Equivalence of inverse Sturm-Liouville problems with boundary conditions rationally dependent on the eigenparameter II ∗ Paul A. Binding † Department of Mathematics and Statistics University of Calgary Calgary, Alberta, Canada T2N 1N4 Patrick J. Browne † Department of Computer Science University of Saskatchewan Saskatoon, Saskatchewan, Canada S7N 5E6 Bruce A. Watson ‡ School of Mathematics University of the Witwatersrand Private Bag 3, P O WITS 2050, South Africa September 15, 2008 Abstract The three classical inverse problems for a Sturm-Liouville boundary value problem −y ′′ + qy = λy, y(0) cos α = y ′ (0) sin α and y ′ (1) = f (λ)y(1) are considered. It is show that the Weyl m-function uniquely determines α, f and q, and is in turn uniquely determined by either two spectra from different values of α or from a spectrum and norming constants. It is also shown that two spectra uniquely determine the norming constants for the one spectrum. In this paper the definiteness assumptions required in our earlier work are removed. ∗

Keywords: Sturm-Liouville, eigenparameter dependent boundary conditions, Mathematics subject classification (2000): 34B25, 46D05, 47E05. † Research supported in part by grants from the NSERC of Canada ‡ Research conducted while visiting University of Calgary and University of Saskatchewan and supported in part by the Centre for Applicable Analysis and Number Theory.

1

1

Introduction

In this article we continue our treatment of the regular Sturm-Liouville equation ly := −y ′′ + qy = λy on [0, 1],

(1.1)

with q ∈ AC[0, 1], subject to the boundary conditions y(0) cos α = y ′ (0) sin α, α ∈ [0, π),

(1.2)

y′ (1) = f (λ), y

(1.3)

and

where f (λ) =

h(λ) g(λ)

and g and h are polynomials with real coefficients and no common zeros. In addition, deg(g) ≥ deg(h) then we set M = deg(g) and assume that g is monic, and if deg(g) < deg(h) then we set M = deg(h) and assume that h is monic, When f (λ) = ∞, (1.3) is interpreted as a Dirichlet condition y(1) = 0. Eigenvalues are listed by increasing real part and frequently (1.1)-(1.3) will be denoted by the shorthand (α, f, q). Let v and w to be the solutions of (1.1) satisfying the terminal conditions v(1, λ) = g(λ), v ′ (1, λ) = h(λ), and w(0, λ) = sin α, w′ (0, λ) = cos α. When confusion is unlikely, we abreviate v(x, λ) to v(x) and similarly for v ′ , w and w′ . Asymptotic approximations to v, v ′ , w and w′ can be found in the Appendix. If deg(h) ≤ deg(g) = m, let h(λ) = Am λm + ... + A0 where Am ∈ R (it may be zero); while if m = deg(h) > deg(g), let g(λ) = Am−1 λm−1 + ... + A0 where Am−1 ∈ R (it may be zero). Let D(λ, α, f, q) = v ′ (0, λ) sin α − v(0, λ) cos α

2

then the zeros of D(λ, α, f, q) are precisely the eigenvalues of (1.1)-(1.3) with the multiplicity of the zero of D being the same as the algebraic multiplicity of the eigenvalue [Naimark]. From the asymptotics in the Appendix, recalling that Q =

deg(h) ≤ deg(g) D(λ, α, f, q) λm deg(h) > deg(g) D(λ, α, f, q) λm

R1 0

q dt we have

 i  √  h√ √ √   sin α λ sin λ + cos λ Am − Q − cot α + O e|ℑ√ λ| , α 6= 0 λ  2 |ℑ√λ|  √  = √ Q sin e λ  − cos λ + √ A − + O , α = 0. m 2 λ λ

(1.4)  h i  |ℑ√λ|  √  √  sin α cos λ + sin√ λ Am−1 + Q + cot α + O e , α 6= 0 λ λ  2  |ℑ√λ|  √  √ = .  sin√ λ + cos λ −Am−1 − Q + O e 3/2 , α = 0 λ 2 λ λ

Theorem 1.1 Let f (λ) = h(λ) g(λ) where h and g are real polynomials with no common zeros. The eigenvalues of (1.1)-(1.3), repeated according to multiplicity and listed in increasing absolute value, are given asymptotically for n → ∞ by
 (n − m)2 π 2 + 2 cot α − 2Am + Q + o n1 , α 6= 0, deg(h) ≤ deg(g) = m,    
  (n + 1 − m)2 π 2 − 2A + Q + o 1 , α = 0, deg(h) ≤ deg(g) = m, m 2 n λn =
  (n + 21 − m)2 π 2 + 2 cot α + 2Am−1 + Q + o n1 , α 6= 0, m = deg(h) > deg(g),   
 (n + 1 − m)2 π 2 + Q + 2Am−1 + o n1 , α = 0, m = deg(h) > deg(g),

For large n all eigenvalues are simple (algebraically) and real.

Theorem 1.2 For large n the eigenfunctions corresponding to the eigenvalue λn have oscillation count n − m in the case of deg(h) ≤ deg(g) = m and for the case of m = deg(h) > deg(g) the oscillation count is n − m + 1 if lim f (λ) = +∞ and n − m if λ→+∞

lim f (λ) = −∞.

λ→+∞

We begin by recalling the definition of the Weyl m-function.

Definition 1.3 Let v(x, λ) be a non-identically zero solution of (1.1) satisfying (1.3). The Weyl m-function of (1.1)-(1.3) is defined by m(λ) =

v ′ (0, λ) cos α + v(0, λ) sin α . v ′ (0, λ) sin α − v(0, λ) cos α

3

The poles of the m-function are precisely the eigenvalues of (1.1)-(1.3) and the order of the pole coincides with the algebraic multiplicity of the eigenvalue. Asymptotics for m(λ) as λ → −∞ are needed in most of our work on inverse problems. Theorem 1.4 For λ → −∞ we have   ( cot α + O √1λ , α 6= 0 m(λ) = √ α = 0. −i λ + O(1), In the last section we will improve on the above theorem for the case of α = 0.

Theorem 1.5 For f : C → R ∪ {∞}, the m-function uniquely determines α, f and q. The first theorem of this section gives an explicit expression for D(λ, α, f, q) as an infinite product.

Theorem 1.6 Under the conditions on f (λ) stated earlier in  ∞ m Y Y  λn+m − λ   (λ − λ ) sin α , α 6= 0, n   n2 π 2   n=1 n=0   ∞ m−1  Y Y  λn+m − λ   (λ − λn ) 
2 , α 6= 0,  sin α n + 21 π 2 n=0 n=0 D(λ, α, f, q) = ∞ m−1 Y Y  λn+m − λ    α = 0, (λ − λ ) −
2 , n    n + 21 π 2  n=0 n=0  m ∞  Y  λn+m−1 − λ  Y  , α = 0, (λ − λ )  n  n2 π 2 n=0 n=1

this section we have m = deg(g) ≥ deg(h), deg(g) < deg(h) = m, m = deg(g) ≥ deg(h), deg(g) < deg(h) = m

Corollary 1.7 Let λn and µn , n = 0, 1, 2, ... be spectra of (α, f, q) and (β, f, q) where sin(α − β) 6= 0. Then λn , µn , n = 0, 1, 2, ... and β uniquely determine α, f and q. In a later section we shall revisit the relationship between two given spectra and the norming constants of one of the spectra.

4

2

Spectral data

Let Λn denote the eigenvalues of (α, f, q) listed without repetition and let νn ≥ 0 denote the multiplicity of the eigenvalue Λn . Suppose f (λ) =

P (λ) Q(λ)

(2.1)

where P (λ) and Q(λ) are real polynomials with no common zeros N = max{deg(P ), deg(Q)} and P (λ) = aN λN + ... + a1 λ + a0 , Q(λ) = bN λN + ... + b1 λ + b0 We begin with the case of aN 6= 0. We define a Pontrjagin space inner product on H = L2 < Y, Z >=

Z

1

y z¯ + yT J¯z

L

CN +1 by (2.2)

0

where y, z denote the L2 components of Y and Z, and y and z denote the CN +1 components of Y and Z. The matrix J is of the form 

  J = 

J1 J2 .. .

... ...

JN JN +1 .. .

JN

... J2N −1

    

where if we denote by J the vector with components J1 , ..., J2N −1 then J is the solution of   1  0     ..  = HJ  .  0

 H1 with H1 the n×(2n−1) matrix T (bN aN −1 − in which H is the block matrix H = H2 aN bN −1 , bN aN −2 − aN bN −2 , . . . , bN a0 − aN b0 ) and H2 the (n − 1) × (2n − 1) matrix 

5

T (−aN , aN −1 , . . . , a0 ), where the T matrices are the (2n − j) × (2n − 1) matrices given as follows   xj . . . x1 0 · · · . . . . . . 0  0 xj . . . x1 0 . . . . . . 0    T (xj , ..., x1 ) =  . ..  . .  . .  0

...

0

xj

. . . . . . x1

Lemma 2.1 J exists has L real entries and is unique and thus defines a Pontrjagin space 2 inner product on H = L CN +1 . Proof:

Then we can pose the boundary value problem (1.1)-(1.3) in the Pontrjagin space H by considering the operator     ly y  y2 − aN −1 y ′ (1) − bN −1 y(1)         with Y = y =  y1  LY =  ... (2.3)    ...  y   yN − a1 y ′ (1) − b1 y(1) yN +1 −a0 y ′ (1) − b0 y(1)

with domain

D(L) = {Y ∈ H : y, y ′ ∈ AC, ly ∈ L2 , y(0) cos α = y ′ (0) sin α, y1 = aN y ′ (1) + bN y(1)} where we recall that ly = −y ′′ + qy. Theorem 2.2 λ is an eigenvalue of (1.1)-(1.3) with eigenfunction y if and only if λ is an eigenvalue of L and a corresponding eigenvector of L is   y(x)   ?     .. .  . Y =    ?  h  i  P N b 1 ′ k k=1 ck −λ λ y (1) − y(1) b + Proof:

6

We define the bilinear form [X, Y ] =< X, Y¯ >,

for all

X, Y ∈ H,

(an alternative approach to using the bilinear product here is to proceed by considering the adjoint problem). Let wn (x) = w(x, Λn ). and let Wn be the eigenvector corresponding to theP eigenfunction wn (x). At this point we note that for n large, Λn = λn+ν where ν = ∞ n=0 (νn − 1) (0 ≤ ν < ∞, [Russakovski]). ˜ of L to the domain We define the extension L ˜ = {Y ∈ H : y, y ′ ∈ AC, ly ∈ L2 }. D(L) For λ 6= cj , j = 1, ..., N, let v(x, λ) b1 c1 −λ v(1, λ) .. .



   V (x, λ) =    

1 λ

h

bN λ) cN −λ v(1, 

v ′ (1, λ) − v(1, λ) b +



PN

bk k=1 ck −λ

and in same way define W (x, λ) from the function w(x, λ).

      i 

In the case of λ = cj we define 

v(x, cj ) 0 .. .

       ′ 0 V (x, cj ) = lim V (x, λ) =   v (1, cj ) λ→cj  0   ..  .   0 0



       .       

