Global branches of solutions for a class of non uniformly fully nonlinear elliptic equations

TL;DR

This paper investigates the global solution structure of a class of fully nonlinear elliptic equations, establishing bifurcation results and analyzing eigenvalue behaviors under singular perturbations.

Contribution

It introduces new global bifurcation results for non-uniform fully nonlinear elliptic equations and characterizes eigenvalues in singular perturbation contexts.

Findings

Existence of simple eigenvalues for perturbed problems

Asymptotic behavior of eigenvalues analyzed

Global bifurcation branches of positive smooth solutions established

Abstract

We consider singular perturbations of eigenvalue problems. We prove that to these problems correspond simple eigenvalues and we study their asymptotic behavior. As a result, we prove global bifurcation results for non uniformly and fully nonlinear elliptic equations. These equations are defined on smooth enough bounded domains and the solutions belonging in these branches are smooth and positive.

Equations

$$-\Delta u=\lambda u+B(\lambda ,x,u,u_{x_i},u_{x_i{x}_j}),x\in \Omega ,u{|}_{\partial \Omega }=0,i,j=1,...,N,$$

$$-\Delta u$$

$$u{|}_{\partial \Omega }$$

$$-\Delta u$$

$$u{|}_{\partial \Omega }$$

$$\underset{||u||\to 0}{\lim }\frac{||H(\lambda ,u)||}{||u||}=0.$$

$$C=\{(\lambda ,u)\in \mathbb{R}\times X:(\lambda ,u)\text{mbox}solvesu=G(\lambda ,u),u\not\equiv 0\},$$

$$||u|{|}_X^2=||u|{|}_{H^6(\Omega )}^2+||u|{|}_{H_0^1(\Omega )}^2.$$

$$\epsilon <u,\phi {>}_X+{\int }_{\Omega }\nabla u\cdot \nabla \varphi dx=\lambda {\int }_{\Omega }(1+|\det D^{2}u{|}^2)u\varphi dx,\text{mbox}forany\varphi \in X,$$

$$\epsilon <u,\phi {>}_X+{\int }_{\Omega }\nabla u\cdot \nabla \varphi dx=\lambda {\int }_{\Omega }u\varphi dx,\text{mbox}forany\varphi \in X.$$

$$\phi (\epsilon )=\{({\lambda }_{0,\epsilon },u_{0,\epsilon }),0\le \epsilon <s\}\subset \mathbb{R}\times X.$$

$$\epsilon <u,\phi {>}_X+{\int }_{\Omega }\nabla u\cdot \nabla \varphi dx=\lambda {\int }_{\Omega }{g}_{n}u\varphi dx,\text{mbox}forany\varphi \in X.$$

$$-\Delta u$$

$$u{|}_{\partial \Omega }$$

$${\phi }_n(\epsilon )=\{({\lambda }_{n,\epsilon },u_{n,\epsilon }),0\le \epsilon <s\}\subset \mathbb{R}\times X.$$

$$\Phi :=\bigcup _{n>N(\delta )}\bigcup _{0\le \epsilon \le s_{\ast }}{\phi }_n(\epsilon ),$$

$${\mathcal{C}}_0=\{(\lambda ,u)\in \mathbb{R}\times X:(\lambda ,u)\text{mbox}solves(\text{ref}eq1.1),u\not\equiv 0\},$$

$$\underset{n\to \infty }{\mathrm{liminf}}\{A_n\}\not\equiv \varnothing .$$

$${\mathcal{C}}_0=\{(\lambda ,u)\in \mathbb{R}\times X:(\lambda ,u)\text{mbox}solves(\text{ref}generalcase),u\not\equiv 0\},$$

$$<\varphi ,\psi {>}_{H_0^1(\Omega )}={\int }_{\Omega }\nabla \varphi \cdot \nabla \psi dx,$$

$$<\varphi ,\psi {>}_{L_g^2(\Omega )}={\int }_{\Omega }g(x)\varphi \psi dx,$$

$${\lambda }_0=\underset{0\not\equiv u\in C_0^{\infty }(\Omega )}{\inf }\frac{{\int }_{\Omega }|\nabla u{|}^{2}dx}{{\int }_{\Omega }{u}^{2}dx},{\lambda }_{n,0}=\underset{0\not\equiv u\in C_0^{\infty }}{\inf }\frac{{\int }_{\Omega }|\nabla u{|}^{2}dx}{{\int }_{\Omega }{g}_n(x)u^{2}dx},$$

$$<u,v{>}_X=\lambda <u,v{>}_{L_{g_n}^2},$$

$$\epsilon <v_{n,\epsilon },\varphi {>}_X+<v_{n,\epsilon },\varphi {>}_{H_0^1(\Omega )}-{\mu }_{n,\epsilon }<v_{n,\epsilon },\varphi {>}_{L_{g_n}^2(\Omega )}=0,$$

$$v_{n,\epsilon }=a_{n,\epsilon }{u}_{n,0}+{\xi }_{n,\epsilon },a_{n,\epsilon }=<v_{n,\epsilon },u_{n,0}{>}_{H_0^1(\Omega )}/||u_{n,0}|{|}_{H_0^1(\Omega )},$$

$$\epsilon <{\xi }_{n,\epsilon },\varphi {>}_X+<{\xi }_{n,\epsilon },\varphi {>}_{H_0^1(\Omega )}-{\mu }_{n,\epsilon }<{\xi }_{n,\epsilon },\varphi {>}_{L_{g_n}^2(\Omega )}=$$

$$=-\epsilon a_{n,\epsilon }<u_{n,0},\varphi {>}_X+a_{n,\epsilon }({\mu }_{n,\epsilon }-{\lambda }_{n,0})<u_{n,0},\varphi {>}_{L_{g_n}^2(\Omega )},$$

$$\epsilon ||{\xi }_{n,\epsilon }|{|}_X^2+||{\xi }_{n,\epsilon }|{|}_{H_0^1(\Omega )}^2-{\mu }_{n,\epsilon }||{\xi }_{n,\epsilon }|{|}_{L_{g_n}^2(\Omega )}^2=-\epsilon a_{n,\epsilon }<u_{n,0},{\xi }_{n,\epsilon }{>}_X,$$

$$||{\xi }_{n,\epsilon }|{|}_X^2+\frac{1}{\epsilon }\left(||{\xi }_{n,\epsilon }|{|}_{H_0^1(\Omega )}^2-{\mu }_{n,\epsilon }||{\xi }_{n,\epsilon }|{|}_{L_{g_n}^2(\Omega )}^2\right)=-a_{n,\epsilon }<u_{n,0},{\xi }_{n,\epsilon }{>}_X.$$

$$<v_{n,\epsilon },u_{n,0}{>}_{H_0^1(\Omega )}\le ||v_{n,\epsilon }|{|}_{H_0^1(\Omega )}||u_{n,0}|{|}_{H_0^1(\Omega )}\le C||v_{n,\epsilon }|{|}_X||u_{n,0}|{|}_X=C.$$

$$||{\xi }_{n,\epsilon }|{|}_{H_0^1(\Omega )}^2-{\mu }_{n,\epsilon }||{\xi }_{n,\epsilon }|{|}_{L_{g_n}^2(\Omega )}^2\ge ({\lambda }_{n,0}^{(2)}-{\mu }_{n,\epsilon })||{\xi }_{n,\epsilon }|{|}_{L_{g_n}^2(\Omega )}^2,$$

$$({\lambda }_{n,0}^{(2)}-{\mu }_{n,\epsilon })||{\xi }_{n,\epsilon }|{|}_{L_{g_n}^2(\Omega )}^2\le -\epsilon a_{n,\epsilon }<u_{n,0},{\xi }_{n,\epsilon }{>}_X\to 0,$$

Full text

1 Introduction

The purpose of this work is to obtain global bifurcation results for a class of nonlinear problems. Our method may be applied to the following problems:

[TABLE]

where $\Omega$ is a bounded domain of $\mathbb{R}^{N}$, $N=2,3,4$, which is smooth enough, at least $C^{6}$. For the operator $B=B(\lambda,x,p,q_{i},r_{ij})$, $B:\mathbb{R}^{N^{2}+N+3}\to\mathbb{R}$ we assume that is smooth enough, at least $C^{6}$ such that $B=o(||u||_{C^{2}(\Omega)})$ uniformly, as $||u||_{C^{2}(\Omega)}\to 0$.

