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Have I understood this correctly: So originally we consider the following partial differential equation:

$$u_t= \frac{u_{xx}}{1+w}-\frac{1}{\epsilon}(1+w)u^3+\frac{wu}{\epsilon(1+w)} \text{ in } \Omega \times (0,\infty)$$ $$w_t=(1+w)^2f^2-w \text{ in } \Omega \times (0,\infty)$$ $$u_x = 0 \text{ on } \partial \Omega$$ $$u(x,0)=u_0(x) \ , \ w(x,0)=w_0(x) \text{ in } \Omega$$

Afterwards it says: For $f, w_0 \in H^2(\Omega)$, the second equation is an ODE in the Banach space $H^2(\Omega)$. There exists a solution of this ODE.

  1. Considering only: $w_t=(1+2)^2f^2-w$, $f, w_0(x) \in C(\Omega)$ is sufficient for Existenz according to Peano's theorem.
  2. However, if we consider the complete pde we need $u \in H^2(\Omega)$ and therefore as we know that $u$ will have the same properties as $f$, we thus assume $f$ and accordingly also $w \in H^2(\Omega)$ and obtain a ODE in $H^2(\Omega)$ and with the theorem of Cauchy, Lipschitz,Picard also the existence of a solution.
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    $\begingroup$ Not to be flip, but why not ask your professor? $\endgroup$
    – LSpice
    Commented May 3, 2022 at 13:21
  • $\begingroup$ I asked him but somehow I didn't really understand his answer or his answer was rather: you have to know. $\endgroup$
    – PeterSo
    Commented May 3, 2022 at 13:34
  • $\begingroup$ Continuity is not enough. Peano's theorem does not hold if your ODE takes values in an infinite-dimensional vector space. mathoverflow.net/questions/163711/… $\endgroup$ Commented May 3, 2022 at 15:38
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    $\begingroup$ Correct, u(t)’=V(t,u(t)) has (unique) local solution for the initial value problems, if V(t,x) is continuous and lipschitz wrto x. $\endgroup$ Commented May 3, 2022 at 17:52
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    $\begingroup$ Funnily though, there is a sense in which the OP is right. One does not really need to use the Banach space machinery here, if the goal is not to demonstrate a solution in $H^2$. Given the stated conditions, $f\in H^2\implies f \in C^{1,1/2}$ and for $w_0\in C^1$ you can also prove that the ODE has solution in $C^1((-\epsilon,\epsilon)\times \Omega)$ just using Picard's theorem. $\endgroup$ Commented May 3, 2022 at 18:56

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Let $X$ be a Banach space, and let $V:\mathbb{R}\times X\rightarrow X$ be continous in its first argument and at least Lipschitz in its second argument: i.e., that $\|{V(t,x)-V(t,y)}||\leq K||x-y||$ for some $K>0$. Differentiation is defined between maps in Banach spaces, and so you are looking for a solution $\gamma:\Omega\rightarrow X$ to the initial value problem $\gamma'(t)=V(t,\gamma(t))$ with $\gamma(0)=x_{0}$, for some interval $x_{0}\in\Omega\subset\mathbb{R}$ .

For this to happen, you have to be able to repeat Picard's theorem for the existence and uniqueness of solutions to initial value problems for ODEs. You need to be able to define a map $F:C(\Omega,X)\rightarrow C(\Omega,X)$ and a sufficiently small interval $x_{0}\in\Omega\subset \mathbb{R}$ such that the solution will be a fixed point of $F$.

As in the classical proof, you define $F(\gamma)(t)=x_0+\int_{0}^{s}V(s,\gamma(s))ds$, where the integral is understood in the Banach sense: $\int_{0}^{t}V(s,\gamma(s))ds$ is the unique element of $X$ that satisfies $\rho(\int_{0}^{t}V(s,\gamma(s))ds)=\int_{0}^{t}\rho (V(s,\gamma(s)))ds$ for all $\rho\in X^{*}$. You can use the theory of Banach spaces to show that there is such an element under the demand of continuity of $V$.

By putting an extra assumption of Lipschitzness on $V$ , you can then shrink $\Omega$ sufficiently to show that $F$ is a contracting mapping between complete Banach spaces --- hence, it follows from the contracting map theorem that there is a fixed point. This will be the solution to the original ODE by using the properties of the Banach-valued integration.

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