2. The set $\mathcal{M}$ and the Pohozaev manifold

We will use the set $\mathcal{M}$ in the minimizing arguments, next we show that it is a nontrivial manifold.

Proof. Observe that for a fixed $\varphi\in C^{\infty}_{0}(\mathbb R^{N})$ , $\varphi\geq0$ , $\varphi \not\equiv 0$ , using (1.2), the function $h(t)=\int_{\mathbb R^{N}}G(t\varphi)dx $ is strictly negative for small t and by (1.5), we have that $h(t)>0$ for t large; this implies that there exists some $\bar t>0$ such that $\bar t\varphi\in \mathcal M.$ Moreover, $\mathcal M$ is a $C^{1}$ manifold. In fact, define the $C^{1}$ functional $J(u) = \int_{\mathbb R^{N}} G(u)dx -1$ , it holds from (1.4) that

(2.1) \begin{equation} J'(u)[u]=\int_{\mathbb R^{N}}(f(u)u - u^{2}-uBu) dx = \int_{\mathbb R^{N}} (f(u) u - 2F(u))dx +2\int_{\mathbb R^{N}}G(u)\geq 2, \ \forall u\in \mathcal M.\end{equation}

Proof. Consider $\zeta \in C^{\infty}_{0}(\mathbb R)$ with $\zeta(x)>0$ and define the function

\begin{equation*}h(t)=\displaystyle\int_{\mathbb R^{2}}G(t\zeta)dx=\displaystyle\int_{\mathbb R^{2}}F(t\zeta) dx-\frac {t^{2}}{2} \displaystyle\int_{\mathbb R^{2}}\zeta B(\zeta)dx-\frac {t^{2}}{2} \displaystyle\int_{\mathbb R^{2}}\zeta^{2}dx.\end{equation*}

From (1.2) for $t>0$ small we have

\begin{equation*}h(t)\leq \frac{\varepsilon -1}{2}t^{2}\displaystyle\int_{\mathbb R^{2}}\zeta^{2}dx.\end{equation*}

For $\varepsilon <1$ , we get $h(t)<0$ for $t>0$ small.

By virtue of (1.5) we obtain

\begin{equation*}h(t) \geq \tau \frac{t^{q}}{q}\displaystyle\int_{\mathbb R^{2}}\zeta^{q}dx-\frac {t^{2}}{2} \displaystyle\int_{\mathbb R^{2}}\zeta^{2}dx -\frac {t^{2}}{2} \displaystyle\int_{\mathbb R^{2}}\zeta B(\zeta)dx\end{equation*}

and

\begin{equation*}h'(t) \geq \lambda t^{q-1}\displaystyle\int_{\mathbb R^{2}}\zeta^{q}dx- t \displaystyle\int_{\mathbb R^{2}}\zeta^{2}dx - t \displaystyle\int_{\mathbb R^{2}}\zeta B(\zeta)dx.\end{equation*}

Then, $h(t)>0$ for $t>0$ large and $h'(t)>0$ for $t>0$ large. Then there is a $\overline{t}>0$ such that

\begin{equation*}\displaystyle\int_{\mathbb R^{2}}G(\overline{t}\zeta)\,dx=h(\overline{t})=0.\end{equation*}

Now we prove that $\mathcal{M}$ is a manifold. Indeed, if $\zeta\in \mathcal{M}$ , then $\zeta \not=0$ . Then, from (1.2) and the fact that $\lim_{|x|\to \infty}\zeta(x)=0 $ and $B\geq 0$ , there exists $ x_0 \in \mathbb R^{2}$ such that $g(\zeta(x_0))<0$ . Thereby, the continuity implies that there is an open interval $B_\delta(x_0)$ such that

\begin{equation*}g(\zeta(x))<0, \quad \forall x \in B_\delta(x_0).\end{equation*}

As a consequence we can always find $\phi \in C_0^{\infty}(\mathbb R^{2}) \subset X$ such that $J'(\zeta)[\phi]=\displaystyle\int_{\mathbb R^{2}}g(\zeta)\phi \,dx < 0$ , where $J(u) = \int_{\mathbb R} G(u)dx$ , showing that $J'(\zeta) \not=0$ .

We will need an estimate for the minimax value b defined by (1.16). For that matter we use the Pohozaev manifold associated to the first equation in (1.1) for $N \geq 3$ , given by

\begin{equation*}\mathcal{P}= \left\{u \in H^{1}(\mathbb{R}^{N})\;:\;\frac{N-2}{2}\displaystyle\int_{\mathbb{R}^{N}}|\nabla u|^{2} \ dx = N \displaystyle\int_{\mathbb{R}^{N}}[F(u) -u^{2}- u v] \ dx, u \neq 0 \right\},\end{equation*}

where $v= Bu$ . Then we can rewrite it as

\begin{equation*}\mathcal{P} = \left\{u \in X\;:\;\frac{N-2}{2} \int_{\mathbb R^N} | \nabla u|^2 \, dx = N \int_{\mathbb R^N} G(u) \, dx, u \neq 0 \right\},\end{equation*}

which, by a direct computation, contains every weak solution of (1.10).

