Bounds on the non-real spectrum of indefinite Sturm-Liouville - - PowerPoint PPT Presentation

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Bounds on the non-real spectrum of indefinite Sturm-Liouville - - PowerPoint PPT Presentation

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Bounds on the non-real spectrum of indefinite Sturm-Liouville operators Operator Theory in Indefinite Inner Product Spaces Philipp Schmitz Page 1 / 15 Singular indefinite Sturm-Liouville operators Let 1 ( pf ) ( x ) + q ( x ) f (

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Bounds on the non-real spectrum of indefinite Sturm-Liouville

  • perators

Operator Theory in Indefinite Inner Product Spaces Philipp Schmitz

Page 1 / 15

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Singular indefinite Sturm-Liouville operators

Let (τf )(x) := 1 r(x)

  • − (pf ′)′(x) + q(x)f (x)
  • ,

x ∈ R with r, 1

p , q ∈ L1 loc(R), p(x) > 0 and r(x) = 0 a. e.

We define a weighted L2-space L2

|r|(R) :=

  • f : R → C : f measurabel, |f |2|r| ∈ L1(R)
  • with an inner product

[f , g] =

  • R

f (x)g(x)r(x) dx. We assume, that τ is in the limit point case at ±∞. Therefore, the maximal

  • perator

A : Dmax → L2

|r|(R),

Af := τf with Dmax =

  • f ∈ L2

|r|(R) : f , pf ′ ∈ AC(R),

τf ∈ L2

|r|(R)

  • is the only self-adjoint realisation of τ in the Krein space
  • L2

|r|(R), [·, ·]

  • .

Page 2 / 15

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Non-real spectrum

The operator Af = sgn(·)

  • − f ′′ − κ(κ + 1) sech2 f
  • ,

dom(A) = Dmax, κ ∈ N, has non-real spectrum [Behrndt, Katatbeh, Trunk ’09].

9900 9900 65 65 Im Re

Page 3 / 15

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Theorem (Behrndt, Philipp, Trunk ’13)

Let Af = sgn(·)

  • − f ′′ + qf
  • ,

dom(A) = Dmax with q ∈ L∞(R). Further assume that ess inf q < 0 holds. Then the non-real spectrum of A consists of isolated eigenvalues and is contained in Σ :=

  • λ ∈ C : dist
  • λ, (−d, d)
  • ≤ 5q∞, | Im λ| ≤ 2q∞
  • ,

where d := −5 ess inf

x∈R q(x) > 0.

Page 4 / 15

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Re Im −d d 5q∞ 2q∞ λ λ ∈ σp(A) \ R Σ Σ :=

  • λ ∈ C : dist
  • λ, (−d, d)
  • ≤ 5q∞, | Im λ| ≤ 2q∞
  • ,

d := −5 ess inf

x∈R q(x) > 0

Page 5 / 15

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Theorem (Behrndt, S., Trunk ’16)

Let Af = sgn(·)

  • − f ′′ + qf
  • ,

dom(A) = Dmax with q ∈ L1(R). Then the non-real spectrum of A consists only of isolated

  • eigenvalues. For each λ ∈ σ(A) \ R the inequality

|λ| ≤ q2

1 ·

  • ln
  • 1

√ 2 + 1 + 1 −2 ≈ 8.3255 · q2

1

holds.

Page 6 / 15

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Proof

Let λ ∈ C \ R: Af = λf , f ∈ dom(A) = Dmax f ′′

+ = −λf+ + qf+ on R+

with f+ ∈ L2(R+), f+, f ′

+ ∈ AC(R+)

f ′′

− = +λf− + qf− on R−

with f− ∈ L2(R−), f−, f ′

− ∈ AC(R−)

and f+(0) = f−(0), f ′

+(0) = f ′ −(0)

Page 7 / 15

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SLIDE 8

Proof

Let λ ∈ C \ R: Af = λf , f ∈ dom(A) = Dmax f ′′

+ = −λf+ + qf+ on R+

with f+ ∈ L2(R+), f+, f ′

+ ∈ AC(R+)

f ′′

− = +λf− + qf− on R−

with f− ∈ L2(R−), f−, f ′

− ∈ AC(R−)

and f+(0) = f−(0), f ′

+(0) = f ′ −(0)

Page 7 / 15

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Proof

Let λ ∈ C \ R: Af = λf , f ∈ dom(A) = Dmax f ′′

+ = −λf+ + qf+ on R+

with f+ ∈ L2(R+), f+, f ′

+ ∈ AC(R+)

f ′′

− = +λf− + qf− on R−

with f− ∈ L2(R−), f−, f ′

− ∈ AC(R−)

and f+(0) = cf−(0), f ′

+(0) = cf ′ −(0),

c = 0

Page 7 / 15

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Proof

Consider the half-axis operator A+ : D+ → L2(R+), A+f = −f ′′ + qf with the domain D+ =

  • f ∈ L2(R+) : f , f ′ ∈ AC(R+),

−f ′′ + qf ∈ L2(R+), f (0) = 0

  • .

This operator is self-adjoint in the Hilbert space L2(R+). If f+ would satisfy −f ′′

+ = −λf+ + qf+ and f+(0) = 0, it would be in D+. Then

we would have A+f+ = λf+ for non-real λ. Hence, f+(0) = 0.

