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Let $U$ be a potential function, and consider the IVP $$ (*) \quad x'' = -U'(x), \qquad x(t_0) = x_0, \quad x'(t_0) = v_0. $$

We suppose the following:

(V) Let $x_0, v_0$ be initial values and let $E := \frac{1}{2} v_0^2 + U(x_0)$, then we suppose there exists an interval $[A,B]$ containing $x_0$ such that

  • $U(A) = U(B) = E$ and $U(x) < E$ for $x \in (A,B)$,
  • $U'(A) \ne 0$
  • $U'(B) \ne 0$.

Define $h(x) := \sqrt{2(E - U(x))}$ on $[A,B]$. This is continuous, on $(A,B)$ continuously differentiable and has no root (here comes my first question, why is it continuously differentiable on $(A,B)$?). Further the improper integral $$ \int_A^B h^{-1}(\xi) d\xi $$ exists (why that?). Now suppose $v_0 > 0$ and consider the IVP $$ x' = \sqrt{2(E - U(x))}, \quad x(t_0) = x_0. $$ Then there exists some time interval $[t_A; t_B]$ (why that?) and a solution $\varphi$ in this interval. This is the inverse function of $H : [A,B] \to [t_A;t_B]$ with $$ H(x) = t_0 + \int_{x_0}^x \frac{d\xi}{\sqrt{2(E-U(x))}}, \quad x \in [A,B], $$ and the following properties:

(10) $\varphi(t_A) = A, \varphi(t_B) = B$ and $\varphi(t) \in (A,B)$ for $t \in (t_A, t_B)$

(10') $\varphi'(t_A) = 0, \varphi'(t_B) = 0$ and $\varphi'(t) > 0$ for $t \in (t_A, t_B)$.

The function $\varphi$ on $(t_A, t_B)$ is also a solution of the IVP (*). We have $$ \varphi'' = \frac{-2U'(\varphi)\varphi'}{2\sqrt{2(E - U(\varphi))}} = -U'(\varphi); $$ (where comes this equation from?) and further $\varphi(t_0) = x_0$ and $\varphi'(t_0) = v_0$; the last one by the definition of $E$. The time interval $[t_A;t_B]$ has the length $$ \frac{T}{2} := \int_A^B \frac{d\xi}{\sqrt{2(E - U(\xi))}}. $$

I have a lot of questions regarding this derivation. I marked the points where I stumbled, hoping you can help me!

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1 Answer 1

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  1. $$ h^\prime_x=\frac{-U^\prime(x)}{\sqrt{2(E-U(x))}} $$ Since $U(x)<E$ on $(A,B)$ then $h^\prime_x$ is continuous on $(A,B)$ under condition that $U^\prime(x)$ is continuous.

  2. To prove that integral exists let's make a trick and parametrize $\zeta=x(t)$ when $x$ is the solution of IVP. Then $E(t)=1/2 x^\prime(t)+U(x(t))$ is constant (it's detrivative $x^\prime x^{\prime\prime}+U^\prime_x x^\prime=0$) Thus $U(x(t))=E-1/2 x^\prime(t)^2$. $$ \int_A^B\frac{dx}{\sqrt{2(E-U(x))}} =\int_{t_A}^{t_B} \frac{x^\prime(t)dt}{\sqrt{x^\prime(t)^2}}=\int_{t_A}^{t_B} sign(x^\prime(t)) dt $$

  3. $H$ is monotonic so the inverse function exists. $H$ is actually $t(x)$ function because you have $$ dt = \frac {dx}{\sqrt{2(E-U(x))}} $$

  4. You can skip notion of $H$ and directly consider $\phi(t)$ . Driect differentiation implies the fact that $\phi$ is the solution of original IVP. Now the fact that $\phi^\prime$ is always positive implies that above integral actually has $1$ instead of $sign$ so the last property is also satisfied.

My prove is not completely rigorous but at least can give you a feeling how energy is connected with dynamics of trajectories and velocities.

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  • $\begingroup$ I do not understand point 3: Why does this relation among the differentials implies that $H$ is the inverse function of $x(t)$? $\endgroup$
    – StefanH
    Aug 30, 2014 at 17:13
  • $\begingroup$ Well, this is the same ODE you can look for function $x(t)$ or $t(x)$ , but this is the same trajectory. $\endgroup$ Aug 30, 2014 at 21:39
  • $\begingroup$ Okay, now I see it, as an equation among differentials it is clear... but I am always a little skeptical about these "seeing differentials as fractions and manipulating"-tricks, but anyway it works. $\endgroup$
    – StefanH
    Aug 31, 2014 at 21:34
  • $\begingroup$ It does not only work this is correct. Learn how the derivative of inverse function are related from the property that $f(f^*(x))=x$ $\endgroup$ Sep 1, 2014 at 0:52
  • $\begingroup$ by chain rule: $f'(f^{*}(x))\cdot (f^{*}(x))' = 1$, and therefore $(f^{*}(x))' = 1 / f'(f^{*}(x))$. In terms of differentials, if $dy/dx = g(x)$, then $dx/dy = 1/g(x) = 1/g(x(y))$, expressed in terms of $y$. $\endgroup$
    – StefanH
    Sep 1, 2014 at 11:10

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