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The Krein signature, Krein eigenvalues, and the Krein Oscillation
Theorem
Todd Kapitula ∗
Department of Mathematics and StatisticsCalvin College
Grand Rapids, MI 49546
June 19, 2009
Abstract. In this paper the problem of locating eigenvalues of negative Krein signature isconsidered for operators of the form JL, where J is skew-symmetric with bounded inverseand L is self-adjoint. A finite-dimensional matrix, hereafter referred to as the Krein matrix,associated with the eigenvalue problem JLu = λu is constructed with the property that if theKrein matrix has a nontrivial kernel for some z0, then ±
√−z0 ∈ σ(JL). The eigenvalues of
the Krein matrix, i.e., the Krein eigenvalues, are real meromorphic functions of the spectralparameter, and have the property that their derivative at a zero is directly related to the Kreinsignature of the eigenvalue. The Krein Oscillation Theorem relates the number of zeros ofa Krein eigenvalue to the number of eigenvalues with negative Krein signature. Because theconstruction of the Krein matrix is functional analytic in nature, it can be used for problemsposed in more than one space dimension. This feature is illustrated in an example for which thespectral stability of the dipole solution to the Gross-Pitaevski equation is considered.
Contents
1. Introduction 1
2. The Krein matrix and Krein eigenvalues 52.1. Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2. Order of zeros and the algebraic multiplicity . . . . . . . . . . . . . . . . . . . . . . . 82.3. The Krein Oscillation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4. Asymptotic results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5. Perturbative results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3. Application: Gross-Pitaevski equation 173.1. Cigar trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2. Pancake trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
References 21
∗E-mail: [email protected]
T. Kapitula 1
1. Introduction
When attempting to understand the dynamics near the critical points of an energy surface fora Hamiltonian system, one must study eigenvalue problems of the form
JLu = λu, (1.1)
where J is a skew-symmetric operator with bounded inverse and L is a self-adjoint operator (see[9] for the case where ker(J ) is nontrivial). The operator L is the linearization of the energy surfaceabout a critical point, and σ(L) corresponds to the local curvature of the surface. If σ(L) ⊂ R+,then under appropriate assumptions the critical point is a minimizer of the energy surface, and ishence stable. If σ(L) 6⊂ R+, then the critical point is not a minimizer. It may be a constrainedminimizer, or it may instead be energetically unstable. In the latter case one wishes to betterunderstand the manner in which being energetically unstable influences σ(JL).
If Im(JL) = 0, then it was seen in [21] that (1.1) is equivalent to the canonical case
J =(
0 1
−1 0
), L =
(L+ 00 L−
), (1.2)
upon setting L+ = L and L− = −JLJ . If Im(JL) 6= 0, then one must be careful when comparingthe two problems; in particular, the reduction of the four-fold eigenvalue symmetry for (1.1) whenIm(JL) = 0 to a two-fold symmetry when Im(JL) 6= 0 causes some difficulties. It will henceforthbe assumed that (1.1) is in the canonical form of (1.2). Furthermore, it will be assumed that:Assumption 1.1. The operators J and L in (1.2) satisfy a variant of [13, Assumption 2.1(b)-(d)],i.e.,
(a) L± are self-adjoint with compact resolvent
(b) σ(L±) ∩ R− is a finite set
(c) there is a self-adjoint operator L0 with compact resolvent such that:
(1) L± = L0 +A±, where A± are L0-compact and satisfy
‖A±u‖ ≤ a‖u‖+ b‖ |L0|ru‖
for some positive constants a, b, and r ∈ [0, 1)
(2) the increasing sequence of nonzero eigenvalues ωj of L0 satisfy for some s ≥ 1,
∞∑j=1
|ωj |−s <∞
(3) there exists a subsequence of eigenvalues {ωnk}k∈N and positive constants c > 0 andr′ > r such that
ωnk+1 − ωnk ≥ c ωr′nk+1.
(d) regarding gker(JL) one has:
(1) ker(L+) ⊥ ker(L−)
(2) dim[gker(JL)] = 2 dim[ker(L)] (see [20, Lemma 3.1] for a simple condition whichensures this dimension count).
Krein signature and the Krein Oscillation Theorem 2
Remark 1.2. The assumption of compactness is not necessary, e.g., see [10, 11, 20, 27] and thereferences therein, but it simplifies the analysis by removing the possibility of having eigenvaluesembedded in the continuous spectrum. The compactness of the operators was recently exploited in[13].
Assumption 1.1, unlike [13, Assumption 2.1], does not assume that L is invertible. However, theinvertibility assumption can be recovered in the following manner. Upon using Assumption 1.1(d:1)let Π : H 7→ [ker(L+)⊕ker(L−)]⊥ be the orthogonal projection, where H is the appropriate Hilbertspace for the problem. It was seen in [20, Section 3] that solving (1.1) for nonzero eigenvalues isequivalent to solving the system
−ΠL+Πu = λv, ΠL−Πv = λu. (1.3)
The operators ΠL±Π are self-adjoint and nonsingular under Assumption 1.1(d:2).Set
R := ΠL+Π, S−1 := ΠL−Π, z := −λ2 (−π/2 < arg λ ≤ π/2), (1.4)
and note that the derivation leading to (1.5) yields that R and S−1 are nonsingular self-adjointoperators with compact resolvent on a Hilbert space H endowed with inner-product 〈·, ·〉. (1.3) isequivalent to the eigenvalue problem
(R− zS)u = 0 (1.5)
in the following sense (the mapping z = −λ2 is illustrated in Figure 1). The nonzero spectrumassociated with (1.2) has the four-fold symmetry {±λ,±λ}. Eigenvalues with positive real part andnonzero imaginary part are mapped in a one-to-one fashion to eigenvalues with nonzero imaginarypart, and instead of the four-fold symmetry there is now only the reflection symmetry {z, z}. Posi-tive real eigenvalues are mapped to negative real eigenvalues. Consequently, all unstable eigenvaluesfor (1.3) are captured by considering (1.5) with
• z ∈ R− if and only if λ ∈ R+
• Im z 6= 0 if and only if Imλ 6= 0 and Reλ > 0.
With respect to the purely imaginary eigenvalues, one has that λ ∈ iR+ is mapped to R+; however,iR− is not included in the domain of the mapping, and hence these eigenvalues are not captured.Since purely imaginary eigenvalues come in the pair ±λ, one has that if there is an eigenvaluez ∈ R+ for (1.5), then for (1.4) there will be the eigenvalue pair ±λ ∈ iR.
Re λ
Im λ
Re z
Im zz=−λ2
Figure 1: (color online) The left panel is the spectral set −π/2 < arg(λ) ≤ π/2 for (1.3),and the right panel is the spectral set 0 < arg(z) ≤ 2π for (1.5). These two sets are relatedby the mapping z = −λ2.
T. Kapitula 3
There has been a great deal of recent work on (1.3). Because of the equivalence between (1.3)and (1.5), the results will be stated in the context of (1.5). Most recently it was shown that underAssumption 1.1,
kr + 2kc + 2k−i = n(R) + n(S) (1.6)
[13, Theorem 2.25]. Here n(A) is the number of strictly negative eigenvalues (counted with algebraicmultiplicities) of the self-adjoint operator A, kr refers to the number of negative real eigenvalues,kc is the number of eigenvalues with Im z < 0, and k−i is the number of positive real eigenvalueswith negative Krein signature.
The Krein signature of an eigenvalue z ∈ R+ with eigenspace Ez says something importantabout the nature of the flow in the direction of Ez along the energy surface. If the signatureis positive, then the energy surface is positive definite in this direction. On the other hand, ifthe signature is negative, then the surface is negative definite in the direction of some nontrivialsubspace of Ez; however, the vector field associated with the linearized flow (and perhaps thenonlinear flow, see [9, Section 4]) is orthogonal to those energetically unstable directions. Thesignature is computed via
k−i (z) = n(S|Ez), k−i =∑
z∈σ(S−1R)∩R+
k−i (z);
in particular, if z ∈ R+ is geometrically and algebraically simple with eigenfunction u, then
k−i (z) =
{0, 〈u,Su〉 > 01, 〈u,Su〉 < 0.
