Global existence and nonexistence for a nonlinear wave equation with damping and source terms

Math. Nachr. 278, No. 11, 1341 – 1358 (2005) / DOI 10.1002/mana.200310310

Global existence and nonexistence for a nonlinear wave equation withdamping and source terms

Yong Zhou∗1

1 Department of Mathematics, East China Normal University, Shanghai 200062, China

Received 12 March 2003, revised 6 August 2003, accepted 9 September 2003Published online 5 August 2005

Key words Nonlinear wave equation, global solution, blow-up, weighted spaces, asymptotic behaviorMSC (2000) 35B30, 35B40, 35B45, 47H20

In this paper we consider a nonlinear wave equation with damping and source term on the whole space. Forlinear damping case, we show that the solution blows up in finite time even for vanishing initial energy. Thecriteria to guarantee blowup of solutions with positive initial energy are established both for linear and nonlineardamping cases. Global existence and large time behavior also are discussed in this work.

c© 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

In this paper, we consider the following initial value problemutt + a |ut|m−1ut − φ∆u = f(x, u) , x ∈ R

n , t > 0 ,

u(x, 0) = u0(x) , ut(x, 0) = u1(x) , x ∈ Rn ,

(1.1)

where a, b > 0, m ≥ 1 are constants and φ(x) > 0 : Rn → R, n ≥ 2, satisfies the following condition

(H) g(x) (φ(x))−1 ∈ C0,γ(Rn) with γ ∈ (0, 1) and g ∈ Ln/2(Rn) ∩ L∞(Rn).

Models of this type are of interest in applications in various areas in mathematical physics (see [1, 12, 17]),as well as in geophysics and ocean acoustics, where for example, the coefficient φ(x) represents the speed ofsound [6]. The force term f , of primary interest is the function

f(x, u) = b |u|p−1u− µu with p > 1 and µ ≥ 0 , (1.2)

see for example [14]. We refer to the case µ = 0 as the mass free case and µ = 0 as the mass case.In [5], it was showed that the local solution with negative initial energy blows up in finite time for linear

damping and mass free case (m = 1 and µ = 0).For Cauchy problem (1.1) with φ(x) = 1 (of course, it does not satisfy (H)), recently the following results was

established (see [8, 15]): If the initial energy is negative and np/(n + p + 1) < m < p, then the local solutionblows up in finite time. When 1 < m < np/(n + p + 1), the solution blows up for sufficiently negative initialenergy and

∫Rn u0u1 dx ≥ 0. Very recently, Levine and Todorova [9] proved that if p > m ≥ 1, then for any

λ > 0 there are choices of initial data from the energy space with initial energyE(0) = λ2, such that the solutionblows up in finite time.

The local existence of u(x, t) ∈ C([0, T );H1(Rn)) for any given u0 ∈ H1(Rn), u1 ∈ L2g(R

n) follows fromthe corresponding theorem in a bounded domain (see [2, 9]). The local existence for (1.1) with µ = 0, m = 1was proved in [5]. In [9, 15], it was proved that the corresponding theorems still hold even for µ = q2(x), whereq(x) is a sufficiently slowly decreasing function.

In this paper we will investigate various blowup criteria both for linear and nonlinear damping, mass caseand free mass case under the condition (H). It is surprising that the blow up criteria only depend on linear and

∗ e-mail: [email protected], Phone: +21 6223 3050, Fax: +21 6223 8105


1342 Zhou: A nonlinear wave equation with damping and source terms

nonlinear damping and are nothing to do with whether it is the free mass case or the mass case. In Section 2,we recall some preliminary results about Eq. (1.1). Then we establish global nonexistence criteria for lineardamping and nonlinear damping in Section 3 and Section 4 respectively. We discuss global existence and largetime behavior both for all possible m ≥ 1 in Section 5. Finally, several applications of our method are mentionedin Section 6.

2 Preliminary results

Just as in [4, 5], the function space for (1.1) is X = D1,2(Rn) × L2g(R

n), with

D1,2(Rn) =f ∈ L

2nn−2 (Rn) : ∇f ∈ L2(Rn)

and the space L2

g(Rn) is defined to be the closure of C∞0 (Rn) functions with respect to the inner product

(f, h)L2g(Rn)

∫Rn

gfh dx .

Clearly, L2g(R

n) is a separable Hilbert space and ‖f‖2L2

g= (f, f)L2

g(Rn). For general 1 < p < ∞, if f is a

measurable function on Rn, define

‖f‖Lpg

(∫

Rn

g |f |p dx)1/p

and let Lpg consist of all f for which ‖f‖Lp

g< ∞. For this weighted space Lp

g(Rn), we have the following

elementary lemma.

Lemma 2.1 (a) Under the assumption (H), the embedding D1,2(Rn) ⊂ L2g(R

n) is compact.(b) Let g satisfies (H), then for any u ∈ D1,2,

‖u‖Lqg≤ ‖g‖Ls‖∇u‖L2 , with s = 2n

2n−qn+2q , 2 ≤ q ≤ 2nn−2 . (2.1)

(c) Assume that g ∈ L1(Rn)∩L∞(Rn), then Lpg(R

n) embeds Lqg(R

n) continuously, for any 1 ≤ q ≤ p <∞.Moreover,

‖f‖Lqg≤ ‖g‖(p−q)/pq

L1 ‖f‖Lpg

(2.2)

for any f ∈ Lpg(R

n).For local existence of (1.1), we have

Theorem 2.2 (Local existence). Suppose (H) holds. Assume that 1 ≤ p ≤ n+2n−2 if n ≥ 2 or 1 ≤ p if n = 2.

Then for any initial data

u0 ∈ D1,2(Rn) , u1 ∈ L2g(R

n) ,

there exist a T > 0 such that the Cauchy problem (1.1) has a unique solution

u ∈ C([0, T );D1,2(Rn)

)and ut ∈ C

([0, T );L2

g(Rn)) ∩ Lp+1

g ([0, T ) × Rn) .

