On the structure of spectra of travelling waves

The linear stability of the travelling wave solutions of a general reaction-diusion system is investigated. The spectrum of the corresponding second order dieren tial operator L is studied. The problem is reduced to an asymptotically autonomous rst order linear system. The relation between the spectrum of L and the corresponding rst order system is dealt with in detail. The rst order system is investigated using exponential dichotomies. A self-contained short presentation is shown for the study of the spectrum, with elementary proofs. An algorithm is given for the determination of the exact position of the essential spectrum. The Evans function method for determining the isolated eigenvalues of L is also presented. The theory is illustrated by three examples: a single travelling wave equation, a three variable combustion model and the generalized KdV equation.


Introduction
We investigate the stability of the travelling wave solutions of the system where u : R + × R → R m , D is a diagonal matrix with positive diagonal elements and f : R m → R m is a continuously differentiable function.The travelling wave solution of this equation has the form u(τ, x) = U (x − cτ ), where U : R → R m .For this function we have The stability of U can be determined by linearization.Putting u(τ, x) = U (x − cτ ) + v(τ, x − cτ ) in (1) the linearized equation for v takes the form According to well known stability theorems (see e.g.[21]) the stability of the travelling wave solution U is determined by the spectrum of the second order differential operator EJQTDE, 2003 No. 15, p. 1 endowed with the supremum norm V = max R |V (t)|.It is assumed that Q : R → C m×m is continuous, and the limits Q ± = lim t→±∞ Q(t) exist.The complex number λ ∈ C is called a regular value of L if the operator L−λI has bounded inverse that is defined in the whole space C 0 (R, C m ).That is for any W ∈ C 0 (R, C m ) there exists a unique solution of LV − λV = W in C 0 (R, C m ) ∩ C 2 (R, C m ), and there exists M > 0 such that for any W ∈ C 0 (R, C m ) we have V ≤ M W .The spectrum σ(L) of L consists of non-regular values.The number λ is called an eigenvalue if L − λI has no inverse.The essential spectrum of L consists of those points of the spectrum which are not isolated eigenvalues with finite multiplicity.It is useful to introduce the first order system corresponding to equation LV − λV = W .Let x = (V, V ) T , y = (0, W ) T , then the first order system is ẋ(t) = A λ (t)x(t) + y(t), where The spectrum of L has been widely investigated.The position of the essential spectrum has been estimated, see e.g.[14], and Weyl's lemma in [21].This can be done using exponential dichotomies for the first order system (5), [5,18].This approach was also generalized to infinite dimensional systems, when A is a bounded operator on a Banach space [6] and also for unbounded operators [4].Fredholm properties are also relevant when determining the spectrum of L [11,18].The relation between these two concepts is dealt with in [16,18].The determination of the isolated eigenvalues requires to solve a linear system with non-constant coefficients, which can be done in general only numerically.For the investigation of the isolated eigenvalues two concepts were introduced.The first was the Evans function [10], which is an analytic function on the complex plane, the zeros of which are the isolated eigenvalues of L. Later a topological invariant was introduced for the study of the spectrum [1].These methods were applied for several systems in physics [3,12,17], chemistry [2,7,13,19] and biology [10,15].
The aim of the paper is to present a self-contained detailed study of the spectrum of L, and to fill the gap between the abstract results (on exponential dichotomies and on topological invariants) and the applications.The novelties of the paper are as follows.
• An algorithm is given for the determination of the exact position of the essential spectrum.The statements concerning the essential spectrum are proved by elementary methods.(Most of the known results give only sufficient conditions for the essential spectrum to lie in the left half plane.) • All the theorems are proved in the finite dimensional case.The presentation does not need abstract techniques, hence for those applying the theory a self-contained method is shown.(According to the author's knowledge a self-contained explanation, including the proofs, is only given for the case of unbounded operators [4].The proof of the finite dimensional case must be compiled from different sources, e.g.[5,8,14,16].) • The relation between the spectrum of L and the corresponding first order system is dealt with in detail.(The standard reference in this context is [14] but the relation is not proved in that book.) 2 Relation between the spectrum of L and the first order system Lemma 1 connects the determination of the spectrum of L with the study of system (5).In order to prove the lemma we will need the following Proposition.
Proof We will show that V → 0 at +∞, it can be verified similarly, that V → 0 at −∞.Let us denote by z the real or imaginary part of the k-th coordinate of V , (k is arbitrary), and let d = D kk .We will prove that lim t→+∞ ż(t) = 0 that implies the desired statement.First we prove in the case c = 0. Since V and W tend to 0 at +∞, for any ε > 0 there exists t 0 such that for all t > t 0 .
Integrating in [t 0 , t] we get In the case c > 0, there exists t 1 > t 0 for which yielding lim +∞ ż = 0, what we wanted to prove.
In the case c < 0 we prove by contradiction.Assume that there exists α > 0 and a sequence t n → ∞, such that | ż(t n )| = α.Let ε = −cα/2 and apply (7) for t 0 = t n when n is large enough.If ż(t n ) = α, then the inequality in the left hand side of (7)  ) → −∞ as t → +∞.This contradicts to the boundedness of z.
Finally, let us consider the case c = 0. Then from the differential equation we get that z tends to zero at infinity.According to the Landau-Kolmogorov inequality (see e.g.[9]) ż ≤ 4 z z .Defining • as the supremum norm on [T, +∞) for T large enough, we get that ż → 0 at infinity.Lemma 1 (i) If system (5) has a unique solution x ∈ C 0 (R, C 2m ) for any y ∈ C 0 (R, C 2m ) and x depends continuously on y, (i.e.there exists M > 0, such that x ≤ M ( y for all y ∈ C 0 (R, C 2m )), then λ is a regular value of L.
(ii) If λ is a regular value of L, then for any differentiable function y ∈ C 0 (R, C 2m ), for which ẏ ∈ C 0 (R, C 2m ), there exists a unique solution x ∈ C 0 (R, C 2m ) of system (5); and there exists M > 0 such that for any y satisfying the above conditions the corresponding unique solution x satisfies x ≤ M ( y + ẏ ).
Then there exists a unique solution x ∈ C 0 (R, C 2m ) of (5).Let V = (x 1 , . . ., x m ) T , U = (x m+1 , . . ., x 2m ) T , then V ∈ C 0 (R, C m ) and U = V , hence V is twice differentiable, and LV − λV = W .The continuity follows from x ≤ M y , namely First we show that for any y ∈ C 0 (R, C 2m ) which is differentiable and ẏ ∈ C 0 (R, C 2m ) there exists a unique solution x ∈ C 0 (R, C 2m ) of (5).Let x = (V, U ) T , y = (y 1 , y 2 ) T , where V, U, y 1 , y 2 : R → C m , then system (5) takes the form The differentiability of y 1 implies that V is twice differentiable and V = U + ẏ1 , hence from the second equation Since λ is a regular value of L, this equation has a unique solution V ∈ C 0 (R, C m ), and there exists Moreover, according to Proposition 1 ) what we wanted to prove.Now we show the continuous dependence of x on y.Let us denote the maximum point of Vk (for some k = 1, . . ., m) by t k .Since Vk → 0 at ±∞, the maximum point is not at infinity, hence Vk (t k ) = 0. Therefore using (8) we get that there exists M 2 > 0 for which Similar inequality can be proved for the minimum of Vk , and for all coordinates of V , hence there exists M 3 > 0 for which V ≤ M 3 ( ẏ1 + y 1 + y 2 ).Since U = V − y 1 , there exists M 4 > 0 for which U ≤ M 4 ( ẏ1 + y 1 + y 2 ).Thus with M = 2(M 1 + M 4 ) we have x ≤ M ( y + ẏ ).Now we turn to the study of general first order systems, the special case of which is system (5).

