MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have some problems proving that a Differential Matrix Equation has a solution. I apologize if the question is too elementary, but I've found this theorem stated everywhere on the web without any reference or clue about how to prove it.

What I exactly mean with a Differential Matrix Equation is: $X'=AX+B$. Where $A$ is a matrix of size $n\times n$ and $B$ is a column vector. Both $A$ and $B$ have coefficients which are holomorfic functions in a convex open set $\Omega$ and continuous on the closure $\bar\Omega$. We also have an initial condition $X(z_0)=U_0$, where $U_0$ is a column vector of complex coefficients.

So far I know how to prove it when $A$ has constant coefficients. However the proof cannot be applied as it uses the trick of $(e^{Az})'=Ae^{Az}$ to find an explicit solution and in general $(e^A)'\neq Ae^A$ if $A$ is nonconstant.

I've also read about Magnus Series, but I don't fully understand them. Also I'm only interested in the existence, so I'd prefer an easier way to prove that there are solutions.

share|cite|improve this question
Please ask on – Deane Yang Jan 31 '11 at 19:50
@Deane I've done so before. Do you know a reference for the proof? – Gabriel Furstenheim Jan 31 '11 at 21:08
If you're only interested in holomorphic parameters, you can write A and B as power series around $z_0$ and look for a power series solution. – Homology Jan 31 '11 at 21:36
I'm not sure whether X' meen $\frac{dX}{dt}$ or $\frac{dX}{dz}$. Either way - problem can be easily reduced to the ODE (in real case) or to the CDE (in complex case) with "sooth enough" right sites... – Michal Oszmaniec Jan 31 '11 at 21:59
@Homology Yes, I'm only interested in holomorphic parameters. The problem I find is that I'm interested in a global solution. With the power series i can only get local solutions and proving they patch to a global solution doesn't seem straightforward as the convex condition (in case it is necessary) looks difficult to recover from a local point of view. – Gabriel Furstenheim Jan 31 '11 at 22:08

This really sounds like homework. Anyway, try this. We work under the assumption that $A,B$ are entire functions, otherwise everything works the same but the results are local.

If $X(z)=u+iv$ is holomorphic, its derivative can be written as $X'(z)=u_x+iv_x$. Now separate real part and imaginary part of your system, what you get is a nice linear system of $2n$ equations for $2n$ unknown functions, in the variable $x$, with $y$ as a parameter. Standard advanced calculus results give you the existence of a smooth global solution, smoothly dependent on the parameter $y$ (don't ask for references, just differentiate the system $y$ and repeat). Now do the same trick writing $X'(z)=v_y-iu_y$: you get another set of $2n$ smooth functions. But of course $X'(z)=AX+B=u_x+iv_x=v_y-iu_y$ which means that $u,v$ satisfy the Cauchy-Riemann conditions and $X$ is holomorphic.

share|cite|improve this answer

This is more a comment than a solution, but I do not have the power. Have you checked in

Daletskii-Krein: Stability of Solutions of Differential Equations in Banach Space, Chapter 6?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.