数学代考|椭圆方程MATH0090 Elliptic Partial Differential Equations

Proof .

Step 1: The first step is to mollify $u$. Given $\sigma>0$, define $\Omega_\sigma:={x \in \Omega \mid d(x, \partial \Omega)>\sigma}$. Now define $u_\sigma: \Omega_\sigma \rightarrow \mathbb{R}$ by

$$u_\sigma(x)=\left(\eta_\sigma * u\right)(x)=\int_{\mathbb{R}^n} \eta_\sigma(x-y) u(y) d y=\int_{B_\sigma(x)} \eta_\sigma(x-y) u(y) d y .$$
We call $u_\sigma$ the $\sigma$ th mollification of $u$. We claim that $u_\sigma \in C^{\infty}\left(\Omega_\sigma\right)$.
Fix $x \in \Omega_\sigma$, and $1 \leq i \leq n$. Then if $h$ is chosen small enough such that $x+h e_i \in \Omega_\sigma$, we have
$$\frac{u_\sigma\left(x+h e_i\right)-u_\sigma(x)}{h}=\frac{1}{\sigma^n} \int_{B_\sigma(x)} \frac{1}{h}\left(\eta\left(\frac{x+h e_i-y}{\sigma}\right)-\eta\left(\frac{x-y}{\sigma}\right)\right) u(y) d y .$$
Since
$$D_i \eta_\sigma(x)=\sigma^{-n+1} D_i \eta\left(\frac{x}{\sigma}\right)$$
we have
$$\frac{1}{h}\left(\eta\left(\frac{x+h e_i-y}{\sigma}\right)-\eta\left(\frac{x-y}{\sigma}\right)\right) \rightarrow \frac{1}{\sigma} D_i \eta\left(\frac{x-y}{\sigma}\right)$$
uniformly on $B_\sigma(x)$, and hence $D_i u_\sigma(x)$ exists and for $x \in \Omega_\sigma$,
$$D_i u_\sigma(x)=\int_{B_\sigma(x)} D_i \eta_\sigma(x-y) u(y) d y$$

A similar argument shows that $D^{(\alpha)} u_\sigma(x)$ exists and for $x \in \Omega_\sigma$,

$$D^{(\alpha)} u_\sigma(x)=\int_{B_\sigma(x)} D^{(\alpha)} \eta_\sigma(x-y) u(y) d y$$
for any multiindex $\alpha$. Thus $u_\sigma \in C^{\infty}\left(\Omega_\sigma\right)$.
Step 2: We now claim that $u_\sigma \rightarrow u$ for a.e. $x \in \Omega_\sigma$. Fix such a point $x \in \Omega_\sigma$. Then using the fact that $\eta_\sigma$ has unit intgeral,
$$\begin{aligned} \left|u_\sigma(x)-u(x)\right| & =\left|\int_{B_\sigma(x)} \eta_\sigma(x-y)(u(y)-u(x)) d y\right| \ & \leq \frac{1}{\sigma^n} \int_{B_\sigma(x)} \eta\left(\frac{x-y}{\sigma}\right)|u(y)-u(x)| d y \ & \leq \frac{C}{\sigma^n} \int_{B_\sigma(x)}|u(y)-u(x)| \rightarrow 0, \end{aligned}$$
by Lebesgue’s Differentiation Theorem.
Step 3: Next, we claim that $u_\sigma$ is classically harmonic in $\Omega_\sigma$. Write $\Delta_x$ to indicate that the differentiation is with respect to $x$ in the Laplacian. Then
$$\begin{aligned} \Delta_x u_\sigma(x) & =\Delta_x\left(\int_{B_\sigma(x)} \eta_\sigma(x-y) u(y) d y\right) \ & =\int_{B_\sigma(x)} \Delta_x\left(\eta_\sigma(x-y)\right) u(y) d y, \end{aligned}$$
as $\eta_\sigma$ is smooth. But by the chain rule,
$$\Delta_x\left(\eta_\sigma(x-y)\right)=\Delta_y\left(\eta_\sigma(x-y)\right)$$
as the $(-1)$ ‘s cancel. But then
$$\int_{B_\sigma(x)} \Delta_y\left(\eta_\sigma(x-y)\right) u(y) d y=0$$
since $u$ is weakly harmonic and $f(y):=\eta_\sigma(x-y) \in C_c^2(\Omega)$.
Step 4: The next thing to prove are the following two statements about mollification. Let $\sigma, \tau>0$. Define $\left(u_\sigma\right)\tau(x)=\eta\tau * u_\sigma$ for $\tau>0$, so $\left(u_\sigma\right)\tau$ is defined in $\Omega{\sigma+\tau}$. Similarly we define $\left(u_\tau\right)\sigma$. We claim for all $x \in \Omega{\sigma+\tau}$ :

