On symmetric positive homoclinic solutions of semilinear p-Laplacian differential equations

Abstract

In this paper we study the existence of even positive homoclinic solutions for p-Laplacian ordinary differential equations (ODEs) of the type ${(u^{\mathrm{\prime }}{|u^{\mathrm{\prime }}|}^{p-2})}^{\mathrm{\prime }}-a(x)u{|u|}^{p-2}+\lambda b(x)u{|u|}^{q-2}=0$, where $2\le p<q$, $\lambda >0$ and the functions a and b are strictly positive and even. First, we prove a result on symmetry of positive solutions of p-Laplacian ODEs. Then, using the mountain-pass theorem, we prove the existence of symmetric positive homoclinic solutions of the considered equations. Some examples and additional comments are given.

MSC:34B18, 34B40, 49J40.

1 Introduction and main results

In this paper we prove the existence of positive homoclinic solutions for p-Laplacian ODEs of the type

$${\left(u^{\mathrm{\prime }}{\left|u^{\mathrm{\prime }}\right|}^{p-2}\right)}^{\mathrm{\prime }}-a(x)u{|u|}^{p-2}+\lambda b(x)u{|u|}^{q-2}=0,x\in \mathbb{R},$$

(1)

where $2\le p<q$ and $\lambda >0$. We assume that

(H) the functions $a(x)$ are $b(x)$ are continuously differentiable, strictly positive, $0<a\le a(x)\le A$ and $0<b\le b(x)\le B$. Let, moreover, $a(x)$ and $b(x)$ be even functions on ℝ, $x{a}^{\mathrm{\prime }}(x)>0$ and $x{b}^{\mathrm{\prime }}(x)<0$ for $x\ne 0$.

By a solution of (1), we mean a function $u:\mathbb{R}\to \mathbb{R}$ such that $u\in C^1(\mathbb{R})$, ${(u^{\mathrm{\prime }}{|u^{\mathrm{\prime }}|}^{p-2})}^{\mathrm{\prime }}\in C(\mathbb{R})$ and Eq. (1) holds for every $x\in \mathbb{R}$. We are looking for positive solutions of (1) which are homoclinic, i.e., $u(x)\to 0$ and $u^{\mathrm{\prime }}(x)\to 0$ as $|x|\to \mathrm{\infty }$.

In the case $p=2$, $q=4$ and $\lambda =1$, similar problems are considered in [1–3] using variational methods. Note that in [2] and [3] the following second-order differential equations are considered:

$$u^{\mathrm{\prime }\mathrm{\prime }}-a(x)u-b(x)u^2+c(x)u^3=0$$

and

$$u^{\mathrm{\prime }\mathrm{\prime }}+a(x)u-b(x)u^2+c(x)u^3=0,$$

where a, b and c are periodic, bounded functions and a and c are positive. These equations come from a biomathematics model suggested by Austin [4] and Cronin [5]. Further results and the phase plane analysis of these equations with constant coefficients are given in [6]. Note that the periodic and homoclinic solutions of p-Laplacian ODEs are considered in [7, 8].

The present work is an extension of these studies to p-Laplacian ODEs. Let $X_T:=W_0^{1,p}(-T,T)$ be the Sobolev space of p-integrable absolutely continuous functions $u:[-T,T]\to \mathbb{R}$ such that

$${\parallel u\parallel }^p={\int }_{-T}^T({|u^{\mathrm{\prime }}(x)|}^p+{|u(x)|}^p)dx<\mathrm{\infty }$$

and $u(-T)=u(T)=0$.

We use a variational treatment of the problem considering the functional $J_T:X_T\to \mathbb{R}$

$$J_T(u)={\int }_{-T}^T(\frac{1}{p}({|u^{\mathrm{\prime }}(x)|}^p+a(x){|u(x)|}^p)-\frac{\lambda }{q}b(x){(u^{+}(x))}^q)dx,$$

where $u^{+}(x)=max\{u(x),0\}$.

Using the well-known mountain-pass theorem, we conclude that the functional $J_T$ has a nontrivial critical point $u_{T,\lambda }\in X_T$, which is a solution of the restricted problem

$$\begin{array}{r}{\left(u^{\mathrm{\prime }}{\left|u^{\mathrm{\prime }}\right|}^{p-2}\right)}^{\mathrm{\prime }}-a(x)u{|u|}^{p-2}+\lambda b(x)u{|u|}^{q-2}=0,x\in (-T,T),\\ u(-T)=u(T)=0.\end{array}$$

(2)

Further, we obtain uniform estimates for the solutions $u_{T,\lambda }$, extended by 0 outside $[-T,T]$. Then, a positive homoclinic solution $u_{\lambda }$ of (1) is found as a limit of $u_{T,\lambda }$, as $T\to \mathrm{\infty }$ in $C_{\mathrm{loc}}^1(\mathbb{R})$. The function $u_{\lambda }$ is also an even function.

To obtain the property, we extend the symmetry lemma of Korman and Ouyang [9] to the p-Laplacian equations. The result is formulated and proved in Section 2.

Our main result is:

Theorem 1 Suppose that $2\le p<q$, $\lambda >0$ and assumptions (H) hold. Then Eq. (1) has a positive solution $u_{\lambda }$ such that $u_{\lambda }(x)\to 0$ and $u_{\lambda }^{\mathrm{\prime }}(x)\to 0$ as $|x|\to \mathrm{\infty }$. Moreover, the solution $u_{\lambda }$ is an even function, $max\{u_{\lambda }(x):x\in \mathbb{R}\}=u_{\lambda }(0)\to +\mathrm{\infty }$ as $\lambda \to 0$ and $u_{\lambda }^{\mathrm{\prime }}(x)<0$ for $x>0$.

