Existence and nonexistence of global solutions for the generalized IMBq equation

Nonlinear Analysis 36 (1999) 961 – 980

Existence and nonexistence of global solutions for the generalized IMBq equation1 Chen Guowang a;∗ , Wang Shubin b a

Department of Mathematics, Zhengzhou University, Zhengzhou 450052, People’s Republic of China b Department of Mathematics and Mechanics, Zhengzhou Institute of Technology, Zhengzhou 450052, People’s Republic of China Received 10 July 1995; accepted 24 June 1997

Keywords: Wave equation; Higher order; Nonlinear; Initial boundary value problem; Existence of global solution; Blow up of solutions

1. Introduction We know that the equation (which we call the Bq-equation) ’t t = ’xx + (’2 )xx + ’xxxx

(1.1)

was derived by J. Boussinesq in 1872 to describe shallow-water waves. The improved Bq-equation (which we call IBq) is ’t t − ’xx − ’xxt t = (’2 )xx :

(1.2)

A modi cation of the IBq equation analogous of the MKdV equation yields ’t t − ’xx − ’xxt t = (’3 )xx ;

(1.3)

which we call the IMBq (see [5]). In the above equations, ’ denotes the unknown function, and subscripts x and t indicate partial derivatives. For the Bq equation and its generalized equation there are many results (see [1, 4, 6–8]), but there are only a few results dealing with the IBq equation (1.2) and the IMBq equation (1.3). In [2] ∗ 1

Corresponding author. This project is supported by National Natural Science Foundation of China and by the Natural Science Foundation of Henan Provice. 0362-546X/99/$ – see front matter ? 1999 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 7 1 0 - 4

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Cauchy problem for generalized IMBq equation with several variables only is studied and the existence and uniqueness of the local classical solution are proved. About the global solution of the IMBq equation there has not been any discussion. In this paper, we are going to study the following initial boundary value problem of the generalized IMBq equation on QT = × [0; T ] ( = (0; 1); T ¿0) : ut t − uxx − uxxt t = f(u)xx ;

(1.4)

u(0; t) = u(1; t) = 0;

(1.5)

u(x; 0) = u0 (x);

ut (x; 0) = u1 (x);

(1.6)

where u(x; t) denotes unknown function, f(s) is the given nonlinear function, u0 (x) and u1 (x) are the given initial value functions. In Section 2 we prove the existence and uniqueness of the generalized local solution of the initial boundary value problem (1.4) – (1.6). The existence and uniqueness of the generalized global solution of the problem (1.4) – (1.6) are proved in Section 3. In Section 4 we prove the existence and uniqueness of the classical global solution of the problem (1.4)–(1.6). In Section 5 blowing up of the solutions for the problem (1.4) – (1.6) is discussed. In the nal section we consider the case f(u) = Ku q for Eq. (1.4). 2. Existence and uniqueness of generalized local solution for problem (1.4) – (1.6) In the following we reduce the problem (1.4) – (1.6) to an integral equation by Green’s function of a bounary value problem for a second-order ordinary dierential equation. We prove the existence and uniqueness of generalized local solution for the integral equation by the contraction mapping principle, i.e., the problem (1.4) – (1.6) has a unique generalized local solution. Let G(x; ) (see [9]) be the Green’s function of the boundary value problem for the ordinary dierential equation y(x) − y00 (x) = 0; i.e.,

( G(x; ) =

y(0) = y(1) = 0;

−1 −1 x 1 2 (e − e ) (e

− e−x )(e1− − e−1 ); 0 ≤ x¡;

−1 −1  1 2 (e − e ) (e

− e− )(e1−x − e x−1 );  ≤ x ≤ 1:

Green’s function G(x; ) satis es the following properties: (1) G(x; ) is de ned and continuous in Q = {0 ≤ x ≤ 1; 0 ≤  ≤ 1}; (2) G(x; ) satis es the homogeneous equation G(x; ) − Gxx (x; ) = 0;

x 6= 

and the homogeneous conditions G(0; ) = 0;

G(1; ) = 0:

(2.1)

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

963

(3) Gx0 (x; ) has the discontinuity point of the rst kind at x =  and satis es the condition Gx0 ( + 0; ) − Gx ( − 0; ) = −1: (4) G(x; ) = G(; x): (5) 0 ≤ G(x; )¡ 27 ; 0 ≤ x ≤ 1; 0 ≤  ≤ 1: Obviously, the properties (1)–(4) hold. We only prove the property (5). In fact, we rst consider G(x; ) on 0 ≤ x¡. We have G(x; ) ≥ 0. Since e x − e−x ¡e − e− , G(x; )¡ 12 (e − e−1 )−1 (e − e2−1 − e−2+1 + e):

(2.2)

From the inequality e2−1 + e1−2 ≥ 2 and Eq. (2.2) it follows that G(x; )¡

e−1 2 e − 2 + e−1 = ¡ : 2(e − e−1 ) 2(e + 1) 7

Using the property (4) we see that the property (5) holds. Suppose that u(x; t) is a classical solution of the problem (1.4) – (1.6). Eqs. (1.4) and (1.5) can be rewritten as follows: [ut t + u + f(u) − f(0)] − [ut t + u + f(u) − f(0)]xx = u + f(u) − f(0)

(2.3)

and ut t (0; t) + u(0; t) + f(u(0; t)) − f(0) = 0; ut t (1; t) + u(1; t) + f(u(1; t)) − f(0) = 0:

