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Volume 2010, Article ID 203248, 11 pagesdoi:10.1155/2010/203248 Research Article A Remark on the Blowup of Solutions to the Laplace Equations with Nonlinear Dynamical Boundary Conditions

Trang 1

Volume 2010, Article ID 203248, 11 pages

doi:10.1155/2010/203248

Research Article

A Remark on the Blowup of Solutions to

the Laplace Equations with Nonlinear Dynamical Boundary Conditions

Hongwei Zhang and Qingying Hu

Department of Mathematics, Henan University of Technology, Zhengzhou 450001, China

Correspondence should be addressed to Hongwei Zhang,wei661@yahoo.com.cn

Received 24 April 2010; Revised 19 July 2010; Accepted 7 August 2010

Academic Editor: Zhitao Zhang

Copyrightq 2010 H Zhang and Q Hu This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We present some sufficient conditions of blowup of the solutions to Laplace equations with semilinear dynamical boundary conditions of hyperbolic type

1 Introduction

LetΩ be a bounded domain of R N , N ≥ 1, with a smooth boundary ∂Ω  S  S1∪S2, where S1

and S2are closed and disjoint and S1possesses positive measure We consider the following problem:

2u

∂t2  k ∂u

a ∂u

where a ≥ 0, b ≥ 0, a  b  1, and k > 0 are constants, Δ is the Laplace operator with respect

to the space variables, and ∂/∂n is the outer unit normal derivative to boundary S u0, u1are

given initial functions For convenience, we take k 1 in this paper

Trang 2

The problem1.1–1.4 can be used as models to describe the motion of a fluid in a container or to describe the displacement of a fluid in a medium without gravity; see 1 5 for more information In recent years, the problem has attracted a great deal of people Lions 6

following problem:

2u

∂t2  k ∂u

Hintermann 2

and uniqueness of the solution for problem 1.5–1.7 Cavalcanti et al 7 11

the existence and asymptotic behavior of solutions evolution problem on manifolds In this direction, the existence and asymptotic behavior of the related of evolution problem

hyperbolic problem with second-order boundary conditions

We will consider the blowup of the solution for problem1.1–1.4 with nonlinear

boundary source term gu Blowup of the solution for problem 1.1–1.4 was considered

by Kirane 3 1, by use of Jensen’s inequality and Glassey’s method 18

et al 19

type dynamical boundary inequality by the test function methods In this paper, we present some sufficient conditions of blowup of the solutions for the problem 1.1–1.4 when Ω is a

bounded domain and S2can be a nonempty set We use a different approach from those ones used in the prior literature 3,19

Another related problem to1.1–1.4 is the following problem:

∂u

∂t ∂u

of the solution of problem1.8–1.10 For more results concerning the related problem 1.8–

1.10, we refer the reader to 3,6,19–31

boundedness, asymptotic behavior, and nonexistence of global solutions for problem1.8–

1.10 were studied

In this paper, the definition of the usually space H1Ω, H s S, L p Ω, and L p S can

Trang 3

2 Blowup of the Solutions

In this paper, we always assume that the initial data u0∈ H s 1/2 S1, u1 ∈ H s S1, s > 1, and

g ∈ C and that the problem 1.1–1.4 possesses a unique local weak solution 2,3,6

u is in the class

u ∈ L∞

0, T; H s1Ω, u t ∈ L0, T, H s S1, u tt ∈ L∞

0, T; L2S1, 2.1 and the boundary conditions are satisfied in the trace sense 2

Lemma 2.1 see 33 t  Ft, u, v t ≥ Ft, v, F ∈ C, t0 ≤ t < ∞, −∞ < u <

∞, and ut0  vt0 Then, vt ≥ ut, t ≥ t0.

Theorem 2.2 Suppose that ux, t is a weak solution of problem 1.1–1.4 and gs satisfies:

1 sgs ≥ KGs, where K > 2, Gs s

0g ρdρ, Gs ≥ β|s| p1, where β > 0, p > 1;

2 E0 u02

S1u12

S1b/au0S2−2S

1G σdσ ≤ −2/ K−2βC1p3−1 2/p−11−

e 1−p/44/p−1 < 0

where C1 mS1p1/p−1 Then, the solution of problem1.1–1.4 blows up in a finite time.

Proof Denote

E t  u t2

S1 ∇u2

a u S2− 2



S1

then from1.1–1.4, we have

d

Hence

Let Ht  ut2

S1t

0

τ

0us2

S1ds dτ Using condition1 ofTheorem 2.2, we have

˙

H t  d

dt H t  2



S1

uu t dσ

t

0

us2

S1ds,

¨

H t  d2

dt2H t  2



S1

u2t dσ 2



S1

uu tt dσ



S1

u2

 2



S1



u2t − u ∂u

∂n  ugu 12u2



≥ 2



S



u2t − u ∂u

∂n  KGu 12u2



dσ.

