- #1
zi-lao-lan
- 4
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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!
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!