# Unique solutions to a nonlinear system of ODEs

Given $A$ is a $n\times n$ matrix, and $f: R\rightarrow R^n$ is continuous and bounded.

(a) If $A$ has no eigenvalues on the imaginary axis, prove that $\dot{x} = Ax + f(t)$ has a unique solution which is bounded on $R$.

(b) Show the counterexample to (a) when $A$ has eigenvalues on the imaginary axis.

My attempt: (a) Since $A$ has no eigenvalues on the imaginary axis, $A$ is infinitesimally hyperbolic. By a well-know theorem, we can rewrite a Jordan canonical form $J$ of $A$ such that $J = [A_s\ 0, 0\ A_u]^T$ where eigenvalues of matrix $A_s$ all have negative real parts and eigenvalues of $A_u$ all have positive real parts. Then the given ODE is equivalent to: $\dot{x} = PJP^{-1}x + f(t)$ where $P$ is matrix whose columns are eigenvectors associated with each eigenvalues of $A_s, A_u$ in the same order.

My question is: how to show that $PJP^{-1}$ transforms $A$ into 2 block matrices: one with all eigenvalues having positive real part, and one with all eigenvalues having negative real parts?

(b) I haven't been able to get this, despite spending several hours trying to find $A$ so that the solutions would have resonance effect. Can someone please give some thought about this?

• (b) what if $A = \left[ \begin {array}{cc} 0&0\\ 1&0\end {array} \right]$ and $f(t) = [ 1 \ \ \ 1]^T$ ? – Smilia Oct 17 '15 at 8:59
• @Guignol: Such a slick example! Although I'm not sure if $\lambda = 0$ counts as "on the imaginary axis," but I guess the origin still counts:) Many thanks! Can you try part (a) as well? – user177196 Oct 17 '15 at 18:10
• nobody wants to help me with part (a), after transforming $A$ into the Jordan form? I just need to prove if there exists a relationship between the solution of $\dot{x} = Ax + f(t)$ and $\dot{x} = A_sx + f(t)$ or $\dot{x} = A_ux + f(t)$. – user177196 Oct 18 '15 at 0:51

The assertion in (a) is wrong. You need to have eigenvalues with negative real part for a bounded solution for all possible bounded $f(t)$.

As for the Jordan form you can write $$A = \begin{bmatrix}P_s & P_u\end{bmatrix} \begin{bmatrix}A_s & 0 \\ 0 & A_u \end{bmatrix} \begin{bmatrix}P_s & P_u\end{bmatrix}^{-1}$$

where columns of $P_s$ and $P_u$ contains the (generalized) eigenvectors of the stable and unstable eigenvalues respectively. You can see this fact from the spectral decomposition of a matrix.

Now define $P = \begin{bmatrix}P_s & P_u\end{bmatrix}$ and $y = P^{-1} x$.

$$\dot{y} = P^{-1} \dot{x} = P^{-1} A x + P^{-1} f(t) = P^{-1} A P y + P^{-1} f(t)$$

$$\begin{bmatrix}\dot{y}_s \\ \dot{y}_u\end{bmatrix} = \begin{bmatrix}A_s & 0 \\ 0 & A_u \end{bmatrix} \begin{bmatrix}y_s \\ y_u\end{bmatrix} + P^{-1} \begin{bmatrix}f_s(t) \\ f_u(t) \end{bmatrix}$$

• I'm not sure why (a) is wrong from your explanation. Can you clarify with a counterexample? Yes, I can see how you rewrite A in that form, but I need to show that from that form, i can break the original ODEs into 2 sub-ODEs, each correspond to either $A_s$ or $A_u$. It's quite hard:( – user177196 Oct 18 '15 at 21:18
• For counter example select $A=1$ and $f(t)=1$. Ok, I understand the question now, see updated answer. – obareey Oct 18 '15 at 21:31
• Absolutely beautiful! The transformation is so simple that I should have thought of it:(( Many thanks for your great help. Btw, is the last term still bounded, as the columns of $P^{-1}$ might contain unstable eigenvalues? – user177196 Oct 18 '15 at 21:57
• Multiplication with a constant never changes the boundedness of a function. The bound might be bigger but still finite and that's what is important. – obareey Oct 18 '15 at 22:01
• I don't know what else to say. Read everything I wrote carefully and do the counter examples. (a) is obviously wrong. That's it. Trying to prove a wrong expression is meaningless. Also, if boundedness means bounded by some function, the claim would be true even if $A$ has eigenvalues on the imaginary axis, i.e. for all $A$. – obareey Oct 22 '15 at 7:46