LARGE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS WITH NONLINEAR GRADIENT TERMS

Vol. 22, No. 4 (1999) 869–883 S 0161-17129922869-4

© Electronic Publishing House

LARGE SOLUTIONS OF SEMILINEAR ELLIPTIC EQUATIONS WITH NONLINEAR GRADIENT TERMS

ALAN V. LAIR and AIHUA W. WOOD (Received 19 June 1998)

Abstract.We show that large positive solutions exist for the equation(P±):∆u±|∇u|^q= p(x)u^γinΩ⊆R^N(N≥3)for appropriate choices ofγ >1,q >0 in which the domainΩis either bounded or equal toR^N. The nonnegative functionpis continuous and may vanish on large parts ofΩ. IfΩ=R^N, thenpmust satisfy a decay condition as|x| → ∞. For(P+), the decay condition is simply_∞

0 tφ(t)dt <∞, whereφ(t)=max|x|=tp(x). For(P−), we require thatt^2+βφ(t)be bounded above for some positiveβ. Furthermore, we show that the given conditions onγ andp are nearly optimal for equation(P+)in that no large solutions exist if eitherγ≤1 or the functionphas compact support inΩ.

Keywords and phrases. Large solution, elliptic equation, existence of solution, semilinear elliptic equation.

1991 Mathematics Subject Classification. Primary 35J25, 35J60.

1. Introduction. We consider existence and nonexistence of large solutions of the equation

∆u±|∇u|^q=p(x)u^γ (P±) in whichqandγare positive constants, the functionpis nonnegative, and the domain Ωis either bounded with smooth boundary or equal toR^N. A solutionu(x)of(P±)is called alargesolution ifu→ ∞asx→∂Ω. In the caseΩ=R^N, x→∂Ωmeans|x| → ∞ and such solutions are calledentirelarge solutions.

Large solutions of semilinear elliptic equations have been studied for decades. Al- most all such studies have dealt with equations of the form

∆u=g(x,u) (1.1)

in which the functiongtakes various forms (see [1, 2, 6, 8, 10, 12] and their references).

Except for [2] and [10], all of these have restricted their attention to bounded domains and functionsgwhich are strictly positive whenu >0.

In [2], Bandle and Marcus proved the existence of large solutions wheng(x,u)= p(x)u^γin whichpis allowed to be zero at finitely many places inΩ. Lair and Shaker [10] proved the existence of large solutions in bounded domains and entire large solutions inR^N forg(x,u)=p(x)f (u), allowingpto be zero on large parts ofΩ.

A few authors considered large solutions of semilinear equations containing non- linear gradient terms (see [1, 7, 11]). Motivation for the present study stems from the work of Bandle and Giarrusso [1] who developed existence and asymptotic behavior

results for large solutions of

∆u±|∇u|^q=f (u) (1.2) on a bounded domain. Our goal here is to develop comprehensive (and nearly optimal) existence results for(P+)whenΩis either bounded or equal toR^N, and at the same time develop somewhat comparable results for problem (P−). In particular, for Ω bounded, we show that problem(P+)has a positive large solution ifγandqsatisfy

γ >max{1,q}, (1.3)

andpsatisfies the following condition.

Condition (P). p(x) ≥ 0,∀x ∈ Ω;p(x) ∈C(Ω); if¯ z ∈Ωandp(z)= 0, then there exists a domainDzcontainingzfor whichDz⊆Ωandp(x) >0 for allx∈∂Dz. Thus,p is allowed to vanish on significant portion of Ω. Indeed, the functionp could be zero on all ofΩexcept for a small (in measure) open set containing∂Ω. For problem(P−)withΩbounded, we do not need inequality (1.3) when we establish the existence of a nonnegative solution (see Theorem 4).

ForΩ=R^N, we prove the existence of a positive entire large solution if p also satisfies

_∞

0 r φ(r )dr <∞, (1.4)

whereφ(r )=max|x|=rp(x)(see Theorem 3). For problem(P−), we require the strong- er decay condition that r^2+βφ(r ) be bounded above for some β >0. In addition, restrictions are placed on the relationship betweenqandγ(see Theorem 5).

Furthermore, for both the bounded and unbounded cases, we show that our con- ditions onγ,q, andp are nearly optimal for(P+)in that ifγ≤1 orphas compact support inΩ, then(P+)has no positive large solutions.

We also note that, among the many open problems related to(P±), the existence of positive large solutions remains unproved for(P+)if 1< γ≤qand for(P−)if the functionpis required to satisfy the weaker decay condition (1.4).

