Invariant Manifolds and Their Zero-Viscosity Limits for Navier-Stokes Equations

Transcription

1 Dynamics of PDE, Vol2, No2, 59-86, 2005 Invariant Manifolds and Their Zero-Viscosity Limits for Navier-Stokes Equations Y Charles Li Communicated by Charles Li, received May, 2005 Abstract First we prove a general spectral theorem for the linear Navier- Stokes (NS) operator in both 2D and D The spectral theorem says that the spectrum consists of only eigenvalues which lie in a parabolic region, and the eigenfunctions (and higher order eigenfunctions) form a complete basis in H l (l = 0,, 2, ) Then we prove the existence of invariant manifolds We are also interested in a more challenging problem, ie studying the zero-viscosity limits (ν 0 + ) of the invariant manifolds Under an assumption, we can show that the sizes of the unstable manifold and the center-stable manifold of a steady state are O( ν), while the sizes of the stable manifold, the center manifold, and the center-unstable manifold are O(ν), as ν 0 + Finally, we study three examples The first example is defined on a rectangular periodic domain, and has only one unstable eigenvalue which is real A complete estimate on this eigenvalue is obtained Existence of an D unstable manifold and a codim stable manifold is proved without any assumption For the other two examples, partial estimates on the eigenvalues are obtained Contents Introduction 59 2 Formulation of the Problems 6 Invariant Manifolds 62 4 Examples 69 References 85 Introduction Navier-Stokes equations define an infinite dimensional dynamical system Turbulence can be represented through a chaotic solution in the infinite dimensional phase spaces For example, one can take the Sobolev spaces H l (l = 0,, 2, ) to 99 Mathematics Subject Classification Primary 5, 76, 7; Secondary 4 Key words and phrases Invariant manifold, zero-viscosity limit, Navier-Stokes equation 59 c 2005 International Press

2 60 Y CHARLES LI be the phase spaces A basic dynamical system question is: Does the phase space foliate under the NS flow? That is, are there invariant manifolds under the NS flow? The answer is yes It seems that under the Euler flow, which can be obtained from the NS flow by setting viscosity to zero, the phase space does not foliate To begin a dynamical system study, one chooses a common steady state of the NS and its corresponding Euler equation, defined on a periodic spatial domain At this steady state, the spectra of the linear NS and the linear Euler are dramatically different The linear NS has the form: ν + relatively compact terms, therefore, it has only a point spectrum via Weyl s essential spectrum theorem On the other hand, the linear Euler has a nontrivial essential spectrum For 2D linear Euler, if the steady state as a vector field has a saddle, then the essential spectrum in the Sobolev space H l (l = 0,, 2, ) is a vertical band symmetric with respect to the imaginary axis [25] The width of the band (= lλ for some Λ > 0) increases with l, representing the fact of cascade and inverse cascade A typical example of such steady states is the cat s eye If the steady state has no saddle, but has at least two centers, then the essential spectrum in the Sobolev space H l is the imaginary axis [25] A typical example of such steady states is the shear [6] [9] For the shear, a sharp upper bound on the number of eigenvalues was also obtained [8] For D linear Euler, the spectrum is an open problem A nice example of the steady states is the ABC flow vector field An interesting open problem is to understand the connection between the spectra of linear NS and linear Euler as ν 0 + Besides the progresses mentioned above in understanding the spectra of linear Euler, progresses were also made in better understanding of the spectra of linear NS, although they were in a different setting For both 2D and D NS posed in L 2 (in terms of weak solutions) on a bounded domain with non-slip boundary condition, Prodi [22] proved that the eigenvalues of the linear NS at a steady state lie inside a parabolic region λ r < a bλ 2 i where a > 0 and b > 0 are some constants, and λ = λ r + iλ i is the spectral parameter Moreover, David Sattinger [24] proved that the linear NS has infinitely many eigenvalues lying in the parabolic region, each of finite multiplicity, which can cluster only at infinity, and the corresponding eigenfunctions (and higher order eigenfunctions) form a complete basis in L 2 Similar results are proved in the current article for a periodic domain and mild solutions in H l (l = 0,, 2, ) The proof involves some deep result from Hilbert-Schmidt operators Together with some deep result from sectorial operator, we prove the existence of invariant manifolds In terms of well-posedness of Navier-Stokes equations (NS), zero-viscosity limit problem has been extensively studied [7] [26] Here we are also interested in the zero-viscosity limit (ν 0 + ) problem for invariant manifolds It is proved in this article that under an assumption, their sizes are of order O( ν) or O(ν) as ν 0 + Existence of invariant manifolds for Euler equations is still open I tend to believe that invariant manifolds for Euler equations do not exist, at least not in the usual sense An intuitive argument is as follows: Kato s technique [7] for proving the existence of a mild solution to Euler equations amounts to writing the solution as u(t) = e R t 0 u(τ) dτ u(0), where u is the velocity, and the Leray projection and the external force are not included, which do not affect the argument Notice that the u(τ) in the exponent depends upon u(0), therefore, one should only expect the evolution operator of Euler equations to be at best C 0 in u(0), just as in t On the other hand, the existence of an invariant manifold is really an at least C in u(0)

3 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 6 phenomenon One can obtain expressions of invariant manifolds around a shear for a Galerkin truncation of the Euler equation [0], which exhibit a lip structure For concrete examples, eg single mode steady states, a continued fraction technique can be designed for calculating the eigenvalues [6] Detailed calculations were conducted by Vincent Liu [] [4] Liu obtained rather detailed information on unstable eigenvalues In some case, Liu showed that as ν 0 +, the unstable eigenvalue does converge to that of linear Euler In this article, besides the example of Liu, we will study two more examples For the first example, a complete eigenvalue information is obtained, and the existence of unstable and stable manifolds is proved without any assumption For the second example, only partial information on the eigenvalues is obtained, as in the example of Liu The article is organized as follows: In section 2, we will present the formulation of the problems Section is on invariant manifolds Examples are studied in Section 4 2 Formulation of the Problems We will study the following form of 2D Navier-Stokes equation (NS) (2) t Ω + {Ψ, Ω} = ν[ Ω + f(x)], where Ω is the vorticity which is a real scalar-valued function of three variables t and x = (x, x 2 ), the bracket {, } is defined as where Ψ is the stream function given by, {f, g} = ( x f)( x2 g) ( x2 f)( x g), u = x2 Ψ, u 2 = x Ψ, the relation between vorticity Ω and stream function Ψ is, Ω = x u 2 x2 u = Ψ, and ν is the viscosity, is the 2D Laplacian, and f(x) is the body force We will pose periodic boundary condition Ω(t, x + 2π, x 2 ) = Ω(t, x, x 2 ) = Ω(t, x, x 2 + 2π), ie the 2D NS is defined on the 2-torus T 2 We require that Ψ and f have mean zero Ψdx = T 2 fdx = 0 T 2 Of course Ω always has mean zero In this case, Ψ = Ω Setting ν = 0 in the 2D NS (2), one gets the corresponding 2D Euler equation We will study the common steady states of 2D NS and 2D Euler given by (22) {Ψ, Ω} = 0, Ω + f(x) = 0 Our three prime examples are: () The shear Ω = cos x 2 defined on the rectangular periodic domain [0, 2π/α] [0, 2π] where /2 < α < (2) The shear Ω = cos(x + x 2 ) where f(x) = 2 cos(x + x 2 ) () The cat s eye Ω = cos x + α cos x 2 where α is a constant and f(x) = cos x + α cos x 2 We will study the following form of D Navier-Stokes equation (2) t Ω + (u )Ω (Ω )u = ν[ Ω + f(x)],