(1.7)
If 〈u,Su〉 = 0 then one knows that the eigenspace has a Jordan block [10, Theorem 2.3].From (1.6) one necessarily has that kr ≥ 1 if n(R) + n(S) is odd (see [12] for a similar result for
(1.1)). Thus, one has a simple instability criterion, which for (1.5) can be refined in the followingmanner. For a self-adjoint operator H define the negative cone C(H) by
C(H) := {u : 〈u,Hu〉 < 0} ∪ {0},
and let dim[C(H)] denote the dimension of the maximal subspace in C(H). First suppose thateach negative real-valued eigenvalue is algebraically simple. By [13, Corollary 2.24] one knows that(1.6) can be refined as
kr = |n(R)− n(S)|+ 2 (min{n(R),n(S)} − dim[C(R) ∩ C(S)])kc + k−i = dim[C(R) ∩ C(S)].
(1.8)
If the assumption that negative real-valued eigenvalues are algebraically simple is removed, thenthe equalities are removed in favor of inequalities:
kr ≥ |n(R)− n(S)|+ 2 (min{n(R),n(S)} − dim[C(R) ∩ C(S)])kc + k−i ≤ dim[C(R) ∩ C(S)].
(1.9)
A close examination of (1.8) yields some important insights concerning the location of thespectra for (1.5). First, if n(S) = 0, then C(S) = {0}, so that kr = n(R) and k−i = kc = 0.Consequently, it will henceforth be assumed in this paper that n(S) ≥ 1. Second, the first line ineither (1.8) or (1.9) yields the lower bound
kr ≥ |n(R)− n(S)| (1.10)
Krein signature and the Krein Oscillation Theorem 4
(for the earliest proofs see [11, 14]). Finally, the inequality in (1.10) is closed upon consideringthe cone intersection C(R) ∩ C(S). If C(R) ∩ C(S) = {0}, then one has that kr = n(R) + n(S),which by (1.6) implies that all of the unstable spectra is purely real. As the dimension of thiscone intersection increases, the number of negative real eigenvalues decreases in favor of either (a)eigenvalues with nonzero imaginary part, or (b) positive real eigenvalues.
As a consequence of (1.8) one has a precise count of kr. It would be useful if one could alsoprecisely count kc, for it would then be the case that one would have an exact count of the totalnumber of spectrally unstable eigenvalues. Alternatively, one could determine k−i , and then use(1.6) to determine kc. As it is seen in the second line of (1.8), the two quantities k−i and kc areclosely related by the cone intersection; furthermore, one has the upper bound
kc + k−i ≤ min{n(R),n(S)} (1.11)
(this upper bound is also valid for (1.9)). It is, and should be, difficult to further refine the secondline of (1.8). It is well-known that if 1 ≤ k−i (z) ≤ dim(Ez)− 1, i.e., there is a collision of a positivereal eigenvalues of opposite sign, then there is generically a Hamiltonian-Hopf bifurcation. Underthis bifurcation a small change of the vector field decreases the count k−i while increasing the countkc. The dimension of the cone intersection is unchanged under small perturbations, so that thetotal count kc + k−i remains unchanged.
In this paper a certain matrix, called the Krein matrix, will be constructed, and it will havethe property that its eigenvalues, called the Krein eigenvalues, can be used to locate all of thoseeigenvalues making up the subcount kc + k−i . These eigenvalues are realized as zeros of the deter-minant of the Krein matrix, which in turn implies that they are realized as zeros of one or moreof the Krein eigenvalues. For positive real zeros which do not coincide with removable singularitiesof the Krein matrix the Krein signature of the eigenvalue will be directly related to the sign ofthe derivative of a Krein eigenvalue. A careful study of the graphical asymptotics of the Kreineigenvalues leads to the Krein Oscillation Theorem, which relates the number of zeros of the Kreineigenvalue in a given interval for which it is real analytic to the number of negative eigenvalues, orto the number of positive eigenvalues with negative signature.
The Evans function is a holomorphic function of the eigenvalue parameter, the zeros of theEvans function are located precisely at the eigenvalues, and the order of the zero is the algebraicmultiplicity of the eigenvalue (see [3, 15, 18, 29] and the references therein). There is one way inwhich the Krein matrix behaves like the Evans function: the zeros of its determinant will correspondto eigenvalues. However, it will differ from the Evans function in many important respects. On theother hand, for the Krein matrix and Krein eigenvalues one has:
(a) the Krein matrix and Krein eigenvalues are meromorphic
(b) the determinant of the Krein matrix does not necessarily vanish for all real eigenvalues
(c) the order of the zero of the Krein eigenvalue will match the algebraic multiplicity of theeigenvalue only for those eigenvalues not located at removable singularities.
A final important point of distinction is that the construction of the Krein matrix is functionalanalytic in nature, and hence it can be formally constructed for problems posed in many spatialdimensions. Alternatively, the Evans function is constructed via ODE techniques, and is thereforeapplicable to only those problems which can be somehow posed on a subset of the line.
Property (a) of the Krein eigenvalues is a necessary condition if one wishes to relate the deriva-tive of the Krein eigenvalue to the signature of the eigenvalue. Regarding (b), it appears at firstglance that this property would seem to preclude the usefulness of the Krein matrix and the Krein
T. Kapitula 5
eigenvalues. However, the matrix will be constructed in such a manner that its determinant willbe zero for all of those eigenvalues for which the associated eigenspace has a nontrivial intersectionwith C(S). As was seen in [13], this then means it will necessarily be zero whenever the eigenvaluehas nonzero imaginary part, or is positive and real with negative signature. Hence, the determinantof the Krein matrix will detect all of those eigenvalues making up the subcount kc + k−i . The onlyeigenvalues that will potentially be missed are either negative real eigenvalues, or the positive realones with positive Krein signature. It will be seen that those eigenvalues which are not realized asa zero of a Krein eigenvalue will correspond to a removable singularity of the Krein matrix.
The paper will be organized as follows. In Section 2 the Krein matrix will be constructed, andthe properties leading to the Krein Oscillation Theorem will be derived. In Section 3 the theorywill be used to study the spectral problem associated with waves of the Gross-Pitaevski equationin both one and two space dimensions.
Acknowledgments. The author gratefully acknowledges the support of the Jack and Lois KuipersApplied Mathematics Endowment, a Calvin Research Fellowship, and the National Science Foun-dation under grant DMS-0806636. The author appreciates illuminating conversations with KeithPromislow and Panayotis Kevrekidis.
2. The Krein matrix and Krein eigenvalues
The construction of the Krein matrix follows the ideas presented in [10, 27]. In contrast to[10, 27] there will be the fundamental difference that R and S−1 are both assumed to have compactresolvent, and hence there will be no issue herein with continuous spectrum. In [27] it was alsoassumed that n(S−1) = 1, which was relevant to the problem solved therein. Those properties ofthe Krein matrix and the Krein eigenvalues which follow directly from either of [10, 27] will beappropriately cited.
2.1. Construction
It will be assumed for all self-adjoint operators that the eigenfunctions form an orthonormalbasis. For the self-adjoint operator S with eigenfunction basis {φj}, let the associated eigenvaluesbe denoted by λj . Since S is nonsingular and compact with 1 ≤ n(S) <∞, one has that σ(S) canbe ordered as λ1 ≤ λ2 ≤ · · ·λn < 0, and λj > 0 for j ≥ n+ 1. Set
N(S) := span{φ1, . . . , φn},
and let P : H 7→ N(S)⊥ and Q = 1− P be orthogonal projections, i.e.,
Pu = u−n∑j=1
〈u, φj〉φj , Qu =n∑j=1
〈u, φj〉φj . (2.1)
Define the operatorsR2 := PRP, S2 := PSP, (2.2)
and note that (a) R2 and S−12 have compact resolvent, and (b) n(S2) = 0 with n(R2) ≤ n(R).
Upon setting p := Pu and writing u = p+∑αjφj , by applying the projection P to (1.5) one gets
the system
(R2 − zS2)p+n∑j=1
αjPRφj = 0. (2.3)
Krein signature and the Krein Oscillation Theorem 6
Applying the projection Q, taking the inner product of the resulting equation with φ` for ` =1, . . . , n, and using the fact that P,Q are commuting self-adjoint operators yields the system ofequations
〈Rp, φ`〉+n∑j=0
αj〈Rφj , φ`〉 = α`zλ`. (2.4)
The system of (2.3) and (2.4) is the one to be studied.In order to close (2.4), one must solve for p in (2.3) and substitute that expression into (2.4).
Since S2 is compact, self-adjoint, and positive definite, the expression S2 = S1/22 S
1/22 is well-defined.
