The Euler-Lagrange first integral

Question

How to calculate $\frac{d\phi}{dz}$ from following equation: $$A\frac{d^2\phi}{dz^2}+\cos(\phi)\sin(\phi)=0,$$ where $A$ is a constant?

Answer 1

$$A\frac{d^2\phi}{dz^2}+\cos(\phi)\sin(\phi)=0,$$ Substitute $$p=\frac {d \phi}{dz} \implies \frac{d^2\phi}{dz^2}=p\frac {dp} {d \phi} $$ $$App'=-\cos(\phi)\sin(\phi)$$ $$A\int pdp=-\frac 12\int \sin(2\phi) d\phi$$ $$Ap^2= \frac 12\cos (2\phi) +K_1$$ $$\frac {d \phi}{dz} =\pm \sqrt {\frac {\cos (2\phi) +K_1} {2A}}$$ $$.......$$

@tgram $$Ap^2= \frac 12\cos (2\phi) +K_1$$ note that $$\cos (2\phi)=2\cos^2 \phi -1=2(1-\sin^2 \phi )-1=1-2\sin^2 \phi$$ The constant $K_1$ absorb all the constants...

And $$p=\frac {d \phi}{dz}$$

Answer 2

Multiply both sides by $\frac{d\phi}{dz}$ and then integrate (using the chain rule to help...).

Answer 3

Since $2A\phi'\phi''+\phi'\sin 2\phi=0$, $C:=A\phi'^2-\frac{1}{2}\cos 2\phi$ is constant. In particular $z=\int\dfrac{d\phi}{\sqrt{(C+\frac{1}{2}\cos 2\phi)/A}}$, which you can reduce to an elliptic integral of the first kind. Inverting this gives $\phi$ in terms of $z$, or if you just want $\phi'=(dz/d\phi)^{-1}$ in terms of $\phi$ no elliptic integrals are needed.

Answer 4

$$ A\phi'\phi'' +\cos\phi\sin\phi\phi' = 0\Rightarrow \frac 12A(\phi'^2)'+\frac 12(\sin^2\phi)' = 0 $$

hence

$$ A\phi'^2+\sin^2\phi = C_1\Rightarrow \phi' = \pm\frac{1}{\sqrt{A}}\sqrt{C_1-\sin^2\phi} $$