Solving a differential equation for simple harmonic motion.

Question

Let $a$ be a positive real number.
Let $m$ be a positive real number.
Let $k$ be a positive real number.

Solve the following differential equation:

$\frac{1}{2} m \frac{dx}{dt}(t)^2 + \frac{1}{2} k x(t)^2 = \frac{1}{2} k a^2$.
$x(0) = a$.
$\frac{dx}{dt}(0) = 0$.

I cannot solve this differential equation rigorously.
My attempt to find the solution is the following:

$\frac{1}{2} m \frac{dx}{dt}(t)^2 + \frac{1}{2} k x(t)^2 = \frac{1}{2} k a^2$
$m \frac{dx}{dt}(t)^2 + k x(t)^2 = k a^2$.
$\frac{dx}{dt}(t)^2 = \frac{k}{m} (a^2 - x(t)^2)$.
$\frac{dx}{dt}(t)^2 = \omega^2 (a^2 - x(t)^2)$, where $\omega := \sqrt{\frac{k}{m}}$.
Since $x(0) = a > 0$ and $\frac{dx}{dt}(0) = 0$, $\frac{dx}{dt}(t) \leq 0$ for small $t \geq 0$.
So, $\frac{dx}{dt}(t) = -\omega \sqrt{a^2 - x(t)^2}$ for a small real number $t \geq 0$.
$\int_{\epsilon}^{t} \frac{1}{\sqrt{a^2 - x(s)^2}} \frac{dx}{ds}(s) ds = -\int_{\epsilon}^{t} \omega ds$, where $\epsilon$ is a small positive real number.
$\arcsin(\frac{x(t)}{a}) - \arcsin(\frac{x(\epsilon)}{a}) = -\omega (t - \epsilon)$.
Since $\lim_{\epsilon \to 0} x(\epsilon) = a$ and $\lim_{\epsilon \to 0} \epsilon = 0$, $\arcsin(\frac{x(t)}{a}) - \arcsin(1) = -\omega t$.
$\arcsin(\frac{x(t)}{a}) - \frac{\pi}{2} = -\omega t$.
$\arcsin(\frac{x(t)}{a}) = \frac{\pi}{2} -\omega t$.
$x(t) = a \sin(\frac{\pi}{2} -\omega t) = a \cos(\omega t)$.
So, $x(t) = a \cos(\omega t)$ is a solution for a small real number $t \geq 0$.

I wanna know a mathematically rigorous solution.

Answer 1

Problems like this have infinitely many solutions by "gluing" constant solution with non-constant solutions. This is primarily because:

• $\dot{x}(0)=0$
• $\dot{x}$ does not approach 0 fast enough

A simple example would suffice to illustrate the idea. Let the differential equation be $$ \dot{x}(t)^2+x(t)^2=1, x(0)=1, \dot{x}(0)=0 $$

Its phase curve is a unit circle, with the starting point located at (1,0). Since $\dot{x}(0)=0$, it can stay there for an arbitrary amount of time ($[0,+\infty]$ to be exact) before moving clockwise on the unit circle. Correspondingly on the extended phase portrait with possible solutions drawn, what you get is a "gluing" of the constant solution with sinusoidal waves, shown below:

Hope this helps. If the language is unfamiliar, read section 2 of "Ordinary Differential Equations" by V.I. Arnold.