Solutions of : $y' = (y^2 + z^2 +1)^{-a} , z' = y(1+z^2)^a$

Question

Study the existence of solutions that are set entirely on $\mathbb R$ for the functions :

$$ y' = (y^2 + z^2 +1)^{-a}, z' = y(1+z^2)^a $$

I came upon this problem while studying for my Dynamical Systems course, but I'm not sure on how to proceed.

One thing I saw was that we could bound the second equation, such as :

$$z' = (1 +z^2)^a \leq (1+y^2 + z^2)^a = y/y'$$

since $y^2 \geq \forall y\in C(\mathbb R).$

So, we have that the second equation is bound between the solution and the derivative of the first equation, since :

$$z \leq \frac{y}{y'}$$

but I cannot see how to use this in order to study the existence of the equation (it can maybe help on proving Lipschitz conditions for the uniqueness but that's not what I need here).

Answer 1

There are multiple ways to go about this, but probably the easiest is to find a situation where all partial derivatives are uniformly bounded. Use for example Corollary 3.14 from this set of lecture notes:

If $\| D f(w)\|$ is uniformly bounded as $\| w\| \to \infty$, then $f$ is globally Lipschitz.

In this case, you can use the supremum norm on the Jacobian of $f(y,z)$. Then, to study the behaviour of the different partial derivatives for $\|(y,z)\| \to \infty$, you can for example revert to polar coordinates and take the limit of large radius. This will give you an interval of $a$-values where the right hand side of your system is globally Lipschitz, yielding global existence of solutions. (I get $-\frac{1}{2} \leq a \leq 0$; of course, I might have made a mistake somewhere)

Answer 2

Of course I don't know if the following considerations are helpful for your course but at least it can be seen that a solution for $\,y\,$ and $\,z\,$ exists.

With $\enspace\displaystyle y=\frac{z'}{(1+z^2)^a}\enspace $ follows $\enspace\displaystyle y'= \frac{z''}{(1+z^2)^a}- \frac{2az(z')^2}{(1+z^2)^{a+1}} \,$ .

With substituting $\,y'\,$ and $\,y\,$ in $\enspace y'(y^2+z^2+1)=1\enspace$ and the equation solved for $\,z''\,$ we get:

$$ z''= \frac{2az(z')^2}{1+z^2} +\frac{(1+z^2)^{3a}}{((z')^2+(1+z^2)^{2a+1})^a} $$

We see now, that one gets the $\,\text{n}^{th}$ derivation $\,z^{(n)}$ by $\,z^{(k+1)}=(z^{(k)})'\,$ in dependence of $\,z\,$ and $\,z'\,$ .

The denominators are always positive (in $\mathbb{R}$) and therefore exists a Taylor series as a solution

for $\,z\,$ . $\,\,$(And $\,y\,$ can be calculated by $\,z\,$ and $\,z'\,$ because of the initial formula above.)

Example for the first values:

$z(0):=0\,$ , $\,z’(0):=0\enspace$ => $\enspace z’’(0)=1\,$ ; $\,\,y(0)=0\,$ , $\,y’(0)=1$