System of differential equations without solution?

Question

Analyzing the certain problem in differential geometry, I obtained the following system of differential equations $$6f'(u)g'(u)-gf''(u)-fg''(u)=0,$$ $$-3g(u)f'(u)^2+3f(u)f'(u)g'(u)+f(u)g(u)f''(u)-g''(u)-f(u)^2g''(u)=0,$$ $$-3f'(u)^2+3g'(u)^2+f(u)f''(u)-g(u)g''(u)=0,$$ $$3g(u)f'(u)g'(u)-3f(u)g'(u)^2-f''(u)-g(u)^2f''(u)+f(u)g(u)g''(u)=0.$$ From geometry point of view, only solution should be constant functions, but are there any arguments from theory of differential equations which can confirm my thoughts.

Answer 1

Put your system in a more systematic form $$ \pmatrix{g&f\\-fg&1+f^2\\-f&g\\1+g^2&-fg} \pmatrix{f''\\g''} = \pmatrix{6f'g'\\-3gf'^2+3ff'g'\\−3f'^2+3g'^2\\3gf'g'−3fg'^2} $$

The first and third equations can be combined to $$ \pmatrix{f&-g\\g&f}\pmatrix{f''\\g''}=3\pmatrix{f'^2-g'^2\\2f'g'} \\\text{or}\\ (f+ig)(f''+ig'')=3(f'+ig')^2 $$ implying by integrating once $$f'+ig'=C(f+ig)^3\tag{*},$$ and twice $$ (f+ig)^{-2}=D-Cu\implies f^2+g^2=|D-Cu|^{-1}. $$

Now combine the 4th and 2nd equation to get similar second derivative terms like the first set of equations, $f(IV)+g(II)$ gives $$ ff''+gg''=-3(f^2g'^2-2ff'gg'+f'^2g^2)=-3(fg'-f'g)^2=-3Im((f-ig)(f'+ig'))^2 $$ But also $$ ff''+gg''=Re((f-ig)(f''+ig''))=3\frac{Re\Bigl((f-ig)(f'+ig'))^2\Bigr)}{f^2+g^2} $$ Now insert the above first-order differential equation (*) to get to the identity $$ -Im(C(f^2+g^2)(f+ig)^2)^2=\frac{Re\Bigl((C(f^2+g^2)(f+ig)^2)^2\Bigr)}{f^2+g^2} \\\iff\\ -(f^2+g^2)\,Im(C(f+ig)^2)^2=Re(C^2(f+ig)^4)=Re(C(f+ig)^2)^2-Im(C(f+ig)^2)^2 \\\iff\\ (1-|Cu-D|^{-1})\,Im(C(Cu-D))^2=Re(C(Cu-D))^2 \\\iff\\ -Im(C(Cu-D))^2=|Cu-D|\cdot Re(C^2(Cu-D)^2). $$

This last should be impossible for $C\ne 0$, starting from the degree inequality giving $Re(C^4)=0$, $Re(C^3D)=0$ etc. Note that there is still a combination of the 2nd and 4th equation open for further restrictions.