Define W (x, λ) analogously when λ = cj is an eigenvalue but when cj not an eigenvalue W (x, λ) has a pole of order 1 at λ = cj and is thus not defined at λ = cj . For λ = Λn define kn by W (x, Λn ) = Wn (x) = kn V (x, Λn ),

7

for all

x ∈ [0, 1]

where kn ∈ C\{0}. We note that kn is well defined as each eigenvalue is geometrically simple and that ˜ (x, λ) = λV (x, λ), for all x ∈ [0, 1], LV ˜ (x, λ) = λW (x, λ), for all x ∈ [0, 1], LW

In the case of νn > 1, the canonical chain of associated vector [Naimark] corresponding to the eigenvector Wn is ∂ 1 W (x, λ) ∂ νn −1 W (x, λ) , ..., . ∂λ1 ∂λνn −1 λ=Λn λ=Λn We thus set

Wnj

∂ j W (x, λ) = , ∂λj λ=Λn

j = 0, ..., νn − 1,

and define the norming constants associated with the eigenvalue Λn by   ρjn = Wnj , Wnνn −1 , j = 0, ..., νn − 1. (j)

(j)

Denoting the L2 [0, 1] components of Wn (x) and V (j) (x, λ) by wn (x) and v (j) (x, λ) (j) (it should be noted that wn (x) and v (j) (x, λ) are also the j th partial derivatives of w(x, λ) and v(x, λ) with respect to λ and evaluated at Λn and λ respectively) we have for λ 6= ci , i = 1, ..., N that   v (j) (x, λ)    X j    b1 (j − k)! (k) j  v (1, λ)  j+1−k   k (c1 − λ)   k=0   . (j)  .. V (x, λ) =      j     X j bN (j − k)! (k)  v (1, λ)    j+1−k   k=0 k (cN − λ) av (j) (1, λ) (j)

and for j = 0, ..., νn − 1, Wn (x) is given by an analogous formula, but with v (j) (x, λ) (j) replaced by wn (x). As a consequence we thus have Z 1 (r) (s) v (r) v (s) dt + av (r) v (s) (1, λ) [V , V ] = 0

  s  N X r X X bk (r − i)!(s − j)! (i) (j) r s + v v (1, λ) i j (ck − λ)s+r+2−j−i j=0 i=0 k=1    r s Z 1 r X s X i j f (i+j+1) (λ) (r−i) (s−j)   v (r) v (s) dt + = v v (1, λ) i+j +1 i+j 0 i=0 j=0 i

8

and for r, s = 0, ..., νn − 1 [Wn(r) , Wn(s) ]

=

Z

0

1

wn(r) wn(s)

dt +

r X s X



i=0 j=0

r i 



 s j f (i+j+1)(Λn ) (r−i) (s−j)  wn wn (1). i+j +1 i+j i

It remains to consider that more challenging case of λ = cj . To simplify the notation we set X bi fj (λ) = aλ + b + ci − λ i6=j

gj (λ) =

g(λ) cj − λ

where gj is extended by continuity to a polynomial on C having gj (cj ) 6= 0. Then (cj − λ)gj (λ) = v(1, λ) giving (k−1)

v (k) (1, cj ) = −kgj

(cj ).

From the definition of v in terms of its value and derivative at x = 1, for all λ ∈ C   1 v ′ (1, λ) = v(1, λ). f (λ) Taking derivatives of the above and noting that limλ→cj 1/f (λ) = 0, we have  s−1  X ∂ s−i (1/f ) s (s) (i) ′ v (1, cj ) = v (1, cj ) i ∂λs−i λ=cj i=0

and thus

(k)

v (k+1) (1, cj ) k+1  k  −1 X k + 1

gj (cj ) = − =

k+1

i=0

i

v

(i) ′

∂ k+1−i (1/f ) (1, cj ) ∂λk+1−i

. λ=cj

These preliminary calculations will enable us to write V (r) (x, cj ) in terms of v (i) (x, cj ), v (i) (1, cj ) ′ and v (i) (x, cj ) for i = 0, ..., r. From the definition of V we have   v(x, λ) b1    c1 −λ v(1, λ)    ..   .     bj−1  cj−1 −λ v(1, λ)     bj gj (λ) V (x, λ) =    b  j+1 v(1, λ)    cj+1 −λ   .   ..     bN  c −λ v(1, λ)  N av(1, λ)

9

which, after differentiating r times with respect to λ and then setting λ = cj gives   v (r) (x, cj )  r    X b1 (r − i)! r   (i)   v (1, c ) j   i (c1 − cj )r+1−i i=0     ..   .     r   X r bj−1 (r − i)!   (i) v (1, c )   j i   (cj−1 − cj )r+1−i i=0     r   −b X ∂ r+1−i (1/f ) r+1 j   (i) ′ (r) (1, c ) v V (x, cj ) =   j i ∂λr+1−i λ=cj   r + 1 i=0    r    X bj+1 (r − i)! r   (i) v (1, c )  j  r+1−i i (c − c )   j+1 j i=0     ..   .     r   X r bN (r − i)! (i)   v (1, cj )   r+1−i i (c − c ) N j   i=0 av (r) (1, cj ) (r)

with the same expression holding for Wn (x), r = 0, ..., νn with v symbolically replaced by wn if Λn = cj and consequently for 0 ≤ r, s ≤ νn − 1 Z 1 (r) (s) wn(r) wn(s) dt [W , W ] = 0    r s (m+p+1) r X s X (cj ) (r−m) (s−p) m p fj   + w wn (1) m+p+1 n m+p m=0 p=0 m   r+1−m r X s  X bj ′ ′ ∂ (1/f ) ∂ s+1−p (1/f ) r+1 s+1 wn(m) wn(p) (1) + r+1−m s+1−p m p (r + 1)(s + 1) ∂λ ∂λ cj m=0 p=0

Thus giving the norming constants in terms of object directly related to the boundary value problem. Our first Lemma in this section will supply the necessary background to express the coefficients of the negative power terms in the Laurent expansion of m(λ) at an eigenvalue Λn in terms to the derivatives of D(λ, α, f, q) at Λn .