(2.4)

From Eqs. (2.3) and (2.4) we get Z ut t (x; t) + u(x; t) + f(u(x; t)) − f(0) =

0

1

G(x; )[u(; t) + f(u(; t)) − f(0)] d:

Integrating the above formula with respect to t twice we have Z u(x; t) = u0 (x) + tu1 (x) − +

Z tZ 0

0

t

0

t

(t − )[u(x; ) + f(u(x; )) − f(0)] d

(t − )G(x; )[u(; ) + f(u(; )) − f(0)] d d:

(2.5)

Hence, the classical solution of the problem (1.4) – (1.6) is the solution of the integral equation (2.5). Deÿnition 2.1. For any T ¿0 if the function u(x; t) ∈ C([0; T ]; C[0; 1]) satis es the integral equation (2.5), then u(x; t) is called the continuous solution of the integral equation (2.5) or the generalized solution of the problem (1.4) – (1.6). If T ¡∞, then

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C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

u(x; t) is called the generalized local solution of the problem (1.4) – (1.6). If T = ∞, then u(x; t) is called the generalized global solution of the problem (1.4) – (1.6). Now we are going to prove the existence and uniqueness of the local continuous solution for the integral equation (2.5) by the contraction mapping principle. For this purpose we de ne the function space X (T ) = {u(x; t) ∈ C([0; 1]; C[0; 1]); u(0; t) = u(1; t) = 0}; equipped with the norm de ned by kukX (T ) = max max |u(x; t)|; 0≤t≤T 0≤x≤1

∀u ∈ X (T ):

It is easy to see that X (t) is a Banach space. Let M = ku0 kC[0; 1] + ku1 kC[0; 1] . We de ne the set K(M; T ) = {u | u ∈ X (T ); kukX (T ) ≤ 2M + 1}: Obviously, K(M; T ) is a nonempty bounded closed convex subset of X (T ) for each M; T ¿0. We de ne the map S as follows: Z t (t − )[w(x; ) + f(w(x; )) − f(0)] d Sw = u0 (x) + tu1 (x) − 0

Z tZ t (t − )G(x; )[w(; ) + f(w(; )) − f(0)] d d; + 0

0

∀w ∈ X (T ): (2.6)

Obviously, S maps X (T ) into X (T ). Our goal is to show that S has a unique xed point in K(M; T ). Lemma 2.1. Assume that u0 (x); u1 (x) ∈ C[0; 1]; u0 (0) = u0 (1) = u1 (0) = u1 (1) = 0 and f(s) ∈C 1 (R); then S maps K(M; T ) into K(M; T ) and S : K(M; T ) → K(M; T ) is strictly contractive if T is appropriately small relative to M . Proof. Assume that w(x; t) ∈K(M; T ). Let us de ne f(Á) : [0; ∞) → [0; ∞) by f(Á) = max {|f(s)|; |f0 (s)|}; |s|≤Á

∀Á ≥ 0:

Observe that f is continuous and nondecreasing on [0; ∞). From Eq. (2.6) we obtain kSwkX (T ) ≤ M + MT +

9 14 [2M

+ 1 + 3f(2M + 1)]T 2 :

If T satis es

(  T ≤ min 1;

14 9(2M + 1 + 3f(2M + 1))

1=2 ) ;

(2.7)

then kSwkX (T ) ≤ 2M + 1. Therefore, if Eq. (2.7) holds, then S maps K(M; T ) into K(M; T ).

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

965

Now we are going to prove that the map S is strictly contractive. Let T ¿0 and w1 ; w2 ∈K(M; T ) be given. We have Z t (t − )[w1 (x; ) − w2 (x; ) + f(w1 (x; )) − f(w2 (x; ))] d Sw1 − Sw2 = − 0

+

Z tZ 0

0

1

(t − )G(x; )[w1 (; ) − w2 (; )

+f(w1 (; )) − f(w2 (; ))] d d: Using the mean value theorem for f we obtain 9 T 2 [1 + f(2M + 1)]kw1 − w2 kX (T ) : kSw1 − Sw2 kX (T ) ≤ 14

If T satis es

(  T ≤ min 1;

14 9(2M + 1 + 3f(2M + 1))

1=2  ;

7 9(1 + f(2M + 1))

1=2 ) ;

(2.8)

then kSw1 − Sw2 kX (T ) ≤ 12 kw1 − w2 kX (T ) . The lemma is proved. Theorem 2.1. Assume that the conditions of Lemma 2.1 hold. Then the problem (1.4) – (1.6) has a unique generalized local solution u(x; t) ∈C([0; T0 ); C(0; 1]); where [0; T0 ) is a maximal time interval. Moreover; if sup (ku(· ; t)kC[0; 1] + kut (· ; t)kC[0; 1] )¡∞;

t∈[0; Ts )

(2.9)

then T0 = ∞. Proof. It follows from Lemma 2.1 and the contraction mapping principle that for appropriately chosen T ¿0; S has a unique xed point u(x; t)∈K(M; T ); which is a generalized solution of the problem (1.4) – (1.6). It is easy to prove that for each T 0 ¿0; Eq. (2.5) has at most one solution which belongs to X (T 0 ). Let [0; T0 ) be the maximal time interval of existence for u ∈X (T0 ). It remains only to show that if Eq. (2.9) is satis ed, then T0 = ∞. Suppose that Eq. (2.9) holds and T0 ¡∞. For any T 0 ∈[0; T0 ); we consider the integral equation Z t (t − )[v(x; ) + f(v(x; )) − f(0)] d v(x; t) = u(x; T 0 ) + ut (x; T 0 )t − +

