bvp4c problem with fewer boundary value than variable

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ulan 2024-5-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2122151-bvp4c-problem-with-fewer-boundary-value-than-variable

评论： Torsten 2024-5-24

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 个评论
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ulan 2024-5-24

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 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.

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