Getting error: Second argument must be a variable or a nonnegative integer specifying the number of differentiations.

Question

Missael Hernandez 2020-10-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/618213-getting-error-second-argument-must-be-a-variable-or-a-nonnegative-integer-specifying-the-number-of

评论： Walter Roberson 2020-10-19

I am trying to symbolically obtain the Euler-Lagrange equations, but I keep getting the "Second argument must be a variable or a nonnegative integer specifying the number of differentiations" error. It occurs with the first term of the equation (dL/dx_dot). I don't what I am doing wrong. Any help is appreciated. Thanks.

syms x(t) theta1(t) theta2(t) F(t) m0 m1 m2 I1 I2 L1 L2 g
T0 = 0.5*m0*diff(x(t),t)^2; %Kinetic energy of mass 
T1 = 0.5*m1*((diff(x(t),t)+0.5*L1*diff(theta1(t),t)*cos(theta1(t)))^2+(L1*diff(theta1(t),t)*sin(theta1(t)))^2)+0.5*I1*diff(theta1(t),t)^2; %Kinetic energy of pendulum 1 
T2 = 0.5*m2*((diff(x(t),t)+L1*diff(theta1(t),t)*cos(theta1(t))+0.5*L2*diff(theta2(t),t)*cos(theta2(t)))^2+(L1*diff(theta1(t),t)*sin(theta1(t))+0.5*L2*diff(theta2(t),t)*sin(theta2(t)))^2)+0.5*I2*diff(theta2(t),t)^2; %Kinetic energy of pendulum 2
T = T0+T1+T2; % Total kinetic energy
U0 = 0; %Potential energy of mass
U1 = m1*g*0.5*L1*cos(theta1(t)); %Potential energy of pendulum 1
U2 = m2*g*(L1*cos(theta2(t))+0.5*L2*cos(theta2(t))); %Potential energy of pendulum 2
U = U0+U1+U2; %Total potential energy
L = T-U; %Lagrangian of system
dL_dx_dot = diff(L,diff(x(t),t));
dL_dx = diff(L,x);
diff(dL_dx_dot,t) - dL_dx == F(t)

3 个评论
显示 1更早的评论隐藏 1更早的评论

Missael Hernandez 2020-10-19

I have R2019a. I didn't think that'd be an issue.

Walter Roberson 2020-10-19

Interesting, I did not notice that in R2020b they improved this.

My answer below applies up to R2020a.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2020-10-19

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/618213-getting-error-second-argument-must-be-a-variable-or-a-nonnegative-integer-specifying-the-number-of#answer_517438

在 MATLAB Online 中打开

It is not possible to differentiate with respect to a function.

You have to make a substitution and hope that the system remains valid.

syms Dx;
dL_dx_dot = subs(diff( subs(L,diff(x(t),t),Dx), Dx), Dx, diff(x(t),t))

This is not guaranteed to be valid in general! In the past I have posted systems that this kind of manipulation is not valid for. It depends upon the rest of the equation being independent of the function whose derivative is being taken, and that is not always the case.

The situation is sort of like making a substitution assuming that you are not dividing by zero -- the substitution would often work, but there are cases where you accidentally introduce nonsense answers when you make such a substitution.