Open Access

Blow-up solution and stability to an inverse problem for a pseudo-parabolic equation

Journal of Inequalities and Applications20122012:274

https://doi.org/10.1186/1029-242X-2012-274

Received: 20 February 2012

Accepted: 14 November 2012

Published: 28 November 2012

Abstract

We consider a two-fold problem for an inverse problem of pseudo-parabolic equations with a nonlinear term. Sufficient conditions for a blow-up solution are derived and a stability result is established.

Keywords

blow-upinverse problempseudo-parabolic equationstability

1 Introduction

Let us consider the following inverse problem for a pseudo-parabolic equation:
(1)
(2)
(3)
(4)
where Ω R n is a bounded domain with a sufficiently smooth boundary Ω, p and a are positive constants, g ( x ) and b i ( x ) are given functions satisfying
ω H 2 ( Ω ) H 0 1 ( Ω ) L p + 2 ( Ω ) , Ω ω ( ω Δ ω ) d x = 1 ,
(A1)
with weight function g ( x ) = ω a Δ ω , and a constant
B 0 = max x Ω ( i = 1 n b i 2 ( x ) ) 1 / 2 , x Ω , b i C ( Ω ¯ ) .
(A2)
The inverse problem consists of finding a pair of functions ( u ( x , t ) , f ( t ) ) satisfying (1)-(4) when
u 0 H 0 1 ( Ω ) L p + 2 ( Ω ) and Ω u 0 ( ω Δ ω ) d x = 1 .
(A3)

Additional information about the solution to the inverse problem is given in the form of the integral overdetermination condition (4). From the physical point of view, this condition may be interpreted as measurements of the temperature u ( x , t ) by a device averaging over the domain Ω [1].

This type of equations arises from a variety of mathematical models in engineering and physical sciences; for example, inverse scattering problems in quantum physics, an inverse problem of interest in geophysics [2].

Existence and uniqueness of solutions to an inverse problem for parabolic and pseudo-parabolic equations are studied in [36]. Stability of solutions is investigated by several authors [1, 7]; but less is known about blow-up solutions. Eden and Kalantarov [8] studied the same problem without a strong damping term Δ u t . Meyvaci [9] established a blow-up result for the pseudo-parabolic equation u t Δ u t Δ u + u x 1 u p = | u | 2 m u , where p 1 is a given integer and m 1 is a number.

Here, we used the following notations:
u = u L 2 ( Ω ) , ( u , v ) = Ω u v d x , u p = u L p ( Ω )
are the arithmetic-geometric inequality and Young’s inequality for a , b > 0 respectively;
a b ε 2 a 2 + 1 2 ε b 2 , a b β a p + C ( p , β ) b q ,
(5)
with 1 / p + 1 / q = 1 , C ( p , β ) = 1 q ( β p ) q / p and the Poincare-Friedrich inequality
λ 1 u 2 u 2 ,
(6)
where λ 1 is the first eigenvalue of the eigenvalue problem
Δ u = λ u , x Ω ; u = 0 , x Ω .
Multiplying both sides of (1) by ω and integrating the resulting equation over Ω lead to the following relation:
f ( t ) = ( u , Δ ω ) + ( ω , i = 1 n b i u x i ) ( ω , | u | p u ) ,
(7)

where conditions (2), (3) and (A1) are used. Substituting (7) into (1), problem (1)-(3) yields a direct problem given by [4].

2 Blow-up result

Firstly, let us note the following lemma known as ‘generalized concavity lemma’ or ‘Ladyzhenskaya-Kalantarov lemma’. It is an important tool to obtain the blow-up solutions to parabolic- and hyperbolic-type equations.

Lemma 1 Let α > 0 , C 1 , C 2 0 and C 1 + C 2 > 0 . Suppose that a positive, twice differentiable function F ( t ) satisfies the inequality
F ( t ) F ( t ) ( 1 + α ) ( F ( t ) ) 2 2 C 1 F ( t ) F ( t ) C 2 F 2 ( t ) , t 0 .
(8)
If
F ( 0 ) > 0 and F ( 0 ) + γ 2 α 1 F ( 0 ) > 0 ,
(9)
then F ( t ) goes to infinity as
t t 1 t 2 = 1 2 C 1 2 + C 2 ln γ 1 F ( 0 ) + α F ( 0 ) γ 2 F ( 0 ) + α F ( 0 ) .
(10)

Here, γ 1 = C 1 + C 1 2 + α C 2 and γ 2 = C 1 C 1 2 + α C 2 .

