Strong instability of standing waves for Hartree equation with harmonic potential

Physica D 237 (2008) 998–1005 www.elsevier.com/locate/physd

Strong instability of standing waves for Hartree equation with harmonic potential Yanjin Wang Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba, Meguro, Tokyo, 153-8914, Japan Received 6 August 2007; received in revised form 8 November 2007; accepted 28 November 2007 Available online 3 December 2007 Communicated by J. Bronski

Abstract This paper is concerned with the standing waves eiωt φ of Hartree equation with harmonic potential. Via construction of a cross-constrained invariant set for Hartree equation with harmonic potential, it is shown that if the initial data is in the invariant set then the corresponding solution blows up in finite time. Under an appropriate assumption on the frequency ω, the strong instability of standing waves is investigated by the above blow-up result. Furthermore, without the assumption we also study the property of the standing waves for Hartree equation with harmonic potential. c 2007 Elsevier B.V. All rights reserved.

Keywords: Hartree equation; Harmonic potential; Standing waves; Blow-up; Strong instability

1. Introduction The nonlinear Schr¨odinger equation arises in various physical contexts in the description of nonlinear waves [20] such as propagation of a laser beam in a medium whose index of refraction is sensitive to the wave amplitude, water waves at the free surface of an idea fluid, and plasma waves. In this paper we consider the following nonlinear Schr¨odinger equation with nonlocal interaction and harmonic potential  iu t + 1u = |x|2 u − (V ∗ |u|2 )u (1.1) u(0, x) = u 0 (x) where 1 is Laplacian operator on Rn (n ≥ 3), u : R + × Rn → C is a complex-valued function, and V ∗ |u|2 =

Z

|u(t, y)|2 dy, λ Rn |x − y|

(1.2)

where 0 < λ < n. The above equation with λ = 1 and n = 3 is equivalent to the Schr¨odinger–Poisson system with harmonic

E-mail addresses: [email protected], wang [email protected]. c 2007 Elsevier B.V. All rights reserved. 0167-2789/$ - see front matter doi:10.1016/j.physd.2007.11.018

potential,  iu t + 1u = |x|2 u − V (x)u, 1V = |u|2 ,  u(0, x) = u 0 , which arises typically if we consider the quantum mechanical time evolution of electrons in the mean field approximation of the many body effects, modeled by the Poisson equation, with a confinement modeled by the quadratic potential of the harmonic oscillator. Recently the classical limit of Eq. (1.1) was studied by Carles, Mauser and Stimming [3]. And if the harmonic potential vanishes then this equation models the classical limit of the field equations describing quantum mechanical nonrelativistic many boson system [11]. If we replace (V ∗ |u|2 )u by |u| p−1 u then Schr¨odinger equation (1.1) has been extensively studied, and there exists a large amount of literature on the Cauchy problem of the nonlinear Schr¨odinger equation (see, e.g., [2,4,15] and the references therein). Especially, the blow-up and global existence of solutions of Schr¨odinger equation were studied in [5,17,23]. And the stability and instability of standing waves were also studied in [7–9]. Now we return to the Hartree equation (1.1). For Eq. (1.1) without harmonic potential, there are lots of works on the Cauchy problem and asymptotic behavior of the solutions

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Y. Wang / Physica D 237 (2008) 998–1005

