Math Theory of Partial Differential Equations Lecture 29: SturmLiouville eigenvalue problems (continued).


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1 Mth Theory of Prtil Differentil Equtions Lecture 29: SturmLiouville eigenvlue problems (continued).
2 Regulr SturmLiouville eigenvlue problem: d ( p dφ ) + qφ + λσφ = 0 ( < x < b), dx dx β 1 φ() + β 2 φ () = 0, β 3 φ(b) + β 4 φ (b) = 0. Here β i R, β 1 + β 2 0, β 3 + β 4 0. Functions p, q, σ re continuous on [, b], p > 0 nd σ > 0 on [, b].
3 6 properties of regulr SturmLiouville problem Eigenvlues re rel. Eigenvlues form n incresing sequence. nth eigenfunction hs n 1 zeros in (, b). Eigenfunctions re orthogonl with weight σ. Eigenfunctions nd eigenvlues re relted through the Ryleigh quotient. Piecewise smooth functions cn be expnded into generlized Fourier series of eigenfunctions.
4 Regulr SturmLiouville eqution: d ( p dφ ) + qφ + λσφ = 0 ( < x < b). dx dx Consider liner differentil opertor L(f ) = d ( p df ) + qf. dx dx Now the eqution cn be rewritten s L(φ) + λσφ = 0.
5 Lgrnge s identity: gl(f ) f L(g) = d dx ( ) p(gf fg ) Integrting over [, b], we obtin Green s formul: b ( ) gl(f ) f L(g) dx = p(gf fg ) b Clim If f nd g stisfy the sme regulr boundry conditions, then the righthnd side in Green s formul vnishes.
6 Suppose φ n nd φ m re eigenfunctions of the SturmLiouville problem corresponding to eigenvlues λ n nd λ m : L(φ n ) + λ n σφ n = 0, L(φ m ) + λ m σφ m = 0. Since φ n nd φ m stisfy the sme regulr boundry conditions, Green s formul implies tht b ( ) φ m L(φ n ) φ n L(φ m ) dx = 0 = b If λ n λ m, then (λ m λ n )φ n (x)φ m (x)σ(x) dx = 0 b φ n (x)φ m (x)σ(x) dx = 0.
7 Suppose φ is complexvlued eigenfunction corresponding to complex eigenvlue λ: L(φ) + λσφ = 0, β 1 φ() + β 2 φ () = 0, β 3 φ(b) + β 4 φ (b) = 0. We re going to show tht λ R. Any complex number z = x + iy is ssigned its complex conjugte z = x iy. Let us pply the complex conjugcy to the Sturmliouville eqution nd the boundry conditions.
8 L(φ) + λσφ = 0, β 1 φ() + β 2 φ () = β 3 φ(b) + β 4 φ (b) = 0. It is known tht z 1 + z 2 = z 1 + z 2 nd z 1 z 2 = z 1 z 2. L(φ) + λ σ φ = 0, β 1 φ()+β 2 φ () = β 3 φ(b)+β 4 φ (b) = 0. If z is rel then z = z. L(φ) + λσφ = 0, β 1 φ()+β 2 φ () = β 3 φ(b)+β 4 φ (b) = 0.
9 Let φ denote the complex conjugte function of φ, i.e., φ(x) = φ(x) for x b. We hve tht φ = f + ig, where f nd g re relvlued functions. Then φ = f ig. Note tht φ = (f ig) = f ig = f + ig = φ. It follows tht L(φ) = (pφ ) + qφ = (pφ ) + qφ = ( pφ ) + qφ = ( pφ ) + qφ = L(φ).
10 L ( φ ) + λσφ = 0, β 1 φ() + β 2 φ () = β 3 φ(b) + β 4 φ (b) = 0. If φ is n eigenfunction belonging to n eigenvlue λ, then φ is n eigenfunction belonging to the eigenvlue λ. Assume tht λ λ. Then But b b φ(x)φ(x)σ(x) dx = 0. φ(x)φ(x)σ(x) dx = Thus λ = λ = λ R. b φ(x) 2 σ(x) dx > 0.
