Algebraic topological techniques for elliptic problems involving fractional Laplacian

Abstract

We prove the existence of infinitely many solutions to an elliptic problem by borrowing the techniques from algebraic topology. The solution(s) thus obtained will also be proved to be bounded.

Appendix

Theorem 4.1

(Theorem 6.2.42 of Papagiorgiou et al. [33]) If X is a Banach space, $I\in C^1(X)$, I satisfies the Ce-condition, $K_I$ is finite with $0\in K_I$, and for some $k\in {\mathbb {N}}$ we have $C_k(I,0)\ne 0$, $C_k(I,\infty )=0$, then there exists a $u\in K_I$ such that $\bar{(}u)<0$, $C_{k-1}(I,u)\ne 0$ or $I(u)>0$, and $C_{k+1}(I,u)\ne 0$.

Theorem 4.2

Any solution to (3.2) is in $L^{\infty }(\Omega )$.

Proof

The argument sketched here is a standard one, and hence we shall only show that an improvement in integrability is possible up to $L^{\infty }$ assuming an integrability of certain order, say $p>1$. The boundedness follows from a bootstrap argument. Without loss of generality we can consider the set $\Omega '=\{x\in \Omega :u(x)>1\}$, and thus by the positivity of a fixed solution (refer Remark 3.2), say, u we have $u=u^+>0$ a.e. in $\Omega$. Let $u\in L^{p}(\Omega )$ for $p>1$. Let $a=\max \{\lambda ,|\mu |\}$. On testing with $u^p$ to obtain the following:

$$\begin{aligned} \begin{aligned} C\Vert u^{\frac{p+1}{2}}\Vert _{2_s^*}^{2}&\le \Vert u^{\frac{p+1}{2}}\Vert \\&\le \left( \lambda \int _{\Omega '}|u|^{p-\gamma }dx+\mu \int _{\Omega '}|u|^{2_s^*-1+p}dx\right) \frac{(p+1)^2}{4p}\\&\le a\left( \int _{\Omega '}|u|^{p}(1+|u|^{2_s^*-1})dx\right) \frac{(p+1)^2}{4p};~\text {since in}~\Omega '~\text {we have}~u>1\\&\le 2a\left( \int _{\Omega '}|u|^{p}|u|^{2_s^*}dx\right) \frac{(p+1)^2}{4p}\\&\le 2a C''\Vert u\Vert _{\beta ^*}^{2_s^*}\Vert u^p\Vert _{{\mathfrak {t}}};~\text {since by using H}\ddot{\mathrm{o}}\text {lder's inequality}. \end{aligned} \end{aligned}$$

(4.1)

Here, ${\mathfrak {t}}=\frac{\beta ^*}{\beta ^*-2_s^*}$ for some $\beta ^*>1$. Thus, we also have

$$\begin{aligned} \begin{aligned} C'\Vert u^{\frac{p}{2}}\Vert _{\beta ^*}^2&\le C'\Vert u^{\frac{p+1}{2}}\Vert _{\beta ^*}^2\le \Vert u^{\frac{p+1}{2}}\Vert _{2_s^*}^2. \end{aligned} \end{aligned}$$

(4.2)

From the story so far, we know the following

$$\begin{aligned} C'\Vert u^{\frac{p}{2}}\Vert _{\beta ^*}^2&\le 2a C''\Vert u\Vert _{\beta ^*}^{2_s^*}\Vert u^p\Vert _{{\mathfrak {t}}}. \end{aligned}$$

(4.3)

For the fixed $\beta ^*>1$, we set $\eta =\frac{\beta ^*}{2{\mathfrak {t}}}>1$ for a suitable choice of t, and $\tau ={\mathfrak {t}}p$ to get

$$\begin{aligned} \Vert u\Vert _{\eta \tau }&\le C^{\frac{{\mathfrak {t}}}{\tau }}\Vert u\Vert _{\tau };~\text {where}~C=2a C''\Vert u\Vert _{\beta ^*}^{\alpha ^+}~\text {is a fixed quantity for a fixed solution}~u. \end{aligned}$$

(4.4)

Let us now iterate with $\tau _0={\mathfrak {t}}$, $\tau _{n+1}=\eta \tau _n=\eta ^{n+1}{\mathfrak {t}}$. After n iterations, the inequality (4.4) yields

$$\begin{aligned} \Vert u\Vert _{\tau _{n+1}}&\le C^{\sum \limits _{i=0}^{n}\frac{{\mathfrak {t}}}{\tau _i}}\prod \limits _{i=0}^{n}\left( \frac{\tau _i}{{\mathfrak {t}}}\right) ^{\frac{{\mathfrak {t}}}{\tau _i}}\Vert u\Vert _{{\mathfrak {t}}}. \end{aligned}$$

(4.5)

By using $\eta >1$ and the method of iteration, i.e. $\tau _0={\mathfrak {t}}$, $\tau _{n+1}=\eta \tau _n=\eta ^{n+1}{\mathfrak {t}}$, we have

$$\begin{aligned} \sum \limits _{i=0}^{\infty }\frac{{\mathfrak {t}}}{\tau _i}=\sum \limits _{i=0}^{\infty }\frac{1}{\eta ^i}=\frac{\eta }{\eta -1}, \end{aligned}$$

and

$$\begin{aligned} \prod \limits _{i=0}^{\infty }\left( \frac{\tau _i}{{\mathfrak {t}}}\right) ^{\frac{{\mathfrak {t}}}{\tau _i}}=\eta ^{\frac{\eta ^2}{(\eta -1)^2}}. \end{aligned}$$

Hence, on passing the limit $n\rightarrow \infty$ in (4.5), we end up getting

$$\begin{aligned} \Vert u\Vert _{\infty }&\le C^{\frac{\eta }{\eta -1}}\eta ^{\frac{\eta ^2}{(\eta -1)^2}}\Vert u\Vert _{{\mathfrak {t}}}. \end{aligned}$$

(4.6)

Thus, $u\in L^{\infty }(\Omega )$. $\square$