2
$\begingroup$

Let $\Omega\subset \mathbb{R}^n, \ n\geq 2$ be a bounded Lipschitz domain. Consider the following inhomogenous system (in divergence form) subject to zero Dirichlet boundary conditions: \begin{equation} -\sum_{i=1}^{n}D_ia_i(x, u, Du)+a_0(x, u, Du)=0\quad\text{in } \Omega, \end{equation}where the vector field $a=(a_1, \dots, a_n)$ and the inhomogeneity $a_0$ satisfy the following (natural) growth condition: \begin{align} |a_i(x, u, Du)|&\leq K(1+|z|^{p-1}), \quad i=1, \dots, n,\\ |a_0(x, u, Du)|&\leq K_0(1+|z|^p) \end{align} for all $x\in \Omega,\ u\in \mathbb{R}^N$ and $z\in \mathbb{R}^{nN}$. We suppose that $a_i (i=0, 1, \dots, n)$ are Caratheodory functions.

A weak solution to the above system is a function $u\in W^{1,\ p}(\Omega, \mathbb{R}^N)$ satisfies \begin{equation*} \int_{\Omega}a(x, u, Du)\cdot D\varphi\ \mathrm{d}x+\int_{\Omega}a_0(x, u, Du)\cdot\varphi\ \mathrm{d}x=0\quad\forall\ \varphi\in C_c^{\infty}(\Omega, \mathbb{R}^N). \end{equation*}

In various papers/books about this topic, see for example Frehse and Beck (2013) "Regular and irregular solutions for a class of elliptic systems in critical dimensions", it is said that by approximation, the above identity remains true for all $\varphi\in W_0^{1,\ p}(\Omega, \mathbb{R}^N)\cap L^{\infty}(\Omega, \mathbb{R}^N)$.

Does anyone know of this approximation argument?

$\endgroup$
  • $\begingroup$ I can't understand one thing. In the paper you have cited, the authors works only on the case $p=n$. But in this case we have that $p'=n/(n-1)\leq n$, which implies that $W^{1,p}$ is continuously embedded in $W^{1,p'}$, hence the equality is true for all $\varphi\in W^{1,p}_0$. What I can't understant is why in this case they take $\varphi$ only in $W^{1,p}\cap L^infty$? Do you know the reason? Is my argument wrong? Can you point me out another paper/book which contain this definition? Thank you $\endgroup$ – Tomás Oct 19 '13 at 1:46
  • $\begingroup$ It\s true $W^{1, p}(\Omega)\hookrightarrow W^{1, p'}(\Omega)$ for $p\geq 2$. However, $a_0\notin L^{p'}$ necessarily. See my comments below your answer. I can refer you to Struwe's Variational Methods Appendix C, Chen and Wu's Second order elliptic equations and systems (Defn. 2.1) and Giaquinta's Multiple integrals in the calculus of variations and nonlinear elliptic systems pg 40-41. They've $p=2$ and $u\in L^{\infty}(\Omega)$. I think it's a technical reason (probably other deeper reasons too) so that the variational integral is differentiable in say $H_0^1(\Omega)\cap L^{\infty}(\Omega)$ $\endgroup$ – Nirav Oct 19 '13 at 6:50
  • $\begingroup$ Ok, now I saw my mistake. $\endgroup$ – Tomás Oct 19 '13 at 11:49
1
$\begingroup$

By using the answer here with $m=\|\varphi\|_\infty$, we can assume that there exist a sequence $\varphi_k\in C_0^\infty$ such that $\varphi_k\to \varphi$ in $W^{1,p}$ and $\|\varphi_k\|_\infty\leq M$ where $M$ is a positive constant.

