Math 412-501Theory of Partial Differential EquationsLecture 2-8:Sturm-Liouville eigenvalue problems(continued).Sturm-Liouville differential equation:ddxpdφdx+ qφ + λσφ = 0 (a < x < b),where p = p(x), q = q(x), σ = σ(x) are knownfunctions on [a, b] and λ is an unknown constant.Sturm-Liouville eigenvalue problem == Sturm-Liouville differential equation ++ linear homogeneous boundary conditionsEigenfunction: nonzero solution φ of the boundaryvalue problem.Eigenvalue: corresponding value of λ.ddxpdφdx+ qφ + λσφ = 0 (a < x < b).The equation is regular if p, q, σ are real andcontinuous on [a, b], and p, σ > 0 on [a, b].The Sturm-Liouville eigenvalue problem is regular ifthe equation is regular and boundary conditions areof the formβ1φ(a) + β2φ′(a) = 0,β3φ(b) + β4φ′(b) = 0,where βi∈ R, |β1| + |β2| 6= 0, |β3| + |β4| 6= 0.6 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.Heat flow in a nonuniform rod without sourcesInitial-boundary value problem:cρ∂u∂t=∂∂xK0∂u∂x(0 < x < L),∂u∂x(0, t) =∂u∂x(L, t) = 0, (insulated ends)u(x, 0) = f (x) (0 < x < L).We assume that K0(x), c(x), ρ(x) are positive andcontinuous on [0, L]. Also, we assume that f (x) ispiecewise smooth.Separation of variables: u(x, t) = φ(x)G (t).Substitute this into the heat equation:cρφdGdt=ddxK0dφdxG .Divide both sides by c(x)ρ(x)φ(x)G (t) = cρu:1GdGdt=1cρφddxK0dφdx= −λ = const.The variables have been separated:dGdt+ λG = 0,ddxK0dφdx+ λcρφ = 0.Boundary conditions∂u∂x(0, t) =∂u∂x(L, t) = 0 holdprovided φ′(0) = φ′(L) = 0.Eigenvalue problem:ddxK0dφdx+ λcρφ = 0, φ′(0) = φ′(L) = 0.This is a regular Sturm-Liouville eigenvalue problem(p = K0, q = 0, σ = cρ, [a, b] = [0, L]).There are infinitely many eigenvalues:λ1< λ2< . . . < λn< λn+1< . . .The corresponding eigenfunctions φnare unique upto multiplicative constants.Dependence on t:G′(t) = −λG (t) =⇒ G (t) = C0e−λtSolutions of the boundary value problem:u(x, t) = e−λntφn(x), n = 1, 2, . . .The general solution of the boundary value problemis a superposition of solutions with separatedvariables:u(x, t) =X∞n=1Cne−λntφn(x).Initial condition u(x, 0) = f (x) is satisfied whenf (x) =X∞n=1Cnφn(x).Hence Cnare coefficients of the generalized Fourierseries for f :Cn=ZL0f (x)φn(x)c(x)ρ(x) dxZL0φ2n(x)c(x)ρ(x) dx.Solution: u(x, t) =X∞n=1Cne−λntφn(x).In general, we do not know λnand φn.Nevertheless, we can determine limt→+∞u(x, t).We need to know which λnis > 0, = 0, < 0.ddxK0dφdx+ λcρφ = 0, φ′(0) = φ′(L) = 0.Rayleigh quotient:λ =−K0φφ′L0+ZL0K0(φ′)2dxZL0φ2cρ dx.Since φ′(0) = φ′(L) = 0, the nonintegral termvanishes. It follows that either λ > 0, or else λ = 0and φ = const. Indeed, λ = 0 is an eigenvalue.Solution of the heat conduction problem:u(x, t) =X∞n=1Cne−λntφn(x).Now we know that λ1= 0. Furthermore, we can setφ1= 1. Besides, 0 < λ2< λ3< . . .It follows thatlimt→+∞u(x, t) = C1=ZL0f (x)c(x)ρ(x) dxZL0c(x)ρ(x) dx.Rayleigh quotientConsider a regular Sturm-Liouville equation:ddxpdφdx+ qφ + λσφ = 0 (a < x < b).Suppose φ is a nonzero solution for some λ.Multiply the equation by φ and integrate over [a, b]:Zbaφddxpdφdxdx +Zbaqφ2dx + λZbaσφ2dx = 0.Integrate the first integral by parts:Zbaφddxpdφdxdx = pφdφdxba−Zbapdφdx2dx.It follows thatλ =−pφφ′ba+Zbap(φ′)2− qφ2dxZbaφ2σ dx.We have used only the facts that p, q, σ arecontinuous and that σ > 0.The Rayleigh quotient can be used for anyboundary conditions.Regular Sturm-Liouville equation:ddxpdφdx+ qφ + λσφ = 0 (a < x < b).Consider a linear differential operatorL(f ) =ddxpdfdx+ qf .Now the equation can be rewritten asL(φ) + λσφ = 0.Lemma Suppose f and g are functions on [a, b]such that L(f ) and L(g ) are well defined. ThengL(f ) − f L(g) =ddxp(gf′− fg′).Proof: L(f ) = (pf′)′+ qf , L(g ) = (pg′)′+ qg.Left-hand side:gL(f ) − f L(g) = g(pf′)′+ gqf − f (pg′)′− fqg= g (pf′)′− f (pg′)′.Right-hand side:ddxp(gf′− fg′)=ddxg(pf′) − f (pg′)= g′pf′+ g(pf′)′− f′pg′− f (pg′)′= g (pf′)′− f (pg′)′.Lagrange’s identity:gL(f ) − f L(g ) =ddxp(gf′− fg′)Integrating over [a, b], we obtain Green’s formula:ZbagL(f ) − f L(g )dx = p(gf′− fg′)baClaim If f and g satisfy the same regular boundaryconditions, then the right-hand side in Green’sformula vanishes.Proof: We have thatβ1f (a) + β2f′(a) = 0, β1g(a) + β2g′(a) = 0,where β1, β2∈ R, |β1| + |β2| 6= 0.Vectors (f (a), f′(a)) and (g (a), g′(a)) areorthogonal to vector (β1, β2). Since (β1, β2) 6= 0, itfollows that (f (a), f′(a)) and (g (a), g′(a)) areparallel. Then their vector product is equal to 0:(g(a), g′(a)) × (f (a), f′(a)) = g (a)f′(a) − f (a)g′(a) = 0.Similarly, g(b)f′(b) − f (b)g′(b) = 0.Hencep(gf′− fg′)ba=