2. Existence results

2.1. Equation (P+).In this section, we develop a critical boundedness result (Lemma 1), which will prove very useful in developing our existence theorem for bounded domains (Theorem 2). This result, in turn, provides the critical element for the existence proof when Ω=R^N. Interestingly, Bandle and Giarrusso [1] had no boundedness result comparable to that of Lemma 1. Indeed, their proof of their main existence result simply assumes that such an upper bound exists.

2.1.1.Ωis a bounded domain

Lemma1. Letunbe a solution of the problem

∆un+|∇un|^q=p(x)u^γn x∈Ω,

un(x)=n, x∈∂Ω. (2.1)

Then0< un≤nonΩ. Furthermore, let¯ BRbe a ball of radiusRsuch thatBR⊆Ω. Then

there exists a constantM=M(R,q,γ)such thatun(x)≤MonBR∀n, provided that 0< mo≤p(x)≤MoinΩand (1.3) holds.

Proof. To prove thatun>0 inΩ, without loss of generality, we letn=1. We observe from the maximum principle that 0≤u1≤1. Furthermore, it is clear that, for any 0< o<1, any solution, sayz, to the problem (which exists by [9, Thm. 8.3, p. 301])

∆z+|∇z|^q=p(x)z^γ, x∈Ω,

z=o, x∈∂Ω (2.2)

satisfiesz≤u1 and 0≤z≤o. Hence, if we show thatz >0 inΩfor some choice ofo∈(0,1), we will be done. To do this, letxo∈R^N\Ω. We assume, without loss¯ of generality, thatxo=0. Letr= |x|and chooseRo>0 large so that ¯Ω⊆B(0,Ro).

ChooseMo>0 so thatp(x) < Moon ¯Ω. Now, choose 0< o<1 so that Mo^γoR²o

2N ≤o. (2.3)

Letv(x)=(Mo^γo/2N)r²for|x| ≤Ro. Definew on the ballB(0,Ro)asw(x)=z(x) forx∈Ω¯andw(x)=oonB(0,Ro)\Ω. We show that¯ v≤winB(0,Ro).

Thus, we suppose that max(v−w)in B(0,Ro) is positive. In this case, the point where the maximum occurs must lie inΩsince

v(x)=Moo^γ

2N r²≤Mo^γo

2N Ro²≤o=w(x), x∈B(0,Ro)\Ω. (2.4) Therefore, at the point where max(v−w)occurs, we have

0≥∆(v−w)=∆(v−z)=Mo^γo−pz^γ+|∇z|^q> p ^γo−z^γ

≥0, (2.5) a contradiction. Therefore,v≤winB(0,Ro)which yieldsv≤winΩor(Mo^γo/2N)r²

≤z(x)inΩ. Sincer >0 inΩ, we getz >0 inΩ. Hence,u1>0 inΩ.

Now, letbe a sufficiently small positive number so thatBR+⊆Ωand letvnbe a solution of

∆vn=movn^γ, x∈BR+,

vn=n, x∈∂BR+. (2.6)

A similar argument as above implies thatvn>0 inBR+. By the maximum principle, it is clear thatvn≤vn+1,n=1,2,.... By [4, Thm. A], it is easy to show thatvnis a radial solution. Thus,vnsatisfies

v_n+N−1

r v_n =movn^γ, x∈BR+,

vn=n, x∈∂BR+. (2.7)

It is clear thatv_n(0)=0 andv_n(r )≥0∀nand∀r. Equation (2.7) may be rewritten as r^N−1v_n(r )

=mor^N−1vn^γ, (2.8)

which may be integrated to get _r

0

s^N−1v_n(s)

ds=mo

_r

os^N−1vn^γds. (2.9)

Thus, we have

r^N−1vn(r )=mo

_r

0s^N−1vn^γds≤mor^N−1vn^γ(r ) _r

0 ds=mor^Nvn^γ(r ) (2.10) which implies that

v_n(r )≤mor vn^γ. (2.11) Letvbe a solution of

∆v=mov^γ, x∈BR+,

v→ ∞, x∈∂BR+. (2.12) The existence ofvis justified by [10, Thm. 1]. By the maximum principle,vn≤vin BR+for alln. Thus,vnis bounded above onBRby a constant which is independent ofn. By (2.11),v_n(r )is also bounded above by a constant independent ofn. LetKbe an upper bound for bothvnandv_n onBR.

If we can find a functionwnwhich satisfies

∆wn+|∇wn|^q≤mown^γ, x∈BR+⊆Ω, wn=n, x∈∂BR+, wn≤Ko, x∈B¯R,

(2.13)

whereKo is a constant independent ofn, then by the maximum principle, we have un≤wn≤Ko, and we will be done.