Z tZ 0

0

0

1

(t − )G(x; )[v(; ) + f(v(; )) − f(0)] d d:

(2.10)

By virtue of Eq. (2.9), ku(· ; T 0 )kC[0; 1] + kui (· ; T 0 )kC[0; 1] is uniformly bounded in T ∈[0; T0 ) which allows us to choose T ∗ ∈(0; T0 ) such that for each T 0 ∈[0; T0 ); the integral equation (2.10) has a unique solution v(x; t)∈X (T ∗ ). The existence of 0

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C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

such a T ∗ follows from Lemma 2.1 and the contraction mapping principle. In particular, Eq. (2.8) reveals that T ∗ can be selected independently of T 0 ∈[0; T0 ). Set T 0 = T0 − T ∗ =2; let v denote the corresponding solution of Eq. (2.10), and de ne u(x; ˆ t) : [0; 1] × [0; T0 + T ∗ =2] → R by ( u(x; ˆ t) =

u(x; t);

t ∈[0; T ∗ ];

v(x; t − T 0 );

t ∈[T 0 ; T0 + T ∗ =2]:

(2.11)

By construction, u(x; ˆ t) is a solution of Eq. (2.10) on [0; T0 + T ∗ =2], and by local uniqueness, uˆ extends u. This violates the maximality to [0; T0 ). Hence, if Eq. (2.9) holds, then T0 = ∞. This completes the proof of the theorem.

3. Existence and uniqueness of generalized global solution for the problem (1.4) – (1.6) In this section we prove the existence and uniqueness of the generalized global solution for the problem (1.4)–(1.6). For this purpose we are going to make a priori estimates of the generalized solution for the problem (1.4) – (1.6). Lemma 3.1. Assume that u0 ; u1 ∈C[0; 1]; f(s) ∈C(R) and the following inequality |f(u)| ≤ AF(u) + B

(3.1)

Ru holds; where F(u) = 0 f(s) ds and A; B¿0 are constants. Then the generalized solution u(x; t) ∈C([0; T ]; C[0; 1]) has the estimate Z 0

1

Z |f(u(x; t))| dx +

1

0

u2 (x; t) dx ≤ M1 (T );

0 ≤ t ≤ T;

(3.2)

where M1 (T ) denotes a constant dependent on T. Proof. Dierentiating Eq. (2.5) with respect to t, we obtain Z ut (x; t) +

0

t

Z u(x; ) d +

= u1 (x) + f(0)t +

Z tZ 0

t

0

0

1

f(u(x; ) d G(x; )[u(; ) + f(u(; )) − f(0)] d d:

(3.3)

Observe that Z tZ 0

0



u(x; s) ds u(x; ) d =

1 2

Z 0

t

d d

Z 0



2 u(x; s) ds

d =

1 2

Z 0

t

2 u(x; ) d

:

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

967

Multiply Eq. (3.3) by u and integrate the product with respect to t, a direct calculation shows 1 2 1 u + 2 2

Z

t

0

2 u(x; ) d Z

1 = u02 + u1 (x) 2 + +

0

Z tZ Z 0

0

1

0

Z tZ Z 0

0

−f(0)

1

0

Z tZ 0

t

0



0

f(u(x; s)) ds u(x; ) d Z

u(x; ) d + f(0)

0

t

u(x; ) d

G(x; )u(; s)u(x; ) d ds d G(x; )f(u(; s))u(x; ) d ds d

1

0

+

Z tZ

G(x; )u(x; ) d d:

(3.4)

Multiplying Eq. (3.3) by f(u) and integrating the product with respect to t, we have 1 F(u(x; t)) + 2

Z

t

0

2 f(u(x; )) d Z

= F(u0 ) + u1 (x) + +

Z tZ Z 0

0

0

0

Z =

0

0

=

0 t 0

1

0

Z tZ 0

d d

Z 0



0



f(u(x; ))u(x; s) ds d

f(u(x; )) d + f(0)

0

t

f(u(x; )) d

G(x; )f(u(; s))f(u(x; )) d ds d G(x; )f(u(x; )) d d:

f(u(x; s)) ds

f(u(x; )) d

0

Z

Z tZ 0

Z Z

0



G(x; )u(; s)f(u(x; )) d ds d

1

0

+

f(u(x; s))u(x; ) ds d + t

Z

0

Z tZ Z

−f(0) Observe that Z tZ 

1

0

t

Z tZ

0



0



(3.5)

f(u(x; ))u(x; s) ds d 

u(x; s) ds d

u(x; ) d:

(3.6)

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C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

Adding Eq. (3.4) to Eq. (3.5) and integrating the sum with respect to x over [0; 1], we get Z 2

1

0

Z F(u(x; t)) dx +

Z +

1

Z

0

t

0

Z =2

1

u2 (x; t) dx +

0

f(u(x; )) d

Z + 2f(0) Z +2

1

Z

1

Z Z

0

1

0

0

Z

0

0

0

0

1

1

Z tZ Z

− 2f(0) Z

0

0

0 1

1

Z tZ Z

0

−2

0

t

0

2 u(x; ) d

dx

1

0

Z 0

1

0

Z tZ 0

t

1

0

u02 (x) dx+2

Z

1

0

Z u1 (x)

