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} $

  • $\begingroup$ @LutzL I forget it when I write. Thanks for the link. $\endgroup$ – Tof Jun 13 '18 at 8:07

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.


$\langle \cdot,\cdot \rangle$ 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 $\Vert C \Vert > 0$

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


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


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

  • $\begingroup$ I don't see how you get from $(1/2)(d/dt) \Vert X \Vert^2 = \langle \dot X, C \rangle$ to $\Vert \dot X \Vert = 2\langle X, C \rangle + C_0$; I understand the right-hand side just fine; but I don't see how the integral of the left is anything other than $(1/2)(\Vert X \Vert^2 - \Vert X_0 \Vert^2)$. Wassup? Cheers! $\endgroup$ – Robert Lewis Jun 12 '18 at 21:28
  • $\begingroup$ Oh! Thanks! I will fix this typo. $\endgroup$ – Cesareo Jun 12 '18 at 21:34
  • $\begingroup$ OK, Thanks for fixing the typo. Also, I should have typed "anything other than $(1/2)(\Vert \dot X \Vert^2 - \Vert \dot X_0 \Vert)$" in my other comment, but you seem to have straightened your post out despite my mistake. But now I want to know how you get $\langle X - X_0(t), C \rangle = 0$ from the equation right above it. Can you show me? Thanks! $\endgroup$ – Robert Lewis Jun 12 '18 at 21:59
  • $\begingroup$ My pleasure. Please see the note attached at the answer bottom. $\endgroup$ – Cesareo Jun 12 '18 at 22:24
  • 1
    $\begingroup$ What does $X+X_0(t)$ mean? Should that not be $X(t)-X_0$? And if the reference point $X_0$ is not constant in time, how can you claim that the motion follows a plane? Is it not that this type of Lorenz equation has spirals on the surface of a cylinder as solutions? $\endgroup$ – LutzL Jun 13 '18 at 8:04

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 $A$.

  • $\begingroup$ Just to be sure, when you said an orthogonal rotation, you mean to change my axis along the force then I have just one coordinate ? and you called it $ae_1$ $\endgroup$ – Tof Jun 13 '18 at 9:44
  • $\begingroup$ Just added the extra step for general transformations, for $U \in SO(3)$ we have $\det(U)=1$. Set $v_1=A/\|A\|$, $v_2$ orthonormal to $v_1$, $v_3=v_1\times v_2$ then with $V=(v_1,v_2,v_3)$ you get $U=V^{-1}$, and there is 1 parameter free in that construction. $\endgroup$ – LutzL Jun 13 '18 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.