First order systems
Now for short let C 0 = C 0 (R, C n ) endowed with the supremum norm x = max R |x(t)|, and let A : R → C n×n be a continuous function for which the limits exist.Let us consider the first order system Our aim is to give necessary and sufficient condition for the existence and uniqueness of a solution x ∈ C 0 of (9) for any y ∈ C 0 , and for the continuous dependence of x on y.Since A is continuous, system (9) has solutions for any y ∈ C 0 , that can be given by the variation of constants formula as where Ψ is the fundamental system of solutions of the homogeneous equation satisfying Ψ(0) = I, i.e. the n columns of the matrix Ψ(t) are n independent solutions of the homogeneous system Hence the question is that for a given y ∈ C 0 does there exist a unique x 0 ∈ C n , such that x ∈ C 0 (x is given by ( 10)), and that does x depend continuously on y.
The dimension of the stable, unstable and central subspaces of the matrices A ± play important role.Let us denote the number of eigenvalues (with multiplicity) of A + with positive, negative, zero real part by n + u , n + s , n + c , respectively.We define n − u , n − s , n − c similarly using A − .First we show that the continuous dependence is violated when n + c > 0 or n − c > 0. The main point in these cases is to prove the existence of a bounded solution of the homogeneous equation, which does not tend to zero.
Theorem 1 Let us assume that at least one of the following two conditions holds: Then the solution of ( 9) does not depend continuously on y, in the sense that there is no M > 0 for which x ≤ M ( y + ẏ ) holds for any differentiable function y ∈ C 0 , for which ẏ ∈ C 0 .
Proof We prove in case (a), the other case is similar.According to Theorem 1.10.1 of [8] there exists z 0 ∈ C n , such that the solution t → Ψ(t)z 0 of ( 11) is bounded in [0, +∞) but does not tend to zero as t → +∞.Then there exist a > 0 and a sequence t k → +∞, such that |Ψ(t k )z 0 | = a for all k = 1, 2, . ... Let h k : R → R be continuously differentiable functions satisfying the following conditions Hence ẏk → 0 at +∞, since h k → 0, ḣk → 0 and A(t) is bounded.On the other hand, there exists Then x k is a solution of ( 9) belonging to y k , and In the further considerations we will assume n + c = 0 or n − c = 0.In this case we will use exponential dichotomies to answer the above question.
Definition 1 System (11) possesses an exponential dichotomy in the interval J if there exist a projection P and positive numbers K, L, α, β, such that for s ≥ t, t, s ∈ J We will show that the exponential dichotomy on R implies the existence, uniqueness and continuous dependence of the solution of ( 9).It can be shown that system (11) possesses an exponential dichotomy on R if A(t) is constant and it has no eigenvalues on the imaginary axis, that is when n + c = n − c = 0.For the time dependent case there is no exponential dichotomy on R in general when n + c = n − c = 0.However, we will show that system (11) possesses exponential dichotomies on R + and on R − .If the projections of these two dichotomies are the same, then system (11) possesses an exponential dichotomy on R as well, and the existence, uniqueness and continuous dependence follows.
We will use the following properties of exponential dichotomies.
Lemma 2 (i) Let P 1 and P 2 be projections, for which Im P 1 = Im P 2 .If system (11) possesses an exponential dichotomy in the interval J with the projection P 1 , then it possesses an exponential dichotomy in the interval J with the projection P 2 as well.
(ii) System (11) possesses an exponential dichotomy in any closed and bounded interval [a, b], with any projection.
(iii) If system (11) possesses an exponential dichotomy in the intervals (a, b] and [b, c) with the same projection P , then it possesses an exponential dichotomy in the interval (a, c) (here a can be −∞, and c can be +∞).

