Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The next problem states that given a $m\times m$ non singular matrix $T$, such that $T^2=I$, is called a reflection for $\dot{x}=f(x) \Leftrightarrow f(Tx)=-Tf(x)$ for all $x\ \epsilon\ R^{m}$. Need to prove that $\phi(t,T\xi)\equiv T\phi(-t,\xi)$ where $\phi(t,\xi)$ is the general solution of $x=f(x)$, and also to show that a reflection of a solution in the x1 axis is also a solution.

I know that $\dot{x}=f(x)$ and has a solution of $\phi(t,\xi)$, i also know that $\phi(0,\xi)=\xi$ because is the identity of the flow, so I tought that if we multiply by $-T$ the function and the solution we have $-Tf(x)$ and $-T\phi(t,\xi)$, and this new transformation given by $-T$ should preserve the new function with its new solution because is just a reflection.

So if we have $-T\phi(t,\xi)$, and if we put $t=0$ for the solution, we will have $-T\phi(0,\xi)=-T\xi$, that it would have been the same to express as $-\phi(0,T\xi)=-T\xi$ or $\phi(0,-T\xi)=-T\xi$, in other words $\phi(t,-T\xi)$ or $-\phi(t,T\xi)$ should be valid for $-Tf(x)$, but i need to arrive to $\phi(t,T\xi)\equiv T\phi(-t,\xi)$, not the previous one's. Maybe im approaching wrong and I should use another technique, or i could be missing something.

Any feedback is appreciated. Thanks.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Your goal $\phi(t,T\xi)\equiv T\phi(-t,\xi)$ will be achieved if you can show two things:

  1. $\phi(0,T\xi) = T \phi(0,\xi)$ -- that is, the functions agree at time $t=0$
  2. $\frac{d}{dt}\phi(t,T\xi)\equiv \frac{d}{dt} \left(T\phi(-t,\xi)\right)$ -- that is, the functions have the same $t$-derivative.

Item 1 is really a tautology: both sides are equal to $T\xi$ by the definition of $\phi$.

In 2, the left-hand side is $f(T\xi)$. To calculate the right-hand side, notice that derivative commutes with linear transformation $T$, and then use the chain rule: $$\frac{d}{dt} \left(T\phi(-t,\xi)\right) = T \frac{d}{dt} \phi(-t,\xi) = -T f(\xi)$$ Thus, both sides agree.

share|improve this answer
    
Thank you for sharing your knowledge, :D, really aprreciated –  JHughes Dec 23 '12 at 14:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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