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

  1. Jul 24, 2008 #1
    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
  2. jcsd
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