Proof
(i) This statement is proved in [5] The following Lemma can be proved using the results in [5].
Lemma 3 (i) If n + c = 0, then system (11) possesses an exponential dichotomy in [0, +∞), the projection of which is denoted by P + .Moreover, dim(ker (11) possesses an exponential dichotomy in (−∞, 0], the projection of which is denoted by P − .Moreover, dim(ker Proof We prove only (i), the second statement can be verified similarly.Since n + c = 0, system ẋ = A + x with constant coefficients possesses an exponential dichotomy in [0, +∞) the projection of which, denoted by P + , has an n + s dimensional range and n + u dimensional kernel, see [5] pp.10.For any δ > 0 there exists t 0 > 0, such that A(t) − A + < δ for t > t 0 .Hence according to proposition 1 of [5] on pp.34 system (11) possesses an exponential dichotomy in [t 0 , +∞) with projection P + .Applying statements (ii) and (iii) of Lemma 2 we can complete the proof.
Let us introduce the following subspaces.
According to the next Proposition the subspace E + s consists of those initial conditions x 0 , for which the solution of the homogeneous equation tend to 0 as t → +∞.Similarly, the subspace E − u consists of those initial conditions x 0 , for which the solution of the homogeneous equation tend to 0 as t → −∞.
We will need the following three lemmas.The first is a generalization of Proposition 2 to the inhomogeneous equation.
(i) Assume n + c = 0. Then lim +∞ x = 0 if and only if (ii) Assume n − c = 0. Then lim −∞ x = 0 if and only if Proof First we show that lim t→+∞ Applying ( 12) for P = P + and s = 0 we obtain Now we prove the convergence of the second term.Let ε > 0 be an arbitrary positive number.Since lim +∞ y = 0, there exists t 1 > 0 such that, for t > t Now we show that ( 15) is equivalent to Applying (13) for P = P + and t = 0 we can see that the integral in (15) We have seen that the second term tends to zero as t → +∞, hence lim t→+∞ Ψ(t)(I − P + )(x 0 + x * ) = 0 implying x 0 + x * = 0 according to Proposition 2, hence x 0 = −x * , yielding (15).
According to (17) the first and second terms tend to zero as t → +∞.Hence lim +∞ x = 0 if and only if the sum of the third and fourth term tends to zero as t → +∞.According to (19) this sum tends to zero if and only if that (15) holds, what we had to prove.The proof of statement (ii) of the Lemma follows similarly from (18) and (20).
Lemma 5 Let us assume n + c = n − c = 0.The following two statements are equivalent.(i) For any a ∈ E + u and any b ∈ E − s there exists a unique solution x 0 of u , then (I − P + )z = 0 and P − z = 0, hence applying (i) with a = b = 0 we get z = 0. Now we show that for any z ∈ C n there exist Let z 1 be the solution of (i) with a = 0, b = P − z, that is Let z 2 be the solution of (i) with a = (I − P + )z, b = 0, that is Then (I − P + )z = (I − P + )(z 1 + z 2 ) and P − z = P − (z 1 + z 2 ), hence the uniqueness implies z = z 1 + z 2 .
The following lemma is proved in [16] Prop.2.1 and in [5] p. 19, but in order to make the paper self-contained we present a short proof.

