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Finite time blow-up and global solutions for fourth order damped wave equations
J. Math. Anal. Appl. 418 (2014) 713–733
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Finite time blow-up and global solutions for fourth order damped wave equations Yongda Wang Dipartimento di Matematica, Politecnico di Milano, Piazza Leonardo da Vinci 32, 20133 Milano, Italy
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 18 October 2013 Available online 12 April 2014 Submitted by P. Yao
This work is devoted to a class of fourth order wave equations with linear damping term and superlinear source term. After showing the uniqueness and existence of local solutions to the equations, we give necessary and sufficient conditions for global existence and finite time blow-up of these solutions. Moreover, the potential well depth is estimated. © 2014 Elsevier Inc. All rights reserved.
Keywords: Fourth order wave equation Initial-boundary value problem Damping term Strong source term
1. Introduction The report [1] about the Tacoma Narrows Bridge collapse [25,26] considers . . . the crucial event in the collapse to be the sudden change from a vertical to a torsional mode of oscillation, see also [25, p. 63]. Hence, if one models a suspension bridge by a beam, there is no way to highlight the torsional oscillations. The nonlinear behavior of suspension bridges, which is by now well established, see [2,8,13,23], also plays a crucial role in causing oscillations. Therefore, a reliable model for suspension bridges should be nonlinear and it should have enough degrees of freedom to display torsional oscillations. In this respect, Lazer and McKenna [14, Problem 11] suggested to study the following equation Δ2 u + c2 Δu + h(u) = 0,
in Rn ,
(1.1)
where h(u) ≈ [u + 1]+ − 1 with u+ = max{u, 0}. Subsequently, equations “like” (1.1), namely, Δ2 u + c2 Δu = b (u + 1)+ − 1 ,
in Σ ⊂ Rn ,
(1.2)
under the Navier boundary condition have been considered in several papers. For the case c2 < λ1 ((λk )k≥1 , the sequence of eigenvalues of −Δ in H01 (Σ)), we refer to [15,18,27]. Tarantello [27] proved that if E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jmaa.2014.04.015 0022-247X/© 2014 Elsevier Inc. All rights reserved.
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b ≥ λ1 (λ1 − c2 ), then (1.2) has a nontrivial solution. Lazer and McKenna [15] discussed the case n = 1 when Σ is an interval and found that (1.2) has at least 2k − 1 nontrivial solutions if b > λk (λk − c2 ). Micheletti and Pistoia [18] studied (1.2) for a more general nonlinearity which is “like” b[(u + 1)+ − 1] and obtained the multiplicity existence of nontrivial solutions. When c2 ≥ λ1 , Micheletti and Pistoia considered (1.2) when b[(u + 1)+ − 1] is replaced by a more general nonlinearity in [17], where they showed the existence of a nontrivial solution. Then Micheletti, Pistoia and Saccon [19] improved the result in [17] and gave the multiplicity results for the same problem. In other papers [32–34] this kind of problem was also investigated under the Navier boundary condition. More recently, Ferrero and Gazzola [5] suggested one should consider the boundaries of a plate Ω = (0, π) × (−l, l) which represents the roadway of a suspension bridge as follows. Because the edges x = 0, π connect with the ground, they are assumed to be hinged and then u(0, y) = uxx (0, y) = u(π, y) = uxx (π, y) = 0,
y ∈ (−l, l),
(1.3)
while the edges y = ±l are free and the boundary conditions there become (see [30, 2.40]) uyy (x, ±l) + σuxx (x, ±l) = 0,
uyyy (x, ±l) + (2 − σ)uxxy (x, ±l) = 0,
x ∈ (0, π).
(1.4)
The free boundaries (1.4) yield small stretching energy for the plate, so Ferrero and Gazzola took c = 0 in (1.1) and introduced a model for the stationary suspension bridge Δ2 u + h(x, y, u) = f (x, y),
in Ω
(1.5)
as well as a model for the nonlinear dynamical suspension bridge utt + Δ2 u + μut + h(x, y, u) = f (x, y, t),
in Ω × (0, T ),
(1.6)
here h(x, y, u) is restoring force due to the hangers of the suspension bridge, f (x, y) or f (x, y, t) is the external force including the gravity. Given an open rectangular plate Ω = (0, π) × (−l, l) ⊂ R2 , we consider the following initial value problem ⎧ 2 p−2 ⎨ utt + Δ u + μut + au = |u| u, (x, y, t) ∈ Ω × [0, T ], (x, y) ∈ Ω, u(x, y, 0) = u0 (x, y), ⎩ ut (x, y, 0) = u1 (x, y), (x, y) ∈ Ω,
(1.7)
with the boundary condition
y ∈ (−l, l), u(0, y, t) = uxx (0, y, t) = u(π, y, t) = uxx (π, y, t) = 0, uyy (x, ±l, t) + σuxx (x, ±l, t) = uyyy (x, ±l, t) + (2 − σ)uxxy (x, ±l, t) = 0, x ∈ (0, π),
(1.8)
for every t ∈ [0, T ], where T > 0, μ > 0, 2 < p < ∞, σ ∈ (0, 12 ) and a = a(x, y, t) is a sign-changing and bounded measurable function. The initial data u0 , u1 belong to suitable spaces, which will be specified later on. Here we study the fourth order wave problem (1.7) with the boundary condition (1.8), because it comes from the physical model for the dynamic suspension bridge if we assume that au describes the restoring force because of the hangers of the suspension bridge and |u|p−2 u is the strong source term which represents the other external forces acting on the bridge. We explain this model briefly below, see for more details [5].
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Let u be the displacement of the plate in the vertical direction, assume that f denotes the external force acting on the plate, then for a small displacement u, the total energy of the plate Ω approximately equals to by a suitable normalization ET (u) =
1 (Δu)2 + (1 − σ) u2xy − uxx uyy − f u dxdy. 2
(1.9)
Ω
Due to the hangers of the bridge, there exists another force h with a potential energy given by then the total energy becomes by adding this potential energy to (1.9): ET (u) =
Ω
Hdxdy,
1 (Δu)2 + (1 − σ) u2xy − uxx uyy + H − f u dxdy. 2
(1.10)
Ω
If the external force f also depends on time, the deformation u has a kinetic energy given as follows by a suitable scaling:
1 2 u dxdy, 2 t
Ω
which should be added to the nonlinear static energy (1.10): Eu (t) = Ω
1 2 u dxdy + 2 t
2 1 2 (Δu) + (1 − σ) uxy − uxx uyy + H(x, y, u) − f u dxdy. 2
(1.11)
Ω
This is the total energy of a nonlinear dynamic bridge. As for the action, we take the difference between the kinetic energy and the potential energy and integrate on [0, T ]: T A (u) = 0
Ω
1 2 u dxdy − 2 t
1 (Δu)2 + (1 − σ) u2xy − uxx uyy + H(x, y, u) − f u dxdy dt. 2
Ω
Taking the critical points of the functional A (u), we have Eq. (1.6) by adding a damping term μut due to internal friction. Eq. (1.6) also appears in other contexts, see [6, (17)], and is sometimes called the Swift–Hohenberg equation. In the present paper, our purpose is to investigate the initial-boundary value problem (1.7)–(1.8). We first set up the existence and uniqueness of local solution to (1.7)–(1.8), then the behavior of the local solution is analyzed by utilizing the potential well theory developed by Sattinger [24] in 1968 to solve the second order evolution equations. We give necessary and sufficient conditions for the global existence and finite time blow-up of solutions to (1.7)–(1.8). For the related results on second order evolution problems, there are works by Ebihara, Nakao and Nambu [3], Ikehata and Suzuki [12], Levine and Serrin [16], Payne and Sattinger [22], Tsutsumi [29,28], Gazzola [7], Gazzola and Squassina [9], Gazzola and Weth [10]. This paper is organized as follows. In Section 2 we prepare several preliminary definitions and recall some basic facts. Also, some important lemmas are proved. We state the main results including local existence, global existence and finite time blow-up of solutions to (1.1)–(1.2) as well as the estimation on the potential well depth in Section 3, see Theorems 3.1, 3.2, 3.3 and 3.4. The last four sections are devoted to the proofs of the main results.
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2. Preliminaries Letting Ω = (0, π) × (−l, l) ⊂ R2 , we denote the standard Lq (Ω) norm by · q for 1 ≤ q < ∞ and the H (Ω) norm by · H 2 = ( · 22 + D2 · 22 )1/2 . In addition, we introduce another function space defined in [5], 2
H∗2 = H∗2 (Ω) := u ∈ H 2 (Ω) : u = 0 on {0, π} × (−l, l) , the dual space of which is defined by H(Ω) and the corresponding duality between them is denoted by ·,· . Clearly, H∗2 satisfies H02 (Ω) ⊂ H∗2 (Ω) ⊂ H 2 (Ω) and is a Hilbert space when endowed with the inner product (u, v)
H∗2
ΔuΔv + (1 − σ)
= Ω
2uxy vxy − uxx vyy − uyy vxx ,
for every u, v ∈ H∗2 .
Ω
This inner product induces a norm uH∗2 =
1/2
|Δu|2 + 2(1 − σ)
Ω
u2xy − uxx uyy
,
for every u ∈ H∗2 ,
Ω
which is equivalent to · H 2 for 0 < σ < 1/2, see for the proof [5, Lemma 7]. Moreover, we have a Sobolev embedding inequality for this case. Lemma 2.1. Assume that 1 ≤ q < ∞. Then for any u ∈ H∗2 , the inequality uq ≤ Sq uH∗2 π holds, where Sq = ( 2l +
(2.1)
√
2 1 (q+2)/2q ( 1−σ )1/2 . 2 )(2πl)
Remark 2.1. Here Sq is not less than the best Sobolev embedding constant. How to obtain the best constant is still open. ¯ and for any (x, y) ∈ Ω, there results Proof of Lemma 2.1. Take any u ∈ H∗2 ⊂ C 1 (Ω) x π π 1/2
2 √ u(x, y) = ux (t, y)dt ≤ ux (x, y)dx ≤ π ux (x, y) dx 0
≤
√
π
0
≤
0
−uxx (x, y)u(x, y)dx
π 0
√
1/2
π π
uxx (x, y)2 dx
1/4 π
0
u(x, y)2 dx
0
which yields that π 0
u(x, y)2 dx ≤ π 4
π 0
uxx (x, y)2 dx.
1/4 ,
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Then it follows that from the two inequalities above u(x, y) ≤ π 3/2
π
uxx (x, y)2 dx
1/2 .
(2.2)
0
Furthermore, we have x π uy (x, y) = uyx (t, y)dt ≤ uyx (x, y)dx, 0
(2.3)
0
and y l u(x, y) = u(x, s) + uy (x, r)dr ≤ u(x, s) + uy (x, y)dy,
for any s ∈ (−l, l).
