Solving a particular system of differential equations

Question

The problem I'm trying to solve is this:

$X'(t) \in \mathbb{R}^3 \,, \, \omega = (\omega_1,\omega_2,\omega_3) $ Find the general solution for

$$X'(t) = \omega \times X(t)$$

After doing the cross product and rearranging a bit I got to

$$\begin{pmatrix} x_1' (t) \\ x_2' (t) \\ x_3' (t) \end{pmatrix} = \begin{pmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 &-\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{pmatrix} \begin{pmatrix} x_1 (t) \\ x_2 (t) \\ x_3 (t) \end{pmatrix}$$

Then I looked for the eigenvalues of the matrix, which are 0, $\sqrt{-\omega_1 ^2 -\omega_2 ^2 -\omega_3 ^2}$ and $-\sqrt{-\omega_1 ^2 -\omega_2 ^2 -\omega_3 ^2}$.

Once I have the eigenvectors, I know how to proceed to find the general solution, but finding the eigenvectors of the second and third eigenvalues lead me to really weird looking stuff, which makes me think that there's another way of solving this.

I've been thinking about this all day, and I can't think of anything else, any help would be greatly appreciated.

Answer 1

Let $A$ and $X(t)$ be the two matrices $$ A = \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 &-\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix} \quad\text{ and }\quad X(t) = \begin{bmatrix} x_1 (t) \\ x_2 (t) \\ x_3 (t) \end{bmatrix}$$

and $\omega = \sqrt{\omega_1^2 + \omega_2^2 + \omega_3^2} \ge 0$. When $A$ is constant, the ODE for $X(t)$

$$\frac{d}{dt}X(t) = A X(t)$$

has solution of the form

$$X(t) = e^{tA} X(0)\quad\text{ where }\quad e^{tA} = \sum_{k=0}^\infty \frac{t^k}{k!} A^k\tag{*1} $$ When $A$ is a skew symmetric matrix, it satisfy an interesting identity:

$$A^3 = -\omega^2 A$$

A consequence of this is for any even function $f(z)$ whose power series expansion converges rapidly enough, you will find

$$f(A)A = f(i\omega)A\quad\text{ and }\quad f(A)A^2 = f(i\omega)A^2$$

Apply this to $e^{tA}$, one can convert the series into a quadratic polynomial in $A$!

$$\begin{align} e^{tA} &= I_n + \left(\sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}(A^2)^k\right)A + \left(\sum_{k=0}^\infty\frac{t^{2k+2}}{(2k+2)!}(A^2)^k\right)A^2\\ &= I_n + \left(\sum_{k=0}^\infty\frac{t^{2k+1}}{(2k+1)!}(i\omega)^{2k}\right)A + \left(\sum_{k=0}^\infty\frac{t^{2k+2}}{(2k+2)!}(i\omega)^{2k}\right)A^2\\ &= I_n + \frac{\sin(t\omega)}{\omega} A + \frac{1-\cos(t\omega)}{\omega^2} A^2 \end{align}\tag{*2} $$ Please note that during practical calculation, there is no need to manipulate power series explicitly like this. Instead, one can formally manipulate $(*2)$ as follows

$$ e^{tA} = I_n + \frac{\sinh(tA)}{A} A + \frac{\cosh(tA) - I_n}{A^2} A^2 = I_n + \frac{\sin(t\omega)}{\omega} A + \frac{1-\cos(t\omega)}{\omega^2} A^2 $$

Please keep in mind in above manipulation, expression like $\displaystyle\;\frac{\sinh(tA)}{A}\;$ doesn't mean compute $\sinh(tA)$ by a power series expansion and then divide it by $A$. Instead, it means evaluate the power series associated with $\displaystyle\;\frac{\sin(tz)}{z}$ "at" $z = A$.

Finally, let us rephrase the solution $(*1)$ in terms of vectors. Let $\hat{e}_i, i = 1,2,3$ be the canonical basis of $\mathbb{R}^3$ as a vector space. Let $\vec{\omega}$ and $\vec{X}(t)$ be the two vectors

$$ \vec{\omega} = \omega_1 \hat{e}_1 + \omega_2 \hat{e}_2 + \omega_3 \hat{e}_3 \quad\text{ and }\quad \vec{X}(t) = x_1(t) \hat{e}_1 + x_2(t) \hat{e}_2 + x_3(t) \hat{e}_3 $$ Let $\hat{\omega}$ be the unit vector $\displaystyle\;\frac{\vec{\omega}}{\omega}\;$. As mentioned in the question, application of $A$ to the column vector $X(t)$ is equivalent to a cross product between $\vec{\omega}$ and $\vec{X}(t)$.

