Solution of a coupled gradient system? Suppose $U \times V \subset \mathbb R^n \times \mathbb R^n$ is an open set and $\phi, \psi: U \times V \to \mathbb R$ are two $C^{\infty}$-smooth functions. Furthermore, for each $y \in V$, $\phi_y: U \to \mathbb R$ has the property: all sublevel sets are compact, i.e., 
$$S_a = \{x \in U: \phi_y(x) = \phi(x, y) \le a\}$$
is compact for every $a \in \mathbb R$. For $\psi$, we have that for each $x$, the function $\psi_x: V \to \mathbb R$ has compact sublevel sets. We define a system of ODEs as follows
$$\begin{align*}
\dot{x}(t) = -\partial_x \phi(x, y), \\
\dot{y}(t) = -\partial_y \psi(x, y).
\end{align*}$$
Since the partial derivatives are smooth and thus locally Lipschitz, for any initial condition, there should be a local unique solution. 
My questions:


*

*Is it possible to infer the solutions are defined for all time?

*Is there a name (and reference) for such defined systems? It seems to resemble a gradient system however they are coupled together.
 A: The answer to question 1 is no.
Consider $U = V = \mathbb{R}$, $\phi(x,y) = \frac{1}{2}(x-y^2)^2$ and $\psi(x,y) = \frac{1}{2}(y-x^2)^2$.
Then the sublevel sets of $\phi_y$ and $\psi_x$ for given $a \in \mathbb{R}$ are just the compact intervals $[y^2-\sqrt{2a}, y^2+\sqrt{2a}]$ and $[x^2-\sqrt{2a}, x^2+\sqrt{2a}]$, respectively.
The system turns out to be
\begin{align*}
\dot x = y^2 - x \\
\dot y = x^2 - y.
\end{align*}
A solution to this system is given by $x(t) = y(t) = \frac{1}{1-e^{t-1}}$, since
\begin{align*} \dot x(t) = \frac{e^{t-1}}{(1-e^{t-1})^2} = \frac{e^{t-1}}{(1-e^{t-1})}
\frac{1}{(1-e^{t-1})} = (x(t)-1)x(t) = x(t)^2-x(t) = y(t)^2-x(t), \end{align*}
and vice versa for $y$. This can not be defined for all time, because $\lim_{t \rightarrow 1} x(t) = \infty$.
Another, similar example is given by $\phi = \frac{1}{2}(x-y-y^3)^2, \psi = \frac{1}{2}(y-x-x^3)^2$, here all solutions with $x_0 \neq -y_0$ blow up in finite time. One can show this by adding both differential equations, which yields, after some completing of squares,
\begin{align*}
\partial_t (x+y) = (x+y)(\frac{1}{4}(x+y)^2 + \frac{3}{4}(x-y)^2) > \frac{1}{4}(x+y)^3.
\end{align*}
This means that $x+y$ becomes unbounded in finite time, as long as $x-y \neq 0$.