For the sake of the representation and simplicity reasons, we first assume that the space dimension is $N=3$ and $B(\lambda,x,u,...,D^{2}u)=\lambda\,|\det D^{2}u|^{2}\,u$ and later on this paper we consider the general case of $B(\lambda,x,u,...,D^{2}u)$. Hence, we are going prove first the existence of a global bifurcation of solutions for the problem

[TABLE]

bifurcating from the principal eigenvalue $\lambda_{0}$ of the corresponding linear equation

[TABLE]

In order to state the main result of this paper we give the following definition of what we mean with the existence of a branch of solutions of an operator equation ([6]).

THEOREM 1.1

Assume that $X$ is a Banach space with norm $||\cdot||$ and consider $G(\lambda,\cdot)=L(\lambda,\cdot)+H(\lambda,\cdot)$, where $L$ is a compact linear map on $X$ and $H(\lambda,\cdot)$ is compact and satisfies

[TABLE]

If $\lambda$ is a simple eigenvalue of $L$ then the closure of the set

[TABLE]

possesses a maximal continuum (i.e. connected branch) of solutions, $\mathcal{C}_{\lambda}$, such that $(\lambda,0)\in\mathcal{C}_{\lambda}$ and $\mathcal{C}_{\lambda}$ either:

(i) meets infinity in $\mathbb{R}\times X$ or,

(ii) meets $(\lambda^{*},0)$, where $\lambda^{*}\neq\lambda$ is also an eigenvalue of $L$.

One possible way to study problems of the form (1.2), is to approximate them by suitably elliptic problems, for which the classical methods apply. We introduce the Hilbert space $X=H^{6}(\Omega)\cap H_{0}^{1}(\Omega)$, with norm given by

[TABLE]

Note that $H^{6}(\Omega)$ denotes the classic Sobolev space $W^{6,2}(\Omega)$. Let also $<.,.>_{X}$ be the inner product in $X$. For $\varepsilon$ small enough positive number, we assume the problem:

[TABLE]

and the corresponding linear eigenvalue problem:

[TABLE]

The choice of the space $X$ had been made in order to use the imbedding theorems; any function belonging to $H^{6}$ must also belong to $L^{\infty}(\Omega)$. In particular, $H^{6}$ is imbedded compactly, at least, in $C^{3}(\bar{\Omega})$, $N=2,3,4$, for sufficient smooth domains. In the sequel, we will frequently use that for $N=3$, the compact embedding is up to $C^{4}(\bar{\Omega})$.

Our intention is to prove that for each $\varepsilon$ small enough, there exists a branch of solutions of (1.4) bifurcating from a certain simple eigenvalue of (1.5), such that these branches converge to the global branch of (1.2), as $\varepsilon\downarrow 0$. In this direction, we face two main difficulties. The first is, that the properties of the eigenvalues of (1.5) are not known; it is unclear if they are close enough to $\lambda_{0}$, and what is their multiplicity. The second difficulty, since we have a singular perturbation, is to prove that the branches of solutions of (1.4) converge in $X$; we cannot expect that the branches of solutions of (1.4) converge uniformly. However, a careful application of Whyburn’s Lemma, is sufficient to give the desired (nonuniform) convergence.

In Section 2, we deal with the eigenvalue problem (1.5). We prove that for any $\varepsilon>0$ small enough, admits a positive eigenvalue $\lambda_{0,\varepsilon}$, which is simple with the associated eigenfunction being positive. The proof is not straightforward. Problem (1.3) (and (1.7 below), admits principal eigenvalue which is simple in the sense that the null space of the corresponding operator, $P_{0}$, is spanned by the eigenfunction and the range of $P_{0}$ is the orthogonal complement of the null space in $H_{0}^{1}(\Omega)$. In other words, $P_{0}$ is Fredholm with index 0, or equivalently $\lambda_{0}$ has algebraic multiplicity one. In our case, we do not have this: considering problem (1.3) in the space $X$, although $X\subset H_{0}^{1}(\Omega)$, the situation is different. In the context of the space $X$, the operator $P_{0}$ is no longer Fredholm with index 0. This means that the principal eigenvalue $\lambda_{0}$ has multiplicity one (since there exists only one function satisfying (1.3)) but has not algebraic multiplicity one. This is the reason why the Theorem of perturbed eigenvalues (see for instance [1, 5]) cannot be applied in the case of (1.3) and (1.5). However, the eigenvalues of the singular perturbations, that remain close enough to $\lambda_{0}$, are proved to be algebraic simple.

THEOREM 1.2

There exists $s>0$ small enough, such that for every $\varepsilon$, with $0<\varepsilon<s$, problem (1.5) admits an eigenvalue $\lambda_{0,\varepsilon}$ which is (algebraic) simple in $X$ and the associated normalized eigenfunction $u_{0,\varepsilon}\in X$ is positive. The perturbed eigenpairs form a continuous curve:

[TABLE]

Moreover, for $0\leq\varepsilon<s$, (1.5) has no other eigenpair, close enough to $(\lambda_{0},u_{0})$, than that belonging in $\varphi(\varepsilon)$.

Let $g_{n}$ be a smooth function, $g_{n}\in C^{4}(\bar{\Omega})$, such that the measure of its positive part $|g^{+}_{n}|$ is not zero in $\Omega$. Then, Theorem 1.2 is also applicable in the case of the linear problems

[TABLE]

For such $g_{n}$, there exists a principal eigenvalue $\lambda_{n,0}$ for the problem

[TABLE]

and the corresponding normalized eigenfunction $u_{n,0}$ belongs in $X$. Then, as in Theorem 1.2, there exist a local, to $(\lambda_{n,0},u_{n,0})$, curve of eigenpairs for the perturbed problems (1.6). Denote this curve by

[TABLE]

Our last result concerning eigenvalue problems is the following Proposition. For the proof we refer to Subsection 2.2.

PROPOSITION 1.1

Let $g_{n}$ and $g_{0}$ be $C^{4}(\bar{\Omega})$ functions, and let $(\lambda_{n,0},u_{n,0})$, $(\lambda_{0},u_{0})$ be the principal eigenpairs of (1.7) with weights $g_{n}$, $g_{0}$, respectively. Then, $u_{n,0}\to u_{0}$ in $X$ if and only if the set

[TABLE]

is compact in $X\times\mathbb{R}$. The positive real numbers $s_{*}$ and $\delta$, depend only on (1.7) with weight $g_{0}$.

Based on these results, we prove the existence of a branch of solutions, $\mathcal{C}_{\varepsilon}$, for the problem (1.4), for any $\varepsilon$ small enough, bifurcating from $\lambda_{0,\varepsilon}$. We are not in the position to exclude the second alternative of Theorem 1.1. The standard argument, based on the maximum principle, which ensures that the branches are global and the solutions belonging in these branches are positive, cannot be applied. However, we prove that if the second alternative of Theorem 1.1 holds, then $\lambda^{*}_{\varepsilon}\to\infty$, as $\varepsilon\downarrow 0$.

In Section 3, we leave $\varepsilon\downarrow 0$, in order to obtain a global branch of solutions for the problem (1.2). The main result for our model equation is the following theorem.

THEOREM 1.3

The principal eigenvalue $\lambda_{0}>0$ of the problem (1.3) is a bifurcation point of the perturbed problem (1.2). More precisely, the closure of the set

[TABLE]

possesses a maximal continuum (i.e. connected branch) of solutions, $\mathcal{C}_{0}$, such that $(\lambda_{0},0)\in\mathcal{C}_{0}$ and $\mathcal{C}_{0}$ is unbounded in $\mathbb{R}\times X$. Moreover, there are maximal connected subsets $\mathcal{C}^{+}_{0}$, $\mathcal{C}^{-}_{0}$ of $\mathcal{C}_{0}$ containing $(\lambda_{0},0)$ in their closure, such that $u(x)>(<)0$, for every $x\in\Omega$, if $(\lambda,u)\in\mathcal{C}^{+(-)}_{0}$, respectively.

This is done with the use of Whyburn’s Lemma [7]:

Let $G$ be any infinite collection of point sets. The set of all points $x$ such that every neighborhood of $x$ contains points of infinitely many sets of $G$ is called the superior limit of $G$ ($\limsup G$). The set of all points $y$ such that every neighborhood of $y$ contains points of all but a finite number of sets of $G$ is called the inferior limit of $G$ ($\liminf G$).

THEOREM 1.4

Let $\{A_{n}\}_{n\in\mathbb{N}}$ be a sequence of connected closed sets such that

[TABLE]

Then, if the set $\bigcup_{n\in\mathbb{N}}A_{n}$ is relatively compact, $\limsup_{n\to\infty}\{A_{n}\}$ is a closed, connected set.