4 62 Y CHARLES LI where u = (u, u 2, u ) is the velocity, Ω = (Ω, Ω 2, Ω ) is the vorticity, = ( x, x2, x ), Ω = u, u = 0, ν is the viscosity, is the D Laplacian, and f(x) = (f(x ), f(x 2 ), f(x )) is the body force We also pose periodic boundary condition of period (2π, 2π, 2π), ie the D NS is defined on the -torus T We require that u, f and Ω all have mean zero In this case, u can be uniquely determined from Ω by Fourier transform: ũ (k) = i k 2 [k 2 Ω (k) k Ω2 (k)], ũ 2 (k) = i k 2 [k Ω (k) k Ω (k)], ũ (k) = i k 2 [k Ω2 (k) k 2 Ω (k)] Setting ν = 0 in the D NS (2), one gets the corresponding D Euler equation We will study the common steady states of D NS and D Euler given by (24) (u )Ω (Ω )u = 0, Ω + f(x) = 0 Our prime example is: The ABC flow u = A sin x + C cos x 2, u 2 = B sin x + A cos x, u = C sin x 2 + B cos x Here Ω = u = f Invariant Manifolds Centering around a C steady state Ω, u, Ψ given by (22) or (24), one can rewrite the NS as () t Ω = LΩ + N(Ω), where for 2D and for D LΩ = ν Ω {Ψ, Ω} {Ψ, Ω }, N(Ω) = {Ψ, Ω}, LΩ = ν Ω (u )Ω (u )Ω + (Ω )u + (Ω )u, N(Ω) = (u )Ω + (Ω )u Theorem The linear Euler operator L is a closed operator Its spectrum consists of infinitely many discrete eigenvalues, each of finite multiplicity, which can cluster only at infinity All the eigenvalues lie inside the parabolic region λ r a bλ 2 i where a > 0 and b > 0 are some constants, and λ = λ r + iλ i is the spectral parameter The corresponding eigenfunctions (and higher order eigenfunctions) form a complete basis in H l (l = 0,, 2, ) Proof The proof for L being a closed operator is trivial We can write L as L = ν + A where A is relatively compact with respect to ν By Weyl s essential spectrum theorem [2], L and ν have the same essential spectrum Since the essential spectrum of ν is empty, so is L Therefore, L has only eigenvalues It is easy to see that for large enough λ > 0, (L λ) is compact, thus each eigenvalue can be of only finite multiplicity and all the eigenvalues can cluster only at infinity [2]

5 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 6 Next we show that large enough λ > 0 is not an eigenvalue Assume the contrary, then there exists nonzero Ω H l+2 such that Applying ( ) l/2, one gets λω ν Ω = AΩ λ( ) l/2 Ω ν ( ) l/2 Ω = ( ) l/2 (AΩ) Multiplying by ( ) l/2 Ω, one obtains λ ( ) l/2 Ω 2 [ ] [ ] + ν ( ) l/2 Ω ( )( ) l/2 Ω [ ] [ ] = ( ) l/2 Ω ( ) l/2 (AΩ) Thus = ( ) λ l/2 Ω 2 ( ) + ν (l+)/2 Ω [ [ ] ( ) Ω] l/2 ( ) l/2 (AΩ) ( ) C l/2 Ω ( ) (l+)/2 Ω ( C ɛ 2 ( ) (l+)/2 Ω 2 + ( ) l/2 ɛ 2 Ω 2) 2 Let ɛ 2 = ν 2C 2C2 and λ = ν +, we have ( ) Ω l/2 2 + ( ) 2 ν (l+)/2 Ω 2 0, which is a contradiction to Ω being an eigenfunction Thus λ = 2C2 ν + is in the resolvent set Next let λ be a complex number in the resolvent set Naturally, we need to consider Ω as a complex-valued function For any h H l, there is a Ω H l+2 such that Applying ( ) l/2, one gets ν Ω + λω AΩ = h ν ( ) l/2 Ω + λ( ) l/2 Ω ( ) l/2 (AΩ) = ( ) l/2 h Multiplying by ( ) l/2 Ω, one obtains [ ] [ ] ν ( )( ) l/2 Ω ( ) l/2 Ω [ ] [ ] = ( ) l/2 (AΩ) ( ) l/2 Ω + [ + λ [ ( ) l/2 h ] ( ) l/2 Ω ] [ [ ( ) l/2 Ω ] ( ) l/2 Ω ] Thus (2) = ( ) ν (l+)/2 Ω 2 ( ) + λ l/2 Ω 2 [ [ ] [ ] [ ] ( ) (AΩ)] l/2 ( ) l/2 Ω + ( ) l/2 h ( ) l/2 Ω

6 64 Y CHARLES LI Taking the real part, one gets Let ɛ 2 = ν 2C, one has () ν Ω 2 l+ + λ r Ω 2 l C + 2 ( ɛ 2 Ω 2 l+ + ) ɛ 2 Ω 2 l ( h 2 l + Ω 2 ) l 2 ν Ω 2 l+ + λ r Ω 2 l 2 h 2 l + C Ω 2 l Taking the imaginary part of (2), one gets that is Then we have that is (4) Adding () and (4), we have ( λ r + Let a = C, b = λ i Ω 2 l C 2 Ω l+ Ω l + h l Ω l, λ i Ω l C 2 Ω l+ + h l λ 2 i Ω 2 l C Ω 2 l+ + C 4 h 2 l, ν 2C λ 2 i Ω 2 l 2 ν Ω 2 l+ + νc 4 2C h 2 l ν 2C λ 2 i C ν 2C, and C 5 = 2 Thus if λ r a + bλ 2 i > 0, (5) (L λ) ) Ω 2 l ( + νc ) 4 h 2 l 2 C ( + νc4 C ), (λ r a + bλ 2 i ) Ω 2 l C 5 h 2 l C /2 5 (λ r a + bλ 2 i )/2 (L λ) is holomorphic in λ and can be continued analytically as long as (L λ) is bounded As shown above, (L λ) exists at large λ > 0 Then from (5), (L λ) exists everywhere outside the parabolic region λ r a bλ 2 i Thus the eigenvalues of L lie inside this region To prove the rest of the theorem, we need a known lemma [Corollary, pp042, []] Lemma 2 Let T be a densely defined unbounded linear operator in a Hilbert space H, with the property that for some λ 0 in the resolvent set of T, (T λ 0 ) is a Hilbert-Schmidt operator Let γ,, γ 5 be non-overlapping differentiable arcs having limiting directions at infinity, and such that no adjacent pair of arcs form an angle as great as π/2 at infinity Suppose that the resolvent (T λ) satisfies an inequality (T λ) = O( λ N ) as λ along each arc γ j for some natural number N Then the eigenfunctions (and higher order eigenfunctions) form a complete basis in H