We set the notation

\begin{equation*}p_0 = \inf_{u \in \mathcal{P}} I(u).\end{equation*}

Since B is linear and arguing as in [Reference Jeanjean and Tanaka25, proof of Lemma 3.1], we can see that

(2.2) \begin{equation}p_0 = \frac {1}{N} \left(\frac{N-2}{2N} \right)^{(N-2)/2} (2D)^{N/2},\end{equation}

where D is as in (1.13).

We prove next some results involving the Pohozaev manifold.

Lemma 2.3. It holds

\begin{equation*}\frac {1}{N} \left(\frac{N-2}{2N} \right)^{(N-2)/2} (2D)^{N/2} \leq b,\end{equation*}

where b is the minimax level of I defined in (1.16).

Proof. Indeed, since B is linear and arguing as [Reference Jeanjean and Tanaka25, step 3 of Lemma 1.4], for each $\overline \gamma \in \Gamma$ , it results $\overline{\gamma}([0, 1]) \cap \mathcal{P} \neq \emptyset$ . Then, there exists $t_0 \in [0, 1]$ such that $\overline{\gamma}(t_0) \in \mathcal{P}$ . The property $\int_{\mathbb{R}^{N}}u B(u) dx \geq 0$ , implies

\begin{equation*}p_0 \leq I(\overline{\gamma}(t_0))\leq \max_{t \in [0, 1]} I(\overline{\gamma}(t)).\end{equation*}

Hence, $p_0 \leq b$ and the result follows from (2.2).

3. Minimizing sequences and Lagrange multipliers

We establish some preliminary results which will be useful in order to prove Theorem 1.1. Along the paper, we will take advantage of the following rephrased form of (1.13)–(1.14), namely

(3.1) \begin{equation} D = \inf_{u\in \mathcal M} T(u), \qquad T(u) =\displaystyle\frac{1}{2}\displaystyle\int_{\mathbb{R}^{N}}|\nabla u|^{2} \ dx .\end{equation}

The next steps consist in proving the boundedness of the minimizing sequences in $H^{1}(\mathbb{R}^N)$ for D.

Proof. Let $\{ u_n \}\subset \mathcal M $ be a minimizing sequence for T, then

\begin{equation*}T(u_{n}) =\frac 12 \int_{\mathbb R^N} |\nabla u_n|^2 dx \longrightarrow D \quad \hbox{as $n \rightarrow + \infty$} \end{equation*}

and

\begin{equation*} \int_{\mathbb R^N} G(u_n) dx = 1, \quad \hbox{that is}, \quad \int_{\mathbb R^N} \left( F(u_n) -\frac{1}{2}u_n^{2} - \frac{1}{2} u_nB(u_n) \right) dx =1. \end{equation*}

Then

(3.2) \begin{equation} \frac 12\int_{\mathbb R^N} |\nabla u_{n}|^2 \, dx \leq C \quad \hbox{for all $n \in \mathbb N$ and for some constant $C > 0$ } \end{equation}

and

\begin{equation*}\int_{\mathbb R^N} F(u_n) \,dx = 1 +\frac{1}{2} \int_{\mathbb R^N} u_n^{2} \, dx+\frac{1}{2} \int_{\mathbb R^N} u_n B(u_n) \, dx.\end{equation*}

By using (1.7) with $\varepsilon = 1/2$ , we get

\begin{equation*} 1 + \frac{1}{2} \int_{\mathbb R^N} u_n^{2} \, dx+\frac{1}{2} \int_{\mathbb R^N} u_nB(u_n) \, dx \leq \frac 14 \int_{\mathbb R^N} u_n ^2 \, dx + C_{1/2} \int_{\mathbb R^N} |u_n|^{2^{*}} \, dx. \end{equation*}

Then, for every $n \in \mathbb N$ , by using (3.2), it follows

\begin{equation*}\frac 12 \int_{\mathbb R^N} u_n^2 dx \leq C_{1/4} \, \int_{\mathbb R^N} |u_n|^{2^{*}} dx \leq C_{1/4} C \int_{\mathbb R^N} |\nabla u_{n}|^2 \, dx \leq \overline C. \end{equation*}

Consequently $\{ u_n \} $ is bounded also in $L^2(\mathbb R^N)$ . Moreover, $ \displaystyle\int_{\mathbb R^N} u_n B(u_n) \, dx \leq \|B\|_{X^{-1}} \int_{\mathbb R^N} u_n^2 dx$ and this ensures its boundedness in X (here $X^{-1}$ is the dual space of X).

By the Ekeland Variational Principle (see [Reference Ekeland20]), we can assume that the minimizing sequence $\{u_{n}\}$ is also a Palais–Smale sequence, that is, using (2.1) there exists a sequence of Lagrange multipliers $\{ \lambda_n \} \subset \mathbb R$ such that

(3.3) \begin{equation}\frac 12 \int_{\mathbb R^N} |\nabla u_n|^2 dx \longrightarrow D \qquad \hbox{as $n \rightarrow + \infty$}\end{equation}

and

(3.4) \begin{equation}T'(u_n) - \lambda_n J'(u_n) \longrightarrow 0 \ \hbox{in } X^{-1}\qquad \text{ as } n \rightarrow + \infty.\end{equation}

In the remaining part of this section, $\{\lambda_{n}\}$ will be the associated sequence of Lagrange multipliers. At this point, it is useful to establish some properties of the levels D and b.

