Linearization of Ricatti equations

[mathnerd15]

In a Riccati equation y'=qo+q1y+q2y^2, if q2 is nonzero then you can make a substitution
v=yq2, S=q2q0, R=q1+(q2'/q2) which satisfies a Riccati, v'=v^2+R(x)v+S(x)=(yq2)'=q0q2+(q1+q2'/q2)v+v^2
with double substitution v=-u'/u, u now satisfies a linear 2nd ODE:
u''-R(x)u'+S(x)u=0, v'=-(u'/u)'=-(u''/u)+v^2, u''/u=v^2-v'=-S+Ru'/u, therefore we have reduced the Ricatti to an equation
u''+Su-Ru'=0, solving this equation will lead to a solution y=-u'/(q2u)

were these substitutions simply invented or is there a reasoning process behind the proof?

 

[bolbteppa]

Apparently you can use Lie theory to understand when a Riccati equation is integrable. That link might help, but that's about all I know at present unfortunately.

 

[pasmith]

In a Riccati equation y'=qo+q1y+q2y^2, if q2 is nonzero then you can make a substitution
v=yq2, S=q2q0, R=q1+(q2'/q2) which satisfies a Riccati, v'=v^2+R(x)v+S(x)=(yq2)'=q0q2+(q1+q2'/q2)v+v^2
with double substitution v=-u'/u, u now satisfies a linear 2nd ODE:
u''-R(x)u'+S(x)u=0, v'=-(u'/u)'=-(u''/u)+v^2, u''/u=v^2-v'=-S+Ru'/u, therefore we have reduced the Ricatti to an equation
u''+Su-Ru'=0, solving this equation will lead to a solution y=-u'/(q2u)

were these substitutions simply invented or is there a reasoning process behind the proof?

Since $q_2 \neq 0$, we can simplify things by changing variables so that $q_2 = 1$.

If we start with $y' = q_0 + q_1 y + q_2 y^2$ and substitute $v = fy$, then we end up with
$$v' = \frac{q_2}{f}v^2 + \left(q_1 + \frac{f'}{f}\right)v + fq_0$$
and we can make the coefficient of $v^2$ equal to 1 by taking $f = q_2$, so that we have
$$v' = v^2 + Rv + S$$
where $R = q_1 + q_2'/q_2$ and $S = q_2q_0$. Now we want to get rid of the $v^2$ entirely, and we might consider the substitution $v = gh$. That yields
$$g'h + h'g = g^2h^2 + Rgh + S.$$
Now we must eliminate the $g^2h^2$ term by setting it equal to either $g'h$ or $h'g$. It is clear from the symmetry that doing one gives the same as doing the other with g and h swapped. So taking the second we have
$$h'g = g^2h^2 \Rightarrow h' = gh^2 \Rightarrow g = \frac{h'}{h^2} = -\left(\frac1h\right)'.$$
The natural thing to do now is to set $h = -1/u$, so that $g = u'$ and $v = -u'/u$. Substituting those back we find
$$g'h = Rgh + S \Rightarrow -u''/u = -Ru'/u + S$$
and thus
$$u'' - Ru' + Su = 0$$
which is linear.