# Bounded solutions of ODE system.

Find $a \in \mathbb{R}$, such that solutions of the system $$\dot x = 3x + y + 1; \dot y = 6x+2y+a$$ are bounded when $-\infty < t< \infty$ and determine if they are stable or not.

My attempt:

Well, unfortunately I can not solve the problem. I only have several vague ideas. First of all, I guess that for solutions to be bounded $\dot x=0, \dot y=0$ must holds. I can not prove it. (btw I think it is clear that converse is true). From $\dot x=0, \dot y=0$ one easily get $a=2$. Of course it is not justified. As for stability again, I feel the solution (I feel there is only one solution, but I can not prove it) is stable since derivatives are zero, but I am not sure it is strong argument. Of course this idea fails as long the idea $\dot x=0, \dot y=0$ fails.

Thanks a lot for your help!

Update:

LutzL provided a killer hint (thanks a lot) and I am trying to expand his hint into readable argument. My try:

Assume $x$ and $y$ are bounded solutions of the system. Consider $z=2x-y$. Cleary, $z$ must be bounded. From the system we can compute $z'=2x'-y'=6x+2y+2-6x-2y-a=2-a$. So since $z$ is bounded we conclude $a=2$ since otherwise $z$ is clearly unbounded.

Ok, I am happy with this part. My idea about $\dot x=0, \dot y=0$ was wrong, but the answer was correct, funny.

Now please advice how to handle the part of the question about the stability of the solution.

Update 2: Ok, I got another idea for stability part. Maybe it is possible to find all bounded solutions explicitly and therefore conclude they are stable? What are the bounded solution (or may solutions) for this system? I still think there should be only one solution, but I can not justify this.

Update 3: Ok, it looks like I have the proof about explicit form of bounded solution:

Since we know that bounded solution are only possible when $a=2$ let us consider the system $$\dot x = 3x + y + 1; \dot y = 6x+2y+2.$$ Let us divide second equation on first, we get $y'_x=2$, so $y=2x+C'$. Plugging into first equations we get $\dot x - 5x = C_1$, which has solutions $x(t) = \frac{C_1}{5} + C_2 e^{-5 t}$. Clearly, this solution is bounded only when $C_2=0$. But then $x=c, \dot x =0, y = -1-3c$, so we proved that bounded solutions have the form $x=c, y=-1-3c$, where $c$ is constant.

Now I think that I can tell that bounded solutions are stable directly from definition of stability, since initial conditions fully determine any bounded solution and any bounded solution with close initial conditions will be closed for any $t$, since they do not depend on $t$ . Is this argument valid?

• Is $a$ a constant? – High GPA May 28 '17 at 9:51
• Yes, a is real number. – Hedgehog May 28 '17 at 9:52

Consider $z=2x-y$. Then $\dot z=2-a$ needs to have a bounded solution if $x$ and $y$ are to be bounded.

• Thanks a lot! I've a bit expanded your answer, can you please check my expansion of your argument? Also I still can not handle part of the problem about stability, but may it can be solved by looking at explicit form of the solution? Can you give a hint about how the solution(s) of the system when $a=2$ look? Thanks again! – Hedgehog May 28 '17 at 10:47
• Yes. Essentially, your system matrix has an eigenvalue $0$ which is a critical case for stability. As the other eigenvalue is $5$, there are no stationary points that are stable in both time directions, you only find this one-parameter family of bounded solutions. – LutzL May 28 '17 at 10:48
• Can you please elaborate? What is the theorem you use to claim that the system is stable? Is it true that as long as we have eigenvalue $0$ the system is stable, no matter what other eigenvalues are? What is the role of eigenvalue 5 in this particular case? Why can we ignore constant terms $1$ and $a$? Can you please give brief explanation about this - I am not particularly interested in this concrete question, but I am preparing for the test and I have to solve similar questions myself. Thanks a lot! – Hedgehog May 28 '17 at 10:55
• Sorry again for spamming you, but this drives me crazy. I know about stability conditions for linear system with constant coefficients: as long as we have $\lambda_i$ such that $Re \lambda_i >0$ the zero solution is not stable. But how can it be stable in this case with eigenvalue $5$? – Hedgehog May 28 '17 at 11:42
• I did not claim that it is stable, at least not in both time directions. In direction $-\infty$ it is stable, but not asymptotically stable. – LutzL May 28 '17 at 11:55

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