Elliptic Equations and Systems with Subcritical and Critical Exponential Growth Without the Ambrosetti–Rabinowitz Condition

Abstract

In this paper, we prove the existence of nontrivial nonnegative solutions to a class of elliptic equations and systems which do not satisfy the Ambrosetti–Rabinowitz (AR) condition where the nonlinear terms are superlinear at 0 and of subcritical or critical exponential growth at ∞. The known results without the AR condition in the literature only involve nonlinear terms of polynomial growth. We will use suitable versions of the Mountain Pass Theorem and Linking Theorem introduced by Cerami (Istit. Lombardo Accad. Sci. Lett. Rend. A, 112(2):332–336, 1978 Ann. Mat. Pura Appl., 124:161–179, 1980). The Moser–Trudinger inequality plays an important role in establishing our results. Our theorems extend the results of de Figueiredo, Miyagaki, and Ruf (Calc. Var. Partial Differ. Equ., 3(2):139–153, 1995) and of de Figueiredo, do Ó, and Ruf (Indiana Univ. Math. J., 53(4):1037–1054, 2004) to the case where the nonlinear term does not satisfy the AR condition. Examples of such nonlinear terms are given in Appendix A. Thus, we have established the existence of nontrivial nonnegative solutions for a wider class of nonlinear terms.

Appendix A: Examples of Weaker Nonlinearity

In this section, we will discuss and compare the conditions (H2) in [17] and (L3) in our work and also give some examples to illustrate that our condition (L3) is weaker than (H2) in [17] and the strict inequality of (A.2) of [1]: $f^{\prime}(x, u)>\frac{f(x, u)}{u}$. Therefore, it is worthwhile to study the existence of nontrivial solutions to problem (P) under our condition (L3).

Let us see what the condition (H2) in [17] really means. First, we recall the (H2) condition:

• (H2) ∃t ₀>0,∃M>0 such that ∀|u|≥t ₀,∀x∈Ω,

  $$0<F(x,u)=\int_{0}^{u}f(x,t)dt\leq M \bigl \vert f ( x,u ) \bigr \vert . $$

From

$$0<F(x,u)\leq M \bigl \vert f ( x,u ) \bigr \vert , $$

we have

$$0<\frac{1}{M}\leq\frac{\vert f ( x,u ) \vert }{F(x,u)},$$

which is

$$ \biggl[ \ln \biggl(\frac{F(x,u)}{e^{u/M}} \biggr) \biggr]^{\prime}\geq0 $$

(A.1)

when f(x,u) is nonnegative. Thus the function $\frac{F(x,u)}{e^{Cu}}$ is nondecreasing for some small positive constant C when u is big enough. So, if $F(x,u)=P(u)e^{\alpha u^{2}}$ where P is a polynomial; the condition (H2) is satisfied since both terms P(u) and $e^{\alpha u^{2}-cu}$ are increasing when u is big enough. However, if we have periodic terms or decreasing terms in the nonlinear terms as the following example shows, (A.1) may not be satisfied and thus the condition (H2) may not hold anymore.

Example 1

The nonlinearity $f(x,u)=e^{u}\cos u+(\sqrt{2}+\sin u)e^{u}$ doesn’t satisfy the condition (H2) in [17]. Indeed, it’s easy to see that in this case, $F(u)=(\sqrt{2}+\sin u)e^{u}$. So if there exists a constant M>0 such that

$$(\sqrt{2}+\sin u)e^{u}\leq M(\sqrt{2}+\sin u+\cos u)e^{u},$$

then

$$\sqrt{2}+\sin u\leq M(\sqrt{2}+\sin u+\cos u), $$

i.e.,

$$(1-M)\sin u-M\cos u\leq\sqrt{2}(M-1). $$

However, we can choose u such that

which is a contradiction. This shows that f(x,u) does not satisfy condition (H2) in [17].

Next, let us discuss what our condition (L3) means. We recall that

• (L3) There are C _∗≥0,θ≥1 such that H(x,t)≤θH(x,s)+C _∗ for all 0<t<s, ∀x∈Ω, where H(x,u)=uf(x,u)−2F(x,u).

The condition (L3) suggests a sort of “weak” nondecreasing property of the function H(x,t). In particular, a nondecreasing function H(x,t) in t variable satisfies our condition (L3) (with θ=1 and C _∗=0). Now suppose that f′ (in terms of u) exists, then H(x,t) being nondecreasing is equivalent to (uf(x,u)−2F(x,u))′≥0, which is in turn equivalent to

$$ f^{^{\prime}}(x,u)\geq\frac{f(x,u)}{u}\quad \text{for all }0<u,\quad \forall x\in \varOmega. $$

(A.2)

This kind of condition was assumed in the work of Adimurthi [1] with strict inequality in (A.2) in order to get the existence of positive solutions of the semilinear Dirichlet problem with critical exponential growth. Indeed, as mentioned in [34], Adimurthi assumed that f is C ¹ and satisfies $f^{^{\prime}}(x,u)>\frac{f(x,u)}{u}$for all u≠0,∀x∈Ω in his paper [1]. In other words, our condition (L3) (even with θ=1 and C _∗=0) is weaker than the condition of Adimurthi. In the following example, we will give an example of a nonlinearity which satisfies our condition (L3) but does not satisfy Adimurthi’s condition.

