I have to show that the solution of a differential equation is an arc of a great circle. The differential equation is as follows $($in spherical coordinates$)$: $$\frac{\sin ^2\theta\phi '}{(1+\sin ^2\theta (\phi ')^2)^{\frac{1}{2}}}=C$$ where $C$ is an arbitrary constant and $\phi '$ denotes the derivative of $\phi$ with respect to $\theta$.

My reasoning:

By setting $\phi (0)=0$, any arc of a great circle will have no change in $\phi$ with respect to $\theta$, so with this initial condition the answer follows by proving that $\phi '=0$. My issue is that upon working this round I end up with $$(\phi ')^2=\frac{C^2}{\sin ^4\theta -C^2\sin ^2\theta}$$ From this I can see no way forward.

Where do i go from here/ what should I do instead?

  • $\begingroup$ Yes, my mistake. Thanks. $\endgroup$ Jun 7 '18 at 17:20

Let $u=\cot \theta$, then $1+u^2=\csc^2 \theta$ and $du=-\csc^2 \theta \, d\theta$.

\begin{align} \frac{d\phi}{d\theta} &= \frac{d\theta}{\sin \theta \sqrt{\sin^2 \theta-C^2}} \\ &= \frac{C\csc^2 \theta}{\sqrt{1-C^2\csc^2 \theta}} \\ d\phi &= -\frac{C\,du}{\sqrt{1-C^2(1+u^2)}} \\ &= -\frac{C\,du}{\sqrt{(1-C^2)-C^2u^2}} \\ &= -\frac{du}{\sqrt{\tan \alpha^2-u^2}} \tag{$C=\cos \alpha$} \\ \phi &=\cos^{-1} \left( \frac{u}{\tan \alpha} \right)+\beta \\ \cos (\phi-\beta) &= \frac{\cot \theta}{\tan \alpha} \\ \cot \theta &= \tan \alpha \cos (\phi-\beta) \\ \end{align}


$$(\sin \theta \cos \phi)(\sin \alpha \cos \beta)+ (\sin \theta \sin \phi)(\sin \alpha \sin \beta)= (\cos \theta)(\cos \alpha)$$

which lies on the plane

$$x\sin \alpha \cos \beta+y\sin \alpha \sin \beta-z\cos \alpha=0$$

  • $\begingroup$ Thanks, that’s a really elegant solution. Is there anything that lead you to that substitution or just experience? $\endgroup$ Jun 7 '18 at 20:23

We have

$$ \phi' = \frac{C\sin^{-2}\theta}{\sqrt{1-C^2\sin^{-2}\theta}} $$

now changing variable

$$ u = C\cot\theta \Rightarrow du = -\sin^{-2}\theta d\theta $$

and then

$$ d\phi = -\frac{du}{\sqrt{1-u^2}} $$

and integrating

$$ \phi = C_1 -\sin^{-1}(u) = C_1-\sin^{-1}(C\cot\theta) $$

and then

$$ C\cot\theta = \sin(C_1-\phi) $$


$$ C\cos\theta =\sin(C_1)\sin\theta\cos\phi-\cos(C_1)\sin\theta\sin\phi $$

or changing to cartesian coordinates

$$ C z - \sin(C_1)x+\cos(C_1)y = 0 $$

which is the equation of the plane intersecting the sphere and containing the great circle.

  • $\begingroup$ Thanks, that’s really neat. As above, is there anything that lead you to that substitution $\endgroup$ Jun 7 '18 at 20:30
  • $\begingroup$ The differential equation deduction using variational procedures, helps a lot in their comprehension and also in it's solution. $\endgroup$
    – Cesareo
    Jun 7 '18 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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