When solving a physics problem, I had the following system of differential equations ($\rho$ and $\varphi$ are functions of $t$, $c$ and $v$ are constants)

$$ \begin{cases} \rho\,\dot{\varphi} = c \cdot \cos \varphi \\ \dot{\rho} = c \cdot \sin\varphi - v \end{cases}$$

I integrated the second equation wrt $t$

$$\rho = \int(c \cdot \sin\varphi - v) dt = - \frac{c \cos \varphi} {\dot\varphi} - vt + \gamma$$

I wanted to find out the constant $\gamma$, so I substituted $c \cos \varphi = \rho \dot\varphi$ and got

$$\rho = -\rho - vt + \gamma$$

But this formula contradicts with the physical interpretation of the problem, since $\rho$ must depend on $c$ outside the constant.

What have I done wrong?

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    $\begingroup$ $\dot\varphi$ is not constant, it depends on $\varphi$. $\endgroup$ – Intelligenti pauca Nov 2 '15 at 19:25
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    $\begingroup$ After you have integrated, try to derivate the RHS to check if it is correct. $\endgroup$ – A.Γ. Nov 2 '15 at 19:28
  • $\begingroup$ You're right, the derivative is not the same. So how should I solve this eqn properly? $\endgroup$ – marmistrz Nov 2 '15 at 19:54

I do not know the physics of this DE, but if we will try to search for the solution in the form $$ \rho(t)=\eta(\phi(t)). $$ then we get $\dot\rho=\eta'(\phi)\dot\phi$ and dividing the second equation by the first one gives $$ \frac{\eta'}{\eta}=(\ln\eta)'=\frac{c\sin\phi-v}{c\cos\phi}=\tan\phi-\frac{v}{c}\frac{1}{\cos\phi} $$ which is possible to integrate $$ \ln\eta(\phi)=-\ln|\cos\phi|-\frac{v}{c}\ln\Big|\frac{\sin\phi+1}{\cos\phi}\Big|+\gamma. $$ Does it make sense from the physical interpretation?

P.S. Quantity $\frac{v}{c}$ makes me think that it is kind of relativity equation.

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  • $\begingroup$ Well, it isn't contradictory at least. I did $$\ln \eta = - \ln \left| \delta \cos\varphi \cdot \left( \frac{1 + \sin\varphi}{\cos\varphi}\right)^{\frac vc}\right|$$ $$\rho(\varphi) = \eta = - \frac{1}{ \delta \cos\varphi \cdot \left( \frac{1 + \sin\varphi}{\cos\varphi}\right)^{\frac vc}}$$ So it is possible for $\rho$ to be $0$. But it would be easier for me to find it out if I could have $\rho(t)$. Well, it may be connected with some relativistic problem, but in its essence it's not. I have to find out the track of two bodies with given velocities. $\endgroup$ – marmistrz Nov 3 '15 at 21:57
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    $\begingroup$ @marmistrz $\rho=\eta(\phi)$ is the relation between the coordinates, the trajectory. To find the particular motion one can find $\phi(t)$ from the first equation $\dot\phi=\frac{c\cos\phi}{\eta(\phi)}$ as $\eta$ is already known, and then substitute to $\rho(t)=\eta(\phi(t))$. $\endgroup$ – A.Γ. Nov 3 '15 at 22:34

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