Attempts to solve Riccati recurrence equations using various substitutions.

A Riccati recurrence equation in y(k) is one of the form $y\left(k+1\right)y\left(k\right)+A\left(k\right)y\left(k+1\right)+B\left(k\right)y\left(k\right)=C\left(k\right)$ where A(k), B(k), and C(k) are independent of y(k).  If the equation is homogeneous ($C\left(k\right)=0$), then we try the substitution $x\left(k\right)=\frac{1}{y\left(k\right)}$, which makes the equation first order linear and if not, then we try $y\left(k\right)=\frac{x\left(k\right)-B\left(k\right)x\left(k+1\right)}{x\left(k+1\right)}$, which makes the equation second order linear. Finally, there is the substitution $y\left(k\right)=\frac{x\left(k+1\right)-A\left(k+1\right)x\left(k\right)}{x\left(k\right)}$, which makes the equation second order linear. If rsolve can solve these new equations, then we back-substitute to obtain solutions to the original problems.

If A(k) is undefined for some k, then a set of equations may be returned, giving values of y(k) for specific k as well as the general formula.

Since it calls rsolve, this procedure can be expensive; because of the back-substitution, the answers may be overly complicated.

See the help page for LREtools[REcreate] for the definition of the format of a problem.

$\mathrm{with}\left(\mathrm{LREtools}\right)\:$

$\mathrm{prob}≔\mathrm{REcreate}\left(y\left(k+1\right)y\left(k\right)+2y\left(k+1\right)-3y\left(k\right)=1\,y\left(k\right)\,\left\{y\left(0\right)=\frac{1}{2}\right\}\right)$

$\mathrm{prob}≔\mathrm{RESol}\left(\left\{\left(y\left(k+1\right)-3\right)y\left(k\right)+2y\left(k+1\right)=1\right\}\,\left\{y\left(k\right)\right\}\,\left\{y\left(0\right)=\frac{1}{2}\right\}\,\mathrm{INFO}\right)$

$\mathrm{riccati}\left(\mathrm{prob}\right)$

$\frac{{\left(-5-\sqrt{5}\right)}^k\sqrt{5}-{\left(-5+\sqrt{5}\right)}^k\sqrt{5}+{\left(-5-\sqrt{5}\right)}^k+{\left(-5+\sqrt{5}\right)}^k}{2\left({\left(-5+\sqrt{5}\right)}^k+{\left(-5-\sqrt{5}\right)}^k\right)}$

$\mathrm{riccati}\left(y\left(k+1\right)y\left(k\right)+2y\left(k+1\right)-ky\left(k\right)=0\,y\left(k\right)\,\varnothing \right)$

${\mathrm{charfcn}}_0\left(k\right)y\left(0\right)$