Math Theory of Partial Differential Equations Lecture 3-2: Spectral properties of the Laplacian. Bessel functions.

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1 Math Theory of Partial ifferential Equations Lecture 3-2: Spectral properties of the Laplacian. Bessel functions.

2 Eigenvalue problem: 2 φ + λφ = 0 in, ( αφ + β φ ) n = 0, where α, β are piecewise continuous real functions on such that α + β 0 everywhere on. We assume that the boundary is piecewise smooth.

3 6 spectral properties of the Laplacian Property 1. All eigenvalues are real. Property 2. All eigenvalues can be arranged in the ascending order λ 1 < λ 2 <... < λ n < λ n+1 <... so that λ n as n. This means that: there are infinitely many eigenvalues; there is a smallest eigenvalue; on any finite interval, there are only finitely many eigenvalues.

4 Property 3. An eigenvalue λ n may be multiple but its multiplicity is finite. Moreover, the smallest eigenvalue λ 1 is simple, and the corresponding eigenfunction φ 1 has no zeros inside the domain. Property 4. Eigenfunctions corresponding to different eigenvalues are orthogonal relative to the inner product f, g = f (x, y)g(x, y) dx dy.

5 Property 5. Any eigenfunction φ can be related to its eigenvalue λ through the Rayleigh quotient: φ φ n ds + φ 2 dx dy λ =. φ 2 dx dy

6 Property 6. There exists a sequence φ 1, φ 2,... of pairwise orthogonal eigenfunctions that is complete in the Hilbert space L 2 (). Any square-integrable function f L 2 () is expanded into a series f (x, y) = n=1 c nφ n (x, y), that converges in the mean. The series is unique: c n = f, φ n φ n, φ n. If f is piecewise smooth then the series converges pointwise to f at points of continuity.

7 Rayleigh quotient Suppose that 2 φ = λφ in the domain. Multiply both sides by φ and integrate over : φ 2 φ dx dy = λ φ 2 dx dy. Green s formula: ψ 2 φ da = ψ φ n ds ψ φ da This is an analog of integration by parts. Now φ φ n ds φ 2 dx dy = λ φ 2 dx dy.

8 It follows that λ = φ φ n ds + φ 2 dx dy. φ 2 dx dy If φ satisfies the boundary condition φ = 0 or φ n = 0 (or mixed), then the one-dimensional integral vanishes. In particular, λ 0. If φ n + αφ = 0 on, then φ φ n ds = α φ 2 ds. In particular, if α 0 everywhere on, then λ 0.

9 Self-adjointness ψ 2 φ dx dy = ψ φ n ds ψ φ dx dy (Green s first identity) ( (φ 2 ψ ψ 2 φ) dx dy = φ ψ n ψ φ ) ds n (Green s second identity) If φ and ψ satisfy the same boundary condition ( αφ + β φ ) ( n = αψ + β ψ ) n = 0 then φ ψ n ψ φ n = 0 everywhere on.

10 If φ and ψ satisfy the same boundary condition then (φ 2 ψ ψ 2 φ) dx dy = 0. If φ and ψ are complex-valued functions then also (φ 2 ψ ψ 2 φ) dx dy = 0 (because 2 ψ = 2 ψ and ψ satisfies the same boundary condition as ψ). Thus 2 φ, ψ = φ, 2 ψ, where f, g = f (x, y)g(x, y) dx dy.

11 Eigenvalue problem: 2 φ + λφ = 0 in, ( αφ + β φ ) n = 0. The Laplacian 2 is self-adjoint in the subspace of functions satisfying the boundary condition. Suppose φ is an eigenfunction belonging to an eigenvalue λ. Let us show that λ R. Since 2 φ = λφ, we have that 2 φ, φ = λφ, φ = λ φ, φ, φ, 2 φ = φ, λφ = λ φ, φ. Now 2 φ, φ = φ, 2 φ and φ, φ > 0 imply λ R.

