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Limit Branch of Solutions as p → ∞ for aFamily of Sub-Diffusive Problems Relatedto the p-LaplacianFernando Charro a & Ireneo Peral aa Departamento de Matematicas , U. Autonoma de Madrid , Madrid,SpainPublished online: 10 Dec 2007.

To cite this article: Fernando Charro & Ireneo Peral (2007) Limit Branch of Solutions as p → ∞ for aFamily of Sub-Diffusive Problems Related to the p-Laplacian, Communications in Partial DifferentialEquations, 32:12, 1965-1981, DOI: 10.1080/03605300701454792

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Communications in Partial Differential Equations, 32: 1965–1981, 2007Copyright © Taylor & Francis Group, LLCISSN 0360-5302 print/1532-4133 onlineDOI: 10.1080/03605300701454792

Limit Branch of Solutions as p → �for a Family of Sub-Diffusive Problems

Related to the p-Laplacian

FERNANDO CHARRO AND IRENEO PERAL

Departamento de Matematicas, U. Autonoma de Madrid, Madrid, Spain

We characterize the limit as p → � of the branches of solutions to−div���u�p−2�u� = �ur�p� in � ⊂ �n

u > 0 in �

u = 0 on ��

with � > 0 and limp→�r�p�

p−1 = R, where R < 1. We show that the limit set is a curveof positive viscosity solutions of the equation

min{��v�x�� −�vR�x��−�v�x�

} = 0 in �� v��� = 0�

where �u ≡ ∑ni�j=1

�u�xj

�2u�xj�xi

�u�xi

and � > 0. The key result is a comparisonprinciple for the limit equation from which we deduce uniqueness for the Dirichletproblem and hence the existence of the curve of solutions.

Keywords Comparison principle; Infinity laplacian; Fully nonlinear.

Mathematics Subject Classification 35J15; 35J60; 35J70.

1. Introduction

Our main interest is to study the behavior as p → � of the sequence of positivesolutions of the problems

−div���u�p−2�u� = �ur�p� in � ⊂ �n

u > 0 in �

u = 0 on ��

(1)

Received June 1, 2006; Accepted April 1, 2007Address correspondence to Ireneo Peral, Departamento de Matematicas, U. Autonoma

de Madrid, 28049 Madrid, Spain; E-mail: ireneo.peral@uam.es

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where � > 0 and limp→�r�p�

p−1 = R with R < 1 and � is a bounded domain. Withoutloss of generality we shall suppose r�p� < p− 1 hereafter, this is the sense in whichwe call problem (1) sub-diffusive. The case r�p� = p− 1, the eigenvalues case, hasbeen studied in Juutinen et al. (1999).

We observe that for fixed p, the expression

u��p�x� = �1

p−1−r�p� u1�p�x� (2)

links the solutions of the problem−div���u1�p�p−2�u1�p� = u

r�p�1�p in �

u1�p > 0 in �

u1�p = 0 on ��

(3)

with those of the problem (1). Since r�p� < p− 1 for p large enough, problem (3)has a unique solution for each fixed �. The proof is an adaptation of a result byBrezis and Oswald (1986), see for instance Abdellaoui and Peral (2003) and thereferences therein for the details in this framework. Thus, we deduce that the set ofpositive solutions for fixed p,

p = ���� u��p� � u��p is the nontrivial solution of (1) � (4)

is a continuous branch, in fact a smooth curve in �0���×W 1��0 ���, given by (2).

We are interested in studying both the convergence of those curves of solutionsas p → � and the existence of a limit equation satisfied by the elements of the limitset. First of all, we shall state the precise notion of limit set in this setting.

Definition 1. The limit set of the branches p is defined as

� ={��� u�� � ∃ sequence

{��p� u�p�p

�}p

s.t. ��p� u�p�p� ∈ p and

limp→���p�

1/p = �� limp→� u�p�p

= u� uniformly}� (5)

Remark 2. It is possible to prove that for fixed �p ≡ � the limit of ��� u��p� is acouple �1� v� where v is a function independent of �. The condition lim��p�

1/p = �behaves as a scaling factor for each p that allow a nontrivial limit set to come up.The arguments below work in the same way defining the limit set via the condition

limp→���p�

1r�p� = ��

Both constants � and � are related by the expression � = �1/R.