Theorem 1 is proved in Section 3. From its proof we have

$$max\{u_{\lambda }(x):x\in \mathbb{R}\}=u_{\lambda }(0)\ge {\left(\frac{a(0)}{\lambda b(0)}\right)}^{1/(q-p)}>0,$$

from which it follows that $u_{\lambda }(0)\to +\mathrm{\infty }$ as $\lambda \to 0$. Observe that if $\lambda =0$, the problem

has a unique solution $u=0$. Indeed, multiplying the equation by u and integrating by parts over ℝ, we obtain

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}({|u^{\mathrm{\prime }}(x)|}^p+a(x){|u(x)|}^p)dx=0,$$

which implies that $u\equiv 0$.

A simplified method can be applied to the equations

$$u^{\mathrm{\prime }\mathrm{\prime }}-a(x)u{|u|}^{p-2}+\lambda b(x)u{|u|}^{q-2}=0,x\in \mathbb{R},$$

(3)

under assumptions (H) and $2\le p<q$, $\lambda >0$. Note that in this case, the even homoclinic solution $u_{\lambda }$ of Eq. (3) satisfies

$$max\{u_{\lambda }(x):x\in \mathbb{R}\}=u_{\lambda }(0)\ge {\left(\frac{a(0)}{\lambda b(0)}\right)}^{1/(q-p)},$$

and again $u_{\lambda }(0)\to +\mathrm{\infty }$ as $\lambda \to 0$. If a and b are constants, Eq. (3) is a conservative system and one can plot the phase curves ${(\frac{v}{2})}^2-a\frac{{|u|}^p}{p}+\lambda b\frac{{|u|}^q}{q}=C$ in the phase plane $(u,v)=(u,u^{\mathrm{\prime }})$. An example is given at the end of Section 3.

2 Preliminary results

Let ${\phi }_p(t)=t{|t|}^{p-2}$, $p\ge 2$ and ${\mathrm{\Phi }}_p(t)=\frac{{|t|}^p}{p}$. It is clear that ${\mathrm{\Phi }}_p(t)$ is a differentiable function and ${\mathrm{\Phi }}_p^{\mathrm{\prime }}(t)={\phi }_p(t)$. Moreover, ${\phi }_p^{\mathrm{\prime }}(t)$ exists and ${\phi }_p^{\mathrm{\prime }}(t)=(p-1){|t|}^{p-2}$ for $p\ge 2$.

Let $L^p(a,b)$, $1<p<\mathrm{\infty }$ be the space of Lebesgue measurable functions $u:(a,b)\to \mathbb{R}$ such that the norm ${|u|}_p^p={\int }_a^b{|u(x)|}^{p}dx<\mathrm{\infty }$.

The dual space of $L^p(a,b)$ is $L^{p^{\mathrm{\prime }}}(a,b)$, where $\frac{1}{p}+\frac{1}{p^{\mathrm{\prime }}}=1$. Let $〈\cdot ,\cdot 〉$ be the duality pairing between $L^{p^{\mathrm{\prime }}}(a,b)$ and $L^p(a,b)$. By the Hölder inequality, $|〈v,u〉|\le {|v|}_{p^{\mathrm{\prime }}}{|u|}_p$ for any $v\in L^{p^{\mathrm{\prime }}}(a,b)$ and $u\in L^p(a,b)$. We will use the following lemmata in further considerations.

Lemma 2 For any $u,v\in L^p(a,b)$, the following inequality holds:

$$〈{\phi }_p(u)-{\phi }_p(v),u-v〉\ge ({|u|}_p^{p-1}-{|v|}_p^{p-1})({|u|}_p-{|v|}_p).$$

Proof of Lemma 2. Note that for $u\in L^p(a,b)$, ${\phi }_p(u)\in L^{p^{\mathrm{\prime }}}(a,b)$. From the Hölder inequality, we have

 □

Lemma 3 Let $p\ge 2$, $u\in C^1([a,b])$ and ${(u^{\mathrm{\prime }}{|u^{\mathrm{\prime }}|}^{p-2})}^{\mathrm{\prime }}\in C([a,b])$. Then

$${\int }_a^b{\left(u^{\mathrm{\prime }}{\left|u^{\mathrm{\prime }}\right|}^{p-2}\right)}^{\mathrm{\prime }}{u}^{\mathrm{\prime }}dx=\frac{p-1}{p}({|u^{\mathrm{\prime }}(b)|}^p-{|u^{\mathrm{\prime }}(a)|}^p).$$

The statement of Lemma 3 follows simply from the identity

$${\left({\left|u^{\mathrm{\prime }}\right|}^p\right)}^{\mathrm{\prime }}=\frac{p}{p-1}{\left(u^{\mathrm{\prime }}{\left|u^{\mathrm{\prime }}\right|}^{p-2}\right)}^{\mathrm{\prime }}{u}^{\mathrm{\prime }}.$$

The one-dimensional p-Laplacian operator $L_p$ for a differentiable function u on the interval I is introduced as $L_p(u):={({\phi }_p(u^{\mathrm{\prime }}))}^{\mathrm{\prime }}$. Let us consider the problem

$$\{\begin{array}{c}{L}_p(u)+f(x,u)=0,x\in (-T,T),\hfill \\ u(-T)=u(T)=0,\hfill \end{array}$$

(4)

where $f\in C^1([-T,T]\times {\mathbb{R}}^{+})$ and satisfies

$$\begin{array}{r}f(-x,u)=f(x,u),x\in (-T,T),u>0,\\ x{f}_x(x,u)<0,x\in (-T,T)\mathrm{\setminus }\{0\},u>0.\end{array}$$

(5)

A function $u:[-T,T]\to \mathbb{R}$ is said to be a solution of the problem (4) if $u\in C^1([-T,T])$ with $u(-T)=u(T)=0$ is such that $u^{\mathrm{\prime }}{|u^{\mathrm{\prime }}|}^{p-2}$ is absolutely continuous and $L_{p}u(x)+f(x,u(x))=0$ holds a.e. in $(-T,T)$.

