(Evans) Incomplete proof of the existence and uniqueness of weak solutions to a reaction-diffusion system

Question

In Example 1 in section 9.2 (Fixed Point Methods), Evans employs Banach's Fixed Point Theorem to prove the existence and uniqueness of a weak solution to a system of reaction-diffusion equations. A weak solution $u$ is such that it satisfies the given variational formulation and at the same time $u \in L^{2}\left(0, T; H_{0}^{1} (U; \mathbb{R}^m) \right)$, $u' \in L^{2}\left(0, T; H^{-1} (U; \mathbb{R}^m) \right)$. However, Evans uses an operator $A$ defined on the space $X := C \left( [0, T]; L^2 (U; \mathbb{R}^m)\right)$, showing that there exists a unique solution from $X$ satisfying the variational formulation. My question is: how can we deduce that this solution (together with its derivative) also belongs to the right spaces required above?

Answer 1

Let $\mathscr U = \{ \mathbf w \in L^2(0,T; H^1_0(U;\mathbb R^m)) \text{ s.t. } \mathbf w' \in L^2(0,T;H^{-1}(U;\mathbb R^m)\}.$ By Theorem 3 in Section 5.9.2 of Evans, $\mathscr U \subset C([0,T];L^2(U;\mathbb R^m))$. Recall that $A:X\to X$ is defined by $\mathbf u \mapsto \mathbf w$ where $\mathbf w \in \mathscr U $ is the unique weak solution of the linear PDE given in equation (8) of the example in Evans. In particular, this means that the image of $A$ is a subset of $\mathscr U$. Thus, if $\mathbf u$ is a fixed point of $A$ then $\mathbf u$ is in the image of $A$, so $\mathbf u \in \mathscr U$.