Electronic Journal of Qualitative Theory of Differential Equations Spec. Ed. I, 2009 No. 26, 1–8; http://www.math.u-szeged.hu/ejqtde/

CONVEX SOLUTIONS OF SYSTEMS ARISING FROM MONGE-AMPÈRE EQUATIONS

Haiyan Wang

Division of Mathematical and Natural Sciences Arizona State University, Phoenix, AZ 85069, USA e-mail: wangh@asu.edu

Honoring the Career of John Graef on the Occasion of His Sixty-Seventh Birthday

Abstract

We establish two criteria for the existence of convex solutions to a boundary value problem for weakly coupled systems arising from the Monge-Ampère equations. We shall use fixed point theorems in a cone.

Key words and phrases: Convex solutions, Monge-Ampère equations, cone. AMS (MOS) Subject Classifications:34B15, 35J60

1 Introduction

In this paper we consider the existence of convex solutions to the Dirichlet problem for the weakly coupled system

\begin{aligned}((u′_{1}(t))^{N})′ & = Nt^{N−1}f(−u_{2}(t)) in 0 < t < 1, \\ ((u′_{2}(t))^{N})′ & = Nt^{N−1}g(−u_{1}(t)) in 0 < t < 1, \\ u′_{1}(0) = u′_{2}(0) & = 0, u_{1}(1) = u_{2}(1) = 0, \end{aligned}  (1.1)

where N ≥ 1. A nontrivial convex solution of (1.1) is negative on [0,1). Such a problem arises in the study of the existence of convex radial solutions to the Dirichlet problem for the system of the Monge-Ampère equations

\{ \begin{matrix}detD^{2}u_{1} = f(−u_{2}) in B, \\ detD^{2}u_{2} = g(−u_{1}) in B, \\ u_{1} = u_{2} = 0 on ∂B, \end{matrix}  (1.2)

where B = {x ∈ ℝ^{N} : |x| < 1} and detD^{2}u_{i} is the determinant of the Hessian matrix (\frac{∂^{2}u_{i}}{∂x_{m}∂x_{n}}) of u_{i}. For how to reduce (1.2) to (1.1), one may see Hu and the author [5].

The Dirichlet problem for a single unknown variable Monge-Ampère equations

\{ \begin{matrix}detD^{2}u = f(−u) in B, \\ u = 0 on ∂B, \end{matrix}  (1.3)

in general domains in ℝ^{n} may be found in Caffarelli, Nirenberg and Spruck [1]. Kutev [7] investigated the existence of strictly convex radial solutions of (1.3) when f(−u) = (−u)^{p}. Delano [3] treated the existence of convex radial solutions of (1.3) for a class of more general functions, namely λ exp f(|x|, u, |∇u|).

The author [10] and Hu and the author [5] showed that the existence, multiplicity and nonexistence of convex radial solutions of (1.3) can be determined by the asymptotic behaviors of the quotient \frac{f(u)}{u^{N}} at zero and infinity.

In this paper we shall establish the existence of convex radial solutions of the weakly coupled system (1.1) in superlinear and sublinear cases. First, introduce the notation

f_{0} = lim_{x→0^{+}} \frac{f(x)}{x^{N}}, f_{∞} = lim_{x→∞} \frac{f(x)}{x^{N}},

and

g_{0} = lim_{x→0^{+}} \frac{g(x)}{x^{N}}, g_{∞} = lim_{x→∞} \frac{g(x)}{x^{N}}.

We shall show that if (1.1) is superlinear, or f_{0} = g_{0} = 0 and f_{∞} = g_{∞} = ∞, (1.1) is sublinear, or f_{0} = g_{0} = ∞ and f_{∞} = g_{∞} = 0, then (1.1) has a convex solution.

Our main results are:

Theorem 1.1 Assume f, g : [0, ∞) → [0, ∞) are continuous. (a). If f_{0} = g_{0} = 0 and f_{∞} = g_{∞} = ∞, then (1.1) has a convex solution. (b). If f_{0} = g_{0} = ∞ and f_{∞} = g_{∞} = 0, then (1.1) has a convex solution.

2 Preliminaries

With a simple transformation v_{i} = −u_{i}, i = 1, 2 (1.1) can be brought to the following equation

\{ \begin{matrix}(( − v′_{1}(r))^{N})′ & = Nr^{N−1}f(v_{2}), 0 < r < 1, \\ (( − v′_{2}(r))^{N})′ & = Nr^{N−1}g(v_{1}), 0 < r < 1, \\ v′_{i}(0) = v_{i}(1) & = 0, i = 1, 2. \end{matrix}  (2.4)

Now we treat positive concave classical solutions of (2.4).

