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TAMU MATH 412 - Lecture3-2web

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Math 412-501Theory of Partial Differential EquationsLecture 3-2:Spectral properties of the Laplacian.Bessel functions.Eigenvalue problem:∇2φ + λφ = 0 in D,αφ + β∂φ∂n∂D= 0,where α, β are piecewise continuous real functionson ∂D such that |α| + |β| 6= 0 everywhere on ∂D.We assume that the boundary ∂D is piecewisesmooth.6 spectral properties of the LaplacianProperty 1. All eigenvalues are real.Property 2. All eigenvalues can be arranged inthe ascending orderλ1< λ2< . . . < λn< λn+1< . . .so that λn→ ∞ as n → ∞.This means that:• there are infinitely many eigenvalues;• there is a smallest eigenvalue;• on any finite interval, there are only finitelymany eigenvalues.Property 3. An eigenvalue λnmay be multiplebut its multiplicity is finite.Moreover, the smallest eigenvalue λ1is simple,and the corresponding eigenfunction φ1has no zerosinside the domain D.Property 4. Eigenfunctions corresponding todifferent eigenvalues are orthogonal relative to theinner producthf , gi =ZZDf (x, y)g (x, y ) dx dy.Property 5. Any eigenfunction φ can be relatedto its eigenvalue λ through the Rayleigh quotient:λ =−I∂Dφ∂φ∂nds +ZZD|∇φ|2dx dyZZD|φ|2dx dy.Property 6. There exists a sequence φ1, φ2, . . . ofpairwise orthogonal eigenfunctions that is completein the Hilbert space L2(D).Any square-integrable function f ∈ L2(D) isexpanded into a seriesf (x, y) =X∞n=1cnφn(x, y ),that converges in the mean. The series is unique:cn=hf , φnihφn, φni.If f is piecewise smooth then the series convergespointwise to f at points of continuity.Rayleigh quotientSuppose that ∇2φ = −λφ in the domain D.Multiply both sides by φ and integrate over D:ZZDφ ∇2φ dx dy = −λZZD|φ|2dx dy .Green’s formula:ZZDψ ∇2φ dA =I∂Dψ∂φ∂nds −ZZD∇ψ · ∇φ dAThis is an analog of integration by parts. NowI∂Dφ∂φ∂nds −ZZD|∇φ|2dx dy = −λZZD|φ|2dx dy .It follows thatλ =−I∂Dφ∂φ∂nds +ZZD|∇φ|2dx dyZZD|φ|2dx dy.If φ satisfies the boundary condition φ|∂D= 0 or∂φ∂n∂D= 0 (or mixed), then the one-dimensionalintegral vanishes. In particular, λ ≥ 0.If∂φ∂n+ αφ = 0 on ∂D, then−I∂Dφ∂φ∂nds =I∂Dα|φ|2ds.In particular, if α ≥ 0 everywhere on ∂D, then λ ≥ 0.Self-adjointnessZZDψ ∇2φ dx dy =I∂Dψ∂φ∂nds −ZZD∇ψ · ∇φ dx dy(Green’s first identity)ZZD(φ ∇2ψ − ψ ∇2φ) dx dy =I∂Dφ∂ψ∂n− ψ∂φ∂nds(Green’s second identity)If φ and ψ satisfy the same boundary conditionαφ + β∂φ∂n∂D=αψ + β∂ψ∂n∂D= 0then φ∂ψ∂n− ψ∂φ∂n= 0 everywhere on ∂D.If φ and ψ satisfy the same boundary condition thenZZD(φ ∇2ψ − ψ ∇2φ) dx dy = 0.If φ and ψ are complex-valued functions then alsoZZD(φ∇2ψ − ψ ∇2φ) dx dy = 0(because ∇2ψ = ∇2ψ and ψ satisfies the sameboundary condition as ψ).Thus h∇2φ, ψi = hφ, ∇2ψi, wherehf , gi =ZZDf (x, y)g(x, y) dx dy.Eigenvalue problem:∇2φ + λφ = 0 in D,αφ + β∂φ∂n∂D= 0.The Laplacian ∇2is self-adjoint in the subspace offunctions satisfying the boundary condition.Suppose