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

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 ( x ) � � ( τ f )( x ) := , x ∈ R r ( x ) with r , 1 p , q ∈ L 1 loc ( R ), p ( x ) > 0 and r ( x ) � = 0 a. e. We define a weighted L 2 -space � � L 2 f : R → C : f measurabel , | f | 2 | r | ∈ L 1 ( R ) | r | ( R ) := with an inner product � [ f , g ] = f ( x ) g ( x ) r ( x ) d x . R We assume, that τ is in the limit point case at ±∞ . Therefore, the maximal operator A : D max → L 2 | r | ( R ) , Af := τ f with | r | ( R ) : f , pf ′ ∈ AC ( R ) , � � f ∈ L 2 τ f ∈ L 2 D max = | r | ( R ) � L 2 � is the only self-adjoint realisation of τ in the Krein space | r | ( R ) , [ · , · ] . Page 2 / 15

Non-real spectrum The operator � − f ′′ − κ ( κ + 1) sech 2 f � Af = sgn( · ) , dom( A ) = D max , κ ∈ N , has non-real spectrum [Behrndt, Katatbeh, Trunk ’09]. Im 65 Re 9900 9900 65 Page 3 / 15

Theorem (Behrndt, Philipp, Trunk ’13) Let − f ′′ + qf � � Af = sgn( · ) dom( A ) = D max , 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 ) ≤ 5 � q � ∞ , | Im λ | ≤ 2 � q � ∞ , where d := − 5 ess inf x ∈ R q ( x ) > 0 . Page 4 / 15

λ ∈ σ p ( A ) \ R Im 2 � q � ∞ λ Σ Re − d d 5 � q � ∞ � � � � Σ := λ ∈ C : dist λ, ( − d , d ) ≤ 5 � q � ∞ , | Im λ | ≤ 2 � q � ∞ , d := − 5 ess inf x ∈ R q ( x ) > 0 Page 5 / 15

Theorem (Behrndt, S., Trunk ’16) Let − f ′′ + qf � � Af = sgn( · ) dom( A ) = D max , with q ∈ L 1 ( R ) . Then the non-real spectrum of A consists only of isolated eigenvalues. For each λ ∈ σ ( A ) \ R the inequality �� − 2 � � 1 | λ | ≤ � q � 2 ≈ 8 . 3255 · � q � 2 1 · ln √ + 1 1 2 + 1 holds. Page 6 / 15

Proof Let λ ∈ C \ R : Af = λ f , f ∈ dom( A ) = D max f ′′ + = − λ f + + qf + on R + f ′′ − = + λ f − + qf − on R − with f + ∈ L 2 ( R + ), f + , f ′ + ∈ AC ( R + ) with f − ∈ L 2 ( R − ), f − , f ′ − ∈ AC ( R − ) and f ′ + (0) = f ′ f + (0) = f − (0) , − (0) Page 7 / 15

Proof Let λ ∈ C \ R : Af = λ f , f ∈ dom( A ) = D max f ′′ + = − λ f + + qf + on R + f ′′ − = + λ f − + qf − on R − with f + ∈ L 2 ( R + ), f + , f ′ + ∈ AC ( R + ) with f − ∈ L 2 ( R − ), f − , f ′ − ∈ AC ( R − ) and f ′ + (0) = f ′ f + (0) = f − (0) , − (0) Page 7 / 15

Proof Let λ ∈ C \ R : Af = λ f , f ∈ dom( A ) = D max f ′′ + = − λ f + + qf + on R + f ′′ − = + λ f − + qf − on R − with f + ∈ L 2 ( R + ), f + , f ′ + ∈ AC ( R + ) with f − ∈ L 2 ( R − ), f − , f ′ − ∈ AC ( R − ) and f ′ + (0) = cf ′ f + (0) = cf − (0) , − (0) , c � = 0 Page 7 / 15

Proof Consider the half-axis operator A + f = − f ′′ + qf A + : D + → L 2 ( R + ) , with the domain f ∈ L 2 ( R + ) : f , f ′ ∈ AC ( R + ) , − f ′′ + qf ∈ L 2 ( R + ) , � � D + = f (0) = 0 . This operator is self-adjoint in the Hilbert space L 2 ( R + ). If f + would satisfy − f ′′ + = − λ f + + qf + and f + (0) = 0, it would be in D + . Then we would have A + f + = λ f + for non-real λ . Hence, f + (0) � = 0. Page 8 / 15

Proof Let λ ∈ C \ R : Af = λ f , f ∈ dom( A ) = D max f ′′ + = − λ f + + qf + on R + f ′′ − = + λ f − + qf − on R − with f + ∈ L 2 ( R + ), f + , f ′ + ∈ AC ( R + ) with f − ∈ L 2 ( R − ), f − , f ′ − ∈ AC ( R − ) and f ′ + (0) = cf ′ f + (0) = cf − (0) , − (0) , c � = 0 Page 9 / 15

Proof Let λ ∈ C \ R : Af = λ f , f ∈ dom( A ) = D max f ′′ − = + λ f − + qf − on R − f ′′ + = − λ f + + qf + on R + with f + ∈ L 2 ( R + ), f + , f ′ + ∈ AC ( R + ) with f − ∈ L 2 ( R − ), f − , f ′ − ∈ AC ( R − ) and f ′ f + (0) = f ′ + (0) − (0) f − (0) Page 9 / 15

