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Harmonic functions - complex analysis

  1. Apr 11, 2006 #1
    so .. if f (z) = u + iv is analytic on D, then u and v are harmonic on D...
    now ...
    if f (z) never vanishes on the domain ...
    then show log |f (z)| is harmonic on the domain ...
    Recall: harmonic means second partial derivative of f with respect to x + second partial derivative of f with respect to y = 0

    umm? did they mean to say that harmonic means second partial derivative of log |f (z)| with respect to x + second partial derivative of log |f (z)| with respect to y = 0
     
    Last edited: Apr 11, 2006
  2. jcsd
  3. Apr 11, 2006 #2
    i mean in order to show log |f (z)| is harmonic on the domain, dont I need to prove second partial derivative of log |f (z)| with respect to x + second partial derivative of log |f (z)| with respect to y = 0?



    I hope my question makes sense ...
     
  4. Apr 11, 2006 #3
    umm bump ...
     
  5. Apr 11, 2006 #4

    HallsofIvy

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    Since that is the definition of "harmonic", yes, that's what they mean when they say "log |f(z)| is harmonic". Of course, you don't necessarily have to verify the definition to prove it. Since " if f (z) = u + iv is analytic on D, then u and v are harmonic on D... " you could instead find a v such that f(z)= log|f(z)|+ iv is analytic or a u such that f(z)= u+ log|f(z)|i is analytic. You might try to do that by using the Cauchy-Riemann conditions: u(x,y)+ iv(x,y) is analytic on D if and only if [itex]\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}[/itex] and [itex]\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}[/itex].
     
  6. Apr 11, 2006 #5
    see .. .what i was tryina to do ... is ... prove that second partial derivative of log |f (z)| with respect to x + second partial derivative of log |f (z)| with respect to y = 0 ... of course it wasnt giving me happy result lol ...although it is technically right ...
     
  7. Apr 11, 2006 #6
    ahhhhhh i still dunno how to do this problem ...

    like i know ... but i dont ... basically i am confused
     
  8. Apr 11, 2006 #7
    umm i still am confused

    isnt it supposed to be a very simple problem?
    why am i getting so confused?
     
  9. Apr 11, 2006 #8

    shmoe

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    So was it zero or not? If you post your work we will have an easier time troubleshooting.
     
  10. Apr 12, 2006 #9
    nope it wasnt zero... thats the problem!

    btw it's kinda hard to post my work, did you get a zero for it?

    i assure you it's not a hw problem ... lol

    if i had time, i would scan my work, but i need to know how to do it by tomorrow.
     
  11. Apr 12, 2006 #10

    shmoe

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    I don't see why it should be hard. If you aren't comfortable with latex:

    https://www.physicsforums.com/showthread.php?t=8997

    you can just use the usual ascii characters- be carefull with using enough parenthesis to make it unambiguous though.

    Just post what you had for the first and second derivatives for x and y.
     
  12. Apr 15, 2006 #11
    According to "Mathworld"

    http://mathworld.wolfram.com/AnalyticFunction.html

    A complex valued function is analytic in D if it is complex differentiable at every point in D.

    Note that the log function is complex differentiable for the absolute value of every complex number not equal to 0. That is,

    Log(z) in undefined when z = 0 and hence not differentiable when z = 0.

    You are given the fact that |f(z)| is never 0, so |f(z)| is always a positive real number.

    Log|f(z)| = Log|u + i*v| = Log[(u^2+v^2)^(1/2)]

    Maybe you could use the fact that u and v are are harmonic functions, take the 2nd partial derivitives of Log[(u^2+v^2)^(1/2)], add them together to get 0 hence completing the proof.

    Not sure if it will work, just a thought.

    Best Regards,

    Edwin G. Schasteen
     
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