why matlab take a lot of time in the symbolic calculation
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merwan behar
el 4 de Jul. de 2021
Comentada: merwan behar
el 5 de Jul. de 2021
I wrote a code to calculate the fundamental frequencies of a piezoelectric nanotube, I used the state space analytical method to solve my differential equation which based on the computation of the exponential of a matrix but the problem takes a lot for the execution and I did not know for I will be very happy if you clarify the problem for me
the code is
clear all
clc
%introduire des variable
h=1e-9;
L=10e-9;
C11=132e9;
e31=-4.1;
mu=0;
E11=5.841e-9;
E33=7.124e-9;
ro=7500;
B=(pi/h);
r1=49.5e-9;
r2=50.5e-9;
A=pi*((r2^2)-(r1^2));
I=pi*(((r2^4)/4)-((r1^4/4)));
fun1=@(x,r) e31.*(r.^2).*sin(x).*B.*sin(B.*r.*sin(x));
fun2=@(x,r) E11.*r.*(cos(B.*r.*sin(x))).^2;
fun3=@(x,r) E33.*(B^2).*r.*(sin(B.*r.*sin(x))).^2;
F31 = integral2(fun1,0,2*pi,49.5,50.5);
X11 = integral2(fun2,0,2*pi,49.5,50.5);
X33 = integral2(fun3,15,2*pi,49.5,50.5);
NE=0;
%introduire la matrice A
syms w
p1=((ro*A*w^2)/(C11*I));
p2=((-(mu^2*ro*A*w^2)+(F31^2/X11))/(C11*I));
p3=((F31*X33)/(X11*C11*I));
p4=(F31/X11);
p5=(X33/X11);
a=[0 1 0 0 0 0;0 0 1 0 0 0;0 0 0 1 0 0;...
(p1) 0 (p2) 0 (p3) 0;0 0 0 0 0 1;0 0 (p4) 0 (p5) 0];
[V,D]=eig(a);
H=eye(6);
M=V*diag(exp(diag(D*L)))/V;
%condition aux limite simplement appuyé
l1=H(1,:);
l3=H(3,:);
l5=H(5,:);
m1=M(1,:);
m3=M(3,:);
m5=M(5,:);
%la matrice final
K=[l1;l3;l5;m1;m3;m5];
DA=det(K);
%fréquence fondamental
for n = 1:5
F = vpasolve(DA,w,[0.4 3],'Random',true)
end
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Respuesta aceptada
Sulaymon Eshkabilov
el 4 de Jul. de 2021
The problem with integral2 calculation part not with vpasolve(). You need to check the limit values for xmin, xmax, ymin, ymax in interal2(). This below given syntax controls tolerances and finds the solutions, BUT
...
F31 = integral2(fun1,0,2*pi,49.5,50.5, 'Method','iterated','AbsTol',0,'RelTol',1e-10); % Solves the problem partially
X11 = integral2(fun2,0,2*pi,49.5,50.5, 'Method','iterated','AbsTol',0,'RelTol',1e-10);
X33 = integral2(fun3,15,2*pi,49.5,50.5, 'Method','iterated','AbsTol',0,'RelTol',1e-10);
NE=0;
As it is given, the polynomial coeffs p2, p3 .. p5 = nan.
Therefore, it does not produce any solutions.
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