Chapter 30 Lecture 34 — Sturm Liouville Theory 30.1 Properties of SL Problems: 1. Eigenvalue Properties: • (a) the eigenvalues λ are all real. • (b) there are an infinite number of eigenvalues λj with λ1 < λ2 < . . . λj → ∞ as j → ∞. • (c) λj > 0 provided α1 α2 < 0, β1 β2 > 0 and q(x) > 0. 2. Eigenfunction Properties: Corresponding to each eigenvalue λj there is an eigenfunction φj (x) that is unique up to multiplication by a constant, and which satisfy: • (a) φj (x) are real and can be normalized 0 r(x)φ2j (x) dx = 1. • (b) the eigenfunctions corresponding to different eigenvalues are orthogonal with respect to the weight function r(x): r(x)φj (x)φk (x) dx = 0 if j = k. (30.1) 0 • (c) φj (x) has exactly j − 1 zeros on 0 < x < . 171 Lecture 34 — Sturm Liouville Theory 3. Expansion Property: The eigenfunctions φj (x) form a complete set so that any piecewise smooth function f (x) can be expanded as a generalized Fourier Series: f (x) = ∞ cn φn (x) (30.2) n=1 by orthogonality: cn = 0 r(x)f (x)φn (x) dx 0 30.2 . (30.3) r(x)φ2n (x) dx Lagrange’s Identity: Lagrange’s Identity (vLu − uLv) dx = −p(x)u v 0 + p(x)uv 0 0 . is fundamental to the development of S-L Theory. Proof: Let u and v be any sufficiently differentiable functions; then vLu dx = v −(pu ) + qu dx 0 (30.4) 0 = −vpu 0 + u pv dx + 0 = uqv dx (30.5) 0 −vpu 0 + upv 0 + u −(pv ) + qv dx(30.6) 0 vLu dx = Therefore −pvu 0 0 + puv 0 + uLv dx. (30.7) 0 Now suppose that u and v both satisfy the SL boundary conditions. I.E. α1 u(0) + α2 u (0) = 0 α1 v(0) + α2 v (0) = 0 172 β1 u() + β2 u () = 0 β1 v() + β2 v () = 0 (30.8) 30.3. PROOFS TO SELECTED PROPERTIES: then vLu dx − 0 uLv dx = −p()u ()v() + p()u()v () 0 (30.10) +p(0)u (0)v(0) − p(0)u(0)v (0) β1 β1 = p() + u()v() + u() − v() (30.11) β2 β2 α1 α1 (30.12) +p(0) − u(0)v(0) − u(0) − v(0) α2 α2 = 0. (30.13) Thus vLu dx = 0 condition. Note: uL?v dx whenever u and v satisfy the SL boundary 0 • If L and BC are such that uLv dx then L is said to be vLu dx = 0 0 self-adjoint. Notation if we define (f, g) = f (x)g(x) dx then we 0 may write (v, Lv) = (u, Lv). 30.3 (30.9) Proofs to selected properties: (1a) λ is real: Let Ly = λry (1) α1 y(0) + α2 y (0) = 0 β1 y() + β2 y () = 0. Take the conjugate of (1) Lȳ = λ̄rȳ. By Lagrange’s Identity: 0 = (ȳ, Ly) − (y, Lȳ) (30.14) = (ȳ, rλy) − (y, rλ̄ȳ) = ȳ(x)rλy(x) dx − y(x)r(x)λ̄ȳ(x) dx 0 (30.15) (30.16) 0 = (λ − λ̄) 2 r(x)y(x) dx (30.17) 0 173 Lecture 34 — Sturm Liouville Theory Since r(x)|y(x)|2 ≥ 0 it follows that λ = λ̄ ⇒ λ is real. (1c) λj > 0 provided α1 /α2 < 0 β1 /β2 > 0 and q(x) > 0. Consider Ly = −(py ) + qy = λry (SL) and multiply (SL) by y and integrate from 0 to : (y, Ly) = 2 −(py ) y + qy dx = λ 0 Therefore λ = 0 2 r(x) y(x) dx (30.18) 0 −(py ) y + qy 2 dx this is known as Rayleigh’s Quotient. ry 2 dx 0 [−py y]0 + = p(y )2 + qy 2 dx 0 (30.19) ry 2 dx 0 β1 2 2 +p() β2 y() − p(0) αα12 y(0) + p(y )2 + qy 2 dx = 0 . (30.20) ry 2 dx 0 Therefore λ > 0 since the RHS is all positive. Note: If q(x) ≡ 0 and α1 = 0 = β1 then with y (0) = 0 = y () we have nontrivial eigenfunction y(x) = 1 and eigenvalue λ = 0. (2b) Eigenfunctions corresponding to different eigenvalues are orthogonal. Consider two distinct eigenvalues λj = λk λj : Lφj = rλj φj and λk : Lφk = rλk φk . Then 0 = (φK , Lφj ) − (φj , LφK ) by Lagrange’s Identity = (φK , rλj φj ) − (φj , rλK φK ) = (λj − λK ) r(x)φK (x)φj (x) dx (30.21) (30.22) (30.23) 0 now λj = λk?K implies that r(x)φk (x)φj (x) dx = 0. 0 174 (30.24) 30.3. PROOFS TO SELECTED PROPERTIES: (3) Expansion Property It is difficult to prove the convergence of the eigenfunction series expansion for f (x) that is piecewise smooth. However, if we assume the expansion converges then it is a simple matter to use orthogonality to determine the coefficients in the expansion: Let f (x) = ∞ cn φn (x). n=1 f (x)φm (x)r(x) dx = ∞ n=1 0 cn r(x)φm (x)φn (x) dx (30.25) 0 orthogonality implies cm = 0 r(x)f (x)φm (x) dx 0 2 r(x) φm (x) dx . (30.26) 175