Equations and Boundary conditions are Unequal

Question

MINATI PATRA 2023-8-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2008132-equations-and-boundary-conditions-are-unequal

评论： Torsten 2023-8-15

采纳的回答： Walter Roberson

Model#02.pdf

在 MATLAB Online 中打开

%%% THe present code is of the attached (Model#02) pdf, need modification to run.

%%% REFERENCE: This type of work was done in DOI: 10.1002/num.22672 (More in size, so cant be uploaded)

main()
Unrecognized function or variable 'S'.

Error in solution>main/ode (line 21)
        figure(10),plot(x,S(2,:),'-b','LineWidth',1.5),hold on

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

Error in bvp5c (line 129)
    bvparguments(solver_name,ode,bc,solinit,options);

Error in solution>main (line 6)
xa = 0; xb = 5; solinit = bvpinit(linspace(xa,xb,100),[0 1 0 1 0 1 0 1 0 1 0 1]); sol = bvp5c(@ode,@bc,solinit); x = linspace(xa,xb,100);  S = deval(sol,x);
function main
We = 2; n = 0.5; M = 1; Pr = 7; Rd = 0.5; Tw = 1.5; Nt = 0.3; Nb = 0.5; Ec = 0.3; Q = 0.5; Le = 2; K = 0.1; D1 = 0.1; 
m = 0.5; E = 0.1; Lb = 0.5;  Pe = 2; D2 = 1; S1 = 0.2; S2 = 0.2; Om = 0.5; 
xa = 0; xb = 5; solinit = bvpinit(linspace(xa,xb,100),[0 1 0 1 0 1 0 1 0 1 0 1]); sol = bvp5c(@ode,@bc,solinit); x = linspace(xa,xb,100);  S = deval(sol,x);
    function BC = bc(ya,yb)
        BC = [ya([1,4]); ya(2) - S1; ya([6,8,10]) - 1; yb([6,8,10]); yb(2) - S2; yb(4) - Om];        
    end 
    function EQ = ode(x,y) 
        Af = (1+(We*y(3))^2)^((n-1)/2) + (n-1)*(We*y(3))^2*(1+(We*y(3))^2)^((n-3)/2);    Ag = (1+(We*y(5))^2)^((n-1)/2) + (n-1)*(We*y(5))^2*(1+(We*y(5))^2)^((n-3)/2);
        At = 4*Rd*(Tw-1)*y(7)^2*(1+(Tw-1)*y(6))^2;      X = - E/(1+D1*y(6));
        EQ = [ -2*y(2);  
            y(3);      (M*y(2) + y(2)^2 - y(4)^2 + y(1)*y(3))/Af; 
            y(5);      (M*y(4) + 2*y(2)*y(4) + y(1)*y(5))/Ag; 
            y(7);      (Pr/At)*( y(1)*y(7) - Nt*y(7)^2 - Nb*y(7)*y(9) - Ec*(y(3)^2 + y(5)^2) - M*Ec*(y(2)^2 + y(4)^2) - Q*y(6) );
            y(9);      Pr*Le*( y(1)*y(9) + K*(1+D1*y(6))^m*y(8)*exp(X) - (Nt/Nb)*((Pr/At)*( y(1)*y(7) - Nt*y(7)^2 - Nb*y(7)*y(9) - Ec*(y(3)^2 + y(5)^2) - M*Ec*(y(2)^2 + y(4)^2) - Q*y(6) ))   );
            y(11);     Lb*y(1)*y(11)+ Pe*( y(9)*y(11) + (D2 + y(10)) )*Pr*Le*( y(1)*y(9) + K*(1+D1*(y(6))^m)*y(8)*exp(X) - (Nt/Nb)*((Pr/At)*( y(1)*y(7) - Nt*y(7)^2 - Nb*y(7)*y(9) - Ec*(y(3)^2 + y(5)^2) - M*Ec*(y(2)^2 + y(4)^2) - Q*y(6) ) ))
            ];
        
        figure(10),plot(x,S(2,:),'-b','LineWidth',1.5),hold on
    end
figure(2),plot(x,S(2,:));hold on
end

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Answer 1

Walter Roberson 2023-8-14

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2008132-equations-and-boundary-conditions-are-unequal#answer_1287142

移动：Walter Roberson 2023-8-14

在 MATLAB Online 中打开

S is the output of the deval() and so is not available until after the bvp5c has been run. But the ode function wants to plot S, which requires S be defined but it is not defined until after the ode is finished.

