Temperature distribution on a circular plate

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SUMMARY

The discussion centers on solving LaPlace's equation in plane polar coordinates for a circular plate's temperature distribution. The equation presented is (1/r)(d/dr(r.df/dr)) + 1/r² . d²/dφ² = 0, leading to a general solution of f = A₀ + Σ(r^m)(Aₘ cos(mφ) + Bₘ sin(mφ)). The critical boundary condition at r = 0 reveals that negative powers of r are not permissible as they lead to singularities, specifically R(0) = (0)^{-m}, which diverges when m is positive.

PREREQUISITES
  • Understanding of LaPlace's equation
  • Familiarity with polar coordinates
  • Knowledge of boundary conditions in differential equations
  • Basic concepts of series solutions in mathematical physics
NEXT STEPS
  • Study the implications of boundary conditions in solving partial differential equations
  • Explore the method of separation of variables in more complex geometries
  • Learn about the physical interpretations of temperature distributions in circular domains
  • Investigate the behavior of solutions near singular points in differential equations
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Students preparing for exams in mathematical physics, particularly those focusing on differential equations and boundary value problems, as well as educators teaching these concepts.

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Hi, I have a maths exam tomorrow (!) 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 tomorrow!
 
Physics news on Phys.org
Negative powers of r not m. The negative powers of r gives:

R(0) = (0)^{-m}
with m positive.
BOOM! It blows up at 0.
 
Oh
Thats really obvious, sorry :blushing:
 

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