The application of Whyburn’s Lemma in bifurcation theory is rather standard; We refer in [2] and the references therein. In our case, we prove that $\mathcal{C}_{\varepsilon}\to\mathcal{C}_{0}$ in $\mathbb{R}\times X\cap B_{R}(\lambda_{0},0)$, for any $R\in\mathbb{R}$, where $B_{R}(\lambda_{0},0)$ denotes the ball of $\mathbb{R}\times X$ with center $(\lambda_{0},0)$ and radius $R$. This means that the branches $\mathcal{C}_{\varepsilon}$ tend to $\mathcal{C}_{0}$ locally in $\mathbb{R}\times X$.

Finally we state as a theorem the existence of global branches for the general case (1.1).

THEOREM 1.5

Assume that $B$ is smooth enough, at least $C^{6}$ such that $B=o(||u||_{C^{2}(\Omega)})$, $B=o(\Delta u)$ uniformly, as $||u||_{C^{2}(\Omega)}\to 0$. Then, the principal eigenvalue $\lambda_{0}>0$ of the operator $-\Delta$ is a bifurcation point of the problem (1.1). More precisely, the closure of the set

[TABLE]

possesses a maximal continuum (i.e. connected branch) of solutions, $\mathcal{C}_{0}$, such that $(\lambda_{0},0)\in\mathcal{C}_{0}$ and $\mathcal{C}_{0}$ is unbounded in $\mathbb{R}\times X$. Moreover, there are maximal connected subsets $\mathcal{C}^{+}_{0}$, $\mathcal{C}^{-}_{0}$ of $\mathcal{C}_{0}$ containing $(\lambda_{0},0)$ in their closure, such that $u(x)>(<)0$, for every $x\in\Omega$, if $(\lambda,u)\in\mathcal{C}^{+(-)}_{0}$, respectively.

The proof of the above Theorem is given in Section 4.

The only work, up to our knowledge, obtained bifurcation results for fully nonlinear elliptic problems, was [3] and based on this existence result, the authors in [4] proved positivity. In both works the uniform ellipticity condition is assumed.

Notation Throughout this work we will consider the Sobolev space $H^{1}_{0}(\Omega)$ and the weighted space $L^{2}_{g}(\Omega)$, with inner products

[TABLE]

and

[TABLE]

for any $\phi,\;\psi\in C^{\infty}_{0}(\Omega)$, respectively.

2 Eigenvalue problems

Standard regularity results imply the following:

LEMMA 2.1

Let $u\in X$, be a solution of the problem (1.4). Then, $u$ is at least a $C^{6}(\bar{\Omega})$-function.

Another well known result concerns the eigenvalue problems (1.3) and (1.7):

THEOREM 2.1

Problems (1.3) and (1.7) admit principal eigenvalues $\lambda_{0}$ and $\lambda_{n,0}$, respectively, given by

[TABLE]

which are simple and the corresponding normalized eigenvalues $u_{0}$ and $u_{n,0}$, respectively, belong to $H^{1}_{0}(\Omega)$ and they are the only positive eigenfunctions of (1.3) and (1.7). Since $\partial\Omega\in C^{6}$ and $g_{n}\in C^{4}(\bar{\Omega})$ we have that $u_{0}$ and $u_{n,0}$ belong also to $X$.

Next we consider the eigenvalue problem

[TABLE]

for $u$, $v$ in $X$. We denote the eigenpairs of (2.2) as $(\tilde{\lambda}^{(i)}_{n},\tilde{u}^{(i)}_{n})$, $i=1,2,...$. Standard spectral theory for compact self-adjoint operators imply the existence of these eigenpairs in $\mathbb{R}\times X$ and state their properties. In the sequel, we assume that $\{\tilde{u}^{(i)}_{n}\}$ consists an orthonormal basis of $L^{2}_{g_{n}}(\Omega)$.

2.1 Proof of Theorem 1.2

We give the proof for the more general case of the problem (1.6). Throughout, this Subsection we assume that $n$, thus $g_{n}$, is fixed.

LEMMA 2.2

Assume that there exist $(v_{n,\varepsilon},\mu_{n,\varepsilon})$ eigenpairs of (1.6) in $X$, i.e.

[TABLE]

for any $\phi\in X$, such that $\mu_{n,\varepsilon}\to\lambda_{n,0}$, as $\varepsilon\to 0$. Then, $v_{n,\varepsilon}\to u_{n,0}$, in $X$.

Proof Without loss of generality we assume that $v_{n,\varepsilon}$ are normalized in $X$. Observe that $v_{n,\varepsilon}\in H^{1}_{0}(\Omega)$, so we may decompose it as

[TABLE]

where $\xi_{n,\varepsilon}\perp u_{n,0}$ in $H^{1}_{0}(\Omega)$, hence $\xi_{n,\varepsilon}\perp u_{n,0}$ also in $L^{2}_{g_{n}}(\Omega)$. From (2.3), we have that $\xi_{n,\varepsilon}$ satisfy

[TABLE]

for any $\phi\in X$. Setting $\phi=\xi_{n,\varepsilon}$, we get that

[TABLE]

or

[TABLE]

Observe that $a_{n,\varepsilon}$ is a bounded sequence in $\mathbb{R}$, since

[TABLE]

Also, $\xi_{n,\varepsilon}=v_{n,\varepsilon}-a_{n,\varepsilon}u_{n,0}$ implies that $\xi_{n,\varepsilon}$ is also bounded in $X$. Moreover, since $\xi_{n,\varepsilon}\perp u_{n,0}$ in $H^{1}_{0}(\Omega)$ and $\mu_{n,\varepsilon}$ remain close enough to $\lambda_{n,0}$, we have

[TABLE]

where $\lambda^{(2)}_{n,0}$ denotes the second eigenvalue of (1.7). Assume now that $\xi_{n,\varepsilon}$ converge weakly to some $\xi_{*}$, in $X$, as $\varepsilon\downarrow 0$. If $\xi_{*}\not\equiv 0$, then from (2.6) and (2.8,) we obtain that

[TABLE]

as $\varepsilon\downarrow 0$, which is a contradiction. Then $\xi_{*}\equiv 0$ and in this case, from (2.7), we have that

[TABLE]

Thus, $\xi_{n,\varepsilon}$ is strongly convergent to zero in $X$. Finally, we conclude that $v_{n,\varepsilon}\to u_{n,0}$, in $X$. $\blacksquare$

We do the following observation: Assume that $u_{n,0}\equiv\tilde{u}^{i}_{n}$, where $\tilde{u}^{i}_{n}$ is an eigenfunction of (2.2), associated with some eigenvalue $\tilde{\lambda}^{i}_{n}$. We are not in the position to exclude this case, but if this happens can be characterized as exceptional and should hold for special weights $g_{n}$. In this case, the curve of eigenpairs $\phi(\varepsilon)$ is given directly by $\phi(\varepsilon)=\{(\lambda_{n\varepsilon},u_{n,\varepsilon})=(\lambda_{0,n}+\varepsilon\,\tilde{\lambda}^{(i)}_{n},\tilde{u}^{(i)}_{n})\}$, for every $\varepsilon$. The properties of this curve, are given by Theorem 1.2, and may be proved by a similar way (see Remark 2.1).

Thus, we will consider the general case; $u_{n,0}$ cannot coincide to any $\tilde{u}^{i}_{n}$. Assume that

[TABLE]

for some fixed $i$ and $\kappa_{n,0}$ denotes the ratio

[TABLE]

since $u_{n,0}$ is normalized in $X$. From (2.9) we have that $u_{n,0}$ does not belong to $span\{\tilde{u}^{(1)}_{n},...\tilde{u}^{(i)}_{n}\}$; if we assume the opposite then

[TABLE]

thus $\kappa_{n,0}\leq\tilde{\lambda}^{(i)}_{n}$, which is a contradiction. From (2.9) we also have that $u_{n,0}$ is not orthogonal to $span\{\tilde{u}^{(1)}_{n},...\tilde{u}^{(i)}_{n}\}$; if the opposite holds then $\kappa_{n,0}\geq\tilde{\lambda}^{(i+1)}_{n}$, which is also a contradiction.

We note that there exists functions which are orthogonal to $u_{n,0}$ in $H_{0}^{1}(\Omega)$, but cannot be orthogonal to $u_{n,0}$ in $X$. The only case that this may happen is when $u_{n,0}$ coincides with an eigenfunction of (2.2), i.e., in the exceptional case. To see this, assume the opposite; all functions orthogonal to $u_{n,0}$ in $H_{0}^{1}(\Omega)$ are orthogonal also in $X$. Then for some eigenfunction $\tilde{u}^{(i)}_{n}$ we have that

[TABLE]

where $a\neq 0$ and $\zeta$ is orthogonal to $u_{n,0}$ in both $X$ and $H_{0}^{1}(\Omega)$. From (2.2) we have that

[TABLE]

for any $\phi\in X$. Setting now $\phi=u_{n,0}$, we obtain that $u_{n,0}$ must satisfy

[TABLE]

which implies $u_{n,0}\equiv\tilde{u}^{(i)}_{n}$. From the point of view of (2.9) this is a contradiction.