Heuristically, if ��� u�� ∈ �, we can take limits as p → � in the expression (2)and conclude that

u� = limp→� u�p�p

= limp→�

(�

1p−1−r�p�p u1�p�x�

)= �

11−R u1� (6)

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Thus, it seems that the limit problem should satisfy a scaling property analogous to(2) in problem (1).

In order to make rigorous this formal computation, we show (Sections 2 and 3)that there exists a convergent subsequence �u1�pi

i of solutions to (3) and a limitfunction u1 ∈ �, and then we use such a subsequence to deduce that u1 is a viscositysolution to the limit problem

min{��u1�x�� − uR

1 �x��−�u1�x�} = 0 in ��

u1 > 0 in �

u1 = 0 on ���

(7)

where �u ≡ ∑Ni�j=1

�u�xj

�2u�xj�xi

�u�xi.

Next, we prove in Section 4 the main result, a comparison principle for the limitproblem (7) and, as a consequence, uniqueness of positive solutions to (7). Noticethat the comparison and uniqueness results extend to

min{��u��x�� −�uR

��x��−�u��x�} = 0 in ��

u� > 0 in �

u� = 0 on ���

(8)

through the re-scaling suggested by (6).This result could also be read as an extension to the fully nonlinear setting of

the uniqueness result by Brezis–Oswald quoted above.Moreover, we deduce from the uniqueness result that the set

� = {��� u�� � u� is the nontrivial solution of (8)

}�

defines a smooth curve of positive solutions of the limit problem given by (6) whichverifies (Section 5) the estimate

�u��L���� =(� ·max

x∈�dist�x� ���

) 11−R

�

This is the sense in which we mean the limit of branches.We point out the similarities between the behavior of problem (1) for p < �

and that of the limit problem (8).We also provide explicit solutions of problem (7) for a certain class of domains

(including the ball and the annulus among others) in Section 6.Finally, we recall for the reader’s convenience the following lemma, stating that

weak solutions of our problem are also viscosity solutions. The proof, which weomit here, follows in an analogous way to that of Lemma 1.8 in Juutinen et al.(1999) (see also Bhattacharya et al., 1989).

Lemma 3. If u is a continuous weak solution of (1), then it is a viscosity solution ofthe same problem, rewritten as{

Fp��u�D2u� = �ur�p� in �

u = 0 on ���(9)

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where

Fp��� X� = −trace

((Id + �p− 2�

�⊗ �

���2)X

)· ���p−2�

In the sequel we shall always consider the more suitable form of our problembetween (1) and (9) without any further reference.

2. A Priori Bounds

The following result is an easy consequence of Morrey’s estimates.

Lemma 4. Consider p > n fixed. There exist constants C1� C2 independent of p suchthat the solution u1�p of (3) satisfies

�u1�p�L���� ≤ C1� (10)

and for every m > n,

�u1�p�x�− u1�p�y���x − y�1− n

m

≤ C2 ∀x� y ∈ � (11)

if p > m.

Proof. 1. Multiplying (1) by u1�p and integrating by parts, we get∫���u1�p�p dx =

∫��u1�p�r+1 dx� (12)

Morrey’s estimates imply that there exists a constant C > 0 independent of psuch that

�u1�p�L���� ≤ C( ∫

���u1�p�p dx

)1/p

Combining both expressions above we have

�u1�p�L���� ≤ C( ∫

��u1�p�r+1 dx

)1/p ≤ C��� 1p �u1�p�r+1p

L�����

from which we deduce (10) since limp→�p

p−r�p�−1 = 11−R

.

2. Since p > m, combining the Hölder inequality and Morrey estimates wehave

�u1�p�x�− u1�p�y���x − y�1− n

m

≤ C( ∫

���u1�p�m dx

)1/m ≤ C��� 1m− 1

p

( ∫���u1�p�p dx

)1/p

= C��� 1m− 1

p

( ∫��u1�p�r+1dx

)1/p ≤ C��� 1m �u1�p�

r+1p

L�����

where C > 0 is a constant independent of m�p. Then, we obtain (11) from (10). �

We then have the following compactness result.

