General Solution of Vector ODE in Given Form

Question

The ODE in question, given below, is straightforward to solve: $$\mathbf{\ddot r} = -\omega^2 \mathbf r$$ for $\mathbf r \in \Bbb R^2$.

However, I seek to show that the general solution may be written as, $$\mathbf r = \mathbf a \sin (\omega t + \alpha) + \mathbf b \cos (\omega t + \alpha)$$ where $\mathbf a$ and $\mathbf b$ are constant and are orthogonal to each other (and $\alpha$ is a real constant).

Answer 1

The most general form of a solution to the ODE would be :

$\vec{\mu}=\begin{pmatrix} x_1\sin({\omega t + \alpha_1})+x_2\cos({\omega t + \alpha_2}) \\ x_3\sin({\omega t + \alpha_3})+x_4\cos({\omega t + \alpha_4}) \end{pmatrix} $

Using :
$ \sin(\alpha \pm \beta) = \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta)$
and :
$ \cos(\alpha \pm \beta) = \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta)$

We manipulate this into a form with two 'oscillating' vectors :

$\vec{\mu}=\begin{pmatrix} x_1\sin({\omega t +\alpha_5+ \gamma_1})+x_2\cos({\omega t +\alpha_5+ \gamma_2}) \\ x_3\sin({\omega t +\alpha_5+ \gamma_3})+x_4\cos({\omega t +\alpha_5+ \gamma_4}) \end{pmatrix} = \begin{pmatrix} x_1\sin(\omega t +\alpha_5)\cos(\gamma_1) + x_1\cos(\omega t +\alpha_5)\sin(\gamma_1)+x_2\cos(\omega t +\alpha_5)\cos(\gamma_2) - x_2\sin(\omega t +\alpha_5)\sin(\gamma_2) \\ x_3\sin(\omega t +\alpha_5)\cos(\gamma_3) + x_3\cos(\omega t +\alpha_5)\sin(\gamma_3)+x_4\cos(\omega t +\alpha_5)\cos(\gamma_4) - x_4\sin(\omega t +\alpha_5)\sin(\gamma_4) \end{pmatrix} $

$ \implies $
$\vec{\mu}=\begin{pmatrix} x'_1\sin({\omega t + \alpha_5})+x'_2\cos({\omega t + \alpha_5}) \\ x'_3\sin({\omega t + \alpha_5})+x'_4\cos({\omega t + \alpha_5}) \end{pmatrix} =\begin{pmatrix} x'_1\\ x'_3 \end{pmatrix}\sin({\omega t + \alpha_5})+ \begin{pmatrix} x'_2 \\ x'_4 \end{pmatrix}\cos({\omega t + \alpha_5}) $

With :
$x'_1:=x_1\cos(\gamma_1)-x_2\sin(\gamma_2) =x_1\cos(\alpha_1-\alpha_5)-x_2\sin(\alpha_2-\alpha_5)$
$x'_2:=x_1\sin(\gamma_1)+x_2\cos(\gamma_2)=x_1\sin(\alpha_1-\alpha_5)+x_2\cos(\alpha_2-\alpha_5)$
$x'_3:=x_3\cos(\gamma_3)-x_4\sin(\gamma_4)=x_3\cos(\alpha_3-\alpha_5)-x_4\sin(\alpha_4-\alpha_5)$
$x'_4:=x_3\sin(\gamma_3)+x_4\cos(\gamma_4)=x_3\sin(\alpha_3-\alpha_5)+x_4\cos(\alpha_4-\alpha_5)$

The vectors $\begin{pmatrix} x'_1\\ x'_3 \end{pmatrix}$ and $\begin{pmatrix} x'_2\\ x'_4 \end{pmatrix}$ are generally not yet orthogonal.
Now we show they can be made orthogonal with the right choice of $\alpha_5$ :

What we need for the vectors to be orthogonal is : $x'_1x'_2 +x'_3x'_4=0$
Solving this equation for $\alpha_5 $ gives the desired result.

It is clear that the equation can have a solution. I guess that is enough to realize it always must have a solution. Rotating the vector or changing its size can not make the above unsolvable.

For the trigonometric identities used see here. . For the main identity used below see 'Arbitrary phase shift' : $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$ with $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$ and $\phi=atan2(b+\sin(\theta),a+b\cos(\theta))$.

