How to solve a system of 4 coupled non-linear differential equations?

Question

The system of equations I have is: $$\frac{dx_1}{dt} = k_2 \left( \frac{y_2 - y_1}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$ $$\frac{dy_1}{dt} = k_2 \left( \frac{x_1 - x_2}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$

$$\frac{dx_2}{dt} = k_1 \left( \frac{y_1 - y_2}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$ $$\frac{dy_2}{dt} = k_1 \left( \frac{x_2 - x_1}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$

where $k_1,k_2$ are constants. How would I go about solving this for $x_1(t),y_1(t),x_2(t),y_2(t)$?

For some context, these differentials equations determine how 2 vortex velocity fields of arbitrary circulation ($k_1, k_2$) interact with each other. The x-y time derivatives represent the component-wise speed of each vortex due to the other vortex.

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My attempt:

So far, I multiplied the first and second equation by $(x_1−x_2)$ and $(y_1−y_2)$ and added them to get $(x_1−x_2)(dx_1/dt)+(y_1−y_2)(dy_1/dt)=0$. Doing the same thing for the third and fourth equations, I got $(x_1−x_2)(dx_2/dt)+(y_1−y_2)(dy_2/dt)=0$. Don't know where to go from here.

Answer 1

Try to substitute $\Delta=x_1-x_2$ and $\Psi=y_1-y_2$. Then substract equation $1$ and equation $3$. And also equation $2$ and equation $4$. With two differential equations it will be easier to solve. $$\Delta'= \dfrac {A\Psi}{\Psi^2+\Delta^2}$$ $$\Psi'= \dfrac {B\Delta}{\Psi^2+\Delta^2}$$ Where $A=-(k_1+k_2),B=k_1+k_2$.

Then try to solve: $$\dfrac {d \Delta}{d\Psi}=-\dfrac { \Psi}{ \Delta}$$

Answer 2

Your system can be written more compact in complex numbers as $$ \dot z_1=ik_1\frac{z_1-z_2}{|z_1-z_2|^2}\\ \dot z_2=ik_2\frac{z_2-z_1}{|z_1-z_2|^2}\\ $$ It follows immediately that $k_2z_1+k_1z_2=const.$ As the right side only depends on differences, the system is translation invariant, thus one can shift it so that the constant is zero. Now set $z_1=k_1w$, $z_2=-k_2w$ so that the dynamic reduces to one equation $$ \dot w=i\frac{w}{(k_1-k_2)|w|^2} $$ Write in polar coordinates $w(t)=r(t)e^{i\varphi(t)}$ so that $$ \dot r+ir\dot φ=\frac{i}{r(k_1-k_2)} $$ which implies $\dot r=0$, the motion of $w$ is circular, and $\dot\phi=\frac1{r^2(k_1-k_2)}$ is the constant angular velocity.

From here on one can substitute back to find the full solution.