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Hi i am trying to solve the following two equations: $$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$ $$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$ where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that is $\phi(\theta)$. Is there any way to do this?

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  • $\begingroup$ Do you need the general solution or are there some initial conditions? $\endgroup$
    – Yuriy S
    Jan 28 at 13:49
  • $\begingroup$ The first one is separable $\endgroup$
    – Empy2
    Jan 28 at 13:50
  • $\begingroup$ @YuriyS I am hoping for a general solution. But not sure if a general solution exist for this case of equations. Also i have been provided with no initial conditions but if you can solve it for any arbitrary initial values i would be glad to know. $\endgroup$
    – Jason Liam
    Jan 28 at 13:58
  • $\begingroup$ A first step should be to define a new function: $\dot{\phi}=f(t) $, then you can easily solve the first equation and find $f$ in terms of $\theta$ so to speak. $\endgroup$
    – Yuriy S
    Jan 28 at 14:32
  • $\begingroup$ How was this equation obtained, out of curiousity? If it's from Euler-Lagrange equations then it'd be helpful to indicate the Lagrangian. $\endgroup$ Jan 28 at 19:17
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$$\frac{d^2\phi}{dt^2} + 2\cot(\theta)\frac{d\theta}{dt}\frac{d\phi}{dt} =0$$ This separable ODE is directly integrable. $$\ln(\frac{d\phi}{dt})+2\ln(\sin(\theta))=\text{constant}$$ $$\sin^2(\theta)\frac{d\phi}{dt}=c_1\quad\implies\quad \frac{d\phi}{dt}=\frac{c_1}{\sin^2(\theta)}$$ Second equation : $$\frac{d^2\theta}{dt^2}-\sin(\theta)\cos(\theta)\left(\frac{d\phi}{dt}\right)^2=0$$

$$\frac{d^2\theta}{dt^2}-\sin(\theta)\cos(\theta)\left(\frac{c_1}{\sin^2(\theta)}\right)^2=0$$ $$2\frac{d\theta}{dt}\frac{d^2\theta}{dt^2}-2(c_1)^2\frac{\cos(\theta)}{\sin^3(\theta)}\frac{d\theta}{dt}=0$$ $$\left(\frac{d\theta}{dt}\right)^2+(c_1)^2\frac{1}{\sin^2(\theta)}=(c_2)^2$$ $$\frac{d\theta}{dt}=\pm\sqrt{(c_2)^2-(c_1)^2\frac{1}{\sin^2(\theta)}}$$ $$t=\int \frac{\sin(\theta)}{\pm\sqrt{(c_2)^2\sin^2(\theta)-(c_1)^2 }}d\theta$$ $$t=\pm \sin^{-1}\left(\frac{c_2}{(c_1)^2+(c_2)^2}\cos(\theta) \right)+c_3$$ $$\frac{c_2}{(c_1)^2+(c_2)^2}\cos(\theta)=\pm\sin(t-c_3)$$ $$\theta(t)=\cos^{-1}\left(\pm\frac{(c_1)^2+(c_2)^2}{c_2}\sin(t-c_3) \right)$$

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  • $\begingroup$ The fact that $\cos \theta(t)$ is nice makes me think that working in cylindrical coordinates would be nicer than spherical coordinates. $\endgroup$ Jan 28 at 19:00
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I think your system can be reduced to a Bernoulli Differential Equation: $y'(x)+p(x)\,y(x) = q(x) \, [y(x)]^n$.

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Rewrite $\dot{\phi}=\phi'(t)$ and $\ddot{\phi}=\phi''(t)$ by applying the chain rule for $\phi(\theta(t))$ and define $u(\theta)=\phi'(\theta)$ to get

$$u'(\theta) + 2 \cot(\theta) \, u(\theta) + \sin(\theta) \cos(\theta) \, [u(\theta)]^3 = 0 $$

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I will double-check this partial solution and write it down here step-by-step later.

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EDIT:

Considering the chain $\phi - \theta - t$, it's possible to write:

$$\frac{d\phi}{dt} = \frac{d\phi}{d\theta} \frac{d\theta}{dt}$$

$$\frac{d}{dt} \left[\frac{d\phi}{d\theta}\right] = \frac{d^2\phi}{dtd\theta} = \frac{d^2\phi}{d\theta^2} \frac{d\theta}{dt}$$

$$\frac{d^2\phi}{dt^2} = \frac{d}{dt} \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] = \frac{d^2\phi}{dtd\theta} \, \frac{d\theta}{dt} + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} = \frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2}$$

Returning those results to your original system:

$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$

$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$

$$$$

$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} + 2\cot{\theta} \, \frac{d\theta}{dt} \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] =0$$

$$\frac{d^2\theta}{dt^2}- \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] ^2=0$$

$$$$

$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} + 2\cot{\theta} \, \frac{d\theta}{dt} \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] =0$$

$$\frac{d^2\theta}{dt^2} = \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2$$

$$$$

$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2 + 2\cot{\theta} \, \frac{d\phi}{d\theta} \, \left[\frac{d\theta}{dt}\right]^2 =0$$

$$$$

So you have a trivial equation, $[\theta'(t)]^2=0$, and

$$\frac{d^2\phi}{d\theta^2} + \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^3 + 2\cot{\theta} \, \frac{d\phi}{d\theta} =0$$

$$$$

Define $u(\theta)=\phi'(\theta)$ to get a Bernoulli Differential Equation,

$$u'(\theta) + 2 \cot(\theta) \, u(\theta) + \sin(\theta) \cos(\theta) \, [u(\theta)]^3 = 0 \text{,}$$

which presumably has analytical solutions.

