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I have a matrix equation system and I need help to solve it

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mahdi
mahdi el 24 de Mayo de 2023
Editada: Torsten el 10 de Jun. de 2023
So I have these 3, 7*1 variable matrices in those system of equations and C{1,1} and C{1,2} are 7*7 matrices and they are constant, also S_1 to S_15 are 1*7 constant matrices. Problem is, it gives me U=0, W=0 and PHI=0 and it doesnt calculate the variables in those U, W and PHI matrices.
syms U [7 1]
U;
syms W [7 1]
W;
syms PHI [7 1]
PHI;
syms U W PHI
eq1 = (S_1*C{1,2}*U)+(S_2*C{1,1}*U)-(S_3*U)+(S_4*C{1,2}*W)+(S_5*C{1,1}*W) == 0;
eq2 = (S_6*C{1,2}*PHI)+(S_7*C{1,1}*PHI)+(S_8*PHI)-(S_9*C{1,1}*W) == 0;
eq3 = (S_10*C{1,2}*W)+(S_11*C{1,1}*W)+(S_12*C{1,1}*PHI)+(S_13*PHI)+(S_14*C{1,1}*U)...
-(S_15*U) == 0;
[U, W, PHI] = solve([eq1, eq2, eq3],[U, W, PHI])
  44 comentarios
mahdi
mahdi el 10 de Jun. de 2023
Editada: mahdi el 10 de Jun. de 2023
Ok thanks, can you explain this line please:
eq1(7) = subs(eq1(7),lhs(eq1(7)),lhs(eq1(7))+sym('100'));
Is this like the line b(7) = -100; ?
Also I must write this whole process after the end of this right?
{}; % for example only
C{1,1} = rand(7);
C{1,2} = rand(7);
Nu = 0.285; A_11 = 64; A_55 = 37; D_11 = 2; w0c = 0.005; R = 0.5;
A11 = 6.4*10^4; A55 = 3.7*10^4; D11 = 2*10^3;
N = 7; a = 0; b = 0.5;
X1 = a+(b-a)*(1-cos(((1:N)-1)*pi/(N-1)))/2;
syms r
w0 = w0c*cos(pi*r/(2*R));
w0onX1 = double(subs(w0,r,X1));
A = C{1,1}*w0onX1.';
B = C{1,2}*w0onX1.';
B1 = C{1,2}*w0onX1.';
C_1 = cell2mat(C(1));
al = zeros(N);
al(2:N,1:N) = C_1(2:N,1:N);
bl = zeros(N);
bl(1:N-1,1:N) = C_1(1:N-1,1:N);
syms U [N,1]
syms W [N,1]
syms PHI [N,1]
z1 = X1;
z2 = B.';
z3 = A.';
z4 = B1.';
Ug1 = 0; Ug2 = 0; Ug3 = 0;
atm=((2*A11*z1.^2).*(C{1,2}))+(((2*A11*z1).*(C{1,1}))-(2*A11));
atm2=(((A11.*z2).*(z1.^2)).*(C{1,2}))+(((((A11-(Nu*A11)).*z3).*z1).*(al)));
atm3=((D11*z1.^2).*C{1,2})+((D11*z1).*C{1,1})-(D11+(A55*z1.^2));
atm4=((A55*z1.^2).*(al));
atm5=((((2*A55*z1)+((2*A11*Ug2).*z1)+((A11*z2.^2).*z1)+(2*Nu*A11*Ug3)).*C{1,2})+...
((2*A55)+(2*A11*Ug2)+((2*A11*Ug1).*z1)+(A11*z3.^2)+(((A11*z4).*z3).*z1).*(al)));
atm6=(((2*A55*z1).*C{1,1})+(2*A55));
atm7=((((2*Nu*A11)+((2*A11*z4).*z1)).*C{1,1})+(2*Nu*A11*z4));
eq1 = (atm*U)+(atm2*W) == 0;
eq2 = (atm3*PHI)-(atm4*W) == 0;
eq3 = (atm5*W)+(atm6*PHI)+(atm7*U) == 0;
vars = [U;W;PHI];
[Omega,b] = equationsToMatrix([eq1,eq2,eq3],vars);
b(7) = -100;
sol = Omega\b;
JabejayiU = double(sol(1:N));
KhizW = double(sol(1+N:N+N));
TaghirzaviyePHI = double(sol(1+(2*N):N+(2*N)));
Then again for example after I wrote all these and the nonlinear system, I must write the iteration procces? Or I can just combine the loop with the nonlinear system?
Torsten
Torsten el 10 de Jun. de 2023
Editada: Torsten el 10 de Jun. de 2023
Is this like the line b(7) = -100; ?
Yes.
And the code I gave you is complete - no modifications are necessary.
Of course I don't know whether "fsolve" returns a valid and reasonable solution. Nonlinear equations can have none or multiple solutions in contrast to linear systems.
Instead of
f = matlabFunction([lhs(eq1);lhs(eq2);lhs(eq3)],"Vars",{[U;W;PHI]})
[sol,fval] = fsolve(f,zeros(21,1))
you can also try
sol = vpasolve([eq1;eq2;eq3],[U;W;PHI])
to see if you get a different solution.

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