(see, e.g., [4,10] and the references therein). For the Cauchy problem of the Hartree equation with harmonic potential, we refer the readers to book [4] and the references therein. And there are some results on standing waves of Hartree equation with Coulomb type potential (see, e.g., [13]). More recently, the stability and instability of the standing waves of the Eq. (1.1) were studied without harmonic potential in [12,6]. But as we know, there is no result on the standing waves for Hartree equation with harmonic potential as Eq. (1.1). The main purpose of this paper is to investigate the strong instability of standing waves for Eq. (1.1). The tool of the proof is variational calculus, which originates in Berestycki and Cazenave [1], Weinstein [19], Zhang [22]. We first obtain the existence of standing waves of Eq. (1.1) with the ground state by a compact lemma and a standard argument of variational calculus. Then we establish another auxiliary variational problem, which helps us to construct a crossconstrained invariant set under the flow generated by Cauchy problem (1.1). We next derive a blow-up result with the initial data in the cross-constrained invariant set. At last, under an appropriate assumption on the frequency, we obtain the main result on strong instability of standing waves for Eq. (1.1). Without the assumption on the frequency we also obtain a property of the standing waves for the Eq. (1.1). This paper is organized as follows: In the next section, we introduce some basic results on the local existence of the Eq. (1.1) and some functional spaces. In Section 3, we establish several variational problems, in particular we obtain the existence of standing waves of Eq. (1.1). In Section 4, an invariant set is first introduced under the flow generated by Eq. (1.1), then a blow-up result is derived with the initial data in the invariant set. In Section 5, under an appropriate assumption we investigate the strong instability of standing waves by the blowup result obtained in Section 4, the property of the standing waves without the assumption on the frequency is also studied. In the last section, we make some remarks on strong instability of standing waves for Eq. (1.1). 2. Preliminaries We define the following space for Eq. (1.1)   Z Σ := φ ∈ H 1 (Rn ); |x|2 |φ(x)|2 dx < ∞ , Rn

where H 1 (Rn ) is the Sobolev space of complex-valued functions. Here Σ is a Hilbert space endowed with the inner product Z ¯ ¯ (φ, ψ)Σ := (∇φ(x) · ∇ ψ(x) + |x|2 φ(x)ψ(x) Rn

¯ + φ(x)ψ(x))dx, whose norm we denote by k · kΣ , and it is continuously embedded in H 1 (Rn ). And we define the Lebesgue space L q as   Z q q 0 q L = φ ∈ S ; kφkq = |φ(x)| dx < ∞ Rn

for q ≥ 1.

Then we have the following compactness lemma (see [16] and [21]). Lemma 2.1. The embedding Σ ,→ L q+1 is compact, where 1 ≤ q < 1 +

4 n−2 .

The following lemma is well-known as Hardy–Littlewood– Sobolev inequality (see, e.g., [18]). Lemma 2.2. Let 1 < r, s < ∞ and 0 < λ < n satisfying that 1 λ 1 r + n = 1 + s . Then

Z

f (y)

| · −y|λ dy ≤ Ck f kr s for some constant C > 0. From Theorem 4.4.6 and Remark 4.4.8 of [4] we have the following local well-posedness for the Cauchy problem (1.1) in the space Σ . Theorem 2.3. Let u 0 ∈ Σ and 0 < λ < min{4, n}. Then there exists a unique solution u of the Cauchy problem (1.1) in C([0, T ); Σ ) for time T ∈ (0, ∞] (maximal existence time) with the following property: either T = ∞ or else T < ∞ and limt→T − k∇u(t, ·)k2 = ∞, where t → T − means that t → T and t < T . Moreover, we have the conservation laws of the mass Z Z 2 |u(t, x)| dx = |u 0 (x)|2 dx (2.1) Rn

Rn

and energy E(t) = E(0) for every t ∈ [0, T ), where Z  1 E(t) = |∇u(t, x)|2 + |x|2 |u(t, x)|2 2 Rn  1 2 2 − (V ∗ |u| )|u(t, x)| dx. 2

(2.2)

(2.3)

We also have the following identities, which are satisfied by the solution of Eq. (1.1). Lemma 2.4. Let u be a solution of the Hartree equation (1.1) with the initial data u 0 as in Theorem 2.3. If the initial data u 0 also satisfies | · |u 0 ∈ L 2 , then, for 0 < λ < min{4, n} and n ≥ 3, the corresponding solution u of Eq. (1.1) satisfies the following identities Z Z d |x|2 |u(t, x)|2 dx = 4= u(t, ¯ x)∇u(t, x) · xdx (2.4) dt Rn Rn and Z d2 |x|2 |u(t, x)|2 dx dt 2 Rn Z =8 (|∇u(t, x)|2 − 2|x|2 |u(t, x)|2 )dx Rn

 Z λ |u(t, x)|2 |u(t, y)|2 − dydx , 4 Rn |x − y|λ

(2.5)

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where =φ means the imaginary part of the complex-valued function φ. By a similar way as in [14], we can prove this lemma. Here we omit the proof. Remark 2.5. From (2.2) and (2.5), when the initial data with E(0) < 0 satisfies the conditions in Lemma 2.4, we see that the corresponding solution of Eq. (1.1) with 2 < λ < min{n, 4} and n ≥ 3 blows up in finite time T , that is to say, lim k∇u(t, ·)k22 = ∞.

