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Self-Adjoint Operators Self-Adjoint Operators Sturm-Liouville Properties Sturm-Liouville Properties Outline Math 531 - Partial Diﬀerential Equations Sturm-Liouville Problems Part B 1 Self-Adjoint Operators Lagrange’s Identity Joseph M. Mahaﬀy, Green’s Formula and Self-adjointness [email protected]

2 Sturm-Liouville Properties Department of Mathematics and Statistics Dynamical Systems Group Orthogonality of Eigenfunctions Computational Sciences Research Center Real Eigenvalues San Diego State University Unique Eigenfunctions San Diego, CA 92182-7720 http://jmahaﬀy.sdsu.edu

Spring 2020

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (1/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (2/16)

Self-Adjoint Operators Lagrange’s Identity Self-Adjoint Operators Lagrange’s Identity Sturm-Liouville Properties Green’s Formula and Self-adjointness Sturm-Liouville Properties Green’s Formula and Self-adjointness Regular Sturm-Liouville Problem Linear Operator

The Regular Sturm-Liouville problem satisﬁes: d dφ Linear Operator: Let L be the linear diﬀerential operator: p(x) + q(x)φ + λσ(x)φ = 0, dx dx d d L = p(x) + q(x), with the homogeneous BCs: dx dx 0 d dy β1φ(a) + β2φ (a) = 0, L(y) = p(x) + q(x)y. 0 dx dx β3φ(b) + β4φ (b) = 0,

where (i) βi are real, (ii) The functions p(x), q(x), and σ(x) are The Sturm-Liouville diﬀerential equation is written: real, continuous functions for x ∈ [a, b] with p(x) > 0 and σ(x) > 0. L(φ) + λσ(x)φ = 0, The following theorems will NOT be proved:

1 There are inﬁnitely many eigenvalues. where λ is an eigenvalue and φ is an eigenfunction. 2 Any piecewise smooth function can be expanded by the Note: The Linear Operator does NOT have to act on an eigenfunctions. eigenvalue problem. 3 Each succeeding eigenfunction has an additional zero.

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (3/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (4/16) Self-Adjoint Operators Lagrange’s Identity Self-Adjoint Operators Lagrange’s Identity Sturm-Liouville Properties Green’s Formula and Self-adjointness Sturm-Liouville Properties Green’s Formula and Self-adjointness Lagrange’s Identity Linear Operator Proof: This identity is readily shown

d dv d du uL(v) − vL(u) = u p(x) + q(x)v − v p(x) + q(x)u , Theorem (Lagrange’s Identity) dx dx dx dx d dv d du Let L be the Linear Operator: = u p(x) − v p(x) . dx dx dx dx d d L = p(x) + q(x). However, from the product rule we see that dx dx d dv d dv du dv u p = u p(x) + · p , The following formula: dx dx dx dx dx dx d du d du dv du d dv du v p = v p(x) + · p . uL(v) − vL(u) = p u − v , dx dx dx dx dx dx dx dx dx Subtracting these equations, we can insert into the expression for is known as the diﬀerential form of Lagrange’s identity. uL(v) − vL(u) to obtain Lagrange’s identity: d dv du uL(v) − vL(u) = p u − v . q.e.d. dx dx dx

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (5/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (6/16)

Suppose that u and v are any two functions with the additional Lagrange’s identity relates to the ﬁrst part of the linear restriction that the boundary terms vanish: diﬀerential operator from the Sturm-Liouville problem. b dv du Theorem (Green’s Formula) p u − v = 0. dx dx a The integration of Lagrange’s identity give’s Green’s formula:

Z b b Theorem (Self-adjoint) dv du [uL(v) − vL(u)]dx = p u − v If u and v are any two functions satisfying the same set of a dx dx a homogeneous BCs of the regular Sturm-Liouville problem, then for u and v continuously diﬀerentiable for x ∈ (a, b). Z b [uL(v) − vL(u)]dx = 0. This linear diﬀerential operator has important properties when a there are homogeneous BCs. The linear operator L satisfying this condition is self-adjoint.

