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LARGE SOLUTIONS OF QUASILINEAR ELLIPTIC EQUATION OF MIXED TYPE†
LARGE SOLUTIONS OF QUASILINEAR ELLIPTIC EQUATION OF MIXED TYPE†
Journal of Applied Mathematics & Informatics. 2014. Sep, 32(5_6): 721-736
Copyright © 2014, Korean Society of Computational and Applied Mathematics
  • Received : July 19, 2013
  • Accepted : May 21, 2014
  • Published : September 30, 2014
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About the Authors
YUAN ZHANG
ZUODONG YANG

Abstract
We consider the equation △ mu = p ( x ) uα + q ( x ) uβ on R N ( N ≥ 2), where p, q are nonnegative continuous functions and 0 < α β . Under several hypotheses on p ( x ) and q ( x ), we obtain existence and nonexistence of blow-up solutions both for the superlinear and sublinear cases. Existence and nonexistence of entire bounded solutions are established as well. AMS Mathematics Subject Classification : 35J65, 35J50.
Keywords
1. Introduction
We consider the existence and nonexistence of large solutions of the equation
PPT Slide
Lager Image
where 1 < p < N , 0 < α β , and the nonnegative functions p and q are locally Hölder continuous, having the property that min{ p ( x ), q ( x )} is c-positive in Ω (i.e.,if min{ p ( x ), q ( x )} vanishes at any point x 0 , then there is an open set Ω 0 containing x 0 for which
PPT Slide
Lager Image
and it is positive for all x on the boundary of Ω 0 .
By a positive large solution of (1),we mean a positive function u C 1 (Ω) which satisfy (1) at every point of Ω and u → ∞ as x → ∂Ω. In the case Ω = R N , if lim r →∞ u ( r ) = ∞, we call it a positive entire large solution (PELS) of (1).
Such problems arise in Riemannian geometry when studying conformal deformation of a metric with prescribed scalar curvature [1] and in the study of large solutions of elliptic systems. Our purpose is to establish conditions on p and q which ensure the existence and nonexistence of positive solutions of (1).
Equations of the above form are mathematical models occuring in the studies of the p-Laplace equation,generalized in reaction-diffusion theory,non-newtonian fluid theory [3 , 4] , non-Newtonian filtration [5] and the turbulent flow of a gas in porous medium [6] . In the non-Newtonian fluid theory, the quantity p is characteristic of the medium. Media with p > 2 are called dilatant fluids and those with p < 2 are called pseudoplastics. If p = 2, they are Newtonian fluids.
Large solutions of the problem
PPT Slide
Lager Image
where Ω is a bounded domain in RN have been extensively studied. A problem with f ( u ) = eu and N = 2 was first considered by Bieberbach [16] . He showed that if Ω is a bounded domain in R 2 such that ∂Ω is a C 2 submanifold of R 2 , then there exists a unique u C 2 (Ω) such that Δ u = eu in Ω and | u ( x ) − ln ( d ( x )) −2 | is bounded on Ω. Here d ( x ) denotes the distance from a point x to ∂Ω. The result was extended by Rademacher [20] to smooth bounded domain in R 3 . Keller [17] and Osserman [19] provided necessary and sufficient conditions of f for the existence of solutions of Eq (2). The existence, but not uniqueness of solutions of the problem (2) with f monotone was studied by Keller [17] . For f ( u ) = − ua with a > 1, problem (2) is of interest in the study of the sub-sonic motion of a gas when a = 2 and is related to a problem involing super-diffusion, particularly for 1 < a ≤ 2 (see [18] ).
Quasilinear elliptic problem with boundary blow-up
PPT Slide
Lager Image
have been studied, see [11 , 13] and the references therein. Diaz and Letelier [13] proved the existence and uniqueness of large solutions to the problem (3) both for f ( u ) = uγ , γ > m − 1 (super-linear case) and ∂Ω being of the class C 2 . Recently, Lu et al. [9] proved the existence of large solutions to the problem (3) both for f ( u ) = uγ , γ > m − 1, Ω = R N or Ω being a bounded domain (super-linear case), and γ m − 1,Ω = R N (sub-linear case), respectively.
The vast majority of papers studying nonnegative entire large solutions for quasi-linear elliptic equations consider equations of the form
PPT Slide
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where f is nondecreasing. (See, for example, [7 , 8 , 9 , 10] and references therein)
Recently, Caisheng Chen et al. [12] studied the following problem
PPT Slide
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where h ( x ), H ( x ): R N → (0,∞) are the locally Hölder continuous function, and 0 < m p −1 < n . They use the test function method to prove the nonexistence of nontrival solution of the problem (5).
