Past exam of the methods of applied functional analysis course of the University of Oxford / 2010 / B / 5

Here $T$ is a nonlinear compact operator: it is continuous and sends bounded subsets of $M$ to relatively compact subsets of $X$. Equivalently, the images of any bounded input sequence have a norm-convergent subsequence.

The Schauder fixed-point theorem states that if $M$ is a nonempty closed bounded convex set in a Banach space and $T:M\to M$ is continuous and compact, then $\boxed{T\text{ has a fixed point in }M}$. Linearity, a contraction estimate, and uniqueness are not assumptions or conclusions of this theorem.

A weak solution is $u\in H_0^1(\Omega )$ such thatLet $M_f=\Vert f{\Vert }_{\infty }$ and define the Nemytskii operator $N(v)=f\circ v$ on $L^2(\Omega )$. It satisfies $\Vert N(v){\Vert }_2\le M_f|\Omega {|}^{1/2}$. It is continuous in $L^2$: from any sequence $v_j\to v$ in $L^2$, every subsequence has a further subsequence converging almost everywhere; continuity and boundedness of $f$, followed by the dominated convergence theorem, give convergence of the corresponding $N(v_j)$ in $L^2$. This subsequence criterion proves continuity of the whole sequence.

For $h\in L^2$, the Lax-Milgram theorem defines the inverse Dirichlet solution map $Sh\in H_0^1$ by $\int \nabla (Sh)\cdot \nabla \phi =\int h\phi$. Testing with $Sh$ and using the Poincaré inequality yieldsThe Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. Therefore $T=S\circ N$ is a nonlinear compact operator and its image lies in the closed $L^2$ ball of radius $R=1+C_P^2{M}_f|\Omega {|}^{1/2}$. This ball is nonempty, closed, bounded and convex. The Schauder fixed-point theorem gives $u=Tu$. Since $Tu\in H_0^1$, its defining identity is exactly the required weak formulation. Thus a weak solution exists. This is the bounded semilinear Dirichlet existence theorem.

The displayed nonlinearity is singular at $s=-1$, so it cannot be used directly as the globally bounded function in part (b). On $s\ge 0$, however,Use nonnegative truncation of a semilinear source: define $F(s)=f(\max \{s,0\})$. This is continuous and bounded on all of $\mathbb{R}$, with $F\ge 1/2$. Part (b) gives a weak solution of $-\Delta u=F(u)$, $u\in H_0^1$.

The weak maximum principle in the needed form says that if $u\in H_0^1$ and $\int \nabla u\cdot \nabla \phi \ge 0$ for every nonnegative $\phi \in H_0^1$, then $u\ge 0$ almost everywhere. Here this also follows directly by testing with the negative part $u_{-}=\max \{-u,0\}\in H_0^1$:The Poincaré inequality gives $u_{-}=0$. Hence $u\ge 0$, $F(u)=f(u)$, and the denominator $1+u$ stays at least one almost everywhere. Thus the original equation has a nonnegative weak solution, with bounded right side.