The textbook(Ordinary Differential Equations and Dynamical Systems, Gerald Teschl) says:

For the differential equation

$\dot x$ = $f(x)$, $\;x(0) = x_0$ where $f \in C^k(M, R^n)$ where $M$ is open subset of $R^n$,

Let $\phi(t, x_0)$ be solution of the equation.

Then, linearized equation of $\phi(t,x_0)$ is $\dot \phi(t,x_0) = A(t)\,\phi(t,x_0)$ where $A(t) = df_{\phi(t,x_0)}$.

However, I can't understand the notation of $df_{\phi(t,x_0)}$ and confused when calculating example.

For example, when $\phi(t, (1,0)) = (\cos(t), \sin(t))$ and $\dot x_1 = -x_2 + x_1(1-x_1^2-x_2^2),\; \dot x_2 = x_1 + x_2(1-x_1^2-x_2^2)$

Then, $f(\phi(t,(1,0)) = (-\sin(t), \cos(t))$.

Textbook says, in this case, $A(t) = \begin{bmatrix} -2\cos^2(t) &-1+\sin^2(2t) \\ 1-\sin(2t) & \cos(t)\end{bmatrix}$

How do I derive this equation? I thought $df_{\phi(t,x_0)}$ means the gradient of $f(\phi(t,x_0))$, but in this case, $\nabla f(\phi(t,x_0)) = \begin{bmatrix} -\cos(t)\frac{\partial t}{x_1} &-\cos(t)\frac{\partial t}{x_2} \\ -\sin(t)\frac{\partial t}{x_1} & -\sin(t)\frac{\partial t}{x_2} \end{bmatrix} = \begin{bmatrix} \cot(t) &-1 \\ 1 & -\tan(t)\end{bmatrix}$

Is it, $df_{\phi(t,x_0)} \neq \nabla f(\phi(t,x_0))$? or I calculated wrong?

I have very poor understanding about general topic about vector calculus and having trouble with this. Help me please...

  • 2
    $\begingroup$ You've done the operations in the wrong order, and the answer you get doesn't make sense (what is the meaning of $\frac{\partial t}{x_1}$?). You need to calculate the gradient of $f(x_1, x_2) = (-x_2+x_1(1-x_1^2-x_2^2), x_1+x_2(1-x_1^2-x_2^2))$ and then plug in $\phi(t, (1, 0))$. $\endgroup$ – Michael Lee May 15 '18 at 17:31
  • 1
    $\begingroup$ Also, it is not a good idea to move on to the study of ODEs and dynamical systems if you feel that your understanding of calculus is poor. Calculus is extremely important to this field. $\endgroup$ – Michael Lee May 15 '18 at 17:35
  • 1
    $\begingroup$ Oh, that is it.. I was confused about that.. the function takes $t$, but I have to calculate the gradient of $f$. The gradient should be calculated before putting $t$ in, and just putting $t$, we get gradient, but I thought it was gradient about the function defined by $t$, and it led to compute like this. my English is bad that if you understood why I was confused. Anyway, Thanks a lot. $\endgroup$ – Jeongmin Park May 15 '18 at 17:38
  • $\begingroup$ The gradient of a function $\mathbb{R}\to \mathbb{R}^2$ would not even be a $2\times 2$ matrix, but a vector in $\mathbb{R}^2$. $\endgroup$ – Michael Lee May 15 '18 at 17:44

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.