7 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 65 From (5), we see that along any ray λ r = cλ i (c finite), (L λ) = O( λ ) as λ Next we show that (L λ) is a Hilbert-Schmidt operator for large enough λ > 0 Notice that (L λ) = [ I + (ν λ) A ] (ν λ) Since we have (ν λ) Ω 2 l = k = ν 2 k 2 + λ/ν k k 2(l ) Ω k 2, νλ k k 2l (ν k 2 + λ) 2 Ω k 2 k 2 k 2 + λ/ν k 2(l ) Ω k 2 (ν λ) Ω l νλ Ω l It is easy to see that Thus AΩ l C Ω l (ν λ) AΩ l C νλ Ω l Choose λ 0 = 2C2 ν, then (ν λ 0 ) A 2 We have [ I + (ν λ 0 ) A ] being a bounded operator Recall that a bounded linear operator T is a Hilbert-Schmidt operator if T ek 2 is finite where {e k } is a complete orthonormal basis [] By using the Fourier basis for H l, one sees that (ν λ 0 ) is a Hilbert-Schmidt operator A product of a bounded linear operator with a Hilbert-Schmidt operator is another Hilbert- Schmidt operator Thus (L λ 0 ) is a Hilbert-Schmidt operator By Lemma 2, the corresponding eigenfunctions (and higher order eigenfunctions) of L form a complete basis in H l (l = 0,, 2, ) We will study the case as described by the following set-up Set-Up Let there be m unstable eigenvalues λ u λ u m, n center eigenvalues λ c λ c n, and of course the rest infinitely many stable eigenvalues λ s λ s 2 Let the real parts of λ u and λ s have the smallest absolute values among the unstable and stable eigenvalues, respectively Remark For specific examples, often there is no center eigenvalue, here we put them in for generality We need H l (l = 0,, 2, ) to be a Banach algebra [] Thus for 2D and D, l 2 Our argument requires that we work with H l where l for both 2D and D

8 66 Y CHARLES LI We have the splitting H l = Hu l + Hc l + Hs l where Hz l (z = u, c, s) are spanned by the corresponding eigenfunctions Projections onto Hz l are denoted by P z (z = u, c, s) Denote by L z = P z L, N z = P z N, Ω z = P z Ω, (z = u, c, s) Then () can be rewritten as (6) t Ω z = L z Ω z + N z (Ω), (z = u, c, s) Since in this case the spectral mapping theorem is trivially true, one has the trichotomy: (7) (8) (9) e tlu Ce αt, t 0 ; e tlc C t n, t R ; e tls Ce βt, t 0 ; where α = Re{λ u } ɛ, β = Re{λ s } ɛ, n n, and 0 < ɛ Re{λ s } Our argument depends upon the following two crucial facts: () One has a stronger inequality than (9) [[5], Theorem 54] due to the fact that L s is sectorial, (0) e tls Ω s l+ C(βt) /2 e βt Ω s l, t 0 (2) Due to finite dimensionality, one has () Ω z l+ C Ω z l, (z = u, c) Remark 4 For (), take z = u for example, since Ω u Ω u m belong to H l for any l, we can take } (2) C = max { Ω uj l+ / Ω uj l j m Theorem 5 In a neighborhood of the fixed point Ω in the Sobolev space H l (T d ) (l, d = 2, ), there exist a m-dimensional C unstable manifold, a n-dimensional C center manifold, a (m + n)-codimensional C stable manifold, a (m + n)-dimensional C center-unstable manifold, and a m-codimensional C center-stable manifold Proof We will take the center-stable manifold as the example The arguments for the others are similar Apply the method of variation of parameters to (6), one gets the integral equations () Ω z (t) = e (t t0)lz Ω z (t 0 ) + t e (t τ)l z t 0 N z (Ω)dτ, (z = u, c, s) To prove the existence of a center-stable manifold, we start with the Banach space { } B γ,l = Ω(t) Ω γ,l = sup e γt Ω z (t) l < +, ( t 0 4 α < γ < α) z=u,c,s For such Ω(t) in B γ,l lim Ω u (t t e(t t0)lu 0 ) 0 0 +

9 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 67 In the equation (), for z = u, let t 0 + ; for z = c, s, let t 0 = 0; then one gets (4) (5) Ω u (t) = t + Ω z (t) = e tlz Ω z (0) + e (t τ)lu N u (Ω)dτ, t 0 e (t τ)lz N z (Ω)dτ, (z = c, s) The right hand side of (4)-(5) defines a map Γ = Γ u + Γ c + Γ s on B γ,l First we show that Γ contracts, then Γ acting is an easier argument From (4), one has e γt Γ u Ω Γ u Ω 2 l C Using (), one gets e γt Γ u Ω Γ u Ω 2 l C + t + t e γt e α(t τ) N u (Ω ) N u (Ω 2 ) l dτ e γt e α(t τ) N u (Ω ) N u (Ω 2 ) l dτ, in this article, all constants are denoted by the same C Since H l is a Banach algebra N u (Ω ) N u (Ω 2 ) l C( Ω l + Ω 2 l ) Ω Ω 2 l Finally one gets e γt Γ u Ω Γ u Ω 2 l C + t e (α γ)(t τ) ( Ω l + Ω 2 l )e γτ Ω Ω 2 l dτ Now we need to replace N(Ω) by its cut-off χ( Ω l /δ)n(ω) for some δ > 0, where (6) χ(r) =, r [0, ] ; χ(r) = 0, r [, + ) ; χ (r) <, r [0, + ) ; χ C 0 (R, R) If both χ( Ω l /δ) and χ( Ω 2 l /δ) are zero, then of course χ( Ω l /δ)n(ω ) χ( Ω 2 l /δ)n(ω 2 ) = 0 If one of them is nonzero, without loss of generality, say χ( Ω l /δ) 0, then χ( Ω l /δ)n(ω ) χ( Ω 2 l /δ)n(ω 2 ) l [χ( Ω l /δ) χ( Ω 2 l /δ)]n(ω ) l + χ( Ω 2 l /δ)[n(ω ) N(Ω 2 )] l χ( Ω l /δ) χ( Ω 2 l /δ) N(Ω ) l + χ( Ω 2 l /δ) N(Ω ) N(Ω 2 ) l δ Ω Ω 2 l N(Ω ) l + χ( Ω 2 l /δ) N(Ω ) N(Ω 2 ) l If χ( Ω 2 l /δ) = 0, then the second part disappears; otherwise, Ω 2 l < δ In any case, one gets Thus, one has χ( Ω l /δ)n(ω ) χ( Ω 2 l /δ)n(ω 2 ) l Cδ Ω Ω 2 l + e γt Γ u Ω Γ u Ω 2 l Cδ Ω Ω 2 γ,l e (α γ)(t τ) dτ = C δ α γ Ω Ω 2 γ,l t

10 68 Y CHARLES LI Thus (7) sup t 0 For z = c in (5), one has e γt Γ u Ω Γ u Ω 2 l C δ α γ Ω Ω 2 γ,l e γt Γ c Ω Γ c Ω 2 l C t 0 e γt e ɛ(t τ) N c (Ω ) N c (Ω 2 ) l dτ, here notice that Ω c and Ω c 2 have the same initial condition Ω c (0) = Ω c 2(0) The same argument as above leads to Thus (8) sup t 0 t e γt Γ c Ω Γ c Ω 2 l Cδ Ω Ω 2 γ,l e (γ ɛ)(t τ) dτ For z = s in (5), one has e γt Γ s Ω Γ s Ω 2 l C C δ γ ɛ Ω Ω 2 γ,l e γt Γ c Ω Γ c Ω 2 l C δ γ ɛ Ω Ω 2 γ,l t The same argument as before leads to Thus 0 0 e γt [β(t τ)] /2 e β(t τ) N s (Ω ) N s (Ω 2 ) l dτ t e γt Γ s Ω Γ s Ω 2 l Cδ Ω Ω 2 γ,l [β(t τ)] /2 e (β+γ)(t τ) dτ C δ β Ω Ω 2 γ,l (9) sup e γt Γ s Ω Γ s Ω 2 l C δ Ω Ω 2 γ,l t 0 β By choosing (20) δ = min (7)-(9) imply that { 6 C(α γ), 6 C(γ ɛ), } 6 C β ΓΩ ΓΩ 2 γ,l 2 Ω Ω 2 γ,l Thus Γ contracts The claim that Γ acts, ie Γ : B γ,l B γ,l ; follows similarly, and is an easier argument By the contraction mapping theorem, Γ has a fixed point Ω in B γ,l, which satisfies (4)-(5), of course, with N replaced by its cut-off Equations (4)-(5) define a map which maps Ω z (0) (z = c, s) into Ω u (0) = e τlu N u (Ω)dτ This map defines the center-stable manifold Due to the cut-off, only in the δ neighborhood of Ω, this center-stable manifold corresponds to the NS flow Smoothness can be proved through the standard argument [] This proves the theorem,