Proof See [10]. □

Theorem 1 Assume that (A1)-(A3) are satisfied and suppose that the initial condition u 0 satisfies the following condition:
2 ( 2 p + 3 ) p + 2 u 0 p + 2 p + 2 > 2 4 a 2 + 2 ( p + 1 ) 2 a p 2 ( 2 B 0 2 + K 0 2 ) ( u 0 2 + a u 0 2 ) ( 1 + B 0 2 ) u 0 2 D 1 ,
(A4)

where K 0 > 0 and D 1 = Δ ω 2 + B 0 2 ω 2 + 2 p + 2 ω p + 2 p + 2 . Then the solution of the problem (1)-(4) with the weight function g ( x ) = ( ω a Δ ω ) ( x ) blows up in a finite time.

Proof Multiplying (1) by u and integrating over Ω give
1 2 d d t ( u 2 + a u 2 ) + u 2 + ( u , i = 1 n b i u x i ) u p + 2 p + 2 = f ( t ) .
(11)
Also, multiplying (1) by u t and integrating over Ω, we obtain
u t 2 + a u t 2 = 1 2 d d t u 2 ( u t , i = 1 n b i u x i ) + 1 p + 2 d d t u p + 2 p + 2 .
(12)
Now, let us consider the following function:
F ( t ) = u 2 + a u 2 + D 0 ,
(13)
where D 0 is a nonnegative parameter to be chosen later. It is clear that
F ( t ) = 2 ( u , u t ) + 2 a ( u , u t ) .
(14)
Using the Cauchy-Schwarz inequality, we have
( F ( t ) ) 2 4 F ( t ) ( u t 2 + a u t 2 ) .
(15)
Substituting (11) into (12), we obtain
u t 2 + a u t 2 = 1 2 ( p + 2 ) F ( t ) p 2 ( p + 2 ) d d t u 2 ( u t , i = 1 n b i u x i ) + 1 p + 2 d d t ( u , i = 1 n b i u x i ) 1 p + 2 d d t f ( t ) .
(16)
We take the derivative of (7) with respect to t
d d t f ( t ) = ( u t , Δ ω ) + ( ω , i = 1 n b i u t x i ) ( p + 1 ) ( ω , u p u t ) .
(17)
Rewrite (16) in view of (17)
u t 2 + a u t 2 = 1 2 ( p + 2 ) F ( t ) p 2 ( p + 2 ) d d t u 2 p + 1 p + 2 ( u t , i = 1 n b i u x i ) + 1 p + 2 ( u , i = 1 n b i u t x i ) + 1 p + 2 ( u t , Δ ω ) 1 p + 2 ( ω , i = 1 n b i u t x i ) + p + 1 p + 2 ( ω , | u | p u t ) .
(18)
After applying the arithmetic-geometric inequality to estimate the terms on the right-hand side of (18), we obtain
(19)
(20)
(21)
(22)
(23)
For q = 2 n n 2 , n 3 , the following inequality is satisfied for some K 1 > 0 :
( Ω | u | q d x ) 1 / q K 1 ( Ω | u | 2 d x ) 1 / 2 .
(24)
We apply the Hölder inequality, with q 1 = n , q 2 = 2 , q 3 = 2 n n 2 , to the last term in (18),
| Ω ω | u | p u t d x | ( Ω | ω | n d x ) 1 / n ( Ω | u t | 2 d x ) 1 / 2 ( Ω | u | 2 n n 2 d x ) n 2 2 n .
(25)
It follows from (24) and (25) with ω n K 2
K 0 u t u p 4 ( p + 1 ) u t 2 + p + 1 p K 0 2 u 2 ,
(26)
where K 0 = K 1 K 2 . Substituting the estimates (19)-(23) and (26) into (18), we write
p + 4 2 ( p + 2 ) ( u t 2 + a u t 2 ) 1 2 ( p + 2 ) F ( t ) + 2 B 0 2 a p ( p + 2 ) u 2 + 2 p a 1 + p 1 ( p + 1 ) 2 ( 2 B 0 2 + K 0 2 ) a ( p + 2 ) a u 2 + 1 p ( p + 2 ) ( 2 Δ ω 2 + B 0 2 a ω 2 ) .
(27)
Since coefficients of the term a u 2 are greater than those of u 2 on the right-hand side of (27), multiplying both sides of (27) by 2 ( p + 2 ) , we get
(28)
where D 2 = 4 p Δ ω 2 + 2 B 0 2 a p ω 2 . From (15) and (28), we have
( 1 + p 4 ) F 1 ( t ) ( F ( t ) ) 2 F ( t ) + β F ( t ) + ( D 2 β D 0 ) ,
(29)
where β = 4 p a 2 + 2 ( p + 1 ) 2 a p ( 2 B 0 2 + K 0 2 ) . We choose D 0 = β 1 D 2 in the last inequality and multiply both sides of (29) by F ( t ) , which gives
F ( t ) F ( t ) ( 1 + p 4 ) ( F ( t ) ) 2 β ( F ( t ) ) 2 .
(30)

