why matlab take a lot of time in the symbolic calculation

Question

merwan behar 2021-7-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/871723-why-matlab-take-a-lot-of-time-in-the-symbolic-calculation

评论： merwan behar 2021-7-5

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 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=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

Sulaymon Eshkabilov 2021-7-4

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https://ww2.mathworks.cn/matlabcentral/answers/871723-why-matlab-take-a-lot-of-time-in-the-symbolic-calculation#answer_739798

在 MATLAB Online 中打开

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;