We formulate an extension of Lemma 1 of [9] for p-Laplacian nonlinear equations. The result of Korman and Ouyang is one-dimensional analogue of the result of Gidas, Ni and Nirenberg [10] for symmetry of positive solutions of semilinear Laplace equations. In the case of p-Laplacian equations, the symmetry of solutions in higher dimensions is discussed by Reihel and Walter [11].

Theorem 4 Assume that $f\in C^1([-T,T]\times {\mathbb{R}}^{+})$ satisfies (5). Then any positive solution u of (4) is an even function such that $max\{u(x):-T\le x\le T\}=u(0)$ and $u^{\mathrm{\prime }}(x)<0$ for $x\in (0,T]$.

Remark 1 Let us note that if the function f satisfies (5), but u is not a positive solution of (4), then u is not necessarily an even function. A simple counter example in the case $p=2$ is the problem

$$\{\begin{array}{c}{u}^{\mathrm{\prime }\mathrm{\prime }}+u-x^2+{\pi }^2-2=0,-\pi <x<\pi ,\hfill \\ u(-\pi )=u(\pi )=0.\hfill \end{array}$$

The term $f(x,u)=u-x^2+{\pi }^2-2$ satisfies (5) in the interval $(-\pi ,\pi )$, but the solution of the problem $u(x)=x^2-{\pi }^2+sinx$ is negative in $(-\pi ,\pi )$ and not an even function. Its graph is presented in Figure 1. It would be more interesting to show an example for the case $p>2$ and f satisfying the additional assumption $f(x,0)=0$.

Figure 1

Graph of the functions $\mathit{u}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{\pi }}^{\mathbf{2}}\mathbf{+}\mathbf{sin}\mathit{x}$ .

Sketch of Proof of Theorem 4 Suppose that the function u has only one global maximum on $[-T,T]$.

Assume that the function $u(x)$ has a finite number of local minima in the interval $[0,T]$, and let $x_1$ be the largest local minimum. Let $\overline{x}\in [x_1,T]$ be the local maximum and $\tilde{x}\in [\overline{x},T]$ be such that $u(x_1)=u(\tilde{x})$. Denote $u_1=u(x_1)=u(\tilde{x})$ and $u_2=u(\overline{x})$, and let $x=\alpha (u)$ and $x=\beta (u)$ be the inverse functions of the function $u=u(x)$ in the intervals $[x_1,\overline{x}]$ and $[\overline{x},T]$, respectively. Multiplying the equation in (4) by $u^{\mathrm{\prime }}$ and integrating in $[x_1,\tilde{x}]$, we obtain by Lemma 3 and (5):

$$\begin{array}{rcl}0& =& {\int }_{x_1}^{\tilde{x}}(L_p(u)u^{\mathrm{\prime }}+f(x,u)u^{\mathrm{\prime }})dx\\ =& \frac{p-1}{p}{\left|u^{\mathrm{\prime }}\right|}^p(\tilde{x})+{\int }_{x_1}^{\overline{x}}f(x,u)u^{\mathrm{\prime }}dx+{\int }_{\overline{x}}^{\tilde{x}}f(x,u)u^{\mathrm{\prime }}dx\\ =& \frac{p-1}{p}{\left|u^{\mathrm{\prime }}\right|}^p(\tilde{x})+{\int }_{u_1}^{u_2}(f(\alpha (u),u)-f(\beta (u),u))du\\ >& 0,\end{array}$$

which leads to contradiction. One can prove the last fact using other arguments; see, for instance, Theorem 2.1 of [12]. Suppose now that u has infinitely many local minima in $[-T,x^{\ast }]$. Further, we can follow the steps of the proof of Lemma 1 of [9] with corresponding modifications based on Lemma 3. □

3 Proof of the main result

Let $X_T=W_0^{1,p}(-T,T)$ be the Sobolev space of p-integrable absolutely continuous functions $u:[-T,T]\to \mathbb{R}$ such that

$${\parallel u\parallel }_T^p={\int }_{-T}^T({|u^{\mathrm{\prime }}(x)|}^p+{|u(x)|}^p)dx<\mathrm{\infty }$$

and $u(-T)=u(T)=0$. Note that if $a(x)$ is strictly positive and bounded, i.e., there exist a and A such that $0<a\le a(x)\le A$, then ${\parallel u\parallel }_{a,T}^p={\int }_{-T}^T({|u^{\mathrm{\prime }}(x)|}^p+a(x){|u(x)|}^p)dx$ is an equivalent norm in $X_T$.

We need an extension to the p-case of the following proposition by Rabinowitz [13].

Proposition 5 Let $u\in W_{\mathrm{loc}}^{1,p}(\mathbb{R})$. Then:

• (i) If $T\ge 1$, for $x\in [T-1/2,T+1/2]$,

  $$\underset{x\in [T-1/2,T+1/2]}{max}|u(x)|\le 2^{\frac{p-1}{p}}{\left({\int }_{T-1/2}^{T+1/2}({|u^{\mathrm{\prime }}(t)|}^p+{|u(t)|}^p)dt\right)}^{1/p}.$$

  (6)

• (ii) For every $u\in W_0^{1,p}(-T,T)$,

  $${\parallel u\parallel }_{L^{\mathrm{\infty }}(-T,T)}\le 2^{\frac{p-1}{p}}{\parallel u\parallel }_T.$$

  (7)

