bvp4c problem with fewer boundary value than variable
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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
8 个评论
Torsten
2024-5-24
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.
Syed Sohaib Zafar
2024-9-3
移动:Walter Roberson
2024-9-3
Do you find the answer to this? I'm also solving a problem similar to this. I Considered some extra boundary conditions but results are not quite good.
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