Notes for AKT-140117/0:21:24

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

Let $f(\epsilon )={\mathcal {L}}(x_{c}+\epsilon x_{q})$. This now becomes a single variable minimum/maximum problem. We set ${\frac {d}{d\epsilon }}f(\epsilon )\mid _{\epsilon =0}=0$, and solve for $x_{c}$. First, simplifying $f(\epsilon )$, we compute

$f(\epsilon )=\int _{0}^{T}dt({\frac {1}{2}}({\dot {x}}_{c}+\epsilon {\dot {x}}_{q})^{2}-V(x_{c}+\epsilon x_{q}))$

$f(\epsilon )=\int _{0}^{T}dt({\frac {1}{2}}{\dot {x}}_{c}^{2}+\epsilon {\dot {x}}_{c}{\dot {x}}_{q}-V(x_{c})-\epsilon x_{q}V'(x_{c})$ + higher order terms).

Thus,

${\frac {d}{d\epsilon }}f(\epsilon )\mid _{\epsilon =0}=\int _{0}^{T}dt({\dot {x}}_{c}{\dot {x}}_{q}-x_{q}V'(x_{c}))$

Integrating by parts with $u={\dot {x}}_{c},v=x_{q}$, this is equal to

${\dot {x}}_{c}{\dot {x}}_{q}\mid _{0}^{T}-\int _{0}^{T}x_{q}{\ddot {x}}_{c}dt-\int _{0}^{T}x_{q}V'(x_{c})dt$

The first term is equal to 0 by boundary conditions of $x_{q}$, so we obtain the equality

$\int _{0}^{T}-x_{q}({\ddot {x}}_{c}+V'(x_{c}))dt=0$

$\Rightarrow {\ddot {x}}_{c}+V'(x_{c})=0$, 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 $V(x)={\frac {1}{2}}x^{2}$) is

$x_{c}(t)=Acos(t)+Bsin(t)$ Plugging in $t=0$ and $t=\pi /2$, we have

$x_{0}=x_{c}(0)=A$ and $x_{n}=x_{c}(\pi /2)=B$, implying that $x_{c}(t)=x_{0}cost+x_{n}sint$, as claimed.