On a biharmonic elliptic problem with slightly subcritical non-power nonlinearity

Abstract

We study the following biharmonic elliptic problem with slightly subcritical non-power nonlinearity:

$$\begin{aligned} \left\{ \begin{array}{lll} \Delta ^2 u =\frac{|u|^{2^*-2}u}{[\ln (e+|u|)]^\varepsilon }\ \ &{}\textrm{in}\ \Omega , \\ u=\Delta u= 0 \ \ &{} \textrm{on}\ \partial \Omega , \end{array} \right. \end{aligned}$$

where $2^*=\frac{2n}{n-4}$, $\Omega$ is a bounded smooth domain in $\mathbb {R}^n$ with $n\ge 5$, $\varepsilon$ is a small positive parameter. By finite-dimensional Lyapunov–Schmidt reduction, we construct a single bubble solution, which concentrates at the non-degenerate critical point of the Robin function.

Appendix

Now, we prove the estimates in Lemma 2.6.

Proof of Lemma 2.6

(i):

From (2.3), we obtain

$$\begin{aligned} \frac{\partial f_\varepsilon (u)}{\partial \varepsilon } =-\frac{|u|^{p-1}u}{[\ln (e+|u|)]^\varepsilon }\,\ln \ln (e+|u|), \end{aligned}$$

which follows that:

$$\begin{aligned} \left| \frac{\partial f_\varepsilon (u)}{\partial \varepsilon }\right| \le |u|^{p}\ln \ln (\textrm{e}+|u|), \end{aligned}$$

and by the mean value theorem, we get the proof of (i).

(ii):

There holds

$$\begin{aligned} f_\varepsilon '(u)=\frac{|u|^{p-1}}{[\ln (\textrm{e}+|u|)]^\varepsilon } \left( p-\frac{\varepsilon |u|}{(\textrm{e}+|u|)\ln (\textrm{e}+|u|)} \right) . \end{aligned}$$

(4.1)

Moreover, since $0\le \frac{|u|}{\varepsilon +|u|}\le 1,$ and $0\le \frac{1}{\ln (\textsf{e}+|u|)}\le 1$, we get the result (ii).

(iii):

From (4.1), we have

$$\begin{aligned} \frac{\partial f_\varepsilon ^{'}(u)}{\partial \varepsilon } =&-\frac{|u|^{p-1}\ln \ln (\textrm{e}+|u|)}{[\ln (\textrm{e}+|u|)]^\varepsilon } \left( p-\frac{\varepsilon |u|}{(\textrm{e}+|u|)\ln (\textrm{e}+|u|)} \right) \\&- \frac{|u|^{p}}{(\textrm{e}+|u|)[\ln (\textrm{e}+|u|)]^{\varepsilon +1}}. \end{aligned}$$

Thus, for $\varepsilon$ small enough, one has

$$\begin{aligned} \left| \frac{\partial f_{\varepsilon }^{\prime }(u)}{\partial \varepsilon }\right| \le&p\frac{|u|^{p-1}\ln \ln (e+|u|)}{[\ln (e+|u|)]^{\varepsilon }} +\frac{|u|^{p-1}}{[\ln (e+|u|)]^{\varepsilon +1}} \\ \le&|u|^{p-1}\left( p\ln \ln (\textrm{e}+|u|) +\frac{1}{\ln (\textrm{e}+|u|)}\right) . \end{aligned}$$

Finally, the result follows by the mean value theorem.

(iv):

We have

$$\begin{aligned} f_{\varepsilon }^{\prime \prime }(u) =&\frac{\varepsilon |u|^{p-2}u}{[\ln (e+|u|)]^{\varepsilon }} \left( \frac{|u|-e\ln (e+|u|)}{(e+|u|)^{2}[\ln (e+|u|)]^{2}}\right) \\&+ \frac{|u|^{p-2}u}{[\ln (\textrm{e}+|u|)]^\varepsilon } \left( p-1-\frac{\varepsilon |u|}{(\textrm{e}+|u|)\ln (\textrm{e}+|u|)}\right) \\&\left( p-\frac{\varepsilon |u|}{(\textrm{e}+|u|)\ln (\textrm{e}+|u|)}\right) . \end{aligned}$$