This result follows from the weighted-norm space version of the corresponding theorem for φ = 1 (see [2, 9]).The supremum of all T ’s for which the solution exists on [0, T ) × R

n is called the lifespan of the solution of(1.1). The lifespan is denoted by T ∗. If T ∗ = ∞, we say the solution is global, while it is nonglobal if T ∗ <∞,and we say that the solution blows up in finite time.

From now on, we take a = b = 1 just for convenience. One can see that this restriction does not affect ourargument below.

Multiplying equation (1.1) by gut, after integration by parts, one has

12d

dt‖ut‖2

L2g

+ ‖ut‖m+1

Lm+1g

+12d

dt‖∇u‖2

L2 =1

p+ 1d

dt‖u‖p+1

Lp+1g

− 12d

dt

∫Rn

g(x)µ |u|2 dx .


Math. Nachr. 278, No. 11 (2005) / www.mn-journal.com 1343

So the corresponding energy to problem (1.1) is defined as

E(u)(t) = E(t) 12‖ut‖2

L2g

+12‖∇u‖2

L2 − 1p+ 1

‖u‖p+1

Lp+1g

+12

∫Rn

g(x)µ |u|2 dx , (2.3)

and one can find that E(t) ≤ E(0) easily from

dE(t)dt

= −‖ut(., t)‖m+1

Lm+1g

≤ 0 . (2.4)

Note that the above energy inequality (2.4) is derived only formally, but one can use standard mollifying techniqueto prove that it holds for the solution. Actually, in the proof of existence of solution, the energy inequality isproved for the approximate solutions, then (2.4) holds for the solution by taking the limit for the approximatesolution. We would like to skip the detailed proof, just for concision, for details, please see [2] for a similar case.

Lemma 2.3 Let 1 ≤ p ≤ n+2n−2 if n ≥ 2, 1 ≤ p if n = 2. Then there exists a positive constant C > 1

depending only on g(x) (throughout this paper, C denotes a generic positive constant depending on g(x) only, itmay be different from line to line), such that

‖u‖sLp+1

g≤ C

(‖∇u‖2

L2 + ‖u‖p+1

Lp+1g

), with 2 ≤ s ≤ p+ 1

on [0, T ) for any u being a solution to (1.1). And consequently,

‖u‖sLp+1

g≤ C

(|H(t)| + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

+∫

Rn

µg |u|2 dx)

(2.5)

with 2 ≤ s ≤ p+ 1 on [0, T ), where H(t) −E(t).

P r o o f. If ‖u‖Lp+1g

≤ 1, then ‖u‖sLp+1

g≤ ‖u‖2

Lp+1g

≤ C ‖∇u‖2L2 by (2.1). If ‖u‖Lp+1

g> 1, then ‖u‖s

Lp+1g

≤‖u‖p+1

Lp+1g

. (2.5) follows from the definition of energy corresponding to the solution.

If we let f(t) 12‖u(., t)‖2

L2g, where u is a solution construct in Theorem 2.2. One can see that the derivative

of f(t) with respect to time

f ′(t) =∫

Rn

gutu dx (2.6)

is well defined and Lipschitz continuous. Moreover, one can get

f ′′(t) = ‖ut‖2L2

g− (|ut|m−1ut, u

)L2

g− ‖∇u‖2

L2 + ‖u‖p+1

Lp+1g

−∫

Rn

g(x)µ |u|2 dx (2.7)

almost every in [0, T ).

3 Global nonexistence for linear damping

First, we should prove the following technical lemma

Lemma 3.1 Suppose that Ψ(t) is a twice continuously differential satisfyingΨ′′(t) + Ψ′(t) ≥ C0Ψ1+α(t) , t > 0 , C0 > 0 , α > 0 ,

Ψ(0) > 0 , Ψ′(0) ≥ 0 .(3.1)

Then Ψ(t) blows up in finite time. Moreover the blowup time can be estimated explicitly (see (3.3) below).



P r o o f. It is easy to see that we can assume Ψ′(0) > 0. In fact, from Eq. (3.1), one can find that thereexists some t0 such that Ψ′(t0) > 0, then we consider (3.1) for t > t0 with shifting initial data Ψ(t0) > 0 andΨ′(t0) > 0.

Set

Φ′(t) = δΦ1+α/2(t) , Φ(0) = Ψ(0) > 0 , (3.2)

where δ is a constant to be determined later. Then (3.2) can be represented as

Φ(t) =(

Ψ(0)−α/2 − δα

2t

)−2/α

.

It is obviously that Φ(t) is increasing and it goes to infinity as t tends to

T0 =2δα

Ψ(0)−α/2 . (3.3)

On the other hand,

Φ′′(t) + Φ′(t) = δ2(1 +

α

2

)Φ1+α(t) + δΦ1+α/2(t)

≤(δ2(1 + α/2) + δΨ(0)−α/2

)Φ1+α(t) ≤ C0Φ1+α(t) ,

(3.4)

provided that δ > 0 is sufficiently small such that δ2(1 + α/2) + δΨ(0)−α/2 ≤ C0. Moreover we can choose δis small enough such that

Φ′(0) = δΨ(0)1+α/2 < Ψ′(0) . (3.5)

Now we claim that

Ψ′(t) > Φ′(t) , for all 0 ≤ t ≤ T0 . (3.6)

Suppose not, by the continuity of the solution for (3.1), due to (3.5), one has Ψ′(t) > Φ′(t), for t > 0 smallenough, then there exists a t0, 0 < t0 < T0, such that

Ψ′(t) > Φ′(t) , 0 ≤ t < t0 and Ψ′(t0) = Φ′(t0) . (3.7)

So we have the equation for Ψ − Φ on 0 < t ≤ t0 as

Ψ′′(t) − Φ′′(t) + Ψ′(t) − Φ′(t) ≥ C0

(Ψ1+α(t) − Φ1+α(t)

) ≥ 0 ,

which can be solved as

Ψ′(t0) − Φ′(t0) ≥ e−t0(Ψ′(0) − Φ′(0)) > 0 .

