Fujita Exponent for a Nonlinear Degenerate Parabolic Equation with Localized Source

where p > 2, q > 0, and the initial data u 0 (x) is nonnegative continuous and compactly supported. We take the coefficient a(x) ≥ 0 which is a compactly supported function, which means that the reaction term acts only locally. Thus, the model may be used to describe a chemical reaction-diffusion process in which, due to the effect of the catalyst, the reaction takes place only at some local sites [1]. As a representative example of quasilinear reactiondiffusion equation, classical p-Laplacian equation with no source term (a(x) ≡ 0 in (1)) and its variants had been extensively studied in the past few decades (see [2–6] and references therein). It is well known that p-Laplacian equation in general does not allow for classical solvability due to the nonlinearity and degeneracy.Therefore, the “solution” to a pLaplacianmeans the weak solution in the usual integral sense (see, e.g., [4]). Among other things, previous studies show that blow-up happens for the problem with homogeneous reaction; that is, a(x) ≡ 1:


Introduction
This short paper is devoted to the Fujita critical exponent of the following -Laplacian equation: −2   )  +  ()   ,  ∈ R,  > 0,  (, 0) =  0 () ,  ∈ R, where  > 2,  > 0, and the initial data  0 () is nonnegative continuous and compactly supported.We take the coefficient () ≥ 0 which is a compactly supported function, which means that the reaction term acts only locally.Thus, the model may be used to describe a chemical reaction-diffusion process in which, due to the effect of the catalyst, the reaction takes place only at some local sites [1].
As a representative example of quasilinear reactiondiffusion equation, classical -Laplacian equation with no source term (() ≡ 0 in (1)) and its variants had been extensively studied in the past few decades (see [2][3][4][5][6] and references therein).It is well known that -Laplacian equation in general does not allow for classical solvability due to the nonlinearity and degeneracy.Therefore, the "solution" to a -Laplacian means the weak solution in the usual integral sense (see, e.g., [4]).Among other things, previous studies show that blow-up happens for the problem with homogeneous reaction; that is, () ≡ 1:   = (           −2   )  +   ,  ∈ R,  > 0,  (, 0) =  0 () ,  ∈ R. ( It is shown that if 0 <  ≤ 1, every solution is global in time, if 1 <  ≤ 2 − 1, all solutions blow up, and if  > 2 − 1, both global in time solutions and blowing up solutions exist.In this case, the numbers  0 = 1 and   = 2 − 1 are called the global existence exponent and the Fujita exponent, respectively.On the other hand, we remark that there is a close connection of problem (1) with the problem of diffusion with flux conditions on the boundary: In fact, if we take a sequence of reaction coefficients converging to a Dirac delta at the origin (i.e., if   () →  0 ()), the corresponding solutions   of (1) should converge to a solution of problem (3).In [7], authors proved that the global existence exponent of ( 3) is  0 = 2( − 1)/ and the Fujita exponent is   = 2( − 1).After that, there are plenty of works that extended their study to doubly degenerate equation or fast diffusion case (see, e.g., [8][9][10] and references therein).The motivation of this short paper is the wish to understand the blow-up properties of reaction-diffusion equations which combine a nonlinear diffusion and localized reaction term, and this is the main difference with the existing studies of blow-up for similar reaction-diffusion equations.For the porous medium equation case   = (  )  + ()  , Ferreira et al. [11] proved that  0 = ( + 1)/2,   =  + 1.
In our previous work, we has obtained the global existence exponent of (1) is  0 = 2( − 1)/.In this note, we will prove that the Fujita exponent of (1) is   = 2( − 1).Thus, it is the same as that of (3), and this is coincident with our intuitive judgement.We will modify several techniques developed in [11] to prove our main result.

Some Preliminary Lemmas
In this section, we will give some preliminary lemmas, whose proofs may be independent and interesting.We first have the following lemma from [3].Lemma 1.There exists a positive constant   < ∞ such that problem has a unique nonnegative solution ().
When  >  − 1, we need to introduce the energy functional of (1) as follows: Lemma 2. When  >  − 1, if there exists  0 > 0 such that ( 0 ) < 0, then all solutions to problem (1) blow up in finite time.
It follows from Since ( 0 ) < 0, we have () < − ∫ , Therefore, by using Hölder inequality, we could obtain Advances in Mathematical Physics 3 Due to the fact that   () → ∞ as  → ∞ and since ( + 1)/2 > 1, there exists a constant  > 0 such that when  is large enough.This is equivalent to the proposition that  − is a concave function.
Furthermore, we claim that where  2 is a positive constant.
Integrating the inequality twice over the interval [0, ], we obtain that The right side is bounded in [0, ], and so is (), which contradicts the fact that () → ∞ as  → .Hence, in the limit  → .
has a global supersolution.
Proof.Consider the following Cauchy problem: The existence and uniqueness of the solution to this problem was established in [5,6,12].The solution has the form where   is a positive, bounded, and decreasing function and satisfies the following boundary value problem: Set (, ) = (,  −2  +  0 ), where  > 0 is to be determined later;  0 is a constant between 0 and 1.

Main Results and Their Proofs
We first deal with the case of  > 2(−1), which is easier than the other cases.
Proof.Lemma 3 states that when  > 2( − 1) and  < −1, equation has a global supersolution.Note that the reaction coefficient of this equation is bigger than (), multiplied by a constant if necessary.So, we conclude that problem (1) has global solutions if  > 2( − 1) and  < −1.
On the other hand, it is well known (see, e.g., [13]) that problem admits a blowing-up solution, which is a subsolution to our problem for suitable .We thus complete our proof.
Our remainder objective is to prove that if 2( − 1)/ <  ≤ 2( − 1), then all solutions blow up in finite time.And this result is composed of several theorems.
To simplify the exposition, we may take (), a characteristic function; () =  [−,] for some 0 <  < +∞ in the remainder of this paper.Note that such assumption imposing on () is not essential since it can be modified to a more general compactly function.Theorem 6.If 2( − 1)/ <  <  − 1, then all the solutions to problem (1) blow up in finite time.
Proof.We will prove this result by the similar idea of Lemma 10 in [11].We firstly construct here a blow-up subsolution matching of a self-similar function with a blowing-up parabola.
Fix a point 0 <  0 ≤  and consider the even function obtained by the reflection of where () and () are to be determined later and  is a selfsimilar solution of the problem It is known from [7] that problem (33) admits blowing-up self-similar solution if  > 2( − 1)/, and they have the form where  = ( − 1)/( ⋅  − 2( − 1)).
In order to have a  1 function (, ), we need To this end, we take (37) Notice that if  ≤  − 1, (, ) ∼ const.In order to see that  is a subsolution to problem (1), we only have to look at the interval (0,  0 ).Direct calculations show Therefore, when  is small,  is a subsolution to problem (1) if the inequality is valid.
We now check that the above self-similar solution constructed can be put below any solution if we let pass enough time.In fact, the self-similar solution has small initial value if  > 0 is large, but its support is not small, since the length  0 is not small.We then use the penetration property of the solutions to the -Laplacian equation to guarantee that there exists  0 > 0 such that the support of (⋅,  0 ) contains the interval [− 0 ,  0 ].Therefore, we could get (,  0 ) ≥ (, 0) by taking  which is large enough.
By comparison,  must blow up in finite time.Now, we turn to the case of  >  − 1.
Proof.From Lemma 2, we know that if there exists  0 > 0, such that energy functional   ( 0 ) < 0, then all solutions to (1) blow up in finite time.

Theorem 7 .
If  =  − 1, then every solution to problem(1) blows up in finite time.