Subscripted assignment dimension mismatch.
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function main
D=1; %L=0;
Pr=1;R=0.1;Sc=1;
xa=0;xb=6;
Lv = [-2.5:0.025:0];
p = [];
for i=1:length(Lv)
L = Lv(i);
fODE = @(x,y) [y(2); y(3); y(2)^2-y(3)*y(1)-1; y(5); -3*Pr*y(1)*y(5)/(3+4*R); y(7); -Sc*y(1)*y(7)];
BCres= @(ya,yb)[ya(1); ya(2)-L-D*ya(3); ya(4)-1; ya(6)-1; yb(2)-1; yb(4);yb(6)];
xint=linspace(xa,xb,101);
solinit=bvpinit(xint,[0 1 0 1 0 1 0]);
sol=bvp4c(fODE,BCres,solinit);
sxint=deval(sol,xint);
%%WE NEED TO PLOT for
S(i,1)=sxint(3,:);
end
figure(1)
plot(Lv,S,'-','Linewidth',1.5);
xlabel('\bf \lambda');
ylabel('\bf C_{f}');
hold on
end
%%While running the code following ERROR occurs:
Subscripted assignment dimension mismatch.
Error in (line 17)
S(i,1)=sxint(3,:);
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采纳的回答
Walter Roberson
2019-5-19
function all_sxint = main
D=1; %L=0;
Pr=1; R=0.1; Sc=1;
xa=0;xb=6;
Lv = [-2.5:0.025:0];
nLv = length(Lv);
all_sxint = cell(nLv, 1);
S = zeros(nLv, 7);
for i=1:nLv
L = Lv(i);
fODE = @(x,y) [y(2); y(3); y(2)^2-y(3)*y(1)-1; y(5); -3*Pr*y(1)*y(5)/(3+4*R); y(7); -Sc*y(1)*y(7)];
BCres= @(ya,yb)[ya(1); ya(2)-L-D*ya(3); ya(4)-1; ya(6)-1; yb(2)-1; yb(4);yb(6)];
xint=linspace(xa,xb,101);
solinit=bvpinit(xint,[0 1 0 1 0 1 0]);
sol=bvp4c(fODE,BCres,solinit);
sxint=deval(sol,xint);
all_sxint{i} = sxint;
S(i,:) = sxint(3,:);
end
figure(1)
plot(Lv, S, '-', 'Linewidth', 1.5);
xlabel('\bf \lambda');
ylabel('\bf C_{f}');
legend({'bc1', 'bc2', 'bc3', 'bc4', 'bc5', 'bc6', 'bc7'})
end
Assign the output to a variable so that you can examine all of the time points for all of the Lv values afterwards, as you indicate that you need to be able to do that.
11 个评论
更多回答(1 个)
Matt J
2019-5-18
编辑:Matt J
2019-5-18
sxint(3,:) is not a scalar, but the left hand side S(i,1) is a scalar location.
8 个评论
Walter Roberson
2019-5-19
Is there a reason you need to calculate at all of the time points, and then to store data for only the third time point? Is there a particular reason why my suggestion to calculate only at the third time point will not work for you?
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