Proof. Clearly by definition $D \geq 0$ . Suppose, by contradiction, that $D =0$ . If $\{ u_n \} $ is a minimizing sequence for $D =0$ , then

\begin{equation*}\frac 12 \int_{\mathbb R^N} |\nabla u_n|^2 dx \to 0 \quad \hbox{as} \quad n \to + \infty\end{equation*}

and

\begin{equation*} 1 = \int_{\mathbb R^N} G(u_n) \, dx = \int_{\mathbb R^N} \left( F(u_n) - \frac 12 u_n^2 - \frac 12 u_nB(u_n) \right) dx. \end{equation*}

Then, for any $\varepsilon>0$ , see (1.8),

\begin{equation*}1 + \frac 12\int_{\mathbb R^N} u_n^2 \, dx + \frac 12\int_{\mathbb R^N} u_nB(u_n) \, dx = \int_{\mathbb R^N} F(u_n) \, dx \leq \frac{\varepsilon}{2} \int_{\mathbb R^N} u_n^2 \, dx+ {C_{\varepsilon}} \int_{\mathbb R^N} |u_n|^{2^{*}} \, dx \end{equation*}

so that

\begin{equation*} 1+ \frac 12(1 - \varepsilon) \int_{\mathbb R^N} u_n^2 \, dx \leq C_{\varepsilon} \int_{\mathbb R^N} |u_n|^{2^{*}} \, dx \leq C_{\varepsilon} C \int_{\mathbb R^N} |\nabla u_n|^2 dx.\end{equation*}

By choosing $\varepsilon = 1/2$ , we obtain

\begin{equation*}1 \leq C_{ 1/2} C \int_{\mathbb R^N} |\nabla u_n|^2 dx \longrightarrow 0 \qquad \hbox{ as $n \longrightarrow + \infty$}.\end{equation*}

This contradiction concludes the proof that $D > 0$ .

Lemma 3.3. The sequence of Lagrange multipliers $\{ \lambda_n \}$ associated to the minimizing sequence $\{u_{n}\}$ is bounded. More precisely, we have that

\begin{equation*}0<\liminf_{n \to +\infty}\lambda_n \leq \limsup_{n \to + \infty} \lambda_n \leq D.\end{equation*}

Hence, for some subsequence, still denoted by $\{\lambda_n\}$ , we can assume that $\lambda_n \to \lambda^{*}$ , for some $\lambda^{*} \in (0, D]$ .

Proof. By (3.4),

(3.5) \begin{equation}2T(u_{n}) =T'(u_n)[u_n] = \lambda_n J'(u_n)[u_n] +o_{n}(1).\end{equation}

Then, from (2.1)

\begin{equation*}2T(u_n) \geq 2\lambda_{n} +o_{n}(1)\end{equation*}

which implies, taking into account (3.3),

\begin{equation*}\limsup_{n \rightarrow + \infty} \lambda_n \leq \frac 12 \limsup_{n \rightarrow + \infty}\|u_{n}\|^2 =D.\end{equation*}

Since $\{u_{n}\}$ is a bounded minimizing sequence, it is easy to see that $|J'(u_{n})[u_{n}] |= |\int_{\mathbb R^{N}} g(u_{n}) u_{n}| \leq C$ , and then by (3.5) and the fact that $2T(u_{n})\to 2D>0$ , we infer that

\begin{equation*}\liminf_{n \to +\infty}\lambda_n >0.\end{equation*}

The proof is thereby completed.

In the sequel, we will show that a minimizing sequence for (3.1) can be chosen to be nonnegative and radially symmetric around the origin. Note that for our proof we do not need to consider the “odd extension” of the nonlinearity, as it is usually done in the literature to show that the minimizing sequence can be replaced by the sequence of the absolute values. In fact we will prove that the minimizing sequence can be replaced, roughly speaking, by the sequence of the positive parts.

Proof. To begin with, we recall that $F(s)=0$ for all $s \leq 0$ . Thus, $F(u_n)=F(u_n^{+})$ for all $n \in \mathbb{N}$ with $u_n^{+}=\max\{0,u_n\}$ . From this, the equality

\begin{equation*}\int_{\mathbb R^N}G(u_n)\,dx=1, \quad \forall n \in \mathbb{N}\end{equation*}

leads to

\begin{equation*}\int_{\mathbb R^N}G(u_n^+)\,dx \geq 1, \quad \forall n \in \mathbb{N}.\end{equation*}