Lemma 6 Let us assume n
Then system (11) possesses an exponential dichotomy in R with a projection P * , for which Im P * = E + s , ker P * = E − u .
Proof Assumption E + s ⊕ E − u = C n implies that there exists a projection P * , for which Im P * = E + s , ker P * = E − u .Namely, for an arbitrary vector z ∈ C n we can define P * z as Since Im P * = Im P + , Lemma 2 and Lemma 3 imply that system (11) possesses an exponential dichotomy in [0, +∞) with the projection P * .On the other hand Im (I − P * ) = Im (I − P − ), hence Lemma 2 and Lemma 3 imply that system (11) possesses an exponential dichotomy in (−∞, 0] with the projection P * .Since system (11) possesses an exponential dichotomy both in [0, +∞) and in (−∞, 0] with the projection P * , according to Lemma 2 it possesses an exponential dichotomy in R.

Theorem 2 Let us assume n
(i) If for any differentiable function y ∈ C 0 , for which ẏ ∈ C 0 there exists a unique solution x ∈ C 0 of (9), then 9) has a unique solution x ∈ C 0 for any y ∈ C 0 , and there exists M > 0, such that for any y ∈ C 0 and for the corresponding solution x ∈ C 0 the inequality x ≤ M y holds.
Proof (i) Let us assume that system (9) has a unique solution x ∈ C 0 for any differentiable function y ∈ C 0 , for which ẏ ∈ C 0 .Let a ∈ E + u and b ∈ E − s be arbitrary vectors.We will show that there exists a differentiable function y ∈ C 0 , with ẏ ∈ C 0 , such that Namely, let h : R → R be a continuously differentiable function satisfying with some r > 0, and where q ∈ R is chosen to satisfy A(t) ≤ q for all t ∈ R. It can be easily shown that there exists k > 0, such that Ψ(t) ≤ ke qt for all t ∈ R. Let Then y and ẏ are continuous (also in zero, because their limits are zero from left and from right), and y, ẏ ∈ C 0 , because for t > 1 we have Let x ∈ C 0 be the solution of ( 9) belonging to y.Then according to Lemma 4 for x 0 = x(0) we have given by ( 21).According to Lemma 5 there exists a unique x 0 ∈ C n satisfying (22), hence Lemma 4 implies x ∈ C 0 .If x * ∈ C 0 is another solution of ( 9), then x * − x ∈ C 0 is a solution of (11).However, according to Corollary 1 x * − x ≡ 0, that is x * ≡ x.
Finally we prove the continuous dependence.According to Lemma 6 there exist a projection P * , and positive numbers K, L, α, β, for which Repeating the proof of Lemma 4 replacing P + and P − with P * we get Therefore from the variation of constant formula ( 10)

