# Nonlinear first order system of ODEs

While solving some physical problem, I have obtained the following system of differential equations with boundary conditions:

$$\left\{\begin{matrix} \frac{d\phi_1}{dz}=\frac{m^2}{\lambda}- \lambda\phi_1^2-\alpha\phi_2^2 \\ \frac{d\phi_2}{dz}=-2\alpha\phi_1\phi_2 \\ \phi_2(\pm\infty)=0 \\ \phi_1(\pm\infty)=\pm\frac{m}{\lambda} \end{matrix}\right.$$

where $\phi_1(z),\phi_2(z)$ are just functions of real variables, $m,\lambda,\alpha\in \mathbb{R}$

As far as I know, to solve this problem I should solve the system of differential equations and after that use boundary conditions.

I see that it's easy to solve this system when $\lambda=\alpha$:

Just sum this two equations to obtain $$\frac{d(\phi_1+\phi_2)}{dz}=\frac{m^2}{\lambda}-\lambda(\phi_1+\phi_2)$$ This DE easily solved $$(\phi_1+\phi_2)=\frac{m}{\lambda}\tanh(mz-m\lambda C_1)$$ where $C_1$ is an integration constant. After that I can find exact solutions for $\phi_1$ and $\phi_2$. And boundary conditions are satisfied automaticaly.

But I would like to obtain the solution for any $\alpha$ and $\lambda$ at least by quadrature.

My attempt was to reproduce the aproach as in case $\lambda=\alpha$:

I've obtained

$$\frac{1}{\sqrt{\alpha}}\frac{(\sqrt{\alpha}\phi_1+\lambda\phi_2)}{dz}=\frac{m^2}{\lambda}-(\sqrt{\lambda}\phi_1+\sqrt{\alpha}\phi_2)^2$$

So, this attempt was a fail.

Also I have noticed that these two equations look like as Riccati equation

And I have no any ideas how to solve it. Any help will be appreciated.

## 1 Answer

It is possible to transform the system of non-linear ODEs so that a first order linear ODE is obtained (in attachment). This linear ODE is analytically solvable.

However, in order to fully solve the problem, the further calculus are arduous. The theoretical process is skeched in attachment, but not realistic in practice because the integration and inversion of function cannot probably be carried out.

That is why a numerical method of solving is certainly more convenient. 