WriteR2 − zS2 = S1/2
2 (R − z1)S1/22 , R := S−1/2
2 R2S−1/22 : N(S)⊥ 7→ N(S)⊥, (2.5)
and note that R is self-adjoint with compact resolvent and satisfies n(R2) = n(R). Upon substi-tuting (2.5) into (2.3) one sees that
p = S−1/22 k−
n∑j=1
αjS−1/22 (R − z1)−1S−1/2
2 PRφj , (2.6)
wherek ∈ ker(R − z1), S−1/2
2 PRφj ∈ [ker(R − z1)]⊥.
Since R is self-adjoint with a compact resolvent, the real-valued eigenvalues, say µj , can be orderedas a monotone increasing sequence with µj → +∞ as j → ∞. Clearly (2.6) makes sense for allz /∈ σ(R), and care must be taken only when z = µj for some j.
Substitution of (2.6) into (2.4) yields the algebraic system for ` = 1, . . . , n,
〈RS−1/22 k, φ`〉+
n∑j=1
αj〈Rφj , φ`〉 −n∑j=1
αj〈S−1/22 (R − z1)−1S−1/2
2 PRφj ,Rφ`〉 = α`zλ`. (2.7)
If z 6= µj for all j, then k = 0 and the first term on the left-hand side in (2.7) disappears. If z = µj ,and if S−1/2
2 PRφ` ∈ [ker(R − µj1)]⊥, then since S−1/22 and R are self-adjoint one has
〈RS−1/22 k, φ`〉 = 〈S−1/2
2 k,Rφ`〉 = 〈S−1/22 k, PRφ`〉 = 〈k,S−1/2
2 PRφ`〉 = 0. (2.8)
The second equality follows from the fact that S2 : N(S)⊥ 7→ N(S)⊥. If S−1/22 PRφ` /∈ [ker(R −
µj1)]⊥, then (2.3) cannot be solved, and hence it makes no sense to consider (2.7). In conclusion,(2.7) can be reduced to
n∑j=1
αj〈Rφj , φ`〉 −n∑j=1
αj〈S−1/22 (R − z1)−1S−1/2
2 PRφj ,Rφ`〉 = α`zλ`, z /∈ σ(R). (2.9)
As a consequence of (2.5) one has
〈S−1/22 (R − z1)−1S−1/2
2 PRφj ,Rφ`〉 = 〈(S−1/22 (R − z1)−1S−1/2
2 PRφj , PRφ`〉.
If one defines the matrices R,D ,C (z) by
Rj` = 〈φj ,Rφ`〉, D = diag(λ1, . . . , λn), C (z)j` = 〈(S−1/22 (R − z1)−1S−1/2
2 PRφj , PRφ`〉,(2.10)
T. Kapitula 7
then upon setting a := (α1, . . . , αn)T (2.9) is equivalent to
K (z)a = 0 , K (z) := R − zD −C (z). (2.11)
In (2.11) note that K (z) is symmetric, i.e, K (z)T = K (z) for all z ∈ C. In particular, K (z) issymmetric for z ∈ R.Definition 2.1. For (1.5) the Krein matrix K (z) is given in (2.11).
In order to better understand the implications associated with (2.11), it is first necessary torewrite C (z). For µj ∈ σ(R) let vj represent the corresponding normalized eigenfunction, so that{vj} is an orthonormal basis. Upon using the fact that S−1/2
2 is self-adjoint, a standard expansionyields that the entries of C (z) can be rewritten as
C (z)j` =∞∑i=1
〈S−1/22 PRφj , vi〉〈S−1/2
2 PRφ`, vi〉µi − z
. (2.12)
Upon settingci := (ci1, c
i2, . . . , c
in)T, cij := 〈S−1/2
2 PRφj , vi〉, (2.13)
one has the compact form
C (z) =∞∑i=1
ci(ci)H
µi − z, (2.14)
where vH refers to the Hermitian of the vector v .Consider solutions to (2.11) near a simple pole µi. Suppose that µi = µi+1 = · · · = µi+` for some
finite ` ∈ N0. Upon examining (2.14) one sees that a Laurent expansion of C (z) in a neighborhoodof z = µi is given by
C (z) =
∑`j=0 ci+j(ci+j)H
µi − z+ C (0) + . . . .
Now, (2.3) is solvable for a 6= 0 at z = µi if and only if ci+j = 0 for all 0 ≤ j ≤ `; in otherwords, (2.3) is solvable at z = µi if and only if z = µi is a removable singularity for K (z). Ifthe singularity is removable, then (2.3) has a nontrivial solution with a = 0 ; furthermore, (2.4) isautomatically satisfied with a = 0 as a consequence of (2.6) and (2.8). Since a = 0 one sees that〈u,Su〉 = 〈u,S2u〉 > 0; hence, if z ∈ R+ then by (1.7) the eigenvalue has positive signature. Inorder to see if there are solutions with a 6= 0 , one must consider the matrix
K (µi) := limz→µi
K (z) = R − µiD −C (0).
If detK (µi) 6= 0, then one has (2.11) is solved only with a = 0 , so that all solutions lie inspan{S−1/2
2 k : k ∈ ker(R − µi1)}. If detK (µi) = 0, then in addition to the (` + 1)-dimensionalsolution set found with a = 0 there will be a set of solutions with a 6= 0 . If in (2.11) a = 0 , thenupon examining (2.3) this implies that for a nontrivial solution one needs z ∈ σ(R). This scenariohas been covered above. Thus, one can conclude that for solutions with z /∈ σ(R) it is necessarilytrue that a 6= 0 . In conclusion, if an eigenvalue exists for which (2.11) is solved with a = 0 , then(a) it is a real eigenvalue, (b) coincides with an eigenvalue of R and is a removable singularity forK (z), and (c) if positive has positive Krein signature.
For nontrivial a (2.11) has a solution if and only if detK (z) = 0. From the above discussionone has that eigenvalues with nonzero imaginary part, and positive real eigenvalues with negativesignature, will be realized as zeros of detK (z). The following lemma has now been proved:
Krein signature and the Krein Oscillation Theorem 8
Lemma 2.2. All eigenvalues of (1.5) are realized as either (a) solutions of detK (z) = 0, or (b)removable singularities of K (z). An eigenvalue z ∈ C of (1.5) which satisfies either Im z 6= 0, orz ∈ R+ with negative Krein sign, will be realized as a solution of detK (z) = 0.
Remark 2.3. It follows immediately from the construction of K (z) that if detK (z) = 0 withz /∈ σ(R), then the geometric multiplicity of the eigenvalue is bounded above by n(S).Remark 2.4. If σ(R) ⊂ R+, then it is clear that solving detK (z) = 0 will capture all of theeigenvalues associated with the count of (1.6). This assumption will be true for the applicationsconsidered in Section 2.5.Remark 2.5. In a manner similar to that presented above one can construct a Krein matrix forthe eigenvalue problem
(S − ξR)u = 0, ξ =1z. (2.15)
Instead of projecting off N(S), one instead projects off N(R); otherwise, the construction is exactlythe same. For ξ = 1/µj < 0 with a = 0 in (2.11) one will have
sign(〈u,Ru〉) = − sign(〈u,Su〉) < 0;
hence, under the formulation of (2.15) the negative real eigenvalue will be captured as a zero ofthe determinant of the Krein matrix. Consequently, if one wishes to definitively capture all of thenegative real eigenvalues via the determinant of a Krein matrix, one must simultaneously constructa Krein matrix for (1.5) and another one for (2.15) and consider the zero set associated with thedeterminant of each.Remark 2.6. When considering the spectral stability of localized waves, it was shown in [31] thatby constructing a matrix E(λ) ∈ CN×N using the solutions of an N -dimensional linear ODE whichhave certain spatial asymptotic properties, then detE(λ) = 0 if and only if λ was an eigenvalue.It was shown in [7] that detE(λ) is equivalent to the Evans function [3]. When considering thespectral stability of periodic waves with large period for systems which possessed m symmetries,it was shown in [30] that the critical eigenvalues were realized as the zeros of the determinant of aparticular matrix E(λ, γ) ∈ Cm×m, where λ ∈ C is the spectral parameter and γ ∈ R is the Floquetmultiplier.
