PDE with functions for coefficients: f_v g_u + f_u g_v = 0

[smallphi]

I know the solution of

f_v g_u - f_u g_v = 0

where f and g are functions of (u,v) and the subscripts _u and _v denote partial derivatives. The equation can be viewed as a PDE for the unknown g with coefficients given by the partial derivatives of the known f. The equation sets the functional determinant of f and g to zero which means f and g are functionally dependent and the solution is g = function(f), for arbitrary differentiable 'function'.

On the other hand, what is the solution of the above equation if I flip the sign to plus:

f_v g_u + f_u g_v = 0

I got some particular solutions but is there a general solution, expressing g in terms of f, like before?
The solution I got so far is:

f(u,v) = a(u) + b(v)
g(u,v) = function (-a(u) + b(v))

where a, b and function are arbitrary differentiable functions of their arguments.

 

[jvicens]

Thank you for your question. The solution to the equation f_v g_u + f_u g_v = 0 is indeed g = function(f), where function is an arbitrary differentiable function. This is because the equation still sets the functional determinant of f and g to zero, meaning they are functionally dependent. However, the specific form of the solution may vary depending on the functions f and g. Your solution is one possible form of the solution, but there may be others as well.