturm-Liouville problem. Assume that we have an infinite linear combination where y n y m for n m. Then f,y m = f (x = c n y n (x, c n y n,y m = = c m y m,y m = c m y m 2. c n y n,y m Let f be an arbitrary function on [,l]. Then we define the generalized Fourier series for f as where are the generalized Fourier coefficients. S(x = c n y n (x, c n = y m 2 f,y n Let y,y 2,... be a set of orthogonal eigenfunctions of a regular Sturm-Liouville problem, and let f be a piece-wise smooth function in [,l]. Then, for each x in [,l] we have that (a S(x = f (x if f is continuous at x, and (b S(x = ( f (x+ + f (x if f has a discontinuity point at x Some Applications Example 5.9. Consider a rod of length l, with constant density ρ, specific heat c v and thermal conductance κ. Let the temperature of the rod at the time t and the distance x (from, say, the left end point be denoted by u(x,t. Assume that the temperature at the end points of the rod are given by (* u(,t = u(l,t =, t >, and that the temperature distribution in the rod at the initial time t = is given by u(x, = f (x, x l. Determine u(x,t for x l, and t. Solution: We have seen (cf. Chapter that the mathematical formulation of this problem is u t(x,t ku xx(x,t =, x l, t, k = κ c V ρ, u(,t = u(l,t =, t >, u(x, = f (x, x l. We begin by performing the following natural scaling of the problem (cf. Chapter : (5.6. t = k l 2 t, x = x l.

9 5.6. SOME APPLICATIONS 35 (A (B Then we arrive at the following standard problem to solve: ( ũ t (x,t ũ xx (x,t =, x, t, (2 ũ(,t = ũ(,t =, t >, (3 ũ(x, = f (x, x, where f (x = f (xl. We can now use Fourier s method to solve this problem (cf. section 4.8. Step : Try to find solutions of the type If we insert this expression in ( above we get i.e. the two equations ũ(x,t = X(xT (t. T (t T (t = X (x X(x = λ, X (x + λx(x =,and T (t + λt (t =. The function u(x,t = X(xT (t must also satisfy the boundary conditions (2: X(T (t = X(T (t =, t, and if we want a non-trivial solution (T we conclude that X( = X( =. This boundary condition together with (A gives us the Sturm-Liouville problem X (x + λx(x =, (** X( = X( =. Step 2: We get three cases depending on the value of λ : λ <, λ =, and λ >. λ < We get only the trivial solution X(x. λ = We get only the trivial solution X(x. λ > Then X(x = Asin ( λx + Bcos and X( = B =, and X( = Asin Thus the SL-problem (** has the following eigenvalues and the corresponding eigenfunctions λ n = (nπ 2, n Z +, X n (x = sin(nπx. Furthermore, for these values of λ = λ n, the equation (B has the solution T (t = T n (t = e (nπ2t, ( λx, ( λ = A = or λ = nπ, n Z +. and we conclude that the general (separable solution of ( and (2 can be written as ũ n (x,t = c n sin(nπxe (nπ2t. Step 3: The superposition principle (cf. section 4.5 tells us that the function ũ(x,t = b n sin(nπxe (nπ2 t

10 36 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES also satisfy ( and (2. We will now see that we can also assert that this function satisfy the initial condition (3 by choosing appropriate constants b n. It is clear that ũ(x, = b n sin(nπx, and if we choose the b n s as the Fourier coefficients of f, i.e. we actually get We conclude that the function b n = 2 ũ(x, = ũ(x,t = Z f (xsin(nπxdx, b n sin(nπx = f (x. b n sin(nπxe (nπ2t, with b n as above satisfy (, (2 and (3. Final step: By using the scaling from (5.6. we see that the solution to the original problem is given by where u(x,t = b n = 2 l ( nπ b n sin l x e ( nπ l 2 kt, ( nπ f (xsin l x dx. Example 5.. Consider a rod between x = and x = e. Let u(x,t denote the temperature of the rod at the point x and time t. Assume that the end points are kept at the constant temperature, that at the initial time t = the rod has a heat distribution given by u(x, = f (x, < x < e, that no heat is added and that the rod has constant density ρ and specific heat c v. Assume also that the rod has heat conductance K which varies as K(x = x 2. The equation which determines the temperature u(x,t is in this case ( (2 (3 c v ρu t = ( x 2 u x, < x < e, t >. x Determine u(x,t for x e, and t >. Solution: We apply Fourier s method of separating the variables and assume that we can find a solution of the form u(x,t = X(xT (t. Inserting this expression in ( above we get c v ρ T T = d ( x 2 X = λ, X dx where λ is a constant and X satisfies the boundary condition Thus T satisfies the equation X( = X(e =. T = λ c v ρ T,