11 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 69 To study the zero-viscosity limits of invariant manifolds, we need to make an assumption Assumption As ν 0 +, Re{λ u } is of order O(), and Re{λ s } is O(ν), and the constant C in (7)-() is O() Theorem 6 Under Assumption, as ν 0 +, in an order O( ν) neighborhood of the fixed point Ω in the Sobolev space H l (T d ) (l, d = 2, ), there exist an unstable manifold and a center-stable manifold; and in its order O(ν) neighborhood, there exist a stable manifold, a center manifold, and a center-unstable manifold Proof Under the Assumption, the δ in (20) is of order O( ν) since β is O(ν) Thus we have the claim for the center-stable manifold By keeping track of similar estimates as in the proof of Theorem 5, one can quickly see the rest of the claims Due to Assumption, zero-viscosity limits of invariant manifolds become a challenging problem The challenge is further amplified by the elusiveness of the invariant manifolds for Euler equations as discussed in the Introduction As a test, we studied a simpler problem in [2], which reveals some unique features of the zero-viscosity limit of the spectra For instance, the zero-viscosity limit of a discrete spectrum can be a continuous spectrum which is not the spectrum of zero-viscosity 4 Examples Using Fourier series for the 2D NS (2), Ω = ω k e ik x, f = k Z 2 /{0} k Z 2 /{0} f k e ik x, where ω k = ω k and f k = f k, one gets the kinetic form of 2D NS ω k = A(q, r) ω q ω r + ν[ k 2 ω k + f k ], where k=q+r A(q, r) = 2 [ r 2 q 2] q r q 2 r 2 where q 2 = q 2 + q 2 2 for q = (q, q 2 ) T, similarly for r Linearize the 2D NS at the steady state given by a single mode, (4) Ω = Γe ip x + where f = p 2 Ω, one gets Γe ip x ωˆk+np = A(p, ˆk + (n )p) Γ ωˆk+(n )p, (42) + A( p, ˆk + (n + )p) Γ ωˆk+(n+)p ν ˆk + np 2 ωˆk+np, where its right hand side defines the linear NS operator L Let ωˆk+np = e λt ωˆk+np, Γ = Γ e iγ, a = 2 Γ p ˆk p, ρ n = p 2 ˆk + np 2, and z n = ρ n e inγ ωˆk+np, 2 ˆk2 then (4) a n z n + z n z n+ = 0,

12 70 Y CHARLES LI where a n = (aρ n ) (λ + ν ˆk + np 2 ) Let w n = z n /z n, one gets [6] (44) a n + w n = w n+, which leads to the continued fraction, (45) w n () = a n + a n 2 + a n + Rewriting (44) as, (46) w n = a n + w n+, which leads to the other continued fraction (47) w n (2) = a n + a n++ Along the spirit of the finite difference of a second order ordinary differential equation, the difference equation (4) should have two linearly independent solutions Rigorous theory has been well developed This is the so-called Poincaré-Perron system For details, see [2] [8] [9] [20] [7] [4] [6] [5] [9] [4] When ν > 0, a n ãn 2 as n + Then there are a growing solution z n /z n ãn 2 and a decaying solution z n /z n ã n 2, as n + ; and vice versa as n The intuition on this is clear from (44) and (46) The eigenvalues are then determined by matching the two decaying solutions given by the two continued fractions (45) and (47) When ν = 0, a n ã = a p 2 λ as n + When Re{ã} 0, or Re{ã} = 0 ( ã > 2), there are a growing solution z n /z n w +, ( w + > ), and a decaying solution z n /z n w, ( w < ), as n + ; and vice versa as n Here w ± solve the characteristic equation [9] w 2 ãw = 0 Again the eigenvalues are determined by matching the two decaying solutions given by the two continued fractions (45) and (47) Finally, Re{ã} = 0 ( ã 2) corresponds to a continuous spectrum [9] The proof of [9] can be generalized to H l setting for any natural number l For more general result, see [25] In summary, when ν > 0 or ν = 0 [Re{ã} = 0, or Re{ã} = 0 ( ã > 2)], the eigenvalues are determined by (48) a 0 + a + a 2+ = a +, a 2+ which is obtained by matching w () n (45) and w (2) n (47) at n =, ie w () = w (2) The corresponding eigenfunctions belong to H l for any natural number l 4 Example Our first prime example is the shear Ω = cos x 2 defined on the rectangular periodic domain [0, 2π/α] [0, 2π] where /2 < α < As shown below, this example has only one unstable eigenvalue Complete information on this eigenvalue can be obtained Thereby, existence of an unstable manifold and a stable manifold can be confirmed This example is motivated by examples 2 and studied later

13 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 7 In this rectangular domain case, Ω has the Fourier expansion Ω = ω k e i[αkx+k2x2] We have equation (4) with k Z 2 /{0} a n = 4 αˆk (αˆk ) 2 + (ˆk 2 + n) 2 (αˆk ) 2 + (ˆk 2 + n) 2 { [ λ + ν (αˆk ) 2 + (ˆk 2 + n) 2]} Theorem 4 The spectra of the 2D linear Euler operator L have the following properties () (αˆk ) 2 + (ˆk 2 + n) 2 >, n Z When ν 0, there is no eigenvalue of non-negative real part When ν = 0, the entire spectrum is the continuous spectrum [ ] i α ˆk 2, iα ˆk 2 (2) ˆk = 0, ˆk 2 = The spectrum consists of the eigenvalues λ = νn 2, n Z/{0} The eigenfunctions are the Fourier modes ω np e inx2 + cc, ω np C, n Z/{0} As ν 0 +, the eigenvalues are dense on the negative half of the real axis (, 0] Setting ν = 0, the only eigenvalue is λ = 0 of infinite multiplicity with the same eigenfunctions as above () ˆk =, ˆk 2 = 0 (a) ν > 0 For any α (05, 095), there is a unique ν (α), 2 α6 7α (49) 4 6α 2 α 2 4(α 2 + )(α 2 < ν (α) < + 4) 2(α 2, + ) 2 where the term under the square root on the left is positive for α (05, 095), and the left term is always less than the right term When ν > ν (α), there is no eigenvalue of non-negative real part When ν = ν (α), λ = 0 is an eigenvalue, and all the rest eigenvalues have negative real parts When ν < ν (α), there is a unique positive eigenvalue λ(ν) > 0, and all the rest eigenvalues have negative real parts ν λ(ν) is a strictly monotonically decreasing function of ν When α (05, 08469), we have the estimate α 2 ( α 2 ) 8(α 2 + ) α 4 (α 2 + ) 6(α 2 + )(α 2 + 4) ν(α2 + ) < λ(ν) α < 2 ( α 2 ) 8(α 2 + ) να2, where the term under the square root on the left is positive for α (05, 08469) α 2 ( α 2 ) 8(α 2 + ) α 4 (α 2 + ) 6(α 2 + )(α 2 + 4) lim α λ(ν) 2 ( α 2 ) ν 0 + 8(α 2 + )

