A Maximal Monotone Operator on $H^1_0(0,1)$

Question

I am trying to figure out if the operator

$$ \langle N(u), v \rangle_{\mathcal{H}^*, \mathcal{H}}=\int \limits_0^1 u'(x)u(x)v(x)dx $$ is maximal monotone on the Hilbert space $\mathcal{H}=H^1_0(0,1)$, i.e. if

$$ \langle N(u)-N(v), u-v \rangle_{\mathcal{H}^*, \mathcal{H}}\geq 0 $$ for all $u,v\in \mathcal{H}.$

I already found the identity using integration by parts

$$ \langle N(u), u \rangle_{\mathcal{H}^*, \mathcal{H}}=\int \limits_0^1 u^2(x)\cdot u'(x)dx=0 $$ but I simply can't figure out the rest.

I´m starting to think, that it is not true, but couldn't find a counter example. Thanks in advance!

Answer 1

We can write (integration by parts, Taylor expansion): $$ \langle N(u)-N(v),u-v\rangle = \int_0^1 (u'u - v'v)(u-v) dx = \int_0^1 (u-v)\frac12\frac d{dx}(u^2-v^2) dx\\ = -\frac12\int_0^1 (u'-v')(u^2-v^2) dx\\ = -\frac12\int_0^1 (u'-v')(2v(u-v) + (u-v)^2) dx\\ =-\frac12\int_0^1 v\frac d{dx}(u-v)^2 + \frac13\frac d{dx}(u-v)^3\ dx\\ =\frac12\int_0^1 v'(u-v)^2. $$ Define $v(x) := \min(x, 1-x) \in H^1_0(0,1)$ with $v'=-1$ on $(1/2,1)$. Now choose $u$ such that $u(x)=v(x)$ on $(0,1/2)$ and $u(x)\ne v(x)$ on $(1/2,1)$. Then $\langle N(u)-N(v),u-v\rangle<0$.

Hence $N$ is not monotone.

Note that $\langle N(u),u\rangle=0$, hence the choices $u=0$ or $v=0$ do not prove non-monotonicity.