$x'_1x'_2 +x'_3x'_4=0 \implies (x_1\cos(\alpha_1-\alpha_5)-x_2\sin(\alpha_2-\alpha_5))(x_1\sin(\alpha_1-\alpha_5)+x_2\cos(\alpha_2-\alpha_5))+(x_3\cos(\alpha_3-\alpha_5)-x_4\sin(\alpha_4-\alpha_5))(x_3\sin(\alpha_3-\alpha_5)+x_4\cos(\alpha_4-\alpha_5))=0$

$(x_2\sin(\alpha_2-\alpha_5)-x_1\sin(\frac{\pi}{2}+\alpha_1-\alpha_5))(x_1\sin(\alpha_1-\alpha_5)+x_2\sin(\frac{\pi}{2}+\alpha_2-\alpha_5)) +(x_4\sin(\alpha_4-\alpha_5)-x_3\sin(\frac{\pi}{2}+\alpha_3-\alpha_5))(x_3\sin(\alpha_3-\alpha_5)+x_4\sin(\frac{\pi}{2}+\alpha_4-\alpha_5))=0$

$(x_2\sin(\alpha_2-\alpha_5)-x_1\sin(\frac{\pi}{2}+\alpha_1-\alpha_2+\alpha_2-\alpha_5))(x_1\sin(\alpha_1-\alpha_5)+x_2\sin(\frac{\pi}{2}+\alpha_2-\alpha_1+\alpha_1-\alpha_5)) +(x_4\sin(\alpha_4-\alpha_5)-x_3\sin(\frac{\pi}{2}+\alpha_3-\alpha_4+\alpha_4-\alpha_5))(x_3\sin(\alpha_3-\alpha_5)+x_4\sin(\frac{\pi}{2}+\alpha_4 -\alpha_3+\alpha_3-\alpha_5))=0$

$(x_2\sin(\alpha_2-\alpha_5)+x_1\sin(\frac{3\pi}{2}+\alpha_1-\alpha_2+\alpha_2-\alpha_5))(x_1\sin(\alpha_1-\alpha_5)+x_2\sin(\frac{\pi}{2}+\alpha_2-\alpha_1+\alpha_1-\alpha_5)) +(x_4\sin(\alpha_4-\alpha_5)+x_3\sin(\frac{3\pi}{2}+\alpha_3-\alpha_4+\alpha_4-\alpha_5))(x_3\sin(\alpha_3-\alpha_5)+x_4\sin(\frac{\pi}{2}+\alpha_4 -\alpha_3+\alpha_3-\alpha_5))=0$

We define some new constants :

$r_1:=\sqrt{{x_2}^2+{x_1}^2+2x_1x_2\cos(\frac{3\pi}{2}+\alpha_1-\alpha_2)}$
$\phi_1:=atan2(x_1\sin(\frac{3\pi}{2}+\alpha_1-\alpha_2),x_2+x_1\cos(\frac{3\pi}{2}+\alpha_1-\alpha_2))$
$r_2:=\sqrt{{x_2}^2+{x_1}^2+2x_1x_2\cos(\frac{\pi}{2}+\alpha_2-\alpha_1)}$
$\phi_2:=atan2(x_2\sin(\frac{\pi}{2}+\alpha_2-\alpha_1),x_1+x_2\cos(\frac{\pi}{2}+\alpha_2-\alpha_1))$
$r_3:=\sqrt{{x_4}^2+{x_3}^2+2x_3x_4\cos(\frac{3\pi}{2}+\alpha_3-\alpha_4)}$
$\phi_3:=atan2(x_3\sin(\frac{3\pi}{2}+\alpha_3-\alpha_4),x_4+x_3\cos(\frac{3\pi}{2}+\alpha_3-\alpha_4))$
$r_4:=\sqrt{{x_4}^2+{x_3}^2+2x_3x_4\cos(\frac{\pi}{2}+\alpha_4-\alpha_3)}$
$\phi_4:=atan2(x_4\sin(\frac{\pi}{2}+\alpha_4-\alpha_3),x_3+x_4\cos(\frac{\pi}{2}+\alpha_4-\alpha_3))$