Letwn=cv_n^λ, wherevnis a solution of (2.7), the constantscandλ, both independent ofn, are determined later. Since

∆wn+|∇wn|^q−mown^γ

=λcv_n^λ−1∆vn+λ(λ−1)cv_n^λ−2|∇vn|²+c^qλ^qv^(λ−1)q|∇vn|^q−moc^γvn^λγ

=λcv_n^λ−1

∆vn+(λ−1)v_n⁻¹|∇vn|²+c^q−1λ^q−1vn^{(λ−1)(q−1)}|∇vn|^q

−moc^γ−1

λ vn^λγ−λ+1

,

(2.14)

to complete the proof, it suffices to findcandλsuch that

∆vn+(λ−1)v_n⁻¹|∇vn|²+c^q−1λ^q−1vn^{(λ−1)(q−1)}|∇vn|^q−moc^γ−1

λ vn^λγ−λ+1≤0 (2.15) onBR+for alln, for then we haveun≤wn≤cK^λ≡Ko. Sincevnsatisfies (2.7) and (2.11), the left side of the above equals

movn^γ+(λ−1)v_n⁻¹|∇vn|²+c^q−1λ^q−1vn^{(λ−1)(q−1)}|∇vn|^q−moc^γ−1

λ vn^λγ−λ+1

=movn^γ+(λ−1)v_n⁻¹ v_n ²+c^q−1λ^q−1vn^{(λ−1)(q−1)}|v_n|^q−moc^γ−1

λ vn^λγ−λ+1

≤movn^γ+(λ−1)m²o(R+)²vn^2γ−1+c^q−1λ^q−1m^qo(R+)^qv(λ+γ)q−q−λ+1 n

−mo

λ c^γ−1vn^λγ−λ+1.

(2.16) Now, sinceγ > q, we can chooseλlarge so that

λγ−λ+1> γ⇐⇒λ≥1, λγ−λ+1>2γ−1⇐⇒λ≥2, λγ−λ+1> (λ+γ)q−q−λ+1⇐⇒λ≥q(γ−1)

γ−q ,

(2.17)

and

mos^γ+(λ−1)m²_o(R+)²s^2γ−1

+c^q−1λ^q−1m^qo(R+)^qs(λ+γ)q−q−λ+1−mo

λ c^γ−1s^λγ−λ+1<0 (2.18) fors≥2 andc≥1. Since 0< v1≤ ··· ≤vn≤vn+1≤ ··· inBR+, we may findβ >0 such thatvn(r )≥β∀nand∀r. For the above choice ofλ, choose the constantc≥1 so that the following holds.

mo2^γ+(λ−1)m²_o(R+)²2^2γ−1

+c^q−1λ^q−1m^qo(R+)^q2(λ+γ)q−q−λ+1−mo

λ c^γ−1β^λγ−λ+1<0. (2.19) Thus, whethervn(r )≤2 orvn(r )≥2, we get

movn^γ+(λ−1)m²_o(R+)²vn^2γ−1+c^q−1λ^q−1m^qo(R+)^qv(λ+γ)q−q−λ+1 n

<mo

λ c^γ−1vn^λγ−λ+1. (2.20)

Hence,∆wn+|∇wn|^q≤mown^γonBR+.

Theorem2. Assume that (1.3) and condition(P)hold. Suppose thatΩis a bounded domain inR^N,N≥3, with smooth boundary. Then, equation(P+)has a large positive solution inΩ.

Proof. By [9, Thm. 8.3, p. 301], it is easy to prove that, for eachk∈N, the boundary value problem

∆vk+|∇vk|^q=p(x)v_k^γ, x∈Ω,

vk(x)=k, x∈∂Ω (2.21)

has a unique positive classical solution. By the maximum principle it can be shown that vk≤vk+1, k≥1, inΩ. Indeed, suppose that there is a point wherev≡vk+1−vk<0.

Letxo∈R^N\Ω. We assume, without loss of generality, that¯ xo=0.Letr= |x|. Then, for some small >0,v+ε/(1+r )has a negative minimum inΩ. At that minimum, we have

0≤∆ v+

1+r

=p

v_k+1^γ −v_k^γ

−|∇vk+1|^q+|∇vk|^q+ 2

(1+r )³− N−1 r (1+r )²

≤0− N−1 r (1+r )³<0,

(2.22)

a contradiction. Hence,vk≤vk+1,fork=1,2,.... Furthermore, by Lemma 1,v1>0 inΩ.

