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:

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

  2. Is there a name (and reference) for such defined systems? It seems to resemble a gradient system however they are coupled together.

  • $\begingroup$ Optimization and Dynamical Systems may be useful. $\endgroup$ – Rodrigo de Azevedo Jun 1 '19 at 8:41
  • $\begingroup$ When you say the same property holds for $\psi$, do you mean "for each $y\in V,\dots$" or "for each $x\in U,\dots$? $\endgroup$ – Calvin Khor Jun 1 '19 at 16:02
  • $\begingroup$ @Calvin Khor: For $\psi$, it should be for each $x$. I will clarify in the question. $\endgroup$ – MyCindy2012 Jun 1 '19 at 20:55
  • $\begingroup$ @RodrigodeAzevedo: Thanks. I skimmed through the book. It is more about gradient flow. But maybe hidden somewhere the answer I am looking for. $\endgroup$ – MyCindy2012 Jun 1 '19 at 21:06
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    $\begingroup$ This reminds me of an optimization problem a colleague and I encountered: find $(x,y)$ such that $x = \arg\min \phi(x,y)$ and $y=\arg\min \psi(x,y)$. In the literature this is known as a bilevel optimization problem. Your trajectories should converge to a local optimum of this bilevel problem. $\endgroup$ – Hyperplane Jun 7 '19 at 13:23

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 $x-y \neq 0$.

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    $\begingroup$ Oh that's a cool example. Essentially two Rosenbrock-type functions at a right angle. If we start out at a point like $(2,2)$, $-\partial_x\phi>0$ points away from the center since $\phi$ decreases faster in this direction. Similarly $-\partial_y\psi$ points away from the center as well. We end up climbing along the diagonal in an accelerated pace since the gradients keeps increasing in magnitude leading to the blow up. But I guess this example only shows that we cannot expect global convergence. $(x_0,y_0)$ needs to be close enough to the equilibrium. $\endgroup$ – Hyperplane Jun 7 '19 at 14:35
  • $\begingroup$ @Hyperplane The phase portrait of this system (Obtained fastest by typing in streamplot{-x+y^2, -y+x^2} in Wolfram Alpha) suggests that only the zero set of initial values where $x_0=y_0<0$ will lead to convergence. I think that without conditions on the $y$-dependence of $\phi$ and $x$-dependence of $\psi$, there can be no general statement on stability. $\endgroup$ – Josef E. Greilhuber Jun 9 '19 at 7:20
  • $\begingroup$ I was wrong about the phase portrait, when $x_0, y_0$ are close enough to $0,0$, there is convergence in this system - my bad. But I found another example, which features a single codimension-1 instable equilibrium. $\endgroup$ – Josef E. Greilhuber Jun 9 '19 at 7:51
  • $\begingroup$ Actually I think this example could also be interpreted as a continuous version of the prisoner's dilemma $\endgroup$ – Hyperplane Jun 11 '19 at 18:59

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