## New answers tagged oc.optimization-and-control

0

The Extended Monotropic Programming problem deals with a more general extension to all n variables at a time than the two variables at a time extension you are interested in, except that the literature I have seen deals only with the constraint set $C$ being a closed linear subspace, not a general convex set.
Specifically:
$\min_{x \in \text{closed linear ...

0

The classical material derivative $D\varphi/Dt$ of a test function $\varphi \in C_c^\infty(\mathbb{R}_+\times D)$ is obtained by setting
$$
\dfrac{D\varphi}{Dt}(x) := \dfrac{\partial\tilde\varphi}{\partial t}(0,x)\;\;\mbox{with}\;\; \tilde\varphi(t,x) = \varphi(t,T_t(x))
$$
for $x\in D$. Expanding out using the chain rule, we have
$$
\dfrac{D\varphi}{Dt}(x) =...

dg.differential-geometry differential-topology na.numerical-analysis oc.optimization-and-control numerical-analysis-of-pde

2

Brownian motion, i.e. Wiener measure, is a good source of ideas and examples here.
For instance if $W_t$ is 1-dimensional standard Brownian motion at time $t$ and $$P(\forall x\,F(x)=x^2)=1$$ and $Y=W_1$ then $F$ and $Y$ are independent but $$E(F(Y))=1\ne E(F(E(Y)))=E(F(0))=0.$$
(Actually here we just need that $W_1$ is a standard normal random variable.)

2

The answer is yes, at least in the limiting case where the number of points tends to infinity.
Specifically, this is known as the quantizer problem (see Chapter 2 of Sphere Packings, Lattices and Groups by Conway and Sloane). The two-dimensional version of the problem was solved by Fejes Tóth, who showed that the hexagonal lattice is optimal.
László Fejes ...

oc.optimization-and-control convex-optimization global-optimization machine-learning cluster-analysis

2

We have $0\le a\le b\le1$ and $T\in(0,\infty)$. We want to know when there is a positive constant $C$ such that
$$\int_0^T dt\, \int_a^b dx\, u^2(x-t)\geq C\int_0^1 dx\,u^2(x) \tag{1}$$
for all measurable functions $u\colon\mathbb R\to\mathbb R$ such that $u(x)=0$ for $x\notin(a,b)$.
The answer is: never. Indeed, without loss of generality $a<b$. The left-...

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