Lemma 2.3 Let D(λ) and Z(λ) be entire functions. Let D(λ) have a zero of order M j 1 ∂k D at a and let Z(a) 6= 0 and ∂∂λZj (a) = 0 for j = 1, ..., M − 1. Set ak = k! (a), then the ∂λk Laurent expansion of Z/D about a has the form M

X pk Z (λ) = E(λ) + Z(a) D (λ − a)k k=1

10

where E(λ) is analytic in a neighbourhood of a and pM pM −j

=

1 , aM

j−1 1 X pM −k aM +j−k , = − aM

j = 1, ..., M − 1.

k=0

Proof: That Z/D has a Laurent expansion of the form claimed in the lemma is well known, it remains to prove that the coefficients p1 , ..., pM are as stated in the Lemma. As D(λ) is entire (and has a zero of order M at a) it is given by D(λ) =

∞ X

k=M

(λ − a)k ak .

We proceed by induction on the index j in pM −j . For j = 0 we have Z(a)pM

= = =

lim (λ − a)M

λ→a

lim

λ→a

Z(λ) D(λ)

Z(λ) D(λ) (λ−a)M

Z(a) . aM

Assuming the forms of pM , ..., pM −(j−1) to be correct we have Z(a)pM −j =

=

lim (λ − a)M −j

λ→a

Z(λ)−Z(a) (λ−a)j lim D(λ) λ→a (λ−a)M

"

j−1

Z(λ) X Z(a)pM −k − D(λ) (λ − a)M −k k=0 

1 + Z(a) lim λ→a (λ − a)j

   ∞ X 

k=M

=

=

=

j−1 ∞ X X

# 1 ak (λ − a)k−M



j−1 X

  − pM −k (λ − a)   k=0 k

# Z(a) 1 k+r−M lim ar pM −k (λ − a) 1− aM λ→a (λ − a)j r=M k=0   min(j−1,r) ∞ X X Z(a) 1 1 − (λ − a)r aM +r−k pM −k  lim aM λ→a (λ − a)j r=0 k=0   j−1 j−1 ∞ r X X X X Z(a) 1 1 − aM +r−k pM −k  (λ − a)r aM +r−k pM −k − (λ − a)r lim aM λ→a (λ − a)j r=0 "

k=0

11

r=j

k=0

Now from the induction hypotheses r 0 X X (λ − a)r aM +r−k pM −k = 1 − aM pM = 0

1−

r=0

and for r = 1, ..., j − 1 r X k=0

k=0

aM +r−k pM −k = aM pM −r +

r−1 X k=0

aM +r−k pM −k

= aM pM −r − aM pM −r = 0.

Thus we have j−1 ∞ X X Z(a) 1 r = − (λ − a) lim aM +r−k pM −k aM λ→a (λ − a)j

Z(a)pM −j

r=j

= −

Z(a) aM

j−1 X k=0

k=0

aM +j−k pM −k .

Concluding the proof of the lemma.

Assuming Λn to be an eigenvalue of multiplicty νn , we have that the vector fields kn (v(0, λ), v ′ (0, λ)) and (w(0, λ), w′ (0, λ)) have M − 1 order contact at Λn . Hence for each x ∈ [0, 1]  ∂ j v(x,λ)   ∂ j w(x,λ)  j j ∂λ ∂λ kn  ∂ j v′ (x,λ) λ=Λn  =  ∂ j w′ (x,λ) λ=Λn  , j = 0, ..., νn − 1. ∂λj ∂λj λ=Λn

Setting Z(λ) =

v ′ (0, λ) cos α +

λ=Λn

v(0, λ) sin α we have that  1/kn , j = 0, = 0, j = 1, ..., νn − 1 λ=Λn

∂ j Z(λ) ∂λj

and the following corollary follows then as a consequence of the above Lemma. Corollary 2.4 The Laurent expansion of m(λ) about Λn takes the form νn 1 X pk m(λ) = En (λ) + kn (λ − a)k k=1

where E(λ) is analytic in a neighbourhood of Λn and pνn = pνn −j

νn ! , D(νn )

j−1 νn ! X D (νn +j−k) = − (ν ) pνn −k , (νn + j − k)! D n k=0

where D and it derivatives are evaluated at (Λn , α, f, q).

12

j = 1, ..., νn − 1,

The next theorem links the derivatives of D(λ, α, f, q) at Λn to the bilinear products [·, ·] of generalized eigenvectors of the eigenvalue Λn . Theorem 2.5 For j = 0, ..., νn − 1, D

(νn +j)

(Λn , α, f, q) = νn kn



νn + j νn

"

∂ νn −1 V ∂j V , ∂λνn −1 Λn ∂λj



Λn

#

.

Proof: Integration by parts, along with the definitions of V (x, λ) and [·, ·] give (λ − µ)[V (x, λ), V (x, µ)] = [λV (x, λ), V (x, µ)] − [V (x, λ), µV (x, µ)] ˜ (x, λ), V (x, µ)] − [V (x, λ), LV ˜ (x, µ)] = [LV

= [v(x, λ)v ′ (x, µ) − v ′ (x, λ)v(x, µ)]x=1 x=0      N X v(1, λ)v(1, µ) bk ck bk bk ck bk + − bk − − bk bk ck − λ ck − µ ck − µ ck − λ k=1  2  v(1, λ)v(1, µ) a λµ a2 λµ − + a µ λ ′ ′ = [v(1, λ)v (1, µ) − v (1, λ)v(1, µ)] − [v(0, λ)v ′ (0, µ) − v ′ (0, λ)v(0, µ)] N X +v(1, λ)v(1, µ) k=1

bk (λ − µ) (ck − λ)(ck − µ)

+v(1, λ)v(1, µ)[aλ − aµ]

= v(1, λ)v(1, µ)[f (µ) − f (λ)] − [v(0, λ)v ′ (0, µ) − v ′ (0, λ)v(0, µ)] +v(1, λ)v(1, µ)[f (λ) − f (µ)]

= v ′ (0, λ)v(0, µ) − v(0, λ)v ′ (0, µ) thus (λ − µ)[V (x, λ), V (x, µ)] = v ′ (0, λ)v(0, µ) − v(0, λ)v ′ (0, µ).