This contradicts with (3.7). Therefor (3.6) is true, i.e., the solution to (3.1) blows up in finite time.

The first global nonexistence result for linear damping of (1.1) (m = 1) reads

Theorem 3.2 Let condition (H) be satisfied and g(x) ∈ L1(Rn) in addition, for n ≥ 3. Suppose 1 < p ≤ n+2n−2

if n ≥ 3 or 1 < p if n = 2. If the nonzero initial datum satisfies

E(0) ≤ 0 ,∫

Rn

gu0u1 dx ≥ 0 , (3.8)

then the corresponding solution blows up in finite time.

Remark 3.3 Theorem 3.2 shows that the solution blows up even for vanishing initial energy, of course undersome mild condition. In [18], an elegant argument to show that the solution to nonlinear evolution equations withvanishing initial energy is given also. In Section 6, we extend this theorem to more general models. Moreover,we will extend it to the Cauchy problem with φ(x) = 1 in a forthcoming work.



P r o o f. Consider

Ψ(t) =12

∫Rn

gu2(x, t) dx

as a function of one variable t. From (2.6) and (2.7) with m = 1, one obtains

Ψ′′(t) = ‖ut‖2L2

g− (ut, u)L2

g− ‖∇u‖2

L2 + ‖u‖p+1

Lp+1g

−∫

Rn

g(x)µ |u|2 dx

= − 2(

12‖ut‖2

L2g

+12‖∇u‖2

L2 − 1p+ 1

‖u‖p+1

Lp+1g

+12

∫Rn

g(x)µ |u|2 dx)

+ 2 ‖ut‖2L2

g− (ut, u)L2

g+p− 1p+ 1

‖u‖p+1

Lp+1g

= − 2E(t) + 2 ‖ut‖2L2

g− Ψ′(t) +

p− 1p+ 1

‖u‖p+1

Lp+1g

≥ − 2E(0)− Ψ′(t) +p− 1p+ 1

‖u‖p+1

Lp+1g

≥ −Ψ′(t) +p− 1p+ 1

2(1+p)/2 ‖g‖(1−p)/2L1 Ψ(1+p)/2(t) ( by (2.2) ) .

Then by Lemma 3.1, we see that Ψ(t) blows up in finite time. Therefore the lifespan of the solution is finite.

The second one is

Theorem 3.4 Let hypothesis (H) be fulfilled and g(x) ∈ L1 in addition, for n ≥ 3. Suppose 1 < p ≤ n+2n−2 if

n ≥ 2 or 1 < p if n = 2. Then the solution to (1.1) blows up in finite time if the initial energy is negative.

P r o o f. By the definition H(t) = −E(t), and (2.4), we have

0 < H(0) ≤ H(t) ≤ 1p+ 1

‖u‖p+1

Lp+1g

.

We borrow the idea of that in [2] and define

F (t) H1−α(t) + θ

∫Rn

guut dx (3.9)

for 0 < α < p−12(p+1) and δ small to be determined later. It should be mentioned that the weight function serves

the mathematical purpose essentially of reducing the given problem in Rn to a finite domain, so that the method

of [2] can be applied.By direct computation,

F ′(t) = (1 − α)H−αH ′(t) + θ ‖ut‖2L2

g− θ ‖∇u‖2

L2

− θ

∫Rn

guut dx+ θ ‖u‖p+1

Lp+1g

− θ

∫Rn

µg |u|2 dx

≥ (1 − α)H−αH ′(t) + θ ‖ut‖2L2

g− θ ‖∇u‖2

L2 − θδ2

2‖u‖2

L2g

− θδ−2

2‖ut‖2

L2g

+ θ ‖u‖p+1

Lp+1g

− θ

∫Rn

µg |u|2 dx (for any δ2 > 0

)=(

(1 − α)H−α − θδ−2

2

)H ′(t) + θ ‖ut‖2

L2g− θ ‖∇u‖2

L2

+ θ(p+ 1)H(t) + θp+ 1

2

(‖ut‖2

L2g

+ ‖∇u‖2L2 +

∫Rn

µg |u|2 dx)

− θ

∫Rn

µg |u|2 dx − θδ2

2‖u‖2

L2g.



If we let δ2 = kHα, where k > 0 to be determined later, and use

Hα(t)‖u‖2L2

g≤ C

(1

p+ 1

)α

‖u‖2+α(p+1)

Lp+1g

( by (2.2) ) ,

then

F ′(t) ≥(

(1 − α) − θ

2k

)H−α(t)H ′(t)

+ θ

(p+ 3

2

)‖ut‖2

L2g

+ θ

(p− 1

2

)(‖∇u‖2

L2 +∫

Rn

µg |u|2 dx)

+ θ(p+ 1)H(t) − θC1k

(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

+∫

Rn

µg |u|2 dx),

where C1 = 12 C(1/(p+ 1))α. Using the decomposition of p+ 1

p+ 1 =p+ 3

2+p− 1

2,

then one get

F ′(t) ≥(

(1 − α) − θ

2k

)H−α(t)H ′(t) + θ

p− 14

‖∇u‖L2

+ θ

(p+ 3

2− C1k

)H(t) + θ

(p− 1

2(p+ 1)− C1k

)‖u‖p+1

Lp+1g

+ θ

(p+ 7

4− C1k

)‖ut‖2

L2g

+ θ

(p− 1

4− C1k

)∫Rn

µg |u|2 dx .

Take k be small enough such that there exists a constant C2 > 0 such that

p− 12(p+ 1)

− C1k ≥ C2 > 0 .

Then we can choose θ (k is fixed) so small that

1 − α− θ

2k≥ 0 and F (0) = H1−α(0) + θ

∫Rn

gu0u1 dx > 0 .