Proof of Theorem 1.2 We give the proof for the general case where (2.9) holds. Assume that $n$ and $\varepsilon$ are fixed, such that $\varepsilon$ is small enough. We proceed with the proof in four steps.

Existence For every $\varepsilon>0$ small enough, assume the problem (1.5). The compactness properties of the space $X$ imply, that is a well defined eigenvalue problem, and the set of the eigenvalues consists an orthonormal basis of $L^{2}_{g_{n}}(\Omega)$. Moreover, its principal eigenvalue satisfies the variational characterization

[TABLE]

Moreover,

[TABLE]

and

[TABLE]

Since $u_{n,0}$ is normalized in $X$, we obtain that

[TABLE]

This implies that $\lambda_{n,\varepsilon}$ is decreasing, and $\lambda_{n,\varepsilon}\to\lambda_{n,0}$, as $\varepsilon\to 0$. Then, Lemma (2.2) implies that $u_{n,\varepsilon}\to u_{n,0}$, in $X$, as $\varepsilon\downarrow 0$.

As a conclusion we get that, for any $n$ and any $0<\varepsilon$ small enough there exists an eigenpair $(\lambda_{n,\varepsilon},u_{n,\varepsilon})$ of (1.6), which form as $\varepsilon\downarrow 0$, a set of eigenpairs $\phi_{n}(\varepsilon)$ with the eigenfunctions been positive and normalized. Moreover, $\lambda_{n,\varepsilon}$ and $u_{n,\varepsilon}$ maybe written as

[TABLE]

where

[TABLE]

and

[TABLE]

for some $\kappa_{n,\varepsilon}$.

Uniqueness For any $n$ fixed, assume that there are $(\mu_{n,\varepsilon},v_{n,\varepsilon})$ eigenpairs of (1.6), $\mu_{n,\varepsilon}\to\lambda_{n,0}$, $\varepsilon\in[0,s)$, for some $s$ small enough. Assume also that $\mu_{n,\varepsilon}\neq\lambda_{n,\varepsilon}$ and that $v_{n,\varepsilon}$ are normalized in $X$. Lemma (2.2) implies that $v_{n,\varepsilon}\to u_{n,0}$, in $X$, as $\varepsilon\downarrow 0$. Then,

[TABLE]

where $(\lambda_{n,\varepsilon},u_{n,\varepsilon})$ is the eigenpair corresponding to the same $\varepsilon$. Thus, $v_{n,\varepsilon}$ and $u_{n,\varepsilon}$ should be orthogonal in $L^{2}_{g_{n}}(\Omega)$ which is a contradiction, since they both converge to $u_{n,0}$ in $X$, thus also in $L^{2}_{g_{n}}(\Omega)$. As a conclusion we have the uniqueness of $\phi_{n}(\varepsilon)$.

Simplicity For any $n$ fixed, assume that $(\lambda_{n,\varepsilon},v_{n,\varepsilon})$ is an eigenpair of (1.6). We assume that $v_{n,\varepsilon}$ is normalized in $X$. Then, Lemma (2.2) implies that $v_{n,\varepsilon}\to u_{n,0}$, in $X$, as $\varepsilon\downarrow 0$. This is also a contradiction, since $v_{n,\varepsilon}$ and $u_{n,\varepsilon}$ should be orthogonal in $L^{2}_{g_{n}}(\Omega)$.

Algebraic Simplicity Consider the operator $L_{n,\varepsilon}(\varepsilon,u):(0,s)\times X\to X$, defined by

[TABLE]

for any $\phi\in X$. From the above step (simplicity) we have that the null space of $L_{n,\varepsilon}$ is $[u_{n,\varepsilon}]$. It suffices to prove that the range of $L_{n,\varepsilon}$ is $E_{n,\varepsilon}$, the orthogonal complement of ${u_{n,\varepsilon}}$ in $X$. Assume the opposite; Let $v_{n,\varepsilon}\in E_{n,\varepsilon}$, such that $L_{n,\varepsilon}(\varepsilon,v_{n,\varepsilon})=u_{n,\varepsilon}$, i.e.,

[TABLE]

for any $\phi\in V$. However, this is a contradiction if we set $\phi=u_{n,\varepsilon}$. Thus the proof is completed. $\blacksquare$

LEMMA 2.3

For any fixed $n$, the curve of eigenpairs $\phi_{n}$ is continuous in $\mathbb{R}\times V$.

Proof Let $\varepsilon\to\varepsilon_{*}$. If $\varepsilon_{*}=0$, from (2.11), we get that $\lambda_{n,\varepsilon}\to\lambda_{n,0}$ and from Lemma 2.2 we have that $u_{n,\varepsilon}\to u_{n,0}$ in $X$. Assume that $\varepsilon_{*}\neq 0$. Let $\lambda_{n,\varepsilon}\to\lambda_{*}$ in $\mathbb{R}$ and $u_{n,\varepsilon}\to u_{*}$ in $X$. Taking the limit, as $\varepsilon\to\varepsilon_{*}$, in

[TABLE]

we conclude that $(\lambda_{*},u_{*})$ is an eigepair corresponding to $\varepsilon_{*}$. From Theorem 1.2, $(\lambda_{*},u_{*})\in\phi_{n}$ and the proof is completed. $\blacksquare$

REMARK 2.1

Assume the exceptional case where $u_{n,0}\equiv\tilde{u}^{i}_{n}$, where $\tilde{u}^{i}_{n}$ is an eigenfunction of (2.2), associated with some eigenvalue $\tilde{\lambda}^{i}_{n}$. Then, the curve of eigenfunctions $\phi(\varepsilon)=\{(\lambda_{n\varepsilon},u_{n,\varepsilon})=(\lambda_{0,n}+\varepsilon\,\tilde{\lambda}^{(i)}_{n},\tilde{u}^{(i)}_{n})\}$, for every $\varepsilon$, has the properties of Theorem 1.2. More precisely, uniqueness maybe proved exactly as above and simplicity is implied from the simplicity of $u_{n,0}$.

2.2 Proof of Proposition 1.1

We remind that if $\varphi_{n}(\varepsilon)=(\lambda_{n,\varepsilon},u_{n,\varepsilon})$, then $(\lambda_{n,\varepsilon},u_{n,\varepsilon})$ satisfies (2.16). For $\varepsilon=0$, $(\lambda_{n,0},u_{n,0})$ are the principal eigenpairs of (1.7) with weights $g_{n}$ and $(\lambda_{0},u_{0})$ denotes the principal eigenpair of (1.7) with weight $g_{0}$.

Next result concerns the asymptotic behaviour of $\phi_{n}$ as $u_{n,0}\to u_{0}$ in $X$. In this case, $g_{n}\to g_{0}$ in $C^{1}(\bar{\Omega})$, $\lambda_{n,0}\to\lambda_{0}$ and $\lambda^{(2)}_{n,0}\to\lambda^{(2)}_{0}$, where $\lambda^{(2)}_{n,0}$ and $\lambda^{(2)}_{0}$ are the second eigenvalues of (1.7), with weights $g_{n}$ and $g_{0}$, respectively.

Proof of Proposition 1.1 Without loss of generalization, we assume that $u_{n,0}$, $n\geq 1$, cannot coincide to any $\tilde{u}^{i}_{n}$; if there where a subsequence of $u_{n,0}$ such that $u_{n,0}=\tilde{u}^{i}_{n}=u_{0}$, for some $i$, then the corresponding curves are $\phi(\varepsilon)=\{(\lambda_{n\varepsilon},u_{n,\varepsilon})=(\lambda_{0,n}+\varepsilon\,\tilde{\lambda}^{(i)}_{n},\tilde{u}^{(i)}_{n})\}$ and the result follows. However, we do not exclude $u_{0}$ to be equal with any eigenfunction of (2.2).

