Why the seond arugment in the command jacobian is not a vector of variables?

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Tony Cheng 2019-4-9

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https://ww2.mathworks.cn/matlabcentral/answers/455237-why-the-seond-arugment-in-the-command-jacobian-is-not-a-vector-of-variables

编辑： Walter Roberson 2023-4-11

采纳的回答： Stephan

Hi guys, I am using the code Jacobian in Matlab to symbolically compute the Jacobian matrix for two vectors-A and B.

These two vectors are defined by some symbolic variables and they are all functions of the variable t as follows:

syms t a1(t) a2(t) a3(t) b1(t) b2(t) b3(t)

A = [a1(t) ; a2(t) ; a3(t) ] ;

B = [b1(t) ; b2(t) ; b3(t) ] ;

What I did is C = jacobian(A , B ) .

However, Matlab gives an error to me and says the second argument must be a vector of variables.

I am very confused: the second argument B is obviously 3 times 1 symbolic vector. Why the code is not write?

Can anyone offer me some solutions?

Many thanks!

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Answer 1

Stephan 2019-4-9

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编辑：Stephan 2019-4-9

Hi,

you declare b1...b3 as symbolic functions of t - not as symbolic variables. Thats why it doesnt work.

Here is an example:

syms  a1(t) a2(t)  a3(t)  b1 b2 b3
a1(t) = b1*sin(t)/(b2+b3);
a2(t) = 3*t + 5/b2+b3;
a3(t) = 4/t^b3-b1;
A = [a1 a2 a3] ;
B = [b1; b2; b3];
C = jacobian(A ,B)

results in:

C(t) =
 
[ sin(t)/(b2 + b3), -(b1*sin(t))/(b2 + b3)^2, -(b1*sin(t))/(b2 + b3)^2]
[                0,                  -5/b2^2,                        1]
[               -1,                        0,         -(4*log(t))/t^b3]

See the different symbolic types in this simple example:

>> syms  a(t) b
>> whos
  Name      Size            Bytes  Class     Attributes
  a         1x1                 8  symfun              
  b         1x1                 8  sym                 
  t         1x1                 8  sym

Best regards

Stephan

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Stephan 2019-4-9

编辑：Walter Roberson 2023-4-11

Why do you want to do this? Can you explain the context?

Here is a little dirty workaround, that lets you have all the b's as functions of t:

syms a1(t) a2(t) a3(t) b1(t) b2(t) b3(t)
a1(t) = b1(t)*sin(t) + b2 - b3;
a2(t) = b2(t)*cos(t) + 2*b3*b1;
a3(t) = b3(t)*3*t^2 - b2^2;
A(1:3) = [a1 a2 a3]
B(1:3) = [b1 b2 b3]
C_help = sym(zeros(3));
for ii = 1:3
    C_help(ii,:) = functionalDerivative(A(ii),B);
end
C(t) = C_help

gives:

A =
 
[ b2(t) - b3(t) + b1(t)*sin(t), 2*b1(t)*b3(t) + b2(t)*cos(t), 3*t^2*b3(t) - b2(t)^2]
 
B =
 
[ b1(t), b2(t), b3(t)]
 
C(t) =
 
[  sin(t),        1,      -1]
[ 2*b3(t),   cos(t), 2*b1(t)]
[       0, -2*b2(t),   3*t^2]

To check the results we use jacobian - where b1...b3 can not be functions of t, but variables:

syms a11(t) a22(t) a33(t) b11 b22 b33
a11(t) = b11*sin(t) + b22 - b33;
a22(t) = b22*cos(t) + 2*b33*b11;
a33(t) = b33*3*t^2 - b22^2;
AA = [a11 a22 a33]
BB = [b11 b22 b33]
CC = jacobian(AA,BB)

which is:

AA(t) =
 
[ b22 - b33 + b11*sin(t), 2*b11*b33 + b22*cos(t), - b22^2 + 3*b33*t^2]
 
BB =
 
[ b11, b22, b33]
 
CC(t) =
 
[ sin(t),      1,    -1]
[  2*b33, cos(t), 2*b11]
[      0, -2*b22, 3*t^2]

So the given workaround appears to give correct results, including b1...b3 to be functions of t.

Tony Cheng 2019-4-15

Dear Stephan,

Thanks so much for your patient and complete solution!

It is a good workaround and I will use it in the project!

Best Regards

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