Solving detK (z) = 0 is equivalent to solving (for a particular j, and not necessarily all j)
rj(z) = 0, j = 1, . . . , n,
where rj(z) is an eigenvalue of K (z) and will henceforth be referred to as a Krein eigenvalue.Regarding the Krein eigenvalues, the following can be deduced from [25] upon using the fact thatK (z) is meromorphic and symmetric for z ∈ R:Proposition 2.7. Suppose that z /∈ σ(R). Then:
(a) the Krein eigenvalues rj(z) are real analytic, as are the associated eigenvectors v j(z)
(b) for Im z 6= 0 the Krein eigenvalues are analytic as long as they are distinct; furthermore,any branch points correspond to algebraic singularities.
2.2. Order of zeros and the algebraic multiplicity
The immediate goal is to relate the order of a zero of a Krein eigenvalue with the size of theassociated Jordan chain. Since the Krein eigenvalues and associated eigenvectors are real analytic
T. Kapitula 9
by Proposition 2.7, for a given z0 /∈ σ(R) there is an appropriate Taylor series. The coefficients canbe found in the following manner. Suppose that rj(z0) = 0, and assume that |v j(z0)| = 1. Upondifferentiating
K (z)v j(z) = rj(z)v j(z)
with respect to z and evaluating at z0 one sees that
K (z0)v ′j(z0) = r′j(z0)v j(z0)−K ′(z0)v j(z0). (2.16)
(2.16) has the solvability condition
r′j(z0) = v j(z0)HK ′(z0)v j(z0), (2.17)
which fixes r′j(z0), and v ′j(z0) is then uniquely determined by solving (2.16) under the requirementthat v ′j(z0) ∈ [ker(K (z0)−rj(z0)1)]⊥. If one now assumes that r′j(z0) = 0, then upon differentiatinga second time and evaluating at z0 one sees that
K (z0)v ′′j (z0) = r′′j (z0)v j(z0)−K ′′(z0)v j(z0)− 2K ′(z0)v ′j(z0). (2.18)
The solvability condition for (2.18) is
r′′j (z0) = v j(z0)HK ′′(z0)v j(z0) + 2v j(z0)HK ′(z0)v ′j(z0). (2.19)
Of course, one can repeat this process in order to determine the coefficients at any order.Suppose that for (1.5) there is the finite Jordan chain
(R− z0S)ψj = Sψj−1 (ψ0 = 0), j = 1, . . . , k. (2.20)
Solvability and the finiteness of the chain implies that
〈ψk,Sψ1〉 6= 0, 〈ψj ,Sψ1〉 = 0, j = 0, . . . , k − 1. (2.21)
In order to relate (2.21) to the derivatives of the Krein eigenvalues, use the decomposition ψ1 =∑αjφj + p and the expression for p given in (2.6) to write
ψ1 =n∑j=1
α1jφj −
n∑j=1
α1jS−1/22 (R − z01)−1S−1/2
2 PRφj , (2.22)
where a1 = (α11, . . . , α
1n)T satisfies |a1| = 1. Using (2.22) and simplifying yields
〈ψ1,Sψ1〉 = aH1 (D + C ′(z0))a1 = −aH
1 K ′(z0)a1. (2.23)
Since v j(z0) = a1, one sees from (2.17) that
r′j(z0) = −〈ψ1,Sψ1〉; (2.24)
in other words, if z0 ∈ R+ the sign of the derivative of the Krein eigenvalue at a zero is the negativeof the Krein signature.
Now suppose that k ≥ 2 in (2.20), which by (2.21) and (2.24) implies that r′j(z0) = 0. Writingψ2 =
∑α2jφj + p2 in (2.20) and projecting with P yields
(R2 − z0S2)p2 +n∑j=1
α2jPRφj = PSψ1.
Krein signature and the Krein Oscillation Theorem 10
Substituting the expression for ψ1 given in (2.22) into the above, solving for p2 using (R2−z0S2)−1 =S−1/2
2 (R − z01)−1S−1/22 , and simplifying yields
ψ2 =n∑j=1
α2jφj −
n∑j=1
α2jS−1/22 (R − z01)−1S−1/2
2 PRφj −n∑j=1
α1jS−1/22 (R − z01)−2S−1/2
2 PRφj .
In order to solve for the α2j one uses the projection Q in (2.20) (see (2.4)) and eventually finds that
these coefficients satisfy
K (z0)a2 = (C ′(z0) + D)a1 = −K ′(z0)a1, a2 = (α21, . . . , α
2n)T. (2.25)
(2.25) can be solved if and only if aH1 K ′(z0)a1 = 0, which, by (2.17) with v(z0) = a1 yields
r′j(z0) = 0; furthermore, the solution is v ′j(z0) = a2. Continuing in this fashion one finds that forψ3 =
∑α3jφj + p the linear system to be solved is
K (z0)a3 = −12
(K ′′(z0)a1 + 2K ′(z0)a2), a3 = (α31, . . . , α
3n)T. (2.26)
An examination of (2.19) reveals that (2.26) can be solved if and only if r′′j (z0) = 0; furthermore,v ′′j (z0) = 2a3. Thus, k ≥ 3 if and only if rj(z0) = r′j(z0) = r′′j (z0) = 0. An induction argumenteventually leads to the conclusion that
rj(z0) = r′j(z0) = · · · = r(k−1)j (z0) = 0, r(k)j (z0) 6= 0. (2.27)
The next goal is to determine an upper bound on the algebraic multiplicity of the zero. Supposethat (2.20) is satisfied for z0 ∈ R\σ(R). A small perturbation of the form S 7→ S + εSε willgenerically break this eigenvalue of algebraic multiplicity k into k simple eigenvalues (e.g., see [32,Section 3] for a concrete example). Upon using (2.27) and the fact that the coefficients of theTaylor expansion for rj(z) for real z will vary smoothly as a function of ε one has that to leadingorder the equation to solve to locate these eigenvalues is
r(k)j (z0)k!
(z − z0)k + ∂εrj(z0)ε = 0. (2.28)
Solving (2.28) yields
(z − z0)k = ζε, ζ := −k!∂εrj(z0)
r(k)j (z0)
∈ R\{0}, (2.29)
so that the k eigenvalues satisfy z = z0 +O(ε1/k).Without loss of generality suppose that ζ ∈ R+. If z0 ∈ R−, then one sees from (2.30) that
z = zj = z0 + ζ1/kei2πj/kε1/k, j = 0, . . . , k − 1,
from which it can be concluded that kr + 2kc = k. Upon using (1.6) one gets the upper bound forthe algebraic multiplicity is k ≤ n(R) + n(S). Suppose that z0 ∈ R+. If k = 2` is even, then uponusing (2.24) one has for (2.28) that
kc =
{`, k−i = 0`− 1, k−i = 1;
T. Kapitula 11
in other words, kc + k−i = k. If k = 2`+ 1 is odd, then kc = `, and
k−i =
{0, r
(k)j (z0) > 0
1, r(k)j (z0) < 0,
which implies that
kc + k−i =
{`, r
(k)j (z0) > 0
`+ 1, r(k)j (z0) < 0.
Upon using (1.11) one gets the upper bound on the algebraic multiplicity for ` ∈ N:
k = 2` =⇒ ` ≤ min{n(R),n(S)}
k = 2`+ 1 =⇒ ` ≤
{min{n(R),n(S)}, r
(k)j (z0) > 0
min{n(R),n(S)} − 1, r(k)j (z0) < 0.
(2.30)
Finally, it is of interest to determine k−i (z0) for k ≥ 2. If z0 ∈ R+ the following can be shownusing the argument leading to [13, Corollary 2.26]. Let Eεz0 correspond to the subspace formed byprojecting onto the eigenspaces associated with the k eigenvalues satisfying |z − z0| = O(|ε|1/k).One has for |ε| > 0 that
n(S|Eεz0 ) = kc + k−i .
Since the negative index is robust, it is then necessarily true that when ε = 0 with k ∈ {2`, 2`+ 1},
k−i (z0) = n(S|E0z0
) =
`, k = 2``, k = 2`+ 1 with r
(k)j (z0) > 0
`+ 1, k = 2`+ 1 with r(k)j (z0) < 0.
(2.31)
The interested reader should note that [32] contains the first statement of (2.31); however, [32]does not relate k−i (z0) to the sign of r(k)j (z0) when k is odd.
