Solve $ny(x)^2=\sqrt{1+y'(x)^2}$ and determine the range of $x$ where $y(x)$ is real-valued

Question

I have the following differential equation:

$$ny(x)^2=\sqrt{1+y'(x)^2}$$

I know that $n$ is a real number, and that the intial condition is $y(a)=b$, where $a$ and $b$ are also real numbers.

The questions I have are:

1. What is the function $y(x)$?
2. In what range of $x$ is the function $y(x)$ real-valued, and how does that range depend on $n$, $a$, and $b$?

Answer 1

HINT

To start with, notice that \begin{align*} ny^{2} = \sqrt{1+(y')^{2}} & \Longleftrightarrow y' = \pm\sqrt{n^{2}y^{4} - 1} \Longleftrightarrow \int\frac{\mathrm{d}y}{\sqrt{n^{2}y^{4} - 1}} = \pm\int 1\mathrm{d}x \end{align*}

Then WA gives the following result. Hopefully this helps.

Answer 2

Starting from @user1337's answer, the result can simplify as $$x(y)=C \pm \frac{i}{\sqrt{n}} F\left(\left.\sin ^{-1}\left(y\sqrt{n} \right)\right|-1\right)$$ where appears the elliptic integral of the first kind.

Now, compute $C$ from the condition $x(b)=a$.

Answer 3

Given

$$ny^{2} = \sqrt{1+(y')^{2}} \tag1$$

Instead of building variability of profiles through $n,\;$ chosen minimum radius $r_1$ instead. $ n=1/r_1^2 $ for convenience.

Using symbols $(z,r)$ in cylindrical coordinates instead of $(x,y)$. $\tan\phi=\frac{dy}{dx}$.

$$\sec\phi=\dfrac{r^2}{r_1^2} \tag2 $$

Integration gives profiles using elliptic integrals of the first kind as mentioned by others, the numerical solution as follows.

ODE for Shell:

Differentiating with respect to arc length $s$ since $ \frac{dr}{ds}=\sin \phi $

$$\frac{d\phi}{ds}=\dfrac{ 2 r \cos^2 \phi} {r_1^2 }\tag3$$

$$ r=\int \sin \phi\; ds, z=\int \cos \phi\; ds,\; BC: (s=0,r=r_1) \tag4$$

Second order ODE

$$r_1^2 \phi^{''}(s)+6\sin \phi(s) \cos^2\phi(s)=0 \tag5 $$

Profiles obtained:

Unbounded domain and range $r> r_1$

Curvature Relation:

Curvature Ratio of the two shell curvatures $(R_1,R_2)$ by definition

$$\dfrac{R_2}{R_1}= \dfrac{k_1}{k_2}= \dfrac{\phi^{'}}{\cos \phi/r} \tag6 $$

Plug in from (2) and (3)

$$\boxed {\dfrac{R_2}{R_1} = \dfrac{k_1}{k_2}=2 } \tag7 $$

is a very interesting geometry property of the curve. The minor shell curvature is half the major shell curvature i.e., the minor shell radius of curvature $R_2$ is double the major osculating shell radius of curvature $R_1$. This is drawn in the sketch below computed for $r_1=4$.

The differential equation can be put in a simpler $y=f(x)$ form.

Derivation of DE for $y=f(x)$

The two principal shell curvatures are:

$$ k1= \phi^{'} =\dfrac {\dfrac {d^2y}{dx^2}}{(1+ (dy/dx)^2)^{3/2}}$$

$$ k2= \frac{\cos \phi}{y} =\dfrac{1}{y\sqrt{1+(dy/dx)^2 }} \tag 8 $$

Dividing and using curvature ratio formula in (7)

$$ y\cdot \dfrac{d^2y}{dx^2} = 2 [1+(dy/dx)^2] \tag9 $$

which can be integrated with boundary conditions $(x=0, y=r_1, dy/dx =0),\;$ leading exactly to the same results.