Math 412-501Theory of Partial Differential EquationsLecture 2-10: Sturm-Liouville eigenvalueproblems (continued). Hilbert space.Regular Sturm-Liouville eigenvalue problem:ddxpdφdx+ qφ + λσφ = 0 (a < x < b),β1φ(a) + β2φ′(a) = 0,β3φ(b) + β4φ′(b) = 0.Here βi∈ R, |β1| + |β2| 6= 0, |β3| + |β4| 6= 0.Functions p, q, σ are continuous on [a, b],p > 0 and σ > 0 on [a, b].6 properties of a regular Sturm-Liouville problem• Eigenvalues are real.• Eigenvalues form an increasing sequence.• n-th eigenfunction has n − 1 zeros in (a, b).• Eigenfunctions are orthogonal with weight σ.• Eigenfunctions and eigenvalues are relatedthrough the Rayleigh quotient.• Piecewise smooth functions can be expandedinto generalized Fourier series of eigenfunctions.Hilbert spaceHilbert space is an infinite-dimensional analog ofEuclidean space. One realization isL2[a, b] = {f :Rba|f (x)|2dx < ∞}.Inner product of functions:hf , gi =Zbaf (x)g(x) dx.If f and g take complex values, thenhf , gi =Zbaf (x)g(x) dxso thathf , f i =Zba|f (x)|2dx ≥ 0.Norm of a function: kf k =phf , f i.Cauchy-Schwarz inequality: |hf , gi| ≤ kf k · kg k.If f , g are real-valued, then hf , g i = kf k · kgk cos θ,where θ is called the angle between f and g.Convergence: we say that fn→ f in the mean ifkf − fnk → 0 as n → ∞.Lemma If fn→ f in the mean thenhfn, gi → hf , g i for any g ∈ L2[a, b].Proof:|hf , g i − hfn, gi| = |hf − fn, gi| ≤ kf − fnk · kgk.Functions f , g ∈ L2[a, b] are called orthogonal ifhf , g i = 0.Alternative inner product:hf , g iw=Zbaf (x)g(x)w(x) dx,where w > 0 is the weight function.Functions f and g are called orthogonal withweight w if hf , g iw= 0.Alternative inner product means an alternativemodel of the Hilbert space:L2([a, b], w dx) = {f :Rba|f (x)|2w(x) dx < ∞}.A set f1, f2, . . . of pairwise orthogonal nonzerofunctions is called complete if it is maximal, i.e.,there is no nonzero function g such that hg , fni = 0,n = 1, 2, . . . .A complete set forms a basis of the Hilbert space,that is, each function g ∈ L2[a, b] can be expandedinto a seriesg =X∞n=1cnfnthat converges in the mean.Thenhg, hi =X∞n=1cnhfn, hifor any h ∈ L2[a, b].In particular,hg, fmi =X∞n=1cnhfn, fmi = cmhfm, fmi.=⇒ the expansion is unique: cm=hg, fmihfm, fmi.Also,hg, g i =X∞n=1cnhfn, gi =X∞n=1|cn|2hfn, fni.hg, g i =X∞n=1|hg, fni|2hfn, fni(Parseval’s equality)Suppose that f1, f2, . . . is an orthonormal basis,i.e., kfnk = 1. Theng =X∞n=1cnfn, where cn= hg , fni.Parseval’s equality becomes kgk2=X∞n=1|cn|2.If h =X∞n=1dnfn, thenhg, hi =X∞n=1X∞m=1cndmhfn, fmi =X∞n=1cndn.Which sequences c1, c2, . . . are allowed ascoefficients of an expansion?Theorem For any sequence c1, c2, . . . such thatP∞n=1|cn|2< ∞, the seriesX∞n=1cnfnconverges in the mean to some functiong ∈ L2[a, b].This gives rise to another model of the Hilbertspace: ℓ2= {(c1, c2, . . . ) :P∞n=1|cn|2< ∞}.Given c = (c1, c2, . . . ), d = (d1, d2, . . . ) ∈ ℓ2,the inner product ishc, di =X∞n=1cndn.Suppose f1, f2, . . . is a set of pairwise orthogonalnonzero functions in L2[a, b] that is not complete.For any function g ∈ L2[a, b], we can still composea seriesP∞n=1cnfn, where cn=hg, fnihfn, fni.This series converges in the mean to some functiong0∈ L2[a, b]. In general, g 6= g0but g − g0isorthogonal to f1, f2, . . . .Then g =P∞n=1cnfn+ (g − g0) implieskgk2=P∞n=1kcnfnk2+ kg − g0k2≥P∞n=1kcnfnk2.Bessel’s inequality:hg, g i ≥X∞n=1|hg, fni|2hfn, fniL: linear operator in the Hilbert space L2[a, b].In general, L is not defined on the whole space buton a linear subspace H ⊂ L2[a, b] which is dense.Example. L(f ) = (pf′)′+ qf .L is called self-adjoint (or symmetric) ifhL(f ), g i = hf , L(g)i for all f , g ∈ H.If L(f ) = λf for some λ ∈ C and nonzero f ∈ H,then λ is an eigenvalue and f is an eigenfunction.If the operator L is self-adjoint, then• all eigenvalues are real;• eigenfunctions belonging to different eigenvaluesare orthogonal.Regular Sturm-Liouville equation:ddxpdφdx+ qφ + λσφ = 0 (a < x < b).Consider a linear differential operatorL(f ) = (pf′)′+ qf .Now the equation can be rewritten asL(φ) + λσφ = 0.Green’s formula:ZbagL(f ) − f L(g )dx = p(gf′− fg′)baIf f and g satisfy the same regular boundaryconditions, thenZbagL(f ) − f L(g )dx = 0.That is, L is self-adjoint on the set of functionssatisfying particular boundary conditions.L(φ) + λσφ = 0 =⇒ −σ−1L(φ) = λφSo eigenvalues/eigenfunctions of the St urm-Liouvilleproblem are not those of operator L but those ofoperator M = −σ−1L.The operator M is self-adjoint with respect to theinner product h·, ·iσ.Eigenvalue problem:φ′′+ λφ = 0, φ(0) = φ(L) = 0.Eigenvalues: λn= (nπL)2, n = 1, 2, . . .Eigenfunctions: φn(x) = sinnπxL.Since this is a regular Sturm-Liouville problem,eigenfunctions form a complete orthogonal set(a basis) in the Hilbert space L2[0, L].Any function f ∈ L2[0, L] is expanded into a seriesf =X∞n=1cnφnthat converges in the mean.Coefficients:cn=hf , φnihφn, φni=2LZL0f (x) sinnπxLdx.So the Fourier series always converges in t he mean.Parseval’s equality:hf , f i =X∞n=1|hf , φni|2hφn, φni=X∞n=1|cn|2hφn, φni.2LZL0|f (x)|2dx =X∞n=1|cn|2(Parseval’s equality for Fourier sine series)Example. f (x) = 2x, 0 ≤ x ≤ π.f (x) ∼X∞n=1(−1)n+14nsin nxParseval’s equality:2πZπ0(2x)2dx =X∞n=1|cn|2=X∞n=116n2.2π·4π33=X∞n=116n2X∞n=11n2=π26Simplicity of eigenvaluesRegular Sturm-Liouville equation:(pφ′)′+ qφ + λσφ = 0 (a < x < b).Initial value problem φ(a) = C0, φ′(a) = C1alwayshas a unique solution.Suppose φ and ψ are eigenfunctions of a regularproblem corresponding to the same eigenvalue λ.Then β1φ(a) + β2φ′(a) = β1ψ(a) + β2ψ′(a) = 0,where β1, β2∈ R, |β1| + |β2| 6= 0.It follows that (φ(a), φ′(a)) = c(ψ(a), ψ′(a)), c ∈ R.Now φ and cψ are solutions to the same initialvalue problem. Hence φ =