How I can speed up this program with a symbolic calculator ??

Question

merwan behar 2021-7-15

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/879653-how-i-can-speed-up-this-program-with-a-symbolic-calculator

编辑： Walter Roberson 2021-7-16

how I can speed up this program with a symbolic calculator knowing that I am using an intel core i3 fourth generation

here is the code

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,49.5e-9,50.5e-9,'method','iterated');
 X11 = integral2(fun2,0,2*pi,49.5e-9,50.5e-9,'method','iterated'); 
X33 = integral2(fun3,15,2*pi,49.5e-9,50.5e-9,'method','iterated');
 %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]; 
 H=eye(6); 
a=vpa(a); 
M=expm(a*L); 
%condition aux limite simplement appuyÃ©
 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=vpa(det(K)); 
%frÃ©quence fondamental 
for n = 1:5 F = vpasolve(DA,w,[0.4 3],'Random',true) 
end

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Answer 1

Walter Roberson 2021-7-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/879653-how-i-can-speed-up-this-program-with-a-symbolic-calculator#answer_747848

编辑：Walter Roberson 2021-7-16

Symbolic expm of a 6 x 6 matrix is going to take a quite long time. It requires equivalent computing power to finding the eigenvalues of the matrix, but eigenvalues of a 6 x 6 symbolic matrix takes tens of hours (and will probably end up having to be in terms of unresolvable roots of a degree 6 polynomial) .

Then you extract coefficients and put together a new matrix and take det() of that 6 x 6 matrix, which will not be fast but at least is doable in an hour or so.

This is basically not really feasible symoblically.

You will need to only go as far symbolically as a*L, after which use matlabFunction() to create the expression as an anonymous function in w, and make the rest into a series of computation steps in a function that will be invoked from an fzero() call (you need fzero to be able to provide the range constraint.)