In what follows leaving $n\to\infty$ means that $u_{n,0}\to u_{0}$ in $X$. Let $\delta$ be a positive number, such that

[TABLE]

The convergence of $(u_{n,0},\lambda_{n,0})\to(u_{0},\lambda_{0})$, in $X\times\mathbb{R}$, implies that there exists $N(\delta)\in\mathbb{N}$, such that

[TABLE]

for any $n>N(\delta)$. We introduce the following quantity, depending only on $(\lambda_{0},u_{0})$:

[TABLE]

Let $n>N(\delta)$ and $\varepsilon\in[0,s_{*}]$. Then,

[TABLE]

Next, for any $n$, we decompose $u_{n,\varepsilon}$ as in (2.12), (2.13). Decomposition (2.12) implies that

[TABLE]

for every $(n,\varepsilon)$, $0\leq\varepsilon\leq s_{*}$. Next we decompose $\eta_{n,\varepsilon}$ as

[TABLE]

where

[TABLE]

Observe that $<\xi_{n,\varepsilon},u_{n,0}>_{H_{0}^{1}(\Omega)}=0$ implies also $<\xi_{n,\varepsilon},u_{n,0}>_{L^{2}_{g_{n}}(\Omega)}=0$. Moreover, from (2.12) and (2.21), $u_{n,\varepsilon}$ maybe written as

[TABLE]

This equality and (2.12) imply that

[TABLE]

so $\beta_{n,\varepsilon}$ satisfies also

[TABLE]

Using (2.12), (2.16) implies

[TABLE]

since

[TABLE]

for any $\phi\in X$. Setting $\phi=u_{n,0}$ in (2.2) and using (2.12), (2.26) we obtain that

[TABLE]

since $u_{n,0}$ is normalized in $X$. Finally, using (2.22) and setting

[TABLE]

we conclude that

[TABLE]

where $\kappa_{n,0}$ is given by (2.10). Next we prove some estimates for $\kappa_{n,\varepsilon}$, $\alpha_{n,\varepsilon}$ and $\beta_{n,\varepsilon}$. Holds that

[TABLE]

Assume that $\kappa_{n,\varepsilon}\leq 0$, for some $(n,\varepsilon)$. Then,

[TABLE]

which is a contradiction to the variational characterization (see (2.1)) of $\lambda_{n,0}$. Thus, $\kappa_{n,\varepsilon}$ remains positive for $\varepsilon\neq 0$. For $\alpha_{n,\varepsilon}$ we have that, for $\varepsilon\neq 0$, $\alpha_{n,\varepsilon}\neq 0$, since $\kappa_{n,\varepsilon}\neq 0$. Moreover, for any fixed $n$, $\alpha_{n,\varepsilon}\to 1$, as $\varepsilon\to 0$ and from (2.13) and Lemma 2.3 we have that $\alpha_{n,\varepsilon}$ is continuous. Hence,

[TABLE]

Observe now that for any fixed $n$, $\beta_{n,\varepsilon}\to 0$, as $\varepsilon\to 0$. Then, $\kappa_{n,\varepsilon}\to\kappa_{n,0}$, as $\varepsilon\to 0$.

Next, we consider $\beta_{n,\varepsilon}$. First, we prove that, for $\varepsilon\neq 0$, $\beta_{n,\varepsilon}=0$, if and only if $u_{n,\varepsilon}=u_{n,0}=\tilde{u}^{(i)}_{n}$. Let $u_{n,\varepsilon}=u_{n,0}=\tilde{u}^{(i)}_{n}$. From (2.23) we get that $\xi_{n,\varepsilon}=0$, which from (2.24) implies that $\beta_{n,\varepsilon}=0$. Let $\beta_{n,\varepsilon}=0$. From (2.21) we have that $\eta_{n,\varepsilon}\equiv\xi_{n,\varepsilon}$. Setting $\phi=\xi_{n,\varepsilon}$ in (2.2) we get that

[TABLE]

However, the same argument as in Lemma 2.2 (see (2.8)), imply that

[TABLE]

where $\lambda^{(2)}_{n,0}$ is the second eigenvalue of (1.7). Then, (2.31) holds only if $\eta_{n,\varepsilon}\equiv\xi_{n,\varepsilon}\equiv 0$, i.e., $u_{n,\varepsilon}=u_{n,0}=\tilde{u}^{(i)}_{n}$. Thus, $\beta_{n,\varepsilon}=0$, for $\varepsilon\neq 0$, is equivalent to the exceptional case $u_{n,\varepsilon}=u_{n,0}=\tilde{u}^{(i)}_{n}$, which is a contradiction to our assumption. So, $\beta_{n,\varepsilon}\neq 0$, for every $n>N(\delta)$ and $0<\varepsilon\leq s_{*}$.

We prove now that $\beta_{n,\varepsilon}>0$, for every $n>N(\delta)$ and $0<\varepsilon\leq s_{*}$. Using decomposition (2.23) in (2.16), setting also $\phi=\xi_{n,\varepsilon}$, we obtain

[TABLE]

or

[TABLE]

Using now (2.24) we get that

[TABLE]

Observe that $\alpha_{n,\varepsilon}+\beta_{n,\varepsilon}\neq 0$, since if $\alpha_{n,\varepsilon}+\beta_{n,\varepsilon}=0$, (2.23) imply that $u_{n,\varepsilon}=\xi_{n,\varepsilon}$. This is a contradiction to Lemma 2.2. The positivity of $\alpha_{n,\varepsilon}$ and $\kappa_{n,\varepsilon}$, see (2.28), imply that $\alpha_{n,\varepsilon}+\beta_{n,\varepsilon}$ is also positive. Then, from (2.32) and (2.33) we conclude that $\beta_{n,\varepsilon}$ is positive, for any fixed $n$ and $\varepsilon$ small enough. Note that for fixed $n$, from (2.22) and Lemma 2.3 we have that $\beta_{n,\varepsilon}$ is continuous. Hence,

[TABLE]

for every $n>N(\delta)$ and $0\leq\varepsilon\leq s_{*}$. For positive $\beta_{n,\varepsilon}$, we also get that

[TABLE]

and finally, from (2.27), (2.35) and (2.19) we obtain the uniform bound for $\lambda_{n,\varepsilon}$

[TABLE]

for every $n>N(\delta)$ and $0\leq\varepsilon\leq s_{*}$. Our last estimate concerns $\eta_{n,\varepsilon}$ in terms of $\alpha_{n,\varepsilon}$ and $\beta_{n,\varepsilon}$. Setting $\phi=\eta_{n,\varepsilon}$ in (2.2), we have

[TABLE]

Using (2.27), (2.28) and (2.22) we get that

[TABLE]

Using decomposition (2.21) and (2.27)-(2.28) we have that

[TABLE]

Then (2.37) is written as

[TABLE]

or

[TABLE]

Observe now that $\xi_{n,\varepsilon}$ is orthogonal to $u_{n,0}$ in $H_{0}^{1}(\Omega)$ and that the estimate (2.36) is uniform, i.e., holds for any $n>N(\delta)$ and $0\leq\varepsilon\leq s_{*}$, thus

[TABLE]

Finally, from (2.38) we conclude that

[TABLE]

for any $n>N(\delta)$ and $0\leq\varepsilon\leq s_{*}$.

The final step is to prove that every sequence $(u_{n,\varepsilon},\lambda_{n,\varepsilon})\in\phi_{n}$, converges (up to some subsequence) to some $(u_{0,\varepsilon},\lambda_{0,\varepsilon})\in\phi_{0}$, for every $n>N(\delta)$ and every $0\leq\varepsilon\leq s_{*}$, in $X\times\mathbb{R}$. Let $(u_{n,\varepsilon},\lambda_{n,\varepsilon})\in\phi_{n}$. Holds that $u_{n,\varepsilon}$ are normalized, hence bounded and $\lambda_{n,\varepsilon}$ is also bounded (see (2.36)). Then, up to some subsequence, $u_{n,\varepsilon}\rightharpoonup u_{*}$ in $X$ and $\lambda_{n,\varepsilon}\to\lambda_{*}$.

Assume that $\varepsilon\to\varepsilon_{*}$, for some $\varepsilon_{*}\neq 0$. From (2.16) we have that

[TABLE]

or

[TABLE]

since $u_{n,\varepsilon}$ are normalized in $X$. Taking the limit $n\to\infty$ and $\varepsilon\to\varepsilon_{*}$, we get that

[TABLE]

Thus, $(u_{*},\lambda_{*})\neq(0,0)$. Moreover, (2.16) implies that

[TABLE]

for any $\phi\in X$, thus $(u_{*},\lambda_{*})$ is an eigenpair belonging to $\phi_{0}$ and that $(u_{n,\varepsilon},\lambda_{n,\varepsilon})\to(u_{*},\lambda_{*})$ in $X\times\mathbb{R}$.