Proof WKB-approach [Olver ’73]: Consider f ′′ + = − λ f + + qf + on R + . Let f + be defined as f + ( x ) = e −√− λ x (1 + R + ( x )) , x ∈ R + . Then f + is a solution iff R + is satisfies √ R ′′ − λ R ′ + − + = q (1 + R + ) . Variation of constant leads to the integral equation � ∞ √− λ ( x − s ) � q ( s ) � √ R + ( x ) = 1 − e − λ (1 + R + ( s )) d s , 2 x which admits solution R + with x →∞ R + ( x ) = 0 lim ◮ � � � � − 1, | R ′ � q � 1 + ( x ) | � q � 1 ◮ | R + ( x ) | ≤ exp ≤ exp − 1 � � � | λ | | λ | | λ | Page 10 / 15

Proof WKB-approach [Olver ’73]: Consider f ′′ − = + λ f − + qf − on R − . Let f − be defined by f − ( x ) = e + √ + λ x (1 + R − ( x )) , x ∈ R − . Then f − is a solution iff R − is satisfies √ R ′′ + λ R ′ − + − = q (1 + R − ) . Variation of constant leads to the integral equation 1 − e − 2 √− λ ( x − s ) � q ( s ) � x � R − ( x ) = √ (1 + R − ( s )) d s , 2 λ −∞ which admits solution R − with x →−∞ R − ( x ) = 0 lim ◮ � � � � − 1, | R ′ � q � 1 − ( x ) | � q � 1 ◮ | R − ( x ) | ≤ exp ≤ exp − 1 � � � | λ | | λ | | λ | Page 11 / 15

Proof Hence, we have solutions f ± of f ′′ ± = ∓ λ f ± + qf ± with ◮ f ± ( x ) = e ∓√∓ λ x (1 + R ± ( x )), x ∈ R ± x →∞ R + ( x ) = 0, lim x →−∞ R − ( x ) = 0 lim ◮ ◮ f ± ∈ L 2 ( R ± ) � � � � − 1, | R ′ � q � 1 ± ( x ) | � q � 1 ◮ | R ± ( x ) | ≤ exp ≤ exp − 1 � � � | λ | | λ | | λ | The eigenvalue problem reduces to the condition f ′ f + (0) = f ′ + (0) − (0) f − (0) , which is the same as √ √ R ′ R ′ − (0) + (0) − − λ + 1 + R + (0) = λ + 1 + R − (0) . Page 12 / 15

Proof Hence, we have solutions f ± of f ′′ ± = ∓ λ f ± + qf ± with ◮ f ± ( x ) = e ∓√∓ λ x (1 + R ± ( x )), x ∈ R ± x →∞ R + ( x ) = 0, lim x →−∞ R − ( x ) = 0 lim ◮ ◮ f ± ∈ L 2 ( R ± ) � � � � − 1, | R ′ � q � 1 ± ( x ) | � q � 1 ◮ | R ± ( x ) | ≤ exp ≤ exp − 1 � � � | λ | | λ | | λ | The eigenvalue problem reduces to the condition f + (0) = f ′ f ′ + (0) − (0) f − (0) , which is the same as √ √ R ′ R ′ − (0) + (0) − − λ + 1 + R + (0) = λ + 1 + R − (0) . This does not hold for �� − 2 � � 1 | λ | > � q � 2 1 · ln √ + 1 . 2 + 1 � Page 12 / 15

Theorem (Behrndt, S., Trunk) We assume an indefinite Sturm-Liouville operator Af = 1 − ( pf ′ ) ′ + qf � � , dom( A ) = D max r with sgn ( x ) · r ( x ) > 0 a.e. in R and q ∈ L 1 ( R ) . Let ◮ r , p , 1 r , 1 p ∈ L ∞ ( R ) ; ◮ pr , ( p ( pr ) ′ ) ∈ AC ( R \ { 0 } ) with ( rp ) ′ ∈ L 2 ( R ) and ( p ( rp ) ′ ) ′ ∈ L 1 ( R ) ; ◮ | r | , p continuously differentiable in 0 . Then for every λ ∈ σ p ( A ) \ R the inequality �� − 2 � � 1 | λ | ≤ � G � 2 ≈ 8 . 3255 · � G � 2 1 · ln √ + 1 1 2 + 1 holds, where � � − 5 (( rp ) ′ ) 2 ( p ( rp ) ′ ) ′ 1 G = + q . � 4 rp 16 r 2 p | rp | Page 13 / 15

However, there is a catch Let − f ′′ + qf q ∈ L 1 ( R ) , � � Af = sgn ( · ) , dom( A ) = D max ◮ For nonnegativ q the non-real point spectrum of A is empty Page 14 / 15

However, there is a catch Let − f ′′ + qf q ∈ L 1 ( R ) , � � Af = sgn ( · ) , dom( A ) = D max ◮ For nonnegativ q the non-real point spectrum of A is empty ◮ Example: q ( x ) = − 100 · 101 · sech 2 ( x ), � q � 2 1 = (100 · 101 · 2) 2 , our bound: ≈ 3 . 4 · 10 9 Page 14 / 15

However, there is a catch Let − f ′′ + qf q ∈ L 1 ( R ) , � � Af = sgn ( · ) , dom( A ) = D max ◮ For nonnegativ q the non-real point spectrum of A is empty ◮ Example: q ( x ) = − 100 · 101 · sech 2 ( x ), � q � 2 1 = (100 · 101 · 2) 2 , our bound: ≈ 3 . 4 · 10 9 Im 65 Re 9900 9900 65 Page 14 / 15

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