φ is an eigenfunction belonging to aneigenvalue λ. Let us show that λ ∈ R.Since ∇2φ = −λφ, we have thath∇2φ, φi = h−λφ, φi = −λhφ, φi,hφ, ∇2φi = hφ, −λφi = −λhφ, φi.Now h∇2φ, φi = hφ, ∇2φi and hφ, φi > 0 imply λ ∈ R.Suppose φ1and φ2are eigenfunctions belonging todifferent eigenvalues λ1and λ2.Let us show that hφ1, φ2i = 0.Since ∇2φ1= −λ1φ1, ∇2φ2= −λ2φ2, we have thath∇2φ1, φ2i = h−λ1φ1, φ2i = −λ1hφ1, φ2i,hφ1, ∇2φ2i = hφ1, −λ2φ2i = −λ2hφ1, φ2i.But h∇2φ1, φ2i = hφ1, ∇2φ2i, hence−λ1hφ1, φ2i = −λ2hφ1, φ2i.We already know that λ2= λ2. Also, λ16= λ2.It follows that hφ1, φ2i = 0.The main purpose of the Rayleigh quotientConsider a functional (function on functions)RQ[φ] =−I∂Dφ∂φ∂nds +ZZD|∇φ|2dx dyZZD|φ|2dx dy.If φ is an eigenfunction of −∇2in the domain Dwith some boundary condition, then RQ[φ] is thecorresponding eigenvalue.What if φ is not?Let λ1< λ2≤ λ3≤ . . . ≤ λn≤ λn+1≤ . . . beeigenvalues of a particular eigenvalue problemcounted with multiplicities.That is, a simple eigenvalue appears once in thissequence, an eigenvalue of multiplicity two appearstwice, and so on.There is a complete orthogonal system φ1, φ2, . . . inthe Hilbert space L2(D) such that φnis aneigenfunction belonging to λn.Theorem (i) λ1= min RQ[φ], where the minimumis taken over all nonzero functions φ which aredifferentiable in D and satisfy the boundarycondition. Moreover, if RQ[φ] = λ1then φ is aneigenfunction.(ii) λn= min RQ[φ], where the minimum is takenover all nonzero functions φ which are differentiablein D, satisfy the boundary condition, and such thathφ, φki = 0 for 1 ≤ k < n. Moreover, the minimumis attained only on eigenfunctions.Main idea of the proof: RQ[φ] =h−∇2φ, φihφ, φi.(see Haberman 5.6)Spectral properties of the Laplacian in a circleEigenvalue problem:∇2φ + λφ = 0 in D = {(x, y ) : x2+ y2≤ R2},u|∂D= 0.In polar coordinates (r, θ):∂2φ∂r2+1r∂φ∂r+1r2∂2φ∂θ2+ λφ = 0(0 < r < R, −π < θ < π),φ(R, θ) = 0 (−π < θ < π).Additional boundary conditions:|φ(0, θ)| < ∞ (−π < θ < π),φ(r, −π) = φ(r , π),∂φ∂θ(r, −π) =∂φ∂θ(r, π) (0 < r < R).Separation of variables: φ(r, θ) = f (r )h(θ).Substitute this into the equation:f′′(r)h(θ) + r−1f′(r)h(θ) + r−2f (r )h′′(θ) + λf (r )h(θ) = 0.Divide by f (r )h(θ) and multiply by r2:r2f′′(r) + r f′(r) + λr2f (r )f (r )+h′′(θ)h(θ)= 0.It follows thatr2f′′(r) + r f′(r) + λr2f (r )f (r )= −h′′(θ)h(θ)= µ = const.The variables have been separated:r2f′′+ rf′+ (λr2− µ)f = 0,h′′= −µh.Boundary conditions φ(R, θ) = 0 and |φ(0, θ)| < ∞hold if f (R) = 0 and |f (0)| < ∞.Boundary conditions φ(r, −π) = φ(r , π) and∂φ∂θ(r, −π) =∂φ∂θ(r, π) hold if h(−π) = h(π) andh′(−π) = h′(π).Eigenvalue problem:h′′= −µh, h(−π) = h(π), h′(−π) = h′(π).Eigenvalues: µm= m2, m = 0, 1, 2, . . . .µ0= 0 is simple, the others are of multiplicity 2.Eigenfunctions: h0= 1, hm(θ) = cos mθ and˜hm(θ) = sin mθ for m ≥ 1.Dependence on r:r2f′′+ rf′+ (λr2− µ)f = 0, f (R) = 0, |f (0)|


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