Differential equation and substitution

[Pomico]

This is a step from my notes that I don't follow. I have sin\theta\frac{d}{d\theta}(sin\theta\frac{d\Theta}{d\theta}) and that, when substituting u=cos\theta and writing that \Theta(\theta)=P(u), \frac{d}{du}((1-u^{2})\frac{dP}{du}) is obtained.

I can see \frac{d\Theta}{d\theta}=\frac{dP}{du} for the far RHS but can't get the 1-u^{2} to come out. I don't really remember how to do this though I'm sure I have been able to at some point so a starting tip would be much appreciated! The main trap I keep falling in is the temptation to try something with chain rule to get rid of all those d\theta bits in the original equation. I know I'm not supposed to as I'm just supposed to be substituting, not solving, but I'm not sure how else to go about it.

 

[tiny-tim]

Hi Pomico!

(have a theta: θ )

This is a step from my notes that I don't follow. I have sin\theta\frac{d}{d\theta}(sin\theta\frac{d\Theta}{d\theta}) and that, when substituting u=cos\theta and writing that \Theta(\theta)=P(u), \frac{d}{du}((1-u^{2})\frac{dP}{du}) is obtained.

I can see \frac{d\Theta}{d\theta}=\frac{dP}{du} for the far RHS but can't get the 1-u^{2} to come out.

Using the chain rule, sinθ d/dθ = sin²θ d/du.

(but I don't see how to get the whole thing … are you sure it isn't 1/sinθ on the left?)

 

[Mark44]

Using the chain rule, sinθ d/dθ = sin²θ d/du.

tiny-tim,
Maybe I'm missing something here, but the line above doesn't make sense to me.

Pomico,
Is there really a need to use uppercase and lowercase versions of the same letter? I.e., \Theta and \theta.

 

[tiny-tim]

hmm … u = cosθ, so du/dθ = -sinθ,

so d/du = dθ/du d/dθ = -1/sinθ d/dθ,

so sin²θ d/du = -sinθ d/dθ

… ok, i missed out a minus sign ​

 

[Pomico]

(but I don't see how to get the whole thing … are you sure it isn't 1/sinθ on the left?)

Hi!
It's definitely not a reciprocal in the notes, that was one of the things bothering me

Pomico,
Is there really a need to use uppercase and lowercase versions of the same letter? I.e., \Theta and \theta.

Yes there is, this is only part of the problem covered in the notes. Earlier on there was a function F(θ,\phi) which was split into variables as
F(θ,\phi) = \Theta(θ)\Phi(\phi)

The equation in my first post isn't the full equation, just the first term. I don't think the rest will change anything but in case I've added an attachment of the lecture - the derivation I'm working from covers mostly pages 3 and 4 of the document. I don't know how to just include these two pages so sorry for adding the whole thing!

 

[Chrisas]

Try this. Consider the substitution \Theta(\theta)=P(u)

You want to find \frac{d\Theta}{d\theta}

\frac{d\Theta}{d\theta} = \frac{dP(u)}{d\theta} = \frac{dP(u)}{du} \frac{du}{d\theta}

You can get \frac{du}{d\theta} from your substitution of u=cos(\theta). Also sin(\theta) = "something with u", using a well known trig identity.

 

[Mark44]

hmm … u = cosθ, so du/dθ = -sinθ,

so d/du = dθ/du d/dθ = -1/sinθ d/dθ,

so sin²θ d/du = -sinθ d/dθ

… ok, i missed out a minus sign ​

I didn't catch the minus sign, either. I guess what you were doing was equating operators. I think that's what threw me--having the operator without an operand for it to work on.

 

[tiny-tim]

The equation in my first post isn't the full equation, just the first term. I don't think the rest will change anything but in case I've added an attachment of the lecture - the derivation I'm working from covers mostly pages 3 and 4 of the document. I don't know how to just include these two pages so sorry for adding the whole thing!

D'oh! I was right!

You're referring to equations (2.26) and (2.27) …

but you can see that (2.26) is divided by sin²θ before the u-substitution, so then it does start with 1/sinθ !

 

[Pomico]

Aah yes I see, thanks for spotting that :p
I'll have another go!

Edit: Got it, thanks again :)