Assume now that $\varepsilon_{*}=0$. We will prove that $(u_{n,\varepsilon},\lambda_{n,\varepsilon})\to(u_{0},\lambda_{0})$ in $X\times\mathbb{R}$. Assume that $u_{*}\equiv 0$, then

[TABLE]

and (2.39) implies that also

[TABLE]

However, these two limits contradict (2.20). Thus, $u_{*}\not\equiv 0$ and follows directly that $(u_{*},\lambda_{*})\equiv(u_{0},\lambda_{0})$ and since both are normalized in $X$ we deduce that $u_{n,\varepsilon}\to u_{0}$, in $X$.

Finally we conclude that, for every $n>N(\delta)$ and $0\leq\varepsilon\leq s_{*}$, every sequence $(u_{n,\varepsilon},\lambda_{n,\varepsilon})\in\phi_{n}$, has a convergent subsequence to some $(u_{0,\varepsilon},\lambda_{0,\varepsilon})\in\phi_{0}$, in $X\times\mathbb{R}$, i.e., the set $\Phi$ is compact in $X\times\mathbb{R}$.

If now $\Phi$ is compact in $X\times\mathbb{R}$, then it is immediate that $(u_{n,0},\lambda_{n,0})$ tends to $(u_{0},\lambda_{0})$ in $X\times\mathbb{R}$ and the proof is completed. $\blacksquare$

3 Bifurcation Results

In this section, we will prove the existence of a global branch of solutions for the problem (1.2) bifurcating from the principal eigenvalue of the linear problem (1.3). To do this, we will first prove the existence of global branches for the problems (1.4) bifurcating from eigenvalues of the linear problem (1.5). Then we will leave $\varepsilon\to 0$.

From Theorem 1.2 we have the existence of a continuous curve, $\phi_{0}(\varepsilon)=(u_{0,\varepsilon},\lambda_{0,\varepsilon})$, of eigenpairs for the corresponding perturbed problems (1.5).

THEOREM 3.1

Let $\varepsilon$ be a positive real number which is small enough. Then, for each $\varepsilon$, the eigenvalue $\lambda_{0,\varepsilon}$ of the problem (1.5) is a bifurcation point of the perturbed problem (1.4). More precisely, the closure of the set

[TABLE]

possesses a maximal continuum (i.e. connected branch) of solutions, $\mathcal{C}_{\varepsilon}$, such that $(\lambda_{0,\varepsilon},0)\in\mathcal{C}_{\varepsilon}$ and $\mathcal{C}_{\varepsilon}$ either:

(i) meets infinity in $\mathbb{R}\times X$ or,

(ii) meets $(\lambda^{*},0)$, where $\lambda^{*}\neq\lambda$ is also an eigenvalue of $L$.

Proof The proof follows directly from Rabinowitz’s Theorem 1.1. We sketch the proof. For any fixed $\varepsilon>0$, small enough, we consider the operator $G:X\to X$, as

[TABLE]

Observe that $G$ may be written as $G(\lambda,u)=\frac{1}{\varepsilon}L(\lambda,u)+\frac{1}{\varepsilon}H(\lambda,u)$, where $L$ is the linear and compact operator

[TABLE]

while $H$ is the compact nonlinear operator

[TABLE]

satisfying $H(\lambda,u)=o(||u||_{X})$, as $||u||_{X}\to 0$; let $||u_{n}||_{X}\to 0$, then $||u_{n}||_{C^{3}(\bar{\Omega})}\to 0$ and

[TABLE]

for any $v\in X$ and $\tilde{u}_{n}=u_{n}/||u_{n}||_{X}$. In addition, from Theorem 1.2, $\lambda_{0,\varepsilon}$ is a simple eigenvalue of $\frac{1}{\varepsilon}L$. It is clear that the conditions of Rabinowitz’s Theorem are satisfied and the result follows. $\blacksquare$.

Next we prove some results, concerning the asymptotic behavior of the branches $C_{\varepsilon}$. Assume a sequence $(\lambda_{\varepsilon_{n}},u_{\varepsilon_{n}})\in\mathcal{C}_{\varepsilon_{n}}$, and set

[TABLE]

We consider the following eigenvalue problems with weight $g_{\varepsilon_{n}}$:

[TABLE]

where $\psi\in X$. From Lemma 2.1 we have that $u_{\varepsilon_{n}}\in C^{6}(\bar{\Omega})$, so $g_{\varepsilon_{n}}\in C^{4}(\bar{\Omega})$. Then, Theorem 2.1 implies the existence of a principal eigenvalue $\lambda_{n}$ with the associated normalized eigenfunction $v_{n}$ being positive and belonging in $X$. We denote by $g_{0}$ the ”weight” function of (1.3), i.e., $g_{0}\equiv 1$.

LEMMA 3.1

Let $(\lambda_{\varepsilon_{n}},u_{\varepsilon_{n}})\in\mathcal{C}_{\varepsilon_{n}}$ and assume that $u_{\varepsilon_{n}}$ is a bounded sequence in $X$, such that $u_{\varepsilon_{n}}\rightharpoonup 0$ in $X$. Then, $v_{n}\to u_{0}$ in $X$.

Proof Since $u_{\varepsilon_{n}}\rightharpoonup 0$ in $X$, the compact imbedding $X\hookrightarrow C^{4}(\bar{\Omega})$ implies that $u_{\varepsilon_{n}}\to 0$, in $C^{4}(\bar{\Omega})$, hence $g_{\varepsilon_{n}}\to g_{0}$ in $C^{2}(\bar{\Omega})$ and $v_{n}\to u_{0}$ in $C^{2}(\bar{\Omega})$. Moreover, the derivatives of order 5 and 6 of $u_{\varepsilon_{n}}$ tend to zero in $L^{2}(\Omega)$ and the result follows. $\blacksquare$

LEMMA 3.2

Let $\varepsilon_{n}>0$, be a sequence that tends to [math], as $n\to\infty$, and $(\lambda_{\varepsilon_{n}},u_{\varepsilon_{n}})\in\mathcal{C}_{\varepsilon_{n}}$. Assume that $(\lambda_{\varepsilon_{n}},u_{\varepsilon_{n}})$ is a bounded sequence in $\mathbb{R}\times X$, such that $u_{\varepsilon_{n}}$ being nonnegative. Denote by $\lambda_{*}$ and $u_{*}$ the limits (up to a subsequence) $\lambda_{\varepsilon_{n}}\to\lambda^{*}$ and $u_{\varepsilon_{n}}\rightharpoonup u_{*}$ (weakly) in $X$. Then, $u_{\varepsilon_{n}}$ converges strongly in $u_{*}$ in $X$, and either

$(\lambda_{*},u_{*})$* is a nontrivial solution of (1.2), or* 2. 2.

$u_{*}\equiv 0$. In this case, $\lambda_{*}=\lambda_{0}$ and $u_{\varepsilon_{n}}\to 0$ in $X$, such that the normalization of $u_{\varepsilon_{n}}$ converges to $u_{0}$ in $X$.

Proof Since $(\lambda_{\varepsilon_{n}},u_{\varepsilon_{n}})\in\mathcal{C}_{\varepsilon_{n}}$, they satisfy

[TABLE]

for each $n$ and every $\psi\in X$, where $g_{\varepsilon_{n}}(x)=1+|\det D^{2}u_{\varepsilon_{n}}|^{2}$. Since $u_{\varepsilon_{n}}$ is a bounded sequence in $X$, the compact imbedding $X\hookrightarrow C^{3}(\bar{\Omega})$ implies that

[TABLE]

up to a subsequence, denoted again by $u_{\varepsilon_{n}}$.

Assume that, $u_{*}\equiv 0$. Then, we consider the eigenvalue problems (3.2). Since, for each $n$, $g_{\varepsilon_{n}}\in C^{1}(\bar{\Omega})$, from Theorem 1.2 we have the existence of a curve, $\phi_{n}$, of eigenpairs for the corresponding to $g_{\varepsilon_{n}}$ perturbed problems:

[TABLE]

However, $u_{\varepsilon_{n}}\rightharpoonup 0$ in $X$, so Lemma 3.1 implies that $v_{n}\to u_{0}$ in $X$.. Then, Proposition 1.1 implies that there exists an interval $[0,s_{*}]$, depending on $u_{0}$, such that the set $\Phi$ is compact in $\mathbb{R}\times X$.