The spectrum of L
In Section 1 we have introduced the matrix functions A λ , see (6).Since function Q tends to a limit at ±∞, the limits exist.We have seen in Section 2 that the dimension of the stable, unstable and central subspaces of the matrices A ± λ play important role.Let us denote the number of eigenvalues (with multiplicity) of A + λ with positive, negative, zero real part by n + u (λ), n + s (λ), n + c (λ), respectively.We define EJQTDE, 2003 No. 15, p. 10 Theorem 3 Let us assume that at least one of the following two conditions holds: Proof Let us assume the contrary, i.e. that λ is a regular value of L. Then according to (ii) of Lemma 1 for any differentiable function y ∈ C 0 (R, C 2m ), for which ẏ ∈ C 0 (R, C 2m ), there exists a unique solution x ∈ C 0 (R, C 2m ) of system (5); and there exists M > 0 such that for any y the inequality x ≤ M ( y + ẏ ) holds.However, Theorem 1 yields that this M cannot exist, which is a contradiction.
In the further considerations we deal with the case n + c (λ) = 0 = n − c (λ).In this case we introduce the subspaces in the same way as in Section 2. (Now they depend also on λ, because A depends on λ.) (ii) λ is a regular value of L if and only if 0 (otherwise U ≡ 0 and x ≡ 0), and LV = λV .(ii) If λ is a regular value of L, then according to (ii) of Lemma 1 for any differentiable y ∈ C 0 (R, C 2m ) for which ẏ ∈ C 0 (R, C 2m ) there exists a unique solution x ∈ C 0 (R, C 2m ) of (5).Then Theorem 2 implies , then according to Theorem 2 for any y ∈ C 0 (R, C 2m ) there exists a unique solution x ∈ C 0 (R, C 2m ) of ( 5) and it depends continuously on y.Hence by (i) of Lemma 1 λ is a regular value of L.
EJQTDE, 2003 No. 15, p. 11 The dimension of E + s (λ) and E − u (λ) can be determined explicitly, because only the eigenvalues of the matrices A ± λ have to be determined to get these dimensions.However, to get dim(E + s (λ) ∩ E − u (λ)) the time dependent system must be solved numerically.This leads to the definition of the Evans function., and the base of the subspace The assumption dim(E + s (λ) ∩ E − u (λ)) > 0 means that the two bases together give a linearly dependent system of vectors.The Evans function is defined as the determinant formed by these 2m vectors.That is the determinant is zero if λ is an eigenvalue.
We have proved that the eigenvalues are the zeros of the Evans function.It can be also shown that the multiplicity of an eigenvalue is equal to the multiplicity of the zero of the Evans function, and that the Evans function is an analytic function on the domain Ω [1].Hence the zeros of D are isolated, that is in the domain where dim E + s (λ) + dim E − u (λ) = 2m there can be only isolated eigenvalues.This statement together with Corollary 2 enables us to determine the essential spectrum explicitly.

Corollary 3
The essential spectrum of L is The bases of the stable and unstable subspaces can be determined numerically in the following way.We calculate the eigenvalues of A + λ with negative real part, and its corresponding eigenvectors.Let us denote these eigenvalues by µ 1 , . . ., µ k , and the eigenvectors by u 1 , . . ., u k (for short we used the notation k = n + s (λ)).Similarly, let us denote the eigenvalues of A − λ with positive real part by ν 1 , . . ., ν l , and the corresponding eigenvectors by v 1 , . . ., v l (for short we used the notation l = n − u (λ)).Then choosing a sufficiently large number we solve the homogeneous equation ẋ(t) = A λ (t)x(t) in [0, ] starting from the right end point with initial condition x( ) = u i e µi for i = 1, . . ., k. Hence we get k = n + s (λ) linearly independent (approximating) solutions of the differential equations, therefore their values at 0 give a base of E + s (λ).Similarly, solving the differential equation in [− , 0] we get a base of E − u (λ), and the determinant defining the Evans function can be computed.We note that if is very large and there is a significant difference between the real parts of the eigenvalues µ 1 , . . ., µ k , then the solution belonging to the eigenvalue with largest real part will dominate and the solutions starting from linearly independent initial conditions will be practically linearly dependent at zero.(Similar case can occur in [− , 0] as well.)To overcome this difficulty the problem can be extended to a wedge product space of higher dimension [3].Now we show a method to determine the eigenvalues and eigenvectors of A ± λ , which determine the dimensions of E + s (λ) and E − u (λ).We will deal with the two cases together, therefore for short we introduce where Q can be Q + or Q − .Let us denote an eigenvalue of A λ by µ and an eigenvector by u Thus we have proved the following proposition.
Thus the eigenvalues of A ± λ are determined by equation (26) of degree 2m.In the special case when Q is an upper or lower triangular matrix the l.h.s. of the equation is a product of m second degree polynomials, hence the solutions can be computed explicitly [20].