Defining the function $h_n\;:\;[0,1] \to \mathbb R$ by

\begin{equation*}h_n(t)=\int_{\mathbb R^N}G(t u_n^+)\,dx\end{equation*}

the conditions on f yield that h is continuous with $h_n(1) \geq 1$ . Once $u_n^+ \not=0$ for all $n \in \mathbb{N}$ , the condition (1.2) ensures that $h_n(t)<0$ for t close to 0. Thus, there is $t_n \in (0,1]$ such that $h_n(t_n)=1$ , that is,

\begin{equation*}\int_{\mathbb R^N}G(t_n u_n^+)\,dx = 1, \quad \forall n \in \mathbb{N},\end{equation*}

implying that $t_n u_n^+ \in \mathcal{M}$ . We also know that

\begin{equation*}\|u_n^+\|^{2} \leq \|u_n\|^{2}.\end{equation*}

Since $t_n \in (0,1]$ , the last inequality gives

\begin{equation*}D \leq T(t_n u_n^+)\leq T(u_n)=D+o_n(1)\end{equation*}

that is,

\begin{equation*}t_n u_n^{+} \in \mathcal{M} \quad \mbox{and} \quad T(t_n u_n^+) \to D,\end{equation*}

showing that $\{t_n u_n^+\}$ is a minimizing sequence for T. Hence, without lost of generality, we can assume that $\{u_n\}$ is a nonnegative sequence.

Moreover,

\begin{equation*} \int_{\mathbb R^N} |\nabla u_n^{*}|^2 \, dx \leq \int_{\mathbb R^N} |\nabla u_n|^2 \, dx, \quad \forall n \in \mathbb{N} \end{equation*}

(see [Reference Baernstein5, Theorem 3]) and

\begin{equation*} \int_{\mathbb{R}^N}u_{n}^{*}B(u_{n}^{*})\, dx=\int_{\mathbb{R}^N}u_{n}B(u_{n})\, dx, \quad \forall n \in \mathbb{N}, \end{equation*}

\begin{equation*} \int_{\mathbb{R}^N}|u_{n}^{*}|^{2}\, dx=\int_{\mathbb{R}^N}|u_{n}|^{2}\, dx, \quad \forall n \in \mathbb{N}, \end{equation*}

which implies

\begin{equation*}\int_{\mathbb{R}^N}G(u_{n}^{*})\, dx= \int_{\mathbb{R}^N}G(u_{n})\, dx, \quad \forall n \in \mathbb{N},\end{equation*}

where $u_n^{*}$ is the Schwartz symmetrization of $u_n$ . Then, every minimizing sequence can be assumed radially symmetric, nonnegative and decreasing in $r=|x|$ .

Let $X_{rad}$ be the space of radially symmetric functions of X. In what follows, we will use that the embedding

(3.6) \begin{equation} X_{rad} \hookrightarrow L^{p}(\mathbb R^N)\end{equation}

is compact for all $p \in (2,2^{*})$ , recall the similar embedding of $H^1_{rad}(\mathbb R^N)$ and see Lions [Reference Lions29] for more details.

It turns out that the key point in the proof of Theorem 1.1 is to verify that the weak limit of $\{u_{n}\}$ in $X_{rad}$ , denoted hereafter by u, is a solution of the minimizing problem (3.1). Before verifying this in Section 4, some preliminary lemmas are required in order to recover compactness.

In the next result, we use the following constant

\begin{equation*}S= \displaystyle\inf_{u\in \mathcal{D}^{1,2}(\mathbb{R}^{N})\setminus \{0\}}\frac{\displaystyle\int_{\mathbb{R}^{N}}|\nabla u|^{2} dx}{\left(\displaystyle\int_{\mathbb{R}^{N}}|u|^{2^{*}} dx\right)^{2/2^{*}}}.\end{equation*}

Lemma 3.5. Assume that $w_n=u_n-u\rightharpoonup 0$ in $X_{rad}$ and $\| w_n\|^{2} \to L>0$ . Then

\begin{equation*}D \geq 2^{-2 /N}S.\end{equation*}

Proof. First of all, we recall (2.1) and that the limit $T'(u_n) - \lambda_n J'(u_n) \rightarrow 0$ as $n \rightarrow + \infty$ gives

\begin{equation*}T'(u_n)[u_n]-\lambda_{n} J'(u_n)[u_n]=o_n(1).\end{equation*}

Using standard arguments, it is possible to prove that

\begin{equation*}T'(u_n)[u_n]-\lambda_{n} J'(u_n)[u_n]=T'(w_n)[w_n]-\lambda_{n} J'(w_n)[w_n]+T'(u)[u]-\lambda^{*} J'(u)[u] + o_n(1)\end{equation*}

and

\begin{equation*}T'(u)-\lambda^{*} J'(u)=0 \quad \mbox{in} \quad X^{-1},\end{equation*}

where $\lambda^{*}$ is the same which appear in Lemma 3.3. Then $T'(w_n)[w_n]-\lambda_{n} J'(w_n)[w_n]=o_n(1)$ , or equivalently,

\begin{equation*} \int_{\mathbb R^N} |\nabla w_n|^2 \, dx =\lambda_n \, \int_{\mathbb R^N} f(w_n) w_n \, dx - \lambda_n \int_{\mathbb R^N} w_n^2 \, dx - \lambda_n \int_{\mathbb R^N} w_nB(w_n) \, dx + o_n(1).\end{equation*}

Using the growth conditions on f, fixed $q \in (2,2^{*})$ and given $\varepsilon >0$ , there exists $C=C(\varepsilon, q)>0$ such that

\begin{equation*}f(t)t \leq \varepsilon t^{2}+C|t|^{q}+(1+\varepsilon)|t|^{2^{*}}, \quad \forall t \in \mathbb{R}.\end{equation*}

