## Tag Archives: weak derivative

### Sharpness of Sobolev embedding

Consider the Sobolev embedding

Thm 1: Suppose ${u\in W^{1,p}(\Omega)}$, ${\Omega}$ is a bounded domain in ${\mathbb{R}^n}$ with ${C^1}$ boundary, ${p, then

$\displaystyle ||u||_{L^q(\Omega)}\leq C(n,p,q,\Omega)||u||_{W^{1,p}(\Omega)}$

for any ${1\leq q\leq \frac{np}{n-p}}$.

Consider the sharpness of Sobolev embedding, which means ${q}$ can not be bigger than ${\frac{np}{n-p}}$.

WLOG assume ${\Omega=B_1(0)}$. Choose a particular function ${u(x)=\frac{1}{|x|^\alpha}}$, then ${u\in W^{1,p}(B_1)}$ when ${\alpha<\frac{n-p}{p}}$. The reason is the following

${u}$ is smooth away from ${0}$ with

$\displaystyle D_iu=\frac{-\alpha x_i}{|x|^{\alpha+2}}$

For any ${\phi\in C_c^\infty(B_1)}$

$\displaystyle \int_{B_1-B_\epsilon}uD_i\phi=-\int_{B_1-B_\epsilon}D_iu\phi+\int_{\partial B_\epsilon}u\phi\nu_ids$

Let ${\epsilon \rightarrow 0}$, we have

$\displaystyle \left|\int_{\partial B_\epsilon}u\phi\nu_ids\right|\leq C\int_{\partial B_\epsilon}\epsilon^{-\alpha}ds\leq \epsilon^{-\alpha+n-1}\rightarrow 0$

Then

$\displaystyle \int_B uD_i\phi=-\int_BD_iu\phi$

for any ${\phi\in C^\infty_c(B)}$ when ${\alpha. ${u}$ has weak derivative ${D_iu}$, ${Du\in L^p(B)}$ only when ${\alpha<\frac{n-p}{p}}$.

Let us calculate

$\displaystyle ||u||_{L^q}=C\left(\int_0^1r^{-\alpha q+n-1}dr\right)^{1/q}$

${u\in L^q(B_1)}$ if and only if ${\alpha<\frac{n}{q}}$.

So if ${q>\frac{np}{n-p}}$, we can find one ${\alpha}$ such that ${\alpha\in (\frac{n}{q}, \frac{n-p}{p})}$. By the above analysis, such ${u\in W^{1,p}(B_1)}$ but ${u\not\in L^q(B_1)}$.

Thm 2: Suppose ${u\in W^{1,p}(\Omega)}$, ${\Omega}$ is a bounded domain in ${\mathbb{R}^n}$ with ${C^1}$ boundary, ${p, then the embedding ${W^{1,p}(\Omega)\rightarrow L^q(\Omega)}$ is compact when ${q\in [1,\frac{n-p}{np})}$.

Consider the sharpness of compact embedding, ${q}$ must be strictly less than ${\frac{n-p}{np}}$. Actually we can find a sequence of ${u_n\in W^{1,p}(\Omega)}$ but ${u_n}$ does not have convergent subsequence in ${L^{p*}(\Omega)}$.

As before assume ${\Omega=B_1}$. Choose ${u\in C^1_0(B_1)}$, define ${u_{\lambda}(x)=\lambda^\alpha u(\lambda x)}$, for ${\lambda\geq 1}$, then ${u_\lambda\in C^1_0(B_{1/\lambda})}$

$\displaystyle ||u_\lambda||_{L^{p*}(B_1)}=||u_\lambda||_{L^{p*}(\mathbb{R}^n)}=\lambda^{\alpha-n/p*}||u||_{L^{p*}(B_1)}$

$\displaystyle ||Du_\lambda||_{L^p(B_1)}=\lambda^{\alpha+1-n/p}||Du||_{L^p(B_1)}$

Since ${u_\lambda}$ has compact support, then

$\displaystyle ||u||_{W^{1,p}}\leq C||Du||_{L^p}\leq C\lambda^{\alpha+1-n/p}||Du||_{L^p(B_1)}$

Choose ${\alpha=n/p*=n/p-1}$, then ${||u||_{W^{1,p}}}$ is bounded. However, ${u_\lambda}$ has no convergent subsequence. Otherwise as ${\lambda\rightarrow \infty}$, ${u_\lambda(x)\rightarrow 0}$ when ${x\neq 0}$, such subsequence must converge to 0 in ${L^{p*}(\Omega)}$. Since we have ${||u_\lambda||_{L^{p*}}=||u||_{L^{p*}}}$, apparently this can not be true.

