Eliminate u: Solve Nonlinear Diff. Eqns.

[Murilo T]

Homework Statement

Those two equations below appered when I was trying to solve a physics problem.

Relevant Equations

u' = sinφ - 1
uφ' = cosφ

I've began to learn how to solve differential equations to eliminate u in those two equations, but it seems that it is a nonlinear differential equation.

I'm wondering if there is other way to eliminate u, or will I just have to learn how to solve nonlienar differential equations?
The solution to the problem is ϕ' = A cos ϕ (1 + sin ϕ).

I appreciate any help!

 

[etotheipi]

What do you get if you (implicitly) differentiate $u \varphi' = \cos{\varphi}$ with respect to the independent variable? Then, can you eliminate $u$?

 

[Murilo T]

What do you get if you (implicitly) differentiate $u \varphi' = \cos{\varphi}$ with respect to the independent variable? Then, can you eliminate $u$?

I guess that what you are saying is to isolate u = cosϕ/ϕ', and then take the derivative of u with respect to ϕ, right?

 

[etotheipi]

Not quite; here, both $u = u(t)$ and $\varphi = \varphi(t)$ are functions of the independent variable, denoted $t$. Also, the prime notation here denotes differentiation with respect to $t$. So you ought to try evaluating something like$$\frac{d}{dt} \left( u \varphi' \right) = \frac{d}{dt} \left( \cos{\varphi} \right)$$remembering to use the chain rule. If you prefer then you could completely equivalently instead start with $u = \cos{\varphi} / \varphi'$ and then differentiate both sides with respect to $t$.

 

[Murilo T]

$$\frac{d}{dt} \left( u \varphi' \right) = \frac{d}{dt} \left( \cos{\varphi} \right)$$

Is it safe to say that this equals u'φ' + uφ'' = -sin(φ) ?

 

[etotheipi]

Is it safe to say that this equals u'φ' + uφ'' = -sin(φ) ?

Oooh, so close! Almost perfect except you just forgot to apply the chain rule to the right hand side. Remember that $$\frac{d(\cos \varphi)}{dt} = \frac{d(\cos \varphi)}{d\varphi} \frac{d\varphi}{dt} =\frac{d(\cos \varphi)}{d\varphi} \varphi'$$so with that in mind, can you correct your equation? Once you've done that, have a go at eliminating $u$ by substituting your original two equations into this one!

 

[Murilo T]

$$\frac{d(\cos \varphi)}{dt} = \frac{d(\cos \varphi)}{d\varphi} \frac{d\varphi}{dt} =\frac{d(\cos \varphi)}{d\varphi} \varphi'$$

Yees, I realized when I substituted u and u', but it was missing a φ'!

Thank you very very much again! Hahhaha

 

[etotheipi]

Haha it's okay. Did you get a differential equation in $\varphi$ only, and start to think about what the best way of solving that is?

 

[Murilo T]

Yep. I can integrate to get a equation with only φ' and φ. With the initial conditions I can solve it now :D

 

[etotheipi]

Cool, yeah it turns out to be quite nice, you can turn it back into a first order differential equation with a bit of further manipulation:$$\begin{align*}

(\sin \varphi - 1) \varphi'' + \frac{\varphi''}{\varphi'} \cos \varphi &= - \varphi' \sin \varphi \\

\frac{\varphi''}{\varphi'} &= (\sec \varphi - 2 \tan \varphi) \varphi' \\

\frac{d}{dt} \left( \ln{\varphi'} \right) &= \frac{d}{dt} \left( \ln(\sec \varphi + \tan \varphi) - 2 \ln \sec \varphi \right) \\ \\

\varphi'&= c(\sec \varphi + \tan \varphi)(\sec^{-2} \varphi) \\

\varphi'&=c \cos \varphi (1+ \sin \varphi)

\end{align*}$$