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Trouble solving a differential equation

  1. Aug 31, 2011 #1
    Hi,

    How can I solve this equation (finding f(x,y)) ?

    [itex]\left (\frac{\partial f}{\partial x} \right )^2+ \left( \frac{\partial f}{\partial y} \right )^2=0[/itex]

    Thanks a lot.
     
  2. jcsd
  3. Sep 1, 2011 #2
    Can't you factor that with some i's? You know, like if I had:

    [tex](x^2+y^2)=0[/tex] and I wanted to factor it, then I'd write:

    [tex](x+iy)(x-iy)=0[/tex]

    Then that means:

    [tex](x+iy)=0[/tex]

    or:

    [tex]x-iy=0[/tex]

    Ok, same dif then. Now do yours.

    Nothing wrong with complex solutions. Try not to be intimidated by them.
     
  4. Sep 3, 2011 #3

    hunt_mat

    User Avatar
    Homework Helper

    Both term in the equation are greater than or equal to zero, so the equation says that they must both be zero and hence:
    [tex]
    \frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0
    [/tex]
    So what does this say about f?
     
  5. Sep 5, 2011 #4
    Thank you very much jackmell, that is very helpful.


    The partial derivatives vanish, I suppose it means that f is constant ?
     
  6. Sep 6, 2011 #5

    hunt_mat

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    Homework Helper

    Yep.
     
  7. Sep 6, 2011 #6
    What about f(x,y) = exp(x)exp(i*y)

    Then first derivative w.r.t. x is just f(x,y)

    First derivative w.r.t. y is i*f(x,y)

    Sum of their squares is zero, yet they are not constant.

    f is necessarily constant under the constraint that f has first partial derivatives which are functions mapping to the real numbers. Then the two squared terms are necessarily zero or greater, so must be zero. Then f would have to be constant.
     
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