CONVERSION OF ODE TO RECURRENCE RELATION
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
syms x k r f(x) g(x) a b beta b1 M L
syms F(k) G(k)
F(0)=0;F(1)=1;F(2)=a/2;G(0)=0;G(1)=1/2;G(2)=b/2;b1=1/beta;
%%%%dnf=diff(f,x,n)
f=F(k);g=G(k);d1f=(k+1)*F(k+1);d2f=(k+1)*(k+2)*F(k+2);d3f=(k+1)*(k+2)*(k+3)*F(k+3);d1g=(k+1)*G(k+1);
d2g=(k+1)*(k+2)*G(k+2);d3g=(k+1)*(k+2)*(k+3)*G(k+3);
f*d2f=sum((k-r+1)*(k-r+2)*F(r)*F(k-r+2),r,0,k);g*d2g=sum((k-r+1)*(k-r+2)*G(r)*G(k-r+2),r,0,k);
f*d2g=sum((k-r+1)*(k-r+2)*F(r)*G(k-r+2),r,0,k);g*d2f=sum((k-r+1)*(k-r+2)*G(r)*F(k-r+2),r,0,k);
(d1f)^2=sum((k-r+1)*(r+1)*F(r+1)*F(k-r+1),r,0,k);(d1g)^2=sum((k-r+1)*(r+1)*G(r+1)*G(k-r+1),r,0,k);
eqns=simplify((1+b1)*d3f-(d1f)^2+f*d2f+g*d2f-(M+L)*d1f==0,(1+b1)*d3g-(d1g)^2+f*d2g+g*d2g-(M+L)*d1g==0)
Rsolve(eqns,{F(k+3),G(k+2)});
%% I want a recurrence relation in terms of F(k+3) and G(k+2) (k = 0 -> Inf) but unable to code it properly
1 Kommentar
Antworten (0)
Siehe auch
Kategorien
Mehr zu Ordinary Differential Equations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!