On the Brezis–Nirenberg problem for a Kirchhoff type equation in high dimension

Abstract

The present paper deals with a parametrized Kirchhoff type problem involving a critical nonlinearity in high dimension. Existence, non existence and multiplicity of solutions are obtained under the effect of a subcritical perturbation by combining variational properties with a careful analysis of the fiber maps of the energy functional associated to the problem. The particular case of a pure power perturbation is also addressed. Through the study of the Nehari manifolds we extend the general case to a wider range of the parameters.

Appendices

Appendix A: Some topological properties of the Nehari manifolds

We collect some topological properties concerning the Nehari manifold ${\mathcal {N}}^-$. Since the dependency on each parameter will be considered, we will write the full notation $\Phi _{a,b,\lambda }$, $t_{a,b,\lambda }^-(u)$, ${\mathcal {N}}_{a,b,\lambda }^-$ and so on.

Similarly to Proposition 4.2 we can prove:

Lemma A.1

For each $a>0$, $b\in {\mathbb {R}}$, $\lambda \in {\mathbb {R}}$ and $u\in H^1_0(\Omega ){\setminus }\{0\},$ only one of the next $i)-iv)$ occurs.

• (i) The function $\psi _{a,b,\lambda ,u}$ is increasing and has no critical points.

• (ii) The function $\psi _{a,b,\lambda ,u}$ has only one critical point in $]0,+\infty [$ at the value $t_{a,b,\lambda }(u)$. Moreover, $\psi ''_{a,b,\lambda ,u}(t_{a,b,\lambda }(u))=0$ and $\psi _{a,b,\lambda ,u}$ is increasing.

• (iii) The function $\psi _{a,b,\lambda ,u}$ has only two critical points, $0< t^-_{a,b,\lambda } (u) < t^+_{a,b,\lambda } (u)$. Moreover, $t^-_{a,b,\lambda }(u)$ is a local maximum and $t^+_{a,b,\lambda }(u)$ is a local minimum with $\psi _{a,b,\lambda ,u}''(t^-_{a,b,\lambda }(u))<0<\psi _{a,b,\lambda ,u}''(t^+_{a,b,\lambda }(u))$.

• (iv) The function $\psi _{a,b,\lambda ,u}$ has only one critical point in $]0,+\infty [$ at the value $t^-_{a,b,\lambda }(u)$. Moreover, $t^-_{a,b,\lambda }(u)$ is a local maximum and $\psi _{a,b,\lambda ,u}''(t^-_{a,b,\lambda }(u))<0$.

Remark A.1

If $b\le 0$ and $\lambda \ge 0$, then only item iv) of Lemma A.1 occurs. Moreover, if $b>0$, then only one of the items $i)-iii)$ occurs.

Lemma A.2

Fix $u\in H_0^1(\Omega ){\setminus }\{0\}$ and $a>0$. Let $V\subset {\mathbb {R}}^2$ be an open set and assume that $t_{a,b,\lambda }^-(u)$ is defined for all $(b,\lambda )\in V$. Then the function $V\ni (b,\lambda )\mapsto t_{a,b,\lambda }^-(u)$ is $C^1$. Moreover the following holds true.

• (i) The functions $t_{a,b,\lambda }^-(u)$ and $\psi _{a,b,\lambda ,u}(t_{a,b,\lambda }^-(u))$ are increasing with respect to b;

• (ii) The functions $t_{a,b,\lambda }^-(u)$ and $\psi _{a,b,\lambda ,u}(t_{a,b,\lambda }^-(u))$ are decreasing with respect to $\lambda$.

Proof

Denote $t_{b,\lambda }=t_{a,b,\lambda }^-(u)$ and note from the implicit function theorem that $\psi _{a,b,\lambda ,u}'(t_{b,\lambda })=0$ and $\psi _{a,b,\lambda ,u}''(t_{b,\lambda })<0$ implies that $t_{b,\lambda }$ is $C^1$ as a function of $(b,\mu ,\lambda )\in V$. Since

$$\begin{aligned} at_{b,\lambda }^2\Vert u\Vert ^2+bt_{b,\lambda }^4\Vert u\Vert ^4-t_{b,\lambda }^{2^*}\mu \Vert u\Vert _{2^*}^{2^*}-\lambda t_{b,\lambda }^p\Vert u\Vert _p^p=0, \end{aligned}$$