11 5.6. SOME APPLICATIONS 37 (4 and X satisfies d ( x 2 X + λx =, < x < e dx x 2 X + 2xX + λx =, < x < e. The equation (4 together with the boundary condition (2 gives us a regular Sturm-Liouville problem on [,e]. The characteristic equation is with the roots r(r + 2r + λ =, r,2 = 2 ± 4 λ. As in Example 5.6 we get three different cases depending on the value of λ: λ = 4 λ < 4 We have a double root r = 2, and the solutions are given by X(x = Ax 2 +Bx 2 lnx. The boundary condition (2 gives X( = A = and X(e = Be 2 =, i.e. we get only the trivial solution X. The roots are now real and different, r r 2, and the solutions are X(x = Ax r + Bx r 2. The boundary conditions gives us X( = A + B = X(e = Ae r + Be r 2 = A = B, A(e r e r 2 =, and since r r 2 we must have A =, and we only get the trivial solution X. λ > 4 We have two complex roots r = λ 2 ± i, and the general solution is 4 X(x = A sin ( λ 4 lnx + B cos ( λ 4 lnx. x x ( The boundary conditions imply that X( = B = and X(e = Ae 2 sin λ = 4, which us gives that λ 4 = nπ, n Z+. Observe that the case n = is the same as λ =. Hence the eigenvalues of the Sturm-Liouville problem 4 (4 and (2 are and the corresponding eigenfunctions are λ n = 4 + n2 π 2, n Z +, X n (x = x sin(nπlnx, n Z +. For every fixed n, the equation (3 is T n = λ n c v ρ T n,

12 38 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES with the solutions We conclude that the functions T n (t = e λn cvρ t, n Z +. u n (x,t = T n (tx n (x = x sin(nπlnxe λn cvρ t, n Z +, are solutions to the original equation, and they also satisfy the boundary conditions. The superposition principle implies that the function u(x,t = a n x sin(nπlnxe λn cvρ t is also a solution of the equation which satisfy the boundary conditions. Finally, to accommodate the initial values we must choose the constants a n so that u(x, = a n This holds if we choose the constants a n as a n = X n 2 Z e = 2 Z e (Note that X n 2 = x sin2 (nπlnxdx = 2 u(x,t = a n where Example 5.. Solve the problem: x sin(nπlnx = f (x, Z e f (xx n (xdx f (x x sin(nπlnxdx.. The wanted temperature distribution is thus given by x sin(nπlnxe λn cvρ t, Z e f (x a n = 2 sin(nπlnxdx. x ( u u t = 2 x 2, (2 u(,t =, (3 (4 (5 (6 u x(,t = 3u(,t, u(x, = f (x. Solution: We use Fourier s method of separation of variables. Step : Assume that u(x,t = X(xT (t and insert this into (. In the same way as in the previous examples we obtain the equation which gives us the two equations T (t T (t = X (x X(x = λ, T (t λt (t =, X (x λx(x =. Step 2: There are three cases of λ we must study.

13 5.6. SOME APPLICATIONS 39 λ = The solutions to (5 and (6 are then T =constant, and X = Ax +B, i.e. u(x,t = Ax +B for some constants A and B. The boundary value (2 gives u(,t = B =, and (3 gives u x(,t = A = 3u(,t = 3A, i.e. A = and we only get the trivial solution u(x,t. λ > The solutions to (5 are T (t = Ae λt and the solutions to (6 are X(x = λx Be +Ce λx. The boundary value (2 gives u(,t = T (tx( = Ae λt (B+C =, hence either A = (which implies u or B = C. The condition (3 is now equivalent to Ae λt ( λb λ e ( + e λ = 3AB λ e e λ, ( ABe 2 λ 3 + λ = AB ( 3 λ, λ < which is only satisfied if AB = (show this!, and in this case we only get the trivial solution u. If we set λ = p 2 we get (in the same manner as in Example 5.5 the solutions (* u n (x,t = B n e p2 nt sin p n x, n =,2,3,..., where p n are solutions to the equation tan p = p 3. Step 3: All functions defined by (* satisfy (, (2 and (3. According to the superposition principle the function u(x,t = B n e p2 nt sin p n x also satisfies (, (2 and (3. Furthermore, (6 with the corresponding boundary conditions is a regular Sturm-Liouville problem and the theory of generalized Fourier series implies that u(x, t will satisfy (4: if we choose the constants B n as u(x, = B n sin p n x = f (t (** B n = f (x,sin p R nx sin p n x 2 = f (xsin p nxdx R sin2 p n xdx. Thus, the solution to the problem is u(x,t = B n e p2 nt sin p n x, where p n are the positive solutions of tan p = p 3, p < p 2 < (see Fig. 5.2., and B n is defined by (**.