In what follows, it is understood that the maximum principle is applied as above, where the factorε/(1+r )is used whenever the functionpis not strictly positive.

To complete the proof, it suffices to show the following:

(C1) ∀xo∈Ω, there exists M (depending on xo but independent of k) such that vk(x)≤M∀xnearxo;

(C2) limx→∂Ωv(x)= ∞, wherev(x)=limk→∞vk(x)forx∈Ω;

(C3) vis classical solution of(P+).

To prove (C1), we consider two cases.

Case(a). p(xo) >0. Since p is continuous, there exists a ballB(xo,r )such that p(x)≥moinB(xo,r )for somemo>0. (C1) then follows easily from Lemma 1.

Case(b). p(xo)=0. By the hypothesis, there exists a domain Ωo⊆Ωsuch that xo∈Ωoandp(x) >0∀x∈∂Ωo. From Case (a) above, we know that∀x∈∂Ωothere exists a ballB(x,rx)and a positive constant Mx such thatvk≤Mx onB(x,rx/2).

SinceΩis bounded (and henceΩo is bounded),∂Ωo is compact. Thus, there exists a finite number of such balls that cover∂Ωo. Let M=max{Mx1,...,Mxk}, where the ballsB(xi,rx_i/2),i=1,...,k, cover∂Ωo. Clearly,vk≤Mon∂Ωo. Yet another maximum principle argument yieldsvk≤MonΩo.

The proof of (C2) is straightforward. For anyL >0 and any sequencexk→x∈∂Ω, sincevL+1=L+1 on∂Ωand is continuous, there is someK >0 such thatvL+1(xk)≥L fork≥K. Note that, sincev≥vL+1inΩ, we havev(xk)≥L,k≥K. Hence,v(xk)→ ∞ ask→ ∞. Thus, we havev→ ∞asx→∂Ω.

To prove (C3), we letxo∈Ωand letB(xo,r )be the ball of radiusr centered atxo

such that it is contained inΩ. Letψbe aC^∞function which is equal to 1 onB(xo,r /2) and zero offB(xo,r ).

Letf (s)=1/(1+s). Multiplying both sides of equation (2.21) byψ²f (vk)and inte- grating overB(xo,r )yields

B(xo,r )ψ²f (vk)∆vkdx+

B(xo,r )ψ²f (vk)|∇vk|^qdx=

B(xo,r )ψ²f (vk)pv_k^γdx.

(2.23) Integration by parts produces

−

B(xo,r )ψ²f(vk)|∇vk|²dx−

B(xo,r )2ψ∇ψf (vk)·∇vkdx +

B(xo,r )ψ²f (vk)|∇vk|^qdx=

B(xo,r )ψ²f (vk)pv_k^γdx. (2.24) Thus, we have

1 1+Mr

2

B(xo,r )ψ²|∇vk|²dx

≤

B(xo,r )

ψ²

(1+vk)²|∇vk|²dx+

B(xo,r )ψ²f (vk)|∇vk|^qdx

=

B(xo,r )2ψ∇ψ·∇vk

1 1+vk)

dx+

B(xo,r )ψ²f (vk)pv_k^γdx

≤

B(xo,r )(ψ|∇vk|)2|∇ψ|

1+v1 +

B(xo,r )ψ²f (vk)pv_k^γdx

≤

B(xo,r )ψ²|∇vk|²dx+ 1 4

B(xo,r )

2|∇ψ|

1+v1

2

dx+M1

≤

B(xo,r )ψ²|∇vk|²dx+M2,

(2.25)

whereMris an upper bound forvkonB(xo,r ),k=1,2,...,is any positive number, and the constantsM1andM2are independent ofk. Hence, we get

B(xo,r )

ψ∇vk ²dx≤M. (2.26)

That is, theL²(B(xo,r ))-norm of|ψ∇vk|is bounded independently ofk. Thus, the L²(B(xo,r /2))-norm of|∇vk|is bounded independently ofk.

By the standard regularity argument (see [10]), we may find a numberr1>0 such that there is a subsequence of {vk}^∞₁, which we still call {vk}^∞₁, that converges in C^1+α(B(xo,r1))for some positive numberα <1.

Letψbe as before but withrreplaced byr1.

Now, we consider two cases regarding the regularity of the functionp(x).

Case1. p(x)∈C^α(Ω). Note that,∆vk=pv_k^γ−|∇vk|^q and∆(ψvk)=2∇ψ·∇vk+ vk∆ψ+ψ∆vk, k≥ 1, converges in C^α(B(xo,r1)) as k → ∞. By Schauder theory, {ψvk}^∞₁ converges inC^2+α(B(xo,r1))and hence{vk}^∞₁converges inC^2+α(B(xo,r1/2)).

