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I am trying to expand system into two dimensions. Dynamical system with damping proportional to squared velocity is given by Newton's eq: $$ \ddot{x}(t) + \mu \dot{x}^2(t) = 0.$$

Now, I want my system to move in two dimensions. To do that, I wrote $\textbf{r} = xe_1 + ye_2$, so $$ \ddot{\textbf{r}}(t) + \mu \dot{\textbf{r}}^2(t) \cdot e_r= 0. $$

I put there versor to have vector + vector, not vector + scalar. $e_1, e_2$ are standard basis and $e_r=\frac{1}{\sqrt{2}}(e_1+e_2)$.

So, bringing everything together gives $$ \ddot{x}(t)e_1 + \ddot{y}(t)e_2 + \mu[\dot{x}(t)e_1 + \dot{y}(t)e_2]^2 \cdot(e_1+e_2) = 0. $$

Which can be reduced to $$ (\ddot{x} +\mu\dot{x}^2+\mu\dot{y}^2) e_1 + (\ddot{y}+\mu\dot{x}^2+\mu\dot{y}^2)e_2 = 0. $$

Is this method correct? Or maybe, there is better way?

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  • $\begingroup$ Your radial unit vector is not normed: $e_r=\frac{1}{\sqrt{2}}(e_1+e_2)$. $\endgroup$
    – Roger
    Jan 20, 2021 at 15:21
  • $\begingroup$ You're right. Thanks. $\endgroup$
    – user281659
    Jan 20, 2021 at 15:30
  • $\begingroup$ In two dimention, you should have two equations. $\endgroup$
    – ytlu
    Jan 20, 2021 at 15:52
  • $\begingroup$ Your 2nd equation only defines the dynamics in the $\mathbf{r}$ direction, which means it's effectively 1-D still, with $\phi$ const. $\endgroup$
    – stafusa
    Jan 20, 2021 at 22:15

1 Answer 1

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I am not quit sure what is your meaning, but your notation somehow didn't make sense to me, in the $\dot{\vec{r}}^2 \cdot \hat{e_r} $. In the expansion, there is a vector square which is a vector? and without the cross term? Then had it inner product with the 45 degree vectors. This algebraic process makes no sense at all.

Therefore, I try to image the damping force which is proportional to square of speed, and in the opposite direction of the velocity (as the usual case.)

In vector form:

$$ \ddot{\vec{r}} + \mu | \vec{v} |^2 \hat{v} = 0 $$

Where $\vec{r} = x \hat{e}_1 + y \hat{e}_2$ and $\vec{v} = \dot{x} \hat{e}_1 + \dot{y} \hat{e}_2$. A hatted vector indicates an unit vector. The $| \vec{v} |^2 = \dot{x}^2 + \dot{y}^2$. The unit vector of velocity, $\hat{v}$: $$ \hat{v} = \frac{\dot{x} \hat{e}_1 +\dot{y} \hat{e}_2 }{\sqrt{\dot{x}^2 + \dot{y}^2}} . $$

Expand the vector equation in each component:

Along $\hat{e}_1$ :

$$ \ddot{x} + \mu (\dot{x}^2 + \dot{y}^2) \frac{\dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}}= \ddot{x} + \mu \dot{x} \sqrt{\dot{x}^2 + \dot{y}^2} =0 $$

Along $\hat{e}_2$ :

$$ \ddot{y} + \mu (\dot{x}^2 + \dot{y}^2) \frac{\dot{y}}{\sqrt{\dot{x}^2 + \dot{y}^2}}= \ddot{y} + \mu \dot{y} \sqrt{\dot{x}^2 + \dot{y}^2} = 0 $$

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    $\begingroup$ Thank you very much! That's exactly what I mean. My problem was that when I have 1D, the situation is clear. When I expand it into 2D, I got $\ddot{r}$ which is vector and $\dot{r}^2$ which is scalar. $\endgroup$
    – user281659
    Jan 21, 2021 at 14:31
  • $\begingroup$ @blahblah You are welcom. I am glad to be helping. $\endgroup$
    – ytlu
    Jan 21, 2021 at 20:38

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