The motions of the harmonic model are the sinusoids $A\cos (\omega (t-t_0))+B\sin (\omega (t-t_0))$: amplitude and phase, the least period, the conserved energy $k{R}^2\mathrm{/}2$ and the elliptical phase curves

Unit-circle parametrisation. For reals $c,s$ with $c^2+s^2=1$ there is a real $\theta$ with $\cos \theta =c$ and $\sin \theta =s$. Indeed, $c\in [-1,1]$ because $c^2\le c^2+s^2=1$; cosine is continuous on $[0,\pi ]$ with $\cos 0=1$ and $\cos \pi =-1$, so by [L4] there is ${\theta }_1\in [0,\pi ]$ with $\cos {\theta }_1=c$. Then ${\sin }^2{\theta }_1=1-c^2=s^2$ by [L3], and $\sin {\theta }_1\ge 0$ on $[0,\pi ]$ by [L3]; hence $\sin {\theta }_1=\mid s\mid$. If $s\ge 0$ take $\theta ={\theta }_1$; if $s<0$ take $\theta =-{\theta }_1$, so that $\cos \theta =\cos {\theta }_1=c$ and $\sin \theta =-\sin {\theta }_1=-\mid s\mid =s$ by the parity identities of [L3].

The periods of cosine. If $T$ is a real number with $\cos (u+T)=\cos u$ for all $u$, then $T\in 2\pi \mathbb{Z}$. Indeed, taking $u=\pi \mathrm{/}2$ gives $\cos (\pi \mathrm{/}2+T)=0$, so $\pi \mathrm{/}2+T=(m+1\mathrm{/}2)\pi$ for an integer $m$ by [L4], that is $T=m\pi$; taking $u=0$ gives $\cos T=1$. From $\cos (x+\pi )=-\cos x$ and $\cos 0=1$ of [L3], induction on $\mid m\mid$ gives $\cos (m\pi )=(-1{)}^m$ for every integer $m$; hence $(-1{)}^m=1$ and $m$ is even, so $T\in 2\pi \mathbb{Z}$. Conversely $T\in 2\pi \mathbb{Z}$ is a period of cosine by [L4].

Clause 1. Let $x:J\to \mathbb{R}$ be a motion and $t_0\in J$. The shifted curve $y(s):=x(t_0+s)$ is defined on the interval $J-t_0$ with nonempty interior and satisfies $m{y}^{\mathrm{\prime }\mathrm{\prime }}+ky=0$ there (the equation has no explicit time dependence). By [L2] there are $A,B$ with $y(s)=A\cos (\omega s)+B\sin (\omega s)$, that is $x(t)=A\cos (\omega (t-t_0))+B\sin (\omega (t-t_0))$ on $J$. Conversely each function of this form is obtained by shifting a solution of [L2] and is therefore a motion. For the datum $(t_0,x_0,v_0)$ apply [L2] to $y$ with $y(0)=x_0$ and $y^{\mathrm{\prime }}(0)=v_0$: there is exactly one solution on $\mathbb{R}$, namely $A=x_0$, $B=v_0\mathrm{/}\omega$; translating back gives the unique maximal motion, defined on $\mathbb{R}$, and every motion with the datum is a restriction by the uniqueness in clause 1 of Existence, uniqueness and conservation of energy for a one-dimensional conservative model: phase curves lie in energy level sets and the velocity is determined by the position and energy up to sign. This is clause 1.

Clause 2. Write $\theta (t):=\omega (t-t_0)$. For every real $\theta$, $R^2-(A\cos \theta +B\sin \theta {)}^2=A^2{\sin }^2\theta +B^2{\cos }^2\theta -2AB\sin \theta \cos \theta =(A\sin \theta -B\cos \theta {)}^2\ge 0,$ where only the expansion of the squares and ${\sin }^2+{\cos }^2=1$ of [L3] were used; hence $\mid x(t)\mid \le R$. If $R>0$, apply step 1.1 to $(c,s)=(A\mathrm{/}R,B\mathrm{/}R)$, which satisfies $c^2+s^2=1$: there is $\phi$ with $\cos \phi =A\mathrm{/}R$ and $\sin \phi =B\mathrm{/}R$. The addition formula of [L3] gives $\cos (\theta -\phi )=\cos \theta \cos \phi +\sin \theta \sin \phi =\frac{A\cos \theta +B\sin \theta }{R},$ so $x(t)=R\cos (\omega (t-t_0)-\phi )$. If $R=0$ then $A=B=0$ and $x\equiv 0$. This is clause 2.

