I have to prove that if $ f:\mathbb{R^d}\rightarrow\mathbb{R^d} $ is dissipative with respect to the scalar product < . , . > then every fixed point of $x\prime =f(x)$ is stable.
I wanted to use $ \forall~\epsilon>0 ~\exists~\delta>0~ :|dx_{0} |<\delta \implies sup_{t \in [0,\infty)} |x(t)-y(t)|< \epsilon $ and the collocation method but I always come to a point were a specified method is required.
Can anybody give me a hint how to solve it with more general validity?
Thanks!
Xi Tong
Let $x$ and $y$ be two trajectories of the field $f$. Then $$\begin{split} \frac{d}{dt}\|x-y\|^2 &= \frac{d}{dt}\langle x-y,x-y\rangle \\& = 2\langle \dot x-\dot y,x-y\rangle \\& = 2\langle f(x)-f(y) ,x-y\rangle \\& \le 0 \end{split}$$ the last step being the dissipativity of $f$. In the special case where $y$ is a constant solution $y\equiv p$, the inequality implies that for every other solution $x$ the quantity $\|x(t)-p\|$ does not increase in time. Hence, $p$ is stable.