The following well-known result of the fixed point index is crucial in our arguments.

Lemma 2.1 ([2, 4, 6]). Let E be a Banach space and K a cone in E. For r > 0, define K_{r} = {u ∈ K : ∥x∥ < r}. Assume that T : K̄_{r} → K is completely continuous such that T x ≠ x for x ∈ ∂K_{r} = {u ∈ K : ∥x∥ = r}.

(i) If ∥T x∥ ≥ ∥x∥ for x ∈ ∂K_{r}, then

i(T, K_{r}, K) = 0.

(ii) If ∥T x∥ ≤ ∥x∥ for x ∈ ∂K_{r}, then

i(T, K_{r}, K) = 1.

In order to apply Lemma 2.1 to (2.4), let X be the Banach space C[0, 1] × C[0, 1] and, for (v_{1}, v_{2}) ∈ X,

∥(v_{1}, v_{2})∥ = ∥v_{1}∥ + ∥v_{2}∥

where ∥v_{i}∥ = sup_{t∈[0,1]}|v_{i}(t)|. Define K to be a cone in X by

K = \{(v_{1}, v_{2}) ∈ X : v_{i}(t) ≥ 0, t ∈ [0, 1], min_{\frac{1}{4} ≤t≤ \frac{3}{4}} v_{i}(t) ≥ \frac{1}{4}∥v_{i}∥, i = 1, 2\}.

Also, define, for r a positive number, Ω_{r} by

Ω_{r} = {(v_{1}, v_{2}) ∈ K : ∥(v_{1}, v_{2})∥ < r}.

Note that ∂Ω_{r} = {(v_{1}, v_{2}) ∈ K : ∥(v_{1}, v_{2})∥ = r}.

Let T : K → X be a map with components (T^{1}, T^{2}), which are defined by

\begin{aligned}T^{1}(v_{1}, v_{2})(r) & = ∫_{r}^{1} ( ∫_{0}^{s} Nτ^{N−1}f(v_{2}(τ))dτ)^{\frac{1}{n}} ds, r ∈ [0, 1], \\ T^{2}(v_{1}, v_{2})(r) & = ∫_{r}^{1} ( ∫_{0}^{s} Nτ^{N−1}g(v_{1}(τ))dτ)^{\frac{1}{n}} ds, r ∈ [0, 1]. \end{aligned}  (2.5)

It is straightforward to verify that (2.4) is equivalent to the fixed point equation

T(v_{1}, v_{2}) = (v_{1}, v_{2}) in K.

Thus, if (v_{1}, v_{2}) ∈ K is a positive fixed point of T, then (−v_{1}, −v_{2}) is a convex solution of (1.1). Conversely, if (u_{1}, u_{2}) is a convex solution of (1.1), then (−u_{1}, −u_{2}) is a fixed point of T in K.

The following lemma is a standard result due to the concavity of u. We prove it here only for completeness.

Lemma 2.2 Let u ∈ C^{1}[0, 1] with u(t) ≥ 0 for t ∈ [0, 1]. Assume that u′(t) is nonincreasing on [0, 1]. Then

u(t) ≥ min{t, 1 − t}||u||, t ∈ [0, 1],

where ||u|| = sup_{t∈[0,1]} u(t). In particular,

min_{\frac{1}{4} ≤t≤ \frac{3}{4}} u(t) ≥ \frac{1}{4}||u||.

Proof Since u′(t) is nonincreasing, we have for 0 ≤ t_{0} < t < t_{1} ≤ 1,

u(t) − u(t_{0}) = ∫_{t_{0}}^{t} u′(s)ds ≥ (t − t_{0})u′(t)

and

u(t_{1}) − u(t) = ∫_{t}^{t_{1}} u′(s)ds ≤ (t_{1} − t)u′(t),

from which, we have

u(t) ≥ \frac{(t_{1} − t)u(t_{0}) + (t − t_{0})u(t_{1})}{t_{1} − t_{0}}.

Considering the above inequality on [0, σ] and [σ, 1], we have u(t) ≥ t||u|| for t ∈ [0, σ],

and

u(t) ≥ (1 − t)||u|| for t ∈ [σ, 1],

where σ ∈ [0, 1] such that u(σ) = ||u||. Hence,

u(t) ≥ min{t, 1 − t}||u||, t ∈ [0, 1].

□ Lemma 2.3 can be verified by the standard procedures.

Lemma 2.3 Assume f, g : [0, ∞) → [0, ∞) are continuous. Then T(K) ⊂ K and T : K → K is a compact operator and continuous.