Therefore we have the equation for F (t) as

F ′(t) ≥ θC2

(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

+∫

Rn

µg |u|2 dx). (3.10)

On the other hand,

∣∣∣∣∫

Rn

guut dx

∣∣∣∣1/(1−α)

≤ ‖u‖1/(1−α)L2

g‖ut‖1/(1−α)

L2g

≤ C ‖u‖1/(1−α)

Lp+1g

‖ut‖1/(1−α)L2

g

≤ C(‖u‖s

Lp+1g

+ ‖ut‖2L2

g

)≤ C

(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

+∫

Rn

µg |u|2 dx)



where we use Young inequality with s = 21−2α ≤ p+ 1 and (2.5). And then

F 1/(1−α)(t) =(H1−α(t) + θ

∫Rn

guut dx

)1/(1−α)

≤ 21/(1−α)

(H(t) +

∣∣∣∣∫

Rn

guut dx

∣∣∣∣1/(1−α)

)

≤ C

(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

+∫

Rn

µg |u|2 dx).

So (3.10) reduces to

F ′(t) ≥ γF 1/(1−α)(t) (3.11)

where γ is a constant depending only on C, C2, and θ only. For (3.11) with positive initial data, F (t) goes toinfinite as t tends to

1 − α

γαFα/(1−α)(0).

The next theorem is concerning finite time blowup for solutions with positive initial energy. First let us recallthe following lemma in [3]

Lemma 3.5 Suppose that a positive, twice differential function ψ(t) on t ≥ 0 satisfies

ψ′′ψ − (1 + α)(ψ′)2 ≥ 0 , (3.12)

where α > 0. If ψ(0) > 0, ψ′(0) > 0, then

ψ −→ ∞ as t −→ t1 ≤ t2 =ψ(0)αψ′(0)

. (3.13)

Lemma 3.6 Suppose

‖u0‖Lp+1g

> λ0 C−2/(p−1)0 and E(0) < E0 p− 1

2(p+ 1)C

−2(p+1)/(p−1)0 , (3.14)

where

C0 = ‖g‖Ls , with s = 2nn+2−(n−2)p , 1 λ0 and ‖∇u(x, t)‖L2 > C−(p+1)/(p−1)0 , (3.16)

for all t ∈ [0, T ).

P r o o f. Thanks to (2.1) and (2.3), one has

E(t) ≥ 12C2

0

‖u(., t)‖2Lp+1

g− 1p+ 1

‖u(., t)‖p+1

Lp+1g

.

Now if we let

f(ζ) =1

2C20

ζ2 − 1p+ 1

ζp+1 , ζ ≥ 0 ,

then the property of f(ζ) is as follows⎧⎪⎨⎪⎩f(ζ) is strictly increasing on [0, λ0),f(ζ) takes its maximum value E0 at λ0,

f(ζ) is strictly decreasing on (λ0,∞).(3.17)



Since

E0 > E(0) ≥ E(t) ≥ f(‖u(., t)‖Lp+1

g

),

for all time t ≥ 0, it follows from (3.17) that there is no time t∗ such that ‖u(., t∗)‖Lp = λ0. Then by thecontinuity of ‖u(., t)‖Lp-norm with respect to time variable and the initial condition thatE(0) is strictly less thanE0, one has

‖u(., t)‖Lp+1g

> λ0 = C−2/(p−1)0 ,

for all time t ≥ 0, and consequently

‖∇u(., t)‖L2 ≥ C−10 ‖u(., t)‖Lp+1

g> C−1

0 λ0 = C−(p+1)/(p−1)0 .

This finishes the proof.

Theorem 3.7 Let (H) be satisfied, and 1 λ0 and E(0) ≤ E0 .

Where λ0 and E0 defined by (3.14) and (3.15). Then no global solutions of problem (1.1) can exist, i.e., thecorresponding solution blows up in finite time.

Remark 3.8 One can prove that the set of functions in D1,2 × Lp+1g satisfying

‖u0‖Lp+1g

= λ0 = C−2/(p−1)0 and E(0) < E0 =

p− 12(p+ 1)

C−2(p+1)/(p−1)0

is empty.In fact, one can compute directly (see (3.17)) that

E(0) =12‖u1‖2

L2g

+12‖∇u0‖2

L2 − 1p+ 1

‖u0‖p+1

Lp+1g

+12

∫Rn

µg |u0|2 dx

≥(

12− 1p+ 1

)C

−2(p+1)/(p−1)0

as ‖u0‖Lp = λ0.

P r o o f. The method used here is the classical concavity method (see for example [7, 11]). The goal is toconstruct suitable function ψ(t) which satisfies (3.12).

Case 1: E(0) < E0. Define

ψ(t) = ‖u(., t)‖2L2

g+∫ t

0

‖u(., τ)‖2L2

gdτ + (T0 − t) ‖u0‖2

L2g

+ β(t + t0)2 , (3.18)

where t0, T0 > t and β are positive constants to be determined later. Hence by direct computation, we have

ψ′(t) = 2∫

Rn

guut dx+ ‖u(., t)‖2L2

g− ‖u0‖2

L2g

+ 2β(t+ t0)

= 2∫

Rn

guut dx+ 2∫ t

0

∫Rn

guut dx dτ + 2β(t+ t0) ,(3.19)

and

ψ′′(t) = 2 ‖ut‖2L2

g+ 2

∫Rn

guutt dx+ 2∫

Rn

guut dx+ 2β . (3.20)



Then, due to the decreasing of energy and (3.16)

ψ′′(t) = 2 ‖ut‖2L2

g− 2 ‖∇u‖2

L2 + 2 ‖u‖p+1

Lp+1g

− 2∫

Rn

µg |u|2 dx+ 2β

≥ (p− 1) ‖∇u‖2L2 − 2(p+ 1)E(t) + (p+ 3) ‖ut‖2

L2g

+ 2β

= 2(p− 1) ‖∇u‖2L2 − 2(p+ 1)E(0) + 2β

+ (p+ 3) ‖ut‖2L2

g+ 2(p+ 1)

∫ t

0

∫Rn

g(ut(x, τ))2 dx dτ

≥ 2(p− 1)C−2(p+1)/(p−1)0 − 2(p+ 1)E(0) + 2β

+ (p+ 3) ‖ut‖2L2

g+ (p+ 3)

∫ t

0

∫Rn

g(ut(x, τ))2 dx dτ

= 2(p+ 1)(E0 − E(0)) + 2β + (p+ 3) ‖ut‖2L2

g

+ (p+ 3)∫ t

0

∫Rn

g(ut(x, τ))2 dx dτ .

