January 27, 2013 STURM-LIOUVILLE THEORY

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1 January 27, 213 STURM-IOUVIE THEORY RODICA D. COSTIN Contents 1. Examples of separation of variables leading to Sturm-iouville eigenvalue problems Heat conduction in dimension one The vibrating string The wave equation 5 2. Second order linear ordinary differential equations Recall some basic results The selfadjoint form of a linear second order equation Homogeneous boundary conditions Formulation of the homogeneous Sturm-iouville problem Green s Identity and selfadjointness of the Sturm-iouville operator Conclusions 1 3. Eigenfunctions associated to one eigenvalue Regular problems When p(x) vanishes at one endpoint Periodic problems Fourier series An example Problems on infinite intervals Abel s Theorem Sturm s Oscillation Theorems A simple example Sturm s Comparison Theorem Existence of eigenvalues: The Prüfer transformation The Prüfer system Behavior of θ(x) and the zeroes of u(x) Boundary conditions and existence of eigenvalues The distance between two consecutive zeroes of the eigenfunction u n 2 8. Completeness Proof of Theorem If S is dense in D and D is dense in H then S is dense in H More examples of separation of variables The wave equation aplace s equation 25 1

2 2 RODICA D. COSTIN 9.3. The vibrating rod An introduction to the Fourier transform The space 2 (R) The Fourier Transform The Fourier transform diagonalizes the diffferentiation operator An example Examples of separation of variables leading to Sturm-iouville eigenvalue problems Many partial differential equations which appear in physics can be solved by separation of variables. Two examples are illustrated here Heat conduction in dimension one. Consider a thin rod of length, perfectly insulated. The temperature u(x, t) at time t and position x [, ] satisfies the heat equation (1) u t = α u xx where α is a parameter (which depends on the rod). Of course, the heat conduction depends on what happens at the two endpoints of the rod x = and x = (boundary condition), and on the initial temperature distribution the rod (initial condition), which need to be specified Separation of variables. ooking for solutions in the form u(x, t) = y(x)t (t) (with separated variables) and substituting in (1) we obtain y(x)t (t) = αy (x)t (t) therefore T (t) αt (t) = y (x) = constant = λ y(x) and we obtain two ordinary differential equation which can be solved: (2) T (t) = αλt (t) and (3) y + λy = Boundary conditions. 1) Suppose both ends of the rod are kept at constant temperature zero: say u(, t) = and u(, t) = for all t. It follows that y() = and y() =. Equation (3) with these boundary conditions is a Sturm-iouville eigenvalue problem. We saw that the eigenvalues of this problem are λ n = n 2 (n = 1, 2,...) and the eigenfunctions are y n (x) = sin(nπ/ x).

3 STURM-IOUVIE THEORY 3 2) Other constant boundary temperatures can be imposed, but these can be reduced to the case of zero boundary conditions. If, say, u(, t) = and u(, t) = T for all t, then substituting u(x, t) = T x + v(x, t) into (1) we find that v(x, t) satisfies the heat equation and has zero boundary conditions: v(, t) = and v(, t) = for all t. 3) If one end radiates (say, x = ) then the boundary condition at x = is u x (, t) = hu(, t) (h > means heat loss due to radiation, h = means there is no radiation) and u(, t) =. This gives: y() =, y ()+hy() = which together with equation (3) form another Sturm-iouville eigenvalue problem Initial condition. The initial distribution of the temperature needs to be specified as well: u(x, ) = u (x). After finding the eigenvalues λ n and eigenfunctions y n of the appropriate Sturm-iouville eigenvalue problem, equation (2) is solved yielding T n (t) = c n e αλnt. Since the heat equation is linear, then a superposition of solutions with separated variables: u(x, t) = n c ne αλnt y n (x) is again a solution. Now it is the time to require that the initial condition be satisfied: u(x, ) = u (x) = n c ny n (x). If the eigenfunctions y n are complete in 2 [, ] then indeed c n exist, and are uniquely determined for any u 2 [, ]: since the eigenfunctions will be shown to be orthogonal, and assuming they have been normalized to have norm one, then c n = y n, u = y n(x)u (x)dx The vibrating string. Consider a vibrating string with space-dependent tension T (x) and variable linear density ρ(x), assumed to vibrate only due to the restoring tension Derivation of the equation. Denote by y(x, t) the displacement at time t. (For each t, the graph of the function x y(x, t) represents the string.) To deduce the equation of the motion we apply Newton s law on each small piece [x, x + x]. The force at each point x is the vertical component of T (x): T V (x) = T (x) sin α where α is the angle between T (x) and the x-axis. Assuming the oscillations are small then sin α α tan α = y Thus T V (x) T (x) y x. The force of the piece [x, x + x] is T V (x + x) T V (x) T V x x = x ( T (x) y ) x x On the other hand, mass times acceleration is ρ(x) x 2 y, therefore t 2 (4) ρ(x) 2 y t 2 = ( T (x) y ) x x x.

4 4 RODICA D. COSTIN The motion depends on what happens at the endpoints of the string (the boundary conditions) and on its initial state (initial condition) which need to be specified for the equation (4) Separation of variables. Solve the partial differential equation (4) by separation of variables: looking for solutions of the form y(x, t) = u(x)f(t) and plugging it into (4) it follows that therefore so ρ(x)u(x)f (t) = d dx ( T (x) du ) f(t) dx f (t) f(t) = 1 ( T (x)u (x) ) = constant = λ ρ(x)u(x) (5) f = λf and (6) We can rewrite (6) as 1 ρu (T u ) = λ (7) (T u ) + λρu = Boundary conditions. Suppose the endpoints of the string are kept fixed: y(, t) =, y(, t) =. Then this implies (8) u() =, u() = The problem (7), (8) is a Sturm-iouville eigenvalue problem. As noted before, this is an eigenvalue/eigenfunction problem for the operator 1 ρ(x) d dx T (x) d dx, which is selfadjoint on the domain in H = 2 ([, ], ρ(x)dx) {u H u, u H, u() =, u() = } 2) Another boundary conditions could be: while x = is fixed, the endpoint x = is free. Then u() = and u () = (there is no transverse force at x = to produce motion). Equation (7) together with the boundary conditions u() =, u () = is another example of a Sturm-iouville eigenvalue problem. 3) Or, x = is fixed, but the endpoint x = is tied to a spring that vibrates: T u () ku() = (T is the string tension, k is the spring constant, the tail is accelerated up and down but there is no transversal force). In this case, the Sturm-iouville eigenvalue problem consists of equation (7) together with the boundary conditions u() =, T u () + ku() =. The appropriate Sturm-iouville problem is solved, finding the eigenvalues λ n and the corresponding eigenfunctions u n (x).