Motivated by the results of the above cited papers, we shall attempted to treat such equation (1). The results of the semilinear equations are extended to the quasilinear ones. We can find the related results for p = 2 in [14] and [23] .
We first give some important lemmas which will be used in our proves.
Lemma 1.1. If v ( x ) : I R R is a locally integrable nonnegative function, then
PPT Slide
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for all a, b I, a < b and 1 ≤ h < ∞; when 0 < h < 1, the inverse inequality holds .
Proof . This lemma can be easily proved using Jessen’s inequality.
Lemma 1.2 (Weak comparison principle [10] ). Let Ω be a bounded domain in R N ( N ≥ 2) with a smooth boundary ∂Ω and θ : (0, ) → (0 ,∞ ) be a continuous and non-decreasing function. Let u 1 , u 2 w 1,p (Ω) satisfy
PPT Slide
Lager Image
for any non-decreasing function
PPT Slide
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Then the inequality
PPT Slide
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implies that
PPT Slide
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Lemma 1.3. If v ( x ) : I R R is a locally integrable nonnegative function, then
PPT Slide
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for all a, b I, a < b and 1 ≤ h < ∞; when 0 < h < 1, the inverse inequality holds .
Proof . This lemma can be easily proved using Jessen’s inequality.
In the sequel proof, we will use the fact that for any nonnegative a and b ,
PPT Slide
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Denote
PPT Slide
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Combinations of the following conditions on the nonnegative continuous functions p and q will be used:
PPT Slide
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Remark 1.1. Condition ( Mpq ) can be changed into
PPT Slide
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Remark 1.2. If m ≥ 2, condition ( mpq ) can be changed into
PPT Slide
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Proof . In fact, condition (8) implies condition ( mpq ). Let 2 ≤ m < ∞, it follows that
PPT Slide
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we have
PPT Slide
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In the first inequality, we used lemma 1.1.
2. Superlinear/mixed case (0 < α ≤ β, β > m − 1)
First, we consider the following existence results.
Theorem 2.1. Suppose Ω is a bounded domain in R N with smooth boundary. If p and q are c-positive, locally Hölder continuous , 0 < α β, β > m −1, then (1) has a large positive solution in Ω.
Proof . First we consider the existence of positive solutions to
PPT Slide
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From Theorem 2.1 of [10] , there exists a positive solution to the boundary value problem for each k N
PPT Slide
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Clearly,
PPT Slide
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Again from [10] , we have that for each k N there exists a unique nonnegative classical solution wk to the boundary value problem
PPT Slide
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Then
PPT Slide
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From the weak comparison principle, we have wk vk for all k N , and hence vk and wk are ordered upper and lower solutions of (9), respectively. From [9] , we concluded that (9) has a positive solution uk . Using the weak comparison principle again, we have u k−1 uk in Ω. So w 1 u k−1 uk vk . From [10] , the sequence vk converges on Ω to a large solution v, and v satisfies Δ mv = q ( x ) vβ on Ω. It follows that w 1 u k−1 uk v . Thus, uk is bounded. Therefore the sequence uk converges on Ω to some function u. Standard bootstrap argument shows that the function u(x) is indeed a solution to (1).
We are left to show that u is a large solution. To see this, we let x 0 ∈ ∂Ω, and let xj be a sequence in Ω such that xj x 0 as j → ∞. Let k N , since uk is monotone, choose Nk N such that uk ( xj ) > k − 1 for j Nk . Thus, un ( xj ) > k −1 for n k and j Nk . Therefore, given any A > 0, k and Nk can be chosen large enough so that u ( xj ) ≥ A for j Nk . Thus, lim j→∞ u ( xj ) = ∞, and hence, lim xx0 u ( x ) = ∞. Since x 0 is arbitrary, it is now apparent that u is a large solution of (1).
Theorem 2.2. Suppose p, q are c-positive and satisfy ( Mpq ), then (1) has an entire large positive solution.
Proof . From Theorem 2.1, we have that for each k N , there exists a positive solution to the boundary value problem
PPT Slide
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Clearly, for any k and | x | ≥ k, v k+1 vk = ∞. The maximum principle gives that
PPT Slide
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in R N .
To show that vk converges to some v C ( R N ) and that v → ∞ as | x | → ∞, we observe that condition ( Mpq ) implies that
PPT Slide
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Thus,
PPT Slide
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where
PPT Slide
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is the unique positive solution of
PPT Slide
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We claim that
PPT Slide
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on | x | ≤ k . Clearly, when | x | = k ,
PPT Slide
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and thus for
PPT Slide
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Let
PPT Slide
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then
PPT Slide
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PPT Slide
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From the weak comparison principle, we get
PPT Slide
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So
PPT Slide
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if | x | ≤ k .