Moreover, for $\varepsilon$ small enough, one has

$$\begin{aligned} |f_\varepsilon ^{\prime \prime }(u)|\le C|u|^{p-2}. \end{aligned}$$

If $n\le 12$, it holds $p-2=\frac{12-n}{n-4}\ge 0$, for some $t\in (0,1)$, the mean value theorem shows that

$$\begin{aligned} |f_\varepsilon ^{\prime }(u+v)-f_\varepsilon ^{\prime }(u)|=|f_\varepsilon ^{\prime \prime }(u+tv)||v|\le C|u+tv|^{p-2}|v|\le C(|u|^{p-2}+|v|^{p-2})|v|. \end{aligned}$$

If $n\ge 13$, let $q=p-1\in (0,1)$ and

$$\begin{aligned} |f_{\varepsilon }^{\prime }(u+v)-f_{\varepsilon }^{\prime }(u)| \le&p\left| \frac{|u+v|^{q}}{[\ln (\textrm{e}+|u+v|)]^{\varepsilon }} -\frac{|u|^{q}}{[\ln (\textrm{e}+|u|)]^{\varepsilon }}\right| \\&+ \varepsilon \left| \frac{|u+v|^{p}}{(\textrm{e}+|u+v|) [\ln (\textrm{e}+|u+v|)]^{\varepsilon +1}}\right. \\&\left. -\frac{|u|^{p}}{(\textrm{e}+|u|) [\ln (\textrm{e}+|u|)]^{\varepsilon +1}}\right| \\ =&F_1+F_2. \end{aligned}$$

There holds

$$\begin{aligned} F_{2}\le C\varepsilon (|u|^{q}+|v|^{q}). \end{aligned}$$

For $F_1$, for any fixed $v\ne 0$, let $x=\frac{u}{v}$, then

$$\begin{aligned}&\frac{1}{|v|^q}\left| \frac{|u+v|^q}{[\ln (\textrm{e}+|u+v|)]^\varepsilon } -\frac{|u|^q}{[\ln (\textrm{e}+|u|)]^\varepsilon }\right| \\&\quad = \left| \frac{|x+1|^{q}}{[\ln (e+|v||x+1|)]^{\varepsilon }} -\frac{|x|^{q}}{[\ln (e+|v||x|)]^{\varepsilon }}\right| = g(x). \end{aligned}$$

The function g is symmetric with respect to $-\frac{1}{2}$, that is, $g(x)=g(-x-1)$; it is increasing in $[-\frac{1}{2},0]$ and decreasing in $[0,\infty )$, and $g(0)\le 1$. Hence

$$\begin{aligned} F_{1}\le p|v|^{q}. \end{aligned}$$

Combining $F_1$ and $F_2$, we obtain (iv).

(v):

There holds

$$\begin{aligned} \ln \ln \left( \textrm{e}+\delta ^{-r}u\right) =&\ln \ln (\delta ^{-r}(\delta ^{r}\textrm{e}+u))\\ =&\ln \left( \ln \delta ^{-r}+\ln (\textrm{e}^{1-r|\ln \delta |}+u)\right) \\ =&\ln \bigg [\ln \delta ^{-r}\Big (1+\frac{\ln (\textrm{e}^{1-r|\ln \delta |}+u)}{\ln \delta ^{-r}}\Big )\bigg ] \\ =&\ln \ln (\delta ^{-r} ) +\ln \Big (1+\frac{\ln (\textrm{e}^{1-r|\ln \delta |}+u)}{r|\ln \delta |}\Big ). \end{aligned}$$

(vi):

Let

$$\begin{aligned} g(t)=\ln \Big [1+\frac{t}{r}\ln (\textrm{e}^{1-\frac{r}{t}}+u)\Big ],\quad t>0. \end{aligned}$$

Then, from (v) and L’Hôpital’s rule, we have

$$\begin{aligned} {\text {lim}}_{t\rightarrow 0}{\frac{g(t)}{t}}&=\lim _{t\rightarrow 0}g^{\prime }(t)=\lim _{t\rightarrow 0}\frac{\frac{1}{r}\ln (\textrm{e}^{1-\frac{r}{t}}+u)+\frac{1}{t}\textrm{e}^{1-\frac{r}{t}}(\textrm{e}^{1-\frac{r}{t}}+u)^{-1}}{1+\frac{t}{r}\ln (\textrm{e}^{1-\frac{r}{t}}+u)} \\&=\frac{1}{r}\ln u. \end{aligned}$$

Set $t=|\ln \delta |^{-1}$, we finish the proof of (vi).