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Proposition 5. Consider the sequence �u1�p p of solutions of (3). Then, there exists asubsequence pi → � and a limit function u1 such that

limi→�

u1�pi= u1

uniformly. Moreover, u1�x� > 0 in �.

Proof. The existence of u1 as uniform limit is a consequence of the previous lemmaand the Arzelá–Ascoli compactness criteria.

To prove that u1 > 0, we adapt an idea from Boccardo et al. (1995). The pointis to find a subsolution for problem (1) valid for all p large enough.

Consider the eigenvalue problem for the p-laplacian{−div����1�p�p−2��1�p� = �1�p�����1�p�p−2�1�p in �

�1�p = 0 in ���

where �1�p��� and �1�p are the principal eigenvalue and eigenfunction respectively.We take every �1�p normalized in such a way that ��1�p�� = 1.

From Fukagai et al. (1999) and Juutinen et al. (1999) we know that there existsan infinity eigenvalue ����� such that

����� = limp→�

(�1�p���

)1/p = (maxx∈�

dist�x� ���)−1

� (13)

and that there exist a sequence of eigenfunctions ��1�p p with ���1�p��� = 1 thatconverge to an eigenfunction �1�� solving the limit problem

min{���1��� −���1���−��1��

} = 0 in �� �1�� = 0 on ���

in the viscosity sense. Moreover, it is known that �1�� > 0.

Now, fix � > 0 and define �1�p = t�1�p, with t = min{1� �

−11−R� − �

}. Then we have

��1�p�� = t��1�p�� ≤ 1 and, from (13),

t ≤ �−11−R� − � < �1�p���

−1p−1−r�p� for p large enough�

Thus, for that range of p,

−p�1�p = �1�p����p−11�p ≤ �1�p���tp−1−r�p��

r�p�1�p < �

r�p�1�p �

and by Abdellaoui and Peral (2003) we conclude that �1�p ≤ u1�p. Passing to the limit(up to a subsequence)

u1�x� ≥ t�1�� > 0� �

Remark 6. The limit function u1 is Lipschitz continuous since, by letting p → �and then m → � in (11), we obtain

�u1�x�− u1�y���x − y� ≤ C2�

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Remark 7. We will prove below that the limit function u1 is the unique solution ofproblem (15). As a consequence, not only a subsequence but the whole sequence u1�p

converges to u1. Moreover, for every sequence ��p p such that limp→���p�1/p = �,we deduce from (2) that the solution u�p�p

of (1) with � = �p converges to a functionu� uniformly in �, and that

u��x� = �1

1−R u1�x�� (14)

3. The Limit Problem

In the present section, we characterize limits of solutions of (3) (elements of �) assolutions of a PDE. See Fukagai et al. (1999) and Juutinen et al. (1999) for relatedresults in the eigenvalue case.

Proposition 8. The limit function u1 in Proposition 5 is a viscosity solution of theproblem

min{��v�x�� − vR�x��−�v�x�

} = 0 in ��

v > 0 in �

v = 0 on ���

(15)

Proof. If the solutions of our problem (1) were of class �2 the p-laplacian can beexpanded to

pu = div���u�p−2�u� = ��u�p−2u+ �p− 2� ��u�p−4 D2u�u� �u�

= �p− 2���u�p−4

{1

p− 2��u�2u+ D2u�u� �u�

}�

But the solutions of our problem are only �1��, and we have to reinterpret theprevious calculation in the viscosity sense (DiBenedetto, 1983).