14 72 Y CHARLES LI In particular, as ν 0 +, λ(ν) = O() (b) ν = 0 When α (05, 08469), we have only two eigenvalues λ 0 and λ 0, where λ 0 is positive, α 2 ( α 2 ) 8(α 2 + ) α 4 (α 2 + ) 6(α 2 + )(α 2 + 4) < λ 0 < α 2 ( α 2 ) 8(α 2 + ) The rest of the spectrum is a continuous spectrum [ iα/2, iα/2] (c) For any fixed α (05, 08469), (40) lim ν 0 + λ(ν) = λ 0 (4) Finally, when ν = 0, the union of all the above pieces of continuous spectra is the imaginary axis ir From Theorems 5 and 4, we have the corollary Corollary 42 For any α (05, 095), and ν (0, ν (α)) where ν (α) > 0 satisfies (49), in a neighborhood of Ω in the Sobolev space H l (T 2 ) (l ), there are an -dimensional C unstable manifold and an -codimensional C stable manifold Remark 4 In the Theorem 4, (40) verifies part of Assumption That is, Re{λ u } is of order O() as ν 0 + Case 2 of Theorem 4 indicates that Re{λ s } is at least O(ν) as ν 0 + Case indicates that the constant C in (7) and () should be O() as ν 0 + (4) Proof We will give the proof case by case () The case (a) ν > 0 If Re{λ} 0, then all the Re{a n } s are nonzero and have the same sign The real parts of the right and left hand sides of (48) are nonzero but of different signs Thus there is no eigenvalue of non-negative real part The case (b) ν = 0 If Re{λ} 0, then all the Re{a n } s are nonzero and have the same sign Again (48) can not be satisfied If Re{λ} = 0, let ã = lim n a n = 4 αˆk λ, then ãz n = (αˆk ) 2 + (ˆk 2 + n) 2 (αˆk ) 2 + (ˆk 2 + n) 2 (z n+ z n ), where (αˆk ) 2 + (ˆk 2 + n) 2 >, n Z By using l 2 norm of {z n } n Z, one sees that possible eigenvalues have to satisfy ã 2 As mentioned before, Re{λ} = 0 and ã 2 correspond to a continuous spectrum [9], which is the interval (2) In this case, one has The claims follow immediately () In this case, [ i α ˆk 2, iα ˆk 2 [λ + ν(n + ) 2 ] ω (n+)p = 0 a n = 4 α 2 + n 2 α α 2 + n 2 [λ + ν(α2 + n 2 )] ]

15 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 7 Thus a n = a n Equation (48) is reduced to (42) a 0 /2 = (4) The first a few a n s are α 2 a + a 2+ a 0 = 4 α α 2 [λ + να2 ], a = 4 α 2 + α α 2 [λ + ν(α 2 + )], a 2 = 4 α α α 2 + [λ + ν(α2 + 4)] Let f(λ) = a +, g(λ) = a 0 /2 a 2+ (a) ν > 0 First we show that if λ (Re{λ} 0) is a solution of (42), then λ must be real Assume Im{λ} > 0, then Thus But arg( a 0 ) > arg(a ) > arg(a 2 ) > 0 arg(f(λ)) arg(a ) arg(g(λ)) = arg( a 0 ) > arg(f(λ)), a contradiction Similarly for the case Im{λ} < 0, thus λ (Re{λ} 0) is real Next we show that there is a ν 0 > 0 such that for every ν ν 0, there is a unique eigenvalue λ > 0 Since a n > 0, n, we have a + < f(λ) < a 2 a Thus when λ is large enough (44) f(λ) < g(λ) From (4), We know that f(0) > 4ν(α 2 +) 2 α + α(α2 +) 4ν(α 2 +4) 2 g(0) = 2να α 2 We want to choose ν such that Thus (45) ν ν 0 (α) = 4ν(α 2 +) 2 α + α(α2 +) 4ν(α 2 +4) 2 2 α6 7α 4 6α 2 4(α 2 + )(α 2 + 4) where 095 < α 0 < and α 0 satisfies 2να α 2 2 α 6 7α 4 6α 2 = 0 (/2 < α < α 0 )

16 74 Y CHARLES LI For example, ν 0 (05) 0244, ν 0 (095) 0029 For each fixed ν, ν ν 0 (α), we have (46) f(0) > g(0) (47) (48) From (44) and (46), we see that there is a λ > 0 such that (42) is true Next, we want to show that this eigenvalue is unique Notice that (λ + να 2 ) g(λ) = (λ + να 2 ) f(λ) = 2α α 2, (λ + να 2 )a + (λ+να 2 ) a 2+ Since (λ+να 2 )a 2n+ (n 0) is a strictly monotonically increasing function of λ for λ > 0, and (λ + να 2 ) a 2n (n ) is a strictly monotonically decreasing function of λ for λ > 0, we see that (λ+να 2 ) f(λ) is a strictly monotonically decreasing function of λ for λ > 0 Thus the eigenvalue which satisfies (λ + να 2 ) f(λ) = (λ + να 2 ) g(λ) is unique Similarly, να 2 a 2n+ (n 0) is a strictly monotonically increasing function of ν for ν > 0, and (να 2 ) a 2n (n ) is a constant function of ν Then (να 2 ) f(0) is a strictly monotonically decreasing function of ν for ν > 0 We know from above that when ν = ν 0 (α), (να 2 ) f(0) > (να 2 ) g(0) = From (4), we have (να 2 ) f(0) < thus when ν > 0 is large enough 2α, constant in ν α2 α (να 2 = )a 4(α 2 + ) 2 ν 2, (να 2 ) f(0) < (να 2 ) g(0) Therefore, there is a unique ν (α) > ν 0 (α), such that (49) (ν α 2 ) f(0) = (ν α 2 ) g(0) Thus we have shown the following claims: When ν > ν (α), there is no eigenvalue of non-negative real part When ν = ν (α), λ = 0 is an eigenvalue, and all the rest eigenvalues have negative real parts When ν < ν (α), there is a unique positive eigenvalue λ(ν) > 0, and all the rest eigenvalues have negative real parts An estimate for ν can be obtained from (4), We have < 2α 4(α 2 +) 2 α ν 2 + α (α 2 +) α2 < 4(α 2 +4) 2 α ν 0 (α) < ν (α) < 4(α 2 +) 2 ν 2 α 2 2(α 2 + ) 2,

