I keep receiving "Error using / Arguments must be numeric, char, or logical." when I try to find value of sym function

Question

Michal Amar 2021-6-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/865910-i-keep-receiving-error-using-arguments-must-be-numeric-char-or-logical-when-i-try-to-find-v

评论： Michal Amar 2021-6-27

I am trying to solve a system of ODE's using the Runge Kutta Method and I keep receiving the above error when I try to run it. Any help would be much appreciated!

d =  @(t) sqrt(d0^2 -k*t*(d0)^2);
m =  @(t) 1000*(1/6)*d^3;
Fd2 = @(t,y,v) (18*pi*v)/(1000*(d^2)) ;  
F1 = @(t,y,v) v ;%dy/dt = v 
F2 = @(t,y,v) Fd2/m - 9.8  % dv/dt = -g + Fd/m
h = 1;
t = 0:h:15; %Analyzing up to t = 15.
v = zeros(1,length(t));
y = zeros(1,length(t));
v(1) = 0;
y(1) = 2000;
for i = 1:1:length(t) - 1
k1 = F1(t(i),y(i),v(i)) 
p1 = F2(t(i),y(i),v(i))
k2 = F1(t(i)+h/2, y(i)+(h/2)*k1, v(i)+(h/2)*p1)
p2 = F2(t(i)+h/2, y(i)+(h/2)*k1, v(i)+(h/2)*p1)
k3 = F1(t(i)+h/2, y(i)+(h/2)*k2, v(i)+(h/2)*p2)
p3 = F2(t(i)+h/2, y(i)+(h/2)*k2, v(i)+(h/2)*p2)
k4 = F1(t(i)+h, y(i)+(h)*k3, v(i)+(h)*p3)
p4 = F2(t(i)+h, y(i)+(h)*k3, v(i)+(h)*p3)
y(i+1) = y(1) + h * (k1+2*k2+2*k3+k4)/6 
v(i+1) = v(i) + h * (p1+2*p2+2*p3+p4)/6 
end

The following is the full error I receive:

Error using  / 
Arguments must be numeric, char, or logical.
Error in RK>@(t,y,v)Fd2/m-9.8 (line 30)
F2 = @(t,y,v) Fd2/m - 9.8  % dv/dt = -g + Fd/m
Error in RK (line 43)
p1 = F2(t(i),y(i),v(i))

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Scott MacKenzie 2021-6-26

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/865910-i-keep-receiving-error-using-arguments-must-be-numeric-char-or-logical-when-i-try-to-find-v#answer_734270

编辑：Scott MacKenzie 2021-6-26

在 MATLAB Online 中打开

If a function calls another function that takes input variables, you need to define the function to pass the variables through to the other function.

So, change

F2 = @(t,y,v) Fd2/m - 9.8

to

F2 = @(t,y,v) Fd2(t,y,v)/m(t) - 9.8

Also, change

m = @(t) 1000*(1/6)*d^3;

to

m = @(t) 1000*(1/6)*d(t)^3;

3 个评论
显示 1更早的评论隐藏 1更早的评论

Scott MacKenzie 2021-6-27

@Michal Amar You're welcome. So, with the function definitions corrected, it seems your problem has shifted from a syntactic error to a semantic error. Semantic errors are generally trickier to resolve.

There are lot of calculations to navigate in your code. Offhand, I don't see any obvious problem. I suggest you post the new problem (unexpected imaginary numbers) as a new question. A fresh set of eyes might help. Good luck.

Michal Amar 2021-6-27