Case of a single equation, m = 1
Now let U : R → R be a solution of where f : R → R is a continuously differentiable function, U − , U + ∈ R and it is assumed that c ≥ 0. The stability of U is determined by the spectrum of the operator The function q(t) = f (U (t)) is continuous and has limits at ±∞, If U tends to the limits U + and U − exponentially, then the integrals in the assumptions of Theorem 3 are convergent.Now A ± λ are 2-by-2 matrices and according to Proposition 3 their eigenvalues (µ 1,2 ) are determined by the equation The essential spectrum can be determined by calculating the dimensions of E + s (λ) and E − u (λ).These dimensions can be easily determined from the sets where n + c (λ) ≥ 1 and n − c (λ) ≥ 1.These sets are formed by those values of λ to which µ = iω is a solution of (31).Hence the set of λ values where n + c (λ) ≥ 1 is the parabola It is easy to show that on the left hand side of the parabola dim E + s (λ) = 2, and on the right hand side dim E + s (λ) = 1, see Figure 1.Similarly, the set of λ values where n − c (λ) ≥ 1 is the parabola It is easy to show that on the left hand side of the parabola dim E − u (λ) = 0, and on the right hand side dim E − u (λ) = 1, see Figure 1.The parabolas given by ( 32) and (33) for q + > q − in (a) and for q − > q + in (b).The dimensions of the subspaces E + s (λ) and E − u (λ) are shown, the upper numbers correspond to the former and the lower numbers correspond to the latter.Now applying Theorem 3 and Corollary 2 we have the following results concerning the spectrum of L.
• Both parabolas belong to the essential spectrum of L.
• The domain lying on the left hand side of both parabolas consists of regular values of L.
• The domain lying on the right hand side of both parabolas contains all the isolated eigenvalues of L the remaining points of this domain (which are not isolated eigenvalues) are regular values of L.
• If q + > q − , then the domain between the two parabolas is filled with eigenvalues.
• If q + < q − , then the domain between the two parabolas is filled with points belonging to the essential spectrum, but they are not eigenvalues.
In this special case of m = 1 the location of the isolated eigenvalues with respect to the imaginary axis can also be determined.It can be shown that zero is a simple eigenvalue and all other isolated eigenvalues of L are negative (real) if and only if U is strictly monotone and f (U − ) < 0, f (U + ) < 0.Here a is the concentration of the fuel, w is the concentration of an inhibitor species and b is scaled temperature.The travelling wave solution describes a flame propagating with velocity c.The number and stability of travelling waves of this system was investigated in [19].We proved that the solutions a, w and b have limits at +∞, that are denoted by a + , w + and b + .If b + = 0, then we refer to the travelling wave solution as a pulse solution.If b + > 0 then we call it a front solution.In the latter case a + = 0 = w + .It was also shown that a saddle-node bifurcation may occur and there can be 1, 2 or 3 travelling wave solutions.The stability of these solutions can also change through Hopf bifurcation.

Flame propagation in a three variable model
The saddle-node and Hopf bifurcation curves were determined numerically.Here we only show how the method described in the previous Section works for this system to determine the essential spectrum of the corresponding linearizdimen ed operator.The results obtained by the Evans function method will be only cited from [19].

Corollary 1
Let us assume n + c = n − c = 0.The homogeneous equation (11) has a unique solution in C 0 (it is x ≡ 0) if and only if

Definition 2
The Evans function belonging to the operator L is D : Ω

EJQTDE, 2003 Figure 1 .
Figure1.The parabolas given by (32) and (33) for q + > q − in (a) and for q − > q + in (b).The dimensions of the subspaces E + s (λ) and E − u (λ) are shown, the upper numbers correspond to the former and the lower numbers correspond to the latter.