we conclude by differentiating both sides, with respect to b, that

$$\begin{aligned} \frac{\partial {t_{b,\lambda }}}{\partial b}=-\frac{t_{b,\lambda }^4\Vert u\Vert ^4}{\psi _{a,b,\lambda ,u}''(t_{b,\lambda })}>0, \end{aligned}$$

and hence $t_{b,\lambda }$ is increasing in b. Now let $f(b)=\psi _{a,b,\lambda ,u}(t_{a,b,\lambda }^-(u))$ and observe that

$$\begin{aligned} f'(b)=\frac{\partial {t_{b,\lambda }}}{\partial b}\psi _{a,b,\lambda ,u}'(t_{b,\lambda })+\frac{t_{b,\lambda }^4\Vert u\Vert ^4}{4}>0, \end{aligned}$$

which implies that f is increasing and hence (i) is proved. The proof of (ii) is similar. $\square$

Remark A.2

Note that a similar result can also be proved with respect to the functions $t_{a,b,\lambda }^+(u)$ and $\psi _{a,b,\lambda ,u}(t_{a,b,\lambda }^+(u))$.

Denote

$$\begin{aligned} {\mathcal {M}}_{a,b,\lambda }=\left\{ \frac{u}{\Vert u\Vert }:u\in {\mathcal {N}}_{a,b,\lambda }^-\right\} . \end{aligned}$$

Lemma A.3

There holds:

• (i) If $b_1< b_2$, then ${\mathcal {M}}_{b_2}\subset {\mathcal {M}}_{b_1}$.

• (ii) If $\lambda _1< \lambda _2$, then ${\mathcal {M}}_{\lambda _1}\subset {\mathcal {M}}_{\lambda _2}$.

Proof

(i) Take $u\in {\mathcal {M}}_{a,b_2,\lambda }$. Once $\psi _{a,b_1,\lambda }'(t)\le \psi _{a,b_2,\lambda }'(t)$ for all $t>0$, it follows that $\psi _{a,b_1,\lambda }'(t_{a,b_2,\lambda }^-(u))< \psi _{a,b_2,\lambda }'(t_{a,b_2,\lambda }^-(u))=0$ and hence, from Lemma A.1 we conclude that $u\in {\mathcal {M}}_{a,b_1,\lambda }$.

(ii) Take $u\in {\mathcal {M}}_{a,b,\lambda _1}$. Once $\psi _{a,b,\lambda _2}'(t)\le \psi _{a,b,\lambda _1}'(t)$ for all $t>0$, it follows that $\psi _{a,b,\lambda _2}'(t_{a,b,\lambda _1}^-(u))< \psi _{a,b,\lambda _1}'(t_{a,b,\lambda _1}^-(u))=0$ and hence, from Proposition A.1 we conclude that $u\in {\mathcal {M}}_{a,b,\lambda _1}$. $\square$

Proposition A.1

Fix $a>0$ and let I be an interval. Then, the following holds true.

• (i) Fix $b\in {\mathbb {R}}$. If $c^-(a,b,\lambda )$ is defined for all $\lambda \in I$, then it is non-increasing as a function of $\lambda$.

• (ii) Fix $\lambda \in {\mathbb {R}}$. If $c^-(a,b,\lambda )$ is defined for all $b\in I$, then it is non-decreasing as a function of b.

Proof

i) Indeed, fix $\lambda _1<\lambda _2$ and $u\in {\mathcal {M}}_{a,b,\lambda _1}$. Since from Lemma A.3 we have that $u\in {\mathcal {M}}_{a,b,\lambda _2}$, it follows from Lemma A.2 that

$$\begin{aligned} c^-(a,b,\lambda _2)\le \psi _{a,b,\lambda _2}(t_{a,b,\lambda _2}^-(u))<\psi _{a,b,\lambda _1}(t_{a,b,\lambda _1}^-(u)), \forall u\in {\mathcal {M}}_{a,b,\lambda _1}. \end{aligned}$$

(23)

and hence $c^-(a,b,\lambda _2)\le c^-(a,b,\lambda _1)$. The proof of ii) is similar. $\square$

Proposition A.2

Fix $a>0$ and let I be an interval. Then, the following holds true.

