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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.

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    $\begingroup$ Please ask on math.stackexchange.com. $\endgroup$
    – Deane Yang
    Commented Jan 31, 2011 at 19:50
  • $\begingroup$ @Deane I've done so before. Do you know a reference for the proof? $\endgroup$
    – Gabriel Furstenheim
    Commented Jan 31, 2011 at 21:08
  • $\begingroup$ 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. $\endgroup$
    – Homology
    Commented Jan 31, 2011 at 21:36
  • $\begingroup$ 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... $\endgroup$
    – Michał Oszmaniec
    Commented Jan 31, 2011 at 21:59
  • $\begingroup$ @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. $\endgroup$
    – Gabriel Furstenheim
    Commented Jan 31, 2011 at 22:08

2 Answers 2

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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?

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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 w.r.to $y$ 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.

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