Sturm-Liouville Theory

Transcription

1 More on Ryan C. Trinity University Partial Differential Equations April 19, 2012

2 Recall: A Sturm-Liouville (S-L) problem consists of A Sturm-Liouville equation on an interval: (p(x)y ) + (q(x) + λr(x))y = 0, a < x < b, together with Boundary conditions, i.e. specified behavior of y at x = a and x = b. Such a problem is called regular if: The boundary conditions are of the form c 1 y(a) + c 2 y (a) = 0, d 1 y(b) + d 2 y (b) = 0, where (c 1,c 2 ),(d 1,d 2 ) (0,0); p, q and r satisfy certain regularity conditions on [a,b].

3 A nonzero function y that solves an S-L problem is called an eigenfunction, and the corresponding value of λ is called an eigenvalue. Eigenvalues and eigenfunctions of (regular) S-L problems have very nice properties. Theorem The eigenvalues of a regular S-L problem form an increasing sequence of real numbers with λ 1 < λ 2 < λ 3 < lim λ n =. n Moreover, the eigenfunction y n corresponding to λ n is unique (up to a scalar multiple), and has exactly n 1 zeros in the interval a < x < b.

4 Theorem Suppose that y j and y k are eigenfunctions corresponding to distinct eigenvalues λ j and λ k of a (regular) S-L problem. Then y j and y k are orthogonal on [a,b] with respect to the weight function w(x) = r(x). That is y j,y k = b a y j (x)y k (x)r(x)dx = 0. We have put the word regular in parentheses because this result actually holds for certain non-regular S-L problems, too. We will look at the proof of this result to see just where regularity is needed.

5 Proof of orthogonality If (y j,λ j ), (y k,λ k ) are eigenfunction/eigenvalue pairs then (py j + (q + λ j r)y j = 0, (py k ) + (q + λ k r)y k = 0. Multiply the first by y k and the second by y j, then subtract to get (py j ) y k (py k ) y j + (λ j λ k )y j y k r = 0. Moving the λ-terms to one side and adding zero, we get (λ j λ k )y j y k r = (py k ) y j (py j ) y k = (py k ) y j + py k y j py j y k (py j ) y k = (py k y j py j y k) = ( p(y k y j y jy k ) ).

6 If λ j λ k, we can divide by λ j λ k and then integrate to get y j,y k = b a ( ) p(x) y k (x)y j(x) y j (x)y k(x) b y j (x)y k (x)r(x)dx = λ j λ. k This proves the orthogonality of y j and y k whenever the RHS equals zero. This is guaranteed to happen if p(a) ( y k (a)y j(a) y j (a)y k(a) ) = p(b) ( y k (b)y j(b) y j (b)y k(b) ) = 0. These equalities occur when: y k (a)y j(a) y j (a)y k(a) = 0 or p(a) = 0; }{{}}{{} A A y k (b)y j(b) y j(b)y k (b) = 0 }{{} B or p(b) = 0. }{{} B While these conditions are sufficient for orthogonality, it should be pointed out that they are not necessary. a

7 Orthogonality for regular S-L problems If our S-L problem is regular then at x = a we have c 1 y j (a) + c 2 y j(a) = 0, c 1 y k (a) + c 2 y k (a) = 0, or in matrix form ( yj (a) y j (a) y k (a) y k (a) )( c1 c 2 ) = ( 0 0 ). Since (c 1,c 2 ) (0,0) the determinant must be zero, that is y j (a)y k (a) y k(a)y j (a) = 0, which is condition A. Likewise, the boundary condition at x = b gives condition B, which verifies orthogonality.

8 Examples Example Use the preceding results to verify orthogonality of the eigenfunctions of y + λy = 0, 0 < x < L, y(0) = y(l) = 0. This is a regular S-L problem with eigenfunctions y n = sin(nπx/l). Since r(x) = 1, we immediately deduce that for m n. L 0 sin ( mπx ) ( nπx ) sin dx = 0 L L

9 Example If m 0, use the preceding results to verify orthogonality of the eigenfunctions of x 2 y + xy + (λ 2 x 2 m 2 )y = 0, 0 < x < a, y(0) is finite, y(a) = 0. This is a singular S-L problem with eigenfunctions y n = J m (α mn x/a). Since p(x) = x, p(0) = 0. This gives condition A. Since the boundary condition y(a) = 0 is regular, we get also get condition B. With r(x) = x, we immediately deduce that for k l. a 0 J m ( αmk a x ) J m ( αml a x ) x dx = 0

10 Example Use the preceding results to verify orthogonality of the eigenfunctions of y + λy = 0, p < x < p, y( p) = y(p), y ( p) = y (p). This is an S-L problem with 2p-periodic boundary conditions. It is left as an exercise to verify that the eigenvalues are λ n = n2 π 2 p 2 for n = 0,1,2,3,... with eigenfunctions y n = cos nπx p or sin nπx p.