From this,

\begin{equation*} \int_{\mathbb R^N} |\nabla w_n|^2 \, dx \leq\lambda_n \left(\varepsilon \int_{\mathbb R^N} w_n^2\, dx + C \int_{\mathbb R^N} |w_n|^q dx +(1+\varepsilon) \int_{\mathbb R^N} |w_n|^{2^{*}} \, dx\right) + o_n(1).\end{equation*}

Now, using the definition of S, we get

(3.7) \begin{eqnarray} \int_{\mathbb R^N} |\nabla w_n|^2 \, dx &\leq &\lambda_n \left(\varepsilon \int_{\mathbb R^N} w_n^2 dx + C \int_{\mathbb R^N} |w_n|^q dx +(1+\varepsilon) \left( \frac{1}{S} \int_{\mathbb R^N} |\nabla w_n|^{2} \, dx\right)^{2^{*}/ 2} \right)+ o_n(1)\nonumber\\&\leq & \lambda_n \left(\varepsilon \int_{\mathbb R^N} w_n^2 dx + C \int_{\mathbb R^N} |w_n|^q dx +(1+\varepsilon) \left( \frac{1}{S} \int_{\mathbb R^N} |\nabla w_n|^2 \, dx \right)^{2^{*}/ 2} \right)+ o_n(1).\end{eqnarray}

Taking to the limit in (3.7), recalling that $\{w_n\}$ is bounded,

(3.8) \begin{equation} \int_{\mathbb R^N} |\nabla w_n|^2 \, dx \longrightarrow L\end{equation}

and that $w_n \to 0$ in $L^{q}(\mathbb R^N)$ (see (3.6)), we obtain

\begin{equation*}L\leq \lambda^*\left( \varepsilon C_1 + (1+\varepsilon)\left(\frac{L}{S}\right)^{2^{*}/2} \right).\end{equation*}

Since $\varepsilon$ arbitrary and the fact that $\lambda^{*}\leq D$ (recall Lemma 3.3), we derive $L\leq D \left(L/S \right)^{2^{*}/2}$ , or equivalently,

(3.9) \begin{equation}S^{2^{*}/2}\leq D L^{2 /(N-2)}.\end{equation}

On the other, hand (3.8) implies that $L = 2D -\|u\|^2 \leq 2D$ . Hence (3.9) becomes

\begin{equation*}S^{2^{*}/2}\leq 2^{2 /(N-2 )} D^{2^{*}/2},\quad \text{ i.e. } \quad D \geq 2^{-2/N}S\end{equation*}

and the proof is finished.

Now we consider the equation

(3.10) \begin{equation}\left\{\begin{array}{l}-\Delta \omega + \omega + B(\omega)= |\omega|^{q-2} \omega\ \ \mbox{in $\Omega$}\\\omega \in H^{1}_{0}(\Omega)\end{array}\right.,\end{equation}

where $\Omega$ is a smooth bounded domain in $\mathbb{R}^{N}$ and q is the constant which appears in the hypothesis (1.5). The functional $I_q\;:\; H^{1,}_0(\Omega)\rightarrow \mathbb{R}$ associated to (3.10) is given by

\begin{equation*}I_q(u)=\dfrac{1}{2}\displaystyle\int\limits_{\Omega}\left[|\nabla u|^2+|u|^2 + B(u)u \right]dx-\dfrac{1}{q}\displaystyle\int\limits_{\Omega}|u|^q dx .\end{equation*}

Since $\Omega$ is bounded, the expression

\begin{equation*}\| u \|_{\Omega} = \left( \int\limits_{\Omega}\left[|\nabla u|^2+|u|^2 + B(u)u \right]dx \right)^\frac{1}{2}\end{equation*}

is equivalent the usual norm of $H^{1}_{0}(\Omega)$ , by the Mountain Pass Theorem [Reference Ambrosetti and Rabinowitz1], there exists $\omega_q\in H^{1}_{0}(\Omega)$ such that

\begin{eqnarray*}I_q(\omega_q)=c_q, \ \ \|\omega_q\|^{2}_{\Omega}=\displaystyle \int\limits_{\Omega}|\omega_q|^qdx &\text{and}&I'_q(\omega_q)=0.\end{eqnarray*}

Moreover,

(3.11) \begin{equation}c_q= \left(\dfrac{q-2}{2q}\right)\displaystyle\int\limits_{\Omega}|\omega_q|^qdx,\end{equation}

In the next result, the condition $\tau>\tau^{*}$ given in (1.5) plays a crucial role. We fix

\begin{equation*}\tau^{*}= \left[\frac{c_q N(2N/N-2)^{(N-2)/2}}{2^{(N-2)/2}S^{N/2}}\right]^{(q-2)/2}.\end{equation*}

Lemma 3.6. It holds

\begin{equation*}b < \frac {1}{ N} \left(\frac{N-2}{2N} \right)^{(N-2)/2} 2^{(N-2)/2} S^{N/2}, \quad \forall \tau > \tau^{*}.\end{equation*}

