Please help me to handle the following problem:
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?
Thanks for your help!
Best Answer
Consider $z=2x-y$. Then $\dot z=2-a$ needs to have a bounded solution if $x$ and $y$ are to be bounded.