t→T −

(2.6)

The above definition of θ comes from I (φ) = 0 and φ ∈ Σ . But in order to study the strong instability of standing waves, we need the conditions λ > 2 and ω > 0. Thus in the following part of the present paper we always make the assumptions, 2 < λ < min{4, n} and ω > 0. Next in order to obtain the strong instability, we need to introduce the following set: K = {φ ∈ Σ ; I (φ) = 0 and L(φ) = 0, φ 6= 0}, where Z L(φ) = Rn

To conclude this section, we introduce the following definition: Definition 2.6. We say that a standing wave eiωt φ(x) is strongly unstable by blow-up, if for any δ > 0, there exists u 0 ∈ Σ such that ku 0 − φkΣ < δ and the solution u(t) of Eq. (1.1) with the initial data u 0 blows up in finite time. 3. Variational problems By a standing wave, we mean a solution of Eq. (1.1) with the form u(t, x) = eiωt φ(x), where ω ∈ R is a given parameter and φ is a ground state of the following stationary problem:  −1φ + ωφ + |x|2 φ − (V ∗ |φ|2 )φ = 0, x ∈ Rn (3.1) φ ∈ Σ , φ 6= 0. We next define Z 1 J (φ) = (|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx 2 Rn Z |φ(x)|2 |φ(y)|2 1 dydx, (3.2) − 4 Rn ×Rn |x − y|λ Z I (φ) = (|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx RnZ |φ(x)|2 |φ(y)|2 − dydx, (3.3) |x − y|λ Rn ×Rn

(|∇φ(x)|2 − 2|x|2 |φ(x)|2 )dx

Z |φ(x)|2 |φ(y)|2 λ − dydx. 4 Rn ×Rn |x − y|λ We now show the following lemma:

Lemma 3.2. Let 2 < λ < min{4, n} and n ≥ 3. Then, for ω > 0, K is not empty, that is to say, there exists φ ∈ Σ such that I (φ) = 0, L(φ) = 0 and φ 6= 0. Proof. By Theorem 3.1 we see that there exists a nonzero φ ∈ Σ satisfying Eq. (3.1). Multiplying (3.1) by x · ∇φ and integrating over Rn , we then have Z (−1φ + ωφ + |x|2 φ)x · ∇φdx Rn

Z −

|φ(y)|2 φ(x)x · ∇φ(x)dydx = 0. λ Rn ×Rn |x − y|

(−1φ + ωφ + |x|2 φ)x · ∇φdx Z  1 =− (n − 2)|∇φ(x)|2 + ω|φ(x)|2 2 Rn  + (n + 2)|x|2 |φ(x)|2 dx.

Rn

After a direct calculation, we also have 2

M = {φ ∈ Σ ; I (φ) = 0 and φ 6= 0}.

(3.4)

Theorem 3.1. Let 2 < λ < min{4, n} and n ≥ 3. Then for ω > 0 there exists Φ ∈ M such that d0 = J (Φ) = min {J (φ)}. φ∈M

(3.5)

Furthermore, such a Φ is a ground state solution of Eq. (3.1). Theorem 3.1 follows from Lemmas 2.1 and 2.2 and a standard argument of calculus of variations. Here we point out that, the above theorem is also valid, if 0 < λ < min{4, n} and ω > −θ , where θ is defined as: n o k∇φk2 + kxφk2 . θ= min (3.6) kφk2 =1,φ∈Σ

(3.8)

Using integration by parts, we obtain Z

|φ(y)|2 φ(x)x · ∇φ(x)dydx λ Rn ×Rn |x − y| Z |φ(x)|2 |φ(y)|2 dydx = −n |x − y|λ Rn ×Rn Z (x − y)|φ(x)|2 |φ(y)|2 +λ x· , |x − y|2+λ R n ×R n

Z

and let

(3.7)

and similarly Z |φ(x)|2 2 φ(y)y · ∇φ(y)dxdy λ Rn ×Rn |x − y| Z |φ(x)|2 |φ(y)|2 dydx = −n |x − y|λ Rn ×Rn Z (y − x)|φ(x)|2 |φ(y)|2 y· +λ . |x − y|2+λ R n ×R n

(3.9)

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Combining the above two equalities we see that Z |φ(y)|2 φ(x)x · ∇φ(x)dydx λ Rn ×Rn |x − y| Z |φ(x)|2 |φ(y)|2 2n − λ =− dydx. 4 |x − y|λ R n ×R n From (3.8)–(3.10), it follows that L(φ) = 0. Thus we have completed the proof of Lemma 3.2.