Now, let β = 2(E0 − E(0)) > 0, then

ψ′′(t) ≥ (p+ 3)β + (p+ 3) ‖ut‖2L2

g+ (p+ 3)

∫ t

0

∫Rn

g(ut(x, τ))2 dx dτ . (3.21)

It is obviously that⎧⎨⎩ψ(t) > 0 , for all t ≥ 0 ,

ψ′(0) = 2∫

Rn

gu0u1 dx+ 2βt0 > 0 , if t0 > 0 is large enough .(3.22)

And for all (ξ, η) ∈ R2, from (3.18), (3.19) and (3.21),

ψ(t)ξ2 + ψ′(t)ξη +ψ′′(t)p+ 3

η2 ≥ ξ2(‖ut‖2

L2g

+∫ t

0

‖u(., τ)‖2L2

gdτ + β(t+ t0)2

)

+ ξη

(2∫

Rn

guut dx+ 2∫ t

0

∫Rn

guut dx+ 2β(t+ t0))

+ η2

(‖ut‖2

L2g

+∫ t

0

∫Rn

g(ut(x, τ))2 dx dτ + β

)≥ 0 .

Thus

ψ(t)ψ′′(t)p+ 3

−(ψ′(t)

2

)2

≥ 0 ,

that is

ψ(t)ψ′′(t) − p+ 34

(ψ′(t))2 ≥ 0 . (3.23)

Then using Lemma 3.4, one obtains that ψ(t) goes to ∞ as t tends to

2(T0 ‖u0‖2

L2g

+ βt20

)(p− 1)βt0

.

The remaining thing is to choose suitable t0 and T0. Let t0 be any number which depends only on p,E0−E(0),‖u0‖L2

gand ‖u1‖L2

gas

t0 >(p+ 3) ‖u0‖2

L2g

+ (p− 1) ‖u1‖2L2

g

4(p− 1)(E0 − E(0)),



which fulfills the requirement of (3.22) automatically. Then for fixed t0, T0 can be chosen as

T0 =2((T0 + 1) ‖u0‖2

L2 + βt20)

(p− 1)(βt0 − 1

2 ‖u0‖2L2

g− 1

2 ‖u1‖2L2

g

)

=4(2(E0 − E(0))t20 + ‖u0‖2

L2g

)2(p− 1)(E0 − E(0))t0 − (p+ 3) ‖u0‖2

L2 − (p− 1) ‖u1‖2L2

g

.

And the lifespan of the solution is less than T0. This finishes the proof.

Case 2: E(0) = E0. For this case, actually we have the following claim.

Claim 3.9 There exists t > 0 such that E( t ) < E0.

Assume that Claim 3.9 is true for a moment, then the proof of Theorem 3.7 is complete since one can applythe previous case (Case 1) after shifting the time origin to t.

Now, we turn our attention to the proof of Claim 3.9. Just as the argument in [16], suppose Claim 1 is not true,which means that E(t) = E0 for all t ≥ 0. Then by the continuity of ‖u(., t)‖Lp+1

g-norm, there exist a tε, small

enough, such that

E(t) = E0 and ‖u(x, t)‖Lp+1g

> λ0 , for all t ∈ [0, tε] . (3.24)

Then we consider the solution on [0, tε],

0 = E(tε) − E0 = −∫ tε

0

∫Rn

gu2t (x, s) dx ds ,

which turns out

u(x, t) = u0(x) , a.e. on [0, tε] .

And consequently due to Eq. (1.1),

0 = ‖∇u‖2L2 + ‖u‖p+1

Lp+1g

−∫

Rn

µg |u|2 dx , (3.25)

a.e. on (0, tε]. On the other hand, due to Lemma 3.5

E0 = E(t) =12‖∇u|2L2 − 1

p+ 1‖u‖p+1

Lp+1g

+12

∫Rn

µg |u|2 dx

=p− 1

2(p+ 1)‖∇u‖2

L2 +p− 1

2(p+ 1)

∫Rn

µg |u|2 dx

≥ p− 12(p+ 1)

‖∇u‖2L2

> E0

which is a contradiction. So the claim is true.

4 Global nonexistence for nonlinear damping

In the remaining part of this paper, we set µ = 0, for simplicity, since from the proofs of Theorems in Section 3,it is easily to see that the term µu does not affect the proof. So the equation reads

utt + |ut|m−1 ut − φ∆u = |u|p−1 u , x ∈ Rn , t > 0 ,

u(x, 0) = u0(x) , ut(x, 0) = u1(x) , x ∈ Rn ,

(4.1)

with m > 1.Just as the case for linear damping, we can establish finite time blow up with negative initial energy.



Theorem 4.1 Suppose (H) holds and g(x) ∈ L1(Rn) in addition for n ≥ 3. Assume that 1 < m < p ≤ n+2n−2

if n ≥ 3, 1 < m < p if n = 2, and E(0) < 0, then the corresponding solution to (4.1) blows up in finite time.

P r o o f. The proof is similar as that of Theorem 3.3. For completeness of the paper and the convenience ofthe readers, we write it as below briefly. We define

F (t) H1−α(t) + θ

∫Rn

guut dx (4.2)

for θ small to be determined later and

0 < α < min

p− 12(p+ 1)

,p−m

(p+ 1)m

. (4.3)

Then

F ′(t) = (1 − α)H−αH ′(t) + θ ‖ut‖2L2

g− θ ‖∇u‖2

L2 − θ

∫Rn

gu |ut|m−1 ut dx+ θ ‖u‖p+1

Lp+1g

.

