What changes are needed to run the code

Psi = pi/2; L1 = 0.1; L2 = 0.1; L = 0.1; Pr = 1; M = 5;
Ec = 0.5; Nb = 0.5; Nt = 0.1; fw = 0.5; Q = 0.1; Le = 2; Kc = 1;
A = linspace(-1,1,101);
for L = [1 2 3]/10
BC = @(ya,yb)[ ya(1)-fw; ya(2)+1; ya([4;7])-1; yb([2;4;6]) ];
ODE = @(x,y)[ y(2); y(3); ( A.*(y(2)+(1/2)*x*y(3)) - y(1)*y(3) + y(2)*( y(2)+M^2*(sin(Psi))^2) - L1*y(4) - L2*y(6))/(1+(1/L));
y(5); -Pr*( y(1)*y(5) - (1/2)*A.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4) );
y(7); -Le*Pr*( y(1)*y(7)-(1/2)*A.*x*y(7)-Kc*y(6) ) - (Nt/Nb)*(-Pr*( y(1)*y(5) - (1/2)*A.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4) )) ];
xa = 0; xb = 6; xn = linspace(xa,xb,101); x = xn; solinit = bvpinit(x,[0 1 0 1 0 1 0]); sol = bvp5c(ODE,BC,solinit);
S = deval(sol,x); f0 = deval(sol,xa);
figure(1);plot(A,S(2,:),'-','LineWidth',2);xlabel('\bf\eta','color','b'); ylabel('\bff^\prime(\eta)','color','b');hold on,
end
Error using vertcat
Dimensions of arrays being concatenated are not consistent.

Error in solution (line 6)
ODE = @(x,y)[ y(2); y(3); ( A.*(y(2)+(1/2)*x*y(3)) - y(1)*y(3) + y(2)*( y(2)+M^2*(sin(Psi))^2) - L1*y(4) - L2*y(6))/(1+(1/L));

Error in bvparguments (line 105)
testODE = ode(x1,y1,odeExtras{:});

Error in bvp5c (line 129)
bvparguments(solver_name,ode,bc,solinit,options);
%% ERROR is:
Dimensions of matrices being concatenated are not consistent.
%% Help needed to run

Respuestas (1)

Jan
Jan el 16 de Nov. de 2021
Editada: Jan el 16 de Nov. de 2021
ODE = @(x,y) [ ...
y(2); ...
y(3); ...
(A.*(y(2)+0.5*x*y(3)) - y(1)*y(3) + y(2)*( y(2)+M^2*(sin(Psi))^2) - L1*y(4) - L2*y(6))/(1+(1/L)); ...
y(5); ...
-Pr*(y(1)*y(5) - 0.5*A.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4)); ...
y(7); ...
-Le*Pr*(y(1)*y(7) - 0.5*A.*x*y(7)-Kc*y(6)) - (Nt/Nb)*(-Pr*(y(1)*y(5) - 0.5*A.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4)))];
Yes, the 3rd, 5th and 7th element have the size 1x101, while the others are scalars. This comes from the vector A. I cannot guess, what you want to do instead.

