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<n}, 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<n-1}. {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<n}, 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|<x_1\\1+x_1\quad if\quad x_1<0, |x_2|<-x_1\\1-x_2\quad if\quad x_2>0, |x_1|<x_2\\1+x_2\quad if\quad x_2<0, |x_1|<-x_2.\end{cases}

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