Young inequality tells us that∫Rn

gu |ut|m−1 ut dx ≤ δm+1

m+ 1‖u‖m+1

Lm+1g

+m

m+ 1δ−(m+1)/m ‖ut‖m+1

Lm+1g

,

then consequently,

F ′(t) ≥(

(1 − α)H−α − mθ

m+ 1δ−(m+1)/m

)H ′(t) + θ ‖ut‖2

L2g− θ ‖∇u‖2

L2

+ θ(p+ 1)H(t) + θp+ 1

2

(‖ut‖2

L2g

+ ‖∇u‖2L2

)− θδm+1

m+ 1‖u‖m+1

Lm+1g

.

If we let δ−(m+1)/m = KH−α, i.e., δm+1 = K−mHαm, K > 0 to be determined later, and use

Hαm(t) ‖u‖m+1

Lm+1g

≤ C

(1

p+ 1

)αm

‖u‖m+1+α(p+1)m

Lp+1g

( by (2.2) ) ,

then

F ′(t) ≥(

(1 − α) − θm

m+ 1K

)H−α(t)H ′(t)

+ θ

(p+ 3

2

)‖ut‖2

L2g

+ θ

(p− 1

2

)(‖∇u‖2L2

)+ θ(p+ 1)H(t) − θC1K

−m(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

),

where C1 = C(p+1)αm(m+1) . Using the decomposition of p+ 1, then one get

F ′(t) ≥(

(1 − α) − θm

m+ 1K

)H−α(t)H ′(t) + θ

p− 14

‖∇u‖2L2

+ θ

(p+ 3

2− C1K

−m

)H(t) + θ

(p− 1

2(p+ 1)− C1K

−m

)‖u‖p+1

Lp+1g

+ θ

(p+ 7

4− C1K

−m

)‖ut‖2

L2g

+ θ

(p− 1

4− C1K

−m

)∫Rn

µg |u|2 dx .

Take K be large enough such that there exists a constant C2 > 0 and

p− 12(p+ 1)

− C1K−m ≥ C2 > 0 .

Then we can choose θ (K is fixed) so small that

1 − α− θm

m+ 1K ≥ 0 and F (0) = H1−α(0) + θ

∫Rn

gu0u1 dx > 0 .



Therefore we have the equation for F (t) as

F ′(t) ≥ θC2

(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

). (4.4)

Then as the proof of Theorem 3.3, we may have

L1/(1−α)(t) ≤ C(H(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

).

So (4.4) reduces to

F ′(t) ≥ γF 1/(1−α)(t) (4.5)

where γ is a constant depending only on C, C2, and θ only. For (4.5) with positive initial data, F (t) goes toinfinite as t tends to

1 − α

γαFα/(1−α)(0).

Theorem 4.2 Let (H) be satisfied, and 1 < m λ0 C−2/(p−1)0 and E(0) ≤ E0 p− 1

2(p+ 1)C

−2(p+1)/(p−1)0 , (4.6)

where

C0 = ‖g‖Ls , with s = 2nn+2−(n−2)p , 1 < p ≤ n+2

n−2 . (4.7)

Then the corresponding solution to (4.1) blows up in finite time.

P r o o f. Case 1: E(0) < E0. One can see that the proof for Theorem 3.6 does not work any more. We usesome techniques developed in [8] in the sequel. So we define a new quantity,

G(t) = E0 +H(t) ,

then it is clearly that G(t) is an increasing function of t andG(t) > 0 for all t ∈ [0, T ). Moreover, due to Lemma3.5 under the condition (4.6) and (4.7), we have

0 < G(t) = E0 − 12‖ut‖2

L2g− 1

2‖∇u‖2

L2 +1

p+ 1‖u‖p+1

Lp+1g

≤ 1p+ 1

‖u‖p+1

Lp+1g

. (4.8)

Set a function F (t), analogous to that in the proof of Theorem 4.1 as

F (t) G1−α(t) + θ

∫Rn

guut dx . (4.9)

By the property of G(t), it can play a role similar as H(t). In fact one can use the same technique parallel to theproof of Theorem 4.1, and gets

F ′(t) ≥ γ(G(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

)with F (0) > 0 . (4.10)

On the other hand, thanks to Lemma 2.3 and Lemma 3.5,∣∣∣∣∫

Rn

guut dx

∣∣∣∣1/(1−α)

≤ C ‖u‖1/(1−α)

Lp+1g

‖ut‖1/(1−α)L2

g

≤ C(‖u‖s

Lp+1g

+ ‖ut‖2L2

g

)≤ C

(‖∇u‖2

L2 + ‖u‖p+1

Lp+1g

+ ‖ut‖2L2

g

)≤ C

(2G(t) + 2E0 + 2 ‖ut‖2

L2g

+2

p+ 1‖u‖p+1

Lp+1g

)

≤ C(G(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

),



with s = 2/(1 − 2α) ≤ p+ 1. Therefore,

F 1/(1−α)(t) ≤ C(G(t) + ‖ut‖2

L2g

+ ‖u‖p+1

Lp+1g

).

So (4.10) reduces to

F ′(t) ≥ γF 1/(1−α)(t) (4.11)

where γ is a constant depending only on C, p, α, and θ only. For (4.11) with positive initial data, F (t) goes toinfinite as t tends to

1 − α

γαFα/(1−α)(0).

Case 2: E(0) = E0. The proof is completely as the same as that for the linear damping case.

5 Global existence and large time behavior

Theorem 5.1 Suppose (H) holds, and 1 0 with any given

u0 ∈ D1,2(Rn) , u1 ∈ L2g(R

n) .

P r o o f. The argument is standard, say by continuation principle [13]. Define

F (t) =12‖ut(., t)‖2

L2g

+12‖∇u(., t)‖2

L2 +1

p+ 1‖u(., t)‖p+1

Lp+1g

,

as a function of t. Differentiate it, thanks to Holder and Young inequality, then we obtain

F ′(t) = −‖ut‖m+1

Lm+1g

+ 2∫

Rn

g |u|p−1 uut dx

≤ −‖ut‖m+1

Lm+1g

+ C(ε) |u|p+1

Lp+1g

+ ε ‖ut‖p+1

Lp+1g

≤ −‖ut‖m+1

Lm+1g

+ C1ε ‖ut‖m+1

Lm+1g

+ C2ε ‖ut‖2L2

g+ C(ε) |u|p+1

Lp+1g

≤ C2ε ‖ut‖2L2

g+ C(ε) |u|p+1

Lp+1g

≤ C3F (t) ,

provided that C1ε < 1. By Gronwall inequality, F (t) ≤ F (0)eC3t, which implies the conclusion of Theorem5.1.