We use Theorem 4.9. from Brezis to conclude without loss of generality that $\varphi_k\to\varphi$ almost everywhere, hence $a_0(x,u,Du)\varphi_k\to a_0(x,u,Du)\varphi$ almost everywhere and $|a_0(x,u,Du)\varphi_k|\leq M|a_0(x,u,Du)|\in L^1$, therefore we can apply Lebesgue theorem to conclude that

$$\tag{1}\int_\Omega a_0(x,u,Du)\varphi_k\to\int_\Omega a_0(x,u,Du)\varphi$$

On the other hand, the functions $a_i(x,u,Du)$ defines bounded linear functionals in $W_0^{1,p}$, which implies that $$\tag{2}\int_\Omega a(x,u,Du)D\varphi_k\to\int_\Omega a(x,u,Du)D\varphi$$

To conclude, we combine $(1)$ and $(2)$ to get the desired equality.

Remark: Just for the sake of clarity, I will prove here that the operator defined by @user98130 in his answer is continuous from $W_0^{1,p}$ to $W_0^{1,p}$.

First note that $\psi$ is Lipschitz, therefore, we can assume that $$|\psi(x)-\psi(y)|\leq C |x-y|,\ \forall\ x,y\in\mathbb{R}$$

Now, suppose that $u_k\to u$ in $W_0^{1,p}$. Note that $$\int_\Omega |\psi(u_k(x))-\psi(u(x)|^p\leq\int_\Omega C^p |u_k(x)-u(x)|^p\tag{3}$$

On the other hand $$\left\|\psi'(u_k)\frac{\partial u_k}{\partial x_i}-\psi'(u)\frac{\partial u}{\partial x_i}\right\|_p\leq \left\|(\psi'(u_k)-\psi'(u))\frac{\partial u_k}{\partial x_i}\right\|_p+\left\|\psi'(u)\left(\frac{\partial u_k}{\partial x_i}-\frac{\partial u}{\partial x_i}\right)\right\|_p\tag{4}$$

We apply Theorem 4.9. of Brezis and then Lebesgue Theorem to conclude that the first term from the right hand side of $(4)$ converges to $0$. To this end, note that by Theorem 4.9. from Brezis's book, we can assume that $u_k\to u$ almost everywhere, which implies that $\psi'(u_k)\to \psi'(u)$ almost everywhere. Moreover $\psi'(u_k)\leq M$, hence, by Lebesgue Theorem, $\psi'(u_k)\to \psi'(u)$ in $L^q$ for all $q\in [1,\infty)$. To conclude, just use Holder inquality.

The second term converges to $0$, because $\psi'$ is bounded and $u_k\to u$ in $W^{1,p}$.

We conclude from $(3)$ and $(4)$ the continuity of the composition operator. Another way to prove the continuity is: Let $T:W_0^{1,p}\to W_0^{1,p}$ be the composition operator, i.e. $T(u)=\psi\circ u$. You can verify that $T$ is Frechet differentiable and that $$T'(u)v=\psi'(u)v$$

Moreover, although this is not necessary, you can also verify that $T$ is $C^1$..

Remark 1: You can also use this approach to prove that $\varphi_k$ can be choose to be bounded.

$\endgroup$
  • $\begingroup$ Why can we assume that $\varphi_k$ is uniformly bounded in $L^{\infty}(\Omega)$? $\endgroup$ – Nirav Oct 17 '13 at 2:27
  • $\begingroup$ Do you remember how $\varphi_k$ is constructed? $\endgroup$ – Tomás Oct 17 '13 at 11:30
  • $\begingroup$ Not really, in Evans PDE \S 5.2.2 he denotes the closure of $C_c^{\infty}(U)$ in $W^{k, \ p}(U)$ by $W_0^{k,\ p}(U)$ and in $S5.3$ he uses mollifiers in the approximation theorems. Certainly, $\varphi_k\in C_c^{\infty}(\Omega)\ \Rightarrow \|\varphi_k\|_{L^{\infty}(\Omega)}\leq C(k)$. I don't see how to deduce that $\sup_{k} \|\varphi_k\|_{L^{\infty}(\Omega)}<\infty$. $\endgroup$ – Nirav Oct 18 '13 at 5:07
  • $\begingroup$ I have edited my answer, please take a look. It is not a full answer, but maybe can help you devise the full one. $\endgroup$ – Tomás Oct 18 '13 at 19:24
  • $\begingroup$ - I agree with the embedding - We only have $a_0\in L^1$ which doesn't imply $a_0\in L^q$ for $q>1$. See my linked question; $|Du|^2\notin L^2$ - Since $T\varphi=0$ on $\partial\Omega$ it behaves like it has compact support and functions in $W^{1, p}$ with compact support have zero trace so they are in $W_0^{1, \ p}$ - I don't think necessarily though $K$ is a positive distance away from $\partial\Omega$. - and all the mollifier convergence would have to be in $L^p_{\text{loc}}$ etc. $\endgroup$ – Nirav Oct 19 '13 at 7:12
1
$\begingroup$

