We are concerned with a semilinear parabolic differential inequality posed in an exterior domain of , . Some blow-up and existence results are established for the considered problem. The novetly of this paper lies in considering a nontrivial Dirichlet boundary condition, which depends both on time and space. Indeed, to the best of our knowledge, in all articles which deal with the blow-up of solutions in exterior domains, the considered boundary condition is trivial or depends only on space.
In this paper, we study the blow-up of solutions to the problem
where is the closed unit ball in , , , , , , , and , . Moreover, in some special cases, the existence of global solutions is derived. We list below some motivations for studying problem (1.1).
In [1], the authors studied the Dirichlet exterior problem
They showed that
(1.2) admits no global nontrivial non-negative solutions if ;
If , then (1.4) has global solutions for all f with sufficiently small and all such that for any and some sufficiently small .
The main idea of the proof is to show that the -norm of u blows up in a certain selected compact region. Note that the choice of this region depends on the considered problem. Moreover, their proof is based also on the maximum principle and a certain lemma (Lemma 3.2, [2], p. 636), which is valid only if the boundary condition is independent on time.
In [9], using the approach introduced in [2], the author studied the problem
where , , , , , , and h is continuous and satisfies , as , where . It was shown that if and
then problem (1.5) admits no global positive solutions.
Motivated by the above cited works, problem (1.1) is investigated in this paper. Here, the novetly lies in considering a nontrivial Dirichlet boundary condition, which depends both on time and space. Our technique is based on a test function approach and some ideas introduced recently in [5,6].
Let us mention the definition of solutions to problem (1.1).
Let , , , and . We say that is a global weak solution to problem (1.1), if and only if,
for every non-negative function satisfying:
There exists such that , ;
There exists such that , ;
;
, ;
,
where denotes the outward (relative to ) unit normal on .
Let
and
We state below the main results that we are going to prove in the present article. Our first result is the following general blow-up criterion for problem (1.1).
Let,,,and. Ifthen problem (
1.1
) admits no global weak solutions.
Further, some special cases of functions will be discussed. We are first concerned with the set of functions
Let.
If,and, then for allproblem (
1.1
) admits no global weak solutions.
Ifthen problem (
1.1
) admits global solutions for some,and.
Next, we are concerned with the set of functions
Let,,,and.
Ifandthen problem (
1.1
) admits no global weak solutions.
Ifandthen problem (
1.1
) admits no global weak solutions.
If, then for all, problem (
1.1
) admits no global weak solutions.
Let.
Ifor, andthen problem (
1.1
) admits global solutions for some,and.
Ifandthen problem (
1.1
) admits global solutions for some,and.
The rest of the paper is organized as follows. In Section 2, we provide some estimates that will be used later in the proofs of our main results. In Section 3, we prove the general blow-up result given by Theorem 1.2. In Section 4, we discuss the special case when and we prove Theorem 1.3. In Section 5, we prove the blow-up results given by Theorem 1.4 in the case when , . Finally, the proofs of the existence results given by Theorem 1.5 are given in Section 6.
Preliminary estimates
Let and H be the harmonic function in given by
Given , let
Moreover, we introduce the functions
and
where is a function satisfying
Further, let
For, the functionsatisfies conditions (i)–(v) in Definition
1.1
.
Let . By definition of , we have
and
Therefore, conditions (i)–(iii) and (v) are satisfied. On the other hand, for , we have
which yields
Hence, by definition of , condition (iv) is satisfied. □
Let,,,,and. Ifis a global weak solution to problem (
1.1
), then for, we havewhereis a constant (independent of T) and
Let . Using Lemma 2.1 and taking in (1.6), we get
Since , and , , it follows that
On the other hand, using (2.1), we obtain
Since , we have
Hence, since and , we get
where
Combining (2.2) with (2.3), the desired estimate follows. □
We argue by contradiction. Let be a global weak solution to problem (1.1). Using Lemma 2.2, for , we have
Writing
and using the ε-Young inequality, , we get
where is a constant (that depends only on p and ε) and . Similarly, writing
and using the ε-Young inequality, , we get
Further, using (3.1), (3.2) and (3.3), for and , we get
Taking , it follows that
Next, using Lemmas 2.3 and 2.4 with , we obtain
as . Hence, taking , we get
Combining (3.4) with (3.5), for , we get
where is a constant (independent of T). Finally, passing to the supremum limit as and using (1.7), a contradiction follows. This proves Theorem 1.2.
Let , , and
For , we have
where
On the other hand, from (4.1), it follows that
Hence, passing to the limit as in (4.2), we get (1.7). Using Theorem 1.2, we deduce that problem (1.1) admits no global weak solutions. This proves part (I) of Theorem 1.3.
Proof of part (II)
Let
By [11, Prposition 6.1], for some with is sufficiently small, the problem
admits a positive solution. Let
Then u solves problem (1.1) with and . This proves part (II) of Theorem 1.3.
Since , there exists such that
where is a constant. Hence, for , we have
Hence, if and
using (5.1), we have
Using Theorem 1.2, we deduce that problem (1.1) admits no global weak solutions. Next, if and
using (5.1), we get
By Theorem 1.2, it follows that problem (1.1) admits no global weak solutions. This proves part (I) of Theorem 1.4.
Proof of part (II)
Let
Using (5.1), we get
Using Theorem 1.2, we deduce that problem (1.1) admits no global weak solutions. This proves part (II) of Theorem 1.4.
Proof of part (III)
If
then for all , we have
Therefore, for all , (5.2) holds. Hence, the same conclusion as before follows. This proves part (III) of Theorem 1.4.
Let or , and
Let u be the function given by (4.3), where and is sufficiently small. For , we have
Hence, u solves problem (1.1) with
and . Note that , for all . This proves part (I) of Theorem 1.5.
Proof of part (II)
and
Let
where
For , after calculations, one obtains
Hence, u solves problem (1.1) with , , , and , . This proves part (II) of Theorem 1.5.
Footnotes
Acknowledgements
The authors would like to thank the referee for his valuable comments which helped to improve the manuscript. The authors extend their appreciation to the Deanship of Scientific Research at King Saud University for funding this work through research group No. RGP-1435-034.
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