Consider a point x0 ∈ � and a function � ∈ �2��� such that u1 − � has a strictlocal minimum at x0. As u1 is the uniform limit of u1�pi

, there exists a sequence ofpoints xi → x0 such that �u1�pi

− ���xi� is a local minimum for each i. Then, as u1�piis a viscosity solution and so a supersolution, we get

−�pi − 2�����xi��pi−4

{ ����xi��2pi − 2

��xi�+ D2��xi����xi�� ���xi��}

= −pi��xi� ≥ u

r�pi�1�pi

�xi��

Rearranging terms, we obtain

−�pi − 2�

[����xi��

�u1�pi�xi��

r�pi�pi−4

]pi−4{ ����xi��2pi − 2

��xi�+ D2��xi����xi�� ���xi��}≥ 1�

Notice that u1�x0� > 0 by Proposition 5. Then, if we suppose that

����x0�� < uR1 �x0�

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we obtain a contradiction letting i → � in the previous inequality. Thus, it is

����x0�� − uR1 �x0� ≥ 0� (16)

We also have that

−���x0� = − D2��x0����x0�� ���x0�� ≥ 0� (17)

because we would get a contradiction otherwise.Therefore, we can put together (16) and (17) writing

min{����x0�� − uR

1 �x0��−���x0�} ≥ 0�

Hence, we conclude that u1 is a viscosity supersolution of equation (15).It remains to be shown that u1 is a viscosity subsolution of the limit

equation (15), i.e. we have to show that, for each x0 ∈ � and � ∈ �2��� such thatu1 − � attains a strict local maximum at x0 (note that x0 and � are not the samethan before) we have

min{����x0�� − uR

1 �x0��−���x0�} ≤ 0�

We can suppose that

����x0�� > uR1 �x0��

because otherwise, we are done. As we did before, the uniform convergence of u1�pito u1 provides us with a sequence of points xi → x0 which are local maxima ofu1�pi

− �. Recalling the definition of viscosity subsolution we have

−�pi − 2�

[����xi��(

u1�pi�xi�

) r�pi�pi−4

]pi−4{ ����xi��2pi − 2

��xi�+ D2��xi����xi�� ���xi��}≤ 1�

for each fixed p. Letting i → � we obtain −���x0� ≤ 0 because in other case weget a contradiction. �

In the following Lemma we show that u� given by relation (14) is a viscositysolution of a rescaled version of (15).

Lemma 9. The solutions of the problem{min

{��u��x�� −�uR��x��−�u��x�

} = 0 in �

u� = 0 in ���(18)

and those of the problem{min

{��u1�x�� − uR1 �x��−�u1�x�

} = 0 in �

u1 = 0 in ���(19)

are related by the expression u��x� = �1

1−R u1�x��

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We omit the proof since it is standard.

4. Uniqueness of Solutions for the Limit Problem

The main result in the present section is the following Comparison Principle for equa-tion (15), from which we deduce uniqueness of positive solutions of the limit problem.

Notice that then the whole sequence u1�p converges uniformly to u1 as p → �.Hence, we deduce from (2) that the whole sequence u�p�p

converges uniformly tosome u� provided limp→���p�1/p = � (see Remark 7). Lemma 9 then implies thatu� is the unique positive viscosity solution of (18).

Theorem 10. Let � ⊂ �n be a bounded domain and consider a subsolution u and asupersolution v of

min{��w�x�� − wR�x��−�w�x�

} = 0 in �� (20)

Suppose that both, u and v are strictly positive in �, continuous up to the boundary andsatisfy u ≤ v on ��. Then, u ≤ v in �.

Remark 11. Every nontrivial solution u of (19) is �-superharmonic. The Harnackinequality for �-superharmonic functions (see Lindqvist and Manfredi, 1995, 1997)implies u > 0 inside �.

Once we have proved the above Comparison Principle, we deduce fromRemark 11 that there exists a unique solution to problem (19). Hence, fromLemma 9 the set � is a smooth curve of solutions parametrized by �. Noticethat we can suppose R �= 0, since the case R = 0 has been already studied in Jensen(1993) and Juutinen (1998).

Our aim is to study the uniqueness of solutions for the problem (19) followingthe techniques in Crandall et al. (1992). Equation in (19) can be written asF�u� �u�D2u� = 0, where F is given by

F � �×�n ×� n −→ �

�r� p� X� −→ min{�p� − rR�− Xp� p�}

which, in the notation of Crandall et al. (1992), is degenerate elliptic in the sense thatF�r� p�X� ≤ F�r� p� Y� whenever Y ≤ X, but not proper, namely, it is non-increasingin r .