−l
s
Integrating the inequality (2.4) about s on (−l, l), we get l
2lu(x, y) ≤
u(x, s)ds + 2l
−l
l
uy (x, y)dy,
−l
which together with (2.2), (2.3) yields that u(x, y) ≤ 1 π 3/2 2l π 3/2 ≤ (2l)1/2 ≤
l π −l
uxx (x, s)2 dx
1/2
l π ds + −l 0
0
l π
uyx (x, y)dxdy
uxx (x, y)2 dxdy
l π
1/2 + (πl)1/2
−l 0
√ 1/2 2 2 2 π 1/2 D u dxdy + (2πl) . 2l 2
2 2uyx (x, y) dxdy
−l 0
Ω
From Lemma 7 in [5], we know that 2 (1 − σ)D2 u2 ≤ u2H∗2 . Therefore, u(x, y) ≤
√ 1/2 π 1 2 1/2 + (2πl) uH∗2 , 2l 2 1−σ
which implies that, |u|q ≤
√ q q/2 1 π 2 + (2πl)q/2 uqH 2 , ∗ 2l 2 1−σ
for 1 ≤ q < ∞. Integrating the above inequality on Ω, we have
1/2
(2.4)
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uq ≤
√ 1/2 1 π 2 (q+2)/2q + (2πl) uH∗2 , 2l 2 1−σ
which completes the proof. 2 Let {Λi }∞ i=1 be the eigenvalue sequence to the eigenvalue problem ⎧ 2 (x, y) ∈ Ω, ⎨ Δ u = Λu, y ∈ (−l, l), u(0, y) = uxx (0, y) = u(π, y) = uxx (π, y) = 0, ⎩ uyy (x, ±l) + σuxx (x, ±l) = uyyy (x, ±l) + (2 − σ)uxxy (x, ±l) = 0, x ∈ (0, π), which has been solved in [5]. Particularly, Λ1 < 1 and we have an elementary result. Lemma 2.2. Assume that −Λ1 < a1 ≤ a ≤ a2 . Then for any u ∈ H∗2 , there holds A1 u2H∗2 ≤ u2H∗2 + (au, u)2 ≤ A2 u2H∗2 , where (·,·)2 is the L2 inner product and A1 , A2 are given by A1 =
1+ 1,
a1 Λ1 ,
a1 < 0, a1 ≥ 0
and
A2 =
1, 1+
a2 Λ1 ,
a2 < 0, a2 ≥ 0.
Proof. We only consider the left inequality, the right one can be proved similarly. When the constant a1 < 0, since Λ1 (u, u)2 ≤ u2H 2 , then ∗
u2H∗2 + (au, u)2 ≥ u2H∗2 + a1 (u, u)2 ≥ u2H∗2 +
a1 u2H∗2 . Λ1
If a1 > 0, then (au, u)2 ≥ 0, hence u2H∗2 + (au, u)2 ≥ u2H∗2 . Therefore, the inequality A1 u2H∗2 ≤ u2H∗2 + (au, u)2 holds. 2 Now, we define the Nehari functional I and the energy functional J on H∗2 → R, I(u) = u2H∗2 + (au, u)2 − upp , J(u) =
for every u ∈ H∗2 ,
1 1 1 u2H∗2 + (au, u)2 − upp , 2 2 p
for every u ∈ H∗2 ,
which play critical roles in dealing with our problem. Let u be an arbitrary nonzero element in H∗2 and consider a real value function defined by j(λ) = J(λu), then
λ ≥ 0,
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j (λ) = λu2H∗2 + λ(au, u)2 − λp−1 upp , j (λ) = u2H∗2 + (au, u)2 − (p − 1)λp−2 upp . Clearly, j(0) = j (0) = 0 and j (0) = u2H 2 + (au, u)2 > 0 for a > −Λ1 . Thus for any 0 = u ∈ H∗2 , j(λ) is ∗ a convex function for small λ > 0 and has the following behaviors. Lemma 2.3. Assume that −Λ1 < a1 ≤ a ≤ a2 . Then for any nontrivial u ∈ H∗2 , (i) limλ→∞ j(λ) = −∞; ¯ = λ(u) ¯ ¯ = 0; (ii) there exists a unique λ > 0 such that j (λ) ¯ (iii) j (λ) < 0. Proof. The proof is almost similar to that of Lemma 2.2 in [22], the only difference is to deal with the term (au, u)2 , which can be treated by applying Lemma 2.2, so we omit it. 2 Then, we could define the potential well depth of the functional J (also known as mountain pass level) by d=
inf
max J(λu).
u∈H∗2 \{0} λ>0
(2.5)
Denote the set of all nontrivial stationary solutions to the problem (1.7)–(1.8) by N = u ∈ H∗2 \{0} : I(u) = 0 , which is the so-called Nehari manifold, see [20,31]. By considering a map s → I(su) for all u such that u2H 2 = 1 and Lemma 2.3, it is easy to check that each half line starting from the origin of H∗2 intersects ∗ only once the manifold N and N separates the two sets N+ = u ∈ H∗2 : I(u) > 0 ∪ {0}
and N− = u ∈ H∗2 : I(u) < 0 .
Then the stable set W and unstable set U may be defined by W = u ∈ N+ : J(u) < d ,
U = u ∈ N− : J(u) < d .
Lemma 2.4. (See [4,12].) The following properties of W and U hold: (i) W is a neighborhood of 0 ∈ H∗2 ; ¯ (closure in H 2 ). (ii) 0 ∈ /U ∗ As Payne and Sattinger did in [22], the potential well depth d defined in (2.5) can be also characterized as d = inf J(u). u∈N
(2.6)
Finally, we consider an energy functional E : H∗2 (Ω) × L2 (Ω) → R defined by 1 E (v, w) = J(v) + w22 , 2
for every pair (v, w) ∈ H∗2 (Ω) × L2 (Ω),
and the Lyapunov function E(t) = E (u(t), ut (t)), defined for any solution u(t) of the problem (1.7)–(1.8).
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3. Main results In this section, the main results on our problem are listed. To begin with, we explain what does a weak solution mean to our problem (1.7)–(1.8). Definition 3.1. A function u ∈ C([0, T ], H∗2 (Ω)) ∩ C 1 ([0, T ], L2 (Ω)) ∩ C 2 ([0, T ], H(Ω)) with ut ∈ L2 ([0, T ], L2 (Ω)) is called a weak solution to (1.7)–(1.8), if the following conditions hold u(0) = u0 ,
ut (0) = u1
and utt , η + (u, η)H∗2 + μ
ut η +
Ω
|u|p−2 uη
auη = Ω
Ω
for all η ∈ H∗2 (Ω) and a.e. t ∈ [0, T ]. Then we are concerned with the existence and uniqueness of local solution to the problem (1.7)–(1.8). Theorem 3.1. Let μ > 0, 2 < p < ∞ and −Λ1 < a1 ≤ a ≤ a2 . Then for any u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω), there exists T > 0 such that the problem (1.7)–(1.8) has a unique local weak solution u on [0, T ]. Moreover, if Tmax = sup T > 0 : u = u(t) exists on [0, T ] < ∞, then lim u(t)q = ∞,
for q ≥ 1 such that q > (p − 2)/2.
t→Tmax
Remark 3.1. If Tmax = ∞, we say that the weak solution of (1.7)–(1.8) is global, while if Tmax < ∞, then the weak solution blows up and Tmax is the blow-up time. Before we give necessary and sufficient conditions for global existence and finite time blow-up of solutions to (1.7)–(1.8), we estimate the potential well depth. Theorem 3.2. Assume that −Λ1 < a1 ≤ a ≤ a2 , 2 < p < ∞. Then the potential well depth d can be estimated by 0<
p − 2p p − 2 p−2 p−2 A1 Sp p−2 ≤ d ≤ Cp , 2p 2p
where Sp is the Sobolev embedding constant, A1 is defined in Lemma 2.2 and Cp is given by Cp =
π(1 + a2 ) 2sin x2p
p/(p−2) .
In particular, if p is an integral number, then Cp =
2
p
2 p−2
p p−2
p!! · 12 ) p−2 (1 + a2 ) p−2 πl, ( (p−1)!! p!! ( (p−1)!!
here p!! = p · (p − 2) · (p − 4) · · · .
· π4 )
(1 + a2 )
πl,
p is an even number, p is an odd number,
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Remark 3.2. Since we still cannot obtain the best embedding constant in (2.1), we only have a range about the depth d. Now, we state the necessary and sufficient conditions for global existence of solutions to the problem (1.7)–(1.8). Theorem 3.3. Let u be the unique local weak solution to (1.7)–(1.8). Assume that u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω). Then u is a global solution and lim u(t)H 2 + ut (t)2 = 0
t→∞
∗
if and only if there exists a real number t0 ∈ [0, Tmax ) such that u(t0 ) ∈ W p − p−2
where D = min{d, p−2 2p A2
2p − p−2
Sp
and
E(t0 ) < D,
}.
Finally, we come to the blowing up result of the solutions to problem (1.7)–(1.8). Theorem 3.4. Let u be the unique local weak solution to (1.7)–(1.8). Assume that u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω). Then u blows up, that is, Tmax < ∞ if and only if there exists a real number t0 ∈ [0, Tmax ) such that u(t0 ) ∈ U
and
E(t0 ) < d.
4. Proof of Theorem 3.1 We start with some definitions. For every T > 0, set the space
H = C [0, T ], H∗2 (Ω) ∩ C 1 [0, T ], L2 (Ω) with the norm uH =
2 2 1/2
max A1 u(t)H 2 (Ω) + ut (t)2 ,
t∈[0,T ]
∗
where A1 is given in Lemma 2.2. For u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω), denote MT = u ∈ H : u(0) = u0 , ut (0) = u1 and u2H ≤ R2 , where R2 ≥ 2(A2 u0 2H 2 + u1 22 ) and A2 is given in Lemma 2.2. ∗
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Then we consider the initial problem ⎧ 2 p−2 ⎨ vtt + Δ v + μvt + av = |u| u, (x, y, t) ∈ Ω × [0, T ], v(x, y, 0) = u0 (x, y), (x, y) ∈ Ω, ⎩ (x, y) ∈ Ω, vt (x, y, 0) = u1 (x, y),
(4.1)
with the boundary condition
y ∈ (−l, l), v(0, y, t) = vxx (0, y, t) = v(π, y, t) = vxx (π, y, t) = 0, vyy (x, ±l, t) + σvxx (x, ±l, t) = vyyy (x, ±l, t) + (2 − σ)vxxy (x, ±l, t) = 0, x ∈ (0, π),
(4.2)
for every t ∈ [0, T ]. Lemma 4.1. Assume that u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω) and −Λ1 < a1 ≤ a ≤ a2 . Then for every u ∈ H, there exists a unique solution v ∈ H ∩ C 2 ([0, T ], H(Ω)) with vt ∈ L2 ([0, T ], L2 (Ω)) to the problem (4.1)–(4.2). Proof. According to Theorem 6 in [5], there is a unique weak solution v ∈ H ∩ C 2 ([0, T ], H(Ω)) to the problem ⎧ 2 ⎨ vtt + Δ v + μvt + h(x, y, v) = f, (x, y, t) ∈ Ω × [0, T ], (x, y) ∈ Ω, v(x, y, 0) = u0 (x, y), ⎩ vt (x, y, 0) = u1 (x, y), (x, y) ∈ Ω,
(4.3)
with the same boundary condition (4.2), where f = f (x, y, t) ∈ C 0 ([0, T ], L2 (Ω)). Hence, for the uniqueness and existence of solution to (4.1)–(4.2), it is sufficient to check that
|u|p−2 u ∈ C 0 [0, T ], L2 (Ω) . In fact, since u ∈ H, we have p−2 p−1 p−1 max u(t) u(t)2 ≤ max u(t)2p−2 ≤ C max u(t)H 2 ,
t∈[0,T ]
t∈[0,T ]
∗
t∈[0,T ]
which implies that |u|p−2 u ∈ C 0 ([0, T ], L2 (Ω)). Therefore, (4.1)–(4.2) admits a unique solution v ∈ H ∩ C 2 ([0, T ], H(Ω)). To complete we need to prove that vt ∈ L2 ([0, T ], L2 (Ω)). Take vt as a testing function into the equation in (4.1), integrate over Ω × [0, t] ⊂ Ω × [0, T ], that is t
t
vtt (τ ), vt (τ ) dτ +
0
v(τ ), vt (τ )
H∗2
t dτ + μ
0
t =
vt (τ )2 dτ + 2
0
t
av(τ ), vt (τ ) 2 dτ
0
u(τ )p−2 u(τ )vt (τ )dτ.
