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Solving laplaces equation in a 90deg annulus

  • #1
208
0

Homework Statement


Solve laplaces equation inside a 90 degree circular annulus(a<r<b, 0<theta<pi/2) subject to the boundary conditions:
[tex] u(r,0)=0[/tex]
[tex] u(r,\pi/2)=f(r)[/tex]
[tex] u(a,\theta)=0[/tex]
[tex] u(b,\theta)=0[/tex]

Laplaces Equation in 2 dimensions:
[tex]
r \left( {\frac {\partial }{\partial r}}u \left( r,\theta \right) +r{
\frac {\partial ^{2}}{\partial {r}^{2}}}u \left( r,\theta \right)
\right) =-{\frac {\partial ^{2}}{\partial {\theta}^{2}}}u \left( r,
\theta \right)
[/tex]
[tex] u(r,\theata) = \phi(\theta)*G(r)[/tex]
doing the usual method for sep of vars, i sub this and divide to get a separated expression:

[tex]
{\frac {r \left( {\frac {d}{dr}}G \left( r \right) +r{\frac {d^{2}}{d{
r}^{2}}}G \left( r \right) \right) }{G \left( r \right) }}=-{\frac {{
\frac {d^{2}}{d{\theta}^{2}}}\phi \left( \theta \right) }{\phi \left(
\theta \right) }}
[/tex]
I then set both sides equal to [tex]-\lambda[/tex]:
[tex]
r \left( {\frac {d}{dr}}G \left( r \right) +r{\frac {d^{2}}{d{r}^{2}}}
G \left( r \right) \right) +G \left( r \right) \lambda=0
[/tex]
[tex]
{\frac {d^{2}}{d{\theta}^{2}}}\phi \left( \theta \right) =\phi \left(
\theta \right) \lambda
[/tex]

I use Cauchy-Euler method to solve the first ODE:
[tex]G = r^{p}[/tex]
subbing this into the differential equation i get:
[tex]
p(p-1)+p+\lambda=0
p = \pm i \sqrt{\lambda}
[/tex]
[tex]
G = r^{\pm i \sqrt{\lambda}}=e^{\pm i \sqrt{\lambda}*ln(r)}
[/tex]
this allows me to write (unfortunately incorrectly) the general solution the the first problem:
[tex]
G = c_{1}cos(\sqrt{\lambda}*ln(r))+c_{2}sin(\sqrt{\lambda}*ln(r))
[/tex]
I am stuck here -- the book says the general solution of this part should be:
[tex]
G = c_{1}cos(\sqrt{\lambda}*ln(\frac{r}{a}))+c_{2}sin(\sqrt{\lambda}*ln(\frac{r}{a}))
[/tex]
I'm not sure how they got to this (the argument of ln has changed). It makes sense, of course because the solution works but how they got there is unclear to me. If someone could please help me get past this barrier i would appeciate it greatly.
 

Answers and Replies

  • #2
208
0
I actually read the back of the book again and found something weird:

They say that the real solutions are the ones that i got but then they say:
"it is more convenient to use independent solutions which simplify at r = a, cos(root(lambda)ln(r/a)) and sin with the same argument."...

i can see why this would simplify the problem a great deal, but what allows them to do this?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,791
919
r ranges between a and b. You can simplify finding boundary values at a by using ln(r/a) (which is 0 at r= a) instead of ln(r). You can do this because ln(r/a)= ln(r)- ln(a) and sin(ln(r)-ln(a))= cos(ln(a))sin(ln(r)- sin(ln(a))cos(ln(r)) while cos(ln(r)- ln(a))= cos(ln(a))cos(ln(r))+ sin(ln(a))sin(ln(r)) so that changing from ln(r) to ln(r/a) only changes the constants multiplying sin(ln(r)) and cos(ln(r)).
 
  • #4
208
0
Ahh, I hadn't realized that. So this will only change my coefficients at the end of the problem? Thanks a lot for the help!
 

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