Thus, equation (19) is not in the framework of Crandall et al. (1992).Nevertheless, it is possible to transform our equation into a valid one by mean of achange of variables. Following Juutinen et al. (1999), we prove the next lemma.

Lemma 12. Let u be a strictly positive supersolution (subsolution) of problem (19) in�. Then, v�x� = �1− R�−1u1−R�x� is a viscosity supersolution (subsolution) of

min{��v�x�� − 1�−�v�x�−

R

1− R

( ��v�x��4v�x�

)}= 0� (21)

in every �∗ such that �∗ ⊂ �.

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Proof. Let � ∈ �2��� be a function touching v from below at x0 ∈ �. We define��x� = ��1− R���x��

11−R which touches u from below at x0. Notice that ��x� is �2

in a neighborhood of x0 since u > 0 in � implies ��x� > 0 near x0. We can computethe derivatives of ��x� in terms of those of ��x�

���x0� =(�1− R���x0�

) R1−R ���x0��

D2��x0� =(�1− R���x0�

) R1−R D2��x0�+ R

(�1− R���x0�

) 2R−11−R ���x0�⊗ ���x0��

As u is a viscosity solution of (19), we have that

0 ≤ min{����x0�� −�R�x0��− D2��x0����x0�� ���x0��

}≤ min

{��1− R���x0��

R1−R �����x0�� − 1��

− ��1− R���x0��3R1−R

(���x0�+

R

1− R

����x0��4��x0�

)}�

Notice that, ��x0� = v�x0� > 0. We deduce that

��1− R���x0��R

1−R �����x0�� − 1� ≥ 0� and

−��1− R���x0��3R1−R

(���x0�+

R

1− R

����x0��4��x0�

)≥ 0�

It follows that

min{����x0�� − 1�−���x0�−

R

1− R

����x0��4��x0�

}≥ 0�

We have proved that v is a viscosity supersolution of (21). The subsolution case isanalogous. �

Equation (21) is given by the function

F � �+ ×�n ×� n −→ �

�r� p� X� −→ min{�p� − 1�− Xp� p� − R

1− R

�p�4r

}� (22)

which is both degenerate elliptic and proper. Notice that the equation is singular inr = 0 but, as we are dealing with positive solutions, there are no sign changes in theinterior of �, and we can proceed.

As it is pointed out in Crandall et al. (1992, Section 5.C), it is possible toestablish comparison when either u or v are strict. This is not our case, but we aregoing to show that it is possible to construct strict supersolutions of (21) startingfrom any positive supersolution. In Juutinen (1998) and Juutinen et al. (1999), wefind related constructions in the eigenvalue case.

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Lemma 13. Let v�x� be a strictly positive viscosity supersolution of (21) in �∗ suchthat �∗ ⊂ �. Then,

v�x� = a · (v�x�+ �)� (23)

is a strict supersolution of the same equation for any fixed a > 1 and � > 0.

Proof. Let � ∈ �2 be a function touching v�x� from below in some x0 ∈ �∗. Wedefine

��x� = 1a��x�− ��

which clearly touches v�x� from below in x0. We can compute the derivatives of��x� in terms of those of ��x�, this is

���x0� = a−1 ���x0� and D2��x0� = a−1 D2��x0�� (24)

Since v�x� is a viscosity supersolution of (21) in �∗, we deduce

����x0�� − 1 ≥ 0� (25)

and

− D2��x0����x0�� ���x0�� −R

1− R

����x0��4v�x0�

≥ 0� (26)

From (24) and (25), we have

����x0�� − 1 = a ����x0�� − 1 ≥ a− 1� (27)

and from (24) and (26) we get

− D2��x0����x0�� ���x0�� −R

1− R

����x0��4v�x0�

≥ a3 R

1− R����x0��4

(1

v�x0�− 1

v�x0�+ �

)�

All the terms involved in the last expression are positive. From (25) and a > 1 we get

− D2��x0����x0�� ���x0�� −R

1− R

����x0��4v�x0�

≥ a3 R

1− R

(�

v�x0��v�x0�+ ��

)≥ R

1− R

(�

�v����v�� + ��

)� (28)