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  • $\begingroup$ How did you get the following step: $\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2 + 2\cot{\theta} \, \frac{d\phi}{d\theta} \, \left[\frac{d\theta}{dt}\right]^2 =0$ , i think you have missed 2 terms here from the above equations particularly the terms $\frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2}$ and $\frac{d^2\theta}{dt^2}$. Can you explain why are you omitting those terms? Thanks $\endgroup$
    – Jason Liam
    Jan 28 at 16:54
  • $\begingroup$ It's the substitution of the second equation into the first one. $\endgroup$
    – Zalnd
    Jan 28 at 17:10
  • $\begingroup$ Ok got it. Now trying to solve the final equation. $\endgroup$
    – Jason Liam
    Jan 28 at 17:31
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Here is my attempt. Divide the first equation by $\dot\phi$, then we get $$ \frac{d\ln\dot\phi}{dt}+2\frac{d\ln(\sin\theta)}{dt}=0.$$ We then obtain the integral of motion $$\ln\dot\phi+2\ln\sin\theta=C.$$

Hence, we find $\dot\phi\sin^2\theta=e^C=B$. Plugging this in the second equation, we find $$ \ddot\theta-B^2\frac{\cos\theta}{\sin^3\theta}\dot\theta=0.$$ Multiplying this by $\dot\theta$, we find

$$\frac12\frac{d\dot\theta^2}{dt}+\frac b2\frac{d\sin^{-2}\theta}{dt}=0,$$ where $b=B^2$. Then we get another integral of motion, a Hamiltonian $H$. $$\frac{\dot\theta^2}2+\frac b{2\sin^2\theta}=H.$$ I believe this is right, but I don't have time to check it carefully now. I hope this helps.

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First, we must get rid of the middle man, $t$. In other words, we will make the following observations:

$$\dfrac{d\phi}{dt}=\dot\theta\dfrac{d\phi}{d\theta}$$ and $$\dfrac{d^2\phi}{dt^2}=\dfrac{d}{dt}\dot\theta\dfrac{d\phi}{d\theta}=\ddot\theta\phi_{\theta}+\dot\theta^2\phi_{\theta\theta}.$$

Notice now how we have $\ddot\theta$ as $\dot\phi^2\sin\theta\cos\theta$. So the first equation becomes $$\dot\theta^2\sin\theta\cos\theta\cdot\phi^3_{\theta}+\dot\theta^2\phi_{\theta\theta}+2\cot\theta\dot\theta^2\phi_{\theta}=0$$ or $$\sin\theta\cos\theta\cdot\phi^3_{\theta}+\phi_{\theta\theta}+2\cot\theta\phi_{\theta}=0$$ This is the equation for $\phi(\theta)$. If you were not too picky about the parameter $t$, this could have been solved by [other answers on this post]. Now, notice that dividing through by $\phi_\theta$, we have $$\sin\theta\cos\theta\cdot\phi^2_{\theta}+\dfrac{\phi_{\theta\theta}}{\phi_{\theta}}+2\cot\theta=0$$ Because we have a that second term as being $$\dfrac{d}{d\theta}\log(\phi_{\theta})$$ substitute $\phi=\int e^{\beta}\mathrm{d}\theta$. Thus, $$\cos\theta\sin\theta e^{2\beta}+\beta_{\theta}+2\cot\theta=0$$ We're almost done. Divide through by $e^{2\beta}$, and subtract the beta-like terms to obtain $$M+Le^{-2\beta}=\dfrac{d}{d\theta}e^{-2\beta}$$ where $M=2\sin\theta\cos\theta$ and $L=4\cot\theta$. This evidently takes the form $$y'+ay=b$$ which has the general solution $$y=e^{-\int a}(k+\int b e^{\int a})$$ for some constant $k$, and so $$e^{-2\beta}=e^{4\int \cot\theta\mathrm{d}\theta}\left[2\int \sin\theta\cos\theta \cdot e^{-4\int\cot\theta \mathrm{d}\theta}\mathrm{d\theta}+k\right]$$ which we know that $$\phi =\int e^{\beta}\mathrm{d\theta}=\int \dfrac{e^{-2\int \cot\theta\mathrm{d}\theta}}{\sqrt{2\int \sin\theta\cos\theta \cdot e^{-4\int\cot\theta \mathrm{d}\theta}\mathrm{d\theta}+C_0}}\mathrm{d\theta}.$$

Edit A more complete solution would be where one notes that $e^{\int \cot \theta}=|\sin \theta|$ I'm ignoring constants because they're a pain, other than the one obtained by the linear solution. Then, $$\phi =\int \dfrac{1}{\sin^2\theta\sqrt{2\int \dfrac{\cos\theta}{\sin ^3\theta}\mathrm{d\theta}+C_0}}\mathrm{d\theta}$$ And since $$\int \dfrac{1}{\sin^3\theta}d\sin \theta=-\dfrac{1}{2\sin^2\theta}$$ we have simply that $$\phi = \int \dfrac{1}{\sqrt{C_0\sin^4\theta-\sin^2\theta}}d\theta=\arctan\left(\sqrt{C_0\sin^2\theta-1}\right)-\dfrac{\arctan\left(\dfrac{\sqrt{C_0\sin^2\theta-1}}{\sqrt{1-C_0}}\right)}{\sqrt{1-C_0}}$$

Please comment if I made any errors.

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