5 STURM-IOUVIE THEORY 5 Remark. The eigenfunctions u n (x) are the normal modes of the string. Then f n (t) can be found by solving (5): f n (t) = c n sin( λ n t)+d n cos( λ n t) Initial condition. To determine a unique solution the initial position of the string must be given: u(x, ) = g(x) and the initial velocity u t (x, ) = v(x) (the equation is of order two in t!). It is then required that a superposition of the solutions n f n(t)u n (x) satisfy the initial condition: n f n()u n (x) = g(x) and n f n()u n (x) = v(x). If u n (x) form a complete set then c n and d n can be determined The wave equation. In particular, if T (x) and ρ(x) are constant then equation (4) becomes y tt = c 2 y xx (where c 2 = T/ρ) which is the wave equation. The eigenfunctions u n satisfty u n+λ n u = and the appropriate boundary conditions. If these are u() =, u() = then we showed that u n (x) = sin(nπx/) which are the normal modes of the string. 2. Second order linear ordinary differential equations 2.1. Recall some basic results. A second order linear ordinary differential equation (ODE) has the form (9) P (x)u + Q(x)u + R(x)u = Because the equation is linear, any linear combination of solutions is again a solution: if u 1, u 2,..., u k are solutions of (9) and c 1, c 2,..., c k are constants then c 1 u 1 (x) + c 2 u 2 (x) c k u k (x) is also a solution of (9). Assumptions. 1) It is assumed assume that the coefficients P (x), Q(x), R(x) are continuous on an interval (a, b). (However, jump discontinuities can also be accommodated in a similar way, only requiring a bit more work.) 2) It is assumed that P (x) does not vanish in (a, b). (Points where P (x) is zero are singular, and solutions are usually very special at such points.) 3) It can be assumed without loss of generality that P (x) > on (a, b). (Since P (x) is never zero on (a, b), and it is continuous, then P (x) is either positive on (a, b) or negative on (a, b). If P < we multiply the equation by 1.) General solution. Recall that the general solution of (9) has the form u(x) = C 1 u 1 (x) + C 2 u 2 (x) where C 1,2 are arbitrary constants and u 1,2 are two solutions of (9) which are linearly independent, in the sense that the vectors (u 1 (x ), u 1 (x )) and (u 2 (x ), u 2 (x )) are linearly independent at some x (a, b) (equivalently, at any x (a, b)). For example, the solutions with the initial conditions u 1 (x ) = 1, u 1 (x ) = and u 2 (x ) =, u 2 (x ) = 1 are linearly independent.

6 6 RODICA D. COSTIN An equivalent condition for two solutions to be linearly independent is that their Wronskian W [u 1, u 2 ] = u 1u 2 u 1 u 2 satisfies W (x ) at some x (a, b) (equivalently, at any x (a, b)). Recall that the Wronskian satisfies the differential equation (1) W (x) = Q(x) P (x) W (x) and therefore [ W (x) = C exp Q(x) ] P (x) dx Existence and uniqueness of the solution to the initial value problem. For x (a, b) and c, d R equation (9) has a unique solution u(x) for x (a, b) satisfying the initial conditions u(x ) = c, u (x ) = d. This solution u(x) is twice differentiable (moreover, u is continuous, as it is seen from (9)), and it depends continuously on the initial conditions. The general solution depends on two parameters, so it makes sense that two conditions are required to determine these parameters. However, there is no apriori guarantee that solutions satisfying different problems, like boundary conditions, do exist The selfadjoint form of a linear second order equation. Consider again equations (9), but with R(x) replaced by R(x) + λ (where λ is a constant) (11) P (x)u + Q(x)u + (R(x) + λ)u = (λ is singled out just to keep track on how it changes after the substitutions that follow). We are interested in bringing (11) to a selfadjoint form: (12) or, expanded, (13) 1 w(x) ( d dx p(x)du dx + q(x) ( pu ) + ( q + λw)u = ) u = λu where p(x), q(x), w(x) are functions suitably determined. Indeed, expanding the left hand-side of (12) we obtain p w u + p w u + ( q ) w + λ u = which must be (11), therefore p w = P, p w = Q, q w = R

7 STURM-IOUVIE THEORY 7 The first two equations imply that p /p = Q/P therefore [ ] Q(x) (14) p(x) = exp P (x) dx Then since w = p/p and q = wr we obtain (15) w(x) = 1 [ ] Q(x) P (x) exp P (x) dx, q(x) = R(x) P (x) exp [ ] Q(x) P (x) dx 2.3. Homogeneous boundary conditions. These conditions are usually inherited from the PDEs which produced the ODE (11) by separation of variables. If the values on the boundary are not zero, substitutions can often be made to ensure zero values on the boundary: these are called homogeneous boundary conditions. These could have the form: Dirichlet conditions: u(a) =, u(b) =, or Newman conditions: u (a) =, u (b) =, or Mixed Dirichlet-Newman conditions: B (16) a [u] αu(a) + α u (a) = B b [u] βu(b) + β u (b) = The mixed conditions are the most general, as they have the Dirichlet and the Neuman conditions as particular cases (if α = = β we obtain Dirichlet conditions, and if α = = β we obtain Newman conditions). Therefore we work with the general mixed Dirichlet-Newman conditions. It must be assumed that the boundary conditions are nontrivial: at least one of the numbers α, α is not zero (note that this condition can be written as α + α ), and similarly, at least one of the numbers β, β is not zero (i.e. β + β ) Another way of writing B a [u], B b [u]. Clearly if we multiply α and α by the same constant, we obtain the same boundary condition B a [u], and similarly for β and β in B b [u]. It is sometimes convenient (and always possible!) to choose these in the form (17) B a [u] cos(θ a )u(a) sin(θ a )p(a)u (a) = B b [u] cos(θ b )u(b) sin(θ b )p(b)u (b) = which are very suitable for Prüfer coordinates. The transformation which brings (16) in the form (17) is the following: dividing α and α by the quantity ± α 2 + (α /p(a)) 2 with the sign chosen to be opposite to the sign of α, we obtain B a [u] = in the form α 1 u(a) α 2 p(a)u (a) = where α α2 2 = 1 and α 2 therefore there exists θ a [, π) so that α 1 = cos(θ a ) and α 2 = sin(θ a ) (we choose θ a < π/2 if α 1 > and θ a > π/2 if α 1 < ). A similar transformation can be performed on B b. Note that we can choose θ b in [n, (n + 1)π) for any integer n (if n is even, we proceed as for the condition at x = a, while if n is odd we choose the opposite sign in front of ± β 2 + (β /p(b)) 2, namely the sign of β ).