Proof of Proposition 5 Let $x,t\in [T-1/2,T+1/2]$. It follows

$$|u(x)|\le |u(t)|+{\int }_{T-1/2}^{T+1/2}|u^{\mathrm{\prime }}(s)|ds.$$

Integrating with respect to $t\in [T-1/2,T+1/2]$ and using the Hölder and Jensen inequalities, we obtain

$$\begin{array}{rcl}|u(x)|& \le & {\int }_{T-1/2}^{T+1/2}|u(t)|dt+{\int }_{T-1/2}^{T+1/2}|u^{\mathrm{\prime }}(s)|ds\\ \le & {\left({\int }_{T-1/2}^{T+1/2}{|u(t)|}^{p}dt\right)}^{1/p}+{\left({\int }_{T-1/2}^{T+1/2}{|u^{\mathrm{\prime }}(t)|}^{p}dt\right)}^{1/p}\\ \le & 2^{\frac{p-1}{p}}{\left({\int }_{T-1/2}^{T+1/2}({|u^{\mathrm{\prime }}(t)|}^p+{|u(t)|}^p)dt\right)}^{1/p}.\end{array}$$

• (ii) Take $u\in W_0^{1,p}(-T,T)$. Since $W_0^{1,p}(-T,T)\subset C[-T,T]$, there exists $\tau \in [-T,T]$ such that by (i)

  $$\begin{array}{rcl}{\parallel u\parallel }_{L^{\mathrm{\infty }}(-T,T)}& =& {\parallel u\parallel }_{C[\tau -1/2,\tau +1/2]}\le 2^{\frac{p-1}{p}}{\left({\int }_{\tau -1/2}^{\tau +1/2}({|u^{\mathrm{\prime }}(t)|}^p+{|u(t)|}^p)dt\right)}^{1/p}\\ \le & 2\parallel u\parallel .\end{array}$$

 □

We are looking for positive solutions of (1), which are homoclinic, i.e., $u(x)\to 0$ and $u^{\mathrm{\prime }}(x)\to 0$ as $|x|\to \mathrm{\infty }$. Firstly, we look for positive solutions of the problem

A function $u:[-T,T]\to \mathbb{R}$ is said to be a solution of the problem ($P_T$) if $u\in C^1([-T,T])$ with $u(-T)=u(T)=0$ is such that ${\phi }_p(u^{\mathrm{\prime }})$ is absolutely continuous and ${({\phi }_p(u^{\mathrm{\prime }}))}^{\mathrm{\prime }}(x)-a(x){\phi }_p(u)(x)+\lambda b(x){\phi }_q(u)(x)=0$ holds a.e. in $(-T,T)$.

A function $u:[-T,T]\to \mathbb{R}$ is said to be a weak solution of the problem ($P_T$) if

$${\int }_{-T}^T({\left({\phi }_p\left(u^{\mathrm{\prime }}\right)\right)}^{\mathrm{\prime }}{v}^{\mathrm{\prime }}dx+a(x){\phi }_p(u)v-\lambda b(x){\phi }_q(u)v)dx=0,\mathrm{\forall }v\in W_0^{1,p}((-T,T)).$$

Standard arguments show that a weak solution of the problem ($P_T$) is a solution of ($P_T$) (see [14] and [15]). Consider the modified problem

where $u^{+}=max(u,0)$. It is easy to see that solutions of the problem ($P_T^{+}$) are positive solutions of the problem ($P_T$). Indeed, if $u(x)$ is a solution of ($P_T^{+}$) and $u(x)$ has negative minimum at $x_0\in (-T,T)$, since for $p\ge 2$, ${({\phi }_p(u^{\mathrm{\prime }}))}^{\mathrm{\prime }}(x_0)\ge 0$, by the equation ${({\phi }_p(u^{\mathrm{\prime }}))}^{\mathrm{\prime }}-a(x){\phi }_p(u)+\lambda b(x){(u^{+})}^{q-1}=0$, we reach a contradiction

$$0={\left({\phi }_p\left(u^{\mathrm{\prime }}\right)\right)}^{\mathrm{\prime }}(x_0)+a(x_0){(-u(x_0))}^{p-1}>0.$$

Then $u(x)\ge 0$ and u is a solution of ($P_T$). We use a variational treatment of the problem ($P_T^{+}$), considering the functional $J_T:X_T\to \mathbb{R}$

$$J_T(u)={\int }_{-T}^T(\frac{1}{p}({|u^{\mathrm{\prime }}(x)|}^p+a(x){|u(x)|}^p)-\frac{\lambda }{q}b(x){(u^{+}(x))}^q)dx.$$

Critical points of $J_T$ are weak solutions of ($P_T^{+}$), i.e.,

$${\int }_{-T}^T({\phi }_p\left(u^{\mathrm{\prime }}\right)v^{\mathrm{\prime }}+a(x){\phi }_p(u)v-\lambda b(x){\left(u^{+}\right)}^{q-1}v)dx,\mathrm{\forall }v\in W_0^{1,p}(-T,T)$$

and, by a standard way, they are solutions of ($P_T^{+}$). We show that $J_T$ satisfies the assumptions of the mountain-pass theorem of Ambrosetti and Rabinowitz [16].

Theorem 6 (Mountain-pass theorem)

Let X be a Banach space with norm $\parallel \cdot \parallel$, $I\in C^1(X,\mathbf{R})$, $I(0)=0$ and I satisfy the (PS) condition. Suppose that there exist $r>0$, $\alpha >0$ and $e\in X$ such that $\parallel e\parallel >r$

• (i) $I(x)\ge \alpha$ if $\parallel x\parallel =r$,

• (ii) $I(e)<0$. Let $c={inf}_{\gamma \in \mathrm{\Gamma }}\{{max}_{0\le t\le 1}I(\gamma (t))\}\ge \alpha$, where

  $$\mathrm{\Gamma }=\{\gamma \in C([0,1],X):\gamma (0)=0,\gamma (1)=e\}.$$

Then c is a critical value of I, i.e., there exists $x_0$ such that $I(x_0)=c$ and $I^{\mathrm{\prime }}(x_0)=0$.

Next, denote by $C_j$ several positive constants.

Lemma 7 Let $2\le p<q$, $\lambda >0$ and assumptions (H) hold. Then for every $T>0$, the problem ($P_T$) has a positive solution $u_{T,\lambda }$. Moreover, there is a constant $K>0$, independent of T, such that

$${\parallel u_{T,\lambda }\parallel }_T\le K.$$

(8)

Proof Step 1. $J_T$ satisfies the (PS) condition.

