# Proving that $F$ is a contraction for a value $\alpha$

Given the differential equation $dy/dx = f(x,y)$ with initial condition $y(x_0)=y_0$. Let $f$ be a continuous function in $x$ and $y$ and Lipschitz-continuous in $y$ with Lipschitz constant L. Let F be a mapping on the space $C^0(I)$ of continuous functions $u:I\rightarrow \mathbb{R}$. $I = [x_0,M]$. $$Fu(x) = y_0 + \int_{x_0}^x f(s,u(s))ds.$$ Let for $\alpha\geq 0$ the norm $||\cdot||_\alpha$ be given on $C^0(I)$ by $$||u||_\alpha = \max_{x\in I} |u(x)e^{-\alpha x}|,\quad u \in C^0(I).$$

What is an apt value of $\alpha$ such that F is a contraction with respect to the norm $||\cdot||_\alpha$? And how is this $\alpha$ found?

• All you have to do is find a $0\leq q<1$ such that $\left \| F(u(x))-F(g(x))\right \|_\alpha \leq q\cdot \left \| u(x)-g(x) \right \|$ ... and this hint is pretty much the half of the problem. Mar 10, 2016 at 19:16

The Lipschitz condition on $f$ tells you that
\eqalign{\|F(u) - F(v)\|_\alpha &\le L\; \sup_{x \in I} e^{-\alpha x} \int_{x_0}^x |u(s)- v(s)|\; ds \cr &\le L \|u - v\|_\alpha \sup_{x \in I} e^{-\alpha x} \int_{x_0}^x e^{\alpha s}\; ds \cr &= L \|u - v\|_\alpha \dfrac{1 - e^{-\alpha (M-x_0)}}{\alpha} } Thus for a contraction you want $L(1 - e^{-\alpha(M-x_0)})/\alpha < 1$. It suffices to take $\alpha \ge L$.