Choose now, $\varepsilon_{n}$ small enough, such that $\varepsilon_{n}\in[0,s_{*}]$ and denote by $\bar{u}_{\varepsilon_{n}}$ the normalization of $u_{\varepsilon_{n}}$ in $X$. Then, dividing (3.3) by $||u_{\varepsilon_{n}}||_{X}$ we get that $\bar{u}_{\varepsilon_{n}}$ is a solution of (3.5) with $\varepsilon_{n}\in[0,s_{*}]$. We claim that $(\lambda_{\varepsilon_{n}},\bar{u}_{\varepsilon_{n}})\in\phi_{n}$, for each $n$ big enough. Assume the opposite; then problem (3.5) for the same $\varepsilon_{n}$ and $g_{\varepsilon_{n}}$, admits two positive eigenfunctions corresponding to different eigenvalues. These eigenvalues should be orthogonal in $L^{2}_{g_{\varepsilon_{n}}}(\Omega)$, which is impossible, since $g_{\varepsilon_{n}}$ is positive. (See Remark 3.1). Thus, $(\lambda_{\varepsilon_{n}},\bar{u}_{\varepsilon_{n}})$ must belong to $\phi_{n}$, for each $n$ big enough. This means that $\lambda_{*}=\lambda_{0}$ and $\bar{u}_{\varepsilon_{n}}\to u_{0}$ (strongly) in $X$. However, this is true if and only if $||u_{\varepsilon_{n}}||_{X}\to 0$. Thus, $u_{\varepsilon_{n}}$ converges strongly to [math] in $X$. This proves the second alternative of the Lemma.

Assume now that $u_{*}\not\equiv 0$. Taking the limit as $n\to\infty$, (3.3) implies that

[TABLE]

where $g_{*}(x)=1+|\det D^{2}u_{*}|^{2}$. However, $u_{*}\in X$ and is nonnegative. Thus, is the unique positive eigenfunction of the above eigenvalue problem. Suppose now that $\liminf||u_{\varepsilon_{n}}||_{X}>||u_{*}||_{X}$, then there exists a subsequence of $u_{\varepsilon_{n}}$, again denote it by $u_{\varepsilon_{n}}$, such that $||u_{\varepsilon_{n}}||_{X}\to M>||u_{*}||_{X}$. Set $\tilde{u}_{\varepsilon_{n}}=u_{\varepsilon_{n}}/||u_{\varepsilon_{n}}||_{X}$. Is clear that $\tilde{u}_{\varepsilon_{n}}$ converges weakly to some $\tilde{u}_{*}$ in $X$. Then, diving (3.3) by $||u_{\varepsilon_{n}}||_{X}$ and taking the limit, we obtain that $\tilde{u}_{*}$ is also a nonnegative eigenfunction of (3.6), which is impossible. Thus, $\liminf||u_{\varepsilon_{n}}||_{X}=||u_{*}||_{X}$ and $u_{\varepsilon_{n}}\to u_{*}$, strongly, in $X$. This proves the first alternative of the Lemma. Thus, the proof is completed. $\blacksquare$

REMARK 3.1

In the proof of the second alternative, we used that $g_{\varepsilon_{n}}$ is positive. However, this is not something crucial. Above we give a different proof for the second alternative; Assume that $u_{*}\equiv 0$. We will prove that $\lambda_{*}=\lambda_{0}$, thus $(\lambda_{\varepsilon_{n}},\bar{u}_{\varepsilon_{n}})\in\phi_{n}$, (using the notation of the above proof).

Dividing (3.3) by $||u_{\varepsilon_{n}}||^{2}_{H^{1}_{0}(\Omega)}$ and setting

[TABLE]

we obtain that

[TABLE]

Since $\tilde{u}_{\varepsilon_{n}}$ are normalized in $H^{1}_{0}(\Omega)$, they converge weakly in $H^{1}_{0}(\Omega)$ to some function $\tilde{u}_{*}$. The compact imbedding $H^{1}_{0}(\Omega)\hookrightarrow L_{g_{\varepsilon_{n}}}^{2}(\Omega)$ implies that $\tilde{u}_{\varepsilon_{n}}\to\tilde{u}_{*}$, in $L_{g_{\varepsilon_{n}}}^{2}(\Omega)$. Taking the limit as $\varepsilon_{n}\to 0$ in (3.7) we get

[TABLE]

where

[TABLE]

From (3.8) we have that $\tilde{u}_{*}\not\equiv 0$ and the limit of $A_{\varepsilon}$ is finite. Since

[TABLE]

we have that $\varepsilon_{n}\tilde{u}_{\varepsilon_{n}}$ converges in $X$ and this limit must be zero. Then, for any $\phi\in X$, we have that

[TABLE]

which in the limit gives

[TABLE]

However, this makes sense if and only if $(\tilde{u}_{*},\lambda_{*})=(\tilde{u}_{0},\lambda_{0})$, where $\tilde{u}_{0}$ is the normalized in $H^{1}_{0}(\Omega)$ eigenfunction of (1.7) corresponding to the principal eigenvalue $\lambda_{0}$. Thus, $(\lambda_{\varepsilon_{n}},\bar{u}_{\varepsilon_{n}})\in\phi_{n}$ and the rest of the proof follows exactly as in the proof of the above Lemma.

In the next result, we actually prove that for $\varepsilon\to 0$ the nonpositive solutions $(\lambda,u)$ belonging to the branches $\mathcal{C}_{\varepsilon}$, must tend to infinity in $\mathbb{R}\times X$.

LEMMA 3.3

Let $(\lambda_{n},u_{n})$ be a sequence belonging to $C_{\varepsilon_{n}}$, such that $u_{n}(x)\geq 0$ and there exist $\{a^{i}_{n}\}\subset\Omega$, with $u_{n}(a^{i}_{n})=0$. Then, $(\lambda_{n},u_{n})$ is unbounded in $\mathbb{R}\times X$.

Proof Assume the opposite i.e., there exists $M>0$, such that $||(\lambda_{n},u_{n})||_{\mathbb{R}\times X}<M$, for any $n\in\mathbb{N}$. Then, $\lambda_{n}\to\lambda_{*}$ and $u_{n}\rightharpoonup u_{*}$, up to some subsequence. From Lemma 3.2, we have that if $u_{*}\equiv 0$, the normalization of $u_{n}$ must converge to $u_{0}$, in $X$ and thus in $L^{\infty}$, which contradicts the positivity of $u_{0}$. If on the other, $u_{*}\not\equiv 0$, we get that $u_{n}\to u_{*}$, strongly in $X$ and $u_{*}$ is a solution of (1.7), with $g(x)=1+|\det D^{2}u_{*}|^{2}$. The contradiction now follows from maximum principle. $\blacksquare$

LEMMA 3.4

Let $\varepsilon_{n}\to 0$. Then, the branches $\mathcal{C}_{\varepsilon_{n}}$ cannot be uniformly bounded, i.e., we cannot find $M>0$ such that $||(\lambda,u)||_{\mathbb{R}\times X}<M$, for any $(\lambda,u)\in\mathcal{C}_{\varepsilon_{n}}$.

Proof Assume the opposite; for some $\varepsilon_{n}\to 0$ there exists $M>0$, such that $||(\lambda,u)||_{\mathbb{R}\times X}<M$, for any $(\lambda,u)\in\mathcal{C}_{\varepsilon_{n}}$. Then, we have that the branches are compact (i.e., the second alternative of Theorem 3.1 holds), which means that must contain solutions of (1.4) that change sign. The connectness of these branches in $X$ and thus in $L^{\infty}(\Omega)$ implies that there exist $(\lambda_{n},u_{n})\in\mathcal{C}_{\varepsilon_{n}}$ with $u_{n}$ vanishing somewhere in $\Omega$ and being positive elsewhere, with $||u_{n}||_{X}<M$. However, this is impossible from Lemma 3.3. $\blacksquare$

The following result, which is immediate, will be used in the proof of Theorem 1.3.

LEMMA 3.5

Fix $\varepsilon$ small enough. Then the branch $\mathcal{C}_{\varepsilon}$ is a closed set in $\mathbb{R}\times X$.

Proof Let $(\lambda_{n},u_{n})\in\mathcal{C}_{\varepsilon}$, to be a convergent sequence in $\mathbb{R}\times X$, in some $(\lambda_{*},u_{*})\in\mathbb{R}\times X$. Then, for any $\phi\in X$, we have that

[TABLE]

which in the limit gives that

[TABLE]

Thus, $(\lambda_{*},u_{*})\in\mathcal{C}_{\varepsilon}$. $\blacksquare$

Next, we prove the main result of this Section; the existence of a global branch of solutions for (1.2) bifurcating from the principal eigenvalue $\lambda_{0}$ of (1.3).