Let ${(u_k)}_k\subset X_T$ be a sequence, and suppose there exist $C_1$ and $k_0$ such that for $k\ge k_0$

$$|J_T(u_k)|=\left|{\int }_{-T}^T(\frac{1}{p}({|u_k^{\mathrm{\prime }}(x)|}^p+a(x){|u_k(x)|}^p)-\frac{\lambda }{q}b(x){(u_k^{+}(x))}^q)dx\right|\le \frac{C_1}{p},$$

(9)

and

$$\left|〈J_T(u_k),u_k〉\right|=\left|{\int }_{-T}^T({|u_k^{\mathrm{\prime }}(x)|}^p+a(x){|u_k(x)|}^p-\lambda b(x){(u_k^{+}(x))}^q)dx\right|\le {\parallel u_k\parallel }_T.$$

(10)

Let us denote $\hat{a}=min(1,a)$. From (9) and (10), it follows that

$$C_1\ge {\int }_{-T}^T(({|u_k^{\mathrm{\prime }}(x)|}^p+a(x){|u_k(x)|}^p)-\frac{\lambda p}{q}b(x){(u_k^{+}(x))}^q)dx\ge -C_1$$

and

$${\parallel u_k\parallel }_T\ge {\int }_{-T}^T(-{|u_k^{\mathrm{\prime }}(x)|}^p-a(x){|u_k(x)|}^p+\lambda b(x){(u_k^{+}(x))}^q)dx\ge -{\parallel u_k\parallel }_T$$

Then

$$C_1+{\parallel u_k\parallel }_T\ge \lambda \frac{(q-p)b}{p}{\int }_{-T}^T{(u_k^{+}(x))}^{q}dx,$$

and

$$\begin{array}{rcl}\hat{a}{\parallel u_k\parallel }_T^p-C_1& \le & {\int }_{-T}^T({|u_k^{\mathrm{\prime }}(x)|}^p+a(x){|u_k(x)|}^p)dx-C_1\\ \le & \frac{\lambda p}{q}{\int }_{-T}^{T}b(x){(u_k^{+}(x))}^{q}dx\le \frac{\lambda pB}{q}{\int }_{-T}^T{(u_k^{+}(x))}^{q}dx.\end{array}$$

We have

$$\hat{a}{\parallel u_k\parallel }_T^p-C_1\le \frac{B}{q(q-p)b}(C_1+{\parallel u_k\parallel }_T),$$

which implies that the sequence ${(u_k)}_k$ is bounded in $X_T$. By the compact embedding $X_T\subset C([-T,T])$, there exist $u\in X_T$ and the subsequence of ${(u_k)}_k$, still denoted by ${(u_k)}_k$, such that $u_k\rightharpoonup u$ weakly in $X_T$ and $u_k\to u$ strongly in $C([-T,T])$. We will show that $u_k\to u$ strongly in $X_T$ using Lemma 2. By uniform convergence of $u_k$ to u in $C([-T,T])$, it follows that

and

$$〈{\phi }_p(u_k)-{\phi }_p(u),a(x)(u_k-u)〉-〈{\phi }_q(u_k)-{\phi }_q(u),b(x)(u_k-u)〉\to 0,k\to \mathrm{\infty }.$$

Then

$$〈{\phi }_p\left(u_k^{\mathrm{\prime }}\right)-{\phi }_p\left(u^{\mathrm{\prime }}\right),u_k^{\mathrm{\prime }}-u^{\mathrm{\prime }}〉\to 0,k\to \mathrm{\infty },$$

and by Lemma 2,

$$〈{\phi }_p\left(u_k^{\mathrm{\prime }}\right)-{\phi }_p\left(u^{\mathrm{\prime }}\right),u_k^{\mathrm{\prime }}-u^{\mathrm{\prime }}〉\ge ({\left|u_k^{\mathrm{\prime }}\right|}_p^{p-1}-{\left|u^{\mathrm{\prime }}\right|}_p^{p-1})({\left|u_k^{\mathrm{\prime }}\right|}_p-{\left|u^{\mathrm{\prime }}\right|}_p)\ge 0,$$

which implies that ${|u_k^{\mathrm{\prime }}|}_p\to {|u^{\mathrm{\prime }}|}_p$. Then ${\parallel u_k\parallel }_T\to {\parallel u\parallel }_T$ and by the uniform convexity of the space $X_T$, it follows that ${\parallel u_k-u\parallel }_T\to 0$, as $k\to \mathrm{\infty }$.

Step 2. Geometric conditions.

Obviously, $J_T(0)=0$. By assumption (H) it follows

$$\begin{array}{rcl}{J}_T(u)& \ge & \frac{\hat{a}}{2p}{\parallel u\parallel }_T^p+{\int }_{-T}^T(\frac{a(x)}{2p}{|u(x)|}^p-\frac{\lambda p}{q}b(x){(u^{+}(x))}^q)dx\\ \ge & \frac{\hat{a}}{2p}{\parallel u\parallel }_T^p+{\int }_{-T}^T{|u(x)|}^p(\frac{a}{2p}-\frac{\lambda p}{q}b(x){|u(x)|}^{q-p})dx>0\end{array}$$

if ${\parallel u\parallel }_T=\rho :={(\frac{aq}{2\lambda p^2})}^{1/(q-p)}>0$. Then $J_T(u)\ge \frac{\hat{a}{\rho }^p}{2p}>0$.