In order to establish the decay rate for positive initial energy, let us recall the following lemma first

Lemma 5.2 ([10]) Let φ(t) be a nonincreasing and nonnegative function defined on [0, T ], T > 1, satisfying

φ1+r(t) ≤ k0(φ(t) − φ(t+ 1)) ,

for t ∈ [0, T ], k0 > 1 and r ≥ 0. Then we have for each t ∈ [0, T ],φ(t) ≤ φ(0)e−k(t1)+ , r = 0 ,

φ(t) ≤ (φ(0)−r + k0r(t1)+)−1/r, r > 0 ,

where (t− 1)+ = maxt− 1, 0 and k = ln(k0/(k0 − 1)).



Theorem 5.3 Suppose (H) holds, and 1 ≤ m ≤ n+2n−2 , 1 < p ≤ n+2

n−2 . If the initial datum satisfies

‖u0‖Lp+1g

< λ0 = C−2/(p−1)0 , E(0) < E0 =

p− 12(p+ 1)

C−2(p+1)/(p−1)0 ,

where C0 defined by (3.15). Then the Cauchy problem (1.1) has a unique global solution

u ∈ C([0,∞);D1,2(Rn)

)and ut ∈ C

([0,∞);L2

g(Rn)) ∩ Lm+1([0,∞) × R

n) .

Moreover,

E(t) ≤ E(0)e−k(t−1)+ , t ≥ 0 , for m = 1 , (5.1)

E(t) ≤(E(0)(m−1)/2 +

(m− 1)C2

(t− 1)+)−2/(m−1)

, t ≥ 0 , for m > 1 , (5.2)

where k and C are constants depending on g(x), m, p and E(0).

P r o o f. First, by the decreasing of energyE(t), one obtains

E(t) ≤ E(0) < E0 =p− 1

2(p+ 1)C

−2(p+1)/(p−1)0 . (5.3)

We claim that

‖u(., t)‖Lp+1g

< λ0 , for all t ≥ 0 . (5.4)

Suppose not, by the continuity of ‖u(., t)‖Lp+1g

-norm, then there exist a t0 such that ‖u(., t0)‖Lp+1g

= λ0. Butfrom (3.17), we has

E(t0) ≥ 12‖∇u(., t0)‖2

L2 − 1p+ 1

‖u(., t0)‖p+1

Lp+1g

≥ E0 .

This contradicts with (5.3). On the other hand for all t ≥ 0,

‖∇u(., t)‖2L2 = 2E(t) − ‖ut(., t)‖2

L2g

+2

p+ 1‖u(., t)‖p+1

Lp+1g

<p− 1p+ 1

C−2(p+1)/(p−1)0 +

2p+ 1

C−2(p+1)/(p−1)0 = C

−2 p+1p−1

0 .

(5.5)

By continuation argument, we know that the local solution constructed by Theorem 2.2 exists globally.Then we pay attention to large time behavior.By (5.5) and the initial condition, we have, for all t ≥ 0

‖u(., t)‖p+1

Lp+1g

≤ Cp+10 ‖∇u(., t)‖p

L2

= Cp+10 ‖∇u(., t)‖p−2

L2 ‖∇u(., t)‖2L2

< Cp+10

(2(p+ 1)p− 1

E(0))p−1

2

‖∇u(., t)‖2L2

< ‖∇u(., t)‖2L2

(5.6)

and we define 0 ≤ δ < 1 as

δ = Cp+10

(2(p+ 1)p− 1

E(0))(p−1)/2

.

Therefore, if we let

I(t) = ‖∇u(., t)‖2L2 − ‖u(., t)‖p+1

Lp+1g

, (5.7)



then, due to (5.6), for all t ≥ 0,

(1 − δ) ‖∇u(., t)‖2L2 < I(t) . (5.8)

Now we set F (t) as

Fm+1(t) =∫ t+1

t

‖ut(., s)‖m+1

Lm+1g

ds = E(t) − E(t+ 1) .

Thanks to mean value theorem and Holder’s inequality,

14‖ut(., t1)‖2

L2g

+14‖ut(., t2)‖2

L2g

≤∫ t+1

t

‖u(., s)‖2L2

gds

≤ ‖g‖1−2/(m+1)L1

(∫ t+1

t

‖u(., s)‖m+1

Lm+1g

ds

)2/(m+1)

≤ CF 2(t)

holds for some t1 ∈ [t, t+ 1/4] and t2 ∈ [t+ 3/4, t+ 1], here and in what follows, C denotes a generic positiveconstant depending on g, m and p etc., but independent of t, which may be different from lie to line. Hence

‖ut(., ti)‖2L2

g≤ CF 2(t) , i = 1 , 2 . (5.9)

Integrating I(s) on [t1, t2] and taking (5.9) into account, we have

∫ t2

t1

I(s) ds ≤2∑

i=1

‖u(., ti)‖L2g‖ut(., ti)‖L2

g+∫ t2

t1

‖ut(., s)‖2L2

gds

+∫ t2

t1

∫Rn

g |ut(x, s)|m |u(x, s)| dx ds

≤ CF (t) supt1≤s≤t2

E1/2(s) + CF 2(t)

+∫ t2

t1

∫Rn

g |ut(x, s)|m |u(x, s)| dx ds

(5.10)

where we used Sobolev embedding (2.1) and the following inequality

E(t) ≥ 12‖∇u(., t)‖2

L2 − 1p+ 1

‖u(., t)‖p+1

Lp+1g

>p− 1

2(p+ 1)‖∇u(., t)‖2

L2 .