Sincexois arbitrary, it follows thatv∈C^2+α(Ω)and is a solution of(P+).

Case2. p(x)∈C(Ω). We havevk s−C(B(xo,r1))v

→and, consequently,∆vk=pv_k^γ−

∇vk|^{q s−C(B(xo,r1))}→p(x)v^γ−|∇v|^q≡z. That the Laplacian is a closed linear operator implies thatv∈D(∆), ∆v=z. Since xo is arbitrary, we have that v is a classical solution of(P+).

2.1.2.Ω=R^N

Theorem3. Let Ω=R^N. Assume that (1.3), (1.4), and condition (P) hold. Then equation(P+)has a positive entire large solution.

Proof. By Theorem 2, fork=1,2,...,the boundary blow-up problem

∆vk+|∇vk|^q=p(x)v_k^γ, |x|< k,

vk(x)→ ∞, as|x|→k (2.27)

has a classical solution. By the maximum principle, it is clear that

v1≥v2≥ ··· ≥vk≥vk+1≥ ··· (2.28) inR^N. Letv(x)=limk→∞vk(x),x∈R^N. We claim thatvis the desired solution. To prove this, we consider the related problem

∆uk=p(x)u^γ_k, |x|< k,

uk(x)→ ∞, as|x| →k. (2.29)

It is shown in [10] that (2.29) has a unique positive solution for eachk, and that u1≥u2≥ ··· ≥uk≥uk+1≥ ··· ≥w >0 (2.30) for somew→ ∞as|x| → ∞. It follows easily from the maximum principle thatvk≥uk

fork=1,2.... Thus,v(x)→ ∞as|x| → ∞. By a similar argument as (C3) in Theorem 2, we have thatvis the desired solution.

2.2. Equation(P−)

2.2.1.Ωis a bounded domain. The following theorem is our main result for prob- lem(P−)on a bounded domain. Much of the proof is similar to that of Theorem 2 and is, therefore, only outlined.

Theorem4. Suppose thatγ >1,q >0, condition(P)holds, andΩis a bounded do- main inR^N,N≥3, with smooth boundary. Then equation(P−)has a large nonnegative solution inΩ.

Proof. The only significant difference between this proof and that of Theorem 2 lies in obtaining an upper boundM for the sequence{vk}nearx0. There, Lemma 1 is needed, but for(P−) the proof is much easier. Indeed, it is easy to prove that vk(x)≤w(x)for allx∈Ω, wherewis a solution of (See [10, Thm. 1]).

∆w=p(x)w^γ, x <Ω,

w(x) → ∞, x →∂Ω. (2.31)

2.2.2.Ω=R^N. Our main result for equation(P−)is the following theorem.

Theorem5. LetΩ=R^N,γ >1, and assume that condition(P)holds.

If there exist positive numbersβ andR such that 0≤p(x)≤Mr^−2−β, whenever r≡ |x|> R, then equation(P−)has a nonnegative entire large solution provided that max{γ,q}>2ifq≥1, andγ≤1+β(1−q)/(2−q)ifq <1.

To prepare for proving this theorem, we now state and prove three lemmas.

Lemma6. LetMbe any nonnegative number andβany positive number.

Then, forRsufficiently large, there is a nonnegative solution of the problem w+N−1

r w−|w|^q≥Mr^−2−βw^γ, r≥R,

w(r )→ ∞, r→ ∞ (2.32)

provided thatγ >1ifq >2, andγ≤1+β(1−q)/(2−q)ifq <1.

Proof. DenoteL(w)≡w+(N−1/r )w−|w|^q−Mr^−2−βw^γ. Letαbe a positive real number whose value will be made precise later. We have

L(r^α)=α(α−1)r^α−2+N−1

r αr^α−1−α^qr^αq−q−Mr^−2−βr^αγ

=α(α+N−2)r^α−2−α^qr^(α−1)q−Mr^−2−βr^αγ

=r^{−(2+β)+αγ}

α(α+N−2)r^{β+α(1−γ)}−α^qr(α−1)q+2+β−αγ−M

≡r^{−(2+β)+αγ}

α(α+N−2)r^α1−α^qr^α2−M .