$\square$

In the following, we show the proof of $P_3$ in Proposition 3.1.

Lemma 4.1

There holds

$$\begin{aligned} P_3= & {} \int \limits _\Omega \Big (f_0(PU_{ \delta ,\xi })-f_\varepsilon (PU_{ \delta ,\xi }) \Big ) Z^l_{ \delta ,\xi }\text {d}x \\= & {} \left\{ \begin{array}{lll} - \frac{2d}{n-4}M\frac{\varepsilon }{|\ln \delta |} +o\Big (\frac{\varepsilon }{|\ln \delta |}\Big )\ \ &{}\textrm{if}\ l=0, \\ O(\varepsilon \delta ^{\frac{n-4}{2}}\ln |\ln \delta |)\ \ &{}\textrm{if}\ l=1,\ldots ,n, \end{array} \right. \end{aligned}$$

where $M=-\int \limits _{\mathbb {R}^n}U^p(\ln U)Z_0\text {d}z =\frac{(n-4)^2\alpha _n^{2^*}\pi ^{\frac{n-2}{2}}}{2 n}\frac{\Gamma (\frac{n-2}{2})}{\Gamma (n+1)}>0$. Moreover

$$\begin{aligned} \varepsilon \delta ^{\frac{n-4}{2}}\ln |\ln \delta | =&d \delta ^{\frac{3n}{2}-6}|\ln \delta |\ln |\ln \delta | = \left\{ \begin{array}{lll} o( \delta ^{n-4}) \ \ &{}\textrm{if}\ 5\le n\le 6, \\ o( \delta ^{n-3}) \ \ &{}\textrm{if}\ n\ge 7. \end{array} \right. \end{aligned}$$

Proof

By Taylor’s expansion with respect of $\varepsilon$, we have

$$\begin{aligned}&\int \limits _\Omega \Big (f_0(PU_{ \delta ,\xi })-f_\varepsilon (PU_{ \delta ,\xi }) \Big ) Z^l_{ \delta ,\xi }\text {d}x\nonumber \\&\quad = \varepsilon \int \limits _\Omega (PU_{ \delta ,\xi })^p\ln \ln (e+PU_{ \delta ,\xi })Z^l_{ \delta ,\xi }\text {d}x -\varepsilon ^2\nonumber \\&\qquad \int \limits _\Omega (PU_{ \delta ,\xi })^p \Big (\ln \ln (e+PU_{ \delta ,\xi })\Big )^2 Z^l_{ \delta ,\xi } \text {d}x, \end{aligned}$$

(4.2)

where Lemma 2.6 implies that

$$\begin{aligned} \bigg |\int \limits _\Omega (PU_{ \delta ,\xi })^p \Big (\ln \ln (e+PU_{ \delta ,\xi })\Big )^2 Z^l_{ \delta ,\xi } \text {d}x\bigg |&\le \int \limits _\Omega U_{ \delta ,\xi }^p\Big (\ln \ln (e+ U_{ \delta ,\xi })\Big )^2 |Z^l_{ \delta ,\xi }|\text {d}x\nonumber \\&=O\Big ( (\ln |\ln \delta |)^2\Big ). \end{aligned}$$

(4.3)

For the first integral in (4.2), we set $g(u)=u^p\ln \ln (e+u)$, then the mean value theorem gives that

$$\begin{aligned} 0\le g(u)-g(v)\le Cu^{p-1}\Big (\ln \ln (e+u)+1\Big )(u-v)\quad \text{ if }\ 0\le v\le u; \end{aligned}$$

it means that

$$\begin{aligned}&\int \limits _\Omega (PU_{ \delta ,\xi })^p\ln \ln (e+PU_{ \delta ,\xi }) Z^l_{ \delta ,\xi }\text {d}x\\&\quad = \int \limits _\Omega U_{ \delta ,\xi }^p \ln \ln (e+U_{ \delta ,\xi })Z^l_{ \delta ,\xi }\text {d}x\\&\qquad + \int \limits _\Omega U_{ \delta ,\xi }^{p-1}\Big (\ln \ln (e+ U_{ \delta ,\xi } )+1\Big )( U_{ \delta ,\xi }- PU_{ \delta ,\xi } ) Z^l_{ \delta ,\xi }\text {d}x. \end{aligned}$$