Let

Γ = \frac{1}{4} ∫ \frac{3}{4} \frac{1}{4} ( ∫ s_{\frac{1}{4}} Nτ^{N−1}dτ)^{\frac{1}{N}} ds > 0.

Lemma 2.4 Let (v_{1}, v_{2}) ∈ K and η > 0. If

f(v_{2}(t)) ≥ (ηv_{2}(t))^{N} for t ∈ [\frac{1}{4}, \frac{3}{4}],

or

g(v_{1}(t)) ≥ (ηv_{1}(t))^{N} for t ∈ [\frac{1}{4}, \frac{3}{4}],

then

∥T(v_{1}, v_{2})∥ ≥ Γη∥v_{2}∥,

or

∥T(v_{1}, v_{2})∥ ≥ Γη∥v_{1}∥,

respectively.

EJQTDE Spec. Ed. I, 2009 No. 26

Proof Note, from the definition of T(v_{1}, v_{2}), that T^{i}(v_{1}, v_{2})(0) is the maximum value of T^{i}(v_{1}, v_{2}) on [0,1]. It follows that

∥T(v_{1}, v_{2})∥ ≥ sup_{t∈[0,1]}|T^{1}(v_{1}, v_{2})(t)|

≥ ∫ \frac{3}{4} \frac{1}{4} ( ∫ s_{\frac{1}{4}} Nτ^{N−1}f(v_{2}(τ))dτ)^{\frac{1}{N}} ds

≥ ∫ \frac{3}{4} \frac{1}{4} ( ∫ s_{\frac{1}{4}} Nτ^{N−1}(ηv_{2}(τ))^{N}dτ)^{\frac{1}{N}} ds

≥ ∫ \frac{3}{4} \frac{1}{4} ( ∫ s_{\frac{1}{4}} Nτ^{N−1}(\frac{η}{4}∥v_{2}∥)^{N}dτ)^{\frac{1}{N}} ds

= Γη∥v_{2}∥.

Similarly,

∥T(v_{1}, v_{2})∥ ≥ sup_{t∈[0,1]}|T^{2}(v_{1}, v_{2})(t)| ≥ Γη∥v_{1}∥.

□

We define new functions f̂(t), ĝ(t) : [0, ∞) → [0, ∞) by

f̂(t) = max{f(v) : 0 ≤ v ≤ t}, ĝ(t) = max{g(v) : 0 ≤ v ≤ t}.

Note that f̂_{0} = lim_{t→0} \frac{f̂(t)}{t^{N}}, f̂_{∞} = lim_{t→∞} \frac{f̂(t)}{t^{N}} and ĝ_{0}, ĝ_{∞} can be defined similarly.

Lemma 2.5 [9] Assume f, g : [0, ∞) → [0, ∞) are continuous. Then

f̂_{0} = f_{0}, f̂_{∞} = f_{∞},

and

ĝ_{0} = g_{0}, ĝ_{∞} = g_{∞}.

Lemma 2.6 Assume f, g : [0, ∞) → [0, ∞) are continuous. Let r > 0. If there exists an ε > 0 such that

f̂(r) ≤ (εr)^{N}, ĝ(r) ≤ (εr)^{N},

then

∥T(v_{1}, v_{2})∥ ≤ 2ε∥(v_{1}, v_{2})∥ for (v_{1}, v_{2}) ∈ ∂Ω_{r}.

Proof From the definition of T, for (v_{1}, v_{2}) ∈ ∂Ω_{r}, we have

∥T(v_{1}, v_{2})∥ = ∑_{i=1}^{2} sup_{t∈[0,1]}|T^{i}(v_{1}, v_{2})(t)|

≤ (∫_{0}^{1} Nτ^{N−1}f(v_{2}(τ))dτ)^{\frac{1}{N}} + (∫_{0}^{1} Nτ^{N−1}g(v_{1}(τ))dτ)^{\frac{1}{N}}
≤ (∫_{0}^{1} Nτ^{N−1}f̂(r)dτ)^{\frac{1}{N}} + (∫_{0}^{1} Nτ^{N−1}ĝ(r)dτ)^{\frac{1}{N}}
≤ (∫_{0}^{1} Nτ^{N−1}dτ)^{\frac{1}{N}} εr + (∫_{0}^{1} Nτ^{N−1}dτ)^{\frac{1}{N}} εr
≤ 2ε∥(u_{1}, u_{2})∥.

□

3 Proof of Theorem 1.1

Proof Part (a). It follows from Lemma 2.5 that f̂_{0} = 0, ĝ_{0} = 0. Therefore, we can choose r_{1} > 0 so that f̂^{i}(r_{1}) ≤ (εr_{1})^{N}, ĝ^{i}(r_{1}) ≤ (εr_{1})^{N} where the constant ε > 0 satisfies

ε < \frac{1}{2}.