The conditions on the coefficients should imply \begin{align*} a(x,u,Du) &\in L^{p'}(\Omega), \\ a_0(x,u,Du) &\in L^1(\Omega). \end{align*} This yields that the mapping \begin{equation*} \varphi \mapsto a(x,u,Du) \cdot D\varphi + a_0(x,u,Du)\,\varphi \, \mathrm{d}x \end{equation*} is continuous from $W_0^{1,p}(\Omega) \cap L^\infty(\Omega)$ to $\mathbb R$. Moreover, it is zero on the subspace $C_c^\infty(\Omega)$.

Now, if $C_c^\infty(\Omega)$ would be dense in $W_0^{1,p}(\Omega) \cap L^\infty(\Omega)$, you would get your desired result. However, this is only true if $p > n$. One could treat with the case $p \le n$ if one would have the weak density of $C_c^\infty(\Omega)$ in $W_0^{1,p}(\Omega) \cap L^\infty(\Omega)$. I am not sure if this weak density holds.

$\endgroup$
  • $\begingroup$ So by weak density, do you mean that if $\varphi\in W^{1, p}_{0}(\Omega)\cap L^{\infty}(\Omega)$ then there exists a sequence $\varphi_m\in C_c^{\infty}(\Omega)$ such that $\varphi_m\rightharpoonup u$ in $W^{1, p}(\Omega)\cap L^{\infty}(\Omega)$? In particular, \begin{equation} \int g\cdot D\varphi_m\ \mathrm{d}x \rightarrow \int g\cdot D\varphi\ \mathrm{d}x\quad\forall\ g\in L^{p'}(\Omega) \end{equation} and similar for $\varphi_m$ as well as \begin{equation} \int \varphi_m f\ \mathrm{d}x\rightarrow\int\varphi f\ \mathrm{d}x\quad\forall\ f\in L^1(\Omega)? \end{equation} $\endgroup$ – Nirav Oct 14 '13 at 9:00
  • $\begingroup$ @Tomas It is true that $\overline{C_c^{\infty}(\Omega)}=W_0^{1, \ p}(\Omega)$ so given any $\varphi\in W_0^{1, \ p}(\Omega)\cap L^{\infty}(\Omega)$ there exists a sequence $\varphi_m\in C_0^{\infty}(\Omega)$ such that\begin{equation} \|\varphi_m-\varphi\|_{W^{1,\ p}(\Omega)}\rightarrow 0 \end{equation} but we do not have necessarily that\begin{equation}\|\varphi_m-\varphi\|_{L^{\infty}(\Omega)}\rightarrow 0.\end{equation} I had problem with this in this question math.stackexchange.com/questions/497197/… $\endgroup$ – Nirav Oct 15 '13 at 3:40
  • $\begingroup$ @Tomas, I thought that if we want to approximate an element $\varphi\in (X, \|\cdot\|_X)\cap(Y, \|\cdot\|_Y)$ by elements in $(Z, \|\cdot\|_Z)$ (where $Z\subset X\cap Y$), then we want to look for a sequence $\varphi_m\in Z$ such that \begin{equation} \|\varphi_m-\varphi\|_X\rightarrow 0\quad\text{and}\quad \|\varphi_m-\varphi\|_Y\rightarrow 0. \end{equation} Is this wrong? $\endgroup$ – Nirav Oct 16 '13 at 2:49
  • $\begingroup$ @gerw, is my interpretation of weak density the same as yours? $\endgroup$ – Nirav Oct 16 '13 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.