# A Perron-type method

I am now reviewing Perron’s method and it is a fun to go over a similar argument like this. The materials come from the lecture by Prof. Li and T.Aubin’s book.

Theorem On a compact Riemannian manifold without boundary ${(M,g)}$, the equation

$\displaystyle -\Delta_g u + a(x)u = -u^p$

with ${u>0, a(x) \in \mathcal{C}^\infty(M)}$ and ${a(x)<0}$ on ${M}$, ${1 has a solution ${u \in \mathcal{C}^\infty(M)}$.

We will use the method of super & subsolution. This equation enjoys a nice maximum principle as well. Compared to Perron’s method to Poisson equation, we actually work on ${\mathcal{C}^{2,\alpha}}$ functions instead of harmonic functions and use Azera-Ascoli theorem to get the convergence result. The uniform bound of ${\mathcal{C}^{2,\alpha}}$ norm can be achieved by Schauder estimate. And this method doesn’t give uniqueness of the equation.

Proof: First we denote

$\displaystyle \begin{cases} \bar{u} = \frac{1}{\varepsilon}\\ \underline{u} = \varepsilon \end{cases}$

Note that when ${\varepsilon}$ is small enough and fixed, we have

$\displaystyle \begin{cases} - \Delta_g \bar{u} +a\bar{u} \ge - \bar{u}^p \\ - \Delta_g \underline{u} + a \underline{u} \le - \underline{u}^p \end{cases}$

Here is the place where we need ${p>1}$. Then ${\bar{u}, \underline{u}}$ are supersolution and subsolution respectively. We will produce a solution ${u}$ such that ${\underline{u} \le u \le \bar{u}}$.

Fix a large constant ${N}$ such that

$\displaystyle a(x) +N>0 \text{ on M}$

$\displaystyle f(x,u):= -u^p +Nu$

satisfies ${\partial_s f(x,s) >0 \quad \forall \underline{u} \le s \le \bar{u}}$.

Let ${-Lu := -\Delta_g u + a(x)u +Nu}$, then the equation is equivalent to

$\displaystyle -Lu = f(x,u)$

Now set ${u_1 = \underline{u}}$ and define a sequence ${\{ u_i \}}$ by ${-Lu_{i+1} = f(x,u_i)}$. Indeed, given ${u_i}$ fixed, by Schauder estimate, we have ${u_{i+1} \in \mathcal{C}^{2,\alpha}}$. We know that since ${u_1}$ is a constant, and we can do the procedure inductively.

Next we will show

$\displaystyle \underline{u} \le u_2 \le u_3 \le \cdots \le \bar{u}$

Since ${-Lu_2 = f(x,\underline{u})}$ and ${ -L\underline{u} \le f(x,\underline{u})}$, ${-L(u_2 - \underline{u}) \ge 0}$. Since ${M}$ is compact, there exists ${x_{min} \in M}$ such that ${\min (u_2 - \underline{u}) = (u_2 - \underline{u})(x_{min})}$. Since ${-\Delta_g (u_2 - \underline{u})(x_{min}) \le 0}$, we have ${(a(x_{min})+N)(u_2 - \underline{u})(x_{min}) \ge 0}$, which gives ${u_2 - \underline{u} \ge 0}$. Also, ${-L\bar{u} \ge f(x,\bar{u})}$. By the construction of ${f}$ we will have ${-L(u_2 - \bar{u}) \le 0}$ and ${u_2 \le \bar{u}}$ follows. Repeating this procedure with ${u_2}$ and ${u_3}$, we will have ${u_2 \le u_3 \le \bar{u}}$, and so on.

Now we obtain a uniform bound for ${f(x, u_i)}$ independent of ${i}$, since

$\displaystyle |f(x,u_i)| \le \max \{| f(x, \underline{u})| ,| f(x, \bar{u})| \}.$

Now look at the equation

$\displaystyle - \Delta_g u_{i+1} = - (a+N) u_{i+1} + f(x, u_i)$

, in which we have a uniform bound for RHS, hence we obtain a uniform bound ${\|u_i\|_{W^{2,p}} \le C(p), \forall 1. Then by Morrey inequality, ${\|u_i\|_{\mathcal{C}^{1,\alpha}} \le C(\alpha), \forall 0<\alpha<1}$. Finally by Schauder estimate, we get a uniform bound of ${\mathcal{C}^{2,\alpha}}$ norm: ${\|u_i\|_{\mathcal{C}^{2,\alpha}} \le C(\alpha)}$. Applying Azela-Ascoli theorem, there exists a subsequence converges to a function ${u}$ in ${\mathcal{C}^2}$. Note that ${\{u_i\}}$ is a monotone increasing sequence with upper bound, so it has a pointwise limit, and this limit can be nothing but ${u}$.

Therefore ${u \in \mathcal{C}^2}$ is a solution to

$\displaystyle -Lu = f(x,u)$

By Schauder estimate again, ${u \in \mathcal{C}^\infty}$. $\Box$