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

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 ?

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

• @LutzL I forget it when I write. Thanks for the link. – 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.

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

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

• 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! – Robert Lewis Jun 12 '18 at 21:28
• Oh! Thanks! I will fix this typo. – Cesareo Jun 12 '18 at 21:34
• 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! – Robert Lewis Jun 12 '18 at 21:59
• My pleasure. Please see the note attached at the answer bottom. – Cesareo Jun 12 '18 at 22:24
• 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? – 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 $A$.

• 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 $ae_1$ – Tof Jun 13 '18 at 9:44
• 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. – LutzL Jun 13 '18 at 9:51