\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)
(%i1) kill(all)$

part 3 : solve Ricatti equation
[email protected],31/01/2017

(%i1) batch(solve_rec)$
\[\mbox{}\\\mbox{read and interpret file: C:\ensuremath{\backslash}maxima-5.38.1\ensuremath{\backslash}share\ensuremath{\backslash}maxima\ensuremath{\backslash}5.38.1\_ 5\_ gdf93b7b\_ dirty\ensuremath{\backslash}share\ensuremath{\backslash}solve\_ rec\ensuremath{\backslash}solve\_ rec.mac}\mbox{}\\\mbox{(\%{}i2) eval\_ when(batch,ttyoff:true,nolabels:true)}\]

Ricatti equation :

(%i71) eq1:x[i+1]*x[i]+A*x[i+1]+B*x[i]+C = 0;
\[\tag{eq1}\label{eq1}{{x}_{i}}\,{{x}_{i+1}}+A\,{{x}_{i+1}}+B\,{{x}_{i}}+C=0\]

use : translational subst. remove 'Constant term'

(%i73) p1:x[i]=z[i]+delta;
p2:x[i+1]=z[i+1]+delta;
\[\tag{p1}\label{p1}{{x}_{i}}={{z}_{i}}+\delta\] \[\tag{p2}\label{p2}{{x}_{i+1}}={{z}_{i+1}}+\delta\]
(%i75) eq2:subst(rhs(p1),lhs(p1),eq1);
eq3:factor(subst(rhs(p2),lhs(p2),eq2));
\[\tag{eq2}\label{eq2}\left( {{z}_{i}}+\delta\right) \,{{x}_{i+1}}+A\,{{x}_{i+1}}+B\,\left( {{z}_{i}}+\delta\right) +C=0\] \[\tag{eq3}\label{eq3}{{z}_{i}}\,{{z}_{i+1}}+\delta{{z}_{i+1}}+A\,{{z}_{i+1}}+\delta{{z}_{i}}+B\,{{z}_{i}}+{{\delta}^{2}}+B\delta+A\delta+C=0\]
(%i77) eq4:ev(eq3,z[i+1]=0,z[i]=0);
eq5: factor(lhs(eq3)-lhs(eq4)=0);
\[\tag{eq4}\label{eq4}{{\delta}^{2}}+B\delta+A\delta+C=0\] \[\tag{eq5}\label{eq5}{{z}_{i}}\,{{z}_{i+1}}+\delta{{z}_{i+1}}+A\,{{z}_{i+1}}+\delta{{z}_{i}}+B\,{{z}_{i}}=0\]
(%i81) eq6:t=z[i+1]*z[i];
eq7:ratsubst(lhs(eq6),rhs(eq6),eq5);
eq8:coeff(lhs(eq7),z[i+1]);
eq9:coeff(lhs(eq7),z[i]);
\[\tag{eq6}\label{eq6}t={{z}_{i}}\,{{z}_{i+1}}\] \[\tag{eq7}\label{eq7}t+\left( \delta+A\right) \,{{z}_{i+1}}+\left( \delta+B\right) \,{{z}_{i}}=0\] \[\tag{eq8}\label{eq8}\delta+A\] \[\tag{eq9}\label{eq9}\delta+B\]
(%i82) eq10:(lhs(eq7)-t)/t = -t/t;
\[\tag{eq10}\label{eq10}\frac{\left( \delta+A\right) \,{{z}_{i+1}}+\left( \delta+B\right) \,{{z}_{i}}}{t}=-1\]
(%i83) eq11:ratsubst(rhs(eq6),lhs(eq6),lhs(eq10))=rhs(eq10);
\[\tag{eq11}\label{eq11}\frac{\left( \delta+A\right) \,{{z}_{i+1}}+\left( \delta+B\right) \,{{z}_{i}}}{{{z}_{i}}\,{{z}_{i+1}}}=-1\]
(%i86) eq12:part(lhs(eq11),1,1)/rhs(eq6);
eq13:part(lhs(eq11),1,2)/rhs(eq6);
eq14:part(lhs(eq11),2);
\[\tag{eq12}\label{eq12}\frac{\delta+A}{{{z}_{i}}}\] \[\tag{eq13}\label{eq13}\frac{\delta+B}{{{z}_{i+1}}}\] \[\tag{eq14}\label{eq14}{{z}_{i}}\,{{z}_{i+1}}\]
(%i88) eq15:subst(v[i],1/z[i],eq12);
eq16:subst(v[i+1],1/z[i+1],eq13);
\[\tag{eq15}\label{eq15}\left( \delta+A\right) \,{{v}_{i}}\] \[\tag{eq16}\label{eq16}\left( \delta+B\right) \,{{v}_{i+1}}\]
(%i89) eq17:eq15+eq16=rhs(eq11);
\[\tag{eq17}\label{eq17}\left( \delta+B\right) \,{{v}_{i+1}}+\left( \delta+A\right) \,{{v}_{i}}=-1\]

build in method: solve_rec

solution = general+particular

(%i90) eq18:solve_rec(eq17,v[i]);
\[\tag{eq18}\label{eq18}{{v}_{i}}={{\mathit{\%{}k}}_{1}}\,{{\left( -\frac{\delta}{\delta+B}-\frac{A}{\delta+B}\right) }^{i}}+\frac{{{\left( -\frac{\delta+A}{\delta+B}\right) }^{i}}}{2\delta+B+A}-\frac{1}{2\delta+B+A}\]

second difference equation ,nonhomgeous

(%i92) eq19:subst(E*v[i],v[i+1],eq17);
eq20:factor(eq19);
\[\tag{eq19}\label{eq19}E\,\left( \delta+B\right) \,{{v}_{i}}+\left( \delta+A\right) \,{{v}_{i}}=-1\] \[\tag{eq20}\label{eq20}\left( E\delta+\delta+BE+A\right) \,{{v}_{i}}=-1\]

other way to solve ⇒

solution = general+particular

general

(%i96) eq21:part(eq20,1,1);
eq22:coeff(eq21,E);
eq23:ev(eq21,E=0);
eqq:rr = - eq22/eq23;
\[\tag{eq21}\label{eq21}E\delta+\delta+BE+A\] \[\tag{eq22}\label{eq22}\delta+B\] \[\tag{eq23}\label{eq23}\delta+A\] \[\tag{eqq}\label{eqq}\mathit{rr}=\frac{-\delta-B}{\delta+A}\]

particular:

v[i+1]*a1+v[i]*a2=f(i)
f(i) = -1 = -1 * 1^i, E=1,invers operator,found from eq20
rem : f(i) = -1*E^i,E=1

(%i98) eq24:solve(eq20,v[i])[1];
eq25:subst(1,E,eq24);
\[\tag{eq24}\label{eq24}{{v}_{i}}=-\frac{1}{\left( E+1\right) \delta+BE+A}\] \[\tag{eq25}\label{eq25}{{v}_{i}}=-\frac{1}{2\delta+B+A}\]
(%i99) gen1:v[i]=constant*(rhs(eqq))^i+rhs(eq25);
\[\tag{gen1}\label{gen1}{{v}_{i}}=\mathit{constant}\,{{\left( \frac{-\delta-B}{\delta+A}\right) }^{i}}-\frac{1}{2\delta+B+A}\]

general solution : x[i]

(%i100) gen:1/(x[i]-delta)=rhs(gen1);
\[\tag{gen}\label{gen}\frac{1}{{{x}_{i}}-\delta}=\mathit{constant}\,{{\left( \frac{-\delta-B}{\delta+A}\right) }^{i}}-\frac{1}{2\delta+B+A}\]
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