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

1. Jul 24, 2008

### 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.

Last edited: Jul 24, 2008
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