The last term in (5.10) can be estimated as

∫ t2

t1

∫Rn

g |ut(x, s)|m |u(x, s)| dx ds ≤ C

∫ t2

t1

‖ut(., s)‖mLm+1

g‖u(., s)‖Lm+1

gds

≤ C

∫ t2

t1

‖∇ut(., s)‖mLm+1‖∇u(., s)‖L2 ds

≤ C supt1≤s≤t2

E1/2(s)Fm(t) .

(5.11)

Putting (5.11) into (5.10) and due to the decreasing of energy E(t), we obtains

∫ t2

t1

I(s) ds ≤ C(F 2(t) + E1/2(t)F (t) + E1/2(t)Fm(t)

). (5.12)



Due to (5.8)

∫ t2

t1

E(s) ds =12

∫ t2

t1

‖ut(., s)‖2L2

gds+

1p+ 1

∫ t2

t1

I(s) ds

+p− 1

2(p+ 1)

∫ t2

t1

‖∇u(., s)‖2L2 ds

≤ CF 2(t) +(

1p+ 1

+p− 1

2(p+ 1)(1 − δ)

)∫ t2

t1

I(s) ds

≤ C(F 2(t) + E1/2(t)F (t) + E1/2(t)Fm(t)

).

(5.13)

On the hand, from the nonincreasing property of E(t),

∫ t2

t1

E(s) ds ≥ 12E(t2) ,

hence from (5.13)

E(t) = E(t2) +∫ t2

t

‖ut(., s)‖m+1

Lm+1g

ds

≤ 2∫ t2

t1

E(s) ds+∫ t2

t

‖ut(., s)‖m+1

Lm+1g

ds

≤ C(F 2(t) + E1/2(t)Fm(t) + E1/2(t)F (t) + Fm+1(t)

).

(5.14)

Then by Young inequality, we have

E(t) ≤ C(F 2(t) + F 2m(t) + Fm+1(t)

). (5.15)

If m = 1, (5.15) gives

E(t) ≤ CF 2(t) = C(E(t) − E(t+ 1)) , (5.16)

then (5.1) follows from (5.16) and Lemma 5.2.If m > 1, since

Fm+1(t) = E(t) − E(t+ 1) ≤ E(0) ,

(5.15) yields

E(t) ≤ C(1 + F 2(m−1)(t) + Fm−1(t)

)F 2(t)

≤ C(1 + (E(0))2(m−1)/(m+1) + (E(0))(m−1)/(m+1)

)F 2(t)

≤ CF 2(t) ,

which implies

E(m+1)/2(t) ≤ CFm+1(t) ≤ C(E(t) − E(t+ 1)) . (5.17)

Then (5.2) follows from (5.17) and Lemma 5.2. This finishes the proof.



6 Applications to other models

In Section 3, we prove that the solution blows up in finite time even for vanishing initial energy, if∫

Rn gu0u1 dx ≥0. Actually, the method used there can be written as an abstract version and can be applied to many other models.We list some of them as follows.

1. Quasilinear wave equation of Kirchoff type with a dissipative term

utt − φ(x) ‖∇u‖2L2 ∆u+ ut = f(u) , x ∈ R

n , t ≥ 0 , (6.1)

for φ(x) satisfies (H), or in a bounded open domain Ω ⊂ Rn with φ(x) = 1 and Dirichlet boundary condition.

2. A More general model⎧⎪⎨⎪⎩(|ut|l−2ut

)t− div

(a(x) |∇u|q−2 ∇u)+ but = f(x, u) in [0, T )× Ω ,

u(x, t) = 0 on [0, T )× ∂Ω ,u(x, t = 0) = u0(x) , ut(x, t = 0) = u1(x) ,

(6.2)

where Ω ⊂ Rn, is open and bounded, with n ≥ 1, b ≥ 0, a(x) ∈ L∞(Ω) such that a(x) ≥ a0 > 0 almost

everywhere in Ω, and 1 < l < p, 1 < q q, 1 < q < p if n = q, 1 < q < p ≤ ∞ if n < q.

3. A problem in elasticity⎧⎪⎨⎪⎩(x)utt − div(C(x)∇u) + but = f(x, u) in [0, T )× Ω ,

u(x, t) = 0 on [0, T ) × ∂Ω ,

u(x, t = 0) = u0(x) , ut(x, t = 0) = u1(x) ,

(6.3)

where Ω are as before, u ∈ Rn, 0 < (x) ∈ L∞(Ω; R+) and C(x) is a linear symmetric operator from R

2n toitself for all x ∈ Ω, i.e., a tensor of rank 4 which can be represented as C(x) = (cijkl(x)), i, j, k, l = 1, . . . , n,and cijkl ∈ L∞(Ω) and it is uniformly positive-definite in Ω, that is,

(C(x)y, y) ≥ c0 |y|2 , for all x ∈ Ω , y ∈ R2n with c0 > 0 .

4. The polyharmonic operator(0 |ut|l−2ut

)t+ (−∆)Lu+ ut = f(x, u) in [0, T ) × Ω , (6.4)

where Ω as before, 0 ∈ L∞(Ω; R+), and 1 < l < p, 2 ≤ 2L 2L, 1 ≤ L < p/2 if n = 2L,

1 ≤ L < p/2 ≤ ∞ if n < q.

5. The mean curvature operator

utt − div

∇up

1 + |∇u|2

!+ ut = f(x, u) in [0, T ) × Ω . (6.5)

Use the method established in Section 3, we can show that if the initial energy is vanishing and∫Ωu0u1 dx ≥

0, the the corresponding solutions to (6.1), (6.2), (6.3), (6.4) and (6.5) blow up in finite time.

Acknowledgements The author would like to express sincere gratitude to his supervisor Professor Zhouping Xin for en-thusiastic guidance and constant encouragement. Thanks also to the referees for their constructive suggestion. This work ispartially supported by Hong Kong RGC Earmarked Grants CUHK-4219-99P and CUHK-4279-00P.

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Global existence and nonexistence for a nonlinear wave equation with damping and source terms