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (7/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (8/16) Orthogonality of Eigenfunctions Self-Adjoint Operators Lagrange’s Identity Self-Adjoint Operators Real Eigenvalues Sturm-Liouville Properties Green’s Formula and Self-adjointness Sturm-Liouville Properties Unique Eigenfunctions Self-adjoint Orthogonality of Eigenfunctions

Let λn and λm be distinct eigenvalues with corresponding eigenfunctions φn and φm, so The theorem for self-adjointness of the linear operator L extends L(φn) + λnσ(x)φn = 0, to other Sturm-Liouville problems. L(φm) + λmσ(x)φm = 0.

1 HW Exercises examine some regular Sturm-Liouville problems. It follows that Z b Z b 2 It is easy to show for periodic BCs. [φmL(φn) − φnL(φm)]dx = (λm − λn) φmφnσ(x)dx a a 3 Also, easy to show if the Sturm-Liouville problem is singular at x = 0 and has p(0) = 0 with φ(0) bounded. By Green’s Formula:

Z b b dφn dφm (λm − λn) φmφnσ(x)dx = p(x) φm − φn . a dx dx a

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (9/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (10/16)

Orthogonality of Eigenfunctions Orthogonality of Eigenfunctions Self-Adjoint Operators Self-Adjoint Operators Real Eigenvalues Real Eigenvalues Sturm-Liouville Properties Sturm-Liouville Properties Unique Eigenfunctions Unique Eigenfunctions Orthogonality of Eigenfunctions Real Eigenvalues

Suppose that an eigenvalue, λ, is complex and it has a From before corresponding eigenfunction, φ(x), Z b b dφn dφm L(φ) + λσφ = 0. (λm − λn) φmφnσ(x)dx = p(x) φm − φn , a dx dx a Take the complex conjugate, so so if φm and φn satisfy the same homogeneous BCs the right hand side is zero. L(φ) + λσφ = L(φ) + λσφ = 0. Thus, Z b However, L(φ) = L(φ), since the coeﬃcients of L are real, so (λm − λn) φmφnσ(x)dx = 0, a L(φ) + λσφ = 0. which says that eigenfunctions corresponding to diﬀerent eigenvalues are orthogonal with respect to the weighting function, σ(x). Thus, if λ is a complex eigenvalue with corresponding eigenfunction φ, then λ is also an eigenvalue with eigenfunction φ.

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (11/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (12/16) Orthogonality of Eigenfunctions Orthogonality of Eigenfunctions Self-Adjoint Operators Self-Adjoint Operators Real Eigenvalues Real Eigenvalues Sturm-Liouville Properties Sturm-Liouville Properties Unique Eigenfunctions Unique Eigenfunctions Real Eigenvalues Uniqueness

Suppose that φ1 and φ2 are eigenfunctions corresponding to λ, so

By our orthogonality theorem, L(φ1) + λσφ1 = 0 and L(φ2) + λσφ2 = 0.

Z b (λ − λ) φφσdx = 0. Since λ is the same, a φ2L(φ1) − φ1L(φ2) = 0. However, φφ = |φ|2 ≥ 0 and σ > 0. Lagrange’s identity implies: Thus, the integral is zero if and only if φ(x) ≡ 0, which is not an eigenfunction, or λ − λ = 0, which implies λ is real. d dφ1 dφ2 φ2L(φ1) − φ1L(φ2) = p φ2 − φ1 = 0. Thus, eigenvalues of a Sturm-Liouville problem are real. dx dx dx

dφ1 dφ2 Therefore, p φ2 dx − φ1 dx is a constant.

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (13/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (14/16)

Using the Lagrange’s identity, we obtained Earlier we showed that periodic BCs give both sine and cosine dφ1 dφ2 p φ2 − φ1 = C. eigenfunctions, so this case does NOT produce unique dx dx eigenfunctions for a given eigenvalue. The constant C = 0, if we have regular Sturm-Liouville problem The eigenvalue, λ does have a unique eigenfunction. with at least one homogeneous BC. Non-uniqueness can create problems with orthogonality for a It follows that given eigenvalue. The Gram-Schmidt process can be applied to create an orthogonal dφ1 dφ2 d φ2 φ2 − φ1 = 0 or = 0. set of eigenfunctions. dx dx dx φ1 Recall that λm 6= λn always produces orthogonal eigenfunctions. Thus, φ2(x) = cφ1(x) with these BCs, so the eigenfunction is unique.

Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (15/16) Joseph M. Mahaﬀy, [email protected] Sturm-Liouville Problems — (16/16)