ECE 515, week of Dec. 3

\(\def\Real{\mathbb{R}}\)

  • Pontryagin maximum principle
    • Examples
    • Lagrangians to Hamiltonians: Fenchel-Legendre transform.

      Legendre (sometimes called Legendre-Fenchel) transform \(H\) of a convex function \(L:\Real\to\Real\)) is defined as
      \[
      H(p):=\sup_x px-L(x).
      \]
      Both functions, \(L\) and \(H\) can take value \(+\infty\).

      If \(L\) is strictly convex, the gradient mapping
      \[
      G:x\mapsto \frac{dL}{dx}
      \]
      is one-to-one, and it is clear that the argmax in the definition of \(H\) solves \(p=G(x)\). Denote the functional inverse to \(G\) as \(F: p=G(x) \Leftrightarrow x=F(p)\). In this case,
      \[
      H(p)=pF(p)-L(F(p)).
      \]
      Differentiating, we obtain
      \[
      \frac{dH}{dp}=F(p)+\left(p-\frac{dL}{dx}\right)\frac{dF}{dp}=F(p)
      \]

      Hence the graph of the derivative of \(L\) with respect to \(x\), in the \((x,p)\) space is the same as the graph of the derivative of \(H\) with respect to \(p\). It follows immediately that the Legendre transform is involutive:
      \[
      L(x)=\sup_p px-H(p).
      \]

      If \(L\) is convex, one can turn it into a strictly convex function by adding \(\epsilon |x|^2/2\). The graph of the resulting gradient,
      \[
      p=G(x)+\epsilon x
      \]
      converges to the graph of \(p=G(x)\) as \(\epsilon \to 0\). So one can define the graph \(p=G(x)\) for all convex functions.

      One can also define Legendre transform for nonconvex \(L\), using the same formula. The correspondence won’t involutive anymore (as Legendre-Fenchel transforms, as defined above, are always convex). Still, one can reproduce many results, in particular, that the derivative of \(H\) is discontinuous at \(p\) iff \(L-px\) has two competing global maxima, see figure below.

      Increments of \(H\) near \(p\) are and \(x_-\) and \(x_+\), respectively: the function \(H\) is not continuously differentiable.

       

      These considerations are relevant for the Homework 5, as in Problem a,
      \[
      V(x,0)=\inf_X \frac{(x-X)^2}{2T}+\cos(X)=\frac{x^2}{2T}-\frac{1}{T}
      \sup \left(xX-\left(\frac{X^2}{2}+T\cos(X)\right)\right),
      \]
      i.e. is (somewhat modified) Legendre transform of the function \(X^2+T\cos X\).

      For more, see chapter on Legendre transform in Arnold’s Mechanics, p 61ff.

    • Transversality: boundary conditions as Lagrangian manifolds
    • Minimal time linear problems. Bang-bang control.
    • Sketch of the proof of PMP.

4 Responses to ECE 515, week of Dec. 3

  1. Jasvir Virdi December 8, 2018 at 9:54 am #

    Does this result imply that for a non convex function have 2 competing global maxima/minima, the function will be not continuously differentiable?

    • yuliy December 9, 2018 at 5:38 pm #

      Yes, if px-L has two competing maxima, H is not smooth at p.

  2. Anthony S Podkowa December 9, 2018 at 5:09 pm #

    In the last equation, is cos(T) supposed to be cos(x(T))=cos(X)? I don’t see how it is possible to extract the T out. Please let me know.

    • yuliy December 9, 2018 at 5:39 pm #

      Thank, Tony, – typo corrected.

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