• (i) Fix $\lambda \in {\mathbb {R}}$. If $c^-(a,b,\lambda )$ is defined for all $b\in I$, then it is right continuous as a function of b.

• (ii) Fix $b\in {\mathbb {R}}$. If $c^-(a,b,\lambda )$ is defined for all $\lambda \in I$, then it is right continuous as a function of $\lambda$.

Proof

(i) Fix $b_0\in I$. We claim that $\lim _{b\downarrow b_0}c^-(a,b,\lambda )=c^-(a,b_0,\lambda )$. Indeed, once $I\ni b\mapsto c^-(a,b,\lambda )$ is non-decreasing, we can assume that $\lim _{b\downarrow b_0}c^-(a,b,\lambda )=c\ge c^-(a,b_0,\lambda )$. Suppose on the contrary that $c>c^-(a,b_0,\lambda )$. Given $\varepsilon >0$ choose $u\in {\mathcal {M}}_{a,b_0,\lambda }$ such that $\Phi _{a,b_0,\lambda }(t_{a,b_0,\lambda }^-(u)u)\in [c^-(a,b_0,\lambda ),c^-(a,b_0,\lambda )+\varepsilon )$ and $c^-(a,b_0,\lambda )+\varepsilon <c$. From Lemma A.2 we conclude that for small $\delta >0$

$$\begin{aligned} c^-(a,b_0+\delta ,\lambda )\le \Phi _{a,b_0+\delta ,\lambda }(t_{a,b_0+\delta ,\lambda }^-(u)u)< c^-(a,b_0,\lambda )+\varepsilon <c, \end{aligned}$$

which is a contradiction and thus $I\ni b\mapsto c^-(a,b,\lambda )$ is right continuous. The proof of (ii) is similar. $\square$

Appendix B: The case $\lambda =0$

We collect some results concerning the fiber maps $\psi$ when $\lambda =0$. The parameter now is $b>0$, while $a>0$ is fixed. For this reason, we write $\psi _{b,u}$ and $\Phi _b$ instead of $\psi _{0,u}$ and $\Phi _0$ and so on. As we already know, for each $u\in H_0^1(\Omega ){\setminus }\{0\}$ the fiber map $\psi _{b,u}$ has satisfies Proposition 4.2. One can see now that the systems $\psi _{b,u}(t)=\psi _{b,u}'(t)=0$ and $\psi '_{b,u}(t)=\psi _{b,u}''(t)=0$ admits a unique solution, with respect to t, b, which are given respectively by (see [5] and [18])

$$\begin{aligned}&t_{0}(u)=\left( \frac{2^*a}{4-2^*}\frac{\Vert u\Vert ^2}{\Vert u\Vert _{2^*}^{2^*}}\right) ^{\frac{1}{2^*-2}}, \\&b_0(u)=a^{\frac{4-N}{2}}S_N^{\frac{N}{2}}C_1(N)\left( \frac{\Vert u\Vert _{2^*}}{\Vert u\Vert }\right) ^N, \end{aligned}$$

and

$$\begin{aligned}&t(u)=\left( \frac{2a}{4-2^*}\frac{\Vert u\Vert ^2}{\Vert u\Vert _{2^*}^{2^*}}\right) ^{\frac{1}{2^*-2}}, \\&b(u)=a^{\frac{4-N}{2}}S_N^{\frac{N}{2}}C_2(N)\left( \frac{\Vert u\Vert _{2^*}}{\Vert u\Vert }\right) ^N. \end{aligned}$$

As a conclusion of this analysis and similar to Propositions 2.4 and 3.7 we have

Proposition B.1

There holds

• (i) For each $b\ge b_0(u)$ and each $u\in H^1_0(\Omega ){\setminus } \{0\}$, $\inf _{t>0}\psi _{b,u}(t)=0$; for each $b<b_0(u)$ there exists $u\in H^1_0(\Omega ){\setminus } \{0\}$ such that $\Phi _b(u)<0$.

• (ii) For each $b\ge b(u)$, the set ${\mathcal {N}}_b=\emptyset$; for each $b<b(u)$, the sets ${\mathcal {N}}_b^+$, ${\mathcal {N}}_b^-$ and ${\mathcal {N}}_b^0$ are non empty.

