Difference between expressions regarding Lipschitz continuity Let $f:\mathcal{I}\times \mathcal{X} \to\mathbb{R}$ be an arbitrary function, e.g., $f(t,x)=t^2+x$.
What are the differences between the following locally Lipschitz continuity definitions:
"$f$ is locally Lipschitz on $x$; that is there are for each $x^0\in\mathcal{X}\;[\mathcal{X}\subseteq \mathbb{R}^n$  is an open set$]$ a real number $\rho>0$ and a locally integrable function,
$$\alpha:\mathcal{I}\to\mathbb{R}_{\geq 0},$$
$[\mathcal{I}\subseteq \mathbb{R}$  is an interval$]$ such that the ball $B_\rho(x^0)$ of radius $\rho$ centered at $x^0$ is contained in $\mathcal{X}$ and
$$\lVert f(t,x)-f(t,y) \rVert \leq a(t)\lVert x-y\rVert,$$ for each $t\in \mathcal{I}$ and $x,y\in B_\rho(x^0)$."
"$f$ is locally Lipschitz on $x$; that is for all $(t^0,x^0)\in \mathcal{I}\times \mathcal{X}\;\;[\mathcal{I}\times \mathcal{X}$ is a domain$]$ there exists an $L\geq 0$ and a ball $B_\rho(t^0,x^0)\subseteq \mathcal{I}\times \mathcal{X}$ of radius $\rho$ centered at $(t^0,x^0)$ such that,
$$\lVert f(t,x_1)-f(t,x_2) \rVert \leq L\lVert x_1-x_2\rVert,$$
for all $(t,x_1),(t,x_2)\in B_\rho(t^0,x^0)$."
The first definition comes from E. D. Sontag, Mathematical Control Theory [p. 476, Theorem 54] and the second from W. Walter, Ordinary Differential Equations [p.107].
PS. Both definitions imply the continuity of $f$ with respect to the second argument $x$, don't they?
 A: Yes, both conditions imply the local Lipschitz continuity of $f$ with respect to the second variable, provided that in the first one $\alpha$ is not allowed to take on $+\infty$ as a value. 
The second condition is stronger: it implies the first, even with locally bounded $\alpha$. But not conversely: here is a counterexample. 
Example.  Let's take $\mathcal I=(-1,1)$ and $\mathcal X=(-1,1)$. Define $f(t,x)=|t|^{1/3} \sqrt{|x|+|t|}$. For each fixed $t\ne 0$, this  function has Lipschitz constant $|t|^{-1/6}/2$. (For $t=0$, the Lipschitz constant is $0$). Therefore, the first condition holds. The second fails: the point $(0,0)$ has no neighborhood in $\mathcal{I}\times \mathcal{X}$ on which $f$ is   Lipschitz in the second variable with uniform constant. 

Suppose that the first condition fails. Let $x^0$ be a point for which there is no $\rho$, $\alpha$ pair with the required property. 
Fix $\rho$ such that the closure of $B_\rho(x_0)$ is contained in $\mathcal{X}$. There exist  sequences $(t_k)$, $(x_k)$ and $(y_k)$ such that 
$$\frac{|f(t_k,x_k)-f(t_k,y_k)| }{|x_k-y_k|}\to\infty\tag{1}$$ 
$x_k,y_k\in B_\rho(x_0)$, and $t_k$ do not escape to the boundary of $\mathcal{I}$. Indeed, otherwise 
$$\alpha(t)=\sup_{x,y\in B_\rho(x^0)}\frac{|f(t ,x )-f(t ,y )| }{|x -y |} $$ 
would be a required function. 
By passing to a subsequence, we can ensure that  $x_k,y_k$ converge to some point $x'\in\mathcal{X}$, and $t_k$ converge to some $t'\in \mathcal{I}$. Then the second condition fails at $(t',x')$.
