The initial boundary value problem for a nonlinear hyperbolic equation with Lewis function in a bounded domain is considered. In this work, the main result is that the solution blows up in finite time if the initial data possesses suitable positive energy. Moreover, the estimates of the lifespan of solutions are also given.
1. Introduction
Let be a bounded domain in with smooth boundary . We consider the initial boundary value problem for a nonlinear hyperbolic equation with Lewis function which depends on spacial variable:
where , , , and is a continuous function.
The large time behavior of solutions for nonlinear evolution equations has been considered by many authors (for the relevant references one may consult with [1–14].)
In the early 1970s, Levine [3] considered the nonlinear wave equation of the form
in a Hilbert space where are are positive linear operators defined on some dense subspace of the Hilbert space and is a gradient operator. He introduced the concavity method and showed that solutions with negative initial energy blow up in finite time. This method was later improved by Kalantarov and Ladyzheskaya [4] to accommodate more general cases.
Very recently, Zhou [10] considered the initial boundary value problem for a quasilinear parabolic equation with a generalized Lewis function which depends on both spacial variable and time. He obtained the blowup of solutions with positive initial energy. In the case with zero initial energy Zhou [11] obtained a blowup result for a nonlinear wave equation in . A global nonexistence result for a semilinear Petrovsky equation was given in [14].
In this work, we consider blowup results in finite time for solutions of problem (1.1)(1.3) if the initial datas possesses suitable positive energy and obtain a precise estimate for the lifespan of solutions. The proof of our technique is similar to the one in [10]. Moreover, we also show the blowup of solution in finite time with nonpositive initial energy.
Throughout this paper denotes the usual norm of .
The source term in (1.1) with the primitive
satisfies
Let be the best constant of Sobolev embedding inequality
from to .
We need the following lemma in [4, Lemma 2.1].
Lemma 1.1.
Suppose that a positive, twice differentiable function satisfies for the inequality
If , , then
2. BlowUp Results
We set
The corresponding energy to the problem (1.1)(1.3) is given by
and one can find that easily from
whence
We note that from (1.6) and (1.7), we have
and by Sobolev inequality (1.8), , , where
Note that has the maximum value at which are given in (2.1).
Adapting the idea of Zhou [10], we have the following lemma.
Lemma 2.1.
Suppose that and . Then
for all .
Theorem 2.2.
For , suppose that and satisfy
If , then the global solution of the problem (1.1)–(1.3) blows up in finite time and the lifespan
Proof.
To prove the theorem, it suffices to show that the function
satisfies the hypotheses of the Lemma 1.1, where , and to be determined later. To achieve this goal let us observe
Hence,
Let us compute the derivatives and . Thus one has
and
for all . In the above assumption (1.7), the definition of energy functionals (2.2) and (2.4) has been used. Then, due to (2.1) and (2.7) and taking ,
Hence for all and by assumption (2.8) we have
Therefore for all and by the construction of , it is clearly that
whence, . Thus for all , from (2.13), (2.15), and (2.17) we obtain
which implies
Then using Lemma 1.1, one obtain that as
Now, we are in a position to choose suitable and . Let be a number that depends on , , , and as
To choose , we may fix as
Thus, for the lifespan is estimated by
which completes the proof.
Theorem 2.3.
Assume that and the following conditions are valid:
Then the corresponding solution to (1.1)–(1.3) blows up in finite time.
Proof.
Let
then
where the lefthand side of assumption (1.7) and the energy functional (2.2) have been used. Taking the inequality (2.27) and integrating this, we obtain
By using PoincareFriedrich's inequality
and Holder's inequality
where . Using (2.30) and (2.31), we find from (2.28) that
Since as so, there must be a such that
By inequality
and by virtue of (2.33) and using (2.32), we get
where
Therefore, there exits a positive constant
such that
This completes the proof.
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