Let's consider the following system \begin{align} \frac{d}{dt}\begin{pmatrix} u\\v \end{pmatrix}=\begin{pmatrix} f(x,t,u,v,u_x,v_x)\\g(x,t,u,v,u_x,v_x) \end{pmatrix}. \end{align} If there locally exists a invertible mapping, s.t. the system in new coordinates becomes \begin{align*} \frac{d}{d\tilde{t}}\begin{pmatrix} \tilde{u}\\\tilde{v} \end{pmatrix}=\begin{pmatrix} \tilde{f}(\tilde{x},\tilde{t},\tilde{u},\tilde{v},\tilde{u}_{\tilde{x}},\tilde{v}_{\tilde{x}})\\ \tilde{g}(\tilde{x},\tilde{t},\tilde{v},\tilde{v}_{\tilde{x}}) \end{pmatrix}. \end{align*} we call the original system can be decoupled.

Intuitively, most of systems aren't decoupled, but I don't know how to prove it.

  • 1
    $\begingroup$ Do you really mean to allow the derivative of $\tilde u$ to depend on $\tilde v$? $\endgroup$ – LSpice Apr 11 at 14:02
  • 2
    $\begingroup$ Also, before you can hope to prove your statement, you need to define it: what does 'most' mean? $\endgroup$ – LSpice Apr 11 at 14:03

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.