0

t

[u(x; )+f(u(x; ))] d dx

[u(x; t) + f(u(x; ))] d dx

Z tZ Z

0

+2

t

0

0 1

Z

0

0

0

+2

1

Z tZ Z

0

+2

Z

dx

F(u0 (x)) dx+

0

1

2 Z

1

Z

G(x; )u(; s)u(x; ) d ds d dx G(x; )f(u(; s))f(u(x; )) d ds d dx G(x; )f(u(; s))u(x; ) d ds d dx G(x; )u(; s)f(u(x; )) d ds d dx 1

0

G(x; )[u(x; ) + f(u(x; ))] d d dx Z

f(u(x; )) d

t

0

 u(x; ) d dx:

(3.7)

In order to rewrite Eq. (3.7) we change some of its terms by the method used in Eq. (3.6) as follows: Z

1

Z tZ Z

0

0

0

Z =

1

G(x; )

1Z 0

G(x; )u(; s)u(x; ) d ds d dx Z

1

0

Z −

0

Z

0

1

0

1

G(x; )

0

t

Z u(; ) d

Z t Z 0

0



0

t

 u(x; ) d d dx

 u(x; s)u(; ) ds d d ds;

(3.8)

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

Z

1

Z tZ Z

0

0

0

Z =

0

− 1

1

G(x; )

Z

0

1

0

Z tZ Z

0

0

0

Z =

1

0

Z

1

0

−

1

0

G(x; )

1

0

f(u(; )) d

Z t Z 0



0



t

f(u(x; )) d d dx

0

 f(u(x; s))f(u(; )) ds d d dx;

(3.9)

G(x; )f(u(; s))u(x; ) d ds d dx

G(x; )

Z

Z

t

0

Z

1

0

Z

G(x; )f(u(; s))f(u(x; )) d ds d dx Z

1

0

Z

Z

0

Z

1

1

969

G(x; )

Z

t

f(u(; )) d

0

Z t Z 0

0





t

u(x; ) d d dx

0

 u(x; s)f(u(; )) ds d d dx:

(3.10)

Using the symmetry of G(x; ) with respect to x and , from Eq. (3.8) – (3.10) it follows that Z 1Z 1Z tZ  G(x; )u(; s)u(x; ) ds d d dx 0

0

1 2

= Z

1

Z

0

1

Z

1

0

1

0

Z tZ

1 2 1

Z

Z

0

1

0

Z tZ 0

Z

1

1

0

1

Z G(x; )

0

Z 0

1

t

0

Z u(; ) d

t

0

 u(x; ) d d dx;

(3.11)

G(x; )f(u(; s))f(u(x; )) ds d d dx Z G(x; )

t

0

Z f(u(; )) d

0

t

 f(u(x; )) d d dx;

(3.12)

G(x; )f(u(; s))u(x; ) ds d d dx

Z tZ

0

Z



0

Z

0

=



0

1

0

+

Z

0

Z

0

0

1

0

= Z

0

0



G(x; )u(; s)f(u(x; )) ds d d dx Z

G(x; )

0

t

Z u(; ) d

0

t

 u(x; ) d d dx:

(3.13)

970

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

Substituting Eqs. (3.11)–(3.13) into Eq. (3.7) we obtain 2 Z 1 Z 1 Z t Z 1 2 F(u(x; t)) dx + u (x; t) dx + u(x; ) d dx 2 0

0

Z +

1 0

Z =2

Z

t

0

Z

Z

1

Z

0

Z +

1

1

Z

0

1

0

Z

1

1 0

0

+2

0

Z

t

0

G(x; ) 1

Z 0

Z tZ 0

t

Z

u02 (x) dx + 2

Z u(; ) d

1

0

Z u1 (x)

0

t

[u(x; ) + f(u(x; ))] d dx

0

1

 u(x; ) d d dx

t

0

Z f(u(; )) d

Z

0

0

t

0

Z

1

dx

[u(x; ) + f(u(x; ))] d dx

G(x; )

0

1

t

1

0

Z

− 2f(0) Z

Z

G(x; )

Z

0

−2

f(u(x; )) d

F(u0 (x)) dx +

+ 2f(0)

0

2

Z

1

0

+

0

0

t

t

0

 f(u(x; )) d d dx

Z f(u(; )) d

t

0

 u(x; ) d d dx

G(x; )[u(x; ) + f(u(x; ))] d d dx Z

f(u(x; )) d

0

t

 u(x; ) d dx:

(3.14)

Now we make a series of estimations for the right-hand side terms in Eq. (3.14). Using Cauchy’s inequality, Holder’s inequality and the condition (3.1) we can obtain Z t  2 Z Z Z t Z 1 7 1 2 3 1 u1 (x) u(x; ) d dx ≤ u (x) dx + u(x; ) d dx; 2 3 0 1 7 0 0 0 0 (3.15) Z 2

1 0

Z u1 (x)

0

t

 f(u(x; )) d dx ≤

14 3 +

Z

3 14

1

0

u12 (x) dx

Z 0

1

Z 0

t

2 f(u(x; )) d

dx;

(3.16)

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

Z 2f(0)

1

Z

0

t

0

Z 1Z

2f(0)