$\implies r_1r_2\sin(\alpha_2-\alpha_5+\phi_1)\sin(\alpha_1-\alpha_5+\phi_2) +r_3r_4\sin(\alpha_4-\alpha_5+\phi_3)\sin(\alpha_3-\alpha_5+\phi_4)=0$

$\frac{r_1r_2}{2}\cos(\alpha_2 +\phi_1 + \alpha_1+\phi_2-2\alpha_5) +\frac{r_3r_4}{2}\cos(\alpha_4 +\phi_3 + \alpha_3+\phi_4-2\alpha_5)=\frac{r_1r_2}{2}\cos(\alpha_2 +\phi_1 - \alpha_1 -\phi_2) +\frac{r_3r_4}{2}\cos(\alpha_4 +\phi_3 - \alpha_3 -\phi_4)$

$\frac{r_1r_2}{2}\sin(-\frac{\pi}{2}+\alpha_2 +\phi_1 + \alpha_1+\phi_2-2\alpha_5) +\frac{r_3r_4}{2}\sin(-\frac{\pi}{2}+\alpha_2 +\phi_1 + \alpha_1+\phi_2-2\alpha_5 + ( -\alpha_2 -\phi_1 - \alpha_1-\phi_2 +\alpha_4 +\phi_3 + \alpha_3+\phi_4 ) )=\frac{r_1r_2}{2}\cos(\alpha_2 +\phi_1 - \alpha_1 -\phi_2) +\frac{r_3r_4}{2}\cos(\alpha_4 +\phi_3 - \alpha_3 -\phi_4)$

Two more new constants :

$r_5:=\sqrt{{\frac{r_1r_2}{2}}^2+{\frac{r_3r_4}{2}}^2+2\frac{r_3r_4}{2}\frac{r_1r_2}{2}\cos( -\alpha_2 -\phi_1 - \alpha_1-\phi_2 +\alpha_4 +\phi_3 + \alpha_3+\phi_4 )}$
$\phi_5:=atan2(\frac{r_3r_4}{2}\sin(-\alpha_2 -\phi_1 - \alpha_1-\phi_2 +\alpha_4 +\phi_3 + \alpha_3+\phi_4 ),\frac{r_1r_2}{2}+\frac{r_3r_4}{2}\cos(-\alpha_2 -\phi_1 - \alpha_1-\phi_2 +\alpha_4 +\phi_3 + \alpha_3+\phi_4 ))$

Finally the end result :

$ \alpha_5 = -\frac{1}{2}\arcsin \bigg( \frac{r_1r_2}{2r_5}\cos(\alpha_2 +\phi_1 - \alpha_1 -\phi_2) +\frac{r_3r_4}{2r_5}\cos(\alpha_4 +\phi_3 - \alpha_3 -\phi_4) \bigg) +\frac{1}{2}(-\frac{\pi}{2}+\alpha_2 +\phi_1 + \alpha_1+\phi_2 +\phi_5 ) $

Answer 2

We know that the solution can be given as $$ \mathbf r = \mathbf A \sin (\omega t ) + \mathbf B \cos (\omega t) $$ where $\mathbf A$, $\mathbf B$ have no special relationship. Introducing the phase angle $α$ and using trigonometric identities for $\sin((ωt+α)-α)$ etc. one gets $$ \mathbf r=\mathbf a\sin(ωt+α)+\mathbf b\cos(ωt+α) $$ with \begin{align} \mathbf a &= \mathbf B \sin(α) + \mathbf A \cos(α) \\ \mathbf b &= \mathbf B \cos(α) - \mathbf A \sin(α) \end{align} and their scalar product is \begin{align} \langle \mathbf a,\, \mathbf b\rangle &= (\|\mathbf B\|^2-\|\mathbf A\|^2)\sin(α)\cos(α)+⟨\mathbf A,\mathbf B⟩(\cos^2(α)-\sin^2(α)) \\ &= \frac12(\|\mathbf B\|^2-\|\mathbf A\|^2)\sin(2α)+⟨\mathbf A,\mathbf B⟩\cos(2α) \end{align}

Thus one can find the angle $α$ with the desired properties as half the angle of the point $$ (\cos(2α),\sin(2α))=\Bigl(\frac12(\|\mathbf A\|^2-\|\mathbf B\|^2),\,⟨\mathbf A,\mathbf B⟩\Bigr) $$ (relative to origin and positive horizontal axis in a Cartesian plane).