Page 8 / 15

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Proof

Let λ ∈ C \ R: Af = λf , f ∈ dom(A) = Dmax f ′′

+ = −λf+ + qf+ on R+

with f+ ∈ L2(R+), f+, f ′

+ ∈ AC(R+)

f ′′

− = +λf− + qf− on R−

with f− ∈ L2(R−), f−, f ′

− ∈ AC(R−)

and f+(0) = cf−(0), f ′

+(0) = cf ′ −(0),

c = 0

Page 9 / 15

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Proof

Let λ ∈ C \ R: Af = λf , f ∈ dom(A) = Dmax f ′′

+ = −λf+ + qf+ on R+

with f+ ∈ L2(R+), f+, f ′

+ ∈ AC(R+)

f ′′

− = +λf− + qf− on R−

with f− ∈ L2(R−), f−, f ′

− ∈ AC(R−)

and f ′

+(0)

f+(0) = f ′

−(0)

f−(0)

Page 9 / 15

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Proof

WKB-approach [Olver ’73]: Consider f ′′

+ = −λf+ + qf+ on R+. Let f+ be defined as

f+(x) = e−√−λx(1 + R+(x)), x ∈ R+. Then f+ is a solution iff R+ is satisfies R′′

+ −

√ −λR′

+ = q(1 + R+).

Variation of constant leads to the integral equation R+(x) = ∞

x

  • 1 − e

√−λ(x−s) q(s)

2 √ −λ(1 + R+(s)) ds, which admits solution R+ with

◮

lim

x→∞ R+(x) = 0 ◮ |R+(x)| ≤ exp

  • q1
  • |λ|
  • − 1, |R′

+(x)|

  • |λ|

≤ exp

  • q1
  • |λ|
  • − 1

Page 10 / 15

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Proof

WKB-approach [Olver ’73]: Consider f ′′

− = +λf− + qf− on R−. Let f− be defined by

f−(x) = e+√+λx(1 + R−(x)), x ∈ R−. Then f− is a solution iff R− is satisfies R′′

− +

√ +λR′

− = q(1 + R−).

Variation of constant leads to the integral equation R−(x) = x

−∞

  • 1 − e−2√−λ(x−s) q(s)

2 √ λ (1 + R−(s)) ds, which admits solution R− with

◮

lim

x→−∞ R−(x) = 0 ◮ |R−(x)| ≤ exp

  • q1
  • |λ|
  • − 1, |R′

−(x)|

  • |λ|

≤ exp

  • q1
  • |λ|
  • − 1

Page 11 / 15

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Proof

Hence, we have solutions f± of f ′′

± = ∓λf± + qf± with ◮ f±(x) = e∓√∓λx(1 + R±(x)), x ∈ R± ◮

lim

x→∞ R+(x) = 0,

lim

x→−∞ R−(x) = 0 ◮ f± ∈ L2(R±) ◮ |R±(x)| ≤ exp

  • q1
  • |λ|
  • − 1, |R′

±(x)|

  • |λ|

≤ exp

  • q1
  • |λ|
  • − 1

The eigenvalue problem reduces to the condition f ′

+(0)

f+(0) = f ′

−(0)

f−(0), which is the same as − √ −λ + R′

+(0)

1 + R+(0) = √ λ + R′

−(0)

1 + R−(0).

Page 12 / 15

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Proof

Hence, we have solutions f± of f ′′

± = ∓λf± + qf± with ◮ f±(x) = e∓√∓λx(1 + R±(x)), x ∈ R± ◮

lim

x→∞ R+(x) = 0,

lim

x→−∞ R−(x) = 0 ◮ f± ∈ L2(R±) ◮ |R±(x)| ≤ exp

  • q1
  • |λ|
  • − 1, |R′

±(x)|

  • |λ|

≤ exp

  • q1
  • |λ|
  • − 1

The eigenvalue problem reduces to the condition f ′

+(0)

f+(0) = f ′

−(0)

f−(0), which is the same as − √ −λ + R′

+(0)

1 + R+(0) = √ λ + R′

−(0)

1 + R−(0). This does not hold for |λ| > q2

1 ·

  • ln
  • 1

√ 2 + 1 + 1 −2 .

  • Page 12 / 15
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Theorem (Behrndt, S., Trunk)

We assume an indefinite Sturm-Liouville operator Af = 1 r

  • − (pf ′)′ + qf
  • ,

dom(A) = Dmax with sgn(x) · r(x) > 0 a.e. in R and q ∈ L1(R). Let

◮ r, p, 1 r , 1 p ∈ L∞(R); ◮ pr, (p(pr)′) ∈ AC(R \ {0}) with (rp)′ ∈ L2(R) and (p(rp)′)′ ∈ L1(R); ◮ |r|, p continuously differentiable in 0.

Then for every λ ∈ σp(A) \ R the inequality |λ| ≤ G2

1 ·

  • ln
  • 1

√ 2 + 1 + 1 −2 ≈ 8.3255 · G2

1

holds, where G = 1

  • |rp|
  • (p(rp)′)′

4rp − 5 ((rp)′)2 16r 2p + q

  • .

Page 13 / 15

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However, there is a catch

Let Af = sgn(·)

  • − f ′′ + qf
  • ,

q ∈ L1(R), dom(A) = Dmax

◮ For nonnegativ q the non-real point spectrum of A is empty

Page 14 / 15

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However, there is a catch

Let Af = sgn(·)

  • − f ′′ + qf
  • ,

q ∈ L1(R), dom(A) = Dmax

◮ For nonnegativ q the non-real point spectrum of A is empty ◮ Example: q(x) = −100 · 101 · sech2(x), q2 1 = (100 · 101 · 2)2,

  • ur bound: ≈ 3.4 · 109

Page 14 / 15

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However, there is a catch

Let Af = sgn(·)

  • − f ′′ + qf
  • ,

q ∈ L1(R), dom(A) = Dmax

◮ For nonnegativ q the non-real point spectrum of A is empty ◮ Example: q(x) = −100 · 101 · sech2(x), q2 1 = (100 · 101 · 2)2,

  • ur bound: ≈ 3.4 · 109

9900 9900 65 65 Im Re

Page 14 / 15

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Thank you for your attention!

Page 15 / 15