How does this differential equation define an oscillation from a to b?

Question

The differential equation reads: $$ \dot{y}^2+(y^2-a^2)(y^2-b^2)=0 $$

where $a,b\in \mathcal{R}$ and $a<b$.

It is claimed that this differential equation defines an oscillation from $a$ to $b$.

So I did the research: differentiate w.r.t. $x$. Then I get:

$$ \ddot{y}= (a^2+b^2)y-2y^3 $$

Then I looked up in MathWorld, compared with equation(20), I get $k=\sqrt{2-(a^2+b^2)}$.

So I plot the graph using JacobiCN[x,k] in mathematica, it seems not right.

For example:

I know we need the initial value to get the behavior, since the initial condition is not clear in the article. I suppose it would be $y(0)=a$, can this initial condition plus that differential equation justify the claim stated in the title?

ps: I have little background on elliptic functions so it would be nice if you provide an answer with reference to help me understand.

I wonder anyone can get the following function: $$ y=b \mathrm{dn}(x,q) \text{ with } q=1-a^2/b^2 $$ for this function defines an oscillation from $a$ to $b$.

Answer 1

In the OP, there is an erroneous factor of $2$ appearing in the linear term on $y$. To arrive at the coveted result, we begin with the ODE

$$\dot y^2(x)=-(y^2(x)-a^2)(y^2(x)-b^2) \tag 1$$

Differentiating both sides of $(1)$ with respect to $x$ yields

$$2\dot y\ddot y=-2y\dot y(y^2-a^2)-2y\dot y(y^2-b^2) \tag 2$$

Now, dividing both side of $(2)$ by $2\dot y$ we obtain

$$\ddot y=-y(y^2-a^2)-y(y^2-b^2)=-y(2y^2-a^2-b^2)=y(a^2+b^2)-2y^3$$

Can you proceed from here?

Answer 2

At $y=a$ and $y=b$ the derivative of $y$ is zero according to the diff eq.

For $a \lt y \lt b$, the second summand in the diff eq is negative while for $y \lt a$ or $b \lt y$ it is positive.

Assuming a real valued solution, $y\lt a$ is not possible because we would not get zero, i.e. the diff eq is not satisfied. Likewise $y \gt b$ is not possible. Two positive values cannot add up to zero.

But $a \lt y \lt b$ is possible and the derivative can then be either positive or negative (depending whether $y^{-1}(a)$ is smaller or larger than $y^{-1}(b)$). Hence, $y$ is monotonic.

Once $a$ or $b$ is reached, the only way to satisfy the diff eq is that $y$ remains constant.

Conclude by the definition of an oscillation.