Cameron.martin: A derivation of Lemma 3.4

2018-05-14T22:26:17Z

A derivation of Lemma 3.4

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This is a more detailed derivation of the result from Lemma 3.4.

Let <math>f(\epsilon) = \mathcal{L}(x_c + \epsilon x_q)</math>. This now becomes a single variable minimum/maximum problem. We set <math>\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon=0} = 0</math>, and solve for <math>x_c</math>. First, simplifying <math>f(\epsilon)</math>, we compute

<math>f(\epsilon) = \int_0^T dt(\frac{1}{2}
(\dot{x}_c + \epsilon\dot{x}_q)^2 - V(x_c+\epsilon x_q))</math>

<math>f(\epsilon) = \int_0^T dt(\frac{1}{2}\dot{x}_c^2 + \epsilon \dot{x}_c\dot{x}_q - V(x_c) - \epsilon x_q V'(x_c)</math> + higher order terms).

Thus,

<math>\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon=0} = \int_0^T dt(\dot{x}_c\dot{x}_q - x_q V'(x_c))</math>

Integrating by parts with <math>u = \dot{x}_c, v = x_q</math>, this is equal to

<math>\dot{x}_c\dot{x}_q \mid_{0}^{T} - \int_0^T x_q\ddot{x}_c dt - \int_0^T x_q V'(x_c)dt</math>

The first term is equal to 0 by boundary conditions of <math>x_q</math>, so we obtain the equality

<math>\int_0^T -x_q(\ddot{x}_c + V'(x_c))dt = 0</math>

<math>\Rightarrow \ddot{x}_c + V'(x_c) = 0</math>, exactly as stated in the conclusion of Lemma 3.4. Solving this ODE with initial conditions gives the desired result. Explicitly, the solution of this ODE (with <math>V(x) = \frac{1}{2}x^2</math>) is

<math>x_c(t) = Acos(t) + Bsin(t)</math>
Plugging in <math>t = 0</math> and <math>t = \pi/2</math>, we have

<math>x_0 = x_c(0) = A</math> and <math>x_n = x_c(\pi/2) = B</math>, implying that <math>x_c(t) = x_0cos t + x_nsin t</math>, as claimed.

Notes for AKT-140117/0:21:24 - Revision history

Cameron.martin: A derivation of Lemma 3.4