Dynamic system show that it has only one equilibrium point

Question

I have the following dynamic system $A\cdot\ddot{x}+B\cdot\dot{x}+ C\cdot x=0 $ and A,B,C are positive definite matrices. I have to show that it has only one equilibrium point and that this point is asymptotically stable. Any help will be appreciated!

UPDATE Following the comment about treating A,B,C as scalars $\dot{x_1}=x_2 = f_1(x_2)\\ \dot{x_2}=(-B\cdot x_2-C\cdot x_1)/A = f_2(x_1,x_2)$
$\longrightarrow \dot{x}=A(x)|_{x=0} \cdot x \\ $

$ A(x)=\begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \\ \end{bmatrix}=\begin{bmatrix} 0 & 1\\ -C/A&-B/A\\ \end{bmatrix}$ $det(A-\lambda\cdot I)= 0 \longrightarrow det(\begin{bmatrix} -\lambda & 1\\ -C/A& -\lambda -B/A\\ \end{bmatrix}) =0 \longrightarrow \lambda^2+\lambda B/A+C/A=0$\

But I don't seem to get any idea on how to generalize it.

Answer 1

At an equilibrium point you get $\dot x=0$ and $\ddot x=0$, so the remaining condition is $Cx=0$, now apply that $C$ is s.p.d.

For the normalization use $\tilde B=A^{-1/2}BA^{-1/2}$ and $\tilde C=A^{-1/2}CA^{-1/2}$ so that with $y=A^{1/2}x$ you get a system $$ \ddot y+\tilde B\dot y+\tilde Cy=0 $$ with still s.p.d. coefficient matrices. Then try to conclude that any exponential basis solution $ve^{\lambda t}$ needs to have $\lambda$ real and negative, or possibly complex with negative real part.

Answer 2

I would start by calling $x=x_1$ and $\dot{x}=x_2$. Then you can rewrite the ODEs as: \begin{equation} \begin{matrix} \dot{x_1}&=&&x_2\\ A\dot{x_2}&=&-Cx_1&-Bx_2 \end{matrix} \end{equation}

or even: \begin{equation} \dot{\left[ \begin{matrix}x_1\\ x_2 \end{matrix} \right]} = \left[ \begin{matrix} 0&I\\ -A^{-1}C&-A^{-1}B \end{matrix}\right] \left[\begin{matrix} x_1\\ x_2 \end{matrix}\right] \end{equation} Then, the equlibria will be when $\dot{\left[\begin{matrix}x_1\\x_2\end{matrix}\right]}=\mathbf{0}$. That means, $0=x_2=\dot{x}$ and $0=-A^{-1}Cx_1=-A^{-1}Cx$ then you can apply that $A$ and $C$ are positive definite to find a value for $x$.

Now, to prove its stability we need to inspect the eigenvalues of $\left[\begin{matrix} 0&I\\ -A^{-1}C&-A^{-1}B \end{matrix}\right]$.

$\lambda=\mathrm{zeros}(\left|\begin{matrix} sI&-I\\ A^{-1}C&sI+A^{-1}B \end{matrix}\right|)=\mathrm{zeros}(s^2I+A^{-1}Bs+A^{-1}C)$, using the fact that $A$, $B$ and $C$ are positive definite and using from Routh-Hurwitz criterion:

The second-degree polynomial, $P(s)=s^{2}+a_{1}s+a_{0}$ has both roots in the open left half plane (and the system with characteristic equation if and only if both coefficients satisfy $a_{i}>0$.

we can finally tell if its stable or not.

Answer 3

If $x_1$ and $x_2$ are equilibrium points then

$$ \cases{ C\cdot x_1 = 0\\ C\cdot x_2 = 0 }\Rightarrow C\cdot(x_1-x_2)=0 $$

but due to the nature of $C$ follows $x_1-x_2=0\Rightarrow x_1 = x_2$