Proof. Take $\omega_q \in H^{1}_{0}(\Omega)$ a solution of problem (3.10). By (1.5) and the definition of b at (1.16), we get

\begin{eqnarray*}b \leq \max_{ t \geq 0} I(t \omega_q) & \leq &\max_{t \geq 0} \left\{ \frac{t^2}{2}\|\omega_q\|^2_{\Omega} - \tau \frac{t^q}{q} \int_{\mathbb R^N} |\omega_q|^q \, dx \right\} \leq \max_{t \geq 0} \left\{ \frac{t^2}{2} - \tau \frac{t^q}{q} \right\} \int_{\mathbb R^N} |\omega_q|^q \, dx\\& = & \max_{t \geq 0} \left\{ \frac{t^2}{2} - \tau \frac{t^q}{q} \right\} c_q \frac{2q}{q-2} = \frac{c_q}{\tau^{2/(q-2)}}.\end{eqnarray*}

The conclusion of the lemma follows.

Proof. Suppose by contradiction that $u_n \not\to u $ in $D^{1,2}(\mathbb{R}^N)$ and let $w_n = u_n - u$ . Thus $\int_{\mathbb R^N} |\nabla w_n|^{2} \, dx \to L>0$ for some subsequence. Then, by Lemma 3.5,

(3.12) \begin{equation}D \geq 2^{- 2/N} S.\end{equation}

From Lemma 2.3

\begin{equation*}\frac 1{N} \left(\frac{N-2}{2N} \right)^{(N-2)/2} (2D)^{N/2} \leq b,\end{equation*}

from which, using (3.12), it follows that

\begin{equation*}\frac {1}{N} \left(\frac{N-2}{2N} \right)^{(N-2)/2} 2^{(N-2 )/2} S^{N/2} \leq b.\end{equation*}

This contradicts Lemma 3.6 and accomplishes the proof.

4. Proof of Theorem 1.1

The results of the last section will be used next.

Proof of Theorem 1.1. We wish to show that D is attained by u, where u is the weak limit of $\{ u_n \}$ . First of all, we know that

(4.1) \begin{equation}T(u) = \frac 12\int_{\mathbb R^N} |\nabla u|^2 \, dx \leq \liminf_{n \rightarrow + \infty} \frac 12 \int_{\mathbb R^N}|\nabla u_n|^2 \, dx =D\end{equation}

so we just need to prove that $u\in \mathcal M$ .

Since

\begin{equation*}u(R)\leq \biggl(\frac{N}{w_{N-1}}\biggl)^{1/t}R^{-N/t}|u|_{L^{t}(\mathbb{R}^{N})},\end{equation*}

for all $t\geq 1$ , for $R>0$ large, we have

\begin{equation*}\frac{1}{2}u_n^{2}-F(u_n) \geq 0 \quad \forall n \in \mathbb{N} \quad \mbox{in } \mathbb R^{N} \setminus B_{R},\end{equation*}

$B_{R}$ being the ball of radius R centered in $0.$ Since $u_n \to u $ in $L^{2^{*}}(B_R)$ , reasoning with a subsequence if needed, we have $u_{n} \to u $ a.e. in $B_{R}$ and there exists $h\in L^{2^{*}}(B_{R})$ such that $|u_{n}(x)|\leq h(x)$ . Moreover, we have $F(u_{n}(x)) \to F(u(x))$ a.e. and $|F(u_{n})|\leq 2^{-1}\varepsilon u_{n}^{2} + C_{\varepsilon} |u_{n}|^{2^{*}}\leq2^{-1}\varepsilon u_{n}^{2} + C_{\varepsilon} |v|^{2^{*}} $ which, joint with the fact that $\varepsilon$ is arbitrary and the Dominated Convergence Theorem, give $\int_{B_{R}} F(u_{n})dx \to \int_{B_{R}} F(u)dx$ . Then by

\begin{equation*}\int_{B_R}F(u_n)\,dx=\frac{1}{2}\int_{\mathbb{R}^{N}}u_nB(u_n)\,dx+\frac{1}{2}\int_{B_R}u_n^{2}\,dx+\int_{\mathbb{R}^N \setminus B_R}\left(\frac{1}{2}u_n^{2}-F(u_n)\right)\,dx +1,\end{equation*}

taking into account the above considerations and the Fatou Lemma we infer

\begin{equation*}\displaystyle\int_{B_R}F(u)\, dx \geq \frac{1}{2}\int_{\mathbb{R}^{N}}uB(u)\,dx+\frac{1}{2}\int_{B_R}u^{2}dx+\int_{\mathbb{R}^N \setminus B_R}\left(\frac{1}{2}u^{2}-F(u)\right)\,dx +1\end{equation*}

which leads to

\begin{equation*}\displaystyle\int_{\mathbb R^N} G(u) \, dx \geq 1.\end{equation*}

Suppose by contradiction that

\begin{equation*}\int_{\mathbb R^N} G(u) \, dx > 1\end{equation*}

and define $h\;:\;[0, 1] \rightarrow \mathbb R$ by $h(t) = \int_{\mathbb R^N} G(tu) \, dx$ . The growth conditions on f ensure that $h(t) < 0$ for t close to 0 and $h(1) = \int_{\mathbb R^N} G(u) \, dx > 1$ . Then, by the continuity of h, there exists $t_0 \in (0, 1)$ such that $h(t_0)=1$ . Then,

\begin{equation*}\int_{\mathbb R^N} G(t_0 u) \, dx = 1 \text{ if and only if } t_0 u \in \mathcal{M}.\end{equation*}