We note that the j th derivative of the right hand side of above with respect to λ at Λn is zero for j = 1, ..., νn − 1 and is kn−1 D(µ, α, f, q) for j = 0. Calculating the corresponding derivatives of the left hand side we thus obtain for j = 0 (Λn − µ)[V (x, Λn ), V (x, µ)] = kn−1 D(µ, α, f, q) and for j = 1, ..., νn − 1 # " # " ∂ j−1 V (x, λ) ∂ j V (x, λ) , V (x, µ) = j , V (x, µ) . (µ − Λn ) ∂λj ∂λj−1 λ=Λn λ=Λn

Combining the above νn expressions we obtain # " νn −1 V (x, λ) (νn − 1)! ∂ , V (x, µ) = D(µ, α, f, q). (µ − Λn )νn ν −1 n ∂λ kn λ=Λn

13

We now take the j + νn th derivative of this expression at µ = Λn to give #   " νn −1 (νn − 1)! (νn +j) ∂ j V (x, λ) ∂ V (x, λ) νn + j , D (Λn , α, f, q) = νn ! νn kn ∂λνn −1 λ=Λn ∂λj λ=Λn or more concisely D

(νn +j)

(Λn , α, f, q) = νn kn





νn + j νn

Which proves the theorem.

 ∂ νn −1 V ∂ j V . , ∂λνn −1 ∂λj Λn

Combining the corollary and the above theorem we are able to express the coefficients of the Laurent expansion of m(λ) in terms of the norming constants as follows. Corollary 2.6 The Laurent expansion of m(λ) about Λn takes the form m(λ) = En (λ) +

νn X k=1

pkn (λ − a)k

where En (λ) is analytic in a neighbourhood of Λn and pνnn = pnνn −j

3

(νn − 1)! , ρ0n

= −

j−1 X

pnνn −k

k=0

ρnj−k (j − k)!ρ0n

j = 1, ..., νn − 1.

Norming constants

In this and future work we require q ∈ W 1,1 (0, 1) in order to integrate by parts once - this is a more stringent constraint than is essential, as all that we require later is a suitable rate of decay of the fourier coefficients if q. This can be achieved by taking q ∈ Lipǫ for any 0 < ǫ < 1 [Zygmund] or by taking q ∈ H s for any s > 0 [H¨ormander]. We also recall that for large n, each eigenvalues is simple. Thus asymptotics expressions for the norming constants are only concerned with those relating to simple eigenvalues and so the norming constant ρ0n is the only one of asymptotic interest. Theorem 3.1 For q ∈ W 1,1 (0, 1) (i.e. q in AC), the norming constants for (α, f, q) obey the following asymptotics as n → ∞ ( sin2 α
+ O n12  , α= 6 0 2 0 . ρn = 1 1 2Λn + O n2 Λn , α = 0

14

Note that for q ∈ L1 we obtain the above theorem but with O(1/n2 ) replaced by O(1/n) which is sufficient for the later theorem in the case of α 6= 0, but is inaddequate for α = 0 where we need O(1/n1+ǫ ) for some ǫ > 0. Alternative conditions which ensure this as q ∈ H s for some s > 0, or q H¨older continuous for some non-zero H¨older exponent. Proof: We consider the Dirichlet and non-Dirichlet cases separately, as their normalization criteria are different (and the resulting asymptotics for the norming constants also differ). As νn = 1 for large n ρ0n = [Wn0 , Wn0 ] Z 1 N  X 2 wn + = 0

k=1

bk ck − Λn

2

wn2 (1) bk

! !2 N X bk 1 ′ wn (1) + 2 2 wn (1) − b + Λn a ck − Λn k=1  2   ′ 2  Z 1 (1) w (wn ) (1) n 2 wn + O = +O . 2 Λn Λ2n 0

But from the Appendix we get   √   O e2|ℑ λ| , α 6= 0 √   w2 (1, λ) =  O e2|ℑ λ| , α = 0 λ   √   O λe2|ℑ λ| , α 6= 0  √  (w′ )2 (1, λ) =  O e2|ℑ λ| , α = 0  i h √ √ √ √ Rx 2 cos  √ λx cos α sin α sin λx + sin2 α q(t) cos λt sin λ(x − t) dt  0  λ   |ℑ√λx|  √ 2 e 2 2 , α 6= 0 + sin α cos λx + O w (x, λ) = λ   |ℑ√λx|  √ √  √ √ Rx 2 λx  e λx sin 2 sin  , α = 0. + λ3/2 λ 0 q(t) sin λt sin λ(x − t) dt + O λ2

Setting ν =

P∞

j=0 (νj

− 1) as n → ∞ we have

 (ν + n − m)π + (1/n),    p (ν + n + 21 − m)π + O(1/n), Λn =  (ν + n + 21 − m)π + O(1/n),   (ν + n + 1 − m)π + O(1/n),

α 6= 0, deg(h) ≤ deg(g) = m, α = 0, deg(h) ≤ deg(g) = m, α 6= 0, m = deg(h) > deg(g), α = 0, m = deg(h) > deg(g),

thus gives

Z

1 0

cos2

p

Λn t dt = =

Z p 1 1 1 + cos 2 Λn t dt 2 2 0 √ 1 sin 2 Λn √ + 2 2 Λn

15

=



1 n2



1 +O 2



1 +O 2

and similarly Z

0

1

sin

2

p

Λn t dt =

1 n2



.

