# Show differential equation solution is arc of great circle

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?

• Yes, my mistake. Thanks. – CooperCape 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}

Rearrange,

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

which lies on the plane

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

• Thanks, that’s a really elegant solution. Is there anything that lead you to that substitution or just experience? – CooperCape 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

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

or

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

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.

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