MATLAB Answers

Jacobian function and derivative that is a fraction

12 views (last 30 days)
Rick
Rick on 26 Sep 2015
Commented: Walter Roberson on 27 Sep 2015
Hello,
I'm trying to use the Jacobian function to find the matrices to linearize my two equations
However, when I try and take the partial derivative with respect to (F/V) for matrix B, then I get an error.
syms CA CB F V CAi k1 k2 k3
f1 = (F/V)*(CAi-CA)-k1*CA-k3*CA^2;
f2 = -(F/V)*CB + k1*CA - k2*CB;
A = jacobian([f1; f2], [CA CB]);
B = jacobian([f1; f2], [F/V CAi])
Error using mupadmex
Error in MuPAD command: The variable is invalid. [stdlib::diff]
Error in sym/jacobian (line 34)
Jsym = mupadmex('symobj::jacobian',F.s,x.s);
I'm wondering what I can do to fix this. Thanks

Accepted Answer

Matt J
Matt J on 26 Sep 2015
Edited: Matt J on 26 Sep 2015
Although the documentation doesn't say so, I would guess that the Jacobian cannot be taken with respect to a symbolic expression like F/V. It has to be with respect to a named variable. So, replace F/V with its own variable like below,
syms CA CB Fv CAi k1 k2 k3
f1 = (Fv)*(CAi-CA)-k1*CA-k3*CA^2;
f2 = -(Fv)*CB + k1*CA - k2*CB;
A = jacobian([f1; f2], [CA CB]);
B = jacobian([f1; f2], [Fv CAi])
  2 Comments
Walter Roberson
Walter Roberson on 27 Sep 2015
Although that works in this case, in general you need to do the change of variables after you construct the expression. For example,
diff(sin(x)^2 + (1-y^2), sin(x))
would seem to have the obvious change
diff(sinx^2 + 5*(1-y^2), sinx)
but if y happens to be cos(x) then your original expression is
diff(sin(x)^2 + 5*(1-cos(x)^2), sin(x))
or
diff(sin(x)^2 + 5*sin(x)^2, sin(x))
or
diff(6*sin(x)^2, sin(x))
and then it is safe to do the change of variables.
Therefore if you have diff(f(x), v(x)) for some functions f(x) and v(x), you need to use something like
diff(subs(f(x), x, solve(Vx = v(x),x)), Vx)

Sign in to comment.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!