Let
PPT Slide
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and note that vk w for all k . Thus, { vk } converges to some v and v ≥ w in C ( R N ). Since w → ∞ as | x | → ∞, v → ∞ as | x | → ∞.
This completes the proof.
Next we consider the following nonexistence results.
Theorem 2.3. Suppose the equation
PPT Slide
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has no PELS, then Eq.(1) has no PELS.
Proof . Suppose (1) has a PELS u . Let vk be a nonnegative solution of
PPT Slide
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Then the sequence { vk } is decreasing and u vk on | x | ≤ k for all k N . Thus { vk } converges to a function v , on R N and u v on R N . Since u is nonnegative and u ( x ) → ∞ as | x | → ∞, the function v has the same properties. A standard regularity argument can be used to show that the function v is a nonnegative entire large solution of (12), which is a contradiction. This completes the proof.
We now establish some important results for the purely sublinear problem (0 < α β m − 1).
3. Sublinear case (0 < α ≤ β ≤ m − 1)
First, we consider existence results.
Theorem 3.1. Suppose m ≥ 2, 0 < α β m − 1, h ( r ) ≡ ϕ 1 ( r ) + ϕ 2 ( r ) − ψ 1 ( r ) − ψ 2 ( r ) and
PPT Slide
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Then Eq.(1)has a PELS if condi-tion (8) holds.
Proof . The proof hinges on an upper and lower solution argument and a key part of the proof is to show that the equation
PPT Slide
Lager Image
have PELS for which v w . To show that equation (13) have PELS, we show that there exists a number b > 1, such that the integral equations
PPT Slide
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PPT Slide
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have positive solutions valid for all r ≥ 0 with v w . We note that condition (8) means that these entire solutions will be PELS. Let v 1 = 1, define the sequence vk iteratively by
PPT Slide
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The sequence vk is monotonically increasing and vk ≥ 1 for all k ≥ 1, so that we have
PPT Slide
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Next we show that vk ( r ) ≤ eMr for 0 ≤ r R, k ≥ 1 by induction, where
PPT Slide
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When k = 1, it is obviously true. Suppose vk ( r ) ≤ eMr , 0 ≤ r R , then
PPT Slide
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we note that
PPT Slide
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It follows that
PPT Slide
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Thus the sequence { vk } is increasing and locally bounded and therefore converges on R N . Furthermore, its limit v is a solution to (14). A similar argument shows that (15) has a solution w . Then the only thing left is to show the existence of b > 1 such that v w on R N . In fact, we choose
PPT Slide
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and show that this work. To do this, we define
PPT Slide
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and show that R = ∞.
Suppose R < ∞ and note the definition of h to get
PPT Slide
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From the Gronwall inequality,
PPT Slide
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Thus there must exist R > R such that v < w on [0, R ], contradicting the definition of R . Therefore we must have R = ∞ and hence v w on R N .
Then from theorem 1 of [11] , we know there exists an entire solution u ( x ) satisfying w ( x ) ≤ u ( x ) ≤ v ( x ), x R N . This completes the proof.
Theorem 3.2. Suppose p and q are nonnegative and locally Hölder continuous in R N and satisfy ( Mpq ). Then (1) has a nonnegative nontrival entire bounded solution in R N .
Proof . We define
PPT Slide
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and consider the equation
PPT Slide
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Let z 0 = c ≥ 0 for r ≥ 0 and define the sequence
PPT Slide
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for k = 1, 2, · · ·. Radial solutions of (17) will be solutions of (18). It is clear that the sequence { zk } is monotonically increasing. We will show that the sequence { zk } is uniformly bounded. Indeed, for the case m ≥ 2, we have
PPT Slide
Lager Image
Now we split the domain of [0,∞) into two intervals. Note that
PPT Slide
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We choose rk such that
PPT Slide
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It is possible that rk = 0 or rk = ∞. Indeed, if our central value c > 1, then rk = 0. If 0 ≤ c < 1, then either rk = ∞ with zk ( r ) ≤ 1 for all r > 0 or rk < ∞ with zk ( rk ) = 1. Since our goal is to bound { zk }, without loss of generality, we assume that rk < ∞. With the split domains, we have
PPT Slide
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Condition ( Mpq ) shows there exists R > 0, so that
PPT Slide
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Gronwall’s inequality then gives
PPT Slide
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The same argument can be used for the case 1 < m < 2 and we omit it here.
