Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2008 / 3

Using the dual pairing, a monotone operator satisfies $⟨A(u)-A(v),u-v⟩\ge 0$ for all $u,v\in M$. It is a strongly monotone operator if some $m>0$ givesA nonlinear coercive operator with respect to $u_0\in M$ satisfies $⟨A(u),u-u_0⟩/\Vert u-u_0\Vert \to +\infty$ as $\Vert u-u_0\Vert \to \infty$ in $M$.

Strong monotonicity, with $v=u_0$, givesAfter division, the lower bound is $m\Vert u-u_0\Vert -\Vert A(u_0){\Vert }_{X^{\ast }}\to +\infty$, proving coercivity. Reflexivity and separability are not needed for this particular estimate.

Take the zero-boundary Sobolev space $X=H_0^1(\Omega )$ with norm $\Vert u{\Vert }_X=\Vert \nabla u{\Vert }_2$. The Poincaré inequality makes this equivalent to the usual Sobolev norm. DefineThe weak formulation is $u\in X$ and $⟨A(u),v⟩=⟨f,v⟩$ for every $v\in X$.

In dimension three the Sobolev embedding theorem gives $\Vert u{\Vert }_6\le C_6\Vert u{\Vert }_X$, so $u^3\in L^2$. The Cauchy-Schwarz inequality and Poincaré inequality yieldThus $A(u)\in X^{\ast }$ for every $u$, and $f\in X^{\ast }$. Finallysince $u^2+uw+w^2\ge 0$. Hence $A$ is strongly monotone with constant one in the chosen norm.