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Let $U \subset \mathbb{R}^n$ be an open set, and $B$ a smooth matrix valuated function $B:U \to M_n(\mathbb{R})$ .

Let $\frac{\partial}{\partial x^1}$ be the constant vector field defined on $U$. I am trying (with no success) to prove that given $x_0 \in U$, and $A_0 \in M_n(\mathbb{R})$ there exist a smooth matrix valuated function $A:U \to M_n(\mathbb{R})$ , such that:

$\frac{\partial A}{ \partial x^1}= B \cdot A $, with $A(x_0)=A_0$. And that the dependence on $(x_0 , A_0)$ is smooth.

Well i arrived to this problem trying to show the smoothness of a distribution defined on a manifold, and using a chart (reading Spivak introduction to differentiable geometry vol 2, pp 338 2nd edition), and i have been trying with the classical existence theorems, my idea is:

Take $x_0=(x_0^1 , \dots , x_0^n) \in U$ and define the first ODE:

$A'(t)= B(c(t))\cdot A(c(t))$.

Where $c$ is an integral curve for $\frac{\partial}{\partial x^1}$ such that $c(0)= x_0$, this equation is very well know to have a solution, with initial condition $A_0$, and seems natural to try to define $A(c(t)) = A(t)$. If I take curves by $(x_0^1 \pm \epsilon, x_0^2, \dots x_0^n)$ I can define ODE with the same initial condition. The problem is i can't see why all these solutions, glue together! I would appreciate any help or advise. in fact, i don't know if it is true! (one friend told me to use a more general existence of solution theorem which lets $B$ be a smooth function of $\mathbb{R}^n$, but i can't find this theorem anywhere) ...

Any help or advise is welcome. Thank you all!

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Edited Version: Note first that your equation cannot have a unique solution, since you are prescribing only one partial derivative and hence multiplying any solution by a function of the other coordinates is again a solution. What I describe is the simplest solution, which is independent of all other coordinates.

More or less by definition of the matrix exponential, the curve $c:\mathbb R\to M_n(\mathbb R)$ defined by $c(t)=e^{tB}$ is the unique solution of the ODE $c'(t)=B\cdot c(t)$ (where the dot indicates the matrix product) with $c(0)=\mathbb I$. If you want to change the initial condition you simply multiply from the right by an appropriate matrix.

To apply this to your problem, you simply take the first coordinate function $x^1$ and subtract the constant $x_0^1$. Then you see that if you define $\tilde A(x):=e^{(x^1-x_0^1)B}=e^{x^1B}e^{-x_0^1B}$, then this solves the equation $\frac{\partial}{\partial x^1}A(x)=B\cdot A(x)$ by definition of the partial derivative with initial condition $\tilde A(x_0)=\mathbb I$. To get the solution with the initial condition you want you have to take $A(x):=e^{(x^1-x_0^1)B}A_0$.

In general, the fact that the solutions do glue together basically is due to the smooth dependence of the solution on a system of ODEs on the initial condition. But in this case, the explicit formula for the solutions sorts out everything.

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  • $\begingroup$ I see, very good idea to look for explicit solutions!... I was having troblue in showing that $A(x)=e^{(x^1-x_0^1)B}A_0$ satisfies the equation ... it seems to me that it satisfices the equation : $\frac{\partial A}{ \partial x^1}= (x^1 - x_0) B \cdot A$. $\endgroup$ – Alicia Basilio Jun 28 '18 at 13:53
  • $\begingroup$ But using your idea what do you think about $ A(x) = e^{B(x)} /cdot ( e^{B(x_0)^{-1}} /cdot A_0) $ $\endgroup$ – Alicia Basilio Jun 28 '18 at 14:00
  • $\begingroup$ Sorry i wanted to type $ A(x) = e^{B(x)} \cdot ( e^{B(x_0)^{-1}} \cdot A_0) $ $\endgroup$ – Alicia Basilio Jun 28 '18 at 14:08
  • $\begingroup$ Well ... after thinking in this answer i think the idea does not work! ... since solutions of type exponential only work for constant $B$, when $B$ is not constant this type of solutions are not posible en.wikipedia.org/wiki/Derivative_of_the_exponential_map ... any way! thanks for the comment! $\endgroup$ – Alicia Basilio Jun 28 '18 at 21:48
  • $\begingroup$ @AliciaBasilio: I has a small mistake in my answer, I have edited to correct that and add a bit of explanation. The Wikipedia article you link to in your last comment is misleading. There is no need to differential the matrix exponential as a function $M_n(\mathbb R)\to GL(n,\mathbb R)$ in order to compute the derivatives of exponential curves. As the curve $c(t)=e^{tB}$ satsifies $c(s+t)=c(s)c(t)$ it readily follows that $c'(t)=c'(0)c(t)$. $\endgroup$ – Andreas Cap Jun 29 '18 at 11:55

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