Finally, we can put together (27) and (28) in the following way

min{����x0�� − 1�− D2��x0����x0�� ���x0�� −

R

1− R

����x0��4v�x0�

}

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≥ min{a− 1�

R

1− R

(�

�v����v�� + ��

)}= ��a� �� R� �v��� > 0� �

Remark 14. For any a��� > 1 such that a��� → 1, and ���� such that ���� → 0(for example a��� = 1+ � and ���� = �), expression (23) defines a family ofapproximations of the identity as � → 0 since

�v− v�L���∗� ≤ �a���− 1��v�L���∗� + a��������

We shall use the comparison principle for semicontinuous functions, due toCrandall and Ishii, which can be find in Crandall and Ishii (1990) and in theAppendix of Crandall et al. (1992). We include here a simplified version

Lemma 15. Let � > 0. Let u�−v be real-valued, upper-semicontinuous functions in �,and �x�� y�� ∈ �×� be a local maximum point of the function u�x�− v�y�− �

2 �x − y�2.Then, there exist symmetric matrices X� and Y� such that(

��x� − y��� X�

) ∈ J2+u�x�� and

(��x� − y��� Y�

) ∈ J2−v�y���

where J2+u�x�� and J

2−v�y�� are the closures of the upper and lower semijets of u and v,

respectively (see Crandall et al., 1992). We also have

−3�(I 00 I

)≤

(X� 00 −Y�

)≤ 3�

(I −I−I I

)� (29)

We shall also need the following lemma, whose proof can be find in (Crandallet al., 1992, Proposition 3.7).

Lemma 16. Suppose that u and −v are real-valued, upper-semicontinuous functionsin �. For every � > 0, we denote

M� = sup��

(u�x�− v�y�− �

2�x − y�2

)�

If �x�� y�� ∈ �×� is such that M� = u�x��− v�y��− �2 �x� − y��2, then

1. lim� →� ��x� − y��2 = 0.2. lim�→� M� = u�x�− v�x� = supx∈� �u�x�− v�x�� whenever x is a limit point of x�.

Now, we are ready to start with the proof of the Comparison Principle.

Proof of Theorem 10. As u− v ∈ ���� and � is compact, u− v attains a maximumat �. In order to arrive at a contradiction, we suppose that max��u− v� > 0.Consider

u�x� = u1−R�x�

1− Rand v�x� = v1−R�x�

1− R�

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We have proved in Lemma 12 that u and v are respectively a subsolution and asupersolution of (21) in every open �∗ such that �∗ ⊂ �. Moreover, Lemma 13implies that

v��x� = �1+ �� · (v�x�+ �)�

is a strict supersolution of (21) in �∗. Notice that, since �u− v���� ≤ 0

u− v� = u− �1+ ��v− �1+ ��� < 0 on ���

Moreover, Remark 14 implies max��u− v�� > 0 for � small enough. Thus, we cansuppose that �∗ contains all the maximum points of u− v� for � fixed.

Now, for each � > 0, let �x�� y�� be a maximum point of u�x�− v��y�− �2 �x − y�2

in �×�. By the compactness of �, we can suppose that x� → x as � → � for somex ∈ � (notice that also y� → x). Lemma 16 implies that x is a maximum point ofu− v� and, consequently, it is an interior point of �∗. We also have

lim�→�

(u�x��− v��y��−

�

2�x� − y��2

)= u�x�− v��x� > 0�

Thus, for � large enough, we have that both x� and y� are interior points of �∗ and

u�x��− v��y��−�

2�x� − y��2 > 0� (30)

Applying Lemma 15, there exist two symmetric matrices X�, Y� such that

(��x� − y��� X�

) ∈ J2+u�x�� and

(��x� − y��� Y�

) ∈ J2−v��y���

and

X��� �� − Y��� �� ≤ 3���− ��2 ∀�� � ∈ �n� (31)

