substitute value for variable
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I have created a code to solve a differential in terms of one variable (all C variables and B are constants and defined previously). Now I want to plug a value in for that variable into the differential equation, but it won't let me. I tried using the 'subs' function, but that doesn't work. Help?
syms I1 I2
F1 = C1*(exp(C2*(I1-3)-1))-((C1*C2)/2)*(I2-3);
W1 = 0;
F2 = piecewise(x<1,W1,1<x<lambda,W2,x>lambda,W3);
Wo = F1 + F2;
Wo1 = diff(Wo,I1);
Wo2 = diff(Wo,I2);
clear I1 I2
I1 = trace(B);
I2 = .5*((trace(B))^2+trace(B^2));
subs(Wo1,I1,I2);
2 个评论
Birdman
2018-2-21
Share all variables, the entire code.
Stephen23
2018-2-21
Here's the whole code. It all works until the last step.
% Define coefficients for all the fiber families
syms x p Wx I1 I2
pres = [p 0 0
0 p 0
0 0 p];
theta = [2 122 241];
C1 = .00124;
C2 = 1.07;
C3 = [1.03*10^-7 1.67*10^-7 1.59*10^-7];
C4 = [26.6 35 38];
lambda = [2.78 5.6 7.74];
C5 = C3.*C4.*lambda.*(exp(C4.*(lambda-1)));
% Define displacement matrix and B matrix
F = [x 0 0
0 1/sqrt(x) 0
0 0 1/sqrt(x)];
B = F*transpose(F);
% Define F1 as given
F1 = C1*(exp(C2*(I1-3)-1))-((C1*C2)/2)*(I2-3);
W1 = 0;
% Run pipline for each fiber family to get individual strain energy
% functions
for n = 1:3
a = [cos(theta(n))
-sin(theta(n))
0];
W2 = C3(n)*exp(C4(n)*(x-1))-1;
W3 = C5(n);
F2 = piecewise(x<1,W1,1<x<lambda(n),W2,x>lambda(n),W3);
Wo(n) = F1 + F2;
Wo1(n) = diff(Wo(n),I1);
Wo2(n) = diff(Wo(n),I2);
ao(n) = transpose(a(n))*a(n);
end
% Plug in values for a matrix and invariantes
clear I1 I2
I1 = trace(B);
I2 = .5*((trace(B))^2+trace(B^2));
%Substitute in the invariant values
for n=1:3
subs(Wo1(n),I1);
subs(Wo2(n),I2);
end
采纳的回答
Karan Gill
2018-2-21
You shouldn't overwrite I1. Overwriting I1 will not automatically substitute for it in Wo1. Instead, remove I1 = trace(B); and use
>> I1_val = trace(B)
I1_val =
2/x + x^2
>> Wo1 = subs(Wo1,I1,I1_val)
Wo1 =
[ piecewise(x < 1 | 139/50 < x | x in Dom::Interval(1, 139/50), (3317*exp((107*I1)/100 - 421/100))/2500000), piecewise(x < 1 | 28/5 < x | x in Dom::Interval(1, 28/5), (3317*exp((107*I1)/100 - 421/100))/2500000), piecewise(x < 1 | 387/50 < x | x in Dom::Interval(1, 387/50), (3317*exp((107*I1)/100 - 421/100))/2500000)]
This way, you separate the symbolic variable from the value you want to substitute, which keeps the meaning of variables in your code clean.
2 个评论
Carolyn
2018-2-21
Karan Gill
2018-2-21
Great! :) (And I understand overwriting I1 was the intuitive thing to do.)
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