How do I solve?: $$\ddot{\vec{u}} = \vec{u} \times \hat{k}$$
I have tried solving a simpler version of this, $\dot{\vec{u}} = \vec{u} \times \hat{k}$. This one was easy: the head of the vector rotates around the $z$-axis with constant angular speed.
However, I have no idea on the second-order one. It doesn't seem to have any relation to the simpler one.. or does it?..
On
You can spell out the cross-product to obtain \begin{align} \ddot{ \vec u} = \begin{pmatrix} k_2 u_3 - k_3 u_2 \\ k_3u_1 -k_1u_3\\ k_1u_2 - k_2 u_1\end{pmatrix} \end{align} Then, you can do the usual trick to transform this into a larger system of first order ODEs by introducing $\vec z := \vec u, $ (zero'th derivative) and $\vec f := \dot{\vec u }$, (first derivative) which yields the linear system (linear in $\vec z, \vec f$): $$ \dot{\begin{pmatrix} z_1 \\ z_2 \\ z_3 \\ f_1 \\ f_2 \\f_3 \end{pmatrix}} = \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & -k_3 & k_2 & 0 & 0 & 0\\ k_3 & 0 & -k_1 & 0 & 0 & 0\\ -k_2 & k_1 & 0 & 0 & 0 & 0\end{pmatrix} \begin{pmatrix} z_1 \\ z_2 \\ z_3 \\ f_1 \\ f_2 \\f_3 \end{pmatrix}$$ You can then solve this system of ODEs through diagonalization - however, it will be a pain, so maybe it is best to consult an algebraic toolbox for this. This will give (lenghty) expressions for eigenvalues $\lambda_j$ and the transformation matrix $P$ in terms of $k_i$, but it can be done.
On
You should perhaps more explicitly mention that $\bf k$ is the third canonical basis vector.
Then it is immediately clear that the right side is zero in $z$ direction and acts as a $-90°$ rotation in the $xy$-plane, $$ (x{\bf i}+y{\bf j})\times{\bf k}=-x{\bf j}+y{\bf i}. $$ This rotation has 2 square roots, the rotations by $-45°$ and by $135°$. Interpreting the $xy$-plane as complex plane, one gets $$ \ddot x+i\ddot y =-i(x+iy) =\frac12(1-i)^2(x+iy) $$ so that $$ x+iy=c_1e^{\frac{1-i}{\sqrt2}t}+c_2e^{\frac{-1+i}{\sqrt2}t},~~c_{1,2}\in\Bbb C $$
So writing $u = (u_i)$, we have $$ \ddot u_1 = u_2, \quad \ddot u_2 = -u_1, \quad \ddot u_3 = 0 $$ Hence, $u_3(t) = at + b$ for some constants $a,b$, the other two equations give us $$ \ddddot u_1 = -u_1 $$ This has the characteristic equation $\lambda^4 + 1 = 0$, giving us $$ \lambda_{1,2} = \frac{1 \pm i}{\sqrt 2}, \lambda_{3,4} = \frac{-1\pm i}{\sqrt 2} $$ Hence, we have $$ u_1(t) = \exp\left(\frac t{\sqrt 2}\right)\bigl(c\sin(t/\sqrt 2) + d\cos(t/\sqrt 2)\bigr) + \exp\left(-\frac t{\sqrt 2}\right)\bigl(e\sin(t/\sqrt 2) + f\cos(t/\sqrt 2)\bigr) $$ for some constants $c,d,e, f$. By using $u_2 = \ddot u_1$, you can calculate the missing component.