Concentration profile, energy, and weak limits of radial solutions to semilinear elliptic equations with Trudinger–Moser critical nonlinearities

Abstract

We investigate the next Trudinger–Moser critical equations,

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u=\lambda ue^{u^2+\alpha |u|^\beta }&{}\text { in }B,\\ u=0&{}\text { on }\partial B, \end{array}\right. } \end{aligned}$$

where $\alpha >0$, $(\lambda ,\beta )\in (0,\infty )\times (0,2)$ and $B\subset {\mathbb {R}}^2$ is the unit ball centered at the origin. We classify the asymptotic behavior of energy bounded sequences of radial solutions. Via the blow–up analysis and a scaling technique, we deduce the limit profile, energy, and several asymptotic formulas of concentrating solutions together with precise information of the weak limit. In particular, we obtain a new necessary condition on the amplitude of the weak limit at the concentration point. This gives a proof of the conjecture by Grossi et al. (Math Ann, to appear) in 2020 in the radial case. Moreover, in the case of $\beta \le 1$, we show that any sequence carries at most one bubble. This allows a new proof of the nonexistence of low energy nodal radial solutions for $(\lambda ,\beta )$ in a suitable range. Lastly, we discuss several counterparts of our classification result. Especially, we prove the existence of a sequence of solutions which carries multiple bubbles and weakly converges to a sign-changing solution.

Proof of Lemma 2.5

In this appendix we show the proof of Lemma 2.5.

Proof of Lemma 2.5

Without loss of the generality we may assume $u_{i,n}\ge 0$. First we claim that $r_{i,n}/\gamma _{i,n}\rightarrow \infty$. If not, we have a constant $C>0$ such that $r_{i,n}/\gamma _{i,n}\le C$ for all $n\in {\mathbb {N}}$. Then putting $v_{n}(r)=u_{i,n}(r_{i,n}r)$ for $r\in [r_{i-1,n}/r_{i,n},1]$, we get from (2.1) that

$$\begin{aligned} {\left\{ \begin{array}{ll} -v_n''-\frac{1}{r}v_n'=\lambda _n r_{i,n}^2 f_n(v_n),\ (-1)^{i-1}v_n>0\text { in }(r_{i-1,n},r_{i,n}),\\ v_n(1)=0=v_n'(\rho _{i,n}/r_{i,n}),\\ v_n(r_{i-1,n}/r_{i,n})=0\text { if }i\ge 2. \end{array}\right. } \end{aligned}$$

Then the above equation implies

$$\begin{aligned} -v_n''-\frac{1}{r}v_n'\le \lambda _n r_{i,n}^2f_n(\mu _{i,n})=\frac{1}{2\mu _{i,n}}\left( \frac{r_{i,n}}{\gamma _{i,n}}\right) ^2\le \frac{C^2}{2\mu _{i,n}}\rightarrow 0, \end{aligned}$$

uniformly on $[r_{i-1,n}/r_{i,n},1]$. It follows that $v_n\rightarrow 0$ uniformly in $[r_{i-1,n}/r_{i,n},1]$. This contradicts our assumption (2.2). This proves the claim. In particular, we get $\gamma _{i,n}\rightarrow 0$. Next, we claim $\rho _{i,n}/r_{i,n}\rightarrow 0$. This is trivial for $i=1$. Hence we assume $i\ge 2$. Define $v_n$ as above. It follows from our assumption (2.3) and Lemma 2.1 that there exist constants $c,C>0$ such that

$$\begin{aligned} C\ge \int _{r_{i-1,n}/r_{i,n}}^{1} v_n'(r)^2rdr\ge \frac{\mu _{i,n}^2}{2\pi c^2}\left( \frac{\rho _{i,n}}{r_{i,n}}\right) . \end{aligned}$$

Then (2.2) shows the claim. In particular, we get $\rho _{i,n}\rightarrow 0$ and $(r_{i,n}-\rho _{i,n})/\gamma _{i,n}\rightarrow \infty$ by the first claim. Next, by the definition of $z_{i,n}$ and (2.8), we get that

$$\begin{aligned} 0\le -z_{i,n}''-\frac{1}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}}z_{i,n}'\le 1 \text { on }\left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right] . \end{aligned}$$

Then, for any $r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right]$, multiplying the equation by $r+\frac{\rho _{i,n}}{\gamma _{i,n}}$ and integrating over (0, r) if $r\ge 0$ and over (r, 0) if $r<0$ give

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\le -z_{i,n}'(r)\le \ \frac{\frac{r^2}{2}+\frac{\rho _{i,n}}{\gamma _{i,n}}r}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}} \text { if }r\ge 0, \text { and }\\ 0\le z_{i,n}'(r)\le -\frac{\frac{r^2}{2}+\frac{\rho _{i,n}}{\gamma _{i,n}}r}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}} \text { if }r<0. \end{array}\right. } \end{aligned}$$

