we denote by $u^+=\max(u,0)$ and $u^-=\max(-u,0)$ the positive and the negative parts of $u$
we have that $u=u^+-u^-$ my question is : what is $u'$ using $u^+$ and $u^-$ ?
and what is $\int_{\Omega} p(t) u'^2(t) dt$ using $u^+$ and $u^-$ ?
Please
Thank you.
I have that $u^{\pm}=\frac12 (|u|\pm u)$ by definition see: http://en.wikipedia.org/wiki/Positive_and_negative_parts
so whene $u>0$ then $|u|=u$
$(u^+)'=\frac12(|u|'+ u')= \frac12( u'+ u')= u'$
whene $u<0$ then $|u|=-u$ then $(u^-)'=\frac12(|u|'- u')= \frac12(-u' - u')= - u'$
and $u^+\times u^-=\frac14(|u|+u)\times(|u|-u)=\frac14(|u|^2+u|u|-u|u|-u^2)=\frac14(|u|^2-u^2)=0$
So $\int_{\Omega}p(t) u'^2(t) dt=\int_{\Omega}p(t) (u^{+'}(t))^2dt+\int_{\Omega}p(t)(u^{-'}(t))^2dt$