Thus, we have

min{� �x� − y�� − 1�−�2 X��x� − y��� �x� − y��� −

R

1− R

�4�x� − y��4u�x��

}≤ 0�

and

min{��x� − y�� − 1�−�2 Y��x� − y��� �x� − y��� −

R

1− R

�4�x� − y��4v��y��

}≥ min

{��

R

1− R

(�

�v����v�� + ��

)}= ���� R� �v��� > 0�

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Limit Branch of p-Laplacian Solutions 1977

Subtracting the first equation from the second one, we have

0 < ����R� �v���

≤ min{� �x� − y�� − 1� �2 Y��x� − y��� �x� − y��� −

R

1− R

�4�x� − y��4v��y��

}(32)

−min{� �x� − y�� − 1�−�2 X��x� − y��� �x� − y��� −

R

1− R

�4�x� − y��4u�x��

}� (33)

Now, there are four cases to be considered depending on the values attained in theminima (32) and (33). The key point in the sequal is that we have v��y�� ≤ u�x�� andX� ≤ Y� from (30) and (31) respectively.

(1) Both minima are equal to � �x� − y�� − 1 and the difference is 0.(2) The minimum at (33) is −�2 X��x� − y��� �x� − y��� − R

1−R

�4�x�−y��4u�x��

and the onein (32) is � �x� − y�� − 1. Actually, it is impossible for this alternative to holdunless both quantities are equal because

� �x� − y�� − 1 = min�32�

≤ −�2 Y��x� − y��� �x� − y��� −R

1− R

�4�x� − y��4v��y��

≤ −�2 X��x� − y��� �x� − y��� −R

1− R

�4�x� − y��4u�x��

= min�33� ≤ � �x� − y�� − 1�

(3) The minimum in (32) is −�2 Y��x� − y��� �x� − y��� − R1−R

�4�x�−y��4v��y��

and the onein (33) is � �x� − y�� − 1. Using the fact that

−�2 Y��x� − y��� �x� − y��� −R

1− R

�4�x� − y��4v��y��

≤ � �x� − y�� − 1�

we find out that min�32�−min�33� ≤ 0.(4) The minimum in (32) is −�2 Y��x� − y��� �x� − y��� − R

1−R

�4�x�−y��4v��y��

and the one

in (33) is −�2 X��x� − y��� �x� − y��� − R1−R

�4�x�−y��4u�x��

. Therefore

min�32�−min�33� = −�2⟨�Y� − X���x� − y��� �x� − y��

⟩− R

1− R�4�x� − y��4

(1

v��y��− 1

u�x��

)≤ 0�

Thus, all the alternatives lead to

0 < ����R� �v��� ≤ min�32�−min�33� ≤ 0

which is a contradiction. �

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5. Estimates for �u��L�

Our goal in this section is to prove estimates for the limit function u�. To this aim,we consider the family of problems

min{��u�x�� −�� uR�x��−�u�x�

} = 0 in �� u��� = 0� (34)

for R ∈ �0� 1�, where

����� = (maxdistx∈��x� ���

)−1(35)

is the principal �-eigenvalue (see Juutinen et al., 1999). It is known (consult Jensen,1993, and Juutinen, 1998, Lemma 6.10) that the unique solution to problem (34)when R = 0 is

��x� = dist�x� ���

maxdistx∈��x� ����

For R = 1, we consider the maximal �-eigenfunction v with �v�L� = 1. Then, wehave the following result.

Proposition 17. Let � ⊂ �n be a bounded domain and R < 1. Consider � > 0 andu� the nontrivial solution of

min{��u��x�� −�uR

��x��−�u��x�} = 0 in �� u���� = 0�

Then we have(� ·max

y∈�dist�y� ���

) 11−R

v�x� ≤ u��x� ≤(� ·max

y∈�dist�y� ���

) 11−R

��x�� (36)

for every x ∈ �, where v�x� and ��x� are defined above. Indeed,

�u��L���� =(� ·max

x∈�dist�x� ���

) 11−R

� (37)

Lemma 18. For every R ∈ �0� 1�, the functions v�x� and ��x� are respectively a sub-and a supersolution of problem (34).

