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<n\\-n\frac{Dv}{v^2},\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)

\displaystyle vD\tilde{u}_n+\tilde{u}_n Dv=\begin{cases}vDu+uDv,\quad uv<n\,\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)

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.

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