Stability of ODE x'=Ax+f(x) Let $\phi \in C^1([0,\infty); R)$ and $\phi' \in C^{0,1}([0,\infty);R) $ and $\phi'(0)=0$  Define $f:R \rightarrow R$ via $$f(x)=\phi'(|x|)\frac{x}{|x|}$$ Then show
(a) $f \in C^{0,1}(R^n;R^n)$ and $f(0)=0$.
(b) Suppose $\phi$ is twice differentiable in $[0,r)$ for some $r>0$, and there exists $C$,$\alpha >0$ s.t. $|\phi"(s)| \leq C|s|^\alpha$ for some $s \in [0,r)$, show that $$|f(x)| \leq C|x|^{1+\alpha}$$ for all $x \in B(0,r)$
(c) Suppose $A \in R^{n*n}$ be antisymmetric, then find a $f \in C^{0,1} (R^n;R^n)$ s.t. 0 is an unstable equilibrium point for the ODE $$x'=Ax+f(x)$$
(d) Same assumption as (c), find a $f \in C^{0,1} (R^n;R^n)$ s.t. 0 is an asymptotically stable equilibrium point for the ODE $$x'=Ax+f(x)$$
I've been trying to solve this problem for a long time. Any help would be appreciated!
 A: What does $C^{0,1}\big([0, \infty), \mathbb{R}\big)$ stand for? Does it mean continuous and Lipschitz? Which points do you have problems with?
For (c) and (d) you can apply Lyapunov stability with Lyapunov function. Take the smooth, quadratic function $V(x) = \frac{1}{2} |x|^2 = \frac{1}{2}(x\cdot x)$. Then $V$ is a Lyapunov function because it is defined everywhere, $V(0)=0$ and $V(x) > 0$ for all $x \in \mathbb{R} \setminus \{0\}.$ Now let $x(t)$ be a solution of $$x' = A\,x + f(x) = A\,x + \phi'\big(|x|\big) \, \frac{x}{|x|}$$
Then \begin{align}\frac{d}{dt} V\big(x(t)\big) &= \frac{1}{2}\, \frac{d}{dt}\big(x(t) \cdot x(t)\big) = \big(x \cdot x'\big) = \big(x \cdot  (A\,x + f(x))\big)  =   \big(x \cdot  A\,x\big) +  \big(x \cdot f(x)\big)\\
&=  \big(x \cdot  A\,x\big) +  \left(x \, \cdot \, \phi'(|x|) \, \frac{x}{|x|}\right) =  \big(x \cdot  A\,x\big) +   \phi'(|x|)  \, \left(x \, \cdot \, \frac{x}{|x|}\right) \\
&= \big(x \cdot  A\,x\big) +   \frac{\phi'(|x|)}{|x|}  \, \left(x \, \cdot \,  x\right) = \big(x \cdot  A\,x\big) +   \frac{\phi'(|x|)}{|x|}  \, |x|^2\\
&= \big(x \cdot  A\,x\big) +   \phi'(|x|)  \, |x|
\end{align}
Now, since $A$ is antisymmetric, $(x \cdot Ax) = (A^*x \cdot x) = -(A x \cdot x) = -(x \cdot Ax)$ is true for any $x \in \mathbb{R}^n$, which means that $(x \cdot Ax) = 0$. Consequently,
 $$\frac{d}{dt} V\big(x(t)\big) = \phi'\big(\,|x(t)|\,\big)\, |x(t)|$$
(c) If $\phi(s)$ is a strictly increasing function on the interval $[0, \infty)$, then $\phi'(s) > 0$ and thus for a non-equilibrium solution $x(t)$ of $x' = A\,x + f(x)$ 
$$\frac{d}{dt} V\big(x(t)\big) = \phi'\big(\,|x(t)|\,\big)\, |x(t)| > 0$$ which means that the equilibrium point $x_0 = 0$ is unstable since if we invert time $t \mapsto -t$ the point will be asymptotically stable (in other words, if we go back in time, the trajectories near $0$ are attracted to it). 
(d) If $\phi(s)$ is a strictly decreasing function on the interval $[0, \infty)$, then $\phi'(s) < 0$ and thus for a non-equilibrium solution $x(t)$ of $x' = A\,x + f(x)$ 
$$\frac{d}{dt} V\big(x(t)\big) = \phi'\big(\,|x(t)|\,\big)\, |x(t)| < 0$$ which means that the equilibrium point $x_0 = 0$ is asymptotically stable. 