(3.10)

Since I (φ τ ) < 0 for every τ > 1, we see that I (φ τ1 ) < 0. ˆ And for the sake of simplicity we next write φ(x) for φ τ1 (x). n+2−λ µ ˆ Then by a direct Now we put φˆ = µ 2 φ(µx). computation we obtain Z 2 ˆ |∇ φ(x)| dx I (φˆ µ ) = µ4−λ Rn

! 2 |φ(y)| 2 ˆ ˆ |φ(x)| − dydx |x − y|λ R n ×R n !  2 Z 2−λ−n x 2 2 ˆ ˆ ω|φ(x)| + +µ 2 |φ(x)| dx, µ Rn



Z

Now we define the following cross-constrained variational problem d1 = inf J (φ). φ∈K

Since K ⊆ M, we then have the following result:

and

Corollary 3.3. Let 2 < λ < min{4, n}, n ≥ 3 and ω > 0. Then there exists Ψ ∈ K such that

L(φˆ µ ) = µ4−λ

d1 = J (Ψ ) = inf J (φ). φ∈K

In addition, d1 ≥ d0 > 0. Next we consider another cross-constrained variational problem d2 = inf J (φ), φ∈U

where U = {φ ∈ Σ ; I (φ) < 0 and L(φ) = 0}. Lemma 3.4. Let 2 < λ < min{4, n}, n ≥ 3 and ω > 0. Then there exists Φ 0 ∈ Σ such that d2 = J (Φ 0 ) = inf J (φ) > 0. φ∈U

(3.11)

Proof. By Lemma 3.2, we see that there exists nonzero φ ∈ Σ n such that I (φ) = 0 and L(φ) = 0. Let u τ = τ 2 φ(τ x). After a direct calculation we then obtain  Z   x 2 I (φ τ ) = τ 2 |∇φ(x)|2 + ω|φ(x)|2 + |φ(x)|2 dx τ Rn Z 2 |φ(y)|2 |φ(x)| − τλ dydx, (3.12) |x − y|λ Rn ×Rn  Z   x 2 L(φ τ ) = τ 2 |∇φ(x)|2 − 2 |φ(x)|2 dx τ Rn Z |φ(x)|2 |φ(y)|2 λτ λ − dydx. (3.13) 4 Rn ×Rn |x − y|λ Since I (φ τ ) = 0 with τ = 1, by λ > 2 we then see that I (φ τ ) < 0 for every τ > 1. On the other hand, L(φ τ ) < 0 as τ → ∞. Thus, noting the fact φ 6= 0, by the continuity of L(φ τ ) as a function of τ over (0, ∞) we can find an appropriate τ1 ∈ (1, ∞) such that L(φ τ1 ) < 0, Z Z λτ1 λ |φ(x)|2 |φ(y)|2 2 2 τ1 |∇φ(x)| dx − dydx > 0. 4 |x − y|λ Rn Rn ×Rn

Z Rn

2 ˆ |∇ φ(x)| dx

! Z 2 |φ(y)| 2 ˆ ˆ |φ(x)| λ dydx − 4 Rn ×Rn |x − y|λ Z −2−λ−n 2 ˆ −µ 2 |x|2 |φ(x)| dx. Rn