### One example related to weak derivative

$\mathbf{Problem:}$ Suppose $D=\{(x_1,x_2)||x_1|<1, |x_2|<1\}$ is the open square in $\mathbb{R}^2$. $u$ is defined

$\displaystyle u(x)=\begin{cases}1-x_1\quad if\quad x_1>0, |x_2|0, |x_1|

Find the first weak derivative of $u$.

$\mathbf{Proof:}$ Suppose $v=\partial_\alpha u$ is the weak derivative, then for any $\phi\in C^\infty_0(D)$, we have

$\displaystyle \int_D v\phi dx=-\int_D u\partial_\alpha\phi dx.\quad \alpha=1,2\quad (1)$

WLOG, assume $\alpha=1$. Let us denote the four domains in the definition of $u$ as $D_1,D_2,D_3,D_4$ respectively. Note that

$\displaystyle \int_D u\partial_1\phi dx=\int_{D_1} u\partial_1 \phi dx+\int_{D_2} u\partial_1\phi dx+\int_{D_3} u\partial_1\phi dx+\int_{D_4} u\partial_1\phi dx\quad(2)$

While

$\displaystyle \int_{D_1} u\partial_1\phi dx=\iint_{D_1}(1-x_1)\partial_1\phi dx_1dx_2=\int_{0}^1\int_{-x_1}^{x_1}(1-x_1)\partial_1 \phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{-1}^1 \int_{|x_2|}^1(1-x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1-x_1)\phi\big|^1_{|x_2|} +\int_{|x_2|}^1 \phi dx_1\right]dx_2$

$\quad$                       $\displaystyle =-\int_{-1}^1(1-|x_2|)\phi(x_2,x_2)dx_2-\int_{D_1}\partial_1u\,\phi dx\quad (3)$

obtained from the integral by parts and the boundary behavior of $\phi$. Similarly for domain $D_2$

$\displaystyle \int_{D_2} u\partial_1\phi dx=\iint_{D_2}(1+x_1)\partial_1\phi dx_1dx_2=\int_{-1}^0\int_{x_1}^{-x_1}(1+x_1)\partial_1 \phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{-1}^1 \int^{-|x_2|}_{-1}(1+x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1+x_1)\phi\big|_{-1}^{-|x_2|} -\int_{-1}^{-|x_2|} \phi dx_1\right]dx_2$

$\quad$                      $\displaystyle =\int_{-1}^1(1-|x_2|)\phi(-x_2,x_2)dx_2-\int_{D_2}\partial_1u\,\phi dx\quad (4)$
As for the domain $D_3$ and $D_4$, we have
$\displaystyle \int_{D_3} u\partial_1\phi dx=\iint_{D_3}(1-x_2)\partial_1\phi dx_1dx_2=\int_{0}^1(1-x_2)\int_{-x_2}^{x_2}\partial_1\phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{0}^1(1-x_2)[\phi(x_2,x_2)-\phi(-x_2,x_2)] dx_1dx_2\quad (5)$

$\displaystyle \int_{D_4} u\partial_1\phi dx=\iint_{D_4}(1+x_2)\partial_1\phi dx_1dx_2=\int_{-1}^0(1+x_2)\int_{x_2}^{-x_2}\partial_1\phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{-1}^0(1+x_2)[\phi(-x_2,x_2)-\phi(x_2,x_2)] dx_1dx_2\quad (6)$
Substituting $(3-6)$ to $(2)$

$\displaystyle \int_D u\partial_1\phi dx=-\int_{D_1}\partial_1u\,\phi dx-\int_{D_2}\partial_1u\,\phi dx$.

So from $(1)$, we know

$\displaystyle v=\partial_1 u=\begin{cases}-1, x\in D_1\\1,x\in D_2\,\\ 0,x\in D_3\cup D_4. \end{cases}$

As for $\partial_2u$, one can get it in a similar way.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Evans, partial differential equation. 5.3