17 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 75 (420) (42) where ν 0 (α) is given by (45) For example, < ν (05) < , 0029 < ν (095) < 0058 Next we want to show that the unique eigenvalue λ(ν) has the property that ν λ(ν) is a strictly monotonically decreasing function of ν Notice that (λ + να 2 )a 2n+ = 4 α 2 + (2n + ) 2 ( ) λ α α 2 + (2n + ) 2 ν2 ν + α2 ( ) λ ν + [α2 + (2n + ) 2 ], (n 0), [ ] (λ + να 2 ) a 2n = 4 α 2 + (2n) 2 α α 2 + (2n) 2 + (2n)2 λ ν +, (n ) α2 Assume that ν λ(ν) is not a strictly monotonically decreasing function of ν, then there is an interval in which ν λ(ν) is a strictly monotonically increasing or a constant function of ν In that interval, [λ(ν) + να 2 ]a 2n+ (n 0) is strictly monotonically increasing, and [λ(ν)+να 2 ] a 2n (n ) is monotonically decreasing Thus [λ(ν) + να 2 ] f(λ(ν)) is a strictly monotonically decreasing function of ν On the other hand, [λ(ν) + να 2 ] g(λ(ν)) = 2α α 2 is a constant function of ν This contradiction shows that ν λ(ν) is a strictly monotonically decreasing function of ν Next we want to show that λ(ν) = O() as ν 0 + From (42) and (4), we have < 4(α 2 +)[λ+ν(α 2 +)] α(α α + 2 +) 4(α 2 +4)[λ+ν(α 2 +4)] 2α α 2 (λ + α να2 ) < 4(α 2 + )[λ + ν(α 2 + )] Let ν 0 +, we have α 2 ( α 2 ) 8(α 2 + ) α 4 (α 2 + ) 6(α 2 + )(α 2 + 4) λ α 2 ( α 2 ) 8(α 2 + ) For the term under the square root on the left to be positive, we need 59 (422) α < α = (42) Thus when α (/2, α ), λ(ν) = O(), as ν 0 + Also from (420), we have the estimate < α 2 ( α 2 ) 8(α 2 + ) α 4 (α 2 + ) 6(α 2 + )(α 2 + 4) ν(α2 + ) < λ(ν) α 2 ( α 2 ) 8(α 2 + ) να2

18 76 Y CHARLES LI (b) ν = 0 We have a n = 4 α 2 + n 2 α α 2 + n 2 λ As n, a n ã = 4 α λ As before, Re{λ} = 0 and λ α 2 corresponds to a continuous spectrum Next we show that outside the disc λ α 2, there is no eigenvalue Outside the disc, ã > 2, a 0 > 2 α2 α 2 > 2, for α (/2, ), a > 2 ( + α ) 2 > 4, for α (/2, ), (424) and a n > 2, n 2 By a result of continued fraction [5], a 2 + Thus while a + g(λ) >, f(λ) a a 2+ a + < This contradiction proves the claim Thus the possible eigenvalues should lie in the region Re{λ} 0 and λ α 2 Since the eigenvalues are symmetric with respect to the imaginary axis [9], we only need to consider the case Re{λ} > 0 First we show that the possible eigenvalue must be real Assume that Im{λ} > 0, then Thus Then Therefore arg( a 0 ) = arg(a ) = arg(a 2 ) = > 0 ( ) arg a 2 + arg(a ) a + ( ) arg a + a 2 + arg(a ) a + arg(f(λ)) arg(a ) = arg( a 0 ) = arg(g(λ)) This is a contradiction The case Im{λ} < 0 is similar, and the claim is proved When λ > 0, λa + < λ f(λ) <, λ a 2 λa ie 4(α 2 +) α λ 2 + α(α2 +) 4(α 2 +4) < λ f(λ) < 4(α 2 +) α λ 2

19 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 77 (425) Thus lim λ + λ f(λ) = 0, lim λ 0 λ f(λ) 4(α2 + 4) + α(α 2 + ) Notice that λ g(λ) = 2α α 2 We want to choose α such that 2α α 2 < 4(α2 + 4) α(α 2 + ) This condition is equivalent to the fact that the term under the square root on the left of (42) being positive Thus we have 59 α < α = , which is the same with (422) Thus when α (/2, α ), there is a positive eigenvalue λ 0, λ 0 f(λ 0) = λ 0 g(λ 0) = 2α α 2 Since λa 2n+ (n 0) is a strictly monotonically increasing function of λ for λ > 0, and λ a 2n (n ) is a constant function of λ, we see that λ f(λ) is a strictly monotonically decreasing function of λ for λ > 0 Thus the positive eigenvalue λ 0 is unique From (424), we have ie 4(α 2 +) α λ α(α2 +) 4(α 2 +4) < 2α α 2 < 4(α 2 +) α λ 2 0 α 2 ( α 2 ) 8(α 2 + ) α 4 (α 2 + ) 6(α 2 + )(α 2 + 4) < λ α 0 < 2 ( α 2 ) 8(α 2 + ), which is similar to (42) Next we want to show that (426) lim ν 0 + λ(ν) = λ 0 Let F (ν, λ) = f(λ) g(λ), and let F N (ν, λ) be the N-th truncation of F (ν, λ), F N (ν, λ) = a a + a 2+ + an Notice that F (ν, λ(ν) = 0, ν (0, ν (α)), α (05, 08469) ; F (0, λ 0 ) = 0, α (05, 08469) ; where λ(ν) and λ 0 have the estimates (42) and (425) Thus for any fixed α (05, 08469), (ν, λ(ν)) lies in a compact set [0, δ] [c, c 2 ] where δ, c and c 2 are positive constants Assume that (426) is not true, then there is a sequence (ν j, λ(ν j )) such that lim λ(ν j ) = λ 0, ν j 0 +,

20 78 Y CHARLES LI where λ 0 λ 0 and λ 0 [c, c 2 ] By a result of continued fraction [6], lim F N(ν, λ) = F (ν, λ) N uniformly for ν [0, δ] and λ [c, c 2 ] Thus for any ɛ > 0, there is an integer N 0 such that for any N N 0, ie F N (ν j, λ(ν j )) F (ν j, λ(ν j )) < ɛ, F N (ν j, λ(ν j )) < ɛ, j For any fixed N, let j +, we have Let N +, we get F N (0, λ 0 ) ɛ F (0, λ 0 ) ɛ Since ɛ is arbitrarily small, we have F (0, λ 0 ) = 0 By the uniqueness of the eigenvalue, λ 0 = λ 0 This contradiction shows that (426) is true (4) When ν = 0, the union of all the continuous spatra [ i α ˆk 2, iα ˆk 2 ], is ir 42 Example 2 Our second prime example is Ω = cos(x + x 2 ) which corresponds to p = (, ) T and Γ = /2 in (4) Here (427) a n = 8 (ˆk + n) 2 + (ˆk 2 + n) 2 (ˆk 2 ˆk ) (ˆk + n) 2 + (ˆk 2 + n) 2 2 { [ λ + ν (ˆk + n) 2 + (ˆk 2 + n) 2]} This example has two possible unstable eigenvalues For shears, the number of unstable eigenvalues of 2D linear Euler is bounded by the number of lattices points inside the disc of radius p [8], which is even for square lattice In [] [4], Vincent Liu studied another shear which has three possible unstable eigenvalues Liu did a detailed calculation on the eigenvalues, which will be discussed in next subsection Here we will do an even more detailed calculation Theorem 44 The spectra of the 2D linear Euler operator L have the following properties

