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<p<\infty} 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<p<\infty}. 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

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