We have now shown that { zk } is uniformly bounded monotonic sequence. Let z be the pointwise limit of the sequence. It is easy to show that the sequence { zk } is actually equicontinuous. Hence the convergence is uniform and the limit function u in in C 1 [0,∞). Let u 1 = M z u 2 , then
PPT Slide
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That is, u 1 and u 2 are upper and lower solutions of (1). Standard upper/lower solution method then gives a solution of (1) such that u 1 u u 2 .
Theorem 3.3. Suppose 0 < α β m − 1, and suppose that p ( x ) = p (| x |), q ( x ) = q (| x |) ∈ C ( R ) are nonnegative and continuous. Then the equation
PPT Slide
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has a large positive radial solution if and only if mpq holds for p or q.
Proof . To prove the necessity, we assume the contrary. That is
PPT Slide
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We will show that (19) has no large positive radial solution. To do this, suppose (19) does have a positive radial solution u ( r ), then it satisfy
PPT Slide
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for r r 0 > 0. Again, using the fact that (20) implies
PPT Slide
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We can choose r 0 large enough so that
PPT Slide
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Hence
PPT Slide
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So
PPT Slide
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Therefore u cannot be a large solution. This completes the proof of necessity. To proof the sufficiency, we assume that mpq is true for p or q . We show that the equation
PPT Slide
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has a positive solution v such that v ( r ) → ∞ as r → ∞. Again, it suffices to show that for any fixed c > 0, the operator defined by
PPT Slide
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has a fixed point in C [0,∞). We observe that mpq implies
PPT Slide
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Hence, any fixed point of (22) is large. We now show that T has a fixed point in C [0,∞). To do this, we first establish a fixed point in [0, R ] for any R > 0. We consider successive approximation. Let u 0 = c , and define u k+1 = Tuk , k = 0, 1, 2, · · · Notice that uk c,k = 0, 1, 2, · · ·, and u′k ≥ 0. It is clear by induction that this sequence is monotonically increasing. We will show it is uniformly bounded on [0, R ]. Since u′k ( r ) ≥ 0, we may split the domains of the relevant functions into two intervals as in the proof of the previous theorem. Choosing rk such that
PPT Slide
Lager Image
and defining
PPT Slide
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We begin as we did in Theorem 3.3 and choose R > r rk ,
PPT Slide
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Thus, the sequence { uk } is uniformly bounded on [0,R]. It is easy to show that { uk } is also equicontinuous on this interval. Hence by Arzela-Ascoli theorem, { uk } has a uniformly convergent subsequence on [0, R ]. Assuming then that ukj u on [0, R ], it is clear that u C ([0, R ]) and Tu = u on [0, R ]. To prove that T has a fixed point in C ([0, R ]), we let { wk } be defined as follows:
PPT Slide
Lager Image
As in the previous proof, it can be shown that { wk } is bounded and equicontinuous on [0, 1]. Thus { wk } has a subsequence,
PPT Slide
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which converges uniformly on [0, 1]. Let
PPT Slide
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Likewise, the subsequence
PPT Slide
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is bounded and equicontinuous on [0, 2] so that it has a subsequence
PPT Slide
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which converges uniformly on [0, 2]. Let
PPT Slide
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Note that
PPT Slide
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on [0, 1] since
PPT Slide
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is a subsequence of
PPT Slide
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Thus v 2 = v 1 on [0, 1]. Continuing this line of reasoning, we obtain a sequence { vk } with the following properties:
PPT Slide
Lager Image
Therefore, it is clear that { vk } converges to v where
PPT Slide
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and the convergence is uniform on bounded sets. Hence v C ([0,∞)) and satisfy
PPT Slide
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Next we give the following nonexistence results.
Theorem 3.4. Suppose p and q are locally Hölder continuous in R N . If mpq holds for p or q, then (1) has no nonnegative bounded entire solution for 0 < α β m − 1.
Proof . Without loss of generality, assume that mpq holds for p and suppose u is a nonnegative entire bounded solution of (1). Consider the equation
PPT Slide
Lager Image
Note that
PPT Slide
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Let
PPT Slide
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then
PPT Slide
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Thus, by the upper/lower solution method, there exists a nontrival nonnegative entire bounded solution to (23). But this contradicts Theorem 3.1 of [9] . Therefore, u must not exist. This completes the proof.
BIO
Yuan Zhang received M.Sc. from Nanjing Normal University. Her research interest is quasilinear elliptic equation (system).
Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu Nanjing 210023, China.
Zuodong Yang is currently a professor at Nanjing Normal University. His research interests are elliptic and parabolic problems.
a. Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu Nanjing 210023, China.
b. School of Teacher Education, Nanjing Normal University, Jiangsu Nanjing 210097, China.
e-mail:zdyang_jin@263.net
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