8 8 RODICA D. COSTIN Singular boundary conditions. Other type of conditions which appear in applications are: Periodic conditions: if p(a) = p(b) then it can be required that the solutions be periodic: u(a) = u(b), and u (a) = u (b) More generally: α 1 u(a) + α 1 u (a) + β 1 u(b) + β 1 u (b) = α 2 u(a) + α 2 u (a) + β 2 u(b) + β 2 u (b) = If p vanishes at an endpoint, say p(a) = : then the boundary condition at x = a is dropped Formulation of the homogeneous Sturm-iouville problem. We will consider real-valued problems: the functions P, Q, R and the numbers α, α, β, β are real. In case complex valued functions are needed, then equations can be written and separately solved for the real and imaginary parts of these functions. Note that with this reality assumption we have p(x) > and w(x) > (see (14), (15)). Given the functions p, q, w continuous on [a, b] and p, w > on [a, b], and B a [u], B b [u] (nontrivial) find the numbers λ so that the following problem has a nontrivial (i.e. nonzero) solution u(x) on [a, b]: [p(x)u ] + [ q(x) + λw(x)]u = (18) Boundary conditions at x = a and x = b (19) The boundary conditions are one of the following: regular conditions: B a [u] αu(a) + α u (a) = ( α + α ) B b [u] βu(b) + β u (b) = ( β + β ) singular conditions: if p(b) = : (2) B a [u] αu(a) + α u (a) = ( α + α ) or, if p(a) = : (21) B b [u] βu(b) + β u (b) = ( β + β ) (22) periodic conditions: (also singular) if p(a) = p(b) C[u] u(a) u(b) = C [u] u (a) u (b) = The numbers λ are called eigenvalues, and the corresponding solutions - eigenfunctions.

9 STURM-IOUVIE THEORY Green s Identity and selfadjointness of the Sturm-iouville operator. We show here that the problem (18) is indeed selfadjoint. et us first show a general formula: emma 1. Green s identity: (23) b a [ (pu ) v u(pv ) ] = p(u v uv ) b a Relation (23) follows easily using integration by parts: b [ (pu ) v u(pv ) ] b = (pu ) v a a b a u(pv ) = pu v b b a (pu )v upv b b a + u (pv ) = p(u v uv ) b a a a Theorem 2. The operator (24) = 1 w(x) ( d dx p(x) d ) dx + q(x) is selfadjoint in the weighted (real) Hilbert space H = 2 ([a, b], w(x)dx) on the domain (25) D = {u H u, u H, B a [u] =, B b [u] = } Proof. To show that u, v u, v = for all u, v D we use, as usually integration by parts. This work was done by the Green s identity, which gives (noting that the terms containing q cancel each other): u, v u, v = b a 1 [ ( pu ) ] + qu v w dx w b a u 1 [ ( pv ) ] + qv w dx w (26) = p(u v uv ) b a = p(a) (u v uv ) x=a p(b) (u v uv ) x=b We have that (27) p(a)(u v uv ) x=a =, p(b)(u v uv ) x=b = because B a [u] = = B a [v] and B b [u] = = B b [v]. Indeed, if α = then u(a) = = v(a) therefore (u v uv ) x=a =, and if α then u (a) = α/α u(a), v (a) = α/α v(a) which substituted into the first relation of (27) gives again zero. The second relation of (27) follows in a similar way.

10 1 RODICA D. COSTIN 2.6. Conclusions. Since is selfadjoint on D, this implies that its eigenvalues (if any!) are real, and that eigenfunctions corresponding to different eigenvalues are orthogonal. We will show that indeed, there exist infinitely many eigenvalues λ n, and that the eigenfunctions u n form a complete set in the Hilbert space 2 ([a, b], w(x)dx). We will accomplish this program by studying the solutions of the differential equation. It turns out that, in addition, λ n can be ordered increasingly, and λ n, and that eigenfunctions u n oscillate, and the larger n, the more rapid the oscillations. 3. Eigenfunctions associated to one eigenvalue 3.1. Regular problems. emma 3. For regular problems (18), (19) the eigenspaces are one-dimensional: there is a unique (up to a scalar multiple) eigenfunction associated to each eigenvalue. Proof. We show that the eigenspace associated to one eigenvalue of (18) is one dimensional: any two (nonzero) solutions u 1 (x), u 2 (x) of (18) (for the same λ) are linearly dependent. Assume, to get a contradiction, that u 1 and u 2 are linearly independent. Then the general solution of the differential equation in (18) is u = C 1 u 1 + C 2 u 2 with C 1, C 2 arbitrary constants. Since the boundary conditions are linear, it follows that B a [u] = C 1 B a [u 1 ] + C 2 B a [u 2 ] =, B b [u] = C 1 B b [u 1 ] + C 2 B b [u 2 ] = therefore the boundary conditions are satisfied by any solution of the differential equation. For u the solution with u(a) = 1, u(a) = we find that α = and u the solution with u(a) =, u (a) = 1 it follows that α =, which contradicts the nontriviality assumption on B a. (A similiar argument can be made at x = b.) Therefore, u 1 and u 2 must be linearly dependent, hence scalar multiples of each other When p(x) vanishes at one endpoint. Suppose that p(b) =. Consider the the Sturm-iouville problem (18), (21). The Sturm-iouville operator (24) is selfadjoint on the domain (28) D = {u H u, u H, B a [u] = } Indeed, relations (27) hold: at x = b because p(b) = and at x = a with the same proof as for Theorem 24. The eigenspaces are still one-dimensionsl, as the proof of emma 3 works. The singular case with p(a) = is similar.