Clause 4, the energy. With $\theta =\omega (t-t_0)$ the derivative of $x$ is $x^{\mathrm{\prime }}(t)=\omega (-A\sin \theta +B\cos \theta )$ by The derivatives of sine and cosine are cosine and minus sine, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$ and Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$. Using $m{\omega }^2=k$ from [L1], $E=\frac{1}{2}m{x}^{\mathrm{\prime }2}+\frac{1}{2}k{x}^2=\frac{1}{2}k\left[\right(-A\sin \theta +B\cos \theta {)}^2+(A\cos \theta +B\sin \theta {)}^2]=\frac{1}{2}k(A^2+B^2)=\frac{1}{2}k{R}^2,$ expanding the squares and using ${\sin }^2+{\cos }^2=1$ of [L3]; in particular the energy is the constant $\frac{1}{2}k{R}^2$, in agreement with the conservation clause of [L5]. This is the first assertion of clause 4.

Clause 3. If $\omega T\in 2\pi \mathbb{Z}$, then by [L4] the cosine is $2\pi$-periodic, so $\cos (\theta +\omega T)=\cos \theta$ and $x(t+T)=x(t)$ for all $t$; this covers the case $R=0$ as well. Suppose $R>0$ and $T>0$ is a period of $x$. Dividing clause 2 by $R$, $\cos (\omega (t+T-t_0)-\phi )=\cos (\omega (t-t_0)-\phi )$ for all $t$; as $t$ ranges over $\mathbb{R}$, $u:=\omega (t-t_0)-\phi$ ranges over all of $\mathbb{R}$, so $\omega T$ is a period of cosine; by step 1.2, $\omega T\in 2\pi \mathbb{Z}$. Hence the positive periods of a nonconstant harmonic motion are exactly the numbers $2\pi n\mathrm{/}\omega$, $n\ge 1$, and the least one is $2\pi \mathrm{/}\omega$. This is clause 3.

Clause 4, the ellipse. Let $R>0$ and use the phase form of clause 2, $\theta =\omega (t-t_0)$: $x(t)=R\cos (\theta -\phi )$ and $x^{\mathrm{\prime }}(t)=-R\omega \sin (\theta -\phi )$, so $\frac{x(t{)}^2}{R^2}+\frac{x^{\mathrm{\prime }}(t{)}^2}{{\omega }^2{R}^2}={\cos }^2(\theta -\phi )+{\sin }^2(\theta -\phi )=1$ by [L3]: the phase curve is contained in ${\mathrm{\Sigma }}_R$. Conversely let $(x_1,v_1)\in {\mathrm{\Sigma }}_R$, so that $(x_1\mathrm{/}R{)}^2+(v_1\mathrm{/}(\omega R){)}^2=1$; by step 1.1 applied to $(c,s)=(x_1\mathrm{/}R,-v_1\mathrm{/}(\omega R))$ there is ${\theta }_1$ with $\cos {\theta }_1=x_1\mathrm{/}R$ and $\sin {\theta }_1=-v_1\mathrm{/}(\omega R)$. The map $t\mapsto \theta (t)=\omega (t-t_0)$ is a bijection of $\mathbb{R}$ onto $\mathbb{R}$ because $\omega >0$, so the times $t$ with $\theta (t)-\phi ={\theta }_1+2\pi n$ are exactly $t=t_0+({\theta }_1+\phi +2\pi n)\mathrm{/}\omega$, $n\in \mathbb{Z}$; at those times $x(t)=R\cos {\theta }_1=x_1$ and $x^{\mathrm{\prime }}(t)=-R\omega \sin {\theta }_1=v_1$. Hence every point of ${\mathrm{\Sigma }}_R$ lies on the phase curve, at the displayed times. Conversely, attainment of $(x_1,v_1)$ gives both $\cos (\theta -\phi )=\cos {\theta }_1$ and $\sin (\theta -\phi )=\sin {\theta }_1$. The addition identities yield $\cos d=1$ and $\sin d=0$ for $d=\theta -\phi -{\theta }_1$, so $\cos (u+d)=\cos u$ for every real $u$; step 1.2 gives $d\in 2\pi \mathbb{Z}$. Hence these are exactly the attainment times. If $R=0$ then $x\equiv 0$ by clause 2 and the phase curve is $\{(0,0)\}$, which is the level set of energy $0$ because $\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2=0$ forces $x=v=0$. This is clause 4.

Clause 5. The rest points are the equilibria, and by [L6] the equation $U^{\mathrm{\prime }}(x)=kx=0$ has the unique solution $x=0$ because $k>0$; the origin is a strict global minimum of $U$, so [L6] gives Lyapunov stability. No asymptotic stability is asserted, and none follows from the conservation clause of [L5]. This is clause 5.