Once we have proved Lemma 18, Proposition 17 follows easily from theComparison Principle. Indeed, by comparison, for every fixed R ∈ �0� 1� we have

v�x� ≤ u���x� ≤ ��x� ∀x ∈ ��

where u�� is the solution of (34). Formula (14) implies

u��x� =(� ·�−1

�) 1

1−R u���x�� (38)

and then we can combine both expressions above and (35) to get (36), from whichwe deduce (37) since �v�L���� = ���L���� = 1.

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Limit Branch of p-Laplacian Solutions 1979

Proof of Lemma 18. 1. Consider a point x0 ∈ � and a function � ∈ �2 such that�v− �� has a maximum in x0. As v is an �-eigenfunction, it satisfies

min{����x0�� −��v�x0��−���x0�

} ≤ 0 in ��

We can consider −���x0� > 0 and ����x0�� −��v�x0� ≤ 0 since we are doneotherwise. Clearly

����x0�� −��v�x0�R ≤ ��

(v�x0�− v�x0�

R) ≤ 0�

since �v�� = 1, and then

min{����x0�� −��v�x0�

R�−���x0�} ≤ 0 in ��

2. Now, consider a point x0 ∈ � and a function � ∈ �2 such that ��− �� hasa minimum in x0. As � is a solution of problem (34) when R = 0, in particular itsatisfies

min{����x0�� −���−���x0�

} ≥ 0 in ��

Then −���x0� ≥ 0 and ����x0�� ≥ �� and we have

����x0�� −����x0�R ≥ ��

(1− ��x0�

R) ≥ 0

since ���� = 1. Thus, we conclude

min{����x0�� −����x0�

R�−���x0�} ≥ 0 in �� �

6. Explicit Solutions of the Limit Problem

Now, we are going to compute the explicit solution of the limit problem for a classof domains including the ball and the annulus among others. Following Juutinen(1998), we define the ridge set of � as

� = �x ∈ � � dist�x� ��� is not differentiable at x

= �x ∈ � � ∃x1� x2 ∈ ��� x1 �= x2� s.t. �x − x1� = �x − x2� = dist�x� ���

and its subset � = �x ∈ � � dist�x� ��� = maxdistx∈��x� ��� , the set of maximaldistance. We have

Proposition 19. Given R ∈ �0� 1� and � > 0, if � ≡ �, then

u��x� = �1

1−R ·(maxx∈�

dist�x� ���) R

1−R · dist�x� ���� (39)

is the unique positive solution of

min{��u��x�� −�uR

��x��−�u��x�} = 0 in �� u���� = 0� (40)

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Proof. From Proposition 17, we have that(� ·max

y∈�dist�y� ���

) 11−R

v�x� ≤ u��x� ≤(� ·max

y∈�dist�y� ���

) 11−R

��x�� ∀x ∈ ��

where v�x� and ��x� are the same as in the previous section. Notice that we havethis inequality without any hypothesis on �.

It is a well-known result (see Juutinen, 1998, and Juutinen et al., 2001, forexample) that if � = �, then v�x� = ��x� is the maximal solution of the �-eigenvalue problem with �v�L� = 1, from which we deduce (39). �

Remark 20. In fact, (39) defines a branch of solutions of (40) even if R > 1whenever � ≡ �. Indeed, as a consequence of Juutinen (1998, Lemma 6.10), wehave

0 = min{��dist�x� ���� − 1�−�dist�x� ���

}= min

{�

11−R ·max

x∈�dist�x� ���

R1−R · (��dist�x� ���� − 1

)�

�3

1−R ·maxx∈�

dist�x� ���3R1−R · (− �dist�x� ���

)}= min

{��u��x�� −�uR��x��−�u��x�

} ∀x ∈ ��

while u is �-harmonic and satisfies ��u� −�uR > 0 outside � (the proof isanalogous to that of the “�-lemma” in Juutinen et al., 2001).

Acknowledgments

The authors would like to express their gratitude to the anonymous referee forhis/her kind suggestions and helpful advices which had improved the final form ofthe manuscript.

Work partially supported by project MTM2004-02223, MEC, Spain. The firstauthor was also supported by a FPU grant of MEC, Spain.

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