1
$\begingroup$

I have a problem with a system of linear differential equations for a vector-valued function: Let $f$ be a smooth map $\mathbb{R}^2 \ni (x,t) \mapsto f(x,t) \in \mathbb{C}^n$ such that $\partial_x f = Af$, $\partial_t f= Bf$ and inicial condition $f(0,0)=f_0 \in \mathbb{C}^n$, where $A$, and $B$ are matrices smoothly dependent on $x$ and $t$ (not constant!) such that all derivatives of $A,B$ are bounded. In the general case $A$ and $B$ will not commute.

My question is: Suppose, $A$, and $B$ were constant. Will there be a solution at all, when $[A,B]\neq 0$?

On the one side, I think so, simply because the equations are linear. On the other side it appears to me, that in the case of constant $A$, $B$ but not commuting, I could integrate along $x$ and along $t$ to get $f(x,0)=e^{Ax}f_0$ and $f(0,t)=e^{Bt}f_0$. But then I would have $f(x,t)=e^{Bt}e^{Ax}f_0 = e^{Ax}e^{Bt}f_0$. In this situation only $f_0 \in \text{ker}\left([e^A,e^B]\right)$ could make this be possible, however. Is this condition necessary for $f_0$ then?

Comment: I overthought the condition $f_0 \in \text{ker}\left([e^A,e^B]\right)$ and now I think this would not be enough to ensure $e^{Bt}e^{Ax}f_0=e^{Ax}e^{Bt}f_0 \forall x,t \in \mathbb{R}^2$. Instead I would need something like $f_0 \in \bigcap_{x,t} \text{ker}([e^{Ax},e^{Bt}])$ --- I am pretty sure that this set is $=\{0\}$ so that afterall this idea seems not to be of much value. Now I don't see any other way out of this. ALSO: From the application in my mind I do expect a unique solution.

$\endgroup$
  • $\begingroup$ Did you mean to write $\partial_t f = Bf$ rather than $\partial_t = Bf$? Cheers! $\endgroup$ – Robert Lewis Jul 21 '18 at 17:11
  • 1
    $\begingroup$ Sure. Thanks for the remark! $\endgroup$ – Caroline Jul 21 '18 at 17:12
  • $\begingroup$ Perhaps Frobenius theorem? $\endgroup$ – user539887 Jul 22 '18 at 8:36
  • $\begingroup$ Hmm. I did not know about Frobenius Theorem. But apparently it covers only underdetermined solutions. In my case, the rank of $f$ is $\geq 2$. As far as I understood it, what I would need is a statement for the following question: "What condition do $A$ and $B$ and maybe $f_0$ need to satisfy such that a unique solution of the problem above exists?". $\endgroup$ – Caroline Jul 22 '18 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.