LARGE SOLUTIONS OF QUASILINEAR ELLIPTIC EQUATION OF MIXED TYPE

Abstract

We consider the equation ${\Delta}_mu=p(x)u^{\alpha}+q(x)u^{\beta}$ on $R^N(N{\geq}2)$, where p, q are nonnegative continuous functions and 0 < ${\alpha}{\leq}{\beta}$. 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.

1. Introduction

We consider the existence and nonexistence of large solutions of the equation

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 x0, then there is an open set Ω0 containing x0 for which and it is positive for all x on the boundary of Ω0.

By a positive large solution of (1),we mean a positive function u ∈ C1(Ω) which satisfy (1) at every point of Ω and u → ∞ as x → ∂Ω. In the case Ω = RN, if limr→∞ 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

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 R2 such that ∂Ω is a C2 submanifold of R2 , then there exists a unique u ∈ C2(Ω) such that Δu = 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 R3. 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

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 C2. Recently, Lu et al. [9] proved the existence of large solutions to the problem (3) both for f(u) = uγ, γ > m − 1, Ω = RN or Ω being a bounded domain (super-linear case), and γ ≤ m − 1,Ω = RN (sub-linear case), respectively.

The vast majority of papers studying nonnegative entire large solutions for quasi-linear elliptic equations consider equations of the form

where f is nondecreasing. (See, for example,[7,8,9,10] and references therein)

Recently, Caisheng Chen et al.[12] studied the following problem

where h(x),H(x): RN → (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

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 RN(N ≥ 2) with a smooth boundary ∂Ω and θ : (0,∞) → (0,∞) be a continuous and non-decreasing function. Let u1, u2 ∈ w1,p(Ω) satisfy

for any non-decreasing function Then the inequality

implies that

Lemma 1.3. If v(x) : I ⊆ R → R is a locally integrable nonnegative function, then

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,

Denote

Combinations of the following conditions on the nonnegative continuous functions p and q will be used:

Remark 1.1. Condition (Mpq) can be changed into

Remark 1.2. If m ≥ 2, condition (mpq) can be changed into

Proof. In fact, condition (8) implies condition (mpq). Let 2 ≤ m < ∞, it follows that we have

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 RN 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

From Theorem 2.1 of [10], there exists a positive solution to the boundary value problem for each k ∈ N

Clearly, Again from [10], we have that for each k ∈ N there exists a unique nonnegative classical solution wk to the boundary value problem

Then

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 uk−1 ≤ uk in Ω. So w1 ≤ uk−1 ≤ uk ≤ vk. From [10], the sequence vk converges on Ω to a large solution v, and v satisfies Δmv = q(x)vβ on Ω. It follows that w1 ≤ uk−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 x0 ∈ ∂Ω, and let xj be a sequence in Ω such that xj → x0 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, limj→∞ u(xj) = ∞, and hence, limx→x0 u(x) = ∞. Since x0 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

Clearly, for any k and |x| ≥ k, vk+1 ≤ vk = ∞. The maximum principle gives that

in RN.

To show that vk converges to some v ∈ C(RN) and that v → ∞ as |x| → ∞, we observe that condition (Mpq) implies that

Thus,

where

is the unique positive solution of

We claim that on |x| ≤ k. Clearly, when |x| = k, and thus for

Let then

From the weak comparison principle, we get

So if |x| ≤ k.

Let and note that vk ≥ w for all k. Thus, {vk} converges to some v and v ≥ w in C(RN). Since w → ∞ as |x| → ∞, v → ∞ as |x| → ∞.

This completes the proof.

Next we consider the following nonexistence results.

Theorem 2.3. Suppose the equation

has no PELS, then Eq.(1) has no PELS.

Proof. Suppose (1) has a PELS u. Let vk be a nonnegative solution of

Then the sequence {vk} is decreasing and u ≤ vk on |x| ≤ k for all k ∈ N. Thus {vk} converges to a function v, on RN and u ≤ v on RN. 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) ≡ ϕ1(r) + ϕ2(r) − ψ1(r) − ψ2(r) and 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

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

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 v1 = 1, define the sequence vk iteratively by

The sequence vk is monotonically increasing and vk ≥ 1 for all k ≥ 1, so that we have

Next we show that vk(r) ≤ eMr for 0 ≤ r ≤ R, k ≥ 1 by induction, where

When k = 1, it is obviously true. Suppose vk(r) ≤ eMr, 0 ≤ r ≤ R, then

we note that

It follows that

Thus the sequence {vk} is increasing and locally bounded and therefore converges on RN. 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 RN. In fact, we choose

and show that this work. To do this, we define

and show that R∞ = ∞.

Suppose R∞ < ∞ and note the definition of h to get

From the Gronwall inequality,

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 RN.

Then from theorem 1 of [11], we know there exists an entire solution u(x) satisfying w(x) ≤ u(x) ≤ v(x), x ∈ RN. This completes the proof.

Theorem 3.2. Suppose p and q are nonnegative and locally Hölder continuous in RN and satisfy (Mpq). Then (1) has a nonnegative nontrival entire bounded solution in RN.

Proof. We define

and consider the equation

Let z0 = c ≥ 0 for r ≥ 0 and define the sequence

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

Now we split the domain of [0,∞) into two intervals. Note that

We choose rk such that

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

Condition (Mpq) shows there exists R > 0, so that

Gronwall’s inequality then gives

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 C1[0,∞). Let u1 = M ≥ z ≡ u2, then

That is, u1 and u2 are upper and lower solutions of (1). Standard upper/lower solution method then gives a solution of (1) such that u1 ≥ u ≥ u2.

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

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

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

for r ≥ r0 > 0. Again, using the fact that (20) implies

We can choose r0 large enough so that

Hence

So

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

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

has a fixed point in C[0,∞). We observe that mpq implies

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 u0 = c, and define uk+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

and defining

We begin as we did in Theorem 3.3 and choose R > r ≥ rk,

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:

As in the previous proof, it can be shown that {wk} is bounded and equicontinuous on [0, 1]. Thus {wk} has a subsequence, which converges uniformly on [0, 1]. Let

Likewise, the subsequence is bounded and equicontinuous on [0, 2] so that it has a subsequence which converges uniformly on [0, 2]. Let

Note that on [0, 1] since is a subsequence of Thus v2 = v1 on [0, 1]. Continuing this line of reasoning, we obtain a sequence {vk} with the following properties:

Therefore, it is clear that {vk} converges to v where

and the convergence is uniform on bounded sets. Hence v ∈ C([0,∞)) and satisfy

Next we give the following nonexistence results.

Theorem 3.4. Suppose p and q are locally Hölder continuous in RN. 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

Note that

Let then

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.