Since
detK (z) =n∏i=1
ri(z),
the following result has now been proven.Lemma 2.8. For z ∈ R\σ(R) a Krein eigenvalue is zero only if z is an eigenvalue. The order ofthe zero is equal to the size of the Jordan block. The algebraic multiplicity of an eigenvalue is theorder of the zero of the determinant of the Krein matrix. If z ∈ R+ and the zero of the Kreineigenvalue is simple, then the sign of the eigenvalue is the opposite the sign of the derivative of theKrein eigenvalue evaluated at the zero.
Remark 2.9. It is clear that the proof, and consequently the results, of Lemma 2.8 relating theorder of the zero of the Krein eigenvalue to the size of the Jordan block, and the order of the zero ofthe determinant of the Krein matrix to the algebraic multiplicity of the eigenvalue can be extendedto Im z 6= 0 as long as the Krein eigenvalues remain distinct.Remark 2.10. If z ∈ σ(R) is an eigenvalue such that detK (z) = 0, i.e., it corresponds to aremovable singularity of the Krein matrix, then the analysis relating the order of the zero of theKrein eigenvalue to the size of the Jordan chain is still valid, as is (2.31). Since there will beeigenvalues which will not be captured by the Krein matrix, however, the relationship between theorder of the zero of detK (z) and the algebraic multiplicity of the eigenvalue is no longer valid.
Krein signature and the Krein Oscillation Theorem 12
2.3. The Krein Oscillation Theorem
A significant consequence of Lemma 2.8 is the Krein Oscillation Theorem (Theorem 2.12. Beforeit can be stated, however, some graphical properties of the Krein eigenvalues must be derived. Itis helpful to first consider the case of n = 1, which in particular implies that k−i ≤ 1 (see [27] forthis case when the compactness assumption is removed). Upon using the formulation of (2.12) onehas that
r1(z) = 〈φ1,Rφ1〉 − λ1z −∞∑i=1
|〈S−1/22 PRφ1, vi〉|2
µi − z. (2.32)
Differentiation yields
r′1(z) = −λ1 −∞∑i=1
|〈S−1/22 PRφ1, vi〉|2
(µi − z)2, r′′1(z) = −2
∞∑i=1
|〈S−1/22 PRφ1, vi〉|2
(µi − z)3(2.33)
For z < µ1 one has that
r′′1(z) < 0, limz→−∞
r1(z)z
= −λ1 > 0, limz→µ−1
r1(z) = −∞. (2.34)
One can then conclude that r1(z) = 0 has at most two real-valued solutions in (−∞, µ1). If thereis only one solution, say z1 ∈ (−∞, µ1), then as a consequence of (2.34) it must be true thatr′1(z1) = 0; furthermore, from (2.30) and (2.31) one can conclude that the eigenvalue has a Jordanblock of size two with k−i (z0) = 1. For i ∈ N one has that
limz→µ±i
r1(z) = ±∞; (2.35)
hence, in each interval (µi, µi+1) there is an odd number (counting multiplicity) of real-valued zerosof r1(z).
Remark 2.11. The structural form of r1(z) for z ∈ (−∞, µ1) is exactly that detailed in [27]. Thedifference is that here µ1 corresponds to an eigenvalue of R, whereas in [27] µ1 is the edge of thecontinuous spectrum. The presence of the continuous spectrum precluded any additional analysisin [27] of r1(z) for real-valued z > µ1.
A cartoon of the situation is depicted in the left panel of Figure 2. Note by Lemma 2.8 thatthe middle zero of r1(z) in the interval (µ1, µ2) corresponds to the eigenvalue with negative Kreinsignature, and that the other two zeros correspond to eigenvalues with positive signature. From(1.11) one has that k−i ≤ 1; hence, all other zeros of r1(z) will be simple with negative derivative.
Now suppose that n ≥ 2. One has
limz→−∞
K (z)z
= −D ;
consequently,
limz→−∞
rj(z)z
= −λj > 0, j = 1, . . . , n. (2.36)
In considering the behavior near the simple poles µj ∈ σ(R) one has that
limz→µj
(z − µj)K (z) = limz→µj
(µj − z)C (z) = − Resz=µj
C (z).
T. Kapitula 13
r1(z)
r2(z)
r1(z)
z zμ1 μ1 μ2μ2
y y
Figure 2: (color online) The Krein eigenvalue r1(z) for n = 1 (left panel), and the Kreineigenvalues r1(z) and r2(z) for n = 2 (right panel). Note that in the left panel k−i = 1,whereas in the right panel k−i = 2.
If µj is algebraically simple, then by using (2.14) one sees that
Resz=µj
C (z) = −cj(cj)H.
Since cj(cj)H is a rank-one matrix with ker(cj(cj)H) = span{cj}⊥, there will be a 1 ≤ `j ≤ n forwhich one will see the asymptotic behavior
|r`j (z)| = O(|µj − z|−1); |rj(z)| = O(1), j 6= `j .
In other words, even though K (z) has a simple pole, for all of the Krein eigenvalues but one thepole is a removable singularity. Since
trace cj(cj)H =∑i
|cji |2 > 0,
one hasResz=µj
r`j (z) =∑i
|cji |2 > 0,
so thatlimz→µ±j
r`j (z) = ±∞ (2.37)
(compare to (2.35)).Now suppose that µj+1 = µj+2 = · · · = µj+` for some ` ∈ N0. One then has that
Resz=µj
C (z) = −(cj(cj)H + · · ·+ cj+`(cj+`)H
).
Since each cj+i(cj+i)H is a rank-one matrix, for each 1 ≤ i ≤ `, ker(cj+i(cj+i)H) = span{cj+i}⊥.Thus,
ker(Resz=µj
C (z)) ⊂ span{cj , . . . , cj+`}⊥ =⇒ dim[ker(Resz=µj
C (z))] ≥ n− `.
It will then be the case that at least n−` of the eigenvalues of K (z) will have a removable singularityat the pole. This case of multiple eigenvalues is degenerate and can be removed via the application
Krein signature and the Krein Oscillation Theorem 14
of a finite rank perturbation to R; hence, it will be assumed that all of the eigenvalues are simple.A cartoon of the situation is depicted in the right panel of Figure 2 when n = 2.
The Krein Oscillation Theorem requires the following assumptions and definitions. Assumethat σ(R) = {µi}∞i=1 is such that all of the eigenvalues are algebraically simple. Let the sequence{µji}∞i=0 ⊂ σ(R), µj0 = −∞, denote the simple poles of rj(z). For i = 0, . . . ,∞ and j = 1, . . . , ndefine the open intervals Iji = (µji , µ
ji+1). Set
• kr(Iji ) to be the number of negative real eigenvalues detected by rj(z) in Iji
• k−i (Iji ) to be the number of positive real eigenvalues with negative Krein signature detectedby rj(z) in Iji
• #rj(Iji ) to be the number of zeros (counting multiplicity) of rj(z) in Iji .
Theorem 2.12 (Krein Oscillation Theorem). Suppose that not all of the singularities of the Kreinmatrix are removable. For each j = 1, . . . , n, #rj(I
j0) is even, and if n = 1, then #r1(Ij0) ∈ {0, 2}.
If µji+1 < +∞ for i ∈ N, then #rj(Iji ) is odd; furthermore,
Iji ⊂ R− =⇒ kr(Iji ) = #rj(I
ji )
Iji ⊂ R+ =⇒ k−i (Iji ) =#rj(I
ji )− 12
.
Finally, if n = 1 and {0} ⊂ I10 with #r1(I1
0 ) = 2, then either kr(I10 ) = 2 or kr(I1
0 ) + k−i (I10 ) = 1.
Proof: The statement regarding the number of zeros in a given interval is a consequence of theasymptotics given in (2.34) and (2.35) (n = 1), and (2.36) and (2.37) n ≥ 2. The statement thateach real negative zero corresponds to an eigenvalue follows immediately from Lemma 2.8. Thestatement regarding the number of eigenvalues of negative Krein sign follows from both the factthat r′j(z0) > 0 if and only if z0 ∈ R+ is a simple eigenvalue with negative signature, as well as(2.31) in the event of a nontrivial Jordan block. The final statement follows from the fact that ifr1(zj) = 0 for j = 1, 2 with z1 < 0 < z2, then from (2.33) it is necessarily true that r′1(z2) < 0,whereas if 0 < z1 < z2, then r′1(z1) > 0 and r′1(z2) < 0.
Remark 2.13. If Iji ⊂ R+ and all of the zeros of rj(z) are simple on this interval, then for theeigenvalues detected by rj(z) one has the ordering
µji < p < n < p < · · · < n < p < µji+1.