From (3.12) and I (φ) < 0 it follows that Z Z 2 |φ(y)| 2 ˆ ˆ |φ(x)| 2 ˆ |∇ φ(x)| dx − dydx < 0. |x − y|λ Rn Rn ×Rn Therefore, noting the fact λ < min{4, n}, by the continuity of I (φˆ µ ) as a function of µ we see that I (φˆ µ ) < 0 for every ˆ < 0 and L(φˆ µ ) > 0 as µ > 1. Noting the facts L(φ) µ → +∞, and the continuity of L(φˆ µ ) as a function of µ, we can find an appropriate µ1 > 1 such that L(φˆ µ1 ) = 0. Thus we have proved that U 6= ∅. We next claim that d2 > 0. Since φ ∈ U implies that φ 6= 0, by L(φ) = 0 we then have  Z  1 1 − J (φ) = |∇φ(x)|2 + ω|φ(x)|2 2 λ Rn    1 1 2 2 + + |x| |φ| dx > 0, (3.14) 2 λ which means d2 ≥ 0. We next use H¨older inequality and Lemma 2.2 to obtain Z |φ(x)|2 |φ(y)|2 dydx |x − y|λ R n ×R n   2n λ  2nλ Z Z   2n−λ 2 4n |φ(y)| 2n   2n−λ dx ≤C dy dx |φ(x)| λ n n |x − y| R R ≤ Ckφk4 4n 2n−λ

for some constant C > 0. Furthermore, using the compact lemma (Lemma 2.1), we see that Z |φ(x)|2 |φ(y)|2 dydx ≤ Ckφk4 4n |x − y|λ 2n−λ R n ×R n

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Z

2 (|∇φ(x)| + ω|φ(x)| + |x| |φ(x)| )dx . (3.15) 2

≤C Rn

2

2

2

On the other hand, since I (φ) < 0, we see that Z Rn

(|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx

|φ(x)|2 |φ(y)|2 dydx. |x − y|λ Rn ×Rn From (3.15) and (3.16), it follows that

<

Z

Z

(3.16)

As for L(u(t, ·)) < 0 for every t ∈ [0, T ), we also use a contradiction argument to show it. Suppose it does not hold. That is to say, there exists a time t2 ∈ (0, T ) such that L(u(t2 , ·)) = 0. By (4.2) and (4.3) we see that J (u(t2 , ·)) < d and I (u(t2 , ·)) < 0. Thus, u(t2 , ·) ∈ U . By Lemma 3.4 we obtain J (u(t2 , ·)) ≥ d2 , which contradicts J (u(t2 , ·)) < d ≤ d2 . Thus, we have obtained the desired result.

(4.4) 

(|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx ≥ ρ > 0, (3.17)

We are now in a position to state the main result of this section.

where the constant, ρ, is independent of φ(x). Therefore, by (3.14) and (3.17) we see that d2 > 0. Since d2 is bounded below on U , we then see that there exists Φ ∈ Σ such that the statement (3.11) is valid. 

Theorem 4.2. Let 2 < λ < min{n, 4} and n ≥ 3. If u 0 ∈ Γ and | · |u 0 ∈ L 2 (Rn ), then the solution u of Eq. (1.1) blows up in finite time T , that is to say, the following is satisfied

Rn

Remark 3.5. Here we note that, the solution Φ of (3.11) may not be in U , that is to say, it is possible that Φ satisfies I (Φ) = 0 and L(Φ) = 0, which is on ∂ U¯ . Here U¯ is the closure of U and ∂ U¯ is the boundary of U¯ . 4. Blow-up result

lim k∇u(t, ·)k22 = ∞.

t→T −

(4.5)

Proof. Let u(t) be the corresponding solution of Eq. (1.1) with the initial data u 0 ∈ Γ . We firstly define an auxiliary set Γ = {φ ∈ Σ ; J (φ) < d, I (φ) < 0 and L(φ) < −},

To obtain the blow-up result, we introduce a crossconstrained invariant set as follows Γ = {φ ∈ Σ ; J (φ) < d, I (φ) < 0 and L(φ) < 0}, for 2 < λ < min{4, n}, n ≥ 3 and ω > 0, where d = min{d0 , d2 }.

where  is a small positive number. Then we next prove that there exists a small positive number  such that the set Γ is invariant under the flow generated by Eq. (1.1). By the continuity of u(t, x) as a function of t we see that J (u(t, ·)), I (u(t, ·)) and L(u(t, ·)) are continuous as three functions of t on [0, T ). Thus if we can prove that lim L(u(t, ·)) < 0,