1. $\left(u_\sigma\right)\tau(x)=u\sigma(x)$,
2. $\left(u_\sigma\right)\tau(x)=\left(u\tau\right)\sigma(x)$. Observe for any $x \in \Omega{\sigma+\tau}$,
   $$\left(u_\sigma\right)\tau(x)=\int{B_\tau(x)} \eta_\tau(x-y) u_\sigma(y) d y=\frac{1}{\tau^n} \int_{B_\tau(x)} \eta\left(\frac{x-y}{\tau}\right) u_\sigma(y) d y,$$
   and by the coarea formula (4) we have
   $$\left(u_\sigma\right)\tau(x)=\frac{1}{\tau^{n-1}} \int_0^1 \int{\partial B_{\tau \rho}(x)} \eta\left(\frac{x-y}{\tau}\right) u_\sigma(y) d y d \rho .$$
   Now recall that $\eta$ is radial, and thus $\eta(z)=\eta(|z|)$, and since on $\partial B_{\tau \rho}(x)$, we have
   $$\left|\frac{x-y}{\tau}\right|=\rho$$

we can write

$$\left(u_\sigma\right)\tau(x)=\int_0^1 n \omega_n \rho^{n-1} \eta(\rho)\left(\frac{1}{n \omega_n(\tau \rho)^{n-1}} \int{\partial B_{\tau \rho}(x)} u_\sigma(y) d y\right) d \rho,$$
and then applying Theorem (2.4).(1) to the harmonic function $u_\sigma$ we obtain
$$\left(u_\sigma\right)\tau(x)=u\sigma(x) \int_0^1 n \omega_n \rho^{n-1} \eta(\rho) d \rho .$$
Finally,
$$\int_0^1 n \omega_n \rho^{n-1} \eta(\rho) d \rho=\int_0^1 \eta(\rho)\left(\int_{\partial B_\rho(0)} d S\right) d \rho=\int_{B_1(0)} \eta(|y|) d y=1,$$
and thus we conclude $\left(u_\sigma\right)\tau(x)=u\sigma(x)$.
To prove the second statement, let $x \in \Omega_{\sigma+\tau}$ and observe we may take all our integrals to be over $\Omega_{\sigma+\tau}$. We have
$$\begin{aligned} \left(u_\sigma\right)\tau(x) & =\left(\eta\tau * u_\sigma\right)(x) \ & =\int_{\Omega_{\sigma+\tau}} \eta_\tau(x-y) u_\sigma(y) d y \ & =\int_{\Omega_{\sigma+\tau}} \eta_\tau(x-y) \int_{\Omega_{\sigma+\tau}} \eta_\sigma(y-z) u(z) d z d y . \end{aligned}$$
Now set $w=x-y+z$. Then
$$\left(u_\sigma\right)\tau(x)=\int{\Omega_{\sigma+\tau}} \eta_\tau(w-z) \int_{\Omega_{\sigma+\tau}} \eta_\sigma(x-w) u(z) d w d z$$
which upon exchanging the order of integration (which is valid, as $\eta_\sigma$ and $\eta_\tau$ are smooth and $u$ integrable) is equal to $\left(u_\tau\right)_\sigma(x)$.

Step 5: We can now complete the proof. Fix some $\tau>0$. We have shown that for a.e. $x \in \Omega_{\sigma+\tau}$ we have $\left(u_\tau\right)\sigma(x)=\left(u\sigma\right)\tau(x)=u\sigma(x)$. Now let $\sigma \rightarrow 0$. Thus for a.e. $x \in \Omega_\tau$, we have $u_\tau(x)=u(x)$, with $u_\tau$ smooth and classically harmonic. But $\tau$ was arbitrary; it follows there exists a smooth harmonic function $\bar{u}$ defined on all of $\Omega$ such that for a.e. $x \in \Omega, u(x)=\bar{u}(x)$. This completes the proof.