I have to show that the solution of a differential equation is an arc of...

2nalfq8

2nalfq8

Answered

2022-06-30

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):
sin 2 θ ϕ ( 1 + sin 2 θ ( ϕ ) 2 ) 1 2 = C
where C is an arbitrary constant and ϕ denotes the derivative of ϕ with respect to θ.

My reasoning:
By setting ϕ ( 0 ) = 0, any arc of a great circle will have no change in ϕ with respect to θ, so with this initial condition the answer follows by proving that ϕ = 0. My issue is that upon working this round I end up with
( ϕ ) 2 = C 2 sin 4 θ C 2 sin 2 θ
From this I can see no way forward.
Where do i go from here/ what should I do instead?

Answer & Explanation

poquetahr

poquetahr

Expert

2022-07-01Added 18 answers

Let u = cot θ, then 1 + u 2 = csc 2 θ and d u = csc 2 θ d θ.
d ϕ d θ = d θ sin θ sin 2 θ C 2 = C csc 2 θ 1 C 2 csc 2 θ d ϕ = C d u 1 C 2 ( 1 + u 2 ) = C d u ( 1 C 2 ) C 2 u 2 ( C = cos α ) = d u tan α 2 u 2 ϕ = cos 1 ( u tan α ) + β cos ( ϕ β ) = cot θ tan α cot θ = tan α cos ( ϕ β )
Rearrange,
( sin θ cos ϕ ) ( sin α cos β ) + ( sin θ sin ϕ ) ( sin α sin β ) = ( cos θ ) ( cos α )
which lies on the plane
x sin α cos β + y sin α sin β z cos α = 0
fythynwyrk0

fythynwyrk0

Expert

2022-07-02Added 7 answers

We have
ϕ = C sin 2 θ 1 C 2 sin 2 θ
now changing variable
u = C cot θ d u = sin 2 θ d θ
and then
d ϕ = d u 1 u 2
and integrating
ϕ = C 1 sin 1 ( u ) = C 1 sin 1 ( C cot θ )
and then
C cot θ = sin ( C 1 ϕ )
or
C cos θ = sin ( C 1 ) sin θ cos ϕ cos ( C 1 ) sin θ sin ϕ
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.

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