We have by Lemma 2.6 that

∥T(v_{1}, v_{2})∥ ≤ 2ε∥(v_{1}, v_{2})∥ < ∥(v_{1}, v_{2})∥ for (v_{1}, v_{2}) ∈ ∂Ω_{r_{1}}.

Now, since f_{∞} = ∞, g_{∞} = ∞, there is an Ĥ > 0 such that

f(v) ≥ (ηv)^{N}, g(v) ≥ (ηv)^{N}

for v ≥ Ĥ , where η > 0 is chosen so that

\frac{1}{2}Γη > 1.

Let r_{2} = max{2r_{1}, 8Ĥ}. If (v_{1}, v_{2}) ∈ ∂Ω_{r_{2}}, there exists one of i = 1 or i = 2 such that sup_{t∈[0,1]} v_{i} ≥ \frac{1}{2}r_{2}. Without loss of generality, assume that sup_{t∈[0,1]} v_{1} ≥ \frac{1}{2}r_{2}. Then

min_{\frac{1}{4} ≤t≤ \frac{3}{4}} v_{1}(t) ≥ \frac{1}{4} sup_{t∈[0,1]} v_{1} ≥ \frac{1}{8}r_{2} ≥ Ĥ,

which implies that

g(v_{1}(t)) ≥ (ηv_{1}(t))^{N} for t ∈ [\frac{1}{4}, \frac{3}{4}].

It follows from Lemma 2.4 that

∥T(v_{1}, v_{2})∥ ≥ Γη∥v_{1}∥ > \frac{1}{2}Γηr_{2} ≥ r_{2} = ∥(v_{1}, v_{2})∥.

By Lemma 2.1,

i(T, Ω_{r_{1}}, K) = 1 and i(T, Ω_{r_{2}}, K) = 0.

It follows from the additivity of the fixed point index that

i(T, Ω_{r_{2}} \ Ω̄_{r_{1}}, K) = −1.

Thus, i(T, Ω_{r_{2}} \ Ω̄_{r_{1}}, K) ≠ 0, which implies T has a fixed point (v_{1}, v_{2}) ∈ Ω_{r_{2}} \ Ω̄_{r_{1}} by the existence property of the fixed point index. The fixed point (−v_{1}, −v_{2}) ∈ Ω_{r_{2}} \ Ω̄_{r_{1}} is the desired positive solution of (1.1).

Part (b). since f_{0} = ∞, g_{0} = ∞, there is an H > 0 such that

f(v) ≥ (ηv)^{N}, g(v) ≥ (ηv)^{N}

for 0 < v ≤ H , where η > 0 is chosen so that

Γη > 1.

If (v_{1}, v_{2}) ∈ ∂Ω_{r_{1}}, then

f(v_{2}(t)) ≥ (ηv_{2})^{N}, g(v_{1}(t)) ≥ (ηv_{1})^{N}for t ∈ [0, 1].

Lemma 2.4 implies that

∥T(v_{1}, v_{2})∥ ≥ Γη∥(v_{1}, v_{2})∥ > ∥(v_{1}, v_{2})∥ for (v_{1}, v_{2}) ∈ ∂Ω_{r_{1}}.

We now determine Ω_{r_{2}}. It follows from Lemma 2.5 that f̂_{∞} = 0 and ĝ_{∞} = 0. Therefore there is an r_{2} > 2r_{1} such that

f̂^{i}(r_{2}) ≤ (εr_{2})^{N}, ĝ^{i}(r_{2}) ≤ (εr_{2})^{N},

where the constant \frac{1}{2} > ε > 0. Thus, we have by Lemma 2.6 that

∥T(v_{1}, v_{2})∥ ≤ 2ε∥(v_{1}, v_{2})∥ < ∥(v_{1}, v_{2})∥ for (v_{1}, v_{2}) ∈ ∂Ω_{r_{2}}.

By Lemma 2.1,

i(T, Ω_{r_{1}}, K) = 0 and i(T, Ω_{r_{2}}, K) = 1.

It follows from the additivity of the fixed point index that i(T, Ω_{r_{2}} \Ω̄_{r_{1}}, K) = 1. Thus, T has a fixed point (v_{1}, v_{2}) in Ω_{r_{2}} \ Ω̄_{r_{1}}. And (−v_{1}, −v_{2}) is the desired convex solution of (1.1). □

4 Acknowledgement

The author would like thank the reviewer for careful reading of the paper and useful suggestions, which helped to improve the presentation of this paper.

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