11 STURM-IOUVIE THEORY Periodic problems. If p(a) = p(b) the Sturm-iouville eigenvalue problem (18), (22) is also selfadjoint: the operator (24) is selfadjoint on (29) D = {u H u, u H, C[u] =, C [u] = } since the last quantity in (26) is clearly zero for u, v in the domain (29). However, for periodic boundary conditions the eigenspaces may be one, or two-dimensional. Indeed, repeating the argument of 3.1 we find that if for an eigenvalue λ, if there are two independent eigenfunctions then all the solutions of the ODE are periodic (but the space of solutions of a second order ODE is a two-dimensional vector space). This means that the eigenspace is either one-dimensional or two-dimensional, and in the latter case we can choose two orthogonal eigenfunctions. 4. Fourier series It is easy to solve the following periodic Sturm-iouville problem: (3) u + λu =, u( π) = u(π), u ( π) = u (π) It has the eigenvalues λ n = n 2, n =, 1, 2,.... For n > there are two orthogonal eigenfunctions corresponding to λ n = n 2 : sin(nx) and cos(nx). For n = the eigenfunction corresponding to λ = is the constant function. It follows that these eigenfunctions are complete using the general theorem: Theorem 4. Completeness of the eigenfunctions Suppose that is a selfadjoint operator in a Hilbert space H, defined on a domain D dense in H. Assume that has the eigenvalues λ 1 λ 2 λ 3... with λ n, and that each eigenspace is finite dimensional, spanned by a set of orthgonal eigenvectors u n. Then the set of eigenvectors is complete: they form a basis for the Hilbert space H. The proof of the theorem is postponed until 8. Applying Theorem 4 to = d2 on dx 2 in H = 2 [ π, π], which is selfadjoint D = {u H u, u H, u(π) u( π) =, u (π) u ( π) = } having the eigenvalues, 1, 1, 2 2, 2 2,... it follows that its eigenfunctions 1, sin(nx), cos(nx) for n = 1, 2,... form an orthogonal basis for 2 [ π, π], therefore:

12 12 RODICA D. COSTIN Theorem 5. Any function f 2 [ π, π] can be expanded in a Fourier series: (31) f = a 2 + [a n cos(nx) + b n sin(nx)] where the Fourier coefficients a n and b n are given by the formulas a n = 1 π π π f(x) cos(nx) dx, b n = 1 π π π f(x) sin(nx) dx, While the series (31) converges to f in the 2 -norm (that is, in squared average), it is useful to know when the series converges point-wise (that is, for a fixed x, as a series of numbers): 1 Theorem 6. Point-wise convergence of a Fourier series A. If f and f are piecewise continuous on [ π, π] then the series (31) converges for every x. B. et c ( π, π). (i) If f is continuous at x = c then the Fourier series (31) at x = c converges to f(c). (ii) If f has a jump discontinuity at x = c then the Fourier series (31) at x = c converges to [f(c ) + f(c+)]/2. C. The behavior of the Fourier series at the points x = π and x = π is seen in the following way. Continue f(x) outside [ π, π] by 2π-periodicity. If f(π ) = f( π+) then x = ±π are points of continuity, and the series (31) converges to f(±π) for x = ±π. Otherwise, x = ±π are points where there is a jump discountinuity and the series (31) converges to [f(π ) + f( π+)]/2 for x = ±π. The proof of Theorem 6 is not given here. Recall that a function is called even if f( x) = f(x) for all x. For example the functions 1, x 2, x 4, x 8 3x 4, x, cos(x) are even. Recall that a function is called odd if f( x) = f(x) for all x. For example the functions x, x 3, x 5, x 9 2x, sin(x), tan x are odd. Remark. If f(x) is an even function then all b n = and therefore f has a cosine-series. If f(x) is an odd function then all a n = and therefore f has a sine-series. Given any function g(x) for x [, π], then g can be continued to x [ π, π] by 1) requiring that the function on [ π, π] be odd, that is continued as { g(x) if < x < π g odd (x) = g( x) if π < x < in which case g(x) has a sine-series, or by 1 Note that piecewise continuous functions on [ π, π] are necessarily in 2 [ π, π].

13 STURM-IOUVIE THEORY 13 2) requiring that the function on [ π, π] be even, that is continued as { g(x) if < x < π g even (x) = g( x) if π < x < in which case g(x) has a cosine-series. (Recall that functions that differ by the value at one point, such as x =, are considered equal in 2.)

14 14 RODICA D. COSTIN The figure show a function on [, 1] x 1 function on [,1] and its extension to [ 1, 1] as an odd function x extension as odd and its extension to [ 1, 1] as an even function x extended as even

15 STURM-IOUVIE THEORY 15 Recall that the eigenfunctions of the Dirichlet homogeneous problem (32) u + λu =, u() =, u(π) = are sin(nx), for n = 1, 2,... and they correspond to the eigenvalues λ n = n 2. Theorem 4 does apply and we obtain that any function in 2 [, π] can be expanded in a sine-series. This series coincides with the Fourier series of the function continued to an odd function on [π, π] as showed at point 1) above. In conclusion, for a function f on [a, b] we can write an expansion in a sine-series, or a cosine-series, or a general Fourier series. The general Fourier series is more appropriate if f is periodic on [a, b] (i.e. f(a) = f(b)). If f(x) has complex values, then the complex form of the Fourier series may be preferable: c n e 2πin/(b a) n Z Note that this series is real-valued if c n = c n for all n An example. et us find the sine-series of function f(x) = 1 for x [, π]. It can be checked that b n = for n even, and its sine-series is sin (x) + sin (3 x) + sin (5 x) + sin (7 x) + sin (9 x) +... π 3π 5π 7π 9π For each x (, π) the sine-series converges to 1 by Theorem 6 B.(i). To understand the value at which the series converges at the end points x = and x = π we first continue the function to an odd function on [ π, π], which is f odd (x) = 1 for x (, π] and f odd (x) = 1 for x [ π, ). Using Theorem 6 B.(i) the series at x = converges to [f odd ( ) + f odd (+)]/2 = and by Theorem 6 C. at x = π the series converges to [f odd ( π+) + f odd (π )]/2 =. Of course, by substituting directly x = or x = π in the series we see that all its terms are zero. The picture shows the plot of the sum of the first five nonzero terms.