Therefore:

Lemma B.1

The following holds true.

• (i) If $a^\frac{N-4}{2}b<C_1(N)$, then there exists $u\in H_0^1(\Omega ){\setminus }\{0\}$ such that $\Phi _b(u)<0$.

• (ii) If $a^\frac{N-4}{2}b\ge C_1(N)$, then $\psi _{b ,u}(t)>0$ for all $u\in H_0^1(\Omega ){\setminus }\{0\}$ and $t>0$.

• (iii) If $a^\frac{N-4}{2}b<C_2(N)$, then ${\mathcal {N}}^0_b, {\mathcal {N}}^-_b, {\mathcal {N}}^+_b$ are non-empty.

• (iv) If $a^\frac{N-4}{2}b\ge C_2(N)$, then ${\mathcal {N}}_b=\emptyset$.

Remark B.1

Comparing Lemmas B.1 and 2.2 we see that

• (i) $\Phi _b$ is weak lower semi-continuous if, and only, $\Phi _b(u)\ge 0$ for all $u\in H_0^1(\Omega )$.

• (ii) If $\Phi _b'(u)u>0$ for all $u\in H_0^1(\Omega ){\setminus }\{0\}$, then $\Phi _b$ satisfies the Palais-Smale condition. Equivalently ${\mathcal {N}}_b=\emptyset$.

Corollary B.1

If $a^\frac{N-4}{2}b<C_2(N)$, then for all $\lambda >0$ we have ${\mathcal {N}}_{b}^-\ne \emptyset$.

Proof

Indeed, this is a consequence of Lemmas B.1 and A.3. We also refer the reader to [13, Lemma 2.6]. $\square$

The next lemma is an application of Lemma A.2 and Remark A.2:

Lemma B.2

Fix $u\in H_0^1(\Omega ){\setminus }\{0\}$. The following holds true.

• (i) The function $(0,b(u))\ni \mapsto t_b^-(u)$ is continuous and increasing.

• (ii) The function $(0,b(u))\ni \mapsto t_b^+(u)$ is continuous and decreasing.

• (iii)
  $$\begin{aligned} \lim _{b\uparrow b(u)}t_b^-(u)=t(u)=\lim _{b\uparrow b(u)}t_b^+(u). \end{aligned}$$

The following proposition can be found in [18, 19] (with some adaptations). We give an outline of the proof (recall from Lemma B.1 that ${\mathcal {N}}^0_b, {\mathcal {N}}^-_b$ are not empty for all $a,b>0$ satisfying $a^\frac{N-4}{2}b<C_2(N)$):

Proposition B.2

Suppose that $a^\frac{N-4}{2}b<C_2(N)$, then

$$\begin{aligned} \Phi _b(u)=\frac{(2^*-2)^2a^2}{4\cdot 2^*(4-2^*)b}, \forall u\in {\mathcal {N}}^0_b. \end{aligned}$$

Moreover,

$$\begin{aligned} c^-(a,b,0)< \frac{(2^*-2)^2a^2}{4\cdot 2^*(4-2^*)b}=c^0(a,b,0). \end{aligned}$$

Proof

The first part is trivial. Now suppose on the contrary that there exists $u\in {\mathcal {N}}^-_b$ such that

$$\begin{aligned} \Phi _b(u)\ge \frac{(2^*-2)^2a^2}{4\cdot 2^*(4-2^*)b}. \end{aligned}$$

From Lemma B.2 we have that $t^-_b(u)=1<t^-_{b'}(u)<t^+_{b'}(u)<t^+_{b}(u)$ for each $0<b<b'<b(u)$ and hence

$$\begin{aligned} \Phi _{b'}(t^-_{b'}(u)u)> & {} \Phi _{b'}(u)\\> & {} \Phi _{b}(u) \\\ge & {} \frac{(2^*-2)^2a^2}{4\cdot 2^*(4-2^*)b}, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{(2^*-2)^2a^2}{4\cdot 2^*(4-2^*)b}<\lim _{b'\uparrow b(u)} \Phi _{b'}(t^-_{b'}(u)u)=\Phi _{b(u)}(t_b(u)u)=\frac{(2^*-2)^2a^2}{4\cdot 2^*(4-2^*)b(u)}, \end{aligned}$$

a contradiction since $b<b(u)$. $\square$