14 4 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES Example 5.2. (The wave equation A vibrating circular membrane with radius R is described by the following equation together with boundary and initial values: ( (2 (3 (4 ( u tt = c 2 ( u xx + u yy,t >, r = x 2 + y 2 R, u(r,t =,t >, (fixated boundary u(r, = f (r,r R, (initial position u (r, = g(r,r R.(initial velocity t Observe that the initial conditions only depend on r = x 2 + y 2 = the distance from the center of the membrane to the point (x,y, and if we introduce polar coordinates x = r cosθ, y = r sinθ, we see that ( can be written as 2 ( u 2 t 2 = u c2 r 2 + u r r + 2 u r 2 θ 2. If we also make the assumption that u(r,θ,t is radially symmetric (i.e. that u(r,θ,t is independent of the angle θ we can write ( as 2 ( u 2 t 2 = u c2 r 2 + u. r r To solve the problem we continue as before and use Fourier s method to separate the variables. With the function u(r,t = W(rG(t inserted into ( we get the equations (5 W + r W + k 2 W =, r R, (6 G + (ck 2 G =, t >. Furthermore, we get the following boundary values from (2: (7 W(R =, and (5 together with (7 is a regular Sturm-Liouville problem which gives us the eigenfunctions ( αn W n (r = J R r, where α n = k n R are solutions of J (kr = (see Example 5.7. Observe that if we write (5 in the general form we see that we have the weight function= r, i.e. the inner product is given by f,g = Z R f (rg(rrdr. By solving (6 for k = k n and using the superposition principle we see that ( (* u(r,t = A n cos ( cαn R t + B n sin ( cαn R t J ( αn R r is a solution to ( and (2. And we can also choose the constants A n so that (3 is satisfied, i.e. ( αn u(r, = A n J R r = f (r, if (** A n = R R J ( αn R r 2 rdr Z R f (rj ( αn R r rdr.

15 5.7. EXERCISES 4 In the same way we see that (4 is satisfied if we choose B n so that cα n (*** B n R = R R J ( αn R r 2 rdr Z R g(rj ( αn R r rdr. Hence, the answer to the problem is given by (* where A n and B n are chosen as in (** and (*** Exercises 5.. [A] Solve the following S-L problem by determining the eigenvalues and eigenfunctions: ( (a x 2 u (x + λu(x =, < x < e L, u( = u ( e L =, ( (b x 2 u (x + λu(x =, < x < e L, u( = u (e = * Solve the following S-L problem by determining the eigenvalues and eigenfunctions: (a u (x + λu(x =, < x < l, u ( = u (l =, (b u (x + λu(x =, < x < l, u ( = u(l = [A] Use Fourier s method to solve the following problem: u t = u xx, x l, t >, u x(,t = u x(l,t =, t >, u(x, = f (x, < x < l * A rod between x = and x = e has constant temperature at the endpoints, and at the time t = the heat distribution is given by x, < x < e. The rod has a constant density ρ and constant specific heat C, but its thermal conductance varies like K = x 2, < x < e. Formulate an initial and boundary values problem for the temperature of the rod, u(x,t. Then use Fourier s method to solve the problem * (a Solve the problem u t = 4u xx, x, t >, u(,t = t >, u x(,t = cu(,t, t >, u(x, = x, x, x < 2, x 2. (b Give a physical interpretation of the problem in (a.

16 42 5. INTRODUCTION TO STURM-LIOUVILLE THEORY AND THE THEORY OF GENERALIZED FOURIER SERIES 5.6. [A] Consider an ideal liquid, flowing orthogonally towards an infinitely long cylinder by the radius a. Since the problem is uniform in the axial coordinate we can treat the problem in plane polar coordinates. a x The speed of the liquid, v(r,θ is then given by the equation where ψ as a solution of the Laplace equation v(r,θ = gradψ, ψ =. At the surface of the cylinder we have the boundary condition ψ r r=a =, and as r we have the following asymptotic boundary condition ψ lim r x = lim ψ r r cosθ = v, where v is a constant. a Show, using separation of variables, that in polar coordinates, the assumption ψ(r, θ = R(rΘ(θ transforms the Laplace equation to the following two equations Θ (θ + m 2 Θ(θ =, R (r + r R (r m2 R(r r2 =, where m is an integer. b Use a to find ψ and v.