16 16 RODICA D. COSTIN x partial sum 3. Note the large overshoot of the partial sum at the at the jump discontinuities at x = and x = π: this is called Gibbs phenomenon. This behavior of truncates of Fourier series gives rise to artifacts in signal processing Problems on infinite intervals. If we have eigenvalue problems on infinite intervals (like [a, + ) or R = (, + )) a similar theory can be developed, only series need to be replaced by integrals. For example, consider the differentiation operator in 2 (R): = Since d dx eikx = ike ikx then the function e ikx is a generalized eigenfunction of (it must be called generalized because it does not belong to the Hilbert space 2 (R)). However, any f 2 (R) can be developed in terms of these generalized eigenfunctions: f(x) = e ikx F (k) dk (the inverse Fourier transform, up to a multiple) to be compared to the Fourier series for f 2 [ π, π] f = f k e ikx k= 5. Abel s Theorem The following results is (1) in disguise: Theorem 7. Abel s Theorem et u, v be two solutions of the second order linear differential equation (33) [p(x)u ] + [ q(x) + λw(x)]u = Then p(u v uv ) = constant d dx.

17 STURM-IOUVIE THEORY 17 Proof. Equation (33) expanded is pu + p u + ( q + λw)u =. For W = W [u, v] = u v uv relation (1) implies that W = p /p W so ln W = ln p + const which implies pw = const. Note that once a solution u of (33) another independent solution v can be found by solving p(u v uv ) = c, which is a first order differential equation, linear nonhomogeneous for v. 6. Sturm s Oscillation Theorems 6.1. A simple example. It is always useful to consider simple examples (or, toy models ) for the more complicated system studied. A simple example (and exactly solvable!) of a Sturm-iouville problem is obtained for for w(x) = 1, p(x) = 1, q(x) =, and α =, β =, and [a, b] = [, π], namely the problem (32). We studied this equation and we found the eigenvalues λ n = n 2 (n = 1, 2,...) and the eigenfunctions u n = sin(nx) x x x sin(x) sin(2x) sin(3x) sin(2x) The figures show, on the same plot, u n and u n+1 for n = 1, 2, 3. Please note how their zeroes interlace: between two consecutive zeroes on u n there is a zero of u n+1. Sturm s Comparison Theorem shows that this is a general feature. (Please note that the numbers of zeros is linked to the number of oscillations.) sin(3x) sin(4x) 6.2. Sturm s Comparison Theorem. Theorem 8. Consider two solutions u(x) and v(x) of two equations (34) [p(x)u ] + g(x)u = and (35) [p(x)v ] + h(x)v = where p(x) > on [a, b]. (i) If g(x) < h(x) on [a, b] then v(x) oscillates more rapidly than u(x): between any two zeroes of u(x) there is a zero of v(x). (ii) If g(x) < on [a, b] then u(x) does not oscillate: it vanishes at most once on [a, b].

18 18 RODICA D. COSTIN Sturms s Comparison Theorem applied to g(x) = q(x) + λw(x) with w > shows that if λ is negative enough there cannot be solutions of (34) which vanish at both endpoints a, b. However, once there is a solution vanishing at x = a and x = b, then the higher λ the more zeroes solutions have (i.e. more oscillations) and that zeroes of solutions for different λs do interlace. Proof of Sturm s Comparison Theorem (i) Multiplying (34) by v, (35) by u and subtracting it is found that d [ p(u v uv ) ] = (h g)uv dx therefore by integration x2 (36) p(u v uv ) (h g)uv x 2 = x 1 Choose x 1 < x 2 be two consecutive zeroes of u. Assume, to get a contradiction, that v has no zero on (x 1, x 2 ). Then v has a constant sign on (x 1, x 2 ), say v > (otherwise replace v by v). We can also assume u > on (x 1, x 2 ). Then the right-hand side of (36) is positive, while the left-hane side is negative since it equals p(x 2 ) u (x }{{} 2 ) v(x }{{} 2 ) p(x }{{} 1 ) u (x }{{} 1 ) v(x }{{} 1 ) }{{} > > > > which is a contradiction. (ii) For h(x) = : since (pv ) = then the general solution of (35) is x c 1 + c 2 a (1/p). Choosing v = 1 + x a (1/p) we have v > hence v has no zero. Therefore by (i) u has at most one zero.. 7. Existence of eigenvalues: The Prüfer transformation The differential equation (in selfadjoint form) is (37) (p(x)u ) + g(x)u = with g(x) = q(x) + λw(x) Assume that p(x) > and that p, g are continuous. Sturm s Comparison Theorem shows the oscillatory character of the solutions of (37). Polar coordinates in the phase space are then natural The Prüfer system. The Prüfer transformation consists in writing in polar coordinates the (phase-space like) quantities u and v = pu : (38) u(x) = r(x) sin θ(x), u (x) = r(x) p(x) The equation (37) is v + gu =, therefore x 1 cos θ(x) d (39) (r cos θ) + gr sin θ = dx The second equation is obtained from u = r/p cos θ which expanded gives (4) r sin θ + rθ cos θ = r/p cos θ

19 On the other hand, expanding (39) we obtain STURM-IOUVIE THEORY 19 (41) r cos θ rθ sin θ + gr sin θ = Solving (4), (41) for r, θ we obtain (42) r = 1 ( ) 1 2 p g r sin 2θ and (43) θ = g sin 2 θ + 1 p cos2 θ F (x, θ) (multiplying (4) by sin θ, multiplying (41) by cos θ and adding them up we obtain (42), while multiplying (4) by cos θ, multiplying (41) by sin θ and adding them up we obtain (43)). Remarkably, equation (43) is a first order equation for θ(x), independent of r(x)! With θ(x) determined by (43) we can integrate (42) to obtain r(x): r(x) = K exp x a 1 2 ( 1 p(s) g(s) ) sin 2θ(s) ds Note that r(x). Note that we can take the constant K = 1 (otherwise we divide u(x) by K). It is easy to see (from (38)) that the boundary conditions (17) become (44) θ(a) = θ a, θ(b) = θ b 7.2. Behavior of θ(x) and the zeroes of u(x). Choosing λ large enough assume that g(x) > on [a, b], so that oscillations are possible. (A) Note that (from (38)) u(x) = if and only if sin θ = so if and only if θ = kπ, k Z (B) On the other hand θ (x) = F (x, θ) >, so θ(x) is an increasing function. In fact, the larger g, the larger θ (the rate of increase of θ). (C) There can be no accumulation of zeroes of u(x) as the distance between two successive zeroes is no smaller that a positive number d. Why: Denoting M F = max{f (x, θ) x [a, b], θ [, 2π]} we have θ (x) < M F. Therefore, if x 1 < x 2 are two successive zeroes of u(x) then we have θ(x 1 ) = kπ and since θ(x 2 ) = (k+1)π and θ(x 2 ) θ(x 1 ) = θ (c)(x 2 x 1 ) for some c (a, b) it follows that (45) x 2 x 1 = 1 θ (c) (θ(x 2) θ(x 1 )) π M F d