(1.1)

Integrating this again, we get

$$\begin{aligned} \begin{aligned} 0\le -z_{i,n}(r)&\le \int _0^r\frac{\frac{r^2}{2}+\frac{\rho _{i,n}}{\gamma _{i,n}}r}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}}dr\\&= {\left\{ \begin{array}{ll} \frac{r^2}{4}\text { if }i=1,\\ \frac{r^2}{4}+\left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) \frac{r}{2}-\frac{1}{2}\left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) ^2\log {\frac{\frac{\rho _{i,n}}{\gamma _{i,n}}+r}{\frac{\rho _{i,n}}{\gamma _{i,n}}}},\text { if }i\ge 2, \end{array}\right. } \end{aligned} \end{aligned}$$

(1.2)

for any $r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right]$. Then from (1.1) and (1.2), we get that $z_{i,n}$ is uniformly bounded in $C^1_{\text {loc}}((-l,\infty ))$ ($C^1_{\text {loc}}([0,\infty ))$ if $l=0$). Furthermore, since (1.1) implies $|z_{i,n}'(r)/(r+\rho _{i,n}/\gamma _{i,n})|$ is locally uniformly bounded in $(-l,\infty )$ ($[0,\infty )$ if $l=0$), using the equation in (2.8), we get that $z_{i,n}$ is uniformly bounded in $C^2_{\text {loc}}((-l,\infty ))$ ($C^2_{\text {loc}}([0,\infty ))$ if $l=0$). Then it follows from the Ascoli-Arzelà theorem and the equation in (2.8) that there exists a function z such that $z_{i,n}\rightarrow z$ in $C^2_{\text {loc}}((-l,\infty ))$ ($C^2_{\text {loc}}((0,\infty ))\cap C^1_{\text {loc}}([0,\infty ))$ if $l=0$).

Now our final aim is to show $l=m<\infty$. This is clear if $i=1$. Hence after this we assume $i\ge 2$. We first claim $l<\infty$. We suppose $l=\infty$ on the contrary. Then $m=\infty$. It follows from (2.8) that z satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -z''=e^z\hbox { in }{\mathbb {R}},\\ z(0)=0=z'(0). \end{array}\right. } \end{aligned}$$

This implies $z(r)=\log \frac{4e^{\sqrt{2}r}}{\left( 1+e^{\sqrt{2}r}\right) ^2}$ $(r\in {\mathbb {R}})$. Then by (2.3), there exits a constant $C>0$ such that

$$\begin{aligned} C&\ge \lambda _n\int _{\rho _{i,n}}^{r_{i,n}}u_nf_n(u_n)rdr\nonumber \\&\ge \frac{1}{2} \int _{0}^{{\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}}}\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta }-1\right\} }\left( r+\frac{\rho _{i,n}}{\gamma _{i,n}}\right) dr\nonumber \\&\ge \frac{1}{2} \frac{\rho _{i,n}}{\gamma _{i,n}}\int _{0}^{{\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}}}\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta }-1\right\} }dr. \end{aligned}$$

(1.3)

Here the Fatou lemma implies

$$\begin{aligned} \liminf _{n\rightarrow \infty }\int _{0}^{{\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}}}\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta }-1\right\} }dr\ge \int _0^\infty e^zdr. \end{aligned}$$

Since the right hand side of the inequality above is positive value and $m= \infty$, we get that the right hand side of (1.3) diverges to infinity which is a contradiction. This proves the claim. Finally we show $l=m$. Let us assume $l<m$ on the contrary. Then we claim that there exists a constant $C>0$ such that

$$\begin{aligned} |z_{i,n}'(r)|\le C \text { for all }n\in {\mathbb {N}}\text { and }r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},0\right] . \end{aligned}$$

In fact, if $m=\infty$, the claim follows easily by the latter formula in (1.1). If $m<\infty$, using (1.1) again we get a constant $C>0$ such that

$$\begin{aligned} 0\le z_{i,n}'(r)\le \frac{C}{m-l+o(1)} \end{aligned}$$

uniformly for all $r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},0\right]$. This proves the claim. On the other hand, by the mean value theorem, we have a sequence $(\xi _n)\subset \left( \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},0\right)$ such that

$$\begin{aligned} z_{i,n}'(\xi _n)=\frac{-z_{i,n}\left( \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}}\right) }{\frac{\rho _{i,n}-r_{i-1,n}}{\gamma _{i,n}}}=\frac{2\mu _{i,n}^2}{l+o(1)}\rightarrow \infty \end{aligned}$$

since $l\in [0,\infty )$. This is a contradiction. We finish the proof.