0

t

u(x; ) d dx ≤ |f(0)|T 2 + |f(0)|T

f(u(x; )) dx d ≤ 2|f(0)|TA

0

Z 1Z 0

Z

t

0

1

0

Z 0

t

|u(x; t)| 2 dx d;

971

(3.17)

F(u(x; t)) dx d + 2B|f(0)|T 2 ; (3.18)

Z

Z

1 0

Z

G(x; )

0

1

Z

0

1

0

Z

1

Z

0

≤

Z

0 1

0

2 7

u(; ) d

0

G(x; ) 1

Z

t

Z

Z

2 7

≤

2

Z

1

t

0

f(u(; )) d

f(u(x; )) d

G(x; ) Z

0

−2f(0)

Z

1 0

1

t

Z

0

≤

Z

t

0

≤

0

1

Z

0

1

"Z

0

3 14

t

0

Z 0

1

t

0

4 |f(0)|TA 7

Z −2

Z

f(u(x; )) d d dx

dx;

(3.20)

f(u(; )) d

f(u(x; )) d

0

1



Z

Z 0

1

Z 0

t

0

dx +

2 7

Z

1

Z

0

2

t

u(x; ) d

0

dx;

Z

1

0

Z 0

t

|u(x; )| 2 dx d;

4 F(u(x; )) dx d + B|f(0)|T 2 ; 7

f(u(x; )) d t

u(x; ) d d dx

0

(3.21)

(3.22)

G(x; )f(u(x; )) d dx d

Z

Z



t

G(x; )u(x; ) d dx d

2 2 ≤ |f(0)|T 2 + |f(0)|T 7 7 −2f(0)

0

2

t

0

Z

t

2

t

0

1

0

 2 Z Z t 2 1 u(x; ) d d dx ≤ u(x; ) d dx; 7 0 0 (3.19) Z

Z

Z

t

0

1

(3.23)

# u(x; ) d dx 2

f(u(x; )) d

dx +

14 T 3

Z tZ 0

0

1

|u(x; )| 2 dx d:

(3.24)

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Substituting Eqs. (3.15)–(3.24) into Eq. (3.14), we can obtain Z 1 Z 1 F(u(x; )) dx + u 2 (x; t) dx 2 0

0

Z ≤2

1

0

1

0

u02 (x) dx + 7

Z 0

1

u12 (x) dx

  Z tZ 1  14 9 9 18 2 + B |f(0)|T + |f(0)| + T |u(x; )| 2 dx d 7 7 7 3 0 0

 + +

Z F(u0 (x)) dx +

18 |f(0)|TA 7

Z tZ 0

0

1

F(u(x; )) dx d:

(3.25)

Multiplying both sides of Eq. (3.25) by A, adding the product to 2B and using Eq. (3.1), we have Z 1 Z 1 (AF(u(x; )) + B) dx + A u 2 (x; t) dx 2 0

Z

≤ 2A

0

1

Z F(u0 (x)) dx + A

0

0

1

u02 (x) dx + 7A

Z 0

1

u12 (x) dx

 Z tZ 1 9 14 9 18 2 T |u(x; )| 2 dx d + B |f(0)|T + A |f(0)| + +A 7 7 7 3 0 0 Z tZ 1 18 [AF(u(x; )) + B] dx d + 2B: (3.26) + |f(0)|TA 7 0 0 





Then the Gronwall’s inequality yields Z 1 Z 1 |f(u(x; ))| dx + u2 (x; t) dx ≤ M1 (T ); 0

0

0 ≤ t ≤ T:

(3.27)

This proves Lemma 3.1. Lemma 3.2. Under the assumptions of Lemma 3:1; the generalized solution u(x; t) of the problem (1:4)–(1:6) has the estimation ut2 + u2 ≤ M2 (T );

0 ≤ x ≤ 1;

0 ≤ t ≤ 1;

(3.28)

where M2 (T ) is constant dependent on T . Proof. Dierentiating Eq. (2.5) twice with respet to t we get Z 1 G(x; )[u(; t) + f(u(; t)) − f(0)] d: ut t(x; t) + u(x; t) + f(u(x; t))=f(0) + 0

(3.29)

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Multiplying both sides of Eq. (3.29) by Aut and using Eqs. (3.1) and (3.2) we get d [Aut2 + Au2 + 2(AF(u) + B)] dt ( Z = 2A f(0) +

1

0

)

G(x; )[u(; t) + f(u(; t)) − f(0)] d ut (x; t)

  "Z #1=2 Z  1 1 4 9 |f(0)| + u2 (; t) d + |f(u(; t))| d |ut | ≤ A  7 2 0 0 ≤ C1 (T )|ut |: Integrating the above inequality with respect to t we obtain Aut2 + Au2 + 2(AF(u) + B) ≤ Au12 + Au02 + 2(AF(u0 ) + B)C2 (T ) Z 1 t 2 u (x; ) d: + 2 0 t Here C1 (T ) and C2 (T ) are constants dependent on T . Using the Gronwall’s inequality we can obtain Eq. (3.28). The lemma is proved. Theorem 3.1. Suppose that u0 (x); u1 (x) ∈ C[0; 1] satisfy the boundary condition (1:5) and f(s) ∈ C 1 (R) satisÿes the condition (3:1); then the problem (1:4) – (1:6) has a unique generalized global solution u(x; t) ∈ C[0; ∞]; C[0; 1]). Proof. By virtue of Theorem 2.1 we only need to prove that Eq. (2.9) holds. In fact, from Lemma 3.2 it follows that ku(· ; t)kC[0; 1] + kut (· ; t)kC[0; 1] ≤ 2(M2 (T ))1=2 : Therefore sup (ku(· ; t)kC[0; 1] + kut (· ; t)kC[0; 1] )¡∞:

t∈[0; t0 ]