Here p represents an eigenvalue with positive Krein signature, and n is an eigenvalue with negativeKrein signature.Remark 2.14. Suppose that all of the singularities of the Krein matrix are removable. Thisimplies that C (z) ≡ 0, which in turn implies that K (z) = R − zD . Since D is negative definite,detK (z) = 0 will have n(R) positive real solutions and n(S)−n(R) negative real solutions. Lettingz0 ∈ R+ be a solution, since D is negative definite one has that
−aHK ′(z0)a = −aHDa > 0, a ∈ Cn.
Upon using (2.17) and (2.24) this then implies z0 is an eigenvalue with negative Krein signature.Upon using Lemma 2.8 and the count of (1.6) one can conclude that
k−i = n(R), kr = |n(R)− n(S)|+ 2[n(R)− n(R)].
T. Kapitula 15
Note that the Krein matrix detects n(S) − n(R) of the negative real-valued eigenvalues; hence,n(R) − n(R) of the negative real-valued eigenvalues correspond to removable singularities. It isalso interesting to note that
n(R) = n(R|N(S)),
which implies in particular that if n(R) = n(S), then
k−i = n(R|N(S)), kr = 2[n(R)− n(R|N(S))].
Remark 2.15. Suppose that all but a finite number of the singularities of the Krein matrix areremovable, which implies that each Krein eigenvalue is a rational function. Then rj(z) ∼ −λjz > 0as z → +∞, which implies that there is an R0 ∈ R+ such that for z > R0 all eigenvalues correspondto removable singularities of the Krein matrix.
2.4. Asymptotic results
It is possible to make a general statement regarding the location of the eigenvalues in thelimit z → +∞. Since n(R) + n(S) is finite, there is an R0 ∈ R+ such that the only eigenvaluesfor |z| > R0 are positive and real with positive Krein signature. As a consequence of the KreinOscillation Theorem this then implies that for each j there is an i+j such that #rj(I
ji ) = 1 for
i ≥ i+j . For each j,
S−1/22 PRφj =
∞∑i=1
〈S−1/22 PRφj , vi〉vi, 〈S−1/2
2 PRφj ,S−1/22 PRφj〉 =
∞∑i=1
|〈S−1/22 PRφj , vi〉|2.
The finiteness of the sums imply that
limi→∞|〈S−1/2
2 PRφj , vi〉| = 0;
hence, an examination of (2.13) reveals that
limi→∞
ci(ci)H = 0 .
Since the residues of K (z) can be made arbitrarily small, as z increases the effect of the pole willbe seen only in smaller and smaller neighborhoods of σ(R). Consequently, the zero of rj(z) whichis effected by the pole at z = µj` will be in a small neighborhood of µj` . In conclusion, one then hasthat for large z the spectrum of (1.5) is well-approximated by σ(R), and that these large eigenvalueswill all have positive Krein signature.
2.5. Perturbative results
In examples for which the nonlinearity is weak one has the situation that
R = A0 + εR1, S−1 = A0 + εS1, 0 < ε� 1 (2.38)
where A0 is a self-adjoint operator with a compact resolvent and n(A0) = n, and R1,S1 are A0-compact. When ε = 0 the spectral problem is well-understood. Letting λj , j = 1, 2, . . . , representthe eigenvalues of A−1
0 , one has that when ε = 0,
N(R) = N(S), σ(S−1R) = {λ−2j }∞j=1, σ(R) = {λ−2
j }∞j=n+1.
Krein signature and the Krein Oscillation Theorem 16
There are eigenvalues of negative sign at λ−21 ≤ · · · ≤ λ−2
n , and all of the others have positive sign.Since ci(ci)H = 0 for all i ≥ n+1, K (z) has removable singularities for all z ∈ σ(R); consequently,
K (z) = diag(1/λ1 − λ1z, . . . , 1/λn − λnz). (2.39)
Remark 2.16. It can certainly be the case that for ε > 0 one has either or both of n(R), n(S) > n(e.g., see [16, 19]). The increase in the number of negative eigenvalues is related to nonzeroeigenvalues of O(ε1/2) being created by a symmetry-breaking bifurcation. In this case the analysisthat follows will still be valid if one first projects onto the subspace associated with the O(1)eigenvalues of S. In order to simplify the subsequent analysis, it will henceforth be assumed thatn(R) = n(S) = n for all ε > 0 sufficiently small.
(2.39) implies that for j = 1, . . . , n, rj(z) = 0 if and only if z = zj := λ−2j . Furthermore,
r′j(zj) > 0, so by Lemma 2.8 all of these eigenvalues have negative Krein sign. All of the eigenvaluesassociated with (1.5) which have positive sign are not captured by detK (z) = 0 because each ofthese is associated with a removable singularity. Since σ(R) ⊂ R+ when ε = 0, the same will betrue for small ε; hence, following Remark 2.4 one knows that for ε > 0 the zeros of the determinantof the Krein matrix will capture all of the eigenvalues associated with the count of (1.6).
The goal is to understand what happens for ε > 0. Since C(R) = C(S) when ε = 0, by therobustness of the cone intersection one has by (1.8) that kc + k−i = n for ε > 0 sufficiently small.Thus, the question is the manner in which the zeros of the Krein eigenvalues capture this result.If zi /∈ σ(R) when ε = 0, so that the zero is not located on a removable singularity, then by theImplicit Function Theorem one can conclude that the zero persists smoothly with r′i(zi) > 0. Nowsuppose that zi = µ ∈ σ(R) for some 1 ≤ i ≤ n when ε = 0. First suppose that µ ∈ σ(R) isgeometrically and algebraically simple. For ε > 0 one will generically have |ci(ci)H| = O(ε); inparticular, this implies that the effect of the pole on the graph of ri(z) will be felt only within anO(ε) neighborhood of zi. Outside of a small neighborhood of the pole regular perturbation theoryyields
ri(z) ∼ −(λi +O(ε))z + (1/λi +O(ε)),
whereas in the neighborhood of λi one will have the expansion
ri(z) ∼ −(λi +O(ε))z + (1/λi +O(ε)) +O(ε)
1/λ2i +O(ε)− z
. (2.40)
Consequently, ri(z) will be nonzero outside a small neighborhood of µ, whereas in a neighborhoodof µ, ri(z) = 0 will have precisely two solutions (counting multiplicity). If the zeros are simple, thenthey will be either be real-valued and have derivatives of opposite sign, or they will be complex-conjugates with nonzero imaginary parts.
Now suppose that µ is no longer geometrically simple when ε = 0. For ε > 0 one can genericallyexpect that the eigenvalues of R can be ordered as µ0(ε) < · · · < µ`(ε), and that each eigenvalue isgeometrically and algebraically simple. The analogous expansion to (2.40) will read
ri(z) ∼ −(λi +O(ε))z + (1/λi +O(ε)) +∑j=0
Oj(ε)µj(ε)− z
. (2.41)
If all of the residues are nonzero, then there will be ` + 2 solutions to ri(z) = 0 within an O(ε)neighborhood of zi. If 1 ≤ `′ ≤ `+ 1 of the residues are zero, then the number of solutions reducesto (`− `′)+2; furthermore, since the total number of eigenvalues within the O(ε) must be invariantunder small perturbation, there will be `′ eigenvalues which are contained in σ(R).
T. Kapitula 17
3. Application: Gross-Pitaevski equation
The previous results and perspective of Section 2.5 will be applied to solutions of a systemwhich is currently undergoing a great deal of mathematical and experimental study. The governingequations for an N -component Bose-Einstein condensate are given by
i∂tUj + ∆Uj + ωjUj +N∑k=1
ajk|Uk|2Uj = V (x )Uj , j = 1, . . . , N, (3.1)
where Uj ∈ C is the mean-field wave-function of species j, A = (ajk) ∈ RN×N is symmetric,∆ represents the Laplacian, ωj ∈ R are free parameters which represent the chemical potential,and V : Rn 7→ R represents the trapping potential (see [2, 4–6, 8, 17, 22, 24, 26, 28] andthe references therein for further details). If ajk ∈ R+ for all j, k, then the intra-species and inter-species interactions are attractive, whereas if ajk ∈ R− for all j, k the intra-species and inter-speciesinteractions are repulsive.