Your plotting should be moved to before the bc function definition.

Meanwhile, your ode function needs to return a column vector of length 12, same length as your initial condition; at the moment it is only length 11.

main()
ans = 1×2
    11     1
Error using bvparguments
Error in calling BVP5C(ODEFUN,BCFUN,SOLINIT): 
  The derivative function ODEFUN should return a column vector of length 12.

Error in bvp5c (line 129)
    bvparguments(solver_name,ode,bc,solinit,options);

Error in solution>main (line 7)
sol = bvp5c(@ode,@bc,solinit);
function main
We = 2; n = 0.5; M = 1; Pr = 7; Rd = 0.5; Tw = 1.5; Nt = 0.3; Nb = 0.5; Ec = 0.3; Q = 0.5; Le = 2; K = 0.1; D1 = 0.1; 
m = 0.5; E = 0.1; Lb = 0.5;  Pe = 2; D2 = 1; S1 = 0.2; S2 = 0.2; Om = 0.5; 
xa = 0; xb = 5;
solinit = bvpinit(linspace(xa,xb,100),[0 1 0 1 0 1 0 1 0 1 0 1]);
sol = bvp5c(@ode,@bc,solinit);
x = linspace(xa,xb,100);
S = deval(sol,x);
figure(10),plot(x,S(2,:),'-b','LineWidth',1.5),hold on
    function BC = bc(ya,yb)
        BC = [ya([1,4]); ya(2) - S1; ya([6,8,10]) - 1; yb([6,8,10]); yb(2) - S2; yb(4) - Om];        
    end 
    function EQ = ode(x,y) 
        Af = (1+(We*y(3))^2)^((n-1)/2) + (n-1)*(We*y(3))^2*(1+(We*y(3))^2)^((n-3)/2);    Ag = (1+(We*y(5))^2)^((n-1)/2) + (n-1)*(We*y(5))^2*(1+(We*y(5))^2)^((n-3)/2);
        At = 4*Rd*(Tw-1)*y(7)^2*(1+(Tw-1)*y(6))^2;      X = - E/(1+D1*y(6));
        EQ = [ -2*y(2);  
            y(3);      (M*y(2) + y(2)^2 - y(4)^2 + y(1)*y(3))/Af; 
            y(5);      (M*y(4) + 2*y(2)*y(4) + y(1)*y(5))/Ag; 
            y(7);      (Pr/At)*( y(1)*y(7) - Nt*y(7)^2 - Nb*y(7)*y(9) - Ec*(y(3)^2 + y(5)^2) - M*Ec*(y(2)^2 + y(4)^2) - Q*y(6) );
            y(9);      Pr*Le*( y(1)*y(9) + K*(1+D1*y(6))^m*y(8)*exp(X) - (Nt/Nb)*((Pr/At)*( y(1)*y(7) - Nt*y(7)^2 - Nb*y(7)*y(9) - Ec*(y(3)^2 + y(5)^2) - M*Ec*(y(2)^2 + y(4)^2) - Q*y(6) ))   );
            y(11);     Lb*y(1)*y(11)+ Pe*( y(9)*y(11) + (D2 + y(10)) )*Pr*Le*( y(1)*y(9) + K*(1+D1*(y(6))^m)*y(8)*exp(X) - (Nt/Nb)*((Pr/At)*( y(1)*y(7) - Nt*y(7)^2 - Nb*y(7)*y(9) - Ec*(y(3)^2 + y(5)^2) - M*Ec*(y(2)^2 + y(4)^2) - Q*y(6) ) ))
            ];
        size(EQ)
        
    end
figure(2),plot(x,S(2,:));hold on
end

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MINATI PATRA 2023-8-15

I think if first equation is differentiated again to make it a second order ODE, problem can be solved.

Torsten 2023-8-15

You generate an artificial degree of freedom by this differentiation. Depending on the second boundary condition you impose you may or may not reproduce the solution of the original problem (first-order ODE with only one boundary condition).

Test it for a simple example.

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