Proof of Theorem 1.3  We apply Whyburn’s Lemma 1.4 in order to prove that $\mathcal{C}_{\varepsilon}$ converge in $\mathbb{R}\times X$, and this limit $C_{0}$ is the global branch of solutions of (1.2) bifurcating from the principal eigenvalue $\lambda_{0}$ of (1.3). For some $R>0$ and some sequence $\varepsilon_{n}\to 0$, as $n\to\infty$, we define the sets $A_{n}$ as follows:

[TABLE]

For every $n\in\mathbb{N}$, these sets are connected (see Theorem 3.1) and closed (see Lemma 3.5). Next, we claim that $\liminf_{n\to\infty}\{A_{n}\}$ is not empty. To see this we consider the points $(\lambda_{0,\varepsilon_{n}},0)$ belonging to $\mathcal{C}_{\varepsilon_{n}}$. From Theorem 1.2 we have that $(\lambda_{0,\varepsilon_{n}},0)\to(\lambda_{0},0)$, hence,

[TABLE]

It remains to prove that the set $\bigcup_{n\in\mathbb{N}}A_{n}$ is relatively compact i.e., every sequence in $A_{n}$ contains a convergent subsequence. Let $(\lambda_{n},u_{n})\in\bigcup_{n\in\mathbb{N}}A_{n}$, then the sequence $(\lambda_{n},u_{n})$ is bounded in $\mathbb{R}\times X$ and so (up to a subsequence) we have that $\lambda_{n}\to\lambda_{*}$ and $u_{n}\rightharpoonup u_{*}$ in $X$. However, since $u_{n}$ is bounded, Lemma 3.3 implies that $u_{n}$ are positive in $\Omega$ and Lemma 3.2 implies that $(\lambda_{n},u_{n})$ converges strongly in $\mathbb{R}\times X$, either to $(\lambda_{0},0)$ or to $(\lambda_{*},u_{*})$ which satisfy (1.2). Thus, $\bigcup_{n\in\mathbb{N}}A_{n}$ is relatively compact.

Then, we leave $R\to\infty$ in order to obtain that $\mathcal{C}_{\varepsilon_{n}}\to C_{0}$, in $\mathbb{R}\times X$, for any $R\in\mathbb{R}$. In order to prove that $C_{0}$ is unbounded in $\mathbb{R}\times X$, we may use, the sequences $\{(\lambda_{n},u_{n})\in\mathcal{C}_{\varepsilon_{n}}\cap\partial B_{R}(\lambda_{0},0)\}$ which converge to some $(\lambda_{*},u_{*})$ in $\mathbb{R}\times X$, satisfies (1.2), for any $R>0$. $\blacksquare$

REMARK 3.2

Theorem 1.3 is also valid to the cases of space dimension $N=2$ or $N=4$. The proof for $N=2$ follows the same arguments as that for $N=3$. For the case for $N=4$ we have to adapt only the proof of Lemma 3.1; due to the embedding, the derivatives of order 4 of $u_{\varepsilon,n}$ are not tending to zero in $C^{4}(\bar{\Omega})$, however they tend to zero in $L^{p}(\Omega)$, for any $p>1$. This is sufficient for Lemma 3.1 to hold also, in the case of $N=4$.

Finally, we state the global bifurcation result for the case of the problem

[TABLE]

where $\Omega$ is a bounded domain of $\mathbb{R}^{N}$, $N=2,3,4$, $\Omega$ is smooth enough, at least $C^{6}$. Using the same procedure as for the problem (1.2) we may prove the following:

THEOREM 3.2

Assume problem (3.10). Then, the principal eigenvalue $\lambda_{0}$ is a bifurcation point of (3.10), such that the first alternative of Theorem 1.1 holds i.e., we have a global bifurcation.

4 The general case 1.1

Proof of Theorem 1.5. We will give the proof for $N=3$ and $B(x,u,Du,D^{2}u)=|\det D^{2}u|^{2}$. Thus, we will prove the existence of global branches of solutions for the problem:

[TABLE]

bifurcating from the principal eigenvalue $\lambda_{0}$ of $-\Delta$ in $\Omega$. Let $\varepsilon>0$ be a small enough number. We assume the following approximating problems:

[TABLE]

Theorem 3.2 is directly applied for problem (4.2), for any $\varepsilon>0$, hence we have the existence of global branches bifurcating from $\lambda_{0}$. We will prove that for $\varepsilon_{n}\downarrow 0$, these branches converge.

Proof of Theorem 1.5  We apply Whyburn’s Lemma 1.4 in order to prove that $C_{\varepsilon_{n}}\to C_{0}$, where $C_{0}$ will denote the global branch of solutions of (4.1) bifurcating from $\lambda_{0}$. For some $R>0$ and some sequence $\varepsilon_{n}\to 0$, as $n\to\infty$, we define the sets $A_{n}$, as follows:

[TABLE]

For every $n\in\mathbb{N}$, these sets are connected (see Theorem 3.2) and closed (in the same way as in Lemma 3.5). Moreover, we note that $\liminf_{n\to\infty}\{A_{n}\}$ is not empty since $(\lambda_{0},0)\in C_{\varepsilon_{n}}$, for every $n$.

It remains to prove that the set $\cup_{n\in\mathbb{N}}A_{n}$ is relatively compact i.e., every sequence in $A_{n}$ contains a convergent subsequence. Let $(\lambda_{n},u_{n})\in\cup_{n\in\mathbb{N}}A_{n}$. The sequence $(\lambda_{n},u_{n})$ is bounded in $\mathbb{R}\times X$ and so (up to a subsequence), we have that $\lambda_{n}\to\lambda_{*}$ and $u_{n}\rightharpoonup u_{*}$ in $X$. Is sufficient to prove that if $u_{*}\equiv 0$ then $u_{n}\to u_{*}\equiv 0$ in $X$ and $\lambda_{n}\to\lambda_{0}$.

Let $u_{*}\equiv 0$, then $u_{n}\to 0$ in $C^{3}(\bar{\Omega})$ and

[TABLE]

Dividing (4.3) by $-\Delta u_{n}$ we get that

[TABLE]

However,

[TABLE]

uniformly, as $n\to\infty$, thus

[TABLE]

Next we claim that $\lambda_{*}\neq 0$; Assume that $\lambda_{n}\to 0$, then from (4.3) we have that

[TABLE]

Then from (4.5) we get that,

[TABLE]

as $n\to\infty$. Using once again (4.3) we have that

[TABLE]

Setting

[TABLE]

from (4.7) we obtain that

[TABLE]

Observe that $\tilde{u}_{n}\rightharpoonup\tilde{u}_{*}$ in $H_{0}^{1}(\Omega)$, such that

[TABLE]

It is also true from (4.6) that

[TABLE]

Thus if $\lambda_{n}\to 0$, we get a contradiction from (4.8). Thus $\lambda_{*}\neq 0$ and (4.8) implies that $\tilde{u}_{*}\neq 0$, such that

[TABLE]

Since $\tilde{u}_{*}$ is nonnegative, $(\lambda_{*},\tilde{u}_{*})$ should coincide with the principal eigenpair $(\lambda_{0},\tilde{u}_{0})$ ($\tilde{u}_{0}$ denotes the normalization of $u_{0}$ in $H_{0}^{1}(\Omega)$).

Observe now that (4.3) may be seen as

[TABLE]

with

[TABLE]

Denote by $\bar{u}_{n}$ the normalization of $u_{n}$ in $X$. Then, $\bar{u}_{n}$ satisfies (4.9) and for $u_{n}\to 0$, in $C^{3}(\tilde{\Omega})$ and $\lambda_{n}\to\lambda_{0}$, we have that

[TABLE]

in $C^{1}(\bar{\Omega})$. Using the same arguments as in Lemma 3.1 i.e., the compact imbedding $X\hookrightarrow C^{4}(\bar{\Omega})$ implies that $u_{\varepsilon_{n}}\to 0$, in $C^{4}(\bar{\Omega})$, hence $g_{\varepsilon_{n}}\to g_{0}$ in $C^{2}(\bar{\Omega})$. Moreover, the derivatives of order 5 and 6 of $u_{\varepsilon_{n}}$ tend to zero in $L^{2}(\Omega)$. Thus, $\bar{u}_{n}\to u_{0}$ in $X$. However, this is true only if $u_{n}\to 0$, in $X$.

It remains now to leave $R\to\infty$ in order to obtain that $C_{\varepsilon_{n}}\to C_{0}$, in $\mathbb{R}\times X$, for any $R$. In order to prove that $C_{0}$ is unbounded in $\mathbb{R}\times X$, we may use the sequences $\{(\lambda_{n},u_{n})\in C_{\varepsilon_{n}}\cap\partial B_{R}(\lambda_{1},0)\}$, which converge to some $(\lambda_{*},u_{*})$, in $\mathbb{R}\times X$, for any $R>0$. $\blacksquare$