Lemma 4.1. Let 2 < λ < min{4, n}, n ≥ 3 and ω > 0. Then the set Γ is invariant under the flow generated by Eq. (1.1) in the following sense: If u 0 ∈ Γ , then the unique solution u(t) of Eq. (1.1), 0 < t < T , with the initial data u 0 satisfies u(t, ·) ∈ Γ ,

for t ∈ [0, T ),

(4.1)

where T > 0 is the maximum existence time of the solution u(t) as in Theorem 2.3.

t→T −

by the continuity of L(u(t, ·)) as a function of t over [0, T ), we then see that there exists a small number  > 0 such that L(u(t, ·)) < − for every time t ∈ [0, T ). And by (2.1) and (2.2) we see that J (u(t, ·)) = J (u 0 ). Thus by the continuity of J (u(t, ·)) we see that lim J (u(t, ·)) = J (u 0 ) < d.

(4.6)

Proof. Let u 0 ∈ Γ and u(t, x) be the corresponding solution of Eq. (1.1) as in Theorem 2.3. By (2.1) and (2.2) we see that

And from I (u(t, ·)) < 0 for every t ∈ [0, T ) it follows that

J (u(t, ·)) = J (u 0 ) < d

t→T −

(4.2)

for every t ∈ [0, T ). We next claim that I (u(t, ·)) < 0 for every t ∈ [0, T ) by a contradiction argument. Suppose that it is not valid that I (u(t, ·)) < 0 for every t ∈ [0, T ). That is to say, there exists t1 ∈ (0, T ) such that I (u(t1 , ·)) = 0. Since u 0 6= 0, it follows that u(t1 , ·) 6= 0 from (2.1) and (2.2), which implies that u(t1 , ·) ∈ M. By Theorem 3.1 we see that J (u(t1 , ·)) ≥ d0 ≥ d which contradicts (4.2). Thus we have proved that I (u(t, ·)) < 0 for every t ∈ [0, T ).

(4.3)

t→T −

lim I (u(t, ·)) ≤ 0.

(4.7)

We next prove that it is impossible that lim L(u(t, ·)) = 0.

t→T −

(4.8)

We argue the above statement by contradiction. Suppose that (4.8) is valid. Thus it is sufficient to consider the following two cases: Case A. limt→T − I (u(t, ·)) = 0. From (4.6) and limt→T − I (u(t, ·)) = 0, it follows that limt→T − ku(t, ·)kΣ < ∞. Thus by the continuity of u(t, x) as a function of t on [0, T ) we see that there exists Φ(x) such that u(t, ·) → Φ in Σ with I (Φ) = 0

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and J (Φ) < d as t → T − , which implies that Φ ∈ M. From Theorem 3.1 it follows that lim J (u(t, ·)) ≥ d0 = J (Φ) ≥ d,

t→T −

(4.9)

If I (φ) < 0, then we have Z (|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx I (µφ) = µ2 Rn

−µ

4

which contradicts J (Φ) < d. Consequently, it is impossible that limt→T − L(u(t, ·)) = 0 in this case. Case B. limt→T − I (u(t, ·)) < 0. By the continuity of u(t, x) as a function of t, we assume that there exists Φ(x) such that u(t, ·) → Φ in Σ as t → T − and kΦkΣ = limt→T − ku(t, ·)kΣ < ∞ (Otherwise, we have obtained the blow-up result for Eq. (1.1).). Thus we see that Φ ∈ U . By Lemma 3.4 we have limt→T − J (u(t, ·)) ≥ d2 ≥ d which contradicts (4.6). Thus, we have proved that (4.8) is wrong when limt→T − I (u(t, ·)) < 0. Therefore, we have obtained that lim L(u(t, ·)) < 0.

|φ(x)|2 |φ(y)|2 dydx |x − y|λ Rn ×Rn

Z

>0 for a sufficiently small µ > 0. So by the continuity of I (µφ) as a function of µ we see that there exists a µ0 ∈ (0, 1) such that I (µ0 φ) = 0. In addition, we have Z 1 2 J1 (µ0 φ) = µ0 (|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx 4 Rn < J1 (φ), which implies that d3 = inf {J1 (φ)} = d0 .