Suppose that N = 1, and for 0 < δ � 1 consider a potential of the form
V (x ) = |y |2 +1δ4|z |2, (y , z ) ∈ Rdy×dz .
It was shown in [1] that for repulsive interactions, i.e., a11 < 0, then if n = dy + dz < 4 solutionsto (3.1) for x ∈ Rn are well-approximated over long time intervals by solutions of the form
U(t,x ) ∼ π−dz/4e−idzt/δ2u(t,y)e−|z |2/2,
where u(t,y) satisfies (3.1) in the y -variables only, i.e.,
iut + ∆u+ ωu+ a11|u|2u = |y |2u, a11 :=a11
πdz
∫Rdz
e−2|z |2 dz . (3.2)
Thus, the new governing equation is simply the old one with the number of spatial variables beingreduced. The case of dy = 1 is referred to as the cigar trap, and the case dy = 2 is call the pancaketrap. In the case of attractive interactions the reduction is rigorous only if dy = 1; however, inpractice it is still assumed that the dy = 2 reduction is valid. In all of the examples that follow itwill be assumed that N = 1 with a11 < 0.
3.1. Cigar trap
In the case of dy = 1 (3.2) can be rewritten after some rescalings as
iut + uxx + ωu− ε|u|2u = x2u, 0 < ε. (3.3)
For 0 < ε� 1 weakly nonlinear solutions steady-state solutions can be constructed via an elemen-tary Lyapunov-Schmidt reduction procedure. The spectrum and eigenfunctions for the operator
H = − d2
dx2+ x2
are well-known; in particular, the eigenvalues are given by µ` = 1+2` for ` ∈ N0, and the associatedeigenfunctions are given by φ`(x) = H`(x)e−x
2/2, where H`(x) is the Gauss-Hermite polynomial oforder ` is scaled so that ∫
Rφ2` (x) dx = 1, j ∈ N0.
Krein signature and the Krein Oscillation Theorem 18
For 0 < ε� 1 it is known that for each ` ∈ N0 there are weakly nonlinear solutions of the form
U`(x) ∝ φ`(x) +O(ε), 0 < ω − µ` = O(ε).
0 4 8
−3
0
3
z
r 1(z)
0 10
x 10−3
−4
0
4
σ(JL)
Re(λ)
Im(λ
)
Figure 3: (color online) The eigenvalue r1(z) for n = 1 (left panel), and σ(JL) (rightpanel). In the left panel the (red) cross indicates the location of a pole of the Krein matrix.In the right panel the eigenvalue with negative signature is that associated with the filled(red) circle in the insert.
The linear operators associated with the spectral stability problem in the framework of (1.2)are given by
L+ = H− ω + 3εU2` , L− = H− ω + εU2
` . (3.4)
One has that ker(L−) = span{U`}, whereas L+ is nonsingular for ε > 0. When ε = 0 one hasL+ = L− with ker(L−) = span{φ`} and n(L−) = `; hence, upon projecting off the kernel via theprojection operator P` one has in the formulation of (2.38) that A0 = P`L−P` with n(A0) = `.Because the interactions are repulsive, for 0 < ε� 1 one continues to have that n(L±) = `. Uponprojecting off ker(L−) for ε > 0 one sees that n(R) = n(S) = `.
The system is of the form of (2.38). Since the bifurcation arises from ω = µ`, when ε = 0 onehas for i 6= `, λi = (µi−µ`)−1 = [2(i− `)]−1, and each eigenvalue is geometrically and algebraicallysimple. As a consequence one has that when ε = 0, zi ∈ σ(R); furthermore, zi is geometrically andalgebraically simple. Now
ri(z) = 2(i− `)− 12(i− `)
z, i = 0, 1, . . . , `− 1.
The simple zeros occur at zi = 4(i − `)2, which for (1.2) correspond to the purely imaginaryeigenvalues ±i2(i − `). For ε > 0 one will generically have near these simple zeros of ri(z) a pairof purely real zeros which correspond to eigenvalues of opposite sign, or a pair of eigenvalues withnonzero imaginary parts.
Figure 3 gives the spectral output in the case ` = 1, and Figure 4 gives the spectral outputin the case ` = 2. In both cases ε = 0.6. The left panels show a plot of the eigenvalue(s) for
T. Kapitula 19
K (z), while the right panels show the spectrum for (1.2). The purely imaginary eigenvalue in theupper half of the complex plane with negative signature is associated with the filled (red) circle,and the (red) crosses in the left panel denote the location of the poles of K (z). In both figuresthe Hamiltonian-Hopf bifurcation does not occur near z = 4, whereas in the second figure thatbifurcation does occur near z = 16. As a consequence of (1.11) one knows that kc + k−i ≤ `; hence,it is known that the figures capture the location of all of the eigenvalues associated with the countof (1.6).
0 4 8 12 16
−3
0
3
6
9
z
r j(z)
0 3.4757
x 10−3
−4
0
4
σ(JL)
Re(λ)
Im(λ
)
r1(z)
r2(z)
Figure 4: (color online) The Krein eigenvalues rj(z) for j = 1, 2 (left panel), and σ(JL)(right panel). In the left panel the (red) cross indicates the location of a pole of the Kreinmatrix. In the right panel the eigenvalue with negative signature is that associated with thefilled (red) circle in the insert.
3.2. Pancake trap
In the case of dy = 2 (3.1) can be rewritten after some rescalings as
iut + ∆u+ ωu− ε|u|2u = (x2 + y2)u, 0 < ε. (3.5)
For 0 < ε� 1 weakly nonlinear solutions steady-state solutions can be constructed via a Lyapunov-Schmidt reduction procedure. The spectrum and eigenfunctions for the operator
H = −∆ + x2 + y2
are well-known; in particular, the eigenvalues are given by µ`,m = 2 + 2(`+m) for `,m ∈ N0, andthe associated eigenfunctions are given by φ`,m(x, y) = H`(x)Hm(y)e−(x2+y2)/2, where Hi(·) is theGauss-Hermite polynomial as in the previous subsection. The situation for 0 < ε � 1 is morecomplicated than the dy = 1 case, for the eigenvalue µ`,m has multiplicity ` + m + 1; hence, theresulting bifurcation equations are more complicated to solve (see [23] for the case ` + m = 2).
Krein signature and the Krein Oscillation Theorem 20
Figure 5: (color online) A cartoon of r1(z) nearz = 4 for the case 0 < µ3−µ2, µ2−µ1 = O(ε) when0 < ε� 1 under the assumption that the residue foreach pole is nonzero. From this graph there will beone real positive eigenvalue with negative sign, andthree positive real eigenvalues with positive sign.
r1(z)
zμ1 μ2 μ3
Instead of giving a general discussion, the focus instead will be on the easiest case: `+m = 1. Thereal-valued solution is the dipole, and is given in polar coordinates by
U(x, y) ∝√
2πre−r
2/2 cos θ +O(ε), 0 < ω − 4 = O(ε).
(e.g., see [19]).The linear operators associated with the spectral stability problem in the framework of (1.2)
are given byL+ = H− ω + 3εU2
` , L− = H− ω + εU2` . (3.6)
One has that ker(L−) = span{U}, and ker(L+) = span{Uθ}. When ε = 0 one has L+ = L− withker(L−) = span{φ0,1, φ1,0} and n(L−) = 1. For 0 < ε� 1 one will have that for L± there will be aneigenvalue which is O(ε). However, these small eigenvalues do not figure in the spectral calculationbecause dim[gker(JL)] = 4 for all ε > 0, so that no small eigenvalues are created for σ(JL) whenε > 0. After projecting off of the kernels one will eventually see that n(R) = n(S) = 1 for all ε > 0sufficiently small.
The system is of the form of (2.38). Since the bifurcation arises from ω = 4, when ε = 0 onehas for i+ j 6= 1, λi,j = [2(i+ j− 1)]−1. As a consequence one has that when ε = 0, 4 = z1 ∈ σ(R);furthermore, z1 is algebraically simple with geometric multiplicity three. One has
r1(z) = 2− 12z,
and there are removable singularities at z = 4. For ε > 0 it is generically expected that (a) thepoles will become simple, and (b) the residue associated with each pole will be nonzero. Followingthe argument associated with (2.41) with ` = 2 one can conclude that there will be at most fourzeros of r1(z) which are within O(ε) of z = 4 (see the cartoon in Figure 5 for a possible scenario).
