## Product rule or Product formula for weak derivative

Problem: Derive the product formula for weak derivative

$\displaystyle D(uv)=uDv+vDu$

holds for all ${u,v\in W^1(\Omega)}$ such that ${uv}$ and ${uDv+vDu\in L^1_{loc}(\Omega)}$.

Proof: Step 1: prove the case ${u\in W^1(\Omega)}$, ${v\in C^1(\Omega)}$

Step 2: prove the case ${u,v\in W^{1}(\Omega)\cap L^\infty_{loc}(\Omega)}$ by step 1. Readers can also see the proof on page269 of Functional Analysis, Sobolev Spaces and Partial Differential Equations (Haim Brezis)

Step 3: Define ${f\in C^0(\mathbb{R}^n)}$

$\displaystyle f_n(t)=\begin{cases}n,\quad t>n\\ t,\quad |t|\leq n\\ -n,\quad t<-n\end{cases}$

then ${f_n}$ is piecewise smooth in ${\mathbb{R}}$ and ${f_n'\in L^\infty (\mathbb{R})}$. So ${f_n(u)\in W^1(\Omega)}$ by lemma 7.8 in Gilbarg and Trudinger’s bk and

$\displaystyle D(f_n(u))(x)=\begin{cases}Du(x),\quad |u(x)|\leq n,\\ 0,\quad\quad |u(x)|>n.\end{cases}$

Denote ${u_n=f_n(u)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)}$ and ${v_n=f_n(v)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)}$. By step 2 we have ${u_nv_n\in W^1(\Omega)}$ and

$\displaystyle \int_{\Omega} u_nv_n D\phi dx=-\int_{\Omega}(u_nDv_n+v_n Du_n)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)$

Note that by assumption

$\displaystyle |u_nv_n|\leq |uv|\in L^1_{loc}(\Omega)$

$\displaystyle |u_nDv_n|\leq |uDv|\in L^1_{loc}(\Omega)$

$\displaystyle |v_nDu_n|\leq |vDu|\in L^1_{loc}(\Omega)$

By dominating convergence theorem, letting ${n\rightarrow \infty}$

$\displaystyle \int_{\Omega} uv D\phi dx=-\int_{\Omega}(uDv+v Du)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)$

Step 4:Consider the problem with additional assumption ${u\geq 0}$ and ${v\geq 0}$.

Firstly we assume ${v\geq 1}$. Define ${\tilde{u}_n=\min\{u,\frac{n}{v}\}=u+(\frac{n}{v}-u)^+}$. Since ${u,\frac{n}{v} \in W^1(\Omega)}$, then ${\tilde{u}_n\in W^1(\Omega)}$, satisfies ${0\leq \tilde{u}_n v\leq n}$, moreover

$\displaystyle D\tilde{u}_n=\begin{cases}Du,\quad uv

$\displaystyle vD\tilde{u}_n+\tilde{u}_n Dv=\begin{cases}vDu+uDv,\quad uv

Suppose ${\displaystyle f_\epsilon(t)=\frac{t}{1+\epsilon t}}$ defined on ${\mathbb{R}_+^1}$, where ${\epsilon>0}$. Then ${f_\epsilon}$ is a piecewise smooth function in ${\mathbb{R}_+^1}$ and ${f'_\epsilon\in L^\infty(\mathbb{R}_+^1)}$. By lemma 7.8 on Gilbarg and trudinger’s book, ${f_\epsilon(\tilde{u}_n)\in W^1(\Omega)}$. Note that ${f_\epsilon(u)\in L^\infty(\Omega)}$ by the conclusion of step 3,

$\displaystyle \int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi=-\int_{\Omega}[Df_\epsilon(\tilde{u}_n)v+f_\epsilon(\tilde{u}_n)Dv]\phi$

that is

$\displaystyle \int_{\Omega}\frac{\tilde{u}_n}{1+\epsilon \tilde{u}_n}vD\phi=-\int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon \tilde{u}_n}\phi+\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon \tilde{u}_n)^2}\phi\quad (1)$

By dominating convergence theorem, as ${\epsilon\rightarrow 0}$

$\displaystyle \int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi\rightarrow \int_{\Omega}\tilde{u}_nvD\phi$

$\displaystyle \int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon\tilde{u}_n}\phi\rightarrow \int_{\Omega}(\tilde{u}_nDv+vD\tilde{u}_n)\phi$

And

$\displaystyle \left|\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon\tilde{u}_n)^2}\phi\right|\leq \epsilon n\int_{\Omega}|D\tilde{u}_n||\phi|\rightarrow 0\text{ as }\epsilon\rightarrow 0$

Letting ${\epsilon\rightarrow 0}$, ${(1)}$ implies

$\displaystyle \int_{\Omega}\tilde{u}_nvD\phi=-\int_{\Omega}[vD\tilde{u}_n+\tilde{u}_nDv]\phi$

Since for any ${n>0}$

$\displaystyle |\tilde{u}_nv|\leq |uv|$

$\displaystyle |vD\tilde{u}_n+\tilde{u}_nDv|\leq |vDu+uDv|$

by dominating convergence theorem, letting ${n\rightarrow \infty}$

$\displaystyle \int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi$

If we only know ${v\geq 0}$, consider ${u}$ and ${v+1}$, repeat the above proof

$\displaystyle \int_{\Omega}u(v+1)D\phi=-\int_{\Omega}[(v+1)Du+uD(v+1)]\phi$

note that ${u\in W^1(\Omega)}$, this is equivalent to

$\displaystyle \int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi$

Step 5: Consider the most general case with no extra assumption. Since

$\displaystyle u^+v^+, u^+v^-, u^-v^+,u^-v^-\in L^\infty_{loc}(\Omega)$

$\displaystyle v^{\pm}Du^\pm+u^\pm Dv^\pm\in L^\infty_{loc}(\Omega)\text{ respectively }$

step 4 will imply

$\displaystyle u^\pm v^\pm\in W^1(\Omega),\quad D(u^\pm v^\pm)=v^{\pm}Du^\pm+u^\pm Dv^\pm$

So

$\displaystyle uv= u^+v^+-u^+v^-- u^-v^++u^-v^-\in W^1(\Omega)$

and

$\displaystyle D(uv)=uDv+vDu$

$\Box$

Remark: Who can simplify this proof? It is ugly.