2.5

Trang 4

Observing that



S1

u ∂u

∂n 



Ω|∇u|2dxb

a



S2

K



S1

G udσ  −E0 K − 2



S1

G udσ 



S1

u2t dσ b

a



S2

u2



Ω|∇u|2dx, 2.7

we know from2.5–2.7 that

¨

H t ≥ 4



S1

u2t dσ − 2E0



S1

u2dσ  2K − 2



S1

G udσ ≥ −2E0 2K − 2β



S1

|u| p1dσ.

2.8

It follows from2.8 that

˙

H t ≥ −2E0t  2K − 2β

t

0



S1

|u| p1dσ ds ˙H 0, 2.9

H t ≥ −E0t2 2K − 2β

t

0

τ

0



S1

|us| p1dσ ds dτ  t ˙ H 0  H0, 2.10

where H0  u02

S1, ˙ H0  2S1u0u1dσ From2.8 and 2.10, we have

¨

H tHt≥2K−2β



S1

|u| p1

t

0

τ

0



S1

|us| p1dσ ds dτ

 t ˙ H 0 − E0t2 H0 − 2E0.

2.11

Using the inversion of the H ¨older inequality, we obtain



S1

|u| p1

S1

|u|2

p1/2

t

0

τ

0



S

|us| p1dσ ds dτt

0

τ

0



S

|us|2dσ ds dτ

p1/2

1

2t

2mS1

p−1/2

. 2.13

Trang 5

Substituting2.12 and 2.13 into 2.11, we have

¨

H t  Ht

≥ 2K − 2βmS1p1/p−1

×

S1

|u|2

p1/2

1

2t

0

τ

0



S1

|us| p1dσ ds dτ

2/p1

 t ˙ H 0 − E0t2 H0 − 2E0

≥ 2K − 2βmS1p1/p−1

S1

|u|2

p1/2

0

τ

0



S1

|us| p1dσ ds dτ

p1/2

 t ˙ H 0 − E0t2 H0 − 2E0, t ≥ 1.

2.14 Noticing that

a  b n≤ 2n−1a n  b n , a > 0, b > 0, n > 1, 2.15

we have

¨

H t  Ht ≥ 2 3−p/2 K − 2βmS1p1/p−1 H p1/2 t  t ˙ H 0 − E0t2 H0 − 2E0.

2.16

We see from2.9 and 2.10 that ˙H t → ∞, Ht → ∞ as t → ∞ Therefore, there is a

t0≥ 1 such that

˙

Multiplying both sides of2.16 by 2 ˙H t and using 2.9, we get

d

dt



˙

H2t  H2t≥ 1

p 325−p/2 K − 2βmS1p1/p−1 d

dt H

p3/2 t  It, t ≥ t0,

2.18 where

I t −4E0t 2 ˙H0−E0t2 ˙H 0t  H0 − 2E0



From2.18 we have

d dt



˙

H2t  H2t − C2H p3/2 t≥ It, t ≥ t0, 2.20

Trang 6

where C2 1/p  32 5−p/2 K − 2βmS1p1/p−1 Integrating2.20 over t, t0, we arrive at

˙

H2t  H2t − C2H p3/2 t ≥

t

t0

I τdτ  ˙ H2t0  H2t0 − C2H p3/2 t0, t ≥ t0.

2.21

Observe that when t → ∞, the right-hand side of 2.21 approaches to positive infinity

since It > 0 for sufficiently large t; hence, there is a t1 ≥ t0such that the right side of2.21

is larger than or equal to zero when t ≥ t1 We thus have

˙

H2t  H2t ≥ C2H p3/2 t, t ≥ t1. 2.22 Extracting the square root of both sides of2.22 and noticing that ˙H tHt ≥ 0, we obtain

˙

H t  Ht ≥ C3H p3/4 t ≥ C3t 1−p/2 H p3/4 t, t ≥ t1, 2.23

since 1− p < 0, t > t1> t0> 1, where C3C2.

Consider the following initial value problem of the Bernoulli equation:

˙

Z  Z  C3t 1−p/2 Z p3/4 , t ≥ t1, Z t1  Ht1. 2.24 Solving the problem2.24, we obtain the solution

Z t  e −t−t1 



H 1−p/4 t1 −p− 1

4

t

t1

C3τ 1−p/2 e 1−p/4τ−t1 

4/1−p

 e −t−t1 H t1J 4/1−p t, t ≥ t1,

2.25

where Jt  1 − p − 1/4H p−1/4 t1C3

t

t1τ 1−p/2 e 1−p/4τ−t1 dτ Obviously, J t1  1 > 0, and for t > t1 1

δ t  p− 1

p−1/4 t1C3

t

t1

τ 1−p/2 e 1−p/4τ−t1 

p− 1

p−1/4 t1C3

t11

t1

τ 1−p/2 e 1−p/4τ−t1 

p− 1

p−1/4 t1C3t1 11−p/2

t11

t1

e 1−p/4τ−t1 

 H p−1/4 t1C3t1 11−p/2

1− e 1−p/4

.

