Method of continuity and second order elliptic equation

\textbf{Thm:} Suppose L_t: \mathfrak{B}\to \mathfrak{V} is a family of linear operators for t\in[0,1] between two banach spaces. And L_t is continuous for t\in[0,1]. If

(1) (closedness)\displaystyle ||L_tx||_{\mathfrak{V}}\leq C||x||_{\mathfrak{B}}, x\in \mathfrak{B}, Or namely L_t is uniformly bounded.

(2)(openness) \displaystyle C||L_tx||_{\mathfrak{V}}\geq ||x||_{\mathfrak{B}}. Or L_t is invertible.

Here C is independent of t,x. Then L_0 is onto \mathfrak{B} to \mathfrak{V} if and only if L_1 is onto.

How can we use this theorem? Well, consider the Dirichlet problem of Laplacian equation

\displaystyle \begin{cases}\Delta u=f \text{ in }\Omega\\ u=0 \text{ on }\partial \Omega\end{cases}

is solvable for under some condition of f and \Omega. For the general second order elliptic equation

Lu=\displaystyle \begin{cases}a^{ij}(x) u_{ij}(x)+b^i(x)u_i(x)+c(x)u=f \text{ in }\Omega\\ u=0 \text{ on }\partial \Omega\end{cases}

is related to the Dirichlet problem by defining L_t=(1-t)\Delta+tL.

Let us assume \partial \Omega\in C^{2,\alpha}. \mathfrak{B}=\{u\in C^{2,\alpha}(\overline{\Omega})|u=0 \text{ on }\partial \Omega\}, \mathfrak{V}=\{u\in C^{\alpha}(\overline{\Omega})\} and

(I) all the coefficients in L are in \mathfrak{V}

(II) L is strictly elliptic

(III) c\leq 0

Apparently, (1) is satisfied by L_t. For (2), ||x||_{\mathfrak{B}}\leq C||L_tx||_{\mathfrak{V}} is equivalent to the solution of L_tu_t=f satisfies

\displaystyle ||u||_{2,\alpha}\leq C||f||_{\alpha}

this is guaranteed by the schauder estimate.

So we have the following theorem

\textbf{Thm:} L and \Omega satisfies (I)-(III). Then the problem

Lu=f in \Omega and  u=0 on \partial \Omega

 having a solution in \mathfrak{B} is equivalent to

\Delta u=f in \Omega and  u=0 on \partial \Omega

having a solution in \mathfrak{B}.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: