0
$\begingroup$

I'm reading the Theorem2 in UNIFORM ESTIMATES AND BLOW-UP BEHAVIOR FOR SOLUTIONS OF $-\Delta u=V(x) e^u$ IN TWO DIMENSIONS They prove that for the solution of $$ -\Delta u= V(x)\exp u \text { in } \mathbb{R}^2 $$ if $V \in L^{p}(\mathbb{R}^2)$, $\exp u \in L^{p^{\prime}}\left(\mathbb{R}^2\right)$, with $1<p\le+\infty$, then $u \in L^{\infty}\left(\mathbb{R}^2\right)$. (Here $1/p + 1/p^{\prime}=1$)

First fix $0<\epsilon<1 $, note that for any $1 \leq p<q<r \leq \infty, L^q \subseteq L^p+L^r$, so they split $V(x)\exp u $ as $V(x)\exp u=f_1+f_2$ with $\left\|f_1\right\|_{L^1\left(\mathbb{R}^2\right)}<\epsilon$ and $f_2 \in L^{\infty}\left(\mathbb{R}^2\right)$. Let $B_r$ be the ball of radius $r$ centered at $x_0$.

They denote by $C$ various constants independent of $x_0$ (but possibly depending on $\epsilon$ ). Then let $u_i$ be the solution of $$ \left\{\begin{array}{rlll} -\Delta u_i & =f_i & \text { in } & B_1 \\ u_i & =0 & \text { on } & \partial B_1 \end{array}\right. $$

By a lemma they proved before, they proved that


Assume $\Omega \subset \mathbb{R}^2$ is a bounded domain and let $u$ be a solution of

$$ \left\{\begin{array}{ccc} -\Delta u=f(x) & \text { in } \quad \Omega, \\ u=0 & \text { on } \quad \partial \Omega, \end{array}\right. $$ Then for every $\delta \in(0,4 \pi)$ we have $$ \int_{\Omega} \exp \left[\frac{(4 \pi-\delta)] u(x) \|}{\|f\|_1}\right] \mathrm{dx} \leq \frac{4 \pi^2}{\delta}(\operatorname{diam} \Omega)^2 . $$


So applied with $\delta=4 \pi-1$, they have $$ \int_{B_1} \exp \left[\frac{1}{\epsilon}\left|u_1\right|\right] \leq \mathrm{C} $$ and in particular $\left\|u_1\right\|_{L^1\left(B_1\right)} \leq C$. We also have $\left\|u_2\right\|_{L^{\infty}\left(B_1\right)} \leq C$. Let $u_3=u-u_1-u_2$ so that $\Delta u_3=0$ on $B_1$. The mean value theorem for harmonic functions implies that \begin{equation}\tag{1} \left\|u_3^{+}\right\|_{L^{\infty}\left(B_{1 / 2}\right)} \leq \mathrm{C}\left\|u_3^{+}\right\|_{L^1\left(B_1\right)} \end{equation} (This can be simply verified by the fact that let $\Omega \subset \mathbb{R}^n$ open and let $u$ be a harmonic function in $\Omega$. If $K \subset \Omega$ is compact, then $$ \sup _{x \in K}|u(x)| \leq \frac{n}{\omega_n \operatorname{dist}(K, \partial \Omega)^n} \int_{\Omega}|u(x)| \mathrm{d} x .) $$

On the other hand they have $$ u_3^{+} \leq u^{+}+\left|u_1\right|+\left|u_2\right| $$ and since $$ p^{\prime}\int_{R^2} u^{+}\mathrm{d} x \leq \int_{R^2} \exp p^{\prime}u\mathrm{d} x \leq C $$ we see that $\left\|u_3^{+}\right\|_{L^1\left(B_1\right)} \leq \mathrm{C}$. Combining this with (1) they find that $$\left\|u_3^{+}\right\|_{L^{\infty}\left(B_{1 / 2}\right)} \leq C.$$ Finally they write \begin{equation}\tag{2} -\Delta u= V(x)\exp u= V(x)\exp u_1 \exp (u_2+u_3)=g \end{equation} (with $V \in L^p(B_1)$, $\|g\|_{L^{1+\delta}\left(B_{1 / 2}\right)} \leq C$ for some $\delta>0$ (since $e^{u_2+u_3} \in L^{\infty}\left(B_{1 / 2}\right)$, $\exp u_1 \in L^{1 / \epsilon}\left(B_1\right)$ with $\left.1 / \epsilon> p^{\prime}\right)$.


What I'm confused is the last step, they said they use once more the mean value theorem and standard elliptic estimates they deduce from (2) that

$$ \left\|u^{+}\right\|_{L^{\mathbb{\infty}}\left(B_{1 / 4}\right)} \leq C\left\|u^{+}\right\|_{L^1\left(B_{1 / 2}\right)}+C \|u\|_{L^{1+\delta}\left(B_{1 / 2}\right)} \leq C . $$ Since $C$ is independent of $x_0$ they conclude that $u^{+} \in L^{\infty}\left(\mathbb{R}^2\right)$.

I'm very confused about how they use the mean value property again and what kind of regularity they use ?

$\endgroup$

1 Answer 1

1
$\begingroup$

By Caldrron-Zygmund inequality, you may obtain : $$ \|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})} \leq C\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} +C\|g\|_{L^{1+\delta}~~(B_{\frac{1}{2}})}.$$ Since the dimension of space is 2, $$ \|u\|_{L^{\infty}~~(B_{\frac{1}{4}})}\leq C\|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})}.$$ Using the $L^1$- theory, see enter image description here

you can obtain $\|u_1\|_{L^{1+\delta} ~~(B_{\frac{1}{2}})}$ is bounded. By the mean value theorem or harnack inequality you can obtain $\|u_3\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$, combine with the fact that $\|u_2\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$ you may obtain $$\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} \leq C,$$ and finish the proof.

$\endgroup$
3
  • $\begingroup$ Thanks for your answer! I want to know what this $L^1$-theory is. $\endgroup$
    – Elio Li
    Aug 18, 2023 at 9:29
  • $\begingroup$ Thanks very much! Can you tell me which lecture this theorem is in? $\endgroup$
    – Elio Li
    Aug 18, 2023 at 13:52
  • $\begingroup$ @Yuxuan Li 链接:pan.baidu.com/s/10Q8_GmtS854urOFrLn8BFg?pwd=4p91 提取码:4p91 $\endgroup$
    – sorrymaker
    Aug 18, 2023 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.