Let $u_0(x)\in X_T$ be such that $u_0(x)>0$ if $x\in (-T,T)$ and also $u_0(-T)=u_0(T)=0$. Consider the function

$${\hat{u}}_0(x)=\{\begin{array}{cc}\mu u_0(x),\hfill & \text{if }x\in [-1,1],\hfill \\ 0,\hfill & \text{if }x\in [-T,T]\mathrm{\setminus }[-1,1].\hfill \end{array}$$

Then

$$J_T({\hat{u}}_0)={\mu }^p{\int }_{-T}^T\frac{1}{p}({|u_0^{\mathrm{\prime }}(x)|}^p+a(x){|u_0(x)|}^p)dx-{\mu }^q{\int }_{-T}^T\frac{\lambda }{q}b(x){(u_0(x))}^{q}dx<0,$$

for μ large enough.

By the mountain-pass theorem, there exists a solution $u_{T,\lambda }\in X_T$ such that

$$c_T=J_T(u_{T,\lambda })=\underset{\gamma \in {\mathrm{\Gamma }}_T}{inf}\underset{t\in [0,1]}{max}{J}_T(\gamma (t)),J_T^{\mathrm{\prime }}(u_{T,\lambda })=0,$$

(11)

where

$${\mathrm{\Gamma }}_T=\{\gamma (t)\in C([0,1],X_T):\gamma (0)=0,\gamma (1)={\hat{u}}_0(x)\}.$$

Moreover, using the variational characterization (11), we have

$$c_T\ge \frac{\hat{a}{\rho }^p}{2p}>0.$$

Therefore, $u_{T,\lambda }$ is a nontrivial and positive solution of ($P_T$). By Theorem 4, $max\{u_{T,\lambda }:-T\le x\le T\}=u_{T,\lambda }(0)$ and $u_{T,\lambda }^{\mathrm{\prime }}(x)<0$ for $x\in (0,T]$.

Step 3. Uniform estimates.

Let $T_1\ge T\ge 1$. By continuation with zero of a function $u\in X_T$ to $[-T_1,T_1]$, we have $X_T\subset X_{T_1}$and ${\mathrm{\Gamma }}_T\subset {\mathrm{\Gamma }}_{T_1}$. Using the variational characterization (11), we infer that $c_{T_1}\le c_T\le c_1$ and then

$${\int }_{-T}^T(\frac{1}{p}({|u_{T,\lambda }^{\mathrm{\prime }}(x)|}^p+a(x)u_{T,\lambda }^p(x))-\frac{\lambda }{q}b(x)u_{T,\lambda }^q(x))dx\le c_1.$$

(12)

Multiplying the equation of ($P_T$) by $u_T$ and integrating by parts, we have

$${\int }_{-T}^T({\left|u_{T,\lambda }^{\mathrm{\prime }}\right|}^p+a(x)u_{T,\lambda }^p)dx={\int }_{-T}^T\lambda b(x)u_{T,\lambda }^{q}dx.$$

Then by (12),

$$\begin{array}{rcl}{c}_1& \ge & {\int }_{-T}^T(\frac{1}{p}({\left|u_{T,\lambda }^{\mathrm{\prime }}\right|}^p+a(x)u_{T,\lambda }^p)-\frac{\lambda }{q}\lambda b(x)u_{T,\lambda }^q)dx\\ \ge & (\frac{1}{p}-\frac{1}{q}){\int }_{-T}^T({\left|u_{T,\lambda }^{\mathrm{\prime }}\right|}^p+a(x)u_{T,\lambda }^p)dx\ge \frac{\hat{a}(q-p)}{pq}{\parallel u_{T,\lambda }\parallel }_T^p.\end{array}$$

We get (8) with $K=\frac{pq{c}_1}{\hat{a}(q-p)}$, which completes the proof. □

Proof of Theorem 1 Take $T_n\to \mathrm{\infty }$ and let $u_n$ be the solution of the problem ($P_{T_n}$) given by Lemma 2. Consider the extension of $u_n$ to ℝ with zero outside $[-T_n,T_n]$ and denote it by the same symbol.

Claim 1. The sequence of functions ${(u_n)}_n$ is uniformly bounded and equicontinuous.

By (8) and the embedding of $X_{T_n}$ in $C([-T_n,T_n])$, there is $K_1$ such that ${\parallel u_n\parallel }_{L^{\mathrm{\infty }}([-T_n,T_n])}\le K_1$. Then by the equation of ($P_{T_n}$), it follows that

$${\parallel {\left({\phi }_p\left(u_n^{\mathrm{\prime }}\right)\right)}^{\mathrm{\prime }}\parallel }_{L^{\mathrm{\infty }}([-T_n,T_n])}\le K_2.$$

(13)

By the mean value theorem for every natural n and every $t\in \mathbb{R}$, there exists ${\xi }_n\in [t-1,t]$ such that

$$u_n(t)-u_n(t-1)=u_n^{\mathrm{\prime }}({\xi }_k).$$

Then, as a consequence of (13), we obtain

$$\begin{array}{rcl}\left|{\phi }_p(u_n^{\mathrm{\prime }}(t))\right|& =& |{\int }_{{\xi }_k}^t{\left({\phi }_p(u_n^{\mathrm{\prime }}(s))\right)}^{\mathrm{\prime }}ds+{\phi }_p(u_n^{\mathrm{\prime }}({\xi }_k))|\\ \le & {\int }_{t-1}^t\left|{\left({\phi }_p(u_n^{\mathrm{\prime }}(s))\right)}^{\mathrm{\prime }}\right|ds+{|u_n^{\mathrm{\prime }}({\xi }_k)|}^{p-1}\\ \le & K_2+{(|u_n(t)|+|u_n(t-1)|)}^{p-1}\\ \le & K_2+{(2K_1)}^{p-1}=:K_3^{(p-1)/p},\mathrm{\forall }t\in \mathbb{R},\end{array}$$

(14)

from which it follows ${\parallel u_n^{\mathrm{\prime }}\parallel }_{L^{\mathrm{\infty }}([-T_n,T_n])}\le K_3$ and the sequence of functions $(u_n)$ is equicontinuous. Further, we claim that the sequence ${(u_n^{\mathrm{\prime }})}_n$ is also equicontinuous.