0 Ω
Recalling that a1 ≤ a ≤ a2 and by Lemma 2.2, we have vt (t)2 + A1 v(t) 2 + 2μ 2 H
t
∗
0
vt (τ )2 dτ ≤ u1 22 + A2 u0 H 2 + 2 ∗ 2
t 0 Ω
u(τ )p−2 u(τ )vt (τ )dτ. (4.4)
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Notice that t 2
u(τ )p−2 u(τ )vt (τ )dτ ≤ C
0 Ω
t
u(τ )2p−2 dt + μ
0 Ω
t ≤C
t
vt (τ )2 dτ 2
0
u(τ )2p−2 dt + μ 2
t
H∗
0
vt (τ )2 dτ 2
0
t ≤ CT + μ
vt (τ )2 dτ , 2
0
then it follows that from (4.4) T
vt (t)2 dτ ≤ CT, 2
0
which deduces that vt ∈ L2 ([0, T ], L2 (Ω)) and completes the proof of Lemma 4.1.
2
Now, we come to prove Theorem 3.1. By Lemma 4.1, for any u ∈ MT , we could introduce a map Φ : H → H defined by v = Φ(u), here v is the unique solution to (4.1)–(4.2). Claim. Φ is a contract map satisfying Φ(MT ) ⊆ MT , for small T > 0. In fact, assume that u ∈ MT , then the corresponding solution v = Φ(u) satisfies (4.4) for all t ∈ [0, T ]. Thus, as we did in the proof of Lemma 4.1, there holds vt 22 + A1 v2H∗2 ≤ u1 22 + A2 u0 2H∗2 + CR2p−2 T ≤
R2 + CR2p−2 T. 2
If T is small enough, then vH ≤ R, which implies that Φ(MT ) ⊆ MT . Let v1 = Φ(w1 ), v2 = Φ(w2 ) with w1 , w2 ∈ MT . Putting v1 , v2 in Eq. (4.1) and then subtracting the two equations, we obtain by setting v = v1 − v2
vtt , η + (v, η)H∗2 + μ(vt , η)2 + (av, η)2 = |w1 |p−2 w1 − |w2 |p−2 w2 ηdxdy Ω
γ(t)(w1 − w2 )ηdxdy
= Ω
for all η ∈ H∗2 and a.e. t ∈ [0, T ], where γ(t) is estimated by γ(t) ≤ (p − 1)(|w1 | + |w2 |)p−2 . Taking η = vt and arguing similarly as above, we have Φ(w1 ) − Φ(w2 )2 = v2H ≤ CR2p−4 T w1 − w2 2H . H Likely, if T is so small that CR2p−4 T < 1, then there exists a constant 0 < δ < 1 such that Φ(w1 ) − Φ(w2 )2 ≤ δw1 − w2 2H . H Therefore, Φ is a contract map, and the claim is proved.
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Then by the Contracting Mapping Principle, there exists a unique u ∈ MT such that u = Φ(u), which is the unique solution to the problem (1.7)–(1.8). Moreover, u ∈ C 2 ([0, T ], H(Ω)). Finally, by the continue principle (see [11]), we know that if u(t)H < ∞, the solution u(t) should be continued, see also [21, p. 158] for a similar argument. Hence, if Tmax < ∞, it follows lim
2 2 A1 u(t)H 2 + ut (t)2 = ∗
t→Tmax
2 lim u(t)H = ∞.
t→Tmax
(4.5)
Consider the energy function E(t) =
1 u(t)2 2 + 1 ut (t)2 + 1 au(t), u(t) − H∗ 2 2 2 2 2
1 u(t)p , p p
which satisfies that t E(t) + μ
ut (τ )2 dτ = E(s), 2
for every 0 ≤ s ≤ t < Tmax .
(4.6)
s
Clearly, E(t) is nonincreasing. Therefore,
1 u(t)2 2 + 1 ut (t)2 + 1 au(t), u(t) ≤ 1 u(t)p + E(0), H∗ 2 p 2 2 2 2 p
for all t ∈ [0, Tmax ).
By Lemma 2.2, there holds A1 u(t)2 2 + 1 ut (t)2 ≤ 1 u(t)p + E(0), H 2 p ∗ 2 2 p
for all t ∈ [0, Tmax ).
(4.7)
According to (4.5), it implies that lim u(t)p = ∞.
t→Tmax
By (2.1), we have lim u(t)H 2 = ∞.
t→Tmax
(4.8)
∗
Moreover, by (4.7) A1 u(t)2 2 ≤ 1 u(t)p + E(0), H p ∗ 2 p which combined with Gagliardo–Nirenberg inequality yields that 2 p p(1−α) u(t)pα2 , C u(t)H 2 − C ≤ u(t)p ≤ C u(t)q H ∗
∗
for α =
2(p − q) . p(q + 2)
If α ∈ (0, 1) such that pα < 2, that is (p − 2)/2 < q < p, then the above inequality and (4.8) immediately yield lim u(t)q = ∞,
t→Tmax
for q ≥ 1 such that q > (p − 2)/2.
This finally completes the proof of Theorem 3.1.
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5. Proof of Theorem 3.2 We first prove the second inequality, assume that u = c sin x is a stationary solution to the problem (1.7)–(1.8). Then it satisfies that I(u) = 0, namely, u2H∗2 + (au, u)2 = upp , which together with a ≤ a2 yields that cp sin xpp = c2 sin x2H∗2 + c2 (a sin x, sin x)2 ≤ c2 sin x2H∗2 + c2 a2 sin x22 , then
1/(p−2)
c≤
π(1 + a2 ) 2sin xpp
1
1
1 p−2
1 p−2
.
Particularly, if p ∈ N, then c≤
p!! · 12 ) p−2 (1 + a2 ) p−2 , ( (p−1)!! p!! · π4 ) ( (p−1)!!
(1 + a2 )
,
p is an even number, p is an odd number.
Therefore, d = inf J(u) ≤ J(c sin x) u∈N
p−2 p c sin xpp 2p p−2 Cp . ≤ 2p =
On the other hand, consider a real function ϕ(s) defined by ϕ(s) =
1 1 A1 s2 − Spp sp , 2 p
s > 0.
A direct computation entails that max ϕ(s) = s>0
p − 2p p − 2 p−2 A1 Sp p−2 . 2p
Therefore, by Lemmas 2.1, 2.2 d=
inf
max J(λu)
u∈H∗2 \{0} λ>0
1 1 1 λu2H∗2 + (aλu, λu)2 − λupp 2 p u∈H∗ \{0} λ>0 2 1 1 max A1 λu2H∗2 − Spp λupH 2 ≥ inf ∗ p u∈H∗2 \{0} λ>0 2
=
≥ Then we complete the proof.
inf 2
max
p − 2p p − 2 p−2 A1 Sp p−2 . 2p
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6. Proof of Theorem 3.3 First we prove that Tmax = ∞ and limt→∞ (u(t)H∗2 + ut (t)2 ) = 0. Without loss of generality, we assume that t0 = 0. If u(0) ∈ W , E(0) < D ≤ d, then u(t) ∈ W
for every t ∈ [0, Tmax ).
and E(t) < d,
In fact, since E(t) is a nonincreasing function, E(t) ≤ E(0) < D ≤ d for all t ∈ [0, Tmax ). Then suppose that there exists t¯ > 0 such that u(t¯) ∈ N . However, by (2.6), it follows that
d ≤ J u(t¯) ≤ E(t¯) < d, which is impossible. Thus, u(t) ∈ W for all t ∈ [0, Tmax ). Then for all t ∈ [0, Tmax ),
p − 2
I(u(t)) u(t)2 2 + au(t), u(t) J u(t) = + H∗ 2 2p p
p − 2 u(t)2 2 + au(t), u(t) . ≥ H 2 ∗ 2p
(6.1)
By (4.6), we have
1 ut (t)2 + J u(t) + μ 2 2
t
ut (τ )2 dτ = E(0) < d. 2
0
Hence,
1 ut (t)2 + p − 2 u(t)2 2 + au(t), u(t) ≤ C, 2 H 2 ∗ 2 2p
(6.2)
which implies that Tmax = ∞ by the continue principle. Moreover, t
ut (τ )2 dτ ≤ C , 2 μ
for every t ∈ [0, ∞).
(6.3)
0
Note that the following trivial inequality holds d
(1 + t)E(t) ≤ E(t). dt Recalling that pJ(u(t)) = we get
p−2 2 2 (u(t)H∗2
+ (au(t), u(t))2 ) + I(u(t)) and integrating this inequality on [0, t],
t (1 + t)E(t) = E(0) +
1 J u(τ ) dτ + 2
0
= E(0) +
1 p
t
ut (τ )2 dτ 2
0
t 0
1 I u(τ ) dτ + 2
t
ut (τ )2 dτ 2
0
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t
727
u(τ )2 2 + au(τ ), u(τ ) dτ H 2
(6.4)
∗
0
for all t ∈ [0, ∞). A simple computation induces that d utt (t), u(t) = dt
2 ut (t)u(t) − ut (t)2 ,
for a.e. t ∈ [0, ∞).
Ω
Testing the equation in (1.7) with u(t), we have 2 utt (t), u(t) + u(t)H 2 + μ
p au2 (t) = u(t)p ,
ut (t)u(t) +
∗
Ω
for a.e. t ∈ [0, ∞).
(6.5)
Ω
Thus, d dt
2 μ ut (t)u(t) + u(t)2 2
2
= ut (t)2 − I u(t) .