It follows from Theorem 2.1 that T0 = ∞. Theorem 3.1 is proved. 4. Existence of the classical global solution for problem (1.4) – (1.6) In order to prove the existence of the classical global solution for the problem (1.4) – (1.6), we rst study the regularity of the generalized global solution for the problem (1.4)–(1.6). Lemma 4.1. Suppose that the conditions of Theorem 3:1 hold and u0 (x); u1 (x) ∈ C k [0; 1] and f(s) ∈ C k+m (R) (k ≥ 1; m ≥ 0 are arbitrary integers); then the generalized solution u(x; t) for the problem (1:4)–(1:6) belongs to C m+2 ([0; T ]; C k−1 [0; 1]) (∀T ¿0).

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Proof. We apply the method of induction. When m = 1, dierentiating Eq. (2.5) with respect to x we obtain Z ux (x; t) = u0x (x) + tu1x (x) − +

Z tZ 0

1

0

t

0

(t − )[ux (x; ) + f0 (u(x; ))ux (x; )] d

(t − )Gx (x; )[u(; ) + f(u(; )) − f(0)] d d:

(4.1)

Hence kux (· ; t)kC[0; 1] ≤ ku0x kC[0; 1] + T ku1x kC[0; 1] Z t + C3 (T ) kux (· ; )kC[0; 1] d + C4 (T ): 0

Using Gronwall’s inequality we get sup kux (· ; )kC[0; 1] ≤ M3 (T ):

0≤t≤T

(4.2)

When m = 2, dierentiating Eq. (4.1) with respect to x and using G(x; ) = Gxx(x; ) (x 6= ) we obtain Z uxx(x; ) = u0xx + tu1xx − Z − +

t 0

(t − )[uxx(x; ) + f(u(x; ))x2 ] d

(t − )[u(x; ) + f(u(x; )) − f(0)] d

Z tZ 0

0

t

0

1

(t − )G(x; )[u(; ) + f(u(; )) − f(0)] d d:

(4.3)

It follows from Eq. (4.3) that sup kuxx(· ; t)kC[0; 1] ≤ M4 (T ):

0≤t≤T

(4.4)

Now suppose that when m = k − 1, the esitmate sup kuxk−1 (· ; t)kC[0; 1] ≤ M5 (T )

0≤t≤T

(4.5)

holds. We can prove that when m = k, the estimate sup kuxk (· ; t)kC[0; 1] ≤ M6 (T )

0≤t≤T

(4.6)

holds, too. Here Mi (T )(i = 3; 4; 5; 6) are constants dependent on T . When k = 2j− 1(i = 1; 2; : : :), that is, k is an odd number, dierentiating Eq. (2.5) with respect to x

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

975

for (2j − 1)-times, we have ux2j−1 (x; t) = u0x2j−1 (x) + tu1x2j−1 (x) Z −

+

1

0

X

(t − )

[u(x; ) + f(u(x; ))]xi d

i=1; 3; :::; 2j−1

Z tZ 0

1

0

(t − )Gx (x; )[u(; )

+ f(u(; )) − f(0)] d d;

j = 1; 2; : : : ; 2j − 1 = k:

(4.7)

When k = 2j(j = 1; 2; : : :), that is, k is an even number, dierentiating Eq. (2.5) with respect to x for 2j-times, we have ux2j (x; t) = u0x2j (x) + tu1x2j (x) Z −

+

1

0

(t − )

[u(x; t) + f(u(x; )) − f(0)]xi d

i=0; 2; :::; 2j

Z tZ 0

X

1

0

(t − )G(x; )[u(; )

+f(u(; )) − f(0)] d d;

j = 1; 2; : : : ; 2j = k:

(4.8)

It follows from Eq. (4.7) and (4.8) that kuxk (· ; t)kC[0; 1] ≤ ku0xk kC[0; 1] + T ku1xk kC[0; 1] Z + C5 (T )

t

0

kuxk (· ; )kC[0; 1] d + C6 (T );

where C5 (T ) and C6 (T ) are constants dependent on T . Gronwall’s inequality gives Eq. (4.6). From Eq. (4.6) we have u(x; t) ∈ C([0; T ]; C k−1 [0; 1]): Let k − 1 = 2j ( j = 0; 1; : : :) in Eq. (4.8). Dierentiating this result with respect to t we obtain Z t X [u(x; t) + f(u(x; )) − f(0)]xi d ux2jt (x; t) = u1x2j − 0 i=0; 2; :::; 2j

+

Z tZ 0

0

1

G(x; )[u(; ) + f(u(; )) − f(0)] d d:

(4.9)

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C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

It is from Eq. (4.9) that ux2jt (x; t) ∈ C([0:T ]; C[0; 1]). Dierentiating Eq. (4.9) with respect to t we obtain X [u(x; t) + f(u(x; t)) − f(0)]xi ux2jtt (x; t) = − i=0; 2;:::;2j

