Vrftsjtryk

I am trying to understand this Answer: https://math.stackexchange.com/a/2366187/597047 as I am curious about it.

1. I do not fully understand what $Phi$ represents. What is the analogue of $Phi$ in the finite-dimensional case? What is meant by "solution mapping"?


2. I do not know what the motivation is for setting $||A||_k$, $||B||_k$, and $l_1, ; l_2$ the way they are written. Why does /u/fourierwho write them this way? What is the motivation?

I greatly appreciate any help.

1. This is the map of the Picard iteration, for $dot x(t)=f(t,x(t))$ it is
   $$ Phi(x)(t)=x(t_0)+int_{t_0}^tf(s, x(s)),ds $$



2. 
   

   $|cdot|_K$ indicates that the norm is the supremum/maximum over the time segment $K$. $$|F|_K=sup_{tin K}|F(t)|$$ where $|F(t)|$ is the norm of the space that $F(t)$ belongs to.

   
   
   

   As solutions of linear ODE and also their differences may grow exponentially in time, this might be an obstacle in the proof that $Phi$ is a contraction mapping. Thus this kind of proof needs a restriction of the time domain of the function space it considers $Phi$ over. The introduction of $R$ is unnecessary, there is no need for the metric space to be bounded in the Banach fixed-point theorem.

   
   

Another kind of proof counters the (potentially) exponential growth of the solutions (as predicted by the bound of the Grönwall lemma) by a faster falling exponential weight factor in a modified maximum norm, $$|x|_L=max_{tin K}e^{-2L|t-t_0|}|x(t)|,~~~L=|A|_K.$$ In this norm $Phi$ has contraction factor $frac12$ over the space ${scr C}=C(K,E)$ and thus a fixed point in that space that the same way then turns out to also be both in $C^1(K,E)$ and a solution of the ODE, no further assembly necessary.

Why does the modified norm work: For prior detailed computations see https://math.stackexchange.com/a/838568/115115 or Inequality in the proof of unique solution of an ODE.

More generally using the Lipschitz property in the localized form, here $L(t)=|A(t)|$, you get for the local differences of the Picard iteration the inequality
$$|Φ(z)(t)-Φ(y)(t)|leint_{t_0}^t L(s)|z(s)-y(s)|.$$
To bound the local differences in the integrand on the right side against a global constant use some weighted sup norm $$|x|_w=sup_{tin I}frac{|x(t)|}{w(t)}.$$ Then the right side is further bounded by
$$...leint_{t_0}^t L(s)w(s),ds;|z-y|_w.$$

Now the norm estimate would be complete if the last expression were just smaller than $q,w(t);|z-y|_w$ with some $0<q<1$, as then $$|Φ(z)-Φ(y)|_wle q;|z-y|_w.$$ Make $w$ the solution to $qdot w(t)=L(t)w(t)$, $w(t_0)=1$, so that $w(t)=exp(int_{t_0}^tL(s)ds/q)$. Then the integral value is $$int_{t_0}^t L(s)w(s),ds=q(w(t)-1)<qw(t)$$ as required.

Usually one would take $L$ a constant maximizing the individual $L(t)$ values over bounded sub-intervals. However, this is often not necessary, see https://math.stackexchange.com/a/2973201/115115 where a non-constant $L(t)$ is used to get a different, better adapted weight function.