(6.6)
Ω
Integrating this identity on [0, t], by (6.2), (6.3) and Lemmas 2.1, 2.2, we obtain t
I u(τ ) dτ =
0
t
ut (τ )2 dτ + 2
0
t ≤
u0 u1 + Ω
μ u0 22 − 2
ut (t)u(t) −
μ u(t)2 2 2
Ω
ut (τ )2 dτ + u0 22 u1 22 + μ u0 22 + 2 2
1 ut (t)2 + 2 2
1 u(t)2 2 2
0
≤ C(μ)
(6.7)
for every t ∈ [0, ∞). By Lemmas 2.1, 2.2 and (6.1) p/2
upp ≤ Spp upH 2 ≤ Spp A2 ∗
=
p/2
Spp A2 u2H∗2
u2H∗2 + (au, u)2
+ (au, u)2
p/2
(p−2)/2
u2H∗2 + (au, u)2
(p−2)/2
2p
J u(t) u2H∗2 + (au, u)2 p−2 (p−2)/2
2p p p/2 E(0)(p−2)/2 u2H∗2 + (au, u)2 , ≤ S p A2 p−2 p/2
≤ Spp A2
which implies that 2
γ u(t)H 2 + au(t), u(t) 2 ≤ I u(t) , ∗
p/2
2p (p−2)/2 where γ = 1 − Spp A2 ( p−2 ) E(0)(p−2)/2 > 0.
(6.8)
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Then combining (6.3), (6.4), (6.7) and (6.8), we get E(t) ≤
C , t
for every t ∈ [0, ∞).
Consequently, by (6.1) we immediately have
u(t)2 2 + au(t), u(t) + ut (t)2 ≤ C , H∗ 2 2 t
for every t ∈ [0, ∞),
which combining with Lemma 2.2 tells us that
lim u(t)H 2 + ut (t)2 = 0.
t→∞
∗
Conversely, if Tmax = ∞ and u(t)2H 2 + ut (t)22 → 0 as t → ∞, then it follows that by Lemmas 2.1, 2.2 ∗
lim u(t)p = 0
t→∞
and
lim u(t)H 2 + au(t), u(t) 2 + ut (t)2 = 0,
t→∞
∗
which imply that lim E(t) = 0.
t→∞
Therefore, by Lemma 2.4 and the above mentioned results, there must be t0 > 0 such that E(t0 ) < D and u(t0 ) ∈ W . 7. Proof of Theorem 3.4 Firstly, we assume that there exists t0 ≥ 0 such that u(t0 ) ∈ U and E(t0 ) < d. Without loss of generality, we may suppose that t0 = 0, then u(t) ∈ U
and E(t) < d,
for every t ∈ [0, Tmax ).
Indeed, (4.6) entails that E(t) ≤ E(0) < d,
for all t > 0.
Then suppose that there exists t¯ > 0 such that u(t¯) ∈ N , but by (2.6),
d ≤ J u(t¯) ≤ E(t¯) < d, which is a contradiction to (7.1), and therefore u(t) ∈ U for all t ∈ [0, Tmax ). From (2.5), the definition of d, we obtain that
d=
2 p/(p−2) p − 2 (uH∗2 + (au, u)2 ) , u∈H∗ \{0} 2p (upp )2/(p−2)
inf 2
then (u2H 2 + (au, u)2 )p/(p−2) 2pd ∗ ≤ . p−2 (upp )2/(p−2)
(7.1)
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Since u(t) ∈ U for all t ∈ [0, Tmax ), I(u(t)) = u(t)2H 2 + (au(t), u(t))2 − u(t)pp < 0, the above inequality ∗ reads 2
2pd < u(t)H 2 + au(t), u(t) 2 , ∗ p−2
for every t ∈ [0, Tmax ).
(7.2)
Now we prove that Tmax < ∞. We follow closely [9, Theorem 3.11]. Assume by contradiction that Tmax = ∞. For any T > 0, we define a continuous positive function by 2 θ(t) = u(t)2 + μ
t
u(τ )2 dτ + μ(T − t)u0 22 , 2
for all t ∈ [0, T ].
0
With the same spirit (but it needs to be careful for the computations since the presence of the term au) as in [9], we have θ(t)θ (t) −
p+2 2 θ (t) ≥ C > 0, 4
for a.e. t ∈ [0, T ].
Setting y(t) = θ(t)−(p−2)/4 , the above differential inequality becomes y (t) ≤ −
p−2 Cy(t), 4
for a.e. t ∈ [0, T ],
−(p+2)/4 which together with y (t) = − p−2 < 0 proves that y(t) = 0 at some time, say as t = T ∗ 4 θ(t) (independent of T). This tells us that
lim θ(t) = ∞,
t→T ∗
that is t
2 lim∗ u(t)2 = ∞
or
t→T
lim∗
t→T
u(τ )2 dτ = ∞. 2
0
If limt→T ∗ u(t)22 = ∞, by (2.1), we get 2 lim u(t)H 2 = ∞
t→T ∗
If limt→T ∗
t 0
∗
(7.3)
u(τ )22 dτ = ∞ holds, then 2 lim u(t)2 = ∞,
t→T ∗
and we also have (7.3). Therefore, u(t) cannot be a global solution. That is Tmax < ∞. Conversely, assume that there exists no t ≥ 0 such that u(t) ∈ U and E(t) < d, we must have either u(t) ∈ W and E(t) < d (this is impossible by Theorem 3.3) or E(t) ≥ d for all t ≥ 0. Then from (4.6), t μ 0
ut (τ )2 dτ ≤ E(0) − d, 2
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by Jensen’s inequality, for any t ≥ 0 there holds t t
ut (τ )2 dτ ≥ 2
0
t
2 ut (τ )dτ
2 ≥ u(t)2 − u0 22 ,
0
Ω
hence, u(t)2 ≤ CT , 2
for all t ∈ [0, T ),
(7.4)
for any finite T > 0. If Tmax < ∞, we know that from Theorem 3.1 lim u(t)p = ∞,
t→Tmax
hence, by Lemmas 2.1 and 2.2, for every M > E(0), there exists some 0 < t¯ < Tmax such that M<
p − 2 u(t) 2 + au(t), u(t) , H 2 ∗ 2p
for every t ≥ t¯.
(7.5)
Denote V(t) = M − E(t),
for every t ≥ t¯,
(7.6)
ut (τ )2 dτ,
(7.7)
and then from (4.6) t V(t) = V0 + μ
2
0
where V0 = M − E(0). Therefore, by (7.5)–(7.7)
0 < V(t) ≤ M − J u(t) ≤ M −
p 1 p M + u(t)p , p−2 p
then u(t)p ≥ p p
2M + V(t) p−2
> pV(t).
(7.8)
Since p > 2, by the Hölder, Young inequalities and (7.7)–(7.8), we have
μ u(t), ut (t) ≤ Cμu(t)1−k u(t)k ut (t) 2 2 p p 1−k 2k 2 ≤ C u(t)p μν u(t)p + C(ν)−1 ut (t)2 2k
˙ < CV (1−k)/p (t) μν u(t)p + C(ν)−1 V(t) (1−k)/p u(t)p + C(ν)−1 V˙ 1/κ (t), ≤ CνV0 p where ν > 0, k ∈ (1, p/2) and then κ = (1 + (1 − k)/p)−1 ∈ (1, 2p/(p + 2)).
(7.9)
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Next, we estimate I(u(t)). Since V0 < M , then from (7.7)–(7.8) p 2M + (p − 2)V0 , (V0 − M ) u(t) p ≤ (V0 − M )p p−2 that is, 2(V0 − M ) ≥
p − 2 2(V0 − M ) u(t)p . p p 2M + (p − 2)V0
(7.10)
Hence, by (7.6) and (7.10)
p − 2 u(t)p ≥ −2E(t) + p − 2 u(t)p −I u(t) = −2J u(t) + p p p p
p − 2 u(t)p ≥ 2(V0 − M ) + p − 2 u(t)p ≥ 2 V(t) − M + p p p p pV0 p−2 u(t)p . ≥ p p 2M + (p − 2)V0
(7.11)
Now we consider a function defined by
F(t) ≡ V(t)1/κ + ε u(t), ut (t) 2 ,
for every t ≥ t¯.
(7.12)
By (7.9) and (7.11), taking ν > 0 sufficiently small, for (6.6) we have 2
d
u(t), ut (t) 2 = ut (t)2 − I u(t) − μ u(t), ut (t) 2 dt p 2 (1−k)/p u(t)p − C(ν)−1 V˙ 1/κ (t) ≥ ut (t) + C u(t) − CνV 2
0
p
p
2
(1−k)/p u(t)p − C(ν)−1 V˙ 1/κ (t) ≥ ut (t) + C − CνV 0
2
p
2 p ≥ ut (t) + C u(t) − C(ν)−1 V˙ 1/κ (t). 2
p
Therefore, if ε > 0 is small enough, p 2
˙ F(t) ≥ C ut (t)2 + u(t)p > 0. Then choosing ε > 0 even smaller if needed, we get 1/κ
F(t) ≥ F0 ≡ V0
+ ε u(0), ut (0) 2 > 0.
Utilizing (7.8) and by the Hölder, Young inequalities, we have
κ F κ (t) ≤ 2κ−1 V(t) + εκ u(t), ut (t) 2 p 2
≤ C u(t)p + ut (t)2 , which together with (7.13) yields that ˙ F(t) ≥ CF κ (t);
(7.13)
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732
then by Lemma 2.10 in [4], F(t) blows up at T ∗ > t¯ and we have F(t) ≥
C . (T ∗ − t)1/(κ−1)
(7.14)
By (7.9), (7.11) and since E(t) ≥ d, V(t) ≤ M − d, u(t)2 = u(t¯)2 + 2 2 2 ε
t
F(τ ) − V 1/κ (τ ) dτ
t¯
2 2 ≥ u(t¯)2 + ε
t t¯
2 ≥ u(t¯)2 − C + C
C 1/κ dτ − (M − d) (T ∗ − τ )1/(κ−1)
1 T∗ − t
(2−κ)/(κ−1)
−
(2−κ)/(κ−1) 1 . T ∗ − t¯
Consequently, 2 lim∗ u(t)2 = ∞,
t→T
which contradicts (7.4). The proof is complete. Acknowledgment The author wishes to thank the China Scholarship Council for supporting him in research in Italy. References [1] O.H. Ammann, T. von Kármán, G.B. Woodruff, The Failure of the Tacoma Narrows Bridge, Federal Works Agency, 1941. [2] J.M.W. Brownjohn, Observations on non-linear dynamic characteristics of suspension bridges, Earthquake Engrg. Struct. Dyn. 23 (1994) 1351–1367. [3] Y. Ebihara, M. Nakao, T. Nambu, On the existence of global classical solution of initial-boundary value problem for u − u3 = f , Pacific J. Math. 60 (1975) 63–70. [4] J. Esquivel-Avila, The dynamics of a nonlinear wave equation, J. Math. Anal. Appl. 279 (2003) 135–150. [5] A. Ferrero, F. Gazzola, A partially hinged rectangular plate as a model for suspension bridges, preprint. [6] P. Galenko, D. Danilov, V. Lebedev, Phase-field-crystal and Swift–Hohenberg equations with fast dynamics, Phys. Rev. E 79 (2009) 051110 (11 pp.). [7] F. Gazzola, Finite time blow-up and global solutions for some nonlinear parabolic equations, Differential Integral Equations 17 (2004) 983–1012. [8] F. Gazzola, Nonlinearity in oscillating bridges, Electron. J. Differential Equations 2013 (2013), No. 211, 47 pp. [9] F. Gazzola, M. Squassina, Global solutions and finite time blow up for damped semilinear wave equations, Ann. Inst. H. Poincaré Anal. Non Linéaire 23 (2006) 185–207. [10] F. Gazzola, T. Weth, Finite time blow-up and global solutions for semilinear parabolic equations with initial data at high energy level, Differential Integral Equations 18 (2005) 961–990. [11] J.K. Hale, G. Raugel, Convergence in gradient-like systems with applications to PDE, Z. Angew. Math. Phys. 43 (1992) 63–124. [12] R. Ikehata, T. Suzuki, Stable and unstable sets for evolution equations of parabolic and hyperbolic type, Hiroshima Math. J. 26 (1996) 475–491. [13] W. Lacarbonara, Nonlinear Structural Mechanics, Springer, 2013. [14] A.C. Lazer, P.J. McKenna, Large-amplitude periodic oscillations in suspension bridges: some new connections with nonlinear analysis, SIAM Rev. 32 (1990) 537–578. [15] A.C. Lazer, P.J. McKenna, Global bifurcation and a theorem of Tarantello, J. Math. Anal. Appl. 181 (1994) 648–655. [16] H.A. Levine, J. Serrin, Global nonexistence theorems for quasilinear evolution equations with dissipation, Arch. Ration. Mech. Anal. 137 (1997) 341–361. [17] A.M. Micheletti, A. Pistoia, Nontrivial solutions for some fourth-order semilinear elliptic problems, Nonlinear Anal. 34 (1998) 509–523.
