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EDIT2: After some discussion here's the original problem: Let M be a n-D manifold and $\dot x=F(x)u_1, F\in \mathbb{R}^{n\times m}, u_1 \in \mathbb{R}^{m}$ be a control system evolving on M (F is the system matrix i.e. state transition function, and $u_1$ is the input of the system. For all practical purposes $u_1$ is an m-vector from an input space $\mathbb{R}^{m}$). Now let $x=\Psi (y)$ be a coordinate change on M and $u_2=M(y)u_1$ a transformation of the input $u_1$ of the first system. By applying these maps on the system, you get the new equations $\dot y=F(y)u_2$. As you may notice, F is the same in both systems. The problem is why is this happening i.e. for what systems and transformations does this property hold?

EDIT1: A more interesting story is when the d.e. is a matrix equation. For example: $F(y)=DyF(x)G(x)$ where, $F\in \mathbb{R}^{m\times n}, Dy \in \mathbb{R}^{m\times m}$ (the Jacobian matrix of $y=y(x)$) and $G \in \mathbb{R}^{n\times n}$? Apparently $x,y$ are m-vectors.

i have the following d.e. $f(y)={y}'f(x)g(x)$. Does anybody know the solution or a way to solve this d.e. (maybe it is a know form)? Note that $f,g,x,y$ are all real. Thanks in advance!

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Your condition is $J^{-1}F(y)M(y)=F(x)$, being $J$ the Jacobian. This definitely is not a pde. You should clearly state what $u_1$ and $u_2$ are as most people is not expert on control theory but can help in solving your problem. –  Jon Dec 17 '11 at 21:37
    
Jon i've added a comment on u. u is just the input of the system which is an m-vector on $\mathbb{R}^{m}$. As an input vector, it can be freely manipulated (actually defining a feedback law if $u=u(x,t)$, but this is not needed here). –  Jorge Dec 17 '11 at 21:48
    
I think that the answer to your question is just that the system must be linear, this for $F$. This will satisfy the condition between the Jacobian $J$, $F$ and $M$. When $M=J$ you just get $F(x)=J^{-1}F(y)J$ and you are done. –  Jon Dec 17 '11 at 22:27
    
Jon please notice that $J is n \times n$ and $ M is m \times m$ thus $J \neq M$. Furthermore, is nonlinear, for a fact. There is a specific example of this with F being the kinematic equations of a unicycle robot. –  Jorge Dec 18 '11 at 2:06
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1 Answer 1

This equation can be integrated in in the following way

$$\int\frac{dy}{f(y)}=\int dx\frac{1}{f(x)g(x)}+C$$

Once the forms of $f$ and $g$ are known, the integrals could be computed.

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Thanks Jon, however there is a more interesting twist. What is the solution if it is a matrix equation (which actually is what interests me)? For example: $F(y)=DyF(x)G(x)$ where, $F\in \mathbb{R}^{m\times n}, Dy \in \mathbb{R}^{m\times m}$ (the Jacobian matrix of $y=y(x)$) and $G \in \mathbb{R}^{n\times n}$? Apparently $x,y$ are m-vectors. –  Jorge Dec 17 '11 at 18:51
    
Indeed, it appeared somewhat trivial. But now you are claiming that also $x$, the independent variable, is a vector. Is this a pde problem now? –  Jon Dec 17 '11 at 19:22
    
You're right, it is a pde problem now. And it gets even better. Consider that x is a parametrization on a manifold M and y is a change of coordinates on M. You can also associate a dynamical system evolving on M given by $\dot y=F(y)u_1$ and apply the coordinate change. Notice the $F$ matrix on the system (it is the same as the pde). You also apply a transformation on $u_1=M(y)u_2$ (again notice M). After some analysis, you get $\dot x=F(x)u_2$ and the system is invariant under the coordinate change. It all boils down to this condition: F(y)=DyF(x)G(x). The case is why is this happening. –  Jorge Dec 17 '11 at 19:31
    
You can check if $F$ is sympletic. In this case you get a Hamiltonian system and your equation to solve is just a condition for a canonical transformation. –  Jon Dec 17 '11 at 19:35
    
Thanks again Jon, but can you be more specific? Canonical maps preserve the form of Hamilton's equations. How can this be linked to this problem? P.S. The system is actually mechanical. –  Jorge Dec 17 '11 at 19:42
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