Find new first integral of DE System

Question

I am trying to find two first integrals of the following system: $$\begin{cases}x' = x+z\\ y'= y \\ z'=z + y^2 \end{cases}$$ A first integral can be found considering only the system in $\mathbb{R}^2$ for $y$ and $z$. Using the integrating factors $\mu_1 = \frac{1}{y^2}$ and $\mu_2 = \frac{1}{(z-y^2)^2}$ we can obtain the function $$H_1(y,z) = \sqrt{\frac{\mu_1}{\mu_2}}=\frac{z}{y} -y$$ which is a first integral of the $y,z$ system, hence a first integral of the system in $\mathbb{R}^3$.

Now I'm trying to get another first integral, functionally independent to $H_1$, but can't really find a way to do it unless I use PDE's, which I want to avoid.

Any help on how to get it will be appreciated! Thanks.

EDIT: by first integral I refer to a function $H(x,y,z)$ such that $$H' = \left<\nabla H, (x',y',z') \right> = 0.$$

Answer 1

Finally found another first integral. Since this is an actual answer to the question I made, I though it would be better to post it as an answer instead of editing the original question. Here it goes:

It is important to note that $$\left(\frac{x}{y} \right)' = \frac{x'y-xy'}{y^2} = \frac{(x+z)y -xy }{y^2} = \frac{z}{y} $$ hence, $$\left(\frac{x}{y} - y \right)' = \frac{z}{y} - y = H_1(y,z)$$ And we've seen already that this is a first integral. Again, using $y'=y$ we have that $y'' = 1$. Therefore: $$\left(H_1(y,z)\ln|y|\right)' = H_1'\ln|y| + H_1(z,y)\frac{1}{y}y = H_1(z,y) = \frac{z}{y} - y$$ since $H_1' = 0$. Thus we shall consider $$H_2(x,y,z) = \frac{x}{y} - y - \left(\frac{z}{y} - y \right)\ln|y|$$ as another first integral, clearly functionally independent to the previous one -it even has a different variable-.

Just as @Mefitico said, I don't know any systematic ways to actually finding such functions. It's all "trial and error", or playing with the system's functions.

Answer 2

I'm not aware of any systematic method for finding SODE first integrals. I can make a suggestion here if it helps you:

Note that the expression you gave for the first integral :

$$ \langle \nabla H ,(\dot{x},\dot{y},\dot{z})\rangle $$

Can be rewritten as:

$$ [H_x,H_y,H_z] \left[ \begin{array}{ccc} 1 && 0 && 1 \\ 0 && 1 && 0 \\ 0 && 1 && y \\ \end{array} \right] \left[\begin{array}{c} x \\ y \\ z \\ \end{array} \right] $$

Note the transpose of the matrix in the middle: $$ \left[ \begin{array}{ccc} 1 && 0 && 0 \\ 0 && 1 && 1 \\ 1 && 0 && y \\ \end{array} \right] $$

The determinant is $y$, so basically we can look for solutions of the following forms:

$$ \left[\begin{array}{c} H_x \\ H_y \\ H_z \\ \end{array} \right] = \left[ \begin{array}{ccc} 1 && 0 && 0 \\ 1/y && 1 && -1/y \\ -1/y && 0 && 1/y \\ \end{array} \right] v(x,y,z) $$

With $v$ orthogonal to $(x,y,z)$. Examples may be: $v=[-y,x,0]^T$, $v=[0,-z,y]^T$,$v=[-z,0,x]^T$, and any of those times a function $g(x,y,z)$.

We also need that a function $H$ may be found with this gradient. Actually, if assuming that: $$ \left[\begin{array}{c} H_x \\ H_y \\ H_z \\ \end{array} \right] = \nabla \phi $$

Then by Helmholtz decomposition is is possible to find H from a guess $v(x,y,z)$. Not sure if the results found will actually satisfy you.