21 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 79 () The set {ˆk + np} n Z has no intersection with the disc of radius p = 2 When ν > 0, there is no eigenvalue of non-negative real part When ν = 0, the entire spectrum is the continuous spectrum [ i 4 ˆk 2 ˆk, i ] 4 ˆk 2 ˆk (2) ˆk = p = (, ) T The spectrum consists of the eigenvalues λ = ν2n 2, n Z/{0} The eigenfunctions are the Fourier modes ω np e inp x + cc, ω np C, n Z/{0} As ν 0 +, the eigenvalues are dense on the negative half of the real axis (, 0] Setting ν = 0, the only eigenvalue is λ = 0 of infinite multiplicity with the same eigenfunctions as above () ˆk = (, ) T When ν > 0, there is no eigenvalue of non-negative real part When ν = 0, the entire spectrum is the continuous spectrum [ i 2, i 2 ] A special eigenvalue is λ = 2ν ( when ν = 0, this eigenvalue λ = 0 is embedded in the continuous spectrum) (4) ˆk = (0, ) T When ν > 0, in the half plane Re{λ} ν, there is a unique pair of eigenvalues λ and λ such that ν < Re{λ} < 4 ( 8 ) < Im{λ} < (8ν)2 2ν, ( ) + 5 When ν = 0, [ i 4, i 4 ] is a continuous spectrum If there is an eigenvalue of positive real part, then there is a quadruplet (λ, λ λ λ) where 0 < Re{λ} < ( 8 6 ) < Im{λ} < 5 8 5, ( + ) 5 (5) Finally, when ν = 0, the union of all the above pieces of continuous spectra is the imaginary axis ir Remark 45 For Case 4, numerical computation indicates that when ν > 0, the real part of the eigenvalue Re{λ(ν)} c > 0, where c is independent of ν When ν = 0, numerical computation indicates that the eigenvalue λ(0) indeed exists and Re{λ(0)} c > 0 (the same c as above) Numerical computation also indicates that as ν 0 +, λ(ν) λ(0) which can be proved given the above facts Cases 2 and indicate that Re{λ s } in Assumption is at least O(ν) as ν 0 + Proof We will give the proof case by case () The case (a) ν > 0 If Re{λ} 0, then all the Re{a n } s are nonzero and have the same sign The real parts of the right and left hand sides of (48) are nonzero but of different signs Thus there is no eigenvalue of non-negative real part The case (b) ν = 0 If Re{λ} 0, then all

22 80 Y CHARLES LI the Re{a n } s are nonzero and have the same sign Again (48) can not be satisfied If Re{λ} = 0, let ã = lim n a n = 8 ˆk 2 ˆk λ, then ãz n = ˆk + np 2 2 ˆk + np 2 (z n+ z n ), (428) where 0 < ˆk +np 2 2 < ˆk +np 2, n Z By using l 2 norm of {z n } n Z, one sees that possible eigenvalues have to satisfy ã 2 As mentioned before, Re{λ} = 0 and ã 2 correspond to a continuous spectrum [9], which is the interval [ i 4 ˆk 2 ˆk, i ] 4 ˆk 2 ˆk (2) In this case, one has [λ + ν2(n + ) 2 ] ω (n+)p = 0 The claims follow immediately () In this case, denote ωˆk+np simply by ω n, one has (429) [λ + 2ν(n 2 + )]ω n = (n + ) 2 4 (n + ) 2 + ω n+ (n ) 2 4 (n ) 2 + ω n This system decouples at n = 0 A few equations around n = 0 are (λ + 0ν)ω 2 = 8 ω 9 40 ω, (λ + 4ν)ω = 5 ω 2, (40) Notice that (λ + 2ν)ω 0 = 8 ω 8 ω, (λ + 4ν)ω = 5 ω 2, (λ + 0ν)ω 2 = 9 40 ω 8 ω a n = 4 ( + n ) 2 [λ + 2ν(n 2 + )] From (40), one gets (4) w 2 = z 2 /z = 8(λ + 4ν) From (45)and (4), one has (42) 8(λ + 4ν) = a 2 + a + (a) ν > 0 If Re{λ} 0, then Re{a n } > 0 for n 2 Thus the real part of the right hand side of (42) is negative, while the real part of its left hand side is positive Thus there is no eigenvalue of non-negative real part (b) ν = 0 If Re{λ} 0, then Re{a n } (n 2) has the same

23 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 8 fixed sign with λ, again (42) leads to a contradiction If Re{λ} = 0, let ã = lim n a n = 4λ, then ãz = 2 z 2, n 2 ãz n = n 2 + (z n+ z n ), n 2 By using l 2 norm of {z n } n, one sees that possible eigenvalues have to satisfy ã 2 As mentioned before, Re{λ} = 0 and ã 2 correspond to a continuous spectrum [9], which is the interval [ i 2, i 2 ] Similarly from (429), one gets (4) w = z /z 2 = 8(λ + 4ν) From (45)and (4), one has 8(λ + 4ν) = a 2 + a + a 4+ Notice that a n = a n, one gets the same conclusion as above Finally, by choosing ω n = 0 if n 0, one gets the eigenvalue λ = 2ν with the eigenfunction (4) In this case, ωˆke iˆk x + cc, ωˆk C a n = 8 n2 + (n + ) 2 { [ λ + ν n 2 n 2 + (n + ) 2 + (n + ) 2]} 2 Thus a (n+) = a n Equation (48) is reduced to (44) a 0 + The first a few a n s are a + a 2+ = ±i a 0 = 8 ( ) (λ + ν), a = 8 ( 5 ) (λ + 5ν), a 2 = 8 ( ) (λ + ν), a = 8 ( 25 ) (λ + 25ν) 2 (a) ν > 0 Consider the region Re{λ} ν, in which a n 8 λ + ν, n 0 By using l 2 norm of {z n } n Z in (4), one sees that possible eigenvalues in this region have to satisfy λ + ν 4 Next in the region { D = λ Re{λ} ν, λ + ν }, 4

24 82 Y CHARLES LI (45) we will use Rouché s theorem to track the eigenvalues Let f(λ) = a 0 + a + a 2+ g(λ) = a 0 + a + i + i, To apply the Rouché s theorem, one needs to show that (46) f(λ) g(λ) < f(λ) + g(λ), λ D If this is not true, then there is a λ D and a δ (0 δ ) such that ie (47) a 0 + i + f(λ) = δg(λ), +δ a + a 2+ + δ = 0 + δ a On the part of the boundary D: Re{λ} = ν, Re{a 0 } = 0, Re{a n } > 0, n By taking the real part of (47), one sees that (47) can not be satisfied On the other part the boundary D: λ + ν = 4, a 0 = 2, a 0, a 2 26, a n 2, n By a result of continued fraction [5], a + Thus +δ a + a 2+ a 4+ a a 2+ a + a a 2 a + a = 26 The rest of (47) has the estimate: a 0 + i + δ + δ a a 0 2 δ + δ δ + δ a

25 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 8 Therefore, (47) can not be satisfied By Rouché s theorem, inside D, f(λ) and g(λ) have the same number of roots g(λ) has two roots λ = ν + 6 i ± (8ν)2 + i64ν, and one of which is in D (at least when ν is small enough) Thus f(λ) has one root in D From (45), this root satisfies { (48) Re a + } < 0, a 0 + i which leads to ν < Re{λ} < (8ν)2 2ν, ( ) ( < Im{λ} < ) (b) ν = 0 Then a n 8 λ, n 0 Again by using l 2 norm of {z n } n Z in (4), one sees that possible eigenvalues in this region have to satisfy λ 4 In fact, if there is an eigenvalue with positive real part, then (48) is true, which leads to 0 < Re{λ} < 6 5, ( ) ( < Im{λ} < ) As proved in [9], such eigenvalues (if exist) come in quadruplet (λ, λ λ λ) As in Case above, [ i 4, i 4 ] is a continuous spectrum [9] (5) When ν = 0, the union of all the continuous spectra [ i 4 ˆk 2 ˆk, i ] 4 ˆk 2 ˆk, is ir 4 Example An example studied in details by Vincent Liu [] [4] is Ω = 2 π cos(2x 2) which corresponds to p = (0, 2) T and Γ = 2π in (4) Here a n = 4 ˆk 2 + (ˆk 2 + 2n) 2 2π ] ˆk [ˆk2 + (ˆk 2 + 2n) 2 4 { λ + ν [ˆk2 + (ˆk 2 + 2n) 2]} Theorem 46 The spectra of the 2D linear Euler operator L have the following properties