20 2 RODICA D. COSTIN 7.3. Boundary conditions and existence of eigenvalues. We showed that the Sturm-iouville eigenvalue problem (18) becomes, in Prüfer coordinates: determine the values λ so that the following boundary value problem has a nontrivial solution: (46) θ = ( q(x) + λw(x)) sin 2 θ + 1 p(x) cos2 θ (47) θ(a) = θ a [, π) (48) θ(b) = θ b [nπ, (n + 1)π) To solve the problem, first solve the equation (46) with the initial condition (47). We obtain θ(x; λ) depending on the parameter λ, chosen large enough so that q(x) + λw(x) > on [a, b]. The solution θ(x; λ) has the following properties: 1) is continuous in λ (since g depends continuously on the parameter λ) and is increasing in λ (since θ is increasing in λ); 2) for x fixed lim λ θ(x; λ) = + (since θ is increasing in λ, without any bound); 3) for x fixed lim λ θ(x; λ) = (because for large λ equation (43) is approximately θ λw(x) sin 2 θ with the solution cot θ λ x c w. In the limit λ then cot θ ). Therefore, for some λ = λ n, condition (48) is satisfied. We thus find an increasing sequence of eigenvalues λ n, with λ n The distance between two consecutive zeroes of the eigenfunction u n. Using the formula (45) for g(x) = q(x) + λ n w(x) we obtain that the distance between two consecutive zeroes of the eigenfunction u n is no smaller that d n = π/m n where M n = max{[ q(x) + λ n w(x)] sin 2 θ + 1 p(x) cos2 θ x [a, b], θ [, 2π]} (Note that this estimate can be used only when p does not vanish at the endpoints of the interval [a, b].) 8. Completeness Consider the problem (18). We showed that there exists a sequence of eigenvalues λ 1 λ 2 λ 3... with λ n and that the corresponding eigenfunctions u n are orthogonal. The hypothesis of Theorem 4 are thus satisfied. It only remains to prove Theorem 4.

21 STURM-IOUVIE THEORY Proof of Theorem 4. As in finite dimensions, the eigenvalues of this selfadjoint operator can be calculated using the maximin principle. (We cannot speak about a minimax since there is no maximum eigenvalue). Recall that, using the Rayleigh quotient we have then R[u] = u, u u, u λ 1 = min R[u] λ 2 = min{r[u] u 1, u = }, λ 3 = min{r[u] u 1, u =, u 2, u = }... or, with the notation W 1 = Sp(u 1 ), W 2 = Sp(u 1, u 2 ),... W n = Sp(u 1, u 2,..., u n )... then λ 2 = min u W 1 R[u], λ 2 = min u W 2 R[u],..., λ n+1 = min u W n R[u]... et u n be eigenfunctions: u n = λ n u n which by assumption satisfy u n u k if n k (note that this is automatic if λ n λ k since is selfadjoint). We can assume u n = 1. To prove completeness of the eigenfunctions u n we first show that any f in the domain D of the operator can be expanded in terms of u n, in other words that the space S = Sp(u 1, u 2, u 2,...) is dense in D. Then since D was assumed to be dense in H it follows that S is dense in H, therefore u 1, u 2, u 3,... form a basis for the Hilbert space (see details in 8.2 below). To show that S is dense in D for any arbitrary f D, form the series (49) f n u n, with f n = u n, f and show that this series converges to f. Note that the partial sums of the series (49) belong to S: f [N] N u n, f u n S We show that the error when approximating f by partial sums (5) h [N] = f f [N] goes to zero as N, that is Since h [N] W N then λ N+1 = min u W n h [N] as N R[u] R[h [N] ] = h[n], h [N] h [N] 2

22 22 RODICA D. COSTIN and therefore (51) h [N] 2 1 λ N+1 h [N], h [N] Now expand, using (5), N N h [N], h [N] = f, f f n u n, f f n f, u n + = f, f N 2f n λ n u n, f + N n,m=1 N fnλ 2 n = f, f f n f m u n, u m N fnλ 2 n Since lim n λ n = + then the eigenvalues are positive staring with a certain rank p: suppose λ n for n > p. Then (for N > p) (52) N p N p f, f fnλ 2 n = f, f fnλ 2 n fnλ 2 n f, f fnλ 2 n Using (52) in (51) we obtain ( h [N] 2 1 f, f λ N+1 n=p+1 ) p fnλ 2 n as N which completes the proof of the convergence and of completeness If S is dense in D and D is dense in H then S is dense in H. Intuitively: the statement that D is dense in H means that for any f H we can find an f D D as close to f as we wish. Similarly, if S is dense in D then we can find f S S as close to f D as we wish. By the triangle s inequality: if f D is close to f, and f S is close to f D, then f S is close to f. So, let f H. We want (for an arbitrary ɛ > ) to find f S S so that d(f, f S ) < ɛ. But we can certainly find f D D so that d(f, f D ) < ɛ/2 and for that f D we can certainly find f S S so that d(f D, f D ) < ɛ/2. By the triangle s inequality then which proves the claim. d(f, f S ) < d(f, f D ) + d(f D, f S ) < ɛ/2 + ɛ/2 = ɛ 9. More examples of separation of variables 9.1. The wave equation. Vibrations of a string and propagation of waves is modeled by the wave equation (53) u tt = c 2 u xx It can be easily checked that if h 1,2 are two arbitrary twice differentiable functions then (54) u(x, t) = h 1 (x ct) + h 2 (x + ct)

