Show that $\alpha=||x(t)||_2$ where $t> 0$.

Question

Consider the initial value problem:

$x^{'}(t)=Ax(t),x(0)=x_0$ where $t\ge 0$

Suppose that $A$ is a skew symmetric matrix and $\alpha=||x_0||_2$.

Show that $\alpha=||x(t)||_2$ where $t> 0$.

My try:

I got easily the fact that $x(t)=e^{Ax} +x_0$ where $\alpha=||x_0||_2$.

But I am not getting how to show that $\alpha=||x(t)||_2$.

Can someone give some hint on how to apply the fact that $A^T=-A$ and how to get this?

Answer 1

We have, using

$\dot x(t) = Ax(t), \tag 0$

$\dfrac{d}{dt} \Vert x(t) \Vert ^2 = \dfrac{d}{dt}\langle x(t), x(t) \rangle = \langle \dot x(t), x(t) \rangle + \langle x(t), \dot x(t) \rangle$ $= \langle x(t), \dot x(t) \rangle + \langle x(t), \dot x(t) \rangle = 2 \langle x(t), \dot x(t) \rangle = 2\langle x(t), A x(t) \rangle; \tag 1$

Since

$A^T = - A, \tag 2$

$\langle x(t), A x(t) \rangle = \langle A^T x(t), x(t) \rangle = \langle -Ax(t), x(t) \rangle = -\langle x(t), A x(t) \rangle; \tag 3$

thus,

$\langle x(t), A x(t) \rangle = 0, \tag 4$

and we conclude that

$\dfrac{d}{dt} \Vert x(t) \Vert^2 = \dfrac{d}{dt} \langle x(t), x(t) \rangle = 0, \; \forall t \in \Bbb R; \tag 5$

therefore $\Vert x(t) \Vert^2$ is constant; hence

$\Vert x(t) \Vert^2 = \Vert x(0) \Vert^2 \Longrightarrow \Vert x(t) \Vert = \Vert x(0) \Vert , \forall t \in \Bbb R. \tag 6$

Answer 2

Note that $$ 2 x(t)^T x'(t) = \frac{d}{dt} \left( || x(t) ||_{2}^{2} \right) $$ Use your governing equation to study the norm.