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Assume a mapping $X: \mathbb{R} \to \mathbb{R}^d$. We know that the solution to ode $$ d X_t = (\mu - X_t) dt $$ is $X_t = (X_0-\mu) e^{- t} + \mu$, which indicates that $X_t$ converges to $\mu$ as $t \to \infty$, regardless of $X_0\in \mathbb{R}^d$ .

How to solve the following two generalized odes of the above one:

  1. $$ d X_t = \exp(-\|X_t- \mu\|) (\mu - X_t) dt $$ where $\|\|$ is the Euclidean norm.

    If the range $\mathbb R^d$ of $X$ is $\mathbb R$, then according this table, we can apply the separation of variable method $$t = \int_{-\infty}^X \frac{1}{\exp(-\|x- \mu\|) (\mu - x)} dx +C,$$ but how to solve the integral in it then?

    But the range of $X$ is $\mathbb R^d$ in general. So it is a system of nonlinear, separable and autonomous odes. How shall we solve this sytem then?

    Does the solution $X_t$ still converge to $\mu$ as $t \to \infty$?

  2. $$ d X_t = \sum_{i=1}^3 \exp(-\|X_t- \mu_i\|) (\mu_i - X_t) dt? $$ Does $X_t$ converge to one of the three $\mu_i$'s as $t \to \infty$, depending on where $X_0$ is? How is the attraction region of $X_0$ for each of the three $\mu_i$'s decided?

By the way, are their names for the above three odes?

Thanks in advance!

Note: The questions similar to those for sdes in an earlier post but here they are for deterministic odes.

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The equation $X'_t=\exp(-\|X_t-\mu\|)(\mu-X_t)$ tells you that the velocity vector points from $X_t$ to $\mu$. Therefore, the solution stays on the line passing through $\mu$ and the initial point $X_0$. This makes it an essentially one-dimensional problem. Here is the formal reduction to one dimension: write $$X_t=\mu+f(t)\frac{X_0-\mu}{\|X_0-\mu\|}$$ where $f$ is a scalar function such that $f(0)=\|X_0-\mu\|$. Differentiate in $t$ to get $$f'(t)\frac{X_0-\mu}{\|X_0-\mu\|}=-e^{-f(t)} f(t)\frac{X_0-\mu}{\|X_0-\mu\|}$$ hence $$f'(t) =-e^{-f(t)}\,f(t) $$

Both 1 and 2 are equations of the form $X_t'=-\nabla \Phi(X_t)$, where $\Phi$ is some scalar function. These are gradient systems (discussed here, for example).

In case 1, $\Phi(X)=-(1+\|X-\mu\|)e^{-\|X-\mu\|}$. This is a function with the only critical point at $\mu$, to which all trajectories converges.

In case 2, $\Phi(X)=-\sum_{i=1}^3(1+\|X-\mu_i\|)e^{-\|X-\mu_i\|}$. This function may have critical points other than $\mu_i$, and the points $\mu_i$ need not be local minima for $\Phi$. For example, if $\mu_i$ are the vertices of regular triangle $(1,0)$, $(-1/2,\pm \sqrt{3}/2)$, then the only local minimum is at the origin, and this is where the trajectories will converge. On the other hand, if this triangle is made 10 times larger, then each $\mu_i$ is a local minimum, but the origin is still a critical point.

You can find the basins of attraction just by looking at a sufficiently detailed contour plot of $\Phi$. Something like this.

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  • $\begingroup$ Thanks! (1) But in the two cases, why does $ -\nabla \Phi(X) (X_t)$ equal to the functions on the RHSes of my odes? (2)in case 2, why "if μi are the vertices of regular triangle (1,0), (−1/2,±3√/2), then the only local minimum is at the origin, and this is where the trajectories will converge. On the other hand, if this triangle is made 10 times larger, then each μi is a local minimum, but the origin is still a critical point"? $\endgroup$ – Ethan Apr 3 '13 at 5:18

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