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Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Finite time blow-up and global solutions for fourth order damped wave equations Yongda Wang Dipartimento di Matematica, Politecnico di Milano, Piazza Leonardo da Vinci 32, 20133 Milano, Italy
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 18 October 2013 Available online 12 April 2014 Submitted by P. Yao
This work is devoted to a class of fourth order wave equations with linear damping term and superlinear source term. After showing the uniqueness and existence of local solutions to the equations, we give necessary and sufficient conditions for global existence and finite time blow-up of these solutions. Moreover, the potential well depth is estimated. © 2014 Elsevier Inc. All rights reserved.
Keywords: Fourth order wave equation Initial-boundary value problem Damping term Strong source term
1. Introduction The report [1] about the Tacoma Narrows Bridge collapse [25,26] considers . . . the crucial event in the collapse to be the sudden change from a vertical to a torsional mode of oscillation, see also [25, p. 63]. Hence, if one models a suspension bridge by a beam, there is no way to highlight the torsional oscillations. The nonlinear behavior of suspension bridges, which is by now well established, see [2,8,13,23], also plays a crucial role in causing oscillations. Therefore, a reliable model for suspension bridges should be nonlinear and it should have enough degrees of freedom to display torsional oscillations. In this respect, Lazer and McKenna [14, Problem 11] suggested to study the following equation Δ2 u + c2 Δu + h(u) = 0,
in Rn ,
(1.1)
where h(u) ≈ [u + 1]+ − 1 with u+ = max{u, 0}. Subsequently, equations “like” (1.1), namely, Δ2 u + c2 Δu = b (u + 1)+ − 1 ,
in Σ ⊂ Rn ,
(1.2)
under the Navier boundary condition have been considered in several papers. For the case c2 < λ1 ((λk )k≥1 , the sequence of eigenvalues of −Δ in H01 (Σ)), we refer to [15,18,27]. Tarantello [27] proved that if E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jmaa.2014.04.015 0022-247X/© 2014 Elsevier Inc. All rights reserved.
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b ≥ λ1 (λ1 − c2 ), then (1.2) has a nontrivial solution. Lazer and McKenna [15] discussed the case n = 1 when Σ is an interval and found that (1.2) has at least 2k − 1 nontrivial solutions if b > λk (λk − c2 ). Micheletti and Pistoia [18] studied (1.2) for a more general nonlinearity which is “like” b[(u + 1)+ − 1] and obtained the multiplicity existence of nontrivial solutions. When c2 ≥ λ1 , Micheletti and Pistoia considered (1.2) when b[(u + 1)+ − 1] is replaced by a more general nonlinearity in [17], where they showed the existence of a nontrivial solution. Then Micheletti, Pistoia and Saccon [19] improved the result in [17] and gave the multiplicity results for the same problem. In other papers [32–34] this kind of problem was also investigated under the Navier boundary condition. More recently, Ferrero and Gazzola [5] suggested one should consider the boundaries of a plate Ω = (0, π) × (−l, l) which represents the roadway of a suspension bridge as follows. Because the edges x = 0, π connect with the ground, they are assumed to be hinged and then u(0, y) = uxx (0, y) = u(π, y) = uxx (π, y) = 0,
y ∈ (−l, l),
(1.3)
while the edges y = ±l are free and the boundary conditions there become (see [30, 2.40]) uyy (x, ±l) + σuxx (x, ±l) = 0,
uyyy (x, ±l) + (2 − σ)uxxy (x, ±l) = 0,
x ∈ (0, π).
(1.4)
The free boundaries (1.4) yield small stretching energy for the plate, so Ferrero and Gazzola took c = 0 in (1.1) and introduced a model for the stationary suspension bridge Δ2 u + h(x, y, u) = f (x, y),
in Ω
(1.5)
as well as a model for the nonlinear dynamical suspension bridge utt + Δ2 u + μut + h(x, y, u) = f (x, y, t),
in Ω × (0, T ),
(1.6)
here h(x, y, u) is restoring force due to the hangers of the suspension bridge, f (x, y) or f (x, y, t) is the external force including the gravity. Given an open rectangular plate Ω = (0, π) × (−l, l) ⊂ R2 , we consider the following initial value problem ⎧ 2 p−2 ⎨ utt + Δ u + μut + au = |u| u, (x, y, t) ∈ Ω × [0, T ], (x, y) ∈ Ω, u(x, y, 0) = u0 (x, y), ⎩ ut (x, y, 0) = u1 (x, y), (x, y) ∈ Ω,
(1.7)
with the boundary condition
y ∈ (−l, l), u(0, y, t) = uxx (0, y, t) = u(π, y, t) = uxx (π, y, t) = 0, uyy (x, ±l, t) + σuxx (x, ±l, t) = uyyy (x, ±l, t) + (2 − σ)uxxy (x, ±l, t) = 0, x ∈ (0, π),
(1.8)
for every t ∈ [0, T ], where T > 0, μ > 0, 2 < p < ∞, σ ∈ (0, 12 ) and a = a(x, y, t) is a sign-changing and bounded measurable function. The initial data u0 , u1 belong to suitable spaces, which will be specified later on. Here we study the fourth order wave problem (1.7) with the boundary condition (1.8), because it comes from the physical model for the dynamic suspension bridge if we assume that au describes the restoring force because of the hangers of the suspension bridge and |u|p−2 u is the strong source term which represents the other external forces acting on the bridge. We explain this model briefly below, see for more details [5].
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Let u be the displacement of the plate in the vertical direction, assume that f denotes the external force acting on the plate, then for a small displacement u, the total energy of the plate Ω approximately equals to by a suitable normalization ET (u) =
1 (Δu)2 + (1 − σ) u2xy − uxx uyy − f u dxdy. 2
(1.9)
Ω
Due to the hangers of the bridge, there exists another force h with a potential energy given by then the total energy becomes by adding this potential energy to (1.9): ET (u) =
Ω
Hdxdy,
1 (Δu)2 + (1 − σ) u2xy − uxx uyy + H − f u dxdy. 2
(1.10)
Ω
If the external force f also depends on time, the deformation u has a kinetic energy given as follows by a suitable scaling:
1 2 u dxdy, 2 t
Ω
which should be added to the nonlinear static energy (1.10): Eu (t) = Ω
1 2 u dxdy + 2 t
2 1 2 (Δu) + (1 − σ) uxy − uxx uyy + H(x, y, u) − f u dxdy. 2
(1.11)
Ω
This is the total energy of a nonlinear dynamic bridge. As for the action, we take the difference between the kinetic energy and the potential energy and integrate on [0, T ]: T A (u) = 0
Ω
1 2 u dxdy − 2 t
1 (Δu)2 + (1 − σ) u2xy − uxx uyy + H(x, y, u) − f u dxdy dt. 2
Ω
Taking the critical points of the functional A (u), we have Eq. (1.6) by adding a damping term μut due to internal friction. Eq. (1.6) also appears in other contexts, see [6, (17)], and is sometimes called the Swift–Hohenberg equation. In the present paper, our purpose is to investigate the initial-boundary value problem (1.7)–(1.8). We first set up the existence and uniqueness of local solution to (1.7)–(1.8), then the behavior of the local solution is analyzed by utilizing the potential well theory developed by Sattinger [24] in 1968 to solve the second order evolution equations. We give necessary and sufficient conditions for the global existence and finite time blow-up of solutions to (1.7)–(1.8). For the related results on second order evolution problems, there are works by Ebihara, Nakao and Nambu [3], Ikehata and Suzuki [12], Levine and Serrin [16], Payne and Sattinger [22], Tsutsumi [29,28], Gazzola [7], Gazzola and Squassina [9], Gazzola and Weth [10]. This paper is organized as follows. In Section 2 we prepare several preliminary definitions and recall some basic facts. Also, some important lemmas are proved. We state the main results including local existence, global existence and finite time blow-up of solutions to (1.1)–(1.2) as well as the estimation on the potential well depth in Section 3, see Theorems 3.1, 3.2, 3.3 and 3.4. The last four sections are devoted to the proofs of the main results.
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2. Preliminaries Letting Ω = (0, π) × (−l, l) ⊂ R2 , we denote the standard Lq (Ω) norm by · q for 1 ≤ q < ∞ and the H (Ω) norm by · H 2 = ( · 22 + D2 · 22 )1/2 . In addition, we introduce another function space defined in [5], 2
H∗2 = H∗2 (Ω) := u ∈ H 2 (Ω) : u = 0 on {0, π} × (−l, l) , the dual space of which is defined by H(Ω) and the corresponding duality between them is denoted by ·,· . Clearly, H∗2 satisfies H02 (Ω) ⊂ H∗2 (Ω) ⊂ H 2 (Ω) and is a Hilbert space when endowed with the inner product (u, v)
H∗2
ΔuΔv + (1 − σ)
= Ω
2uxy vxy − uxx vyy − uyy vxx ,
for every u, v ∈ H∗2 .
Ω
This inner product induces a norm uH∗2 =
1/2
|Δu|2 + 2(1 − σ)
Ω
u2xy − uxx uyy
,
for every u ∈ H∗2 ,
Ω
which is equivalent to · H 2 for 0 < σ < 1/2, see for the proof [5, Lemma 7]. Moreover, we have a Sobolev embedding inequality for this case. Lemma 2.1. Assume that 1 ≤ q < ∞. Then for any u ∈ H∗2 , the inequality uq ≤ Sq uH∗2 π holds, where Sq = ( 2l +
(2.1)
√
2 1 (q+2)/2q ( 1−σ )1/2 . 2 )(2πl)
Remark 2.1. Here Sq is not less than the best Sobolev embedding constant. How to obtain the best constant is still open. ¯ and for any (x, y) ∈ Ω, there results Proof of Lemma 2.1. Take any u ∈ H∗2 ⊂ C 1 (Ω) x π π 1/2
2 √ u(x, y) = ux (t, y)dt ≤ ux (x, y)dx ≤ π ux (x, y) dx 0
≤
√
π
0
≤
0
−uxx (x, y)u(x, y)dx
π 0
√
1/2
π π
uxx (x, y)2 dx
1/4 π
0
u(x, y)2 dx
0
which yields that π 0
u(x, y)2 dx ≤ π 4
π 0
uxx (x, y)2 dx.