Consequently, by (4.1)

\begin{equation*}D \leq T(t_0 u) = \frac{t_0^2}{2} \int_{\mathbb R^N} |\nabla u|^2 \, dx = t_0^2 \, T(u) \leq t_0^2 \, D < D\end{equation*}

which is an absurd. Thus $ \int_{\mathbb R^N} G(u) \, dx =1 $ , i.e. $ u\in \mathcal M$ . The fact that the solution u of the minimizing problem gives rise to a solution of (1.10), follows by standard arguments; indeed, since u is a solution of the minimizing problem (3.1), i.e. $D =T(u)= \inf_{w\in \mathcal M}T(w)$ , then there exists an associated Lagrange multiplier $\lambda$ such that, in the weak sense,

\begin{equation*}-\Delta u = \lambda g( u).\end{equation*}

Testing the previous equation on the same minimizer u, we deduce

\begin{equation*}2T(u)=\lambda\int_{\mathbb R^{N}}g(u)u dx = \lambda J'(u)[u] \geq 2\lambda .\end{equation*}

so that, by Lemma 3.2, $T(u)\geq\lambda>0$ . Setting $u_{\sigma}(x)=u(\sigma x)$ for $\sigma>0$ , we easily see that

\begin{equation*}-\Delta u_{\sigma} = \lambda \sigma^{2}g(u_{\sigma}).\end{equation*}

Choosing $\sigma=\lambda^{-1/2}$ we obtain a solution of (1.10). Arguing as in [Reference Berestycki and Lions10, Theorem 3], $u_\sigma$ is a solution of (1.10). Rewriting $u_{\sigma}$ as u, and since $v=B(u)$ , we conclude that (u, v) is a solution of (1.1).

5. Proof of Theorem 1.2

We establish preliminary results in the entire plan which will be useful in order to prove Theorem 1.2. Here we point out that, we will deal with minimizing sequences $\{u_{n}\}$ for the minimization problem (1.13)–(1.15), rewritten in the form

(5.1) \begin{equation} D = \inf_{u\in \mathcal M} T(u), \qquad T(u) =\displaystyle\frac{1}{2}\displaystyle\int_{\mathbb{R}^{N}}|\nabla u|^{2} \ dx .\end{equation}

We suppose that $u_{n}$ is nonnegative and radially symmetric.

It is useful to state the next result which is a consequence of [Reference Alves, Montenegro and Souto3, Lemma 5.1]. The proof is based on Trudinger–Moser inequality and the Compactness Lemma of Strauss [Reference Berestycki and Lions10, Theorem A.I].

Lemma 5.1. Assume that f satisfies (1.2), (1.5), and (1.6). Let $\{w_n\}\subset X_{rad}$ be a sequence of radially symmetric functions such that

\begin{equation*}w_n \rightharpoonup w \ \ \mbox{in} \ \ X\end{equation*}

and

\begin{equation*}\displaystyle\sup_{n}|\nabla w_n|^{2}_{2}=\rho \leq \frac 12 \ \ \mbox{and} \ \ \displaystyle\sup_{n}|w_n|^{2}_{2}=M <\infty.\end{equation*}

Then,

\begin{equation*}\displaystyle\int_{\mathbb R^{2}}F(w_n)dx\longrightarrow \displaystyle\int_{\mathbb R^{2}}F(w)dx.\end{equation*}

The relation between the minimum D and the minimax level b defined in (1.16) is stated in the following lemma.

Proof. Arguing as in Lemma 2.2, given $\eta \in X$ with $\eta^{+}=\max\{\eta,0\}\neq 0$ , there is $t_{0}>0$ such that $t_0 \eta^{+} \in \mathcal{M}$ . Then,

\begin{equation*}D\leq \frac{t_{0}^{2}}{ 2} \int_{\mathbb R^{2}} |\nabla \eta^{+}|^2 \, dx = I(t_{0}\eta^{+})\leq \displaystyle\max_{t\geq 0}I(t\eta^{+}).\end{equation*}

Due to the fact that $f(s)=0$ for $s\leq 0$ , for $\eta\in X$ , $\eta\neq 0$ with $\eta^{+}=0$ , we obtain $\max_{t\geq 0}I(tv)=\infty$ . Hence in any case $D\leq b$ .