23 STURM-IOUVIE THEORY 23 satisfies the wave equation and therefore (54) is the general solution of the wave equation. Note that c represents the speed of propagation of the wave. et us solve (53) for x [, ] and t >. For this we need the boundary conditions at x = and at x =, which we take for simplicity to be the homogeneous Dirichlet problem: (55) u(, t) =, u(, t) = and we also need initial conditions, at t =. Since (53) is second order in t we need the initial positions and the initial velocity. Therefore the conditions are (56) u(x, ) = f(x), u t (x, ) = g(x) Separation of variables. ooking for solutions of (53) in the form u(x, t) = X(x)T (t) the PDE becomes T (t)x(x) = c 2 T (t)x (x) therefore T (t) c 2 T (t) = X (x) = constant = λ X(x) (we include c 2 with T just for the convenience of solving the equation for X). The boundary conditions (55) imply that (57) X() =, X() = We solved the Sturm-iouville problem with (57); the solutions are (58) λ = λ n = n2 π 2 Next solve which gives X (x) + λx(x) = 2, n = 1, 2..., X n(x) = sin nπx T (t) + c 2 λ n T (t) = (59) T (t) = T n (t) = a n cos cnπt + b n sin cnπt (note the undetermined constants a n, b n ). We obtained the solutions [ (6) u(x, t) = u n (x, t) = a n cos cnπt + b n sin cnπt ] sin nπx which represent the modes of vibration: the mode u n has frequency ω n = 1 cnπ 2π = c 2 cycles per unit time, and it is called the nth harmonic. The harmonic with n = 1 is called the fundamental frequency, or first harmonic. In the case of the vibrating sting all the harmonic frequency are multiples of the fundamental one.

24 24 RODICA D. COSTIN Superposition. Since the PDE (53) is linear, then any sum of solutions (6) is again a solution, so let [ (61) u(x, t) = u n (x, t) = a n cos cnπt + b n sin cnπt ] sin nπx Initial conditions. We now require that the solution (61) satisfies (56): (62) u(x, ) = a n sin nπx = f(x) and (63) u t (x, ) = b n cnπ nπx sin = g(x) Assuming that f, g 2 [, ] we can expand them as a sine-series (recall that sin nπx are eigenfunctions of a selfadjoint operator satisfying the hypothesis of the Completeness Theorem 4), and Since we obtain a n = sin nπx 2 = (64) a n = 2 sin nπx, f sin nπx 2, b n = cnπ sin 2 nπx dx = 1 2 f(x) sin nπx dx, b n = 2 nπc sin nπx, g sin nπx 2 ( 1 cos 2nπx ) dx = 2 g(x) sin nπx Connection with the general solution. To see how the solution (61), (64) is related to the general solution (54) use the trigonometric formulas 2 sin A cos B = sin(a + B) + sin(a B) and 2 sin A sin B = cos(a B) cos(a + B) to rewrite (61) as where u(x, t) = + b n 2 [ an 2 ( (x + ct)nπt sin + sin ( (x ct)nπt cos + cos (x + ct)nπt ) (x ct)nπt )] 1 [F (x + ct) F (x ct)] 2 ( F (x) = a n sin nxπt + b n cos nxπt ) Note that u(x, ) = F (x) therefore F (x) is the initial condition. Note that the graph of F (x ct) (for any fixed t) is the same as the graph of F (x) (the initial shape), only shifted to the right by ct: it represents dx

25 STURM-IOUVIE THEORY 25 the initial wave traveling to the right with speed c. Similarly, F (x + ct) represents the initial wave traveling to the left with speed c aplace s equation. The two-dimensional heat equation (modeling the temperature distribution in a lamina) is u t = α 2 (u xx + u yy ) The stationary solutions (for which u(x, y, t) u(x, y)) satisfy aplace s equation (65) u xx + u yy = It is clear that in order to solve (65) we need information about the temperature on the boundary of the lamina. Consider as example a lamina in the shape of a seminfinite vertical strip: x and y and the Dirichlet problem: we need the temperature along the segment (x, ) with x, along the half lines (, y), y and (, y), y and for y + : (66) u(x, ) = f(x) for x u(, y) = for y u(, y) = for y lim y + u(x, y) = for x Separation of variables. ooking for solutions of (65) in the form u(x, Y ) = X(x)Y (y) the PDE becomes X (x)y (y) + X(x)Y (y) = therefore X (x) X(x) = Y (y) = λ Y (y) Solving for X(x) we obtain (58) then solving for Y (y): Y (y) = Y n (y) = a n exp( nπy/) + b n exp(nπy/) The last condition in (66) implies that b n = Solving the problem by superposition. Since the equation (65) is linear, then a sum of the solutions with separated variables is again a solution: u(x, y) = X n (x)y n (y) = a n exp( nπy/) sin nπx Requiring that the first condition in (66) be satisfied we obtain u(x, ) = a n sin nπx If f 2 [, ] then a n are found by (64). = f(x)

26 26 RODICA D. COSTIN 9.3. The vibrating rod. Transverse vibrations of a homogeneous rod is described by the equation (67) u xxxx + u tt = Assume that the rest position of the rod is for x π. As before, we solve by separation of variable: u(x, t) = X(x)T (t) which gives X (IV ) (x) = T (t) = λ X(x) T (t) The boundary conditions come from the system studied. For example, they could be 1) free ends: X = X = for x = and x = π, or 2) supported ends: X = X = for x = and x = π, or 3) clamped ends: X = X = for x = and x = π, or 4) X = X = for x = and x = π, or 5) periodicity: X() = X(π), X () = X (π), X () = X (π), X () = X (π). et us solve the problem with free ends: X (IV ) = λx(x) =, X = X = for x =, x = π Can we expect a complete set of eigenfunctions? Consider the operator = d4 in H = 2 [, π] on the domain dx 4 D = {f H f, f, f, f (IV ) H, f () = f () =, f (π) = f (π) = } Is the operator sefladjoint on D? For f, g D calculate f, g = = (68) = π π π f (IV ) (x)g(x) dx = f g π π f (x)g (x) dx f (x)g (x) dx = f g π π + f (x)g (x) dx f (x)g (x) dx = f g π π f (x)g (x) dx π f (x)g (x) dx = fg π π + f(x)g (IV ) (x) dx = f, g therefore is indeed selfadjoint on D. Note also that is positive semidefinite, so the eigenvalues λ are nonnegative. Indeed we have from (68) that f, f = π f (x) 2 dx and f, f = implies f (x) = hence f(x) = a + bx D. To calculate the eigenfunctions we find the general solution of the ODE.

