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First, a motivating example. Suppose $f(x)$ is convex, differentiable, with a single minimum $x^*$. Then the differential equation $$\dot{x}(t) = -\nabla f(x(t))$$ drives $x(t)$ to $x^*$.

Now my question is about a generalization of this. Let $f(x), g(x)$ be two smooth convex functions and let ${\cal G}$ be the set of minima of $g(x)$, which we assume to be nonempty. Consider the differential equation $$ \dot{x}(t) = - \frac{1}{t} \nabla f(x(t)) - \nabla g(x(t))$$ Is it true that this equation drives $x(t)$ to the minimum of $f(x)$ on ${\cal G}$? If not, would it be true if we replaced $1/t$ by a different function, say one which perhaps decays slower or faster?

This statement seems to be true in a few simple examples I tried. For example, taking $g(x)=(x_1+x_2-2)^2$ and $f(x)=x_1^2+x_2^2$ and solving the resulting equation numerically, I do get that solutions seem to approach $(1,1)$.

Update: I asked this question on MO and received a solution. Unfortunately, the solution was given at a level of mathematical maturity that exceeds my own and I am unable to fill in the details of the arguments. Asking the answerer a followup question has not really clarified matters. I am hoping that someone here on could fill in the details.

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    $1/t^p$ with $p>1$ would not work (even if $g\equiv 0$, you don't always get to the minimum of $f$), so it seems that $1/t$ is a borderline case. Perhaps $1/t^p$ with $p<1$ would be more reliable.... I can't prove anything but the questions is interesting.2012-07-01
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    Considering @Leonid's point, it may be better to consider something like $\dot x(t)=-\nabla f(x(t))-t\nabla g(x(t))$ instead.2012-07-01
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    @LeonidKovalev - indeed, it was silly to suggest that something that decays faster would work. I would, of course, be more than satisfied if this turned out to be true for some $p \in (0,1)$ :)2012-07-03

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