11 Although A, A, B and B may not hold, the periodic boundary conditions imply that ( y k (p)y j (p) y j(p)y k (p) ) ( y k ( p)y j( p) y j( p)y k ( p) ) = 0. Since r(x) = 1, this immediately implies the orthogonality relations for m n. p p p p p p sin mπx p cos mπx p sin mπx p sin nπx p cos nπx p cos nπx p dx = 0, dx = 0, dx = 0,

12 Fourier convergence for S-L problems The eigenfunctions of an S-L problem provide a family of orthogonal functions. As with sine and cosine, we can use these to give series expansions for sufficiently nice functions. Theorem Let y 1,y 2,y 3,... be the eigenfunctions of a regular S-L problem on [a,b]. If f is piecewise smooth on [a,b], then f (x+) + f (x ) 2 = A n y n (x), n=1 where A n = f,y n y n,y n = b a b f (x)y n (x)r(x)dx. yn 2 (x)r(x)dx a

13 Remarks The series n=1 A ny n is called the eigenfunction expansion of f. f (x+)+f (x ) Recall that f (x) = 2 anywhere f is continuous. So the eigenfunction expansion is equal to f at most points. Although we have only stated this result for regular S-L problems, it frequently holds for singular problems as well. The original Fourier convergence theorem provides an example of this phenomenon (the S-L problem involved in this case is non-regular).

14 The hanging chain Consider a chain (or heavy rope, cable, etc.) of length L hanging from a fixed point, subject to only to downward gravitational force. x x = L u Chain at rest Displaced chain We place the chain along the (vertical) x-axis, displace the chain from rest, and let u(x,t) = Horizontal deflection of chain from equilibrium at height x and time t.

15 Under ideal assumptions (e.g. planar motion, small deflection, no energy loss due to friction or air resistance, etc.) we obtain the boundary value problem where u tt = g (xu xx + u x ), 0 < x < L, t > 0, u(l,t) = 0, t > 0, u(x,0) = f (x), u t (x,0) = v(x), f (x) is the initial shape of the chain, v(x) is the initial (horizontal) velocity of the chain, g is the acceleration due to gravity.

16 Writing u(x,t) = X(x)T(t), separation of variables (and physical considerations) yields T + λ 2 gt = 0, t > 0, xx + X + λ 2 X = 0, 0 < x < L, X(0) finite, X(L) = 0. The general solution for T is T(t) = A cos ( gλt) + B sin( gλt). The ODE for X can be rewritten as (xx ) + λ 2 X = 0, yielding a singular S-L problem (with p(x) = x, q(x) = 0, r(x) = 1, and parameter λ 2 ).

17 To find the eigenfunctions, we substitute s = 2 x. This yields the parametric Bessel equation of order 0: s 2d2 X ds 2 + s dx ds + λ2 s 2 X = 0, 0 < s < 2 L, X(0) finite, X(2 L) = 0. As we have seen, this means λ = λ n = α n 2 L ( ) αn s X(s) = X n (s) = J 0 2, L where α n is the nth positive zero of J 0. Back-substitution then gives ( ) x X(x) = X n (x) = J 0 α n. L

18 From this we find that T(t) = T n (t) = A n cos ( gλ n t) + B n sin( gλ n t) ( ) ( ) g α n t g α n t = A n cos + B n sin, L 2 L 2 and superposition gives the general solution u(x,t) = = X n (x)t n (t) n=1 n=1 ( )( ( x g J 0 α n A n cos L L ) ( α n t g + B n sin 2 L )) α n t. 2

19 The initial shape condition requires that ( ) x f (x) = u(x,0) = A n J 0 α n. L n=1 }{{} X n(x) According to S-L theory, this means that L ( ) x A n = f,x f (x)j 0 α n dx n 0 L = X n,x n L ( ) x J0 2 α n dx L = 0 1 LJ 2 1 (α n) L 0 ( ) x f (x)j 0 α n dx. L Setting u t (x,0) = v(x) and using similar reasoning yields 2 L ( ) x B n = α n J1 2(α n) v(x)j 0 α n dx. gl L 0