(2.33)

Requiring thatα1≥0 andα1≥α2yield α≤ β

γ−1 (2.34)

and

α≥2−q

1−q forq <1, or α≤q−2

q−1 forq >2, (2.35)

respectively. Thus, forq <1, we require that γ≤1+β

1−q 2−q

. (2.36)

In this case, we takeα=(1−q)/(2−q). Forq >2, we takeγ >1 andα≤min{β/(γ− 1),(q−2)/(q−1)}. Hence, we may chooseRlarge so thatL(r^α)≥0. Consequently, w(r )≡r^α, whereαis as defined above, is the desired solution.

Lemma7. LetM andβ be any positive numbers, and letφ(r )≡2^2+βMr^−2−β. If there is a nonnegative solutionuof

∆u−|∇u|^s≥φ(r )u^γ, |x| ≥R,

u(x)→ ∞, |x| → ∞ (2.37)

that satisfiesu≥r(s−2)/(s−1)for somes >2, then there is a nonnegative solution of

∆w−|∇w|^q≥φ(r )w^γ, |x| ≥R,

w(x)→ ∞, |x| → ∞, (2.38) where1≤q≤2, provided thatγ >2+β.

Proof. Letw=cu^α, whereuis a nonnegative solution of (2.37), 0< c,α <1 are to be determined later. We have

L(w)≡∆w−|∇w|^q−φ(r )w^γ

≥cαu^α−1

|∇u|^s+φ(r )u^γ

−

cα(1−α)u^α−2|∇u|²+(cα)^qu^(α−1)q|∇u|^q+φ(r )c^γu^αγ

≥cαu^α−1|∇u|^s+

cαu^α+γ−1−c^γu^αγ φ(r )

−cα

(1−α)u^α−2|∇u|²+u^(α−1)|∇u|^q .

(2.39)

We choose 0< c,α <1 so thatL(w)≥0. Sinceu→ ∞, we may chooseR >0 so that u≥1.

Hence, we need to consider only two cases:

(a) |u| ≥1 and|∇u| ≥1;

(b) |u| ≥1 and|∇u|<1.

For case (a), it can be easily shown thatL(w)≥0 by appropriately choosingcandα.

For (b), it suffices to have

13cαu^α+γ−1≥c^γu^αγ,

13φ(r )u^α+γ−1≥cα(1−α)u^α−2,

13φ(r )u^α+γ−1≥cαu^α−1.

(2.40)

Thus, we need only to require that

13φ(r )u^γ≥cα. (2.41)

Sinceφ(r )=2^2+βMr^−2−βandu(x)≥r(s−2)/(s−1), it is sufficient to have

132^2+βMr(−2−β)+(γ)(s−2)/(s−1)≥cα. (2.42)

This is true since we can chooseslarge enough so that (−2−β)+γs−2

s−1

>0. (2.43)

Hence, we haveL(w)≥0.

Lemma8. Suppose thatβ >0and0≤p(x)≤2^2+βMr^−2−βfor largeM. Letwbe a nonnegative solution of

∆w−|∇w|^q≥2^2+βM(1+r )^−2−βw^γ, |x| ≥R

w(x)→ ∞, |x| → ∞ (2.44)

for someR≥1. LetMo=max|x|=Rw(x). Then, any solution of

∆vk−|∇vk|^q=p(x)v_k^γ, |x|< k

vk(x)→ ∞, |x| →k (2.45) must satisfy

vk(x)≥w(x)−Mo forR≤ |x|< k, (2.46) for anyk > R.

Proof. Suppose that the conclusion is false. That is, suppose that

R≤x≤kmax

w(x)−Mo−vk(x)

=w(xo)−Mo−vk(xo) >0 (2.47) for somexo. Sincew(x)−Mo−vk(x)≤ −vk(x)≤0 if|x| =Randw(x)−Mo−vk(x)→

−∞ as|x| →k, we know thatR <|xo|< k. Hence, ∆(w−Mo−vk)≤0 atxo and

∇(w−Mo−vk)=0 atxo. Thus,

0≥∆(w−Mo−vk)=∆w−∆vk

≥ |∇w|^q+(1+r )^−2−βM2^2+βw^γ−|∇vk|^q−pv_k^γ

=2^2+β(1+r )^−2−βMw^γ−pv_k^γ

>

2^2+βM(1+r )^−2−β−p v_k^γ.

(2.48)

Hence, we get

0>

2^2+βM(1+r )^−2−β−p

v_k^γ. (2.49)

However, sincep(x)≤Mr^−2−βandr≥R≥1, we get p(x)≤M

1+r r

_2+β

(1+r )^−2−β≤M2^2+β(1+r )^−2−k. (2.50) Hence, we get

2^2+βM(1+r )^−2−β−p≥0, (2.51)

which contradicts (2.49). Hence,w(x)−Mo−vk(x)≤0 inR≤ |x| ≤k.