11 Some fcts bout Eucliden spce Eucliden spce R 3. Let v = (v 1, v 2, v 3 ), u = (u 1, u 2, u 3 ) be two vectors. v u = v 1 u 1 + v 2 u 2 + v 3 u 3 is the dot product. v nd u re orthogonl if v u = 0. v = v v. Vectors e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1) form n orthonorml bsis. v = v 1 e 1 + v 2 e 2 + v 3 e 3 = (v e 1 )e 1 + (v e 2 )e 2 + (v e 3 )e 3.
12 Let v 1,v 2,v 3 be orthogonl nonzero vectors. They form bsis in R 3 so tht for ny u R 3 we hve u = c 1 v 1 + c 2 v 2 + c 3 v 3. Note tht u v n = c n v n v n so tht c n = u v n v n v n. Pythgoren theorem implies tht u 2 = c 1 v c 2 v c 3 v 3 2. Observe tht c n v n 2 = c n 2 v n v n. Hence u u = u v 1 2 v 1 v 1 + u v 2 2 v 2 v 2 + u v 3 2 v 3 v 3 (Prsevl s equlity)
13 Let v 1,v 2 be orthogonl nonzero vectors. Given vector u R 3, let u 0 = u (c 1 v 1 + c 2 v 2 ), where c n = u v n v n v n. It is esy to check tht u 0 v 1 = u 0 v 2 = 0 so tht u 0 (u u 0 ) = 0. Pythgoren theorem implies tht u 2 = c 1 v c 2 v u 0 2 c 1 v c 2 v 2 2. Since c n v n 2 = c n 2 v n v n, we get u u u v 1 2 v 1 v 1 + u v 2 2 v 2 v 2 (Bessel s inequlity)
14 Suppose A nd B re liner opertors in R 3. We sy tht B is djoint to A (denoted B = A ) if Au v = u Bv for ll u,v R 3. Let A = ( ij ) 1 i,j 3, B = (b ij ) 1 i,j 3. Then Ae j = 1j e 1 + 2j e 2 + 3j e 3, hence ij = Ae j e i. Similrly, b ij = Be j e i = e i Be j. It follows tht ij = b ji, i.e., B is the trnspose of A. A is clled selfdjoint if A = A. Selfdjoint opertors hve only rel eigenvlues. Suppose v 1, v 2 re eigenvectors of A belonging to eigenvlues λ 1, λ 2. Then λ 1 v 1 v 2 = Av 1 v 2 = v 1 Av 2 = λ 2 v 1 v 2. If λ 1 λ 2 then v 1 v 2 = 0.
15 From Eucliden spce to Hilbert spce Hilbert spce is n infinitedimensionl nlogue of Eucliden spce. One reliztion is L 2 [, b] = {f : b f (x) 2 dx < }. Inner product of functions: f, g = b f (x)g(x) dx. Since fg 1 2 ( f 2 + g 2 ), the inner product is well defined for ny f, g L 2 [, b]. Norm of function: f = f, f. Convergence: we sy tht f n f in the men if f f n 0 s n.
16 Functions f, g L 2 [, b] re clled orthogonl if f, g = 0. Alterntive inner product: f, g w = b where w is the weight function. f (x)g(x)w(x) dx, Functions f nd g re clled orthogonl with weight w if f, g w = 0.
17 A set f 1, f 2,... of pirwise orthogonl nonzero functions is clled complete if it is mximl, i.e., there is no nonzero function g such tht g, f n = 0, n = 1, 2,.... A complete set forms bsis of the Hilbert spce, tht is, ech function g L 2 [, b] cn be expnded into series g = n=1 c nf n tht converges in the men. The expnsion is unique: c n = g, f n f n, f n.
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