12 Suppose φ 1 and φ 2 are eigenfunctions belonging to different eigenvalues λ 1 and λ 2. Let us show that φ 1, φ 2 = 0. Since 2 φ 1 = λ 1 φ 1, 2 φ 2 = λ 2 φ 2, we have that 2 φ 1, φ 2 = λ 1 φ 1, φ 2 = λ 1 φ 1, φ 2, φ 1, 2 φ 2 = φ 1, λ 2 φ 2 = λ 2 φ 1, φ 2. But 2 φ 1, φ 2 = φ 1, 2 φ 2, hence λ 1 φ 1, φ 2 = λ 2 φ 1, φ 2. We already know that λ 2 = λ 2. Also, λ 1 λ 2. It follows that φ 1, φ 2 = 0.

13 The main purpose of the Rayleigh quotient Consider a functional (function on functions) φ φ n ds + φ 2 dx dy RQ[φ] =. φ 2 dx dy If φ is an eigenfunction of 2 in the domain with some boundary condition, then RQ[φ] is the corresponding eigenvalue. What if φ is not?

14 Let λ 1 < λ 2 λ 3... λ n λ n+1... be eigenvalues of a particular eigenvalue problem counted with multiplicities. That is, a simple eigenvalue appears once in this sequence, an eigenvalue of multiplicity two appears twice, and so on. There is a complete orthogonal system φ 1, φ 2,... in the Hilbert space L 2 () such that φ n is an eigenfunction belonging to λ n.

15 Theorem (i) λ 1 = min RQ[φ], where the minimum is taken over all nonzero functions φ which are differentiable in and satisfy the boundary condition. Moreover, if RQ[φ] = λ 1 then φ is an eigenfunction. (ii) λ n = min RQ[φ], where the minimum is taken over all nonzero functions φ which are differentiable in, satisfy the boundary condition, and such that φ, φ k = 0 for 1 k < n. Moreover, the minimum is attained only on eigenfunctions. Main idea of the proof: RQ[φ] = 2 φ, φ. φ, φ (see Haberman 5.6)

16 Spectral properties of the Laplacian in a circle Eigenvalue problem: 2 φ + λφ = 0 in = {(x, y) : x 2 + y 2 R 2 }, u = 0. In polar coordinates (r, θ): 2 φ r r φ r φ r 2 θ + λφ = 0 2 (0 < r < R, π < θ < π), φ(r, θ) = 0 ( π < θ < π).

17 Additional boundary conditions: φ(r, π) = φ(r, π), φ(0, θ) < φ θ ( π < θ < π), φ (r, π) = θ (r, π) Separation of variables: φ(r, θ) = f (r)h(θ). Substitute this into the equation: (0 < r < R). f (r)h(θ) + r 1 f (r)h(θ) + r 2 f (r)h (θ) + λf (r)h(θ) = 0. ivide by f (r)h(θ) and multiply by r 2 : r 2 f (r) + r f (r) + λr 2 f (r) f (r) + h (θ) h(θ) = 0.

18 It follows that r 2 f (r) + r f (r) + λr 2 f (r) f (r) The variables have been separated: = h (θ) h(θ) = µ = const. r 2 f + rf + (λr 2 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 φ θ (r, π) = θ (r, π) hold if h( π) = h(π) and h ( π) = h (π).

19 Eigenvalue problem: h = µh, h( π) = h(π), h ( π) = h (π). Eigenvalues: µ m = m 2, m = 0, 1, 2,.... µ 0 = 0 is simple, the others are of multiplicity 2. Eigenfunctions: h 0 = 1, h m (θ) = cos mθ and h m (θ) = sin mθ for m 1.

20 ependence on r: r 2 f + rf + (λr 2 µ)f = 0, f (R) = 0, f (0) <. We may assume that µ = m 2, m = 0, 1, 2,.... Also, we know that λ > 0 (Rayleigh quotient!). New variable z = λ r removes dependence on λ: z 2 d2 f dz 2 + z df dz + (z2 m 2 )f = 0. This is Bessel s differential equation of order m. Solutions are called Bessel functions of order m.

MATH 311 Topics in Applied Mathematics Lecture 25: Bessel functions (continued).

MATH 311 Topics in Applied Mathematics Lecture 25: Bessel functions (continued). Bessel s differential equation of order m 0: z 2 d2 f dz 2 + z df dz + (z2 m 2 )f = 0 The equation is considered on the