- #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.