Prove differential equation solution is even function

Question

Let $0\leq c<1$,$u,v,w \in C^1(\mathbb{R})$. Suppose $u'=vw ,\; v'=-uw, \; w'=-c^2 uv$ and $u(0)=0, v(0)=1, w(0)=1$ Then, for all $x\in \mathbb{R} $,$v(x)=v(-x)$ and there is $x_0$ with $v(x_0)=0$

I know that $u^2+v^2=1,\; c^2 u^2+w^2=1$ and I think I use this property in order to prove $v$ is even and has zero point.

Answer 1

The structure $$\pmatrix{u'\\v'}=w\pmatrix{0&1\\-1&0}\pmatrix{u\\v}$$ of your equations provides the parametrization $u(t)=\sin(\alpha(t))$, $v(t)=\cos(α(t))$ for the first two components with angular velocity $α'(t)=w(t)$, $α(0)=0$. Inserting into the third equation gives $$α''(t)=w'(t)=-c^2\sin(α(t))\cos(α(t))=-\frac12c^2\sin(2α(t)).$$ This is an equation for a physical pendulum at angle $2\alpha$ with first integral $$ H(α(t),w(t))=w(t)^2+c^2\sin^2(α(t))=w^2+c^2u^2=1. $$ The amplitude of the oscillation is attained at $α'=w=0$ where $|\sin(α)|=c^{-1}>1$. As that is impossible, the pendulum is in constant rotation, the angle $α$ is continuously increasing and thus also periodically passes the values $\frac\pi2+k\pi$ where $v=\cos(α)$ is zero.

Answer 2

First equality: Multiply the first equation by $u$ and the second by $v$. Adding them together yields $uu' + v v' = 0$, which is equivalent to $\frac{1}{2} \frac{d}{dx}(u^2 + v^2) = 0$. Hence $u^2 + v^2 = C$. Plugging in the initial condition implies that $u^2 + v^2 = 2$.

Second equality: Hint: Multiply first equation by $c^2 u$ and third equation by $w$.

Answer 3

If $x\mapsto \bigl(u_0(x),v_0(x),w_0(x)\bigr)$ is the solution of the original IVP then the triple $x\mapsto \bigl(u_1(x),v_1(x),w_1(x)\bigr)$ defined by $$u_1(x):=-u_0(-x),\quad v_1(x):=v_0(-x),\quad w_1(x):=w_0(-x)$$ is also a solution of the same IVP (check this!). The general existence and uniqueness theorem for systems of ODEs then implies that in fact $$\bigl(u_0(x),v_0(x),w_0(x)\bigr)\equiv\bigl(u_1(x),v_1(x),w_1(x)\bigr)\ ;$$ in particular $x\mapsto v_0(x)$ is an even function.