MHB Complex Variables - Solution of a System

[joypav]

Suppose the polynomial p has all its zeros in the closed half-plane $Re w\le0$, and any zeros that lie on the imaginary axis are of order one.
$$p(z)=det(zI-A),$$
where I is the n x n identity matrix.

Show that any solution of the system

$$\dot{x}=Ax+b$$

remains bounded as $t\to{\infty}$.

Work out in detail the solution of the system with
$$
\begin{bmatrix}
0 & w \\
-w & 0 \\
\end{bmatrix}, w>0
$$
to verify what happens in one special case.

So we have...
$$p(z)=det(zI-A)$$
$$p(z)=det( \begin{bmatrix}
z & 0 \\
0 & z \\
\end{bmatrix}-
\begin{bmatrix}
0 & w \\
-w & 0 \\
\end{bmatrix})=
det(\begin{bmatrix}
z & -w \\
w & z \\
\end{bmatrix})=z^2+w^2=(z+wi)(z-wi)
$$
Both zeros are on the imaginary axis and are both of order one.
Now I am supposed to find the solution of the system?

Do we use that the solution of the system is of the form
$$x_l(t)=\sum_{j=1}^2p_{jl}(t)e^{\lambda_jt}, l=1,2?$$
So
$$\lambda_1=wi,\lambda_2=-wi$$

I have no idea if I'm going in the right direction or not.

I would really like some help with the specific problem, not necessarily the more generalized proof. I think if I knew how to solve the specific problem given then I'd have a much better understanding.

 

[I like Serena]

Now I am supposed to find the solution of the system?

Do we use that the solution of the system is of the form
$$x_l(t)=\sum_{j=1}^2p_{jl}(t)e^{\lambda_jt}, l=1,2?$$
So
$$\lambda_1=wi,\lambda_2=-wi$$

I have no idea if I'm going in the right direction or not.

I would really like some help with the specific problem, not necessarily the more generalized proof. I think if I knew how to solve the specific problem given then I'd have a much better understanding.

Hi joypav!

The solution should look slightly diffferent.

To solve it, we need to diagonalize $A$ into $BDB^{-1}$, where $D$ is a diagonal matrix with the eigenvalues on its diagonal, and where $B$ is a matrix formed of the corresponding eigenvectors.
So:
$$\dot x = Ax + b = BDB^{-1}x+b$$
Let $\tilde x=B^{-1}x$ and $\tilde b = B^{-1}b$, then:
$$\dot{\tilde x} = B^{-1}\dot x = B^{-1}(BDB^{-1}x+b) = D\tilde x + \tilde b$$
The solutions are $\tilde x_i = C_i e^{\lambda_i t} - \frac{\tilde b_i}{\lambda_i}$.
Thus:
$$x=B\tilde x = B\begin{pmatrix}C_1 e^{\lambda_1 t} - \frac{\tilde b_1}{\lambda_1} \\ \dots \\ C_n e^{\lambda_n t} - \frac{\tilde b_n}{\lambda_n}\end{pmatrix}$$

Is it bounded when $t\to \infty$? (Wondering)

 

[joypav]

Yes, because for each n,
$$\lambda_n\le0$$
So the limit as t goes to infinity will go to 0 for each n?

 

[I like Serena]

Yes, because for each n,
$$\lambda_n\le0$$
So the limit as t goes to infinity will go to 0 for each n?

It will go to zero if f $Re\ \lambda_n<0$ for all n.
If $Re\ \lambda_n=0$ for some n, we won't reach 0, but we will be bounded.