7 comentarios

MINATI PATRA
MINATI PATRA el 16 de Nov. de 2021
Hi Jan
Actually I am trying to draw S(2,:) w.r.t A ( varies in x-axis ) for different values of L = [1 2 3]/10
i.e. plot(A,S(2,:))
Jan
Jan el 16 de Nov. de 2021
Editada: Jan el 16 de Nov. de 2021
Then you have to run the integration for all values of A. You cannot find all different trajectories with one integration.
Psi = pi/2;
L1 = 0.1;
L2 = 0.1;
L = 0.1;
Pr = 1;
M = 5;
Ec = 0.5;
Nb = 0.5;
Nt = 0.1;
fw = 0.5;
Q = 0.1;
Le = 2;
Kc = 1;
A = linspace(-1, 1, 10);
xa = 0;
xb = 6;
xn = linspace(xa,xb,101);
x = xn;
figure;
axesH = axes('NextPlot', 'add');
for L = [1 2 3]/10
for iA = A
BC = @(ya,yb)[ ya(1)-fw; ya(2)+1; ya([4;7])-1; yb([2;4;6]) ];
ODE = @(x,y) [ ...
y(2); ...
y(3); ...
(iA.*(y(2)+0.5*x*y(3)) - y(1)*y(3) + y(2)*( y(2)+M^2*(sin(Psi))^2) - L1*y(4) - L2*y(6))/(1+(1/L)); ...
y(5); ...
-Pr*(y(1)*y(5) - 0.5*iA.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4)); ...
y(7); ...
-Le*Pr*(y(1)*y(7) - 0.5*iA.*x*y(7)-Kc*y(6)) - (Nt/Nb)*(-Pr*(y(1)*y(5) - 0.5*iA.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4)))];
solinit = bvpinit(x, [0 1 0 1 0 1 0]);
sol = bvp5c(ODE, BC, solinit);
S = deval(sol, x);
f0 = deval(sol, xa);
plot(axesH, iA, S(2,:), '.', 'LineWidth', 2);
end
end
xlabel('\bf\eta','color','b');
ylabel('\bff^\prime(\eta)','color','b');
I assume you want to modify the oputput. Maybe collect the different output of S(:,2) in an array at first and draw it as plot(A, collectedS)?
MINATI PATRA
MINATI PATRA el 21 de Nov. de 2021
@ Dear Jan
Thanks for your try. But unfortunately Your attempt is not upto mark (I may be wrong).
Please have a look into my trial. It gives so many curves instead of THREE. Can it be modified?
A = -0.50; Psi = pi/2; L1 = 0.1; L2 = 0.1; L = 0.1; Pr = 2; M = 5;
Ec = 0.5; Nb = 0.5; Nt = 0.1; fw = 0.5; Q = 0.1; Le = 2; Kc = 1;
Av = linspace(-1,1,101);
for M = [1 2 3]
for i = 1:length(Av)
A = Av(i);
BC = @(ya,yb)[ ya(1)-fw; ya(2)+1; ya([4;7])-1; yb([2;4;6]) ];
ODE = @(x,y)[ y(2); y(3); ( A.*(y(2)+(1/2)*x*y(3)) - y(1)*y(3) + y(2)*( y(2)+M^2*(sin(Psi))^2) - L1*y(4) - L2*y(6))/(1+(1/L));
y(5); -Pr*( y(1)*y(5) - (1/2)*A.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4) );
y(7); -Le*Pr*( y(1)*y(7)-(1/2)*A.*x*y(7)-Kc*y(6) ) - (Nt/Nb)*(-Pr*( y(1)*y(5) - (1/2)*A.*x*y(5) + Nb*y(5)*y(7) + Nt*y(5)^2 + M^2*Ec*y(2)^2+Q*y(4) )) ];
xa = 0; xb = 6; xn = linspace(xa,xb,101); x = xn; solinit = bvpinit(x,[0 1 0 1 0 1 0]); sol = bvp5c(ODE,BC,solinit);
S = deval(sol,x); f0 = deval(sol,xa); Cf = (1+(1/L))*f0(3); Nu = - f0(5); Sh = - f0(7);
figure(41); plot(Av,S(2,:),'LineWidth',2); xlabel('\bfA','color','b'); ylabel('\bff^\prime(\eta)','color','b'); hold on,
end
end
Jan
Jan el 22 de Nov. de 2021
Please use the tools in the forum to format your code. This improves the readability.
Why do you expect 3 curves only?
MINATI PATRA
MINATI PATRA el 22 de Nov. de 2021
Dear Jan Since I have taken 3 values of the parameter L And variations of A will be in X axis
MINATI PATRA
MINATI PATRA el 6 de Dic. de 2021
Dear Jan
This is still unsolved.
Can it be possible for a last try?
Jan
Jan el 6 de Dic. de 2021
I'm not sure, which problem is still open. The code of my answer is running, and only the diagram looks a little bit strange. I do not know, what you want to get instead of this solution. So please explain again, what is still unsolved.

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Preguntada:

el 16 de Nov. de 2021

Comentada:

Jan
el 6 de Dic. de 2021

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