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If I fix $x \in M$ and $y \in M_0$, how do I show that the mapping $g: \mathbb{R} \rightarrow \mathbb{R}$ given by

$ g(s) := I[x+sy], \hspace{20pt} s \in \mathbb{R}$

is differentiable and what is then the derivative $g'(0)$ ?

I have the following setup:

Let U: $\mathbb{R}^3$ $\rightarrow\mathbb{R}$ be continuously differentiable, $x_1, x_2 \in \mathbb{R}^3 , \left[t_2,t_3\right]$ an interval with $t_2>t_1$ and

$M:= \left[x \in C^2 (\left[t_1,t_2\right], \mathbb{R}^3) | x(t_1)=x_1, x(t_2)=x_2 \right] $

$M_0:= \left[y \in C^2 (\left[t_1,t_2\right], \mathbb{R}^3) | y(t_1)=y(t_2)=0 \right] $

For $x \in M$ we define $I : M \rightarrow \mathbb{R}$ by

$ I\left[x\right] := \int_{t_1}^{t_2}\frac{1}{2}|\dot{x}|^2-U((x(t)) dt $

StudentP
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1 Answers1

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From: \begin{align} I(x+sy)&=\int\limits_{t_1}^{t_2}\frac{1}{2}\langle(x+sy)^\prime(t), (x+sy)^\prime(t)\rangle-U((x+sy)(t))\,\mathrm{d}t \\ &=\int\limits_{t_1}^{t_2}\frac{1}{2}\langle x^\prime(t), x^\prime(t)\rangle-U(x(t)) + s\langle x^\prime(t), y^\prime(t)\rangle -s\mathrm{d}U(x(t))(y(t))+o(s)\,\mathrm{d}t \\ &=I(x) + s\int\limits_{t_1}^{t_2}\langle x^\prime(t), y^\prime(t)\rangle -\mathrm{d}U(x(t))(y(t))\,\mathrm{d}t+o(s) \end{align} And you know that: $$ g^\prime(s)=\lim\limits_{h\to0}\frac{I(x+(s+h)y)-I(x+sy)}{h}=\int\limits_{t_1}^{t_2}\langle(x+sy)^\prime(t), y^\prime(t)\rangle -\mathrm{d}U((x+sy)(t))(y(t))\,\mathrm{d}t $$ So you end up with: $$ g^\prime(0)=\int\limits_{t_1}^{t_2}\langle x^\prime(t), y^\prime(t)\rangle -\mathrm{d}U(x(t))(y(t))\,\mathrm{d}t $$

user581446
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  • 6