Claim 2. The sequence of functions ${(u_n^{\mathrm{\prime }})}_n$ is equicontinuous.

To prove this statement, we follow the method given by Tang and Xiao [7]. For completeness, we present it in details.

Suppose that ${(u_n^{\mathrm{\prime }})}_n$ is not an equicontinuous sequence in $C_{\mathrm{loc}}(\mathbb{R})$. Then there exist an ${\epsilon }_0$ and sequences $(t_k^1)$ and $(t_k^2)$ such that $0<t_k^1-t_k^2<\frac{1}{k}$ and

$$|u_n^{\mathrm{\prime }}\left(t_k^1\right)-u_n^{\mathrm{\prime }}\left(t_k^2\right)|\ge {\epsilon }_0.$$

(15)

By (14), there are numbers $w^1$ and $w^2$ and the subsequence $(u_{n_k}^{\mathrm{\prime }})$ such that $u_{n_k}^{\mathrm{\prime }}(t_k^1)\to w^1$ and $u_{n_k}^{\mathrm{\prime }}(t_k^2)\to w^2$ as $k\to \mathrm{\infty }$. By (15), $|w^1-w^2|\ge {\epsilon }_0$. On the other hand, by (13) we have

$$|{\phi }_p\left(u_{n_k}^{\mathrm{\prime }}\left(t_k^2\right)\right)-{\phi }_p\left(u_{n_k}^{\mathrm{\prime }}\left(t_k^1\right)\right)|\le {\int }_{t_k^1}^{t_k^2}\left|{\phi }_p{(u_{n_k}^{\mathrm{\prime }}(s))}^{\mathrm{\prime }}\right|ds\le \frac{K_2}{k}.$$

Then passing to a limit as $k\to \mathrm{\infty }$, we obtain ${\phi }_p(w^1)={\phi }_p(w^2)$. Hence, $w^1=w^2$ which contradicts $|w^1-w^2|\ge {\epsilon }_0$. Thus, the sequence ${(u_n^{\mathrm{\prime }})}_n$ is equicontinuous.

Let $T>0$. By Claim 1 and Claim 2 and the Arzelà-Ascoli theorem, there is a subsequence of ${(u_n)}_n$, still denoted by ${(u_n)}_n$, and functions $u_{\lambda 1}$ and $v_{\lambda 1}$ of $C([-T,T])$ such that ${\parallel u_n-u_{\lambda 1}\parallel }_{C([-T,T])}\to 0$ and ${\parallel u_n^{\mathrm{\prime }}-v_{\lambda 1}\parallel }_{C([-T,T])}\to 0$. Trivially, it follows that $u_{\lambda 1}\in C^1([-T,T])$, $u_{\lambda 1}^{\mathrm{\prime }}=v_{\lambda 1}$ and ${\parallel u_n-v_{\lambda 1}\parallel }_{C^1([-T,T])}\to 0$. Repeating this procedure as in [7], we obtain that there is a subsequence of ${(u_n)}_n$, still denoted by ${(u_n)}_n$, and $u_{\lambda }$ such that $u_n\to u_{\lambda }$ in $C_{\mathrm{loc}}^1(\mathbb{R})$. The function $u_{\lambda }$ satisfies Eq. (1). Indeed, let $[x_1,x_2]$ be an interval of ℝ and $T_n>0$ such that $[x_1,x_2]\subset [-T_n,T_n]$. By the above considerations, taking a limit as $n\to \mathrm{\infty }$ in the equation

$${\left(u_n^{\mathrm{\prime }}{\left|u_n^{\mathrm{\prime }}\right|}^{p-2}\right)}^{\mathrm{\prime }}-a(x)u_n^{p-1}+\lambda b(x)u_n^{q-1}=0,x\in [x_1,x_2],$$

equivalent to

$$\begin{array}{rcl}{u}_n^{\mathrm{\prime }}{\left|u_n^{\mathrm{\prime }}\right|}^{p-2}(x)& =& u_n^{\mathrm{\prime }}{\left|u_n^{\mathrm{\prime }}\right|}^{p-2}(x_1)+{\int }_{x_1}^x(a(t)u_n^{p-1}(t)-\lambda b(t)u_n^{q-1}(t))dt\\ =& 0,x\in [x_1,x_2],\end{array}$$

we obtain

$$\begin{array}{rcl}{u}_{\lambda }^{\mathrm{\prime }}{\left|u_{\lambda }^{\mathrm{\prime }}\right|}^{p-2}(x)& =& u_{\lambda }^{\mathrm{\prime }}{\left|u_{\lambda }^{\mathrm{\prime }}\right|}^{p-2}(x_1)+{\int }_{x_1}^x(a(t)u_{\lambda }^{p-1}(t)-\lambda b(t)u_{\lambda }^{q-1}(t))dt\\ =& 0,x\in [x_1,x_2],\end{array}$$

and hence

$${\left(u_{\lambda }^{\mathrm{\prime }}{\left|u_{\lambda }^{\mathrm{\prime }}\right|}^{p-2}\right)}^{\mathrm{\prime }}-a(x)u_{\lambda }^{p-1}+\lambda b(x)u_{\lambda }^{q-1}=0,x\in [x_1,x_2].$$

Since $x_1$ and $x_2$ are arbitrary, $u_{\lambda }$ is a solution of (1). Moreover, we have

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}({|u_{\lambda }^{\mathrm{\prime }}(x)|}^p+a(x){|u_{\lambda }(x)|}^p)dx<\mathrm{\infty }.$$

(16)

It remains to show that $u_{\lambda }$ is nonzero and $u_{\lambda }(\pm \mathrm{\infty })=0$ and $u_{\lambda }^{\mathrm{\prime }}(\pm \mathrm{\infty })=0$.