Z +

1

0

G(x; )[u(; t) + f(u(; t)) − f(0)] d:

(4.10)

We see from Eq. (4.10) that ux2jt2 (x; t) ∈ C([0; T ]; C[0; 1]). When k − 1 = 2j − 1 ( j = 1; 2; : : :) we obtain by the same method shown above ux2j−1tt ∈ C([0; T ]; C[0; 1]). We can apply the method of induction to prove that u(x; t) ∈ C m+2 ([0; T ]; C k−1 [0; 1]): The lemma is proved. Theorem 4.1. Suppose that u0 ; u1 ∈ C 3 [0; 1]; u0 (0) = u0 (1) = 0; u1 (0) = u1 (1) = 0; f ∈ C 3 (R) and Z u f(s) ds + B; (4.11) |f(u)| ≤ A 0

where A¿0; B¿0 are constants; then the generalized global solution u(x; t) for the problem (1:4)–(1:6) is the classical global solution for the problem (1:4) – (1:6). Proof. The assumptions of Theorem 4.1 and the conclusion Lemma 4.3 show that u(x; t) ∈ C 3 ([0; ∞); C 2 [0; 1]). It is easy from Eq. (2.5) to see that u(x; t) satis es the initial boundary value conditions (1.5) and (1.6). Now we are going to prove that u(x; t) satis es Eq. (1.4). Dierentiating Eq. (2.5) twice with respect to t, we get ut t(x; t) = −[u(x; t) + f(u(x; t)) − f(0)] Z 1 G(x; )[u(; t) + f(u(; t)) − f(0) d: + 0

(4.12)

Dierentiating Eq. (4.12) twice with respect to x, we get uxxt t (x; t) = −[uxx(x; t) + f(u(x; t))xx] − [u(x; t) + f(u(x; t)) − f(0)] Z 1 Gxx (x; )[u(; t) + f(u(; t)) − f(0)] d: + 0

(4.13)

Observing that G(x; ) = Gxx(x; ) (x 6= ) and substracting Eq. (4.13) from Eq. (4.12) we see that u(x; t) satisi es Eq. (1.4). The theorem is proved. 5. Blow up of the solution for problem (1.4) – (1.6) In this section we are going to consider the blow up of the solution for the problem (1.4) – (1.6). By the above method it is easy to prove the following lemma.

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

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Lemma 5.1. Suppose that u0 (x); u1 (x) ∈ C k [0; 1]; u0 (0) = u0 (1) = u1 (0) = u1 (1) = 0, f(s) ∈ C k (R); then the problem (1.4) – (1.6) has a unique local classical solution u(x; t) ∈ C k ([0; T0 ); C k−1 [0; 1]); where [0; T0 ) is a maximal time interval and k ≥ 3 is a natural number. Lemma 5.2. (Glassey [3]). Let (t) ∈ C 2 satisfy  ≥ h()

(t ≥ 0)

˙ = ÿ¿0. Suppose that h(s) ≥ 0 for all s ≥ . Then with (0) = ¿0; (0) ˙ ˙ exists; and (a) (t)¿0 whereever (t) (b) the inequality −1=2 Z s Z (t)  ÿ2 + 2 h() d ds t≤ 



obtains. “·” denotes the derivative with respect to t. Theorem 5.1. Let u(x; t) be the classical solution of the problem (1.4) – (1.6). Suppose that the following conditions are satisÿed: R1 R1 (a) −(=2) 0 u0 (x) sin x dx = ¿0; −=2 0 u1 (x) sin x dx = ÿ¿0; where (=2) sin x denotes the ÿrst normalized eigenfunction for the problem 00

+  = 0; x ∈ (0; 1); (0) = (1) = 0 and  = 2 denotes the corresponding ÿrst eigenvalue; (b) f(s) ∈ C 2 is an even and convex function satisfying (i) f(s) − s is a nonnegative; nondecreasing function for s ≥ ; (ii) f(s) grows fast enough as s → ∞ so that the integral −1=2 Z s Z ∞ 2 2 2 22 2 2 T=  + ÿ − s + f() d ds 1 + 2 1 + 2 1 + 2  

(5.1)

converges. Then lim sup |u(x; t)| = +∞

t→t0− x∈[0; 1]

for some ÿnite time t0 ≤ T ; where T is given by Eq. (5:1). Proof. Let (t) = −

 2

Z 0

1

u(x; t) sin x dx;

multiply Eq. (1.4) by (=2) sin x and integrate over (0; 1). Integrating by parts with respect to x we obtain Z 1  2 2 sin xf(u) dx: (5.2)  =  + 1 + 2 1 + 2 0 2

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C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

Since f(s) is an even and convex function, by Jensen’s inequality we have ! ! Z 1 Z Z 1  1   sin xf(u) dx ≥ f u sin x dx = f − u sin x dx = f(): 2 2 0 0 2 0 (5.3) Substituting Eq. (5.3) into Eq. (5.2) we get  +

2 2  ≥ f() 1 + 2 1 + 2

˙ = ÿ¿0. Since h(s) = f(s) − s ≥ 0 for all s ≥ , it follows from with (0) = ¿0, (0) Lemma 5.2 that −1=2 Z s Z (t)  2 2 2 22 2 2  +ÿ − s + f() d ds t= 1 + 2 1 + 2 1 + 2   and thus (t) develops a singularity in a nite time t0 ≤ T . Finally, since (t)¿0, we have (t) ≤ sup |u(x; t)| x∈[0;1]

which proves the theorem. Corollary. For each p, 1 ≤ p ≤ ∞, Z ku(· ; t)kLp [0;1] =

0

1

!1=p p

|u(x; t)| dx

blows up in ÿnite time.