Figure 6 gives the numerically generated spectrum when ε = 0.6. The left panel shows a plotof the eigenvalue r1(z), while the right panel shows the spectrum for (1.2). The purely imaginaryeigenvalue in the upper half of the complex plane with negative signature is associated with the filled(red) circle, and the (red) crosses in the left panel denote the location of the poles of r1(z). Whenone compares Figure 6 to the cartoon of Figure 5 it is clear that there is a significant difference:in the cartoon the effect of all three poles is clearly present, whereas in the numerical output theeffect of the second pole at µ2 is missing.
T. Kapitula 21
3 4
−2
0
2
z
r 1(z)
0 10
x 10−3
−4
0
4
σ(JL)
Re(λ)
Im(λ
)Figure 6: (color online) The Krein eigenvalue r1(z) (left panel), and σ(JL) (right panel).In the left panel the (red) cross indicates the location of a pole of the Krein matrix. In theright panel the eigenvalue with negative signature is that associated with the filled (red)circle in the insert.
The missing zero can be explained with the help of Figure 7. Therein one sees σ(S−1R) (bluediamond) overlayed with the singularities of K (z) (red cross). Using the ordering zj < zj+1 forzj ∈ σ(S−1R), note that z4, the only one of the four eigenvalues which is not captured as a zero ofr1(z), practically coincides with µ2; in fact, one has |z4 − µ2| = O(10−12). Recall that for positivereal eigenvalues there are two possibilities: (a) the eigenvalue will coincide with a zero of r1(z), or(b) the eigenvalue will coincide with a removable singularity and be of positive sign. Numericallyone sees that the residue of K (z) at µ2 is O(10−29). Thus, one can conclude that the locationof eigenvalue z4 falls under scenario (b). In conclusion, the residues associated with µ1 and µ3
are nonzero and that associated with µ2 is zero; hence, by the Krein Oscillation Theorem one canexpect that r1(z) will have an odd number of zeros in the interval (µ1, µ3). In fact, there arethree simple zeros, and by Remark 2.13 the middle zero corresponds to an eigenvalue of negativesignature. The fourth eigenvalue is of positive signature and coincides with the pole µ2.
3 3.5 4
0
x 10−4
z
Figure 7: (color online) The eigenvalues of S−1R (blue diamond) and the poles of theKrein matrix (red cross). Note that z4 and µ2 coincide.
Krein signature and the Krein Oscillation Theorem 22
References
[1] N. Abdallah, F. Mehats, C. Schmeiser, and R. Weishaupl. The nonlinear Schrodinger equation with astrongly anisotropic harmonic potential. SIAM J. Math. Anal., 37(1):189–199, 2005.
[2] F. Abdullaev and R. Kraenkel. Coherent atomic oscillations and resonances between coupled Bose-Einstein condensates with time-dependent trapping potential. Phys. Rev. A, 62:023613, 2000.
[3] J. Alexander, R. Gardner, and C.K.R.T. Jones. A topological invariant arising in the stability oftravelling waves. J. Reine Angew. Math., 410:167–212, 1990.
[4] B. Anderson, K. Dholakia, and E. Wright. Atomic-phase interference devices based on ring-shapedBose-Einstein condensates: two-ring case. Phys. Rev. A, 67(3):033601, 2003.
[5] R. Battye, N. Cooper, and P. Sutcliffe. Stable Skyrmions in two-component Bose-Einstein condensates.Phys. Rev. Lett., 88(8):080401, 2002.
[6] R. Bradley, B. Deconinck, and J. Kutz. Exact nonstationary solutions to the mean-field equations ofmotion for two-component Bose-Einstein condensates in periodic potentials. J. Phys. A: Math. Gen.,38:1901–1916, 2005.
[7] T. Bridges and G. Derks. Hodge duality and the Evans function. Phys. Lett. A, 251:363–372, 1999.
[8] T. Busch, J. Cirac, V. Perez-Garcıa, and P. Zollar. Stability and collective excitations of a two-momentBose-Einstein condensed gas: A moment approach. Phys. Rev. A, 56(4):2978–2983, 1997.
[9] B. Deconinck and T. Kapitula. On the orbital (in)stability of spatially periodic stationary solutions ofgeneralized Korteweg-de Vries equations. preprint.
[10] M. Grillakis. Analysis of the linearization around a critical point of an infinite dimensional Hamiltoniansystem. Comm. Pure Appl. Math., 43:299–333, 1990.
[11] M. Grillakis. Linearized instability for nonlinear Schrodinger and Klein-Gordon equations. Comm. PureAppl. Math., 46:747–774, 1988.
[12] M. Grillakis, J. Shatah, and W. Strauss. Stability theory of solitary waves in the presence of symmetry,II. Journal of Functional Analysis, 94:308–348, 1990.
[13] M. Haragus and T. Kapitula. On the spectra of periodic waves for infinite-dimensional Hamiltoniansystems. Physica D, 237(20):2649–2671, 2008.
[14] C.K.R.T. Jones. Instability of standing waves for non-linear Schrodinger-type equations. Ergod. Th. &Dynam. Sys., 8:119–138, 1988.
[15] T. Kapitula. Stability analysis of pulses via the Evans function: dissipative systems, volume 661 ofLecture Notes in Physics, pages 407–427. 2005.
[16] T. Kapitula and P. Keverkidis. Linear stability of perturbed Hamiltonian systems: theory and a caseexample. J. Phys. A: Math. Gen., 37(30):7509–7526, 2004.
[17] T. Kapitula and P. Kevrekidis. Bose-Einstein condensates in the presence of a magnetic trap and opticallattice. Chaos, 15(3):037114, 2005.
[18] T. Kapitula and B. Sandstede. Eigenvalues and resonances using the Evans function. Disc. Cont. Dyn.Sys., 10(4):857–869, 2004.
[19] T. Kapitula, K. Law, and P. Kevrekidis. Interaction of excited states in two-species Bose-Einsteincondensates: a case study. (submitted).
T. Kapitula 23
[20] T. Kapitula, P. Kevrekidis, and B. Sandstede. Counting eigenvalues via the Krein signature in infinite-dimensional Hamiltonian systems. Physica D, 195(3&4):263–282, 2004.
[21] T. Kapitula, P. Kevrekidis, and B. Sandstede. Addendum: Counting eigenvalues via the Krein signaturein infinite-dimensional Hamiltonian systems. Physica D, 201(1&2):199–201, 2005.
[22] T. Kapitula, P. Kevrekidis, and Z. Chen. Three is a crowd: Solitary waves in photorefractive mediawith three potential wells. SIAM J. Appl. Dyn. Sys., 5(4):598–633, 2006.
[23] T. Kapitula, P. Kevrekidis, and R. Carretero-Gonzalez. Rotating matter waves in Bose-Einstein con-densates. Physica D, 233(2):112–137, 2007.
[24] K. Kasamatsu, M. Tsubota, and M. Ueda. Quadrupole and scissors modes and nonlinear mode couplingin trapped two-component Bose-Einstein condensates. Phys. Rev. A, 69(4):043621, 2004.
[25] T. Kato. Perturbation Theory for Linear Operators. Springer-Verlag, Berlin, 1980.
[26] P. Kevrekidis, H. Nistazakis, D. Frantzeskakis, B. Malomed, and R. Carretero-Gonzalez. Families ofmatter-waves in two-component Bose-Einstein condensates. Eur. Phys. J. D, 28:181–185, 2004.
[27] Y. Li and K. Promislow. Structural stability of non-ground state traveling waves of coupled nonlinearSchrodinger equations. Physica D, 124(1–3):137–165, 1998.
[28] M. Matthews, B. Anderson, P. Haljan, D. Hall, C. Wieman, and E. Cornell. Vortices in a Bose-Einsteincondensate. Phys. Rev. Lett., 83(13):2498–2501, 1999.
[29] R. Pego and M. Weinstein. Eigenvalues, and instabilities of solitary waves. Phil. Trans. R. Soc. Lond.A, 340:47–94, 1992.
[30] B. Sandstede and A. Scheel. On the stability of travelling waves with large spatial period. J. Diff. Eq.,172:134–188, 2001.
[31] J. Swinton. The stability of homoclinic pulses: a generalisation of Evans’s method. Phys. Lett. A, 163:57–62, 1992.
[32] V. Vougalter and D. Pelinovsky. Eigenvalues of zero energy in the linearized NLS problem. J. Math.Phys., 47:062701, 2006.