2.26

Trang 7

From2.10, we see that

H p−1/4 tt  1 1−p/2



−E0t2 ˙H 0t  H0

t2 2t  1

p−1/4

−→ −E0p−1/4 2.27

as t → ∞ Take t1sufficiently large such that Hp1/4 t1t1 11−p/2 ≥ 1/2−E0p−1/4 It

follows from2.26 and the condition ofTheorem 2.2that

δ t ≥ 1

2−E0p−1/4 C3

1− e 1−p/4

Therefore,

By virtue of the continuity of Jt and the theorem of the intermediate values, there is a constant t1 <  T ≤ t1 1 such that JT  0 Hence, Zt → ∞ as t → T It follows from

Lemma 2.1that Ht ≥ Zt, t ≥ t1 Thus, H t → ∞ as t → T The theorem is proved.

Theorem 2.3 Suppose that gs is a convex function, g0  0, gs ≥ ls p , where a is a real number

p > 1, and u x, t is a weak solution of problem 1.1–1.4



S1

u0σψ1σdσ  α ≥ λ1

l

1/p−1

> 0,



S1

u1σψ1σdσ  β > 0, 2.30

where ψ1 is the normalized eigenfunction (i.e., ψ1 ≥ 0,S1ψ1σdσ  1) corresponding the smallest

eigenvalue λ1> 0 of the following Steklov spectral problem [ 23 ]:

∂ψ

a ∂ψ

where Ω, S1, S2, k, a, b are defined as in Section 1 Then, the solution of problem1.1–1.4 blows up

in a finite time.

Proof Let

y t 



S

Trang 8

Then, y0  y0 α > 0, y t 0  y1  β > 0 It follows from 1.1–1.4 that yt satisfies

y tt −



S1

∂u

∂n ψ1



S1

Using Green’s formula, we have

0





S

∂u

∂n ψ1



Ω∇u · ∇ψ1dx





S

∂u

∂n ψ1



S

u ∂ψ1

∂n dσ



Ωu Δψ1dx



S1

∂u

∂n ψ1



S1

u ∂ψ1

∂n dσ



S2

∂u

∂n ψ1



S2

u ∂ψ1

∂n dσ





Ωu Δψ1dx

 B1 B2,

2.36

where we have used2.31 and the fact that ψ1is the eigenfunction of the problem1.1–1.4,

B1and B2are denoted as the expressions in the first and the second parenthesis, respectively From2.32, we have

B1



S1

∂u

∂n ψ1dσ − λ1



S1

If a  0, it is clear that B2  0 otherwise, by 1.3 and 2.33,

B2



S2

b

a u ψ1



S2

u

b

Therefore,2.36 implies that B1 0, that is,



S1

∂u

∂n ψ1dσ  λ1



S1

Now,2.35 takes the form

y tt  −λ1y



S1

From Jensen’s inequality and the condition gs ≥ ls p, we have



S

g uψ1dσ ≥ g

S

1

Trang 9

Substituting the above inequality into2.40, we get

y tt  λ1y ≥ ly p , t > 0. 2.42

Since y0  α > 0, y t 0  β > 0, from the continuity of yt, it follows that there is a right

neighborhood0, δ of the point t  0, in which ˙yt > 0, and hence yt > y0 > 0 If there

exists a point t0 such that ˙yt > 0t ∈ 0, t0, but ˙yt0  0, then yt is monotonically

increasing on 0, t0 2.42 that on 0, t0

y tt ≥ yly p−1− λ1



≥ y0



ly0p−1− λ1



and thus y t t is monotonically increasing on 0, t0 0  0 Therefore,

˙yt > 0 and hence yt > y0as t > 0.

Multiplying both sides of2.42 by 2y tand integrating the product over

y2t2l

p 1



y p1− y p1

0



− λ1



y2− y2 0



 y2

1  By

Since By0  y2

1 > 0 and

B

y

 2ly p − 2λ1y > 2y0



ly0p−1− λ1



then By > By0 > 0, ct > 0 Extracting the square root of both sides of 2.44, we have

y t



2l

p 1



y p1− y p1

0



− λ1



y2− y2 0



 y2 1

−1/2

, t > 0. 2.46

Equation2.46

T

∞

y0



2l

p 1



y p1− α p1

− λ1



y2− α2

 β2

1/2

ds < ∞, 2.47

and yt → ∞ as t → T− The theorem is proved

Remark 2.4 The results of the above theorem hold when one considers1.1–1.4 with more general elliptic operator, like

Lu ≡ − divkx∇u  cxu, 0 < k0≤ kx ≤ k1, c x ≥ 0, in Ω × 0, T, 2.48

Trang 10

and the corresponding boundary conditions

2u

∂t2  kx ∂u

∂n  gu, on S1× 0, T,

k x ∂u

∂n  bu  0, bx ≥ 0, on S2× 0, T.

2.49

Acknowledgments

The authors are very grateful to the referee’s suggestions and comments The authors are supported by National Natural Science Foundation of China and Foundation of Henan University of Technology

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The authors are very grateful to the referee’s suggestions and comments The authors are supported by National Natural Science Foundation of China and Foundation of Henan University of Technology

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