Analytical solution to coupled nonlinear ODEs

Question

I am looking to solve several coupled nonlinear ODEs like this one:

$\hspace{20mm} \frac{d x(t)}{dt} = C_1 \cdot x(t) + C_2 \cdot y(t) + C_3\cdot (x(t)^2 + y(t)^2) x(t),$
$\hspace{20mm} \frac{d y(t)}{dt} = C_1 \cdot y(t) - C_2 \cdot x(t) + C_3\cdot (x(t)^2 + y(t)^2) y(t).$

I have tried multiplying the first by $y(t)$ and the second by $x(t)$ and adding them or subtracting them, but with no luck.

I have noticed that each system is almost symmetrical, is there any general procedure for such systems? If not, how do I solve this system?

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Context

I am looking into Rain-Wind induced vibrations of stay cables. I am trying to approximate the model I have derived using a Multiple Time-Scale Perturbation Analysis. During this analysis, many coupled nonlinear ODEs need to be solved, which are almost symmetrical.

Answer 1

We have two coupled ODEs

$$\dot x = c_1 x + c_2 y + c_3 \left( x^2 + y^2 \right) x$$

$$\dot y = c_1 y - c_2 x + c_3 \left( x^2 + y^2 \right) y$$

Multiplying the first ODE by $x$, the second ODE by $y$, and adding them, we get

$$x \dot x + y \dot y = c_1 \left( x^2 + y^2 \right) + c_3 \left( x^2 + y^2 \right)^2$$

Let $r^2 := x^2 + y^2$. Differentiating, we get $2 r \dot r = 2 x \dot x + 2 y \dot y$. Thus, we have the ODE

$$r \dot r = c_1 r^2 + c_3 r^4$$

Dividing both sides by $r$, we get

$$\dot r = c_1 r + c_3 r^3$$

Integrating,

$$r (t) = \sqrt{\frac{c_1}{\left(c_3 + \frac{c_1}{r_0^2}\right) e^{-2 c_1 t} - c_3}}$$

Going back to the original system of ODEs, multiplying the first ODE by $y$, the second ODE by $x$, and subtracting them, we get

$$\dot x y - x \dot y = c_2 \left( y^2 + x^2 \right)$$

Dividing both sides by $x^2$, we get

$$\frac{x \dot y - \dot x y}{x^2} = - c_2 \left(\left(\frac{y}{x}\right)^2 + 1\right)$$

Note that the left-hand side is the derivative of $\frac{y}{x}$. Let $\xi := \frac{y}{x} = \tan (\theta)$. We then obtain the ODE

$$\dot \xi = - c_2 \left( \xi^2 + 1 \right)$$

Integrating,

$$\arctan (\xi) = \arctan(\xi_0) - c_2 t$$

Since $\theta = \arctan (\xi)$, we have

$$\theta (t) = \theta_0 - c_2 t$$

To summarize, we transform the original system of two coupled nonlinear ODEs into a system of two decoupled polynomial ODEs

$$\begin{aligned} \dot r &= c_1 r + c_3 r^3 \\ \dot \xi &= - c_2 \left( \xi^2 + 1 \right) \end{aligned}$$

and then integrate. Once we have $r$ and $\xi$, we have $r$ and $\theta$. We then use $x = r \cos (\theta)$ and $y = r \sin (\theta)$ to go from polar coordinates to Cartesian coordinates.