Let $u \in H^2([0, 1])$ with $u(0) = u'(1) = 0$. Is the following calculus correct? $$\underset{0}{\overset{1}{\int}} u''(x) u(x) dx = u'(x) u(x) \underset{0}{\overset{1}{|}} - \underset{0}{\overset{1}{\int}} u'(x) u'(x) dx = - \underset{0}{\overset{1}{\int}} \frac{(u'(x)^3)'}{3} dx = 0. $$
Thank you!
This is only correct for $u''=0$, since $$ -\frac13\int_0^1 (u'(x)^3)' dx = -\int_0^1 u'(x)^2 u''(x)dx, $$ the latter integral is not equal to $\int_0^1 u''(x)u(x)dx$ (except for special cases $u=0$ or $u''=1$), while the first integral is not zero in general.