26 84 Y CHARLES LI () The set {ˆk + np} n Z has no intersection with the disc of radius p = 2 When ν > 0, there is no eigenvalue of non-negative real part When ν = 0, the entire spectrum is the continuous spectrum [ i ˆk 2 2π, i ˆk 2 2π (2) ˆk = (0, ) T or (0, 2) T The spectrum consists of the eigenvalues λ = ν[ˆk 2 + (ˆk 2 + 2n) 2 ], n Z, where for ˆk = (0, 2) T, n The eigenfunctions are the Fourier modes ωˆk+np e i(ˆk+np) x + cc, ωˆk+np C, n Z As ν 0 +, the eigenvalues are dense on the negative half of the real axis (, 0] Setting ν = 0, the only eigenvalue is λ = 0 of infinite multiplicity with the same eigenfunctions as above () ˆk = (2, 0) T When ν > 0, there is no eigenvalue of non-negative real part When ν = 0, the entire spectrum is the continuous spectrum [ i 2π, i 2π ] A special eigenvalue is λ = 4ν (when ν = 0, this eigenvalue λ = 0 is embedded in the continuous spectrum) (4) ˆk = (, 0) T There is a unique ν, 89 0π 4 < ν < 0π When ν > ν, there is no eigenvalue of non-negative real part When ν = ν, λ = 0 is an eigenvalue, and all the rest eigenvalues have negative real parts When ν < ν, there is a unique positive eigenvalue λ(ν) > 0, and all the rest eigenvalues have negative real parts ν λ(ν) is a strictly monotonically decreasing function of ν λ(ν) has the estimate 89 π 680 5ν < λ(ν) < π 20 ν In particular, as ν 0 +, λ(ν) = O() When ν = 0, we have only two eigenvalues λ 0 and λ 0, where λ 0 is positive 89 π 680 < λ 0 < π 20 ] More- The rest of the spectrum is a continuous spectrum [ i 2, i 2π over, lim λ(ν) = λ 0 ν 0 + ] 2 2π (5) ˆk = (, ) T When ν > 0, in the half plane Re{λ} 2ν, there is a unique pair of eigenvalues λ and λ such that 2ν < Re{λ} < π 40, 5 2 0π < Im{λ} < π

27 INVARIANT MANIFOLDS FOR NAVIER-STOKES EQUATIONS 85 When ν = 0, [ i 2, i 2π 2 ] is a continuous spectrum If there is an 2π eigenvalue of positive real part, then there is a quadruplet (λ, λ λ λ) where 0 < Re{λ} < π 40, π < Im{λ} < 2 0π (6) Finally, when ν = 0, the union of all the above pieces of continuous spectra is the imaginary axis ir Remark 47 Except for Case 4, the rest of the proof is similar to that of Theorem 44 Overall, Liu [] [4] did not realize the continuous spectrum For Case 4, Liu [] [4] gave an elegant proof For Case 5, Liu [] [4] claimed more than what were actually proved In [4], the sketch after line 9, pp472 can be realized as shown in the proof of Case 4 of Theorem 44 above But Equation (9) on the same page does not imply Re{η (ν)} > 0 in Equation (5) on pp467 Arguments between lines 5 and 8 on pp48 are not solid, therefore, the existence of η in Part B of Theorem 2 on pp467 was not proved The sketch between lines 22 and 25 on pp484 is not completed, therefore, (4) in Theorem on pp468 was not proved Proving these claims seems tricky Remark 48 Cases 2 and indicates that Re{λ s } in Assumption is at least O(ν) as ν 0 + Case 4 shows the existence of an unstable eigenvalue In fact, by the property of continued fraction, the corresponding eigenfunction Ω u (ν) also converges as ν 0 + Thus Ω u l+ / Ω u l has a uniform bound as ν 0 + Remark 49 In an effort to minimize the number of unstable modes to one and to have Case 4 not Case 5 in the above theorem, we need to study the rectangular periodic domain [0, 2π/α] [0, 2π] where /2 < α <, and the steady state Ω = cos x 2 This is our Example studied before References [] R Adams, Sobolev Space, Academic Press, New York, 975 [2] L Bers, F John, M Schechter, Partial Differential Equations, Interscience, New York, 964 [] N Dunford, J Schwartz, Linear Operators, Part II, Interscience, New York, 96 [4] W Gautschi, Computational aspects of three-term recurrence relations, SIAM Review 9 (967) [5] D Henry, Geometric Theory of Semilinear Parabolic Equations, Lecture Notes in Mathematics, 840, Springer-Verlag, 98 [6] W B Jones, W J Thron, Continued Fractions, Analytic Theory and Applications, Encyclopedia of Math and Its Appl, 980 [7] T Kato, Quasi-linear equations of evolution, with applications to partial differential equations, Lecture Notes in Math, Springer 448 (975), 25 [8] Y Latushkin, Y Li, and M Stanislavova, The spectrum of a linearized 2D Euler operator, Studies in Applied Mathematics 2 (2004), 259 [9] Y Li, On 2D Euler equations I On the energy-casimir stabilities and the spectra for linearized 2D Euler equations, J Math Phys 4, no2 (2000), 728 [0] Y Li, On 2D Euler equations: II Lax pairs and homoclinic structures, Commun Appl Nonlinear Analysis 0, no (200), [] Y Li, Persistent homoclinic orbits for nonlinear Schrödinger equation under singular perturbations, Dynamics of PDE, no (2004), 87 [2] Y Li, Zero dispersion and viscosity limits of invariant manifolds for focusing nonlinear Schrödinger equations, Submitted to J Math Anal Appl (2005)

28 86 Y CHARLES LI [] V Liu, An example of instability of the Navier-Stokes equations on the 2-dimensional torus, Comm PDE 7 (992), 995 [4] V Liu, On unstable and neutral spectra of incompressible inviscid and viscid fluids on the 2D torus, Quart Appl Math LIII, No (995), 465 [5] L Lorentzen, H Waadeland, Continued Fractions with Applications, North-Holland, Amsterdam, 992 [6] L D Meshalkin and Ya G Sinai, Investigation of the stability of a stationary solution of a system of equations for the plane movement of an incompressible viscous liquid, J Appl Math Mech (PMM) 25 (96), 700 [7] L M Milne-Thomson, The Calculus of Finite Differences, Macmillan, London, 9 [8] O Perron, Über einen Satz der Herren Poincaré, J Reine Angew Math 6 (90), 7 [9] O Perron, Über die Poincarésche lineare Differenzengleichung, J Reine Angew Math 7 (90), 6 [20] O Perron, Uber linear Differenzengleichungen, Acta Math 4 (9), 09 [2] H Poincaré, Sur les équations linéaires aux différentielles ordinaires et aux différences finies, Amer J Math 7 (885), 20 [22] G Prodi, Teoremi di tipo locale per il sistema di Navier-Stokes e stabilita delle soluzioni stazionarie, Rend Sem Mat Univ Padova 2 (962), 74 [2] M Reed, B Simon, Methods Modern Mathematical Physcis, IV, Academic Press, NY, 978 [24] D Sattinger, The mathematical problem of hydrodynamic stability, Journal of Mathematics and Mechanics (Indiana Univ Math J) 9, No9 (970), 797 [25] R Shvidkoy, Y Latushkin, The essential spectrum of the linearized 2D Euler operator is a vertical band, Contemp Math 27 (200), 299 [26] J Wu, The inviscid limits for individual and statistical solutions of the Navier-Stokes equations, PhD Thesis, Chicago University (996) Department of Mathematics, University of Missouri, Columbia, MO 652 address: cli@mathmissouriedu