Difference between "diff(x^2)" and "2*x"

Hi, I can't figure out how are the following two situations different:

syms x1 x2 F;
U=x1^2+(x2-F)^2;
solve('diff(U,x1)=0','diff(U,x2)=0','x1,x2')

vs.

solve('2*x1=0','2*(x2-F)=0','x1,x2')

In the first case MATLAB can't find a solution, although it is a simple linear system (the same one as in the second case which works fine). Any ideas? Thank you:)

Answer 1

[Matt Fig]

I think the problem is you are mixing calling methods. When you pass a string to SOLVE, it looks at the string and determines the unknowns with SYMVAR or looks for your extra arguments. On the other hand, if you have an equation that has symbolic variables that are defined in the workspace, then you should not pass a string to SOLVE.

Here are two equivalent ways to solve your problem. Note carefully the differences.

clear all
syms x1 x2 F   % In this method we use symbolic variables.
U = x1^2+(x2-F)^2;  % And a symbolic expression.
% Note that SOLVE assumes expr is equal to zero...
S = solve(diff(U,x1),diff(U,x2),x1,x2) % NO STRINGS!

Now the other method is to use strings. Here we do not need to define any symbolic variables. This is a different method with different rules.

clear all % NO symbolic variables, we use strings only...
STR1 = 'diff(x1^2+(x2-F)^2,x1)=0'; % These will be evaluated
STR2 = 'diff(x1^2+(x2-F)^2,x2)=0'; % inside SOLVE...
S = solve(STR1,STR2,'x1,x2')  % We use strings here.

So you see what is happening in your code. When you define U as a symbolic expression in the workspace you expect SOLVE to see it when looking at your string, but it doesn't. When evaluating:

solve('diff(U,x1)=0','diff(U,x2)=0','x1,x2') % Your code

SOLVE ends up with this:

solve('0=0','0=0','x1,x2') % U is a constant! dU/dv = 0

which returns the same thing your original code did.

[Matt Fig]

Jan commented:

Oh I see! Thank you Matt, now I understand what was going on...Have a nice day!

Answer 2

[Wayne King]

   syms x1 x2 F;
   U=x1^2+(x2-F)^2;
   sol = solve(diff(U,x1)==0,diff(U,x2)==0,x1,x2);
   sol.x1
   sol.x2

Identical for me to:

    sol2 = solve(2*x1==0,2*(x2-F)==0,x1,x2);
    sol2.x1
    sol2.x2

[Matt Fig]

Jan commented:

Thanks for your answer, I should add that I'm using MATLAB R2010b and the syntax has changed, but that still doesn't explain this behavior to me. Your modified code

     syms x1 x2 F;
     U=x1^2+(x2-F)^2;
     sol = solve('diff(U,x1)=0','diff(U,x2)=0','x1,x2')
     sol.x1
     sol.x2

still generates:

Warning: Could not extract individual solutions. Returning a MuPAD set object.

I will update to a recent version, because I'm almost certain that this is not due to a trivial mistake on my part (although it is possible of course). Thank you for any further comments!