So, inequality (8) is satisfied with α = p 4 > 0 , C 1 = 0 , C 2 = β > 0 . Thus, the desired result is obtained by applying Lemma 1. □

3 Stability of problem

In this part, we consider the following inverse source problem:
(31)
(32)
(33)
(34)
where Ω R n is a bounded domain with a sufficiently smooth boundary Ω and ω, u 0 and φ ( t ) are given functions, p > 0 . Assume that ω satisfies the conditions
Ω ω 2 d x = 1 , ω H 0 1 ( Ω ) L p + 2 ( Ω )
(A5)
and u 0 satisfies
u 0 H 0 1 ( Ω ) L p + 2 ( Ω ) and Ω u 0 ( x ) ω ( x ) d x = φ ( 0 ) .
(A6)
Theorem 2 Suppose that the conditions (A5) and (A6) are satisfied and assume that φ and φ are continuous functions defined on [ 0 , ) which tend to zero as t . Then
lim t ( u 2 + u p + 2 p + 2 ) = 0

with a constant B 0 < 2 ( 1 + λ 1 1 ) λ 1 , where λ 1 is constant in (6).

Proof We multiply (31) by ω, integrate over Ω and use (34) to obtain
f ( t ) = φ ( t ) + ( ω , u t ) + ( ω , u ) ( ω , i = 1 n b i u x i ) + ( ω , | u | p u ) .
(35)
Inserting (35) into (31), we obtain
(36)
Now, let us multiply (36) by u + u t and integrate over Ω:
(37)
Using Cauchy, Poincare and Young inequalities on the right-hand side of (37), we have
(38)
(39)
(40)
(41)
(42)
(43)
Rewriting (37) with estimates (38)-(43), we obtain the following inequality:
(44)
where
D ( t ) = ( | φ | 2 + | φ | 2 ) ( 1 2 ω 2 + 1 ε ω 2 + B 0 2 ε ω 2 ) + ( φ ( t ) ) 2 + | φ ( t ) φ ( t ) | + C ( ε , p ) ω p + 2 p + 2 ( | φ | p + 2 + | φ | p + 2 ) .
We choose ε 0 > 0 such that ε 0 ε < 1 B 0 ( 4 + B 0 λ 1 ) 4 λ 1 and take
K 3 = min { 2 3 ( 1 ε 0 B 0 ( 4 + B 0 λ 1 ) 4 λ 1 ) , 1 ε 0 } .
So, (44) follows
d d t [ 1 2 u 2 + u 2 + 1 p + 2 u p + 2 p + 2 ] + K 3 ( 3 2 u 2 + u p + 2 p + 2 ) D ( t ) .
(45)
The last term on the left-hand side of (45) can be written
3 2 u 2 + u p + 2 p + 2 λ 1 2 u 2 + u 2 + u p + 2 p + 2 K 4 ( 1 2 u 2 + u 2 + 1 p + 2 u p + 2 p + 2 ) ,
(46)
where K 4 = min ( λ 1 , 1 ) . It follows from (45) and (46)
d d t η ( t ) + K 5 η ( t ) D ( t ) .
(47)
Here, K 5 = K 3 K 4 and η ( t ) = 1 2 u 2 + u 2 + 1 p + 2 u p + 2 p + 2 . After solving first-order differential inequality (47), it follows that
u 2 + u p + 2 p + 2 0 as  t .

 □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Sakarya University

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© Yaman; licensee Springer 2012

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