Direct integration gives Z 1 p p 1 √ 2 cos Λn x sin Λn x dx = Λn 0

Z 1 p 1 √ sin 2 Λn x dx Λn 0 √ 1 − cos 2 Λn = 2Λn = O(1/n2 ).

Finally √ Z 1 Z p p 2 cos Λn x x √ q(t) cos Λn t sin Λn (x − t) dt dx Λn 0 0 √ √ Z 1 Z 1 p cos Λn x sin Λn (x − t) p 2q(t) cos Λn t = Λn dx dt i 0 t √ √   Z 1 Z 1 Z 1 p p 1 + cos 2 Λn x sin 2 Λn x √ √ dx − sin 2 Λn t dx dt = q(t) (1 + cos 2 Λn t) 2 Λn 2 Λn t t 0 Z 1 p 1 = − √ (1 − t)q(t) sin 2 Λn t dt + O(1/Λn ) 2 Λn 0   Z 1 p d[(1 − t)q(t)] 1 cos 2 Λn t dt + O(1/Λn ) q(0) + = − 4Λn dt 0 = O(1/Λn ). Similarly Z

0

1

√ Z p p 2 sin Λn x x √ q(t) sin Λn t sin Λn (x − t) dt dx = O(1/Λn ). Λn 0

The Theorem now follows from combining the asymptotics.

Theorem 3.2 For the non-Dirichlet case m(λ) = cot α +

νk ∞ X X k=0 j=1

16

pjk . (λ − Λk )j

Proof: For fixed λ let H(µ) = m(µ)



1 µ − 2 µ−λ µ +1



= m(µ)

1 + λµ . (µ − λ)(µ2 + 1)

Let Γn = {ξ 2 |ξ ∈ γn } where γn is as indicated in figure 1 and  nπ,
α 6= 0 ζn = . n + 21 π, α = 0 6

iζn 

6

ζn -

h

−iζn

-

Figure 1: γn in the ξ-plane We observe that for large n m(µ) = cot α + O(n−1 ) uniformly on Γn and that length(Γn ) ≤ n2 K for some positive constant K, independent of n. Thus Z   2 n | cot α| H(µ) dµ ≤ O A4n Γn → 0 as n → ∞. Let Jn =

νr n X X r=0 j=1

pjr ∂ j−1 (j − 1)! ∂µj−1



 µ µ 2 + 1 Λr

and noting that for large n we have νn = 1 and p1n = O(1) we may let J = limn→∞ Jn . Assuming that m(±i) are both finite (for simplicity) we have for large n that 1 2πi

Z

n+N (f ) νr X X m(i) + m(−i) pjr H(µ) dµ = m(λ) − − Jn+N (f ) − 2 (µ − Λr )j Γn r=0

17

j=1

where N (f ) is fixed and depends only on the function f . Letting n → ∞ we thus obtain ∞

ν

r XX m(i) + m(−i) pjr m(λ) = +J + . 2 (µ − Λr )j r=0

j=1

√ λ = iτ with τ → ∞ we obtain that

Using the fact that m(λ) → cot α for

cot α =

m(i) + m(−i) +J 2

and hence that

∞ X νr X

m(λ) = cot α +

r=0 j=1

pjr . (µ − Λr )j

We now develop a similar representation for the Dirichlet case, but here substantially more care is need. In particular specifying m(λ) up to an additive constant is straightforward but the determination of this remaining constant demands work.

Lemma 3.3 For the Dirichlet case m(λ) = m(0) +

∞ X νn X

pjn

n=0 j=1



 1 1 − . (λ − Λn )j (−Λn )j

Proof: Proceeding as in the non-Dirichlet case, but with H(µ) = and ζn =



as m(λ) = O(n) on Γn we have that

m(µ) µ(µ − λ)

nπ,
α=0 , 1 n + 2 π, α = 6 0

lim

Z

n→∞ Γ n

H(λ) dλ = 0

thus, assuming λ 6= 0, that 0 and λ are not an eigenvalues, we have   ∞ νn ∂ j−1 1 pjn m(λ) m(0) X X − + 0= λ λ (j − 1)! ∂µj−1 µ(µ − λ) µ=Λn n=0 j=1

giving

m(λ) = m(0) +

∞ X νn X

n=0 j=1

∂ j−1 pjn (j − 1)! ∂µj−1

18



 1 1 + µ λ − µ µ=Λn

from which the statement of the lemma follows. For this last lemma to be of any use we need to show that m(0) can be obtained from the spectrum and norming constants.

Lemma 3.4 In the Dirichlet case i.e. α = 0, m(−σ 2 ) = σ + O(1/σ), √ In particular m(λ) + i λ → 0 as λ → −∞.

σ → ∞.

Proof: Bootstrapping the process √ in [Appendix,HH] one more time than appears in the reference one obtains that for λ = iσ with σ → ∞ we have (cf. section 2)   eσ Q −2 s1 (0) = 2 + + O(σ ) 4 σ   σ Q σe 2 + + O(σ −2 ) s′1 (0) = − 4 σ   σ e Q −2 s2 (0) = − 2 + + O(σ ) 4σ σ   σ e Q s′2 (0) = 2 + + O(σ −2 ) 4 σ R1 where Q = 0 q(t) dt. Recall that v = gs1 + hs2 . CASE: deg(h) ≤ deg(g) = m