1/4 ,
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Then it follows that from the two inequalities above u(x, y) ≤ π 3/2
π
uxx (x, y)2 dx
1/2 .
(2.2)
0
Furthermore, we have x π uy (x, y) = uyx (t, y)dt ≤ uyx (x, y)dx, 0
(2.3)
0
and y l u(x, y) = u(x, s) + uy (x, r)dr ≤ u(x, s) + uy (x, y)dy,
for any s ∈ (−l, l).
−l
s
Integrating the inequality (2.4) about s on (−l, l), we get l
2lu(x, y) ≤
u(x, s)ds + 2l
−l
l
uy (x, y)dy,
−l
which together with (2.2), (2.3) yields that u(x, y) ≤ 1 π 3/2 2l π 3/2 ≤ (2l)1/2 ≤
l π −l
uxx (x, s)2 dx
1/2
l π ds + −l 0
0
l π
uyx (x, y)dxdy
uxx (x, y)2 dxdy
l π
1/2 + (πl)1/2
−l 0
√ 1/2 2 2 2 π 1/2 D u dxdy + (2πl) . 2l 2
2 2uyx (x, y) dxdy
−l 0
Ω
From Lemma 7 in [5], we know that 2 (1 − σ)D2 u2 ≤ u2H∗2 . Therefore, u(x, y) ≤
√ 1/2 π 1 2 1/2 + (2πl) uH∗2 , 2l 2 1−σ
which implies that, |u|q ≤
√ q q/2 1 π 2 + (2πl)q/2 uqH 2 , ∗ 2l 2 1−σ
for 1 ≤ q < ∞. Integrating the above inequality on Ω, we have
1/2
(2.4)
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uq ≤
√ 1/2 1 π 2 (q+2)/2q + (2πl) uH∗2 , 2l 2 1−σ
which completes the proof. 2 Let {Λi }∞ i=1 be the eigenvalue sequence to the eigenvalue problem ⎧ 2 (x, y) ∈ Ω, ⎨ Δ u = Λu, y ∈ (−l, l), u(0, y) = uxx (0, y) = u(π, y) = uxx (π, y) = 0, ⎩ uyy (x, ±l) + σuxx (x, ±l) = uyyy (x, ±l) + (2 − σ)uxxy (x, ±l) = 0, x ∈ (0, π), which has been solved in [5]. Particularly, Λ1 < 1 and we have an elementary result. Lemma 2.2. Assume that −Λ1 < a1 ≤ a ≤ a2 . Then for any u ∈ H∗2 , there holds A1 u2H∗2 ≤ u2H∗2 + (au, u)2 ≤ A2 u2H∗2 , where (·,·)2 is the L2 inner product and A1 , A2 are given by A1 =
1+ 1,
a1 Λ1 ,
a1 < 0, a1 ≥ 0
and
A2 =
1, 1+
a2 Λ1 ,
a2 < 0, a2 ≥ 0.
Proof. We only consider the left inequality, the right one can be proved similarly. When the constant a1 < 0, since Λ1 (u, u)2 ≤ u2H 2 , then ∗
u2H∗2 + (au, u)2 ≥ u2H∗2 + a1 (u, u)2 ≥ u2H∗2 +
a1 u2H∗2 . Λ1
If a1 > 0, then (au, u)2 ≥ 0, hence u2H∗2 + (au, u)2 ≥ u2H∗2 . Therefore, the inequality A1 u2H∗2 ≤ u2H∗2 + (au, u)2 holds. 2 Now, we define the Nehari functional I and the energy functional J on H∗2 → R, I(u) = u2H∗2 + (au, u)2 − upp , J(u) =
for every u ∈ H∗2 ,
1 1 1 u2H∗2 + (au, u)2 − upp , 2 2 p
for every u ∈ H∗2 ,
which play critical roles in dealing with our problem. Let u be an arbitrary nonzero element in H∗2 and consider a real value function defined by j(λ) = J(λu), then
λ ≥ 0,
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j (λ) = λu2H∗2 + λ(au, u)2 − λp−1 upp , j (λ) = u2H∗2 + (au, u)2 − (p − 1)λp−2 upp . Clearly, j(0) = j (0) = 0 and j (0) = u2H 2 + (au, u)2 > 0 for a > −Λ1 . Thus for any 0 = u ∈ H∗2 , j(λ) is ∗ a convex function for small λ > 0 and has the following behaviors. Lemma 2.3. Assume that −Λ1 < a1 ≤ a ≤ a2 . Then for any nontrivial u ∈ H∗2 , (i) limλ→∞ j(λ) = −∞; ¯ = λ(u) ¯ ¯ = 0; (ii) there exists a unique λ > 0 such that j (λ) ¯ (iii) j (λ) < 0. Proof. The proof is almost similar to that of Lemma 2.2 in [22], the only difference is to deal with the term (au, u)2 , which can be treated by applying Lemma 2.2, so we omit it. 2 Then, we could define the potential well depth of the functional J (also known as mountain pass level) by d=
inf
max J(λu).
u∈H∗2 \{0} λ>0
(2.5)
Denote the set of all nontrivial stationary solutions to the problem (1.7)–(1.8) by N = u ∈ H∗2 \{0} : I(u) = 0 , which is the so-called Nehari manifold, see [20,31]. By considering a map s → I(su) for all u such that u2H 2 = 1 and Lemma 2.3, it is easy to check that each half line starting from the origin of H∗2 intersects ∗ only once the manifold N and N separates the two sets N+ = u ∈ H∗2 : I(u) > 0 ∪ {0}
and N− = u ∈ H∗2 : I(u) < 0 .
Then the stable set W and unstable set U may be defined by W = u ∈ N+ : J(u) < d ,
U = u ∈ N− : J(u) < d .
Lemma 2.4. (See [4,12].) The following properties of W and U hold: (i) W is a neighborhood of 0 ∈ H∗2 ; ¯ (closure in H 2 ). (ii) 0 ∈ /U ∗ As Payne and Sattinger did in [22], the potential well depth d defined in (2.5) can be also characterized as d = inf J(u). u∈N
(2.6)
Finally, we consider an energy functional E : H∗2 (Ω) × L2 (Ω) → R defined by 1 E (v, w) = J(v) + w22 , 2
for every pair (v, w) ∈ H∗2 (Ω) × L2 (Ω),
and the Lyapunov function E(t) = E (u(t), ut (t)), defined for any solution u(t) of the problem (1.7)–(1.8).
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3. Main results In this section, the main results on our problem are listed. To begin with, we explain what does a weak solution mean to our problem (1.7)–(1.8). Definition 3.1. A function u ∈ C([0, T ], H∗2 (Ω)) ∩ C 1 ([0, T ], L2 (Ω)) ∩ C 2 ([0, T ], H(Ω)) with ut ∈ L2 ([0, T ], L2 (Ω)) is called a weak solution to (1.7)–(1.8), if the following conditions hold u(0) = u0 ,
ut (0) = u1
and utt , η + (u, η)H∗2 + μ
ut η +
Ω
|u|p−2 uη
auη = Ω
Ω
for all η ∈ H∗2 (Ω) and a.e. t ∈ [0, T ]. Then we are concerned with the existence and uniqueness of local solution to the problem (1.7)–(1.8). Theorem 3.1. Let μ > 0, 2 < p < ∞ and −Λ1 < a1 ≤ a ≤ a2 . Then for any u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω), there exists T > 0 such that the problem (1.7)–(1.8) has a unique local weak solution u on [0, T ]. Moreover, if Tmax = sup T > 0 : u = u(t) exists on [0, T ] < ∞, then lim u(t)q = ∞,
for q ≥ 1 such that q > (p − 2)/2.
t→Tmax
Remark 3.1. If Tmax = ∞, we say that the weak solution of (1.7)–(1.8) is global, while if Tmax < ∞, then the weak solution blows up and Tmax is the blow-up time. Before we give necessary and sufficient conditions for global existence and finite time blow-up of solutions to (1.7)–(1.8), we estimate the potential well depth. Theorem 3.2. Assume that −Λ1 < a1 ≤ a ≤ a2 , 2 < p < ∞. Then the potential well depth d can be estimated by 0<
p − 2p p − 2 p−2 p−2 A1 Sp p−2 ≤ d ≤ Cp , 2p 2p
where Sp is the Sobolev embedding constant, A1 is defined in Lemma 2.2 and Cp is given by Cp =
π(1 + a2 ) 2sin x2p
p/(p−2) .
In particular, if p is an integral number, then Cp =
2
p
2 p−2
p p−2
p!! · 12 ) p−2 (1 + a2 ) p−2 πl, ( (p−1)!! p!! ( (p−1)!!
here p!! = p · (p − 2) · (p − 4) · · · .
· π4 )
(1 + a2 )
πl,
p is an even number, p is an odd number,
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Remark 3.2. Since we still cannot obtain the best embedding constant in (2.1), we only have a range about the depth d. Now, we state the necessary and sufficient conditions for global existence of solutions to the problem (1.7)–(1.8). Theorem 3.3. Let u be the unique local weak solution to (1.7)–(1.8). Assume that u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω). Then u is a global solution and lim u(t)H 2 + ut (t)2 = 0
t→∞
∗
if and only if there exists a real number t0 ∈ [0, Tmax ) such that u(t0 ) ∈ W p − p−2
where D = min{d, p−2 2p A2
2p − p−2
Sp
and
E(t0 ) < D,
}.
Finally, we come to the blowing up result of the solutions to problem (1.7)–(1.8). Theorem 3.4. Let u be the unique local weak solution to (1.7)–(1.8). Assume that u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω). Then u blows up, that is, Tmax < ∞ if and only if there exists a real number t0 ∈ [0, Tmax ) such that u(t0 ) ∈ U
and
E(t0 ) < d.
