Exact or Numerical solutions of a system of differential equatios

Question

I need to solve the following system of differential equations:

\begin{align*} x' + 3a \sqrt{ x+ y}x &= -b \sqrt{xy} \\\\ y' + 3c \sqrt{x+y}y &= b \sqrt{xy} \end{align*}

where $a$, $b$, $c$ are constant.

I appreciate any help.

Kaveh

Answer 1

This isn't much of an answer, but I thought that I'd put down a few observations here that you may find helpful.

First of all, if $b=0$, then you have a first integral, in that the ratio $x^{3c}/y^{3a}$ is constant, so you can use this to eliminate, say, $y$ from the first equation, and then the result becomes an autonomous first order equation for $x$ as a function of $t$, which can, in theory, be integrated by quadrature, so let's set this case aside.

When $b\not=0$, you can scale $t$ (the independent variable) so as to make $b=1$, which I'll assume from now on. Also, I'll replace your $3a$ and $3c$ by $a$ and $c$, respectively, because I don't see what good the $3$s do; they just clutter the formulae.

If you have been trying to do numerics, you may have run into a bit of trouble because the equations as you have written them aren't differentiable, which leads to instabilities in the numerics and nonuniqueness in the solutions. One way to deal with this is to 'resolve' the singularities by 'unfolding' the domain. Your equations require that $x{+}y$ and $xy$ both be nonnegative, which can only happen when $x$ and $y$ are nonnegative. The solutions that have either $x$ or $y$ vanishing identically are easy to find, so set those aside. To see what happens to the other solutions, make the substitution $(x,y) = \bigl(4u^2v^2,(u^2-v^2)^2\bigr)$. The differential equations then become polynomial: $$ \begin{align} 4u' &= \phantom{-}v -av(2uv) -cu(u^2{-}v^2),\\\\ 4v' &= -u -au(2uv) +cv(u^2{-}v^2). \end{align} \tag1 $$ (A word of caution: In this parameterization, $\sqrt{xy}=2uv(u^2-v^2)$, which changes sign as one crosses the lines $u=0$, $v=0$, $u=v$ and $u=-v$, so you have to interpret $\sqrt{xy}$ as a signed quantity in your original equations. Of course, $\sqrt{x{+}y}=u^2+v^2$, so it does not change sign.)

The linear terms on the right hand side of $(1)$ describe the rotation vector field in the $uv$-plane, so the solutions near the origin are either a center or a spiral sink or source. Hence, the corresponding curves in the $xy$-plane 'bounce' between $x=0$ and $y=0$ an infinite number of times.)

In fact, $$ \frac{2(u^2+v^2)'}{(u^2{+}v^2)^2} = -c-(a{-}c)\frac{4u^2v^2}{(u^2{+}v^2)^2}, $$ and the right hand side of this expression varies between $-c$ and $-a$, so, unless $a$ and $c$ have opposite signs, a nonzero solution curve always spirals towards (when $a,c>0$) or away from (when $a,c<0$) the origin.

It's also useful to look at things in polar coordinates. Set $$ x=\tfrac12 r^2 (1+\cos\phi)\qquad\text{and}\qquad y = \tfrac12 r^2 (1-\cos\phi). $$ Then, interpreting $\sqrt{xy}$ as $\tfrac12 r^2\sin\phi$ and $\sqrt{x{+}y}$ as $r$, one computes that $$ \begin{align} r' &= -\tfrac14\ r^2\bigl( (a{+}c)+(a{-}c)\cos\phi\bigr),\\\\ \phi' &= b + \tfrac12(a{-}c)\ r\sin\phi. \end{align} \tag2 $$ When $a$ and $c$ are positive, this shows that $r$ is strictly decreasing. In fact, when $r_0>0$ is small, one has $\phi\approx bt$, so integrating in the first equation gives $$ r \approx \left(\frac1{r_0} + \frac{(a{+}c)}4\ t + \frac{(a{-}c)}{4b}\ \sin bt\right)^{-1}. $$ This gives a reasonably good approximation to the solutions when $r_0 = \sqrt{x_0{+}y_0}$ is small.

Meanwhile, if $a < 0 < c$ and $\sqrt{xy}$ is interpreted to be positive (respectively, when $ c < 0 < a$ and $\sqrt{xy}$ is interpreted to be negative), then there is a fixed point in the first quadrant of the $xy$-plane at $$ (x,y) = \left(\frac1{a(a{-}c)},\frac1{c(c{-}a)}\right), $$ and this gives rise to fixed points in the $uv$-plane, somewhat symmetrically arranged around the origin. Stability analysis of these fixed points will probably tell you something useful. They appear to be saddles (I haven't checked this for sure), but I have no idea where the separatrices go.