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
<br /> v&#039; = \frac{q_2}{f}v^2 + \left(q_1 + \frac{f&#039;}{f}\right)v + fq_0<br />
and we can make the coefficient of v^2 equal to 1 by taking f = q_2, so that we have
<br /> v&#039; = v^2 + Rv + S<br />
where R = q_1 + q_2&#039;/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
<br /> g&#039;h + h&#039;g = g^2h^2 + Rgh + S.<br />
Now we must eliminate the g^2h^2 term by setting it equal to either g&#039;h or h&#039;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
<br /> h&#039;g = g^2h^2 \Rightarrow h&#039; = gh^2 \Rightarrow g = \frac{h&#039;}{h^2} = -\left(\frac1h\right)&#039;.<br />
The natural thing to do now is to set h = -1/u, so that g = u&#039; and v = -u&#039;/u. Substituting those back we find
<br /> g&#039;h = Rgh + S \Rightarrow -u&#039;&#039;/u = -Ru&#039;/u + S<br />
and thus
<br /> u&#039;&#039; - Ru&#039; + Su = 0<br />
which is linear.