0
$\begingroup$

Suppose we want to convert, \begin{align*} y_1’’ &= t^2-y_1’-y_2^2 &\quad \quad y_1(0)=0\quad y_1’(0)=1\\ y_2’’ &= t-y_2’-y_1^3 &\quad \quad y_2(0)=1\quad y_2’(0)=0 \end{align*} into a first-order autonomous linear system of ODEs.

To convert to first-order and autonomous you can define, \begin{align*} x_1 &= y_1 \\ x_2 &= y_1’ \\ x_3 &= y_2 \\ x_4 &= y_2’ \\ x_5 &= t \end{align*} and we end up with \begin{align*} x_1’ &= x_2 \\ x_2’ &= x_5^2-x_2-x_3^2 \\ x_3’ &= x_4 \\ x_4’ &= x_5+x_4-x_1^3 \\ x_5’ &= 1 \end{align*} But it remains to convert it to a linear system. How does one accomplish that? Can one simply define $x_6=y_2^2$ and $x_7=y_1^2$ and $x_8=y_1^3$?

$\endgroup$
  • $\begingroup$ Why do you think it is necessary (or even possible) to convert a non-linear system to a linear one? $\endgroup$ – Lutz Lehmann Mar 21 '20 at 14:06
  • $\begingroup$ @LutzLehmann I just wondered whether it could be done. The exercise is related to numerical analysis. I suppose the way to do it is to use linear approximation with Taylor Series. $\endgroup$ – PreSchooler Mar 21 '20 at 19:37
1
$\begingroup$

It is not so easy. The thing with monomials in a function is that it's differential: $$(y(t)^n)' = y'(t)\cdot ny(t)^{n-1}$$

This is product between function and it's derivative. It is non-linear.

I think you need to find some representation where function concatenation with polynomial is linear. This is possible with for example Carleman matrices.

So our function y is represented $\bf M_y$, squaring represented by $\bf M_2$, Differentiation $\bf M_D$

Now we can write $(y(t)^2)'$ as $$\bf M_D M_2M_y$$

But I am not sure that such $\bf M_D$ exists for these Carleman matrices.

$\endgroup$
  • $\begingroup$ Thanks, I figured it cannot possibly be so easy. That is why I asked. $\endgroup$ – PreSchooler Mar 21 '20 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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