For this case g(−σ 2 ) = (−σ 2 )m + O(σ 2m−2 ), h(−σ 2 ) = Aσ 2m + O(σ 2m−2 where A ∈ R. From this and the above asymptotics for sj (0), j = 1, 2 we get   Q − 2A (−1)m σ 2m eσ −2 2 + O(σ ) , 2+ v(0, −σ ) = 4 σ   (−1)m+1 σ 2m+1 eσ Q − 2A v ′ (0, −σ 2 ) = + O(σ −2 ) . 2+ 4 σ Thus v ′ (0, −σ 2 ) + σv(0, −σ 2 ) v(0, −σ 2 )   1 = O . σ

m(−σ 2 ) − σ = −

19

CASE: deg(g) < deg(h) = m For this case h(−σ 2 ) = (−1)m σ 2m + O(σ 2m−2 ), g(−σ 2 ) = A(−1)m−1 σ 2m−2 + O(σ 2m−4 ) where A ∈ R giving   (−1)m−1 σ 2m−1 eσ Q + 2A −2 v(0, −σ ) = + O(σ ) 2+ 4 σ   (−1)m σ 2m eσ Q + 2A ′ 2 −2 v (0, −σ ) = + O(σ ) 2+ 4 σ
from which it follows, as previously, that m(−σ 2 ) − σ = O σ1 . 2

The following theorems express m(λ) in terms of eigenvalues and norming constants.

Theorem 3.5 For α = 0 and M = deg(h) > deg(g)  ∞ νn X X 2 + m(λ) = 1 − 2M + 2ν + n=0

where ν =

P∞

n=0 (νn

j=1

− 1).

 pjn  (λ − Λn )j

Proof: We recall √ that Λn = (n + 1 + ν − M )2 π 2 + O(1) and note that √ −i λ[1 + O(e−|ℑ λ| )] as λ → −∞ and thus that √ √ lim [m(λ) − λ cot λ] = 0.

√

λ cot

√

λ =

λ→−∞

The Mittag-Leffler expansion of √

√ √ λ cot λ [Tichmarsh -Theory of functions pg 113] gives

λ cot

√

λ = −1 −

∞ X k=0

2λ . −λ

n2 π 2

From our previous theorem we have that m(λ) = m(0) + S(λ) +

∞ X

n=0

where S(λ) =

∞ X νn X n=0 j=2

pjn



p1n



1 1 + λ − Λn Λn

1 1 − (λ − Λn )j (−Λn )j

20





which is infact a finite summation, as all but finitely many eigenvalues are simple. Let S∞ = lim S(λ) = − λ→−∞

∞ X νn X

n=0 j=2

pjn . (−Λn )j

With the aid of the Mittag-Leffler expansion, above, MX −ν−2 √ √ m(λ) − λ cot λ = m(0) + S(λ) + 1 + p1n n=0

∞ X



1 1 + λ − Λn Λn



# λ −2 + Λn+M −ν−1 λ − Λn+M −ν−1 λ − n2 π 2 n=0   MX −ν−2 1 1 1 pn = m(0) + S(λ) + 1 + + λ − Λn Λn n=0 " #  ∞ ∞  X X p1n+M −ν−1 2λ 2λ λ + −2 + − . Λn+M −ν−1 λ − Λn+M −ν−1 n=0 λ − Λn+M −ν−1 λ − n2 π 2 n=0 "

p1n+M −ν−1

λ

As p1n /Λn = 2 + O(n−2 ) we have that " " # # ∞ ∞ X X p1n+M −ν−1 p1n+M −ν−1 λ −2 = −2 . lim λ→−∞ Λn+M −ν−1 λ − Λn+M −ν−1 n=0 Λn+M −ν−1 n=0 Noting that Λn+M −ν−1 − n2 π 2 = O(1), we have there exists a constant K, such that for λ large and negative, ∞ X 2λ 2λ λ − Λn+M −ν−1 − λ − n2 π 2 n=0  ∞  X |λ| ≤ K |λ|(n2 π 2 − λ) n=0 √ √ K( λ cot λ + 1) = − 2λ → 0 as λ → −∞. Assembling these pieces we have, that as λ → −∞ m(λ) −

√

" # ∞ MX −ν−2 X √ p1n+M −ν−1 1 1 + −2 pn λ cot λ → m(0) + S∞ + 1 + Λn Λn+M −ν−1 n=0

n=0

from which we obtain m(0) =

∞ X νn X

n=0 j=2

 ∞  1 X pjn pn − 1 − 2(M − ν − 1) − −2 . (−Λn )j Λn n=0

21

Substituting these expressions back into that for m(λ) we get ∞ X νn X

 ∞  1 X pjn pn m(λ) = 1 − 2M + 2ν + − −2 (−Λn )j Λn n=0 j=2 n=0   ∞ X νn X 1 1 j pn + − (λ − Λn )j (−Λn )j n=0 j=1  ∞  ∞ X νn X X p1n 1 j + 2+ pn = 1 − 2M + 2ν + (λ − Λn )j λ − Λn n=0 n=0 j=2

proving the theorem.

Theorem 3.6 For α = 0 and deg(h) ≤ deg(g) = M we have   ∞ νn j X X p n . 2 + m(λ) = −2M + 2ν + (λ − Λn )j n=0

j=1

Proof: Here Λn = (n + ν + 21 − M )2 π 2 + O(1) and we note that √ O(e−2|ℑ λ| )] as λ → −∞ and Mittag-Leffler expansion gives √ ∞ 2 tan λ X √ = . 1 2 2 (k + 2 ) π − λ λ k=0 Thus lim m(λ) +

√

λ→−∞

λ tan

√

√

λ tan

√

√ λ = i λ[1 +

λ = 0.

