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Temperature distribution on a circular plate

  1. Jan 19, 2010 #1
    Hi, I have a maths exam tommorrow (!) and I'm stuck on something:

    Its about solving LaPlace's equation in plane polar coordinates. There is a circular plate witha temp dist that satisfies

    (1/r)(d/dr(r.df/dr)) + 1/r2 . d2/dp2 = 0
    (p is phi)
    so I used a separable eq of the form f=RP

    and solved it to get R = Crm+Dr-m
    when m doesn't equal 0

    and R(r) = A + B ln(r)
    when m=0

    The question is:
    What boundary condition does R(r) satisfy at r = 0? Use this to show that
    the general solution to this problem is

    f= A0 + SUM OF(rm)( Am cos(mp) + Bm sin(mp))

    the explanation is because you can't have negative powers of m when R(r) is finite at the centre. Why can't you have negative powers of m? Negative powers will give fractions but they will still be finite, so why isn't it

    f= A0 + SUM OF(Cmrm+Dmr-m)( Am cos(mp) + Bm sin(mp))

    Thanks if you can help before tommorrow!
     
  2. jcsd
  3. Jan 19, 2010 #2

    jambaugh

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    Science Advisor
    Gold Member

    Negative powers of r not m. The negative powers of r gives:

    [tex]R(0) = (0)^{-m}[/tex]
    with m positive.
    BOOM! It blows up at 0.
     
  4. Jan 19, 2010 #3
    Oh
    Thats really obvious, sorry :blushing:
     
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