By Lemma 2.6 and (2.12), there holds

$$\begin{aligned}&\int \limits _\Omega U_{ \delta ,\xi }^{p-1}(x)\Big [\ln \ln \Big (e+ U_{ \delta ,\xi } (x)\Big )+1\Big ]\Big ( U_{ \delta ,\xi }(x)- PU_{ \delta ,\xi } (x)\Big ) Z^l_{ \delta ,\xi }(x)\text {d}x\\&\quad \le C \delta ^{ \frac{n-4}{2}}(\ln |\ln \delta |+1 )\int \limits _{\mathbb {R}^n} U^{p-1}(z) \Big ( H( \delta z+\xi ,\xi )+O( \delta )\Big )Z_l(z)\text {d}z\\&\quad \le C \delta ^{ \frac{n-4}{2}}\ln |\ln \delta | \Big (H (\xi ,\xi )+O( \delta )\Big )\int \limits _{\mathbb {R}^n} U^{p-1}(z)Z_l(z)\text {d}z \\&\quad = O\Big ( \delta ^{\frac{n-4}{2}}\ln |\ln \delta |\Big ). \end{aligned}$$

From Lemma 2.6 and (3.4), one has

$$\begin{aligned}&\int \limits _\Omega U_{ \delta ,\xi }^p(x)\ln \ln \left( e+U_{ \delta ,\xi }(x)\right) Z^l_{ \delta ,\xi }(x)\text {d}x\nonumber \\&\quad = \delta ^{n-\frac{n+4}{2}-\frac{n-4}{2}}\int \limits _{\frac{\Omega -\xi }{ \delta }} U^p (z)\ln \ln \left( e+ \delta ^{ -\frac{n-4}{2}} U(z)\right) Z_l(z)\text {d}z\nonumber \\&\quad = \ln \ln ( \delta ^{ -\frac{n-4}{2}} )\int \limits _{\frac{\Omega -\xi }{ \delta }}U^p (z)Z_l(z)\text {d}z\nonumber \\&\qquad +\frac{1}{|\ln \delta |} \int \limits _{\frac{\Omega -\xi }{ \delta }}U^p (z)\bigg [|\ln \delta |\ln \bigg (1+\frac{ \ln \Big (e^{1-\frac{n-4}{2}|\ln \delta |}+U(z) \Big )}{\frac{n-4}{2}|\ln \delta |}\bigg )\bigg ]Z_l(z)\text {d}z.\nonumber \\ \end{aligned}$$

(4.4)

Moreover, let $\psi (z)=U^p (z)\bigg [|\ln \delta |\ln \bigg (1+\frac{ \ln \Big (e^{1-\frac{n-4}{2}|\ln \delta |}+U(z) \Big )}{\frac{n-4}{2}|\ln \delta |}\bigg )\bigg ]Z_l(z)$ for $l=1,\ldots ,n$, and it follows that $\int _{\mathbb {R}^n}\psi (z)\text {d}z=0$, since $\psi (z)$ is a odd function. Further, using Lemma 2.6, direct calculations show that

$$\begin{aligned}&\int \limits _{\mathbb {R}^n}\psi (z)\text {d}z-\int \limits _{\frac{\Omega -\xi }{ \delta }}\psi (z)\text {d}z=\int \limits _{\mathbb {R}^n \backslash \frac{\Omega -\xi }{ \delta }}\psi (z)\text {d}z\\&\quad \le C \bigg [|\ln \delta |\ln \bigg (1+\frac{ \ln \Big (e^{1-\frac{n-4}{2}|\ln \delta |}+\alpha _n \Big )}{\frac{n-4}{2}|\ln \delta |}\bigg )\bigg ]\int \limits _{\mathbb {R}^n \backslash \frac{\Omega -\xi }{ \delta }}U^p (z)|Z_l(z)|\text {d}z\\&\quad \le C\Big (\frac{2}{n-4}\ln \alpha _n+o(1)\Big ) \delta ^{ n+1} =O( \delta ^{ n+1}). \end{aligned}$$