Example 2

Consider the function f(x,u)=u(u−1)³ e ^u, which implies

$$\bigl( f(x,u) \bigr)^{\prime}= \bigl[ u(u-1)^{3}+ (u-1)^{3}+3u(u-1)^{2}\bigr] e^{u}. $$

So we can see that $f^{^{\prime}}(x,u)\geq\frac{f(x,u)}{u}$ for all 0<u,∀x∈Ω. Therefore, f satisfies (A.2) and thus our condition (L3). However, when u=1, the equality holds, which means that f does not satisfy Adimurthi’s condition of strict inequality [1].

If we further assume that f(x,u) is positive, then (A.2) gives

$$\frac{f^{^{\prime}}(x,u)}{f(x,u)}\geq\frac{1}{u}\quad\text{for all }0<u,\quad \forall x \in \varOmega, $$

which thus implies that the function $\frac{f(x,u)}{u}$ is nondecreasing for all 0<u,∀x∈Ω. The assumption that the function $\frac {f(x,u)}{u}$ is nondecreasing for all 0<u,∀x∈Ω is also a standard condition and is assumed in many works. In fact, our condition (L3) (even with θ=1 and C _∗=0) is weaker than this standard condition. Indeed, let $g(x,u)=\frac{f(x,u)}{u}$, which is nondecreasing for all 0<u,∀x∈Ω. We get with 0<u, x∈Ω:

$$F(x,u)=\int_{0}^{u} sg(x,s)ds\leq g(x,u)\int _{0}^{u} sds=\frac{u^{2}g(x,u)}{2}= \frac{uf(x,u)}{2},$$

which thus means that H(x,u)≥0. Moreover, with 0<u<v,x∈Ω, we have

from which we can conclude that

$$H(x,u)\leq H(x,v). $$

Example 3

Consider the function $F(x,u)=u^{2}e^{\sqrt{u}}$ and then $f(x,u)= ( 2u+\frac{u\sqrt{u}}{2} ) e^{\sqrt{u}}$. We have $\frac{f(x,u)}{u}= ( 2+\frac{\sqrt{u}}{2} ) e^{\sqrt{u}}$, which is a nondecreasing function. This shows that f(x,u) satisfies our condition (L3). Moreover, for every small positive constant C, then $\frac{F(x,u)}{e^{Cu}}=u^{2}e^{\sqrt{u}(1-C\sqrt{u})}$ is not always increasing when u is big enough. This means that f(x,u) does not satisfy the condition (H2).

In other words, from Example 3, we can see that there exist nonlinearities that satisfy our condition (L3) but do not satisfy the condition (H2).

1.1 A.1 About the Critical Growth

We will finish this paper by analyzing the critical growth of the nonlinearity term f(x,u). We will see that in some cases, we don’t need to assume the condition (H2)-type or (H5)-type as in [17]. More precisely, we consider the following three cases: $\lim_{u\rightarrow+\infty } \frac{\vert f ( x,u ) \vert }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0; \lim_{u\rightarrow+\infty } \frac{\vert f ( x,u ) \vert }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=c\in ( 0,\infty )$, and $\lim_{u\rightarrow+\infty }\frac{\vert f ( x,u ) \vert }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=\infty$.

1.1.1 A.1.1 Case 1

In this subsection, we will discuss the first case,

$$\lim_{u\rightarrow+\infty } \frac{f ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0,\quad\text{uniformly on }x \in\varOmega. $$

This case is easy to study. Indeed, by l’Hôpital’s rule, we also get

$$\lim_{u\rightarrow+\infty } \frac{F ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=\lim_{u\rightarrow +\infty } \frac{f ( x,u ) }{2\alpha_{0}u\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0, \quad\text{uniformly on }x\in\varOmega. $$

Using l’Hôpital’s rule again, we get

$$\lim_{u\rightarrow+\infty } \frac{uF ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=\lim_{u\rightarrow +\infty } \frac{uf ( x,u ) +F(x,u)}{2\alpha_{0}u\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0, \quad\text{uniformly on }x\in \varOmega. $$

So if we have the condition of (H5) type, i.e.,

$$\lim_{u\rightarrow+\infty } \frac{uf ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }\geq\beta>0,\quad\text{uniformly on }x\in \varOmega, $$

we can easily deduce the condition of (H2) type (so we have the AR condition) by noticing that

$$\lim_{u\rightarrow+\infty } \frac{uF ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0<\beta\leq\lim_ {u \rightarrow+\infty } \frac{uf ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) },\quad\text{uniformly on }x \in\varOmega. $$

1.1.2 A.1.2 Case 2

Now we will consider the case

$$\lim_{u\rightarrow+\infty } \frac{f ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=c\in ( 0,\infty ) , \quad\text{uniformly on }x\in\varOmega. $$

In this case, it’s clear that

$$\lim_{u\rightarrow+\infty } \frac{uf ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=\infty,\quad\text{uniformly on }x\in \varOmega, $$

which means that the condition of (H5) type is satisfied automatically. Also, by l’Hôpital’s rule again, we get

$$\lim_{u\rightarrow+\infty } \frac{F ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=\lim_{u\rightarrow +\infty } \frac{f ( x,u ) }{2\alpha_{0}u\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0, \quad \text{uniformly on }x\in\varOmega, $$

so the condition of (H2) type is also satisfied automatically by again noticing that

$$\lim_{u\rightarrow+\infty } \frac{F ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=0<c=\lim_{u \rightarrow +\infty } \frac{f ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) },\quad\text{uniformly on }x\in\varOmega. $$

1.1.3 A.1.3 Case 3

We consider the last case,

$$\lim_{u\rightarrow+\infty } \frac{f ( x,u ) }{\exp ( \alpha_{0}\vert u\vert ^{2} ) }=\infty,\quad\text{uniformly on }x\in \varOmega. $$

In this case, the condition of (H5) type is satisfied automatically.