t→T −

φ∈M

By the continuity of L(u(t, ·)) as a function of t, we see that there exists a small number  > 0 such that L(u(t, ·)) < − for every t ∈ [0, T ). Since Γ is invariant under the flow generated by Eq. (1.1), Γ is also invariant under the same flow. Thus by Lemma 2.4 we have Z Z |x|2 |u(t, x)|2 dx ≤ |x|2 |u 0 (x)|2 dx Rn Rn Z + 4t= u¯ 0 (x)x · ∇u 0 (x)dx − 8t 2 , (4.10) Rn

which implies that Rn |x|2 |u(t, x)|2 dx < 0 as t → ∞, but this is clearly a contradiction. This concludes that the solution u of Hartree equation (1.1) with initial data u 0 ∈ Γ blows up in finite time T , that is, R

lim k∇u(t, ·)k22 = ∞.

t→T −

(5.1)

Therefore, we see that U ⊆ V . Noting the fact that d0 and d2 are dependent on ω, now we can make the following appropriate assumption on d0 and d2 . Assumption 5.1. There exists ω > 0 such that d2 ≥ d0 . Lemma 5.1. Let φ µ (x) = µφ(x). Then there exists a unique µ0 > 0 such that I (φ µ0 ) = 0, I (φ µ ) > 0 when µ ∈ [0, µ0 ), and I (φ µ ) < 0 when µ ∈ (µ0 , ∞). Furthermore, J (φ µ0 ) > J (φ µ ) for every µ ∈ (0, ∞) and µ 6= µ0 . Proof. Substituting φ µ for (3.2) and (3.3), we have Z µ2 µ J (φ ) = |∇φ(x)|2 dx + ω|φ(x)|2 dx 2 Rn  2 2 + |x| |φ(x)| x dx Z |φ(x)|2 |φ(y)|2 µ4 dx, 4 Rn ×Rn |x − y|λ Z I (φ µ ) = µ2 |∇φ(x)|2 dx + ω|φ(x)|2 dx Rn  + |x|2 |φ(x)|2 x dx −

The proof is completed.



5. Strong instability by blow-up In order to obtain the strong instability, we consider the following auxiliary variational problem

− µ4

d3 := inf {J1 (φ)}, φ∈V

Z

|φ(x)|2 |φ(y)|2 dx, |x − y|λ Rn ×Rn

and by a simple calculation we also obtain

where

d J (φ µ ) = I (φ µ ). dµ

1 J1 (φ) = J (φ) − I (φ) 4 Z 1 = (|∇φ(x)|2 + ω|φ(x)|2 + |x|2 |φ(x)|2 )dx 4 Rn

µ

and

Theorem 5.2. Let ω > 0 with Assumption 5.1. Then the standing wave u = eiωt φ is nonlinearly unstable in the following sense: For any δ > 0, there exists T < ∞ and an

V = {φ ∈ Σ ; I (φ) ≤ 0}.

Now this lemma directly follows from the above three equalities. 

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initial data u 0 ∈ Σ with ku 0 − φk < δ, such that the solution u of (1.1) with u 0 satisfies lim k∇u(t, ·)k2 = +∞.

t→T −

(5.2)

Proof. By Assumption 5.1, we see that d = d0 . Since φ is a ground state solution of the stationary equation (3.1), then by Lemma 5.1 we have J (φ µ ) < J (φ) = d

(5.3)

for every µ > 1, where φ µ (x) = µφ(x). In addition, from Lemma 3.2 it follows that L(φ) = 0 and I (φ) = 0. We then obtain that L(φ µ ) < 0

and

I (φ µ ) < 0

(5.4)

for every µ > 1. Thus, by (5.3) and (5.4) we see that φ µ ∈ Γ for every µ > 1, which implies that the solution eiωt φ of Hartree equation (1.1) blows up in finite time. Moreover, it is clear that φ µ → φ in Σ as µ → 1 and µ > 1, which implies for any δ > 0 there exists u 0 such that ku 0 − φkΣ < δ, and the solution of Hartree equation (1.1) with the initial data u 0 blows up in finite time. This completes the proof of Theorem 5.2.  Theorem 5.3. Suppose that ω does not satisfy Assumption 5.1. Then, for the standing wave u = eiωt φ, there exists δ1 > 0 such that, for any δ > 0, there exists T < ∞ and an initial data u 0 ∈ Σ with δ1 < ku 0 − φk < δ1 + δ which satisfy that the corresponding solution of equation (1.1) blows up in finite time T , that is, lim k∇u(t, ·)k2 = +∞.