By Theorem 4, $u_n$ is an even function and attains its maximum at 0. Then by Eq. (1),

$$u_n^{p-1}(0)(-a(0)+\lambda b(0)u_n^{q-p}(0))\ge 0.$$

By assumption (H)

$$u_n(0)\ge {\left(\frac{a(0)}{\lambda b(0)}\right)}^{1/(q-p)}\ge {\left(\frac{a}{\lambda B}\right)}^{1/(q-p)}=C_3>0,$$

independently of n. Hence, passing to a limit as $n\to \mathrm{\infty }$, we obtain

$$u_{\lambda }(0)\ge {\left(\frac{a}{\lambda B}\right)}^{1/(q-p)}>0.$$

Note, that this implies $max\{u_{\lambda }(x):x\in R\}=u_{\lambda }(0)\to +\mathrm{\infty }$ as $\lambda \to 0$.

From (16) and Proposition 5, it follows

(17)

so $u_{\lambda }(\pm \mathrm{\infty })=0$.

Now, we will show that $u_{\lambda }^{\mathrm{\prime }}(\mathrm{\infty })=0$. The arguments for $u_{\lambda }^{\mathrm{\prime }}(-\mathrm{\infty })=0$ are similar.

If $u_{\lambda }^{\mathrm{\prime }}(\mathrm{\infty })\ne 0$, there exist ${\epsilon }_1>0$ and a monotone increasing sequence $x_k\to \mathrm{\infty }$ such that $|u_{\lambda }^{\mathrm{\prime }}(x_k)|\ge {(2{\epsilon }_1)}^{1/(p-1)}$. Then for $x\in [x_k,x_k+\frac{{\epsilon }_1}{K_2}]$,

$$\begin{array}{rcl}{|u_{\lambda }^{\mathrm{\prime }}(x)|}^{p-1}& =& |{\phi }_p(u_{\lambda }^{\mathrm{\prime }}(x_k))+{\int }_{x_k}^x{\phi }_p{(u_{\lambda }^{\mathrm{\prime }}(t))}^{\mathrm{\prime }}dt|\\ \ge & {|u_{\lambda }^{\mathrm{\prime }}(x_k)|}^{p-1}-{\int }_{x_k}^{x_k+\frac{{\epsilon }_1}{K_2}}\left|{\phi }_p{(u_{\lambda }^{\mathrm{\prime }}(t))}^{\mathrm{\prime }}\right|dt\\ \ge & 2{\epsilon }_1-\frac{{\epsilon }_1}{K_2}\cdot K_2={\epsilon }_1,\end{array}$$

which contradicts (16).

Moreover, u is an even function that attains its only maximum at 0, since the same holds for the functions $u_n$. Arguing as in the proof of Theorem 4, we easily obtain that $u^{\mathrm{\prime }}(x)<0$ if $x>0$. □

Remark 2 A simplified method can be applied to the equations

$$u^{\mathrm{\prime }\mathrm{\prime }}-a(x)u{|u|}^{p-2}+\lambda b(x)u{|u|}^{q-2}=0,x\in \mathbb{R},$$

under assumptions (H) and $2\le p<q$, $\lambda >0$. Namely, first one looks for the even positive solutions $u_{T,\lambda }$ of the problem

$$\{\begin{array}{c}{u}^{\mathrm{\prime }\mathrm{\prime }}-a(x){\phi }_p(u)+\lambda b(x){\phi }_q(u)=0,x\in (-T,T),\hfill \\ u(-T)=u(T)=0,\hfill \end{array}$$

considering the functional $I_T:H_0^1(-T,T)\to \mathbb{R}$

$$I_T(u)={\int }_{-T}^T(\frac{1}{2}{u}^{\mathrm{\prime }}{(x)}^2+\frac{1}{p}a(x){|u(x)|}^p-\frac{\lambda }{q}b(x){(u^{+}(x))}^q)dx,$$

where $H_0^1(-T,T)$ is the Sobolev space of square integrable functions such that

$${\parallel u\parallel }^2={\int }_{-T}^T{u}^{\mathrm{\prime }}{(x)}^{2}dx<\mathrm{\infty }.$$

Since $H_0^1(-T,T)$ is a Hilbert space, compactly embedded in $C([-T,T])$ the proof of the (PS)-condition is easier. Similar considerations are made in [1] and [3]. Then, the even homoclinic solution $u_{\lambda }$ is obtained as a $C_{\mathrm{loc}}^1$ limit of the sequence $u_{T,\lambda }$. Note that in this case, the even homoclinic solution $u_{\lambda }$ of Eq. (3) satisfies

$$max\{u_{\lambda }(x):x\in \mathbb{R}\}=u_{\lambda }(0)\ge {\left(\frac{a(0)}{\lambda b(0)}\right)}^{1/(q-p)},$$

and again $u_{\lambda }(0)\to +\mathrm{\infty }$ as $\lambda \to 0$. If a and b are constants, Eq. (3) is a conservative system and one can plot the phase curves ${(\frac{v}{2})}^2-a\frac{{|u|}^p}{p}+\lambda b\frac{{|u|}^q}{q}=C$ in the phase plane $(u,v)=(u,u^{\mathrm{\prime }})$. Consider the equation $u^{\mathrm{\prime }\mathrm{\prime }}-u^3+\lambda u^5=0$. The phase portrait in a $(u,v)$ plane, for $\lambda =0.5$ in the rectangle $\{(u,v):-2\le u\le 2,-1.25\le v\le 1.25\}$, is plotted on Figure 2.

Figure 2

Phase portrait of ${\mathit{u}}^{\mathbf{\prime }\mathbf{\prime }}\mathbf{-}{\mathit{u}}^{\mathbf{3}}\mathbf{+}\mathbf{0.5}{\mathit{u}}^{\mathbf{5}}\mathbf{=}\mathbf{0}$ , in $\mathbf{[}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{]}\mathbf{\times }\mathbf{[}\mathbf{-}\mathbf{1.25}\mathbf{,}\mathbf{1.25}\mathbf{]}$ .