$$A X(t)\quad\leftrightarrow\quad \vec{\omega} \times \vec{X}(t)$$

If we substitute $(*2)$ into $(*1)$ and employ this type of correspondence between matrices and vectors. We will obtain

$$ \vec{X}(t) = \vec{X}(0) + \sin(t\omega)\big( \hat{\omega} \times \vec{X}(0) \big) + (1 - \cos(t\omega)) \big( \hat{\omega} \times ( \hat{\omega} \times \vec{X}(0)) \big) $$ If we split $\vec{X}(0)$ into two orthogonal vectors, one parallel and another perpendicular to $\hat{\omega}$:

$$\vec{X}(0) = \vec{X}_{\parallel}(0) + \vec{X}_{\perp}(0) \quad\text{ where }\quad \begin{cases} \vec{X}_{\parallel}(0) &= ( \vec{X}_(0) \cdot \hat{\omega} ) \hat{\omega}\\ \vec{X}_{\perp}(0) &= -\hat{\omega} \times ( \hat{\omega} \times \vec{X}(0) ) \end{cases} $$ We can re-express $\vec{X}(t)$ in a much more informative form:

$$\vec{X}(t) = \underbrace{\vec{X}_{\parallel}(0)}_{\vec{X}_{\parallel}(t)} + \underbrace{\cos(t\omega)\vec{X}_{\perp}(0) + \sin(t\omega)\big(\hat{\omega} \times \vec{X}_{\perp}(0) \big)}_{\vec{X}_{\perp}(t)}$$

As one can see, the parallel part of $\vec{X}(t)$ remains a constant while the perpendicular part of $\vec{X}(t)$ rotate around the axis in direction of $\hat{\omega}$ in circle.

Answer 2

Recall that $\omega \times X(t)$ is orthogonal to both $\omega$ and $X(t)$. Therefore, $X'(t)$ is orthogonal to $X(t)$ and to $\omega$. In particular, $$ \frac{d}{dt}(X(t)\cdot X(t)) = 2X'(t)\cdot X(t) = 0. $$ That means that $|X(t)|$ is constant in $t$. Similiary $\frac{d}{dt}(X(t)\cdot\omega)=0$, which keeps $X(t)\cdot\omega$ constant in time. So the motion of $X(t)$ remains in the plane $$ X(t)\cdot\omega = X(0)\cdot\omega, $$ and the motion is in a circle because $|X(t)|=|X(0)|$ for all $t$. To be concrete about the representation, project $X(0)$ onto the line through the origin with direction vector $\omega$. That projection is $$ P=X(0)-\frac{X(0)\cdot \omega}{\omega\cdot\omega}\omega. $$ The derivative $X'(t)=\omega\times X(t)$ means that $X(t)$ rotates in a circular direction in the plane perpendicular to $\omega$; the direction is in the direction of your fingers (according to the right-hand rule) when your thumb points along $\omega$. That's why it's best to choose the second orthogonal vector in the plane with normal $\omega$ to be $$ Q = \frac{1}{|\omega|}\omega \times P. $$ Then $|Q|=|P|$ and $$ X(t)= \frac{X(0)\cdot\omega}{\omega\cdot\omega}\omega+\cos(\alpha t)P+\sin(\alpha t)Q $$ for some constant $\alpha$. The constant $\alpha$ is positive and its determined by the length of the derivative vector: $$ |X'(0)|=\alpha|Q| \mbox{ and } |X'(0)| = |\omega\times X(0)|=|\omega\times P|= |\omega||Q|. $$ Now check the above solution where $\alpha=|\omega|$: $$ \begin{align} X'(t) & = |\omega|(-\sin(|\omega|t)P+\cos(|\omega|t)Q),\\ \omega \times X(t) & = \omega\times\left[ \frac{X(0)\cdot\omega}{\omega\cdot\omega}\omega+\cos(|\omega| t)P+\sin(|\omega| t)Q \right] \\ & = |\omega|\left(\frac{1}{|\omega|}\omega\right)\times\left\{\cos(|\omega|t)P+\sin(|\omega|t)Q\right\} \\ & = |\omega|\left\{\cos(|\omega|t)Q-\sin(|\omega|t)P\right\} \end{align} $$ So the equation is satisfied. Plus, by design, $X(0)$ is correct.