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Let $x'=\frac{1}{1+|x|}$ with $x(0)=\xi$

Let $\xi,\nu \in \Bbb R$ and $t \ge0$ show: $|u_\xi(t)-u_\nu(t)| \le e^t|\xi -\nu|$

What I've done so far:

Define $u_\xi(t):=\xi + \int_{t_0}^tf(s,u_\xi(s))ds$ and $u_\nu(t):=\nu + \int_{t_0}^tf(s,u_\nu(s))ds$ and $f(t,x):=\frac{1}{1+|x|}$.

So $\frac{d}{dt}u(t):=f(t,u(t))$ and

$$\begin{align}|u_\xi(t)-u_\nu(t)| &= |\xi + \int_{t_0}^tf(s,u_\xi(s))ds -(\nu +\int_{t_0}^tf(s,u_\nu(s))ds)|\\ &=|(\xi-\nu) + \int_{t_0}^tf(s,u_\xi(s))ds -\int_{t_0}^tf(s,u_\nu(s))ds|\\ &\le|\xi-\nu|+|\int_{t_0}^tf(s,u_\xi(s))ds -\int_{t_0}^tf(s,u_\nu(s))ds|\\ &\le|\xi-\nu|+\int_{t_0}^t1\cdot|u_\xi(s) -u_\nu(s)|ds\\&\le|\xi-\nu|\cdot e^{\int_{t_0}^t1ds}\tag{gronwall}\label{*}\\&=|\xi-\nu|\cdot e^t\end{align} $$

With Lipschitz-constant $L=1$

  • 0
    Where in your proof do you use that $u' = \frac{1}{1+|u|}$? As written your proof seems to work for any function $f$. How do you apply Gronwalls? I would first show $|u_\zeta(t) - u_\nu(t)| \leq |\zeta - \nu| + \int_0^t |u_\nu(s) -u_\zeta(s)|{\rm d}s$ and then apply Gronwalls on this. – 2017-02-20

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