Combining this information with the series representation of m(λ) we have As in the previous theorem m(λ) = m(0) + S(λ) +

∞ X

p1n

n=0

where S(λ) =

∞ X νn X n=0 j=2

pjn





1 1 + λ − Λn Λn

1 1 − (λ − Λn )j (−Λn )j

and S∞ = lim S(λ) = − λ→−∞

22

νn ∞ X X

n=0 j=2

pjn . (−Λn )j





Using the Mittag-Leffler expansion for the tangent function, above, m(λ) +

MX −ν−1 √ √ p1n λ tan λ = m(0) + S(λ) + n=0



1 1 + λ − Λn Λn



" # " # ∞ ∞ X X p1n+M −ν 2λ 2λ λ + . −2 + − Λn+M −ν λ − Λn+M −ν λ − Λn+M −ν λ − (n + 21 )2 π 2 n=0 n=0

As p1n /Λn = 2 + O(n−2 ) we have that " " # # ∞ ∞ X X p1n+M −ν p1n+M −ν λ lim −2 = −2 . λ→−∞ Λn+M −ν λ − Λn+M −ν Λn+M −ν n=0

n=0

Noting that Λn+M −ν − (n + 21 )2 π 2 = O(1), we have there exists a constant K, such that for λ large and negative, ∞ X 2λ 2λ − 1 2 2 λ − Λn+M −ν λ − (n + 2 ) π n=0 " # ∞ X |λ| ≤ K |λ|((n + 21 )2 π 2 − λ) n=0 √ K tan λ √ = 2 λ → 0 as λ → −∞. As λ → −∞ we thus have " # MX −ν−1 ∞ 1 X √ √ p 1 n+M −ν p1n m(λ) + λ tan λ → m(0) + S∞ + + −2 Λ Λ n n+M −ν n=0 n=0 from which we obtain 0 = m(0) + 2(M − ν) + S∞ +

∞  1 X p

n

n=0

Λn

 −2

Substituting back into the expression for m(λ) we get m(λ) = −2(M − ν) + S(λ) − S∞    X ∞ ∞  1 X 1 1 pn 1 pn + + −2 − λ − Λn Λn Λn n=0 n=0  ∞ X νn ∞  X X pjn p1n = −2(M − ν) + + 2+ (λ − Λn )j n=0 λ − Λn n=0 j=2

23

from which the theorem follows. In conclusion it should be noted that we have done more than just prove uniqueness results for the three inverse spectral problems - we have shown that they are equivalent for the case of boundary conditions being rational in the spectral parameter and have given formulae for move from one of the three sets of data to any other.

4

Appendix

We collect here some asymptotic estimates which can be derived from those in [19] by bootstrapping one step further than was done in [19]. Proceeding as in [HH], it can easily be verified that the solutions s1 , s2 of (1.1) with terminal conditions s1 (1) = 1 = s′2 (1) s2 (1) = 0 = s′1 (1) asymptotically for large λ, are given by √ √ Q sin λ √ +O s1 (0) = cos λ + 2 λ

√

e|ℑ λ| λ

!

,

√ ! √ Q e|ℑ λ| √ = , λ sin λ − cos λ + O 2 λ √ ! √ √ sin λ Q e|ℑ λ| , cos λ + O s2 (0) = − √ + 2λ λ3/2 λ √ ! √ |ℑ λ| √ Q e sin λ √ +O s′2 (0) = cos λ + 2 λ λ

√

s′1 (0)

where Q = forms at 0.

R1 0

√

q dt. Hence, as v = gs1 + hs2 , v and v ′ have the following asymptotic

CASE: deg(h) ≤ deg(g) = m

  √  √ Q − Am sin λ + O λm−1 e|ℑ λ| , v(0, λ) = λ cos λ + λ 2    √  √ √ 1 Q ′ m+ 12 m v (0, λ) = λ cos λ + O λm− 2 e|ℑ λ| . sin λ + λ Am − 2 m

√

m− 21



24

CASE: m = deg(h) > deg(g)

√ !#   |ℑ λ| √ √ √ e Q √ v(0, λ) = λm−1 − λ sin λ + cos λ Am−1 + +O , 2 λ " √ !# √   |ℑ λ| √ sin λ e Q . v ′ (0, λ) = λm cos λ + √ +O Am−1 + 2 λ λ

"

Asymptotics for the functions v and v ′ at general points are given by

   √ √  λM cos λ(1 − x) + O λM − 21 e|ℑ λ|(1−x) ,   √ v(x, λ) = √  −λM − 21 sin λ(1 − x) + O λM −1 e|ℑ λ|(1−x) ,    √ √  λM + 21 sin λ(1 − x) + O λM e|ℑ λ|(1−x) , M   √ v ′ (x, λ) = √  λM cos λ(1 − x) + O λM − 21 e|ℑ λ|(1−x) , M

M = deg(g) ≥ deg(h), M = deg(h) > deg(h), = deg(g) ≥ deg(h), = deg(h) > deg(h).

The solutions w(x, λ) have the following asymptotic forms.

√ √ √ Rx  sin √α sin α cos q(t) cos λx + λt sin λ(x − t) dt   λ0 √  √  e|ℑ λx| cos α√ sin λx +O + , α 6= 0 w(x, λ) = λ λ  |ℑ√λx|   √ √ √ R   sin√ λx + 1 x q(t) sin λt sin λ(x − t) dt + O e , α=0 λ 0 λ3/2 λ  √   √ √  − λ sin α sin λx + O e|ℑ λx| , α 6= 0  √  √ . w′ (x, λ) =  cos√ λx + O e|ℑ λx| , α = 0 λ λ

References [1] A. I. Benedek, R. Panzone, On inverse eigenvalue problems for a second-order differential equation with parameter contained in the boundary conditions, Notas de Algebra y Analisis, 9 (1980), 1-13. [2] A. I. Benedek, R. Panzone, On Sturm-Liouville problems with the square-root of the eigenvalue parameter contained in the boundary condition, Notas de Algebra y Analisis, 10 (1981), 1-59.

25

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