In the question, I'm asked to show that \begin{align*} F\begin{pmatrix}x\\y\end{pmatrix}=x^2+y^2 \end{align*} is an integral for the linear map \begin{align*} L(\text{x})= \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} \text{x}. \end{align*}

I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F ∘ L(x) = F(x)$.

My question is how to you compose the two in this case? In $F\begin{pmatrix}x\\y\end{pmatrix}$ Do you set both variables equal $L(x)$ or only one....

  • $\begingroup$ The condition "$F$ is constant along orbits of $L$" is should be written as $F\circ L(x)=F(x)$ $\endgroup$
    – Miel Sharf
    Dec 15 '14 at 14:50
  • $\begingroup$ yes sorry! thanks for pointing that out $\endgroup$
    – Janet
    Dec 15 '14 at 14:54
  • $\begingroup$ The condition that $F ∘ L=F$ is offtopic. $\endgroup$
    – Did
    Dec 15 '14 at 15:00
  • $\begingroup$ The exact question in the textbook asks you to prove that $F$ is an integral for a linear map $L$ by calculating $F ∘ L(x) = F(x)$ $\endgroup$
    – Janet
    Dec 15 '14 at 15:03
  • $\begingroup$ Which textbook is this? Are we considering the (rather poor) dynamical system in discrete time $\mathbf x_{n+1}=L(\mathbf x_n)$, or the dynamical system in continuous time $\dot{\mathbf x}(t)=L(\mathbf x(t))$? $\endgroup$
    – Did
    Dec 15 '14 at 15:05

The following is based on the assumption that one considers a dynamical system in continuous time. It appears that the object of interest is actually a dynamical system in discrete time (aka a recursive sequence...) hence the answer below does not apply. Leaving it for the picture...

Note that the question actually asks why $x^2+y^2=(-y)^2+x^2$ (?!), which makes for a rather poor question, I am sorry to say. Note also that every sequence $(\mathbf x_n)$ defined by $\mathbf x_{n+1}=L(\mathbf x_n)$ has period $4$, as an elementary computation shows.

I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F \text{ o } L(x) = L(x)$.

No. That $F$ is an invariant of the dynamical system associated to $L$ means that, if $$\dot{\mathbf x}(t)=L(\mathbf x(t)),$$ then the quantity $F(\mathbf x(t))$ does not depend on $t$. This is usually done showing that $u'(t)=0$ for every $t$, where $$u(t)=F(\mathbf x(t)),$$ which is implied by the condition that, for every $\mathbf x$, $$\nabla F(\mathbf x)\cdot L(\mathbf x)=0.$$ In your case, $$\nabla F(\mathbf x)=2\mathbf x,$$ hence one is left with $$\nabla F(\mathbf x)\cdot L(\mathbf x)=2\mathbf x\cdot L(\mathbf x)=2\begin{pmatrix}x\\y\end{pmatrix}\cdot\begin{pmatrix}y\\-x\end{pmatrix}=0.$$ Here is a rough sketch of some solutions $(\mathbf x(t))$:

$\hspace2in$enter image description here


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.