4. Proof of Theorem 3.1 We start with some definitions. For every T > 0, set the space
H = C [0, T ], H∗2 (Ω) ∩ C 1 [0, T ], L2 (Ω) with the norm uH =
2 2 1/2
max A1 u(t)H 2 (Ω) + ut (t)2 ,
t∈[0,T ]
∗
where A1 is given in Lemma 2.2. For u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω), denote MT = u ∈ H : u(0) = u0 , ut (0) = u1 and u2H ≤ R2 , where R2 ≥ 2(A2 u0 2H 2 + u1 22 ) and A2 is given in Lemma 2.2. ∗
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Then we consider the initial problem ⎧ 2 p−2 ⎨ vtt + Δ v + μvt + av = |u| u, (x, y, t) ∈ Ω × [0, T ], v(x, y, 0) = u0 (x, y), (x, y) ∈ Ω, ⎩ (x, y) ∈ Ω, vt (x, y, 0) = u1 (x, y),
(4.1)
with the boundary condition
y ∈ (−l, l), v(0, y, t) = vxx (0, y, t) = v(π, y, t) = vxx (π, y, t) = 0, vyy (x, ±l, t) + σvxx (x, ±l, t) = vyyy (x, ±l, t) + (2 − σ)vxxy (x, ±l, t) = 0, x ∈ (0, π),
(4.2)
for every t ∈ [0, T ]. Lemma 4.1. Assume that u0 ∈ H∗2 (Ω), u1 ∈ L2 (Ω) and −Λ1 < a1 ≤ a ≤ a2 . Then for every u ∈ H, there exists a unique solution v ∈ H ∩ C 2 ([0, T ], H(Ω)) with vt ∈ L2 ([0, T ], L2 (Ω)) to the problem (4.1)–(4.2). Proof. According to Theorem 6 in [5], there is a unique weak solution v ∈ H ∩ C 2 ([0, T ], H(Ω)) to the problem ⎧ 2 ⎨ vtt + Δ v + μvt + h(x, y, v) = f, (x, y, t) ∈ Ω × [0, T ], (x, y) ∈ Ω, v(x, y, 0) = u0 (x, y), ⎩ vt (x, y, 0) = u1 (x, y), (x, y) ∈ Ω,
(4.3)
with the same boundary condition (4.2), where f = f (x, y, t) ∈ C 0 ([0, T ], L2 (Ω)). Hence, for the uniqueness and existence of solution to (4.1)–(4.2), it is sufficient to check that
|u|p−2 u ∈ C 0 [0, T ], L2 (Ω) . In fact, since u ∈ H, we have p−2 p−1 p−1 max u(t) u(t)2 ≤ max u(t)2p−2 ≤ C max u(t)H 2 ,
t∈[0,T ]
t∈[0,T ]
∗
t∈[0,T ]
which implies that |u|p−2 u ∈ C 0 ([0, T ], L2 (Ω)). Therefore, (4.1)–(4.2) admits a unique solution v ∈ H ∩ C 2 ([0, T ], H(Ω)). To complete we need to prove that vt ∈ L2 ([0, T ], L2 (Ω)). Take vt as a testing function into the equation in (4.1), integrate over Ω × [0, t] ⊂ Ω × [0, T ], that is t
t
vtt (τ ), vt (τ ) dτ +
0
v(τ ), vt (τ )
H∗2
t dτ + μ
0
t =
vt (τ )2 dτ + 2
0
t
av(τ ), vt (τ ) 2 dτ
0
u(τ )p−2 u(τ )vt (τ )dτ.
0 Ω
Recalling that a1 ≤ a ≤ a2 and by Lemma 2.2, we have vt (t)2 + A1 v(t) 2 + 2μ 2 H
t
∗
0
vt (τ )2 dτ ≤ u1 22 + A2 u0 H 2 + 2 ∗ 2
t 0 Ω
u(τ )p−2 u(τ )vt (τ )dτ. (4.4)
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Notice that t 2
u(τ )p−2 u(τ )vt (τ )dτ ≤ C
0 Ω
t
u(τ )2p−2 dt + μ
0 Ω
t ≤C
t
vt (τ )2 dτ 2
0
u(τ )2p−2 dt + μ 2
t
H∗
0
vt (τ )2 dτ 2
0
t ≤ CT + μ
vt (τ )2 dτ , 2
0
then it follows that from (4.4) T
vt (t)2 dτ ≤ CT, 2
0
which deduces that vt ∈ L2 ([0, T ], L2 (Ω)) and completes the proof of Lemma 4.1.
2
Now, we come to prove Theorem 3.1. By Lemma 4.1, for any u ∈ MT , we could introduce a map Φ : H → H defined by v = Φ(u), here v is the unique solution to (4.1)–(4.2). Claim. Φ is a contract map satisfying Φ(MT ) ⊆ MT , for small T > 0. In fact, assume that u ∈ MT , then the corresponding solution v = Φ(u) satisfies (4.4) for all t ∈ [0, T ]. Thus, as we did in the proof of Lemma 4.1, there holds vt 22 + A1 v2H∗2 ≤ u1 22 + A2 u0 2H∗2 + CR2p−2 T ≤
R2 + CR2p−2 T. 2
If T is small enough, then vH ≤ R, which implies that Φ(MT ) ⊆ MT . Let v1 = Φ(w1 ), v2 = Φ(w2 ) with w1 , w2 ∈ MT . Putting v1 , v2 in Eq. (4.1) and then subtracting the two equations, we obtain by setting v = v1 − v2
vtt , η + (v, η)H∗2 + μ(vt , η)2 + (av, η)2 = |w1 |p−2 w1 − |w2 |p−2 w2 ηdxdy Ω
γ(t)(w1 − w2 )ηdxdy
= Ω
for all η ∈ H∗2 and a.e. t ∈ [0, T ], where γ(t) is estimated by γ(t) ≤ (p − 1)(|w1 | + |w2 |)p−2 . Taking η = vt and arguing similarly as above, we have Φ(w1 ) − Φ(w2 )2 = v2H ≤ CR2p−4 T w1 − w2 2H . H Likely, if T is so small that CR2p−4 T < 1, then there exists a constant 0 < δ < 1 such that Φ(w1 ) − Φ(w2 )2 ≤ δw1 − w2 2H . H Therefore, Φ is a contract map, and the claim is proved.
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Then by the Contracting Mapping Principle, there exists a unique u ∈ MT such that u = Φ(u), which is the unique solution to the problem (1.7)–(1.8). Moreover, u ∈ C 2 ([0, T ], H(Ω)). Finally, by the continue principle (see [11]), we know that if u(t)H < ∞, the solution u(t) should be continued, see also [21, p. 158] for a similar argument. Hence, if Tmax < ∞, it follows lim
2 2 A1 u(t)H 2 + ut (t)2 = ∗
t→Tmax
2 lim u(t)H = ∞.
t→Tmax
(4.5)
Consider the energy function E(t) =
1 u(t)2 2 + 1 ut (t)2 + 1 au(t), u(t) − H∗ 2 2 2 2 2
1 u(t)p , p p
which satisfies that t E(t) + μ
ut (τ )2 dτ = E(s), 2
for every 0 ≤ s ≤ t < Tmax .
(4.6)
s
Clearly, E(t) is nonincreasing. Therefore,
1 u(t)2 2 + 1 ut (t)2 + 1 au(t), u(t) ≤ 1 u(t)p + E(0), H∗ 2 p 2 2 2 2 p
for all t ∈ [0, Tmax ).
By Lemma 2.2, there holds A1 u(t)2 2 + 1 ut (t)2 ≤ 1 u(t)p + E(0), H 2 p ∗ 2 2 p
for all t ∈ [0, Tmax ).
(4.7)
According to (4.5), it implies that lim u(t)p = ∞.
t→Tmax
By (2.1), we have lim u(t)H 2 = ∞.
t→Tmax
(4.8)
∗
Moreover, by (4.7) A1 u(t)2 2 ≤ 1 u(t)p + E(0), H p ∗ 2 p which combined with Gagliardo–Nirenberg inequality yields that 2 p p(1−α) u(t)pα2 , C u(t)H 2 − C ≤ u(t)p ≤ C u(t)q H ∗
∗
for α =
2(p − q) . p(q + 2)
If α ∈ (0, 1) such that pα < 2, that is (p − 2)/2 < q < p, then the above inequality and (4.8) immediately yield lim u(t)q = ∞,
t→Tmax
for q ≥ 1 such that q > (p − 2)/2.
This finally completes the proof of Theorem 3.1.
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5. Proof of Theorem 3.2 We first prove the second inequality, assume that u = c sin x is a stationary solution to the problem (1.7)–(1.8). Then it satisfies that I(u) = 0, namely, u2H∗2 + (au, u)2 = upp , which together with a ≤ a2 yields that cp sin xpp = c2 sin x2H∗2 + c2 (a sin x, sin x)2 ≤ c2 sin x2H∗2 + c2 a2 sin x22 , then
1/(p−2)
c≤
π(1 + a2 ) 2sin xpp
1
1
1 p−2
1 p−2
.
Particularly, if p ∈ N, then c≤
p!! · 12 ) p−2 (1 + a2 ) p−2 , ( (p−1)!! p!! · π4 ) ( (p−1)!!
(1 + a2 )
,
p is an even number, p is an odd number.
Therefore, d = inf J(u) ≤ J(c sin x) u∈N
p−2 p c sin xpp 2p p−2 Cp . ≤ 2p =
On the other hand, consider a real function ϕ(s) defined by ϕ(s) =
1 1 A1 s2 − Spp sp , 2 p
s > 0.
A direct computation entails that max ϕ(s) = s>0
p − 2p p − 2 p−2 A1 Sp p−2 . 2p
Therefore, by Lemmas 2.1, 2.2 d=
inf
max J(λu)
u∈H∗2 \{0} λ>0
1 1 1 λu2H∗2 + (aλu, λu)2 − λupp 2 p u∈H∗ \{0} λ>0 2 1 1 max A1 λu2H∗2 − Spp λupH 2 ≥ inf ∗ p u∈H∗2 \{0} λ>0 2
=
≥ Then we complete the proof.
inf 2
max
p − 2p p − 2 p−2 A1 Sp p−2 . 2p
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6. Proof of Theorem 3.3 First we prove that Tmax = ∞ and limt→∞ (u(t)H∗2 + ut (t)2 ) = 0. Without loss of generality, we assume that t0 = 0. If u(0) ∈ W , E(0) < D ≤ d, then u(t) ∈ W
for every t ∈ [0, Tmax ).
and E(t) < d,
In fact, since E(t) is a nonincreasing function, E(t) ≤ E(0) < D ≤ d for all t ∈ [0, Tmax ). Then suppose that there exists t¯ > 0 such that u(t¯) ∈ N . However, by (2.6), it follows that
d ≤ J u(t¯) ≤ E(t¯) < d, which is impossible. Thus, u(t) ∈ W for all t ∈ [0, Tmax ). Then for all t ∈ [0, Tmax ),
p − 2
I(u(t)) u(t)2 2 + au(t), u(t) J u(t) = + H∗ 2 2p p
p − 2 u(t)2 2 + au(t), u(t) . ≥ H 2 ∗ 2p
(6.1)
By (4.6), we have
1 ut (t)2 + J u(t) + μ 2 2
t
ut (τ )2 dτ = E(0) < d. 2
0
Hence,
1 ut (t)2 + p − 2 u(t)2 2 + au(t), u(t) ≤ C, 2 H 2 ∗ 2 2p
(6.2)
which implies that Tmax = ∞ by the continue principle. Moreover, t
ut (τ )2 dτ ≤ C , 2 μ
for every t ∈ [0, ∞).
