How do I extract a particular value in a 'bvp4c' code solver?
9 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have this bvp4c code for the following equations :
f"=g(g^2+gamma^2)/(g^2+lambda*gamma^2) ------ (1)
g'= (1/3)*f'^2-(2/3)*(f*f")+ Mn*f' ------------------------(2)
t"+Rd*t"+ 2*Pr*f*t'/3+ Nb*t'*p'+Nt*(t')^2= 0------(3)
p"+(2*Lew*f*p')/3+ Nt*t"/Nb= 0 ------------------------(4)
With boundary conditions:
f=0, f'=1, t=1, p=1; at η=0
f'=0, t=0, p=0; at η-->infinity
I have used bvp4c solver to solve it numerically
function sol= proj
clc;clf;clear;
global lambda gama Pr Rd Lew Nb Nt Mn m
%Nb=0.5;
gama=1;
Mn=1;
Rd=0.1;
Pr=1;
Nb=0.3;
Lew=10;
%lambda=1.5;
pp=0.5;
qq=[0.05:0.01:0.5];
%figure(1)
%plot(2,1);hold on
options=bvpset('stats','on','RelTol',1e-9);
m=linspace(0,10);
solinit= bvpinit(m,[1,0,0,0,0,0,0]);
for i=1:numel(pp)
lambda=pp(i);
for i=1:numel(qq);
Nt=qq(i)
sol= bvp4c(@projfun,@projbc,solinit,options);
y1=deval(sol,0)
%y2=deval(y(6,:),0)
solinit= sol;
plot(sol.x,sol.y(6,:),'LineWidth',2);hold on
%legend('N_{b} = 0.1','N_{b} = 0.3','N_{b} = 0.5');
end
end
end
function f= projfun(x,y)
global lambda gama Pr Rd Lew Nb Nt Mn
f= [y(2)
y(3)*(y(3)^2+gama^2)/(y(3)^2+lambda*gama^2)
y(2)^2/3-(2*y(1)*y(3)*(y(3)^2+gama^2))/(3*(y(3)^2+lambda*gama^2))+Mn*y(2)
y(5)
-(2*Pr*y(1)*y(5))/(3*(1+Rd)) - (Nb*y(5)*y(7))/(1+Rd) - (Nt*y(5)^2)/(1+Rd)
y(7)
-((2*Lew*y(1)*y(7))/3)+ (Nt/Nb)*((2*Pr*y(1)*y(5))/(3*(1+Rd)) + (Nb*y(5)*y(7))/(1+Rd) + (Nt*y(5)^2)/(1+Rd))];
end
function res= projbc(ya,yb)
res= [ya(1); ya(2)-1; ya(4)-1.0; ya(6)-1.0; yb(2); yb(4); yb(6)];
end
I want to extract the values of y(6,:) particularly
Using 'deval(sol,0)' gives all the values of the 'y' rows
Is there any command to extract the values of y(6,:) or y(4,:) in particular.
0 commentaires
Réponse acceptée
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Numerical Integration and Differential Equations dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!