smallphi
Jul24-08, 12:42 PM
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.
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.