We now prove Theorem 5.

Proof. By Theorem 4, we have that, fork=1,2,...,the boundary blow-up problem

∆vk−|∇vk|^q=p(x)v_k^γ, |x|< k,

vk(x)→ ∞, as|x| →k (2.52) has a nonnegative classical solution. By the maximum principle, it is clear that

v1≥v2≥ ··· ≥vk≥vk+1≥ ··· (2.53) inR^N. Letv(x)=limk→∞vk(x),x∈R^N, where we assume thatvk= ∞for|x|> kfor allk. Let v(x)=limk→∞vk(x),x∈R^N. We claim thatv is the desired solution. To prove this, we need only to prove thatv satisfies(P−)and thatv→ ∞as|x| → ∞.

To prove the second statement, it suffices to find a nonnegative lower bound, say w(x), for the sequence{vk}^∞₁ such thatw→ ∞as|x| → ∞. Also, by another standard regularity argument (see [10]), we can show thatvis smooth and is a classical solution of(P−).

Ifq >2 orq <1, letwbe a solution of (2.44) which we know to exist by Lemma 6, where M is a constant. Then, by Lemma 11, vk(x) ≥ w(x)−Mo, where Mo = max|x|=1w(x)for 1≤ |x| ≤R.

Thus,v(x)≥w(x)−Mofor 1≤ |x| ≤R. Hence,v(x)→ ∞as|x| → ∞. We conclude thatvis a solution of(P−). In particular, ifq >2, thenv≥w=r(q−2)/(q−1).

Assume that 1≤q≤2. By Lemma 6, there is a solution, sayu, to equation (2.37), wheres >2. Now, letwbe a solution of (2.38). It is clear thatwsolves (2.44) and hence v≥w−Mo, whereMois defined as in Lemma 8. Again, we getv→ ∞as|x| → ∞, and is a classical solution of(P−).

3. Nonexistence results

Theorem9. Suppose that condition(P)and (1.4) hold. If0≤γ≤1, then(P+)has no positive entire large solution inR^N.

Proof. We first show that (1.4) implies that _∞

0 r^1−N _r

0s^N−1φ(s)ds dr <∞. (3.1) Indeed, for anyR >0, we have

_R

0r^1−N _r

0s^N−1φ(s)ds dr

= 1 2−N

_R

0

d dr

r^2−Nr

0s^N−1φ(s)ds dr

= 1 2−Nr^2−N

_r

0s^N−1φ(s)ds|^R₀− 1 2−N

_R

0r^2−Nr^N−1φ(r )dr

= 1 2−NR^2−N

_R

0 s^N−1φ(s)ds− 1 2−N

_R

0r φ(r )dr

≤ 1 N−2

_∞

0 r φ(r )dr <∞.

(3.2)

By (3.1), we can chooserolarge enough so that β≡

_∞

rot^1−N _t

0s^N−1φ(s)ds dt <1. (3.3) Now, suppose that(P+)has a positive large solutionu(x). Using technique similar to that described in [5], we define

¯

u(r )≡ 1 vo

s^N−1r

|x|=ru(x)dσr≡

|x|=ru(x)dσ , (3.4) wherevo(s^N−1r )is the volume of the(N−1)-dimensional sphere andσr is the mea- sure on the sphere. We have

∆u¯=u¯+N−1 r u¯=

|x|=r∆udσ

=

|x|=r

p(x)u^γ−|∇u|^q

dσ≤φ(r )

|x|=ru^γdσ

≤φ(r )

|x|=rudσ _γ

(By Jensen’s Inequality)

=φ(r )u¯^γ(r ).

(3.5)

Thus, we have

¯

u+N−1

r u¯≤φ(r )u¯^γ(r ). (3.6) Integrating the above inequality yields

¯

u(r )≤u(r¯ o)+r_o^N−1u¯(ro)r^2−N 2−N

^r

ro

+ _r

rot^1−N _t

ros^N−1φ(s)u¯^γ(s)ds dt

≤u(r¯ o)+ 1

N−2rou¯(ro)+

_r

rot^1−N _t

ros^N−1φ(s)¯u^γ(s)ds dt

≤u(r¯ o)+ 1

N−2rou¯(ro)+

romax≤r≤Ru(r )¯ _γ

β forro≤r≤R

≡A+

romax≤r≤Ru(r )¯ _γ

β.

