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bvp4c problem with fewer boundary value than variable

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ulan
ulan on 24 May 2024 at 6:52
Commented: Torsten on 24 May 2024 at 17:46
In my problem i have 11 varaible of y but only 9 boundary conditions what can i do here? Another additional problem is equation 3 and 4 both contain θ₂ and φ₂ how do i write y9' and y11' for them? in my code i took one of the constant as zero. I also tried taking some other boundary condition (f'' and f'')=0 to solve it but solution is unstable.
clc;clear;
work=0;
nnn=0;
aa=.5;
bb=10;
cc=30;
name='B';
sol=main(aa,bb,cc,name);
function sol1=main(aa,bb,cc,name)
sol=jobs(aa);
% sol1=jobs(bb);
% sol2=jobs(cc);
figure('Name',name)
hold on
subplot(2,2,1)
plotting(2,"Pimary Velocity(F')",sol,cc)%,sol1,sol2,aa,bb,
subplot(2,2,2)
plotting(5,"Secondary Velocity(G)",sol,cc)
subplot(2,2,3)
plotting(8,"Concentation 1 (\Xi)",sol,cc)
subplot(2,2,4)
plotting(10,"Temparature(\Theta)",sol,cc)
xlim([0 3.2])
hold off
sol1=sol;
end
function plotting(line,titles,sol,aa)%,sol1,sol2,aa,bb,cc
hold on
grid on
a=plot(sol.x,sol.y(line,:),'r');
% b=plot(sol1.x,sol1.y(line,:),'g');
% c=plot(sol2.x,sol2.y(line,:),'b');
xline(0);
yline(0);
% legend([a; b; c],{num2str(aa); num2str(bb) ;num2str(cc)})
title(titles)
hold off
end
function out=jobs(VARABLE)
RC=.1;FW=.5;
M=0.5;BI =1.5;BE = 1.5;AE = 1+BE*BI;R=0.6;GR=8;
GM=6;PR=.71;EC=0.3; S = 0.5;S0 = 1;K=.5;SC=.6;
D = 2;GAMMA = 1;EPS = .8;ST = 5;STT= .5;
NEBLA = 2;LAMB=.8;short=AE^2+BE^2;XX=1+D;W1=1.5;CP=1.5;DF=.8;n=.01;
sol1 = bvpinit(linspace(0,1,500),[1 1 1 1 1 0 0 0 0 0 1]);
sol = bvp4c(@bvp2d, @bc2d, sol1);
sol1=bvpinit(sol,[0 10]);
sol = bvp4c(@bvp2d, @bc2d, sol1);
% sol1=bvpinit(sol,[0 2]);
% sol = bvp4c(@bvp2d, @bc2d, sol1);
out=sol;
function yvector =bvp2d(~,y)
% if y(1)==0
% y(1)=0.01;
% end
yy4=(-y(4)-y(3)*y(1)-W1*(2*y(4)*y(2)+y(3)^2)+K*y(2)+(M/(AE^2+BE^2))*...
(AE*y(2)+BE*y(5))+R*y(5)-GAMMA*y(2)^2+GM*y(10)+GR*y(8))/(-W1*y(1));
yy7=(-y(1)*y(6)-y(7)-2*W1*y(2)*y(7)+K*y(5)-(M/short)*(AE*y(5)+...
BE*y(2))+R*y(2)+GAMMA*y(5)^2)/(W1*y(1));
% yy11=(y(1)*y(9)-y(1)*y(11)*DF*SC+(EC/CP)*(y(3)^2+y(6)^2)+...
% (M*EC)/short*(y(2)^2+y(5)^2)+RC-DF*SC*(y(10)+1))/(1/PR-DF*S0*SC);
% yy9=(y(1)*y(9)+EC/CP*(y(3)^2+y(6)^2)+M*EC/short*(y(2)^2+y(5)^2))/(-PR);
% yy11=(-y(1)*y(11)-S0*yy9-RC*y(10))*SC;
yy11=(-y(1)*y(11)+RC*y(10))*SC;
yy9=(y(1)*y(9)+EC/CP*(y(3)^2+y(6)^2)+M*EC/short*(y(2)^2+y(5)^2)+DF*yy11)/(-PR);
yvector=[y(2);y(3);y(4);yy4;y(6);y(7);yy7;y(9);yy9;y(11);yy11];
end
function residual=bc2d(y0,yinf)
residual=[y0(1)-FW;y0(2)-1;y0(5);y0(8)-1;y0(10)-1;yinf(3);yinf(4);
yinf(2);yinf(5);yinf(8);yinf(10)];
end
end
  7 Comments
ulan
ulan on 24 May 2024 at 17:32
Thats what i tried to do at first but my instructor prohibited me to do it.He said not to mix these two equation instead take one of the constant as 0.Thank you for your valuable time.
Torsten
Torsten on 24 May 2024 at 17:46
He said not to mix these two equation
The two equations are mixed in the state they are now because they both contain θ'' and φ''. If you solve for θ'' and φ'', they become "unmixed" because you explicitly solve for the highest derivatives.

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