Proof. By definition $D\geq 0$ . Assume by contradiction that $D=0$ an let $\{u_n\}$ be a (nonnegative and radially symmetric) minimizing sequence in $X_{rad}$ for T, that is,

\begin{equation*}\displaystyle\int_{\mathbb R^{2}} |\nabla u_n|^2 dx \to 0 \ \ \mbox{and} \ \ \displaystyle\int_{\mathbb R^{2}} G(u_n) \, dx =0.\end{equation*}

For each $\mu_n>0$ , the function $\xi_n(x)=u_n( x / \mu_n)$ satisfies

\begin{equation*}\displaystyle\int_{\mathbb R^{2}} |\nabla \xi_n|^2dx=\displaystyle\int_{\mathbb R^{2}} |\nabla u_n|^2dx \ \ \mbox{and} \ \ \displaystyle\int_{\mathbb R^{2}} G(\xi_n) \, dx =0.\end{equation*}

Since

\begin{equation*}\displaystyle\int_{\mathbb R^{2}} \xi_n^{2}dx+ \displaystyle\int_{\mathbb R^{2}} \xi_nB(\xi_n)dx = \mu_{n}^{2}\biggl[\displaystyle\int_{\mathbb R^{2}} u_n^{2}dx+ \displaystyle\int_{\mathbb R^{2}} u_nB(u_n)dx\biggl],\end{equation*}

we choose $\mu_{n}^{2}=\biggl[\displaystyle\int_{\mathbb R^{2}} u_n^{2}dx+ \displaystyle\int_{\mathbb R^{2}} u_nB(u_n)dx\biggl]^{-1}$ to obtain

\begin{equation*}\displaystyle\int_{\mathbb R^{2}} |\nabla \xi_n|^2 dx\to 0, \ \ \biggl[\displaystyle\int_{\mathbb R^{2}} u_n^{2}dx+ \displaystyle\int_{\mathbb R^{2}} u_nB(u_n)dx\biggl]=1 \ \ \mbox{and} \ \ \displaystyle\int_{\mathbb R^{2}} G(\xi_n) \, dx =0.\end{equation*}

We can assume that there exists $ v \in X_{rad}$ , radially symmetric, such that $\xi_n \rightharpoonup \xi$ in $X_{rad}$ . From Lemma 5.1, we get

\begin{equation*}\displaystyle\int_{\mathbb R^{2}}F(\xi_n)dx\longrightarrow \displaystyle\int_{\mathbb R^{2}}F(\xi)dx.\end{equation*}

Notice that $\int_{\mathbb R^{2}} G(\xi_n) \, dx =0$ implies $\int_{\mathbb R^{2}} F(\xi_n) \, dx =1/2$ and $\int_{\mathbb R^{2}} F(\xi) \, dx = 1/2$ . Then $v \neq 0$ . But

\begin{equation*}\displaystyle\int_{\mathbb R^{2}} |\nabla \xi|^2dx \leq \liminf_{n\to+\infty} \displaystyle\int_{\mathbb R^{2}} |\nabla \xi_n|^2dx\longrightarrow 0,\end{equation*}

implies $\xi=0$ , which is an absurd.

Proof. It is enough to repeat the same argument of the proof of Lemma 3.6, recalling now that $N=2$ and by (1.5) we choose

\begin{equation*}\tau* = \left(\frac{c_q}{4}\right)^{(q-2)/2}.\end{equation*}

Proof of Theorem 1.2. We will show that D is attained by u, where u is the weak limit of $\{ u_n \}$ . Indeed, since $u_n \rightharpoonup u$ in $X_{rad}$ we have

(5.2) \begin{equation}T(u) = \frac 12\int_{\mathbb R^{2}} |\nabla u|^2 \, dx \leq \liminf_{n \rightarrow + \infty} \frac 12 \int_{\mathbb R^{2}}|\nabla u_n|^2 \, dx =D.\end{equation}

Moreover, by Lemma 5.1 we have

\begin{equation*}\displaystyle\int_{\mathbb{R}^{2}}F(u) dx= \displaystyle\liminf_{n\to\infty}\displaystyle\int_{\mathbb{R}^{2}}F(u_n) dx\geq \frac 12 \displaystyle\int_{\mathbb{R}^{2}}u^{2} dx + \frac 12 \displaystyle\int_{\mathbb{R}^{2}}uB(u)dx\end{equation*}

leading to

\begin{equation*}\displaystyle\int_{\mathbb R^{2}} G(u) \, dx \geq 0.\end{equation*}

As in the previous case $N\geq3$ , we just need to prove that $u\in \mathcal M$ , i.e. $\int_{\mathbb R^N} G(u) \, dx = 0$ .

We again argue by contradiction. Suppose that

\begin{equation*}\int_{\mathbb R^{2}} G(u) \, dx > 0.\end{equation*}

As in the proof of Lemma 2.2, we set $h\;:\;[0, 1] \rightarrow \mathbb R$ by $h(t) = \int_{\mathbb R} G(tu) \, dx$ . Using the growth condition of f, we have $h(t) < 0$ for t close to 0 and $h(1) = \int_{\mathbb R} G(u) \, dx > 0$ . Hence, by the continuity of h, there exists $t_0 \in (0, 1)$ such that $h(t_0)=0$ , that is, $t_{0}u\in \mathcal M$ . Consequently, by (5.2)

\begin{equation*}D \leq T(t_0 u) = \frac{t_0^2}{2} \int_{\mathbb R^2} |\nabla u|^2 \, dx = t_0^2 \, T(u) \leq t_0^2 \, D < D\end{equation*}

which is an absurd.

As for the case $N\geq3$ , one shows that the minimizer u of (5.1) gives rise to a solution of (1.10), and then (u, v) with $v=B(u)$ solves (1.1).