straight-line solutions of dynamical system

Question

Given the dynamical system

$\dot x = f(x,y)$ and $\dot y = g(x,y)$

where f and g are homogeneous of degree n (i.e. $G(\alpha x,\alpha y)=\alpha^n G(x,y))$

show that straight line solutions of the form $y(t)=mx(t)$ exist for the system if and only if

$mf(1,m) = g(1,m)$

Starting with a simple example

$\dot x = 2x+y$ and $\dot y = x+y$

Both functions are homogeneous of degree 1 since

$f(\alpha x, \alpha y) = 2\alpha x + \alpha y = \alpha (2x+y) = \alpha f(x,y) $

and

$g(\alpha x, \alpha y) = \alpha x + \alpha y = \alpha (x+y) = \alpha g(x,y) $

If you want straight line solutions, then:

$mf(1,m) = g(1,m)$

$m(2+m) = 1+m$

$m=0.618, -1.618$

Phase portrait of the system showing nullclines and the two straight line solutions y = 0.618x and y=-1.618x

I see how the solutions are generated but how can I go about expressing this more formally for any function that is homogeneous of degree n

Answer 1

There are solutions satisfying $y(t)=mx(t)$ if and only if $$ g(x,mx)=mf(x,mx) $$ for all $x$. This is because solutions of that form exists if and only if the vector field along the corresponding line has the same direction. Being homogeneous of degree $n$ gives you the equation $$ g(1,m)=mf(1,m) $$ after canceling out $x^n$.

Answer 2

If a line through the origin $$t\mapsto p(t)(a,b)$$ is a solution of the given system then $$\eqalign{\dot x(t)=a\dot p(t)&=f\bigl(p(t)(a,b)\bigr)=p^n(t)F(a,b)\cr \dot y(t)=b\dot p(t)&=g\bigl(p(t)(a,b)\bigr)=p^n(t)G(a,b)\cr}$$ for all $t$, hence $$bF(a,b)-aG(a,b)=0\ .$$