t→T −

(5.5)

Proof. In this case we have d = d2 < d0 . Since φ is a ground state solution of the stationary equation, then J (φ µ ) < J (φ) = d0

(5.6) φ µ (x)

for every µ > 1, where is defined as in Theorem 5.2. Noting the fact that J (φ µ ) → −∞ as µ → ∞, by Lemma 5.1 we see that there exists a unique µ1 > 1 such that J (φ µ ) < J (φ µ1 ) = d = d2 with µ > µ1 > 1. In addition, from Lemma 3.2 it follows that L(φ) = 0 and I (φ) = 0. We then obtain that L(φ µ ) < 0 and I (φ µ ) < 0. Thus, we have that φ µ ∈ Γ when µ > µ1 . Now we let δ1 = (µ1 − 1)kφkΣ , then following the same way as in the proof of Theorem 5.2, we can obtain our desired result. 

implies the strong instability of standing wave by blow-up for Eq. (1.1). We next consider Assumption 5.1. Since we do not understand the relation between ω and the solution Φ of problem (3.11), it seems difficult to give a sufficient condition on ω satisfying Assumption 5.1. But we next analyse that there must exist an ω with Assumption 5.1 as λ → 4, λ < 4 and n ≥ 4. Suppose that Φ is a solution of problem (3.11). By Corollary 3.3 and Remark 3.5 we see that if I (Φ) = 0 and L(φ) = 0 then Assumption 5.1 is satisfied. Without loss of generality we next assume that I (Φ) < 0, L(Φ) = 0. n+2

By a direct computation with Φµ = µ 2 Φ(µx), we have Z I (Φµ ) = (µ4 |∇Φ(x)|2 + |x|2 |φ(x)|2 + ωµ2 |Φ(x)|2 )dx Rn Z |Φ(x)|2 |Φ(y)|2 dydx, − µ4+λ |x − y|λ R n ×R n Z 1 (µ4 |∇Φ(x)|2 + |x|2 |φ(x)|2 + ωµ2 |Φ(x)|2 dx) J (Φµ ) = 2 Rn Z µ4+λ |Φ(x)|2 |Φ(y)|2 − dydx. 4 |x − y|λ Rn ×Rn Since I (Φµ ) > 0 as µ → 0, by the continuity of I (Φµ ) as a function of µ we see that there exists 0 < µ0 < 1 such that I (Φµ0 ) = 0. We next consider Z 1 J (Φ) − J (Φµ ) = ((1 − µ4 )|∇Φ(x)|2 2 Rn (1 − µ4+λ ) + ω(1 − µ2 )|Φ(x)|2 dx) − 4 Z |Φ(x)|2 |Φ(y)|2 × dydx n n |x − y|λ Z R ×R 1 ≥ ω(1 − µ2 )|Φ(x)|2 dx 2 Rn   1 − µ4 (1 − µ4+λ ) − + 2 λ Z × |∇Φ(x)|2 dx, (6.1) Rn

where the last inequality comes from L(Φµ ) = 0 as µ = 1. We now let 1 − µ4 (1 − µ4+λ ) − . 2 λ As λ = 4, we see that g(µ) is decreasing on [0, 1]. And since g(1) = 0 and g(0) = 14 , we see that the right-hand side of (6.1) is nonnegative. Thus, as λ → 4, λ < 4 and ω → ∞, the right-hand side of (6.1) may be nonnegative with n ≥ 4. But we do not understand the relation between the solution of problem (3.11) and ω, so here we cannot give an explicit condition on ω with Assumption 5.1. g(µ) =

6. Remarks Let φ be the ground state solution of the stationary equation (3.1). Theorem 5.2 means that for every δ > 0 there exists initial data in the ball at eiωt φ with radius δ such that the corresponding solution blows up in finite time. Theorem 5.3 implies that there exists initial data in a vacuum isolating at eiωt φ such that the corresponding solution of Eq. (1.1) blows up in finite time. By Definition 2.6 we see that Theorem 5.2

Y. Wang / Physica D 237 (2008) 998–1005

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