27 For λ denote ν = λ 1/4 and then STURM-IOUVIE THEORY 27 (69) X(x) = c 1 cos(νx) + c 2 sin(νx) + c 3 e νx + c 4 e νx and for λ = X(x) = d 1 + d 2 x + d 3 x 2 + d 4 x 4 which belongs to D if d 3 = d 4 =. Hence the eigenspace corresponding to λ = is two-dimansional, consisting of linear functions X (x) = d 1 + d 2 x. Imposing the boundary condition in (69) we obtain a linear system of four equations for the four constants c 1,..., c 4. The condition that this system has a nontrivial solution is that its determinant be zero. Calculation of this determinant yields the condition (7) cosh νπ cos νπ = 1 which determines λ. However, (7) is a transcendental equation for ν (we cannot solve it explicitly). We can deduct that there is a sequence of solutions ν tending to + in the following way. Rewrite (7) as (71) cosh x = 1 cos x, (x = νπ) The function cosh x is increasing and greater than 1 for x >. The graph of the right-hand side of (71) has vertical asymptotes. On intervals ( π/2 + 2nπ, 2nπ] (n integer) it decreases from + to 1, and on intervals [2nπ, π/2+ 2nπ) it increases from 1 to +. Therefore, in each of these intervals there is a solution of equation (71). On the other intervals 1/ cos x is negative and equation (71) has no solutions. A sequence of nonnegative eigenvalues λ n do exist and λ n.the Completeness Theorem applies and the eigenfunctions are complete in 2 [π]. The equation is then solved using specified initial conditions. 1. An introduction to the Fourier transform 1.1. The space 2 (R). Consider any function in the vector space of continuous functions which are zero outside a closed interval: C (R) = {f : R C fcontinuous, f(x) = for all x [a, b] for some a < b} (they are called contiuous functions with compact support). If f C vanished outside some [a, b] then clearly it has finite 2 (R)-norm since f 2 = + f(x) 2 dx = b a f(x) 2 dx < The space 2 (R) is defined as the completion of C with respect to the 2 (R)-norm ( + ) 1/2 f = f(x) 2 dx

28 28 RODICA D. COSTIN and it is a Hilbert space with respect to the inner product f, g = + f(x)g(x) dx (which is finite by the Cauchy-Schwarts inequality). Remark If + f(x) 2 dx < then necessarily lim x ± f(x) =. The Hilbert space 2 (R) has many features common to 2 [a, b]; for example, if two functions differ at only a number of points (finitely many, or countably many) are considered equal in 2. One novel feature is that, for a function, even continuous on R, to belong to 2 (R), this function needs to decay to zero fast enough for x ± (so that the improper integral converges). For example, consider functions decaying like a power, say f(x) = 1/(1 + x a ). For which a is such a function in 2 (R)? For large x, f(x) x a and f(x) 2 x 2a which integrated gives a multiple of x 1 2a. The improper integral converges if 1 2a <, therefore for a > 1/2. Other examples of functions belonging to 2 (R) are e x, e x The Fourier Transform. For f C its Fourier transform, Ff, is defined as (72) ˆf(ξ) = (Ff)(ξ) = 1 2π e ixξ f(x) dx It can be shown that ˆf = f (Parseval s identity) and that the Fourier transform can be extended as a unitary operator from 2 (R) to itself, and that its inverse is: (73) f(x) = (F 1 ˆf)(x) = 1 2π e ixξ ˆf(ξ) dξ Note the similarity with Fourier series, only we have an integral instead of a series: recall f = n einx f n. Note also the similarity with matrix multiplication: if U ξ,x = e iξx and f = f x then (72) is like x U ξ,xf x while the inverse of U, which equals its adjoint, is U x,ξ = U ξ,x = e iξx. Remark. Some books define the Fourier transform (72) without the 1 prefactor 2π. In this case, the transform is no longer a unitary operator, and the inverse (73) must have the prefactor 1 2π.

29 STURM-IOUVIE THEORY The Fourier transform diagonalizes the diffferentiation operator. Indeed, consider the linear operator d dx in 2 (R), defined on the domain D = C 1 {f C f C } Just like for finite intervals, the space C 1 is dense in 2 (R). For f, g C 1 we have d f, g = f dx (x)g(x) dx = f(x)g(x) f(x)g d (x) dx = f, dx g and therefore d dx is skew-symmetric (hence its eigenvalues - if any!- are purely imaginary). We can see that e ixξ d (for ξ R) are eigenfunctions of dx. However they are generalized eigenfunctions, since they do not belong to the Hilbert space 2 (R), and we have expansions as integrals (72) rather than series. Indeed, d dx eixξ = iξe ixξ Moreover, integrating by parts we find that (F d dx f)(ξ) = 1 2π or (74) e ixξ f (x) dx = 1 2π df = iξ ˆf dx iξe ixξ f(x) dx = iξ(ff)(ξ) or F d dx F 1 is the operator of multiplication by iξ - hence it is diagonal! An example. Problem: solve the initial value problem for the heat equation on the line (75) u t = αu xx, for x R, t > with (76) u(x, ) = u (x) 2 (R) Take the Fourier transform in x in the heat equation: denoting we obtain which gives using (74) û(ξ, t) = 1 2π û t = αû xx û t = ξ 2 αû which is an ODE in t whose general solution is e ixξ u(x, t) dx (77) û(ξ, t) = F (ξ)e ξ2 αt

30 3 RODICA D. COSTIN Using the initial condition (76) we see that we must have F (ξ) = û (ξ) and taking the inverse Fourier transform in (77) we obtain the solution u(x, t) = 1 e ixξ ξ2 αtû (ξ) dξ 2π where = 1 2π = 1 2π K(x, y, t) = e ixξ ξ2 αt e iyξ u (y) dy dξ K(x, y, t) u (y) dy e ixξ ξ2 αt iyξ dξ which is easily calculated by completing the squares: ( ξ 2 αt i(x y)ξ = ξ α t i x y ) 2 2 α + t therefore 1 K(x, y, t) = e (x y)2 4αt 4παt and therefore the solution to (75), (76) is 1 u(x, t) = e (x y)2 4αt u (y) dy 4παt (x y)2 4αt