Hence, for $\delta$ small enough, we conclude that

$$\begin{aligned} \int \limits _\Omega U_{ \delta ,\xi }^p\ln \ln (e+ U_{ \delta ,\xi } )Z^l_{ \delta ,\xi }\text {d}x = O( \delta ^{ n+1}\ln |\ln \delta |)=o( \delta ^n). \end{aligned}$$

(4.5)

From (4.2), (4.3), and (4.5), we finish the proof for $l=1,\ldots ,n$.

When $l=0$, from (3.4), we get $\int _{\mathbb {R}^n}U^pZ_0\text {d}z=0$, and also, $\int _{\mathbb {R}^n \backslash \frac{\Omega -\xi }{ \delta }}U^pZ_0\text {d}z=0$; according to (4.4) and Lemma 2.6, there holds

$$\begin{aligned}&\int \limits _\Omega U_{ \delta ,\xi }^p(x)\ln \ln (e+ U_{ \delta ,\xi }(x) )Z^0_{ \delta ,\xi }(x)\text {d}x \\&\quad = \frac{1}{|\ln \delta |}\frac{2}{n-4}\int \limits _{\mathbb {R}^n} U^p (z)\ln ( U(z))Z_0(z)\text {d}z +o\Big (\frac{1}{|\ln \delta |}\Big ). \end{aligned}$$

Finally, let $\mathcal {B}(\cdot ,\cdot )$ stands for the common Beta function, then by a fact that $|\partial B_1(0)|=\frac{2\pi ^{\frac{n}{4}}}{\Gamma (\frac{n}{4})}$, one has

$$\begin{aligned} M =&-|\partial B_1(0)|\frac{(n-4)\alpha _n^{2^*}}{2}\int \limits _0^\infty \frac{r^{n-1}}{(1+r^2)^{\frac{n+4}{2}}}\ln \bigg (\frac{1}{(1+r^2)^{\frac{n-4}{2}}}\bigg ) \frac{r^2-1}{(1+r^2)^{\frac{n-2}{2}}}\text {d}r\\ =&\frac{(n-4)^2\alpha _n^{2^*}}{2}\frac{2\pi ^{\frac{n-2}{2}}}{\Gamma (\frac{n-2}{2})} \int \limits _0^\infty \frac{r^{n-1}(r^2-1)}{(1+r^2)^{n+1}} \ln (1+r^2)\text {d}r\\ =&\frac{(n-4)^2\alpha _n^{2^*}\pi ^{\frac{n-2}{2}}}{2 n \Gamma (\frac{n-2}{2})} \int \limits _0^\infty \bigg (\frac{r }{(1+r^2)}\bigg )^n\frac{2r}{1+r^2}\text {d}r\\ =&\frac{(n-4)^2\alpha _n^{2^*}\pi ^{\frac{n-2}{2}}}{2 n\Gamma (\frac{n-2}{2})} \int \limits _0^\infty \frac{s^{\frac{n-4}{2}} }{(1+s)^{n+1}} \text {d}s\\ =&\frac{(n-4)^2\alpha _n^{2^*}\pi ^{\frac{n-2}{2}}}{2 n\Gamma (\frac{n-2}{2})}\mathcal {B}\Big (\frac{n-4}{2}+1,\frac{n-4}{2}\Big )\\ =&\frac{(n-4)^2\alpha _n^{2^*}\pi ^{\frac{n-2}{2}}}{2 n\Gamma (\frac{n-2}{2})}\frac{\Gamma (\frac{n-4}{2}+1)\Gamma (\frac{n-2}{2})}{\Gamma (n+1)} = \frac{(n-4)^2\alpha _n^{2^*}\pi ^{\frac{n-2}{2}}}{2 n}\frac{\Gamma (\frac{n-2}{2})}{\Gamma (n+1)}. \end{aligned}$$

From above all estimations, the proof of this lemma is accomplished. $\square$