(6.3)
0
Note that the following trivial inequality holds d
(1 + t)E(t) ≤ E(t). dt Recalling that pJ(u(t)) = we get
p−2 2 2 (u(t)H∗2
+ (au(t), u(t))2 ) + I(u(t)) and integrating this inequality on [0, t],
t (1 + t)E(t) = E(0) +
1 J u(τ ) dτ + 2
0
= E(0) +
1 p
t
ut (τ )2 dτ 2
0
t 0
1 I u(τ ) dτ + 2
t
ut (τ )2 dτ 2
0
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t
727
u(τ )2 2 + au(τ ), u(τ ) dτ H 2
(6.4)
∗
0
for all t ∈ [0, ∞). A simple computation induces that d utt (t), u(t) = dt
2 ut (t)u(t) − ut (t)2 ,
for a.e. t ∈ [0, ∞).
Ω
Testing the equation in (1.7) with u(t), we have 2 utt (t), u(t) + u(t)H 2 + μ
p au2 (t) = u(t)p ,
ut (t)u(t) +
∗
Ω
for a.e. t ∈ [0, ∞).
(6.5)
Ω
Thus, d dt
2 μ ut (t)u(t) + u(t)2 2
2
= ut (t)2 − I u(t) .
(6.6)
Ω
Integrating this identity on [0, t], by (6.2), (6.3) and Lemmas 2.1, 2.2, we obtain t
I u(τ ) dτ =
0
t
ut (τ )2 dτ + 2
0
t ≤
u0 u1 + Ω
μ u0 22 − 2
ut (t)u(t) −
μ u(t)2 2 2
Ω
ut (τ )2 dτ + u0 22 u1 22 + μ u0 22 + 2 2
1 ut (t)2 + 2 2
1 u(t)2 2 2
0
≤ C(μ)
(6.7)
for every t ∈ [0, ∞). By Lemmas 2.1, 2.2 and (6.1) p/2
upp ≤ Spp upH 2 ≤ Spp A2 ∗
=
p/2
Spp A2 u2H∗2
u2H∗2 + (au, u)2
+ (au, u)2
p/2
(p−2)/2
u2H∗2 + (au, u)2
(p−2)/2
2p
J u(t) u2H∗2 + (au, u)2 p−2 (p−2)/2
2p p p/2 E(0)(p−2)/2 u2H∗2 + (au, u)2 , ≤ S p A2 p−2 p/2
≤ Spp A2
which implies that 2
γ u(t)H 2 + au(t), u(t) 2 ≤ I u(t) , ∗
p/2
2p (p−2)/2 where γ = 1 − Spp A2 ( p−2 ) E(0)(p−2)/2 > 0.
(6.8)
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Then combining (6.3), (6.4), (6.7) and (6.8), we get E(t) ≤
C , t
for every t ∈ [0, ∞).
Consequently, by (6.1) we immediately have
u(t)2 2 + au(t), u(t) + ut (t)2 ≤ C , H∗ 2 2 t
for every t ∈ [0, ∞),
which combining with Lemma 2.2 tells us that
lim u(t)H 2 + ut (t)2 = 0.
t→∞
∗
Conversely, if Tmax = ∞ and u(t)2H 2 + ut (t)22 → 0 as t → ∞, then it follows that by Lemmas 2.1, 2.2 ∗
lim u(t)p = 0
t→∞
and
lim u(t)H 2 + au(t), u(t) 2 + ut (t)2 = 0,
t→∞
∗
which imply that lim E(t) = 0.
t→∞
Therefore, by Lemma 2.4 and the above mentioned results, there must be t0 > 0 such that E(t0 ) < D and u(t0 ) ∈ W . 7. Proof of Theorem 3.4 Firstly, we assume that there exists t0 ≥ 0 such that u(t0 ) ∈ U and E(t0 ) < d. Without loss of generality, we may suppose that t0 = 0, then u(t) ∈ U
and E(t) < d,
for every t ∈ [0, Tmax ).
Indeed, (4.6) entails that E(t) ≤ E(0) < d,
for all t > 0.
Then suppose that there exists t¯ > 0 such that u(t¯) ∈ N , but by (2.6),
d ≤ J u(t¯) ≤ E(t¯) < d, which is a contradiction to (7.1), and therefore u(t) ∈ U for all t ∈ [0, Tmax ). From (2.5), the definition of d, we obtain that
d=
2 p/(p−2) p − 2 (uH∗2 + (au, u)2 ) , u∈H∗ \{0} 2p (upp )2/(p−2)
inf 2
then (u2H 2 + (au, u)2 )p/(p−2) 2pd ∗ ≤ . p−2 (upp )2/(p−2)
(7.1)
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Since u(t) ∈ U for all t ∈ [0, Tmax ), I(u(t)) = u(t)2H 2 + (au(t), u(t))2 − u(t)pp < 0, the above inequality ∗ reads 2
2pd < u(t)H 2 + au(t), u(t) 2 , ∗ p−2
for every t ∈ [0, Tmax ).
(7.2)
Now we prove that Tmax < ∞. We follow closely [9, Theorem 3.11]. Assume by contradiction that Tmax = ∞. For any T > 0, we define a continuous positive function by 2 θ(t) = u(t)2 + μ
t
u(τ )2 dτ + μ(T − t)u0 22 , 2
for all t ∈ [0, T ].
0
With the same spirit (but it needs to be careful for the computations since the presence of the term au) as in [9], we have θ(t)θ (t) −
p+2 2 θ (t) ≥ C > 0, 4
for a.e. t ∈ [0, T ].
Setting y(t) = θ(t)−(p−2)/4 , the above differential inequality becomes y (t) ≤ −
p−2 Cy(t), 4
for a.e. t ∈ [0, T ],
−(p+2)/4 which together with y (t) = − p−2 < 0 proves that y(t) = 0 at some time, say as t = T ∗ 4 θ(t) (independent of T). This tells us that
lim θ(t) = ∞,
t→T ∗
that is t
2 lim∗ u(t)2 = ∞
or
t→T
lim∗
t→T
u(τ )2 dτ = ∞. 2
0
If limt→T ∗ u(t)22 = ∞, by (2.1), we get 2 lim u(t)H 2 = ∞
t→T ∗
If limt→T ∗
t 0
∗
(7.3)
u(τ )22 dτ = ∞ holds, then 2 lim u(t)2 = ∞,
t→T ∗
and we also have (7.3). Therefore, u(t) cannot be a global solution. That is Tmax < ∞. Conversely, assume that there exists no t ≥ 0 such that u(t) ∈ U and E(t) < d, we must have either u(t) ∈ W and E(t) < d (this is impossible by Theorem 3.3) or E(t) ≥ d for all t ≥ 0. Then from (4.6), t μ 0
ut (τ )2 dτ ≤ E(0) − d, 2
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by Jensen’s inequality, for any t ≥ 0 there holds t t
ut (τ )2 dτ ≥ 2
0
t
2 ut (τ )dτ
2 ≥ u(t)2 − u0 22 ,
0
Ω
hence, u(t)2 ≤ CT , 2
for all t ∈ [0, T ),
(7.4)
for any finite T > 0. If Tmax < ∞, we know that from Theorem 3.1 lim u(t)p = ∞,
t→Tmax
hence, by Lemmas 2.1 and 2.2, for every M > E(0), there exists some 0 < t¯ < Tmax such that M<
p − 2 u(t) 2 + au(t), u(t) , H 2 ∗ 2p
for every t ≥ t¯.
(7.5)
Denote V(t) = M − E(t),
for every t ≥ t¯,
(7.6)
ut (τ )2 dτ,
(7.7)
and then from (4.6) t V(t) = V0 + μ
2
0
where V0 = M − E(0). Therefore, by (7.5)–(7.7)
0 < V(t) ≤ M − J u(t) ≤ M −
p 1 p M + u(t)p , p−2 p
then u(t)p ≥ p p
2M + V(t) p−2
> pV(t).
(7.8)
Since p > 2, by the Hölder, Young inequalities and (7.7)–(7.8), we have
μ u(t), ut (t) ≤ Cμu(t)1−k u(t)k ut (t) 2 2 p p 1−k 2k 2 ≤ C u(t)p μν u(t)p + C(ν)−1 ut (t)2 2k
˙ < CV (1−k)/p (t) μν u(t)p + C(ν)−1 V(t) (1−k)/p u(t)p + C(ν)−1 V˙ 1/κ (t), ≤ CνV0 p where ν > 0, k ∈ (1, p/2) and then κ = (1 + (1 − k)/p)−1 ∈ (1, 2p/(p + 2)).
(7.9)
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Next, we estimate I(u(t)). Since V0 < M , then from (7.7)–(7.8) p 2M + (p − 2)V0 , (V0 − M ) u(t) p ≤ (V0 − M )p p−2 that is, 2(V0 − M ) ≥
p − 2 2(V0 − M ) u(t)p . p p 2M + (p − 2)V0
(7.10)
Hence, by (7.6) and (7.10)
p − 2 u(t)p ≥ −2E(t) + p − 2 u(t)p −I u(t) = −2J u(t) + p p p p
p − 2 u(t)p ≥ 2(V0 − M ) + p − 2 u(t)p ≥ 2 V(t) − M + p p p p pV0 p−2 u(t)p . ≥ p p 2M + (p − 2)V0
(7.11)
Now we consider a function defined by
F(t) ≡ V(t)1/κ + ε u(t), ut (t) 2 ,
for every t ≥ t¯.
(7.12)
By (7.9) and (7.11), taking ν > 0 sufficiently small, for (6.6) we have 2
d
u(t), ut (t) 2 = ut (t)2 − I u(t) − μ u(t), ut (t) 2 dt p 2 (1−k)/p u(t)p − C(ν)−1 V˙ 1/κ (t) ≥ ut (t) + C u(t) − CνV 2
0
p
p
2
(1−k)/p u(t)p − C(ν)−1 V˙ 1/κ (t) ≥ ut (t) + C − CνV 0
2
p
2 p ≥ ut (t) + C u(t) − C(ν)−1 V˙ 1/κ (t). 2
p
Therefore, if ε > 0 is small enough, p 2
˙ F(t) ≥ C ut (t)2 + u(t)p > 0. Then choosing ε > 0 even smaller if needed, we get 1/κ
F(t) ≥ F0 ≡ V0
+ ε u(0), ut (0) 2 > 0.
Utilizing (7.8) and by the Hölder, Young inequalities, we have
κ F κ (t) ≤ 2κ−1 V(t) + εκ u(t), ut (t) 2 p 2
≤ C u(t)p + ut (t)2 , which together with (7.13) yields that ˙ F(t) ≥ CF κ (t);
(7.13)
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then by Lemma 2.10 in [4], F(t) blows up at T ∗ > t¯ and we have F(t) ≥
C . (T ∗ − t)1/(κ−1)
(7.14)
By (7.9), (7.11) and since E(t) ≥ d, V(t) ≤ M − d, u(t)2 = u(t¯)2 + 2 2 2 ε
t
F(τ ) − V 1/κ (τ ) dτ
t¯
2 2 ≥ u(t¯)2 + ε
t t¯
2 ≥ u(t¯)2 − C + C
C 1/κ dτ − (M − d) (T ∗ − τ )1/(κ−1)
1 T∗ − t
(2−κ)/(κ−1)
−
(2−κ)/(κ−1) 1 . T ∗ − t¯
Consequently, 2 lim∗ u(t)2 = ∞,
t→T
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