I have the following problem to solve and am not certain of whether my attempt at the solution is correct.

Problem description:
Given an IVP $\dot{y} = \frac{1}{1+|y|},y(0)=y_0\in\mathbb{R}\forall t>0$ Check existence and uniqueness of solutions (local and global).

Attempt at solution
The first thing that comes to mind when hearing EE is the Theorem of Picard, which states, that given an ODE $x'=f(t,x(t)$ the function $f$ is continuous w.r.t. $t$ and Lipschitz-Continuous w.r.t. $x$ in a domain $U \subset \mathbb{R}\times\mathbb{R}$, where $(t_0,x_0)\in U, x(t_0)=x_0$ being the initial condition. Then there exists an interval $I=[t_0-\epsilon, t_0+\epsilon]$ s.t. there exists a unique solution to the IVP $x'(t)=f(t,x(t)),x(t_0)=x_0$.

Now I have $f(t,y(t))=f(y(t))$ and thus an autonomous ODE. Following holds: $||f(y(t))||_\infty \leq 1\quad\forall t\in I$ where $I$ is any interval. Generally $f$ is continuous $\forall t$

Now let $y,z$ be solutions of the ODE, then: $||f(y)-f(z)||=||\frac{|z|-|y|}{(1+|y|)(1+|z|)}||=(*)$

Now assuming $y(t),z(t)\lessgtr 0$ I can certainly get: $(*) \leq \frac{max_I |z(t) -y(t)|}{min_I |(1+|y(t)|)(1+|z(t)|)|}\leq max_I |z(t) -y(t)| = ||z-y||$
where in the first inequality I use $||x|-|y||\leq |x-y|$.

Thus I have obtained a Lipschitz-constant and can use Picard's theorem to assert, that for any $y_0\neq 0$ there exists a unique solution to the IVP. My problem is, that in the equation above I have used the reverse triangle inequality to obtain the needed inequality for the L-constant. But if, say, $z(t)>0,y(t)<0$ then I get $||y+z||$ which does not help.

On the other hand, if I look at $f(y(t))$, then it is obvious, that $y(t)$ is increasing for all $t$ ($\dot{y}=f(y(t))\geq 0$) From this I would deduce, that $y(t)$ is unique given $y(0)=y_0$, globally.

Question: What am I missing to check (global)existence and uniqueness and what (if any) errors are present in the argumentation above.

  • 2
    $\begingroup$ I don't understand your problem. You have $$\bigl\lvert \lvert a\rvert - \lvert b\rvert\bigr\rvert \leqslant \lvert a-b\rvert.$$ Always. If $a$ and $b$ are real and have the same sign, you have equality, if they have opposite sign, you have a strict inequality. Anyway you have $\lvert f(a) - f(b)\rvert \leqslant \lvert a-b\rvert$. $\endgroup$ Jan 27, 2014 at 23:04
  • $\begingroup$ I think that was me being stupid. I've been appending $-$ where it should not be. So now with the reverse triangle inequality I have a Lipschitz-continuity with $L=1$ on any domain $U=[a,b]\times\mathbb{R}$, thus global existence and uniqueness? $\endgroup$
    – Nox
    Jan 28, 2014 at 0:41

1 Answer 1


Another way to check the Lipschitz condition of $F(y)=1/(1+|y|)$ is to use the derivative $F'(y)$. Granted, it does not exist at $0$. But since $F$ is even, we have $$F(a)-F(b) = F(|a|)-F(|b|)$$ which reduces the consideration to nonnegative arguments. And there the mean value theorem only requires differentiability on $(0,\infty)$.

Since $|F'(y)|=1 /(1+y)^2\le 1$ for $y>0$, the function is Lipschitz with $L=1$. Global existence and uniqueness follow from the Picard theorem.


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