(3.7)

Leth(R)=maxro≤r≤Ru(r ).¯ Since 0≤γ≤1, then the last inequality gives h(R)≤A+

h(R)_γ

β≤A+β

1+h(R)

∀R≥ro, (3.8)

or equivalently,

0≤u(R)¯ ≤h(R)≤A+β

1−β. (3.9)

Thus, ¯uis bounded and hence cannot be a large solution. Consequently,ucannot be a large solution.

Combining Theorems 3 and 9, we get the following corollary.

Corollary10. Assume that conditions(P)and (1.4) hold. Letγ > q, then equa- tion(P+)has a positive entire large solution if and only ifγ >1.

Theorem11. Suppose thatΩis bounded and0≤γ≤1. Then(P+)has no positive large solution inΩ.

Proof. Suppose that(P+)has a positive large solutionu(x)inΩ. Assume, without loss of generality, that 0∈Ω. LetBbe a ball of radiusRcentered at 0 and containing Ωsuch that∂B∩∂Ω= ∅. LetM=max_x∈¯Ωp(x).

Letvnbe the unique solution of

∆vn=Mvn^γ, x∈B,

vn=n, x∈∂B. (3.10)

Clearly,vn≤uin B. (We assume thatu= ∞inB\Ω). As shown in Lemma 1,vn is radially symmetric and thus satisfies

vn+N−1

r vn =Mvn^γ, x∈B,

vn=n, x∈∂B. (3.11)

Hence, we have

r^N−1v_n

=Mr^N−1vn^γ. (3.12)

Integrating this inequality yields vn(r )=Mr^1−N

_r

0s^N−1vn^γ(s)ds≥0. (3.13) Choose 0< ro< Rso that

R²−r_o²

2N ≤ 1

2M. (3.14)

Integrating (3.13), using (3.14), and noting thatvnis monotonically increasing gives vn(r )=vn(ro)+M

_r

rot^1−N _t

0s^N−1vn^γ(s)ds dt

≤vn(ro)+Mvn^γ(r )

r²−r_o² 2N

≤vn(ro)+¹₂vn^γ(r ).

(3.15)

Since 0≤γ≤1,we havevn^γ≤1+vn, which gives

vn(r )≤vn(ro)+¹₂+¹₂vn. (3.16) Thus, we get

vn(r )≤1+2vn(ro) for ro≤r≤R. (3.17) In particular, we have

vn(R)≤1+2vn(ro), (3.18)

that is

vn(ro)≥n−1

2 . (3.19)

Thus, for eachxo∈Ωsuch thatro= |xo|satisfies (3.14), we get u(xo)≥vn(ro)≥n−1

2 → ∞ asn → ∞. (3.20)

Hence,u(x)does not exist.

Corollary12. Suppose thatγ > qandΩis a bounded domain inR^N(N≥3). Let p(x)satisfy condition(P). Then equation(P+)has a positive large solution if and only ifγ >1.

4. Condition onpis nearly optimal. We have shown in Theorem 2 that if the non- negative function p is such that each of its zero points is enclosed by a bounded surface of nonzero points, then equation(P+)has a large positive solution. In this section we show that, if the condition does not hold in the sense thatp vanishes in an “outer ring” of the domain, then equation(P+)has no positive large solution.

Theorem13. Suppose thatg(x,0)=0∀x∈Ω.If there exists a domainD⊆Ωsuch thatD¯⊂Ωandg(x,t)=0∀x∈Ω\D,∀t≥0, then there is no positive large solution of

∆u+|∇u|^q=g(x,u), x∈Ω. (4.1) Note that this includes the caseg(x,u)=p(x)u^γ,γ≥0, andp(x)=0inΩ\D.

Proof. Suppose that such a solutionuexists. Letwbe the unique positive solu- tion of

∆w=0, x∈Ω\D, w=0, x∈∂D, w=1, x∈∂Ω.

(4.2)

It should be noted thatusatisfies

∆u≤0, x∈Ω\D, u= ∞, x∈∂Ω, u≥0, x∈∂D.

(4.3)

By the maximum principle, we getkw≤ufor anyk >0 sincekwsatisfies

∆(kw)=0, x∈Ω\D, kw=k, x∈∂Ω, kw=0, x∈∂D.

(4.4)

Therefore, for anyxo∈Ω\D, we havew(xo) >0 andkw(xo)≤u(xo)∀k >0. Hence, u(xo)= ∞, which is a contradiction.

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Lair: Department of Mathematics and Statistics, Air Force Institute of Technol- ogy/ENC,2950P Street, Wright-Patterson AFB, OH45433-7765, USA

E-mail address:[email protected]