Lecture to solve 2nd order differential equation in matrix form.

Question

I have an system of three differential equation coupled. I put this system to matrix form. I think it should be more easy to solve. $$ \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \\ \ddot x_3 \\ \end{bmatrix} = \begin{bmatrix} 0 & a_{12} & -a_{31} \\ -a_{12} & 0 & a_{23} \\ a_{31} & -a_{23} & 0 \\ \end{bmatrix} \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \dot x_3 \\ \end{bmatrix}+\begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \end{bmatrix} $$ All constants are positive or zero.

Could someone tell me what I should read to get the knowledge to solve this differential equation ?

Thanks for your help. Tof

Edit: $a_{23} -> -a_{23} $

Answer 1

Note that the $A$ matrix is a skew symmetric matrix which represents a vectorial product operator $\vec A \times$ so the DE could read

$$ \ddot X = \vec A \times \dot X + C $$

So we have

$$ \langle \dot X, \ddot X \rangle = \langle \dot X, C \rangle $$

because $\langle \dot X, \vec A\times \dot X \rangle = 0$ or

$$ \frac{1}{2}\frac{d}{dt}\Vert\dot X\Vert^2 = \langle \dot X,C \rangle $$

and after integration

$$ \Vert\dot X\Vert^2 = 2 \langle X, C \rangle + C_0 $$

which can be arranged as

$$ \big\langle X - X_0(t),C \big\rangle = 0 $$

so the movement is such that $X-X_0(t)$ is normal to $C$ etc.

NOTE

$\langle \cdot,\cdot \rangle$ represents the scalar product between two vectors. Here $X_0(t)$ represents a vector multiplied by the scalar $||\dot X||^2$ plus a constant vector. Now assuming $\Vert C \Vert > 0$

$$ \Vert\dot X\Vert^2 = \left\langle \Vert\dot X \Vert^2\frac{C}{\Vert C \Vert^2},C \right\rangle $$

and

$$ C_0 = \langle x_0, C \rangle $$

hence

$$ X_0 = \Vert\dot X\Vert^2\frac{C}{\Vert C\Vert^2} + x_0 $$

Answer 2

In $$ \ddot X=A\times \dot X+C $$ you can apply an orthogonal rotation $$ U\ddot X=\det(U)^{-1}(UA)\times (U\dot X)+UC. $$ The rotation $U$ can be chosen so that $UA\sim e_1$. Thus one can assume w.l.o.g. that $A=ae_1$. Then the components of the system read as (set at first $V=\dot X$) $$ \pmatrix{\dot v_1\\\dot v_2\\\dot v_3} =a\pmatrix{0\\-v_3\\v_2}+\pmatrix{c_1\\c_2\\c_3} $$ Integration by standard methods gives \begin{align} v_1&=c_1t+b_1\\ v_2&=ab_2\cos(at)-ab_3\sin(at)-\frac{c_3}a&\implies av_3=c_2-\dot v_2&=c_2+a^2b_2\sin(at)+a^2b_3\cos(at)\\ v_3&=ab_2\sin(at)+ab_3\cos(at)+\frac{c_2}a \end{align}

Now integrate once again to get $X$, \begin{align} x_1&=\frac12c_1t^2+b_1t+d_1\\ x_2&=b_2\sin(at)+b_3\cos(at)-\frac{c_3}at+d_2\\ x_3&=-b_2\cos(at)+b_3\sin(at)+\frac{c_2}at+d_3\\ \end{align} We get as the result a superposition of an accelerated motion in direction $A=ae_1$ and of a circular and a linear motion in the plane perpendicular to $A$.