Notes for AKT-140117/0:21:24

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

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

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ u = \dot{x}_c, v = x_q }[/math], this is equal to

[math]\displaystyle{ \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]\displaystyle{ x_q }[/math], so we obtain the equality

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

[math]\displaystyle{ \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]\displaystyle{ V(x) = \frac{1}{2}x^2 }[/math]) is

[math]\displaystyle{ x_c(t) = Acos(t) + Bsin(t) }[/math] Plugging in [math]\displaystyle{ t = 0 }[/math] and [math]\displaystyle{ t = \pi/2 }[/math], we have

[math]\displaystyle{ x_0 = x_c(0) = A }[/math] and [math]\displaystyle{ x_n = x_c(\pi/2) = B }[/math], implying that [math]\displaystyle{ x_c(t) = x_0cos t + x_nsin t }[/math], as claimed.