6. The case f (u) = Kuq in Eq. (1.4) In this section we are going to discuss the case f(u) = Kuq in Eq. (1.4), which is utt − uxx − uxxtt = (Kuq )xx ;

(1.4a )

where K 6= 0 and q¿1 are constants. Obviously, Eq. (1:4a ) includes IBq equation (1.2) and IMBq equation (1.3). Theorem 6.1. Suppose that u0 ; u1 ∈ C 3 [0; 1] satisfy the boundary condition (1:5); K¿0 is a real number and q ≥ 3 is an odd number. Then the problem (1.4a ), (1.5), (1.6) has a unique global classical solution u(x; t).

C. Guowang, W. Shubin / Nonlinear Analysis 36 (1999) 961 – 980

979

Proof. By virtue of Theorem 4.1 we only need to prove that f(s) = Ks q satis es the condition (4.11). In fact, taking p = (q + 1)=q, p0 = q + 1 and using Young’s inequality we have  qp  Z s 1 |u| K K q q+1 + 0 =K u =q : + Ks q ds + |Kuq | = K|u|q ≤ K p p q+1 q+1 q + 1 0 The theorem is proved. Remark 6.1. f(s) = eq also satisÿes the condition (4:11). Suppose that u0 ; u1 ∈ C 3 [0; 1] satisÿes the boundary condition (1:5). It is easy to see that the initial boundary value problem (1:5); (1:6) for IMBq equation (1:3) has a unique global classical solution u(x; t). According to Lemma 5.1, if K 6= 0, q satis es that q¿1, (−1)q = 1, u0 (x); u1 (x) ∈ C 3 [0; 1], u0 (0) = u0 (1) = u1 (0) = u1 (1) = 0, then the problem (1:4a ), (1.5), (1.6) has a unique local classical solution u(x; t). Theorem 6.2. Let u(x; t) be the classical solution of the problem (1:4a ); (1:5); (1:6). Suppose that K 6= 0 and q¿1; (−1)q = 1. R1 R1 (a) If K¿0; −(=2) 0 u0 (x) sin x dx = ¿( K1 )1=(−1) and −(=2) 0 u1 (x) sin x dx = ÿ¿0; then lim sup |u(x; t)| = +∞

t→t0− x∈[0;1]

for some ÿnite time −1=2 Z ∞ 2 2 22 K 2 2 2 q+1 q+1 (s  + ÿ − s + −  ) ds: t0 ≤ T = 1 + 2 1 + 2 (1 + 2 )(q + 1)  R1 R1 (b) If K¡0; (=2) 0 u0 (x) sin x dx = ¿(−1=K)1=(q−1) and (=2) 0 u1 (x) sin x dx = ÿ¿0; then lim sup |u(x; t)| = +∞

t→t0− x∈[0;1]

for some ÿnite time −1=2 Z ∞ 2 2 2 22 K 2 2 q+1 q+1 (s  +ÿ − s + − ) ds: t0 ≤ T = 1 + 2 1 + 2 (1 + 2 )(q + 1)  Proof. For the case (a) it follows from Theorem 5.1 that the conclusion is valid. R1 Let (t) = (=2) 0 u(x; t) sin x dx. We can easily prove the case (b) by the same method used in Theorem 5.1. The theorem is proved. R1 Remark 6.2. It u0 (x) and u1 (x) satisfy the conditions: −(=2) 0 u0 (x) sin x dx = R1 ¿1 and −(=2) 0 u1 (x) sin x dx = ÿ¿0; then the solution u(x; t) of the problem (1:5) and (1:6) of IBq equation will blow up in ÿnite time.

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References [1] J.L. Bona, R.L. Sachs, Global existence of smooth solutions and stability if solitary waves for a generalized Boussinesq equation, Comm. Math. Phy. 118 (1988) 15–29. [2] G. Chen, J. Xing, Z. Yang, Cauchy problem for generalized IMBq equation with several variables, Nonlinear Anal., to appear. [3] R.T. Glassey, Blow-up theorems for nonlinear wave equations, Math. Z. 132 (1973) 182–203. [4] F. Lineras, Global existence of small solutions for a generalized Boussinesq equation, J. Di. Eq. 106 (1993) 257–293. [5] V.G. Makhankov, Dynamics of classical solitons (in non-integrable systems), Physics Reports, A review section of Phys. Lett. (Section C), 35(1) (1978) 1–128. [6] R.L. Sachs, On the blow up of certain solution of the “good” Boussinesq equation, Appl. Anal. 34 (1990) 145–152. [7] M. Tsutsumi, T. Matahashi, On the Cauchy problem for the Boussinesq type equation, Math. Japonica 26(2) (1991) 371–379.  Stuttgart, ber das homogene Dirichlet-Problem bei nichtlinearen paritiellen [8] G. Warnecke, U. Dierentialgleichungen vom Typ der Boussinesq-Gleichung, Math. Meth. Appl. Sci. 9 (1987) 493–519. [9] E. Zauderer, Partial Dierential Equations of Applied Mathematics, Wiley, New York, 1983.