# Boundary Value Problems/Lesson 7

## Rectangular Domain (R2)

${\displaystyle \displaystyle u_{xx}+u_{yy}=0}$

## Disk Domain (Polar)

For a disk with a radius of "c", let the polar coordinates be ${\displaystyle 0 , and ${\displaystyle -\pi <\theta <\pi }$
${\displaystyle r^{2}u_{rr}+ru_{r}+u_{\theta \theta }=0}$

${\displaystyle u(c,\theta )=f(\theta )}$  , boundary condition.

${\displaystyle u(r,\pi )=u(r,-\pi )}$  continuity of potential.

${\displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )}$  continuity of derivative.

${\displaystyle \displaystyle u(r,\theta )=R(r)\Theta (\theta )}$  The solution as a product of two independent functions. By substitution into the above PDE we have:

${\displaystyle \displaystyle r^{2}R''\Theta +rR'\Theta +R\Theta ''=0}$

Separate,

${\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}+{\frac {\Theta ''}{\Theta }}=0}$
${\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}=-{\frac {\Theta ''}{\Theta }}={\mbox{Constant}}}$

The constant may be greater than , equal to or less than zero.

• ${\displaystyle \displaystyle \lambda ^{2}>0}$

${\displaystyle \displaystyle -{\frac {\Theta ''}{\Theta }}=\lambda ^{2}}$

${\displaystyle \displaystyle \Theta ''+\lambda ^{2}\Theta =0}$

${\displaystyle \displaystyle \Theta (\theta )=Acos(\lambda \theta )+Bsin(\lambda \theta )}$
Use the continuity conditions and try to determine something more about A, B and λ.
${\displaystyle \displaystyle u(r,\pi )=u(r,-\pi )}$  thus ${\displaystyle \displaystyle \Theta (\pi )=\Theta (-\pi )}$  and ${\displaystyle \displaystyle Acos(\lambda \pi )+Bsin(\lambda \pi )=Acos(\lambda -\pi )+Bsin(\lambda -\pi )}$
${\displaystyle \displaystyle Bsin(\lambda \pi )=-Bsin(\lambda \pi )}$
${\displaystyle \displaystyle 2Bsin(\lambda \pi )=0}$
Either ${\displaystyle \displaystyle B=0}$  or ${\displaystyle \displaystyle sin(\lambda \pi )=0}$
Before choosing, apply the second boundary condition:

The continuity of the derivative provides a second condition:
${\displaystyle \displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )}$  thus ${\displaystyle \displaystyle \Theta _{\theta }(\pi )=\Theta _{\theta }(-\pi )}$
${\displaystyle \displaystyle -A\lambda sin(\lambda \pi )+\lambda Bcos(\lambda \pi )=-A\lambda sin(\lambda -\pi )+B\lambda cos(\lambda -\pi )}$
${\displaystyle \displaystyle -A\lambda sin(\lambda \pi )=A\lambda sin(\lambda \pi )}$
${\displaystyle \displaystyle 2A\lambda sin(\lambda \pi )=0}$
Either ${\displaystyle \displaystyle A=0}$  or ${\displaystyle \displaystyle sin(\lambda \pi )=0}$
If either A or B are zero then ${\displaystyle \displaystyle sin(\lambda \pi )=0}$  also must hold. So all we need is ${\displaystyle \displaystyle sin(\lambda \pi )=0}$  which implies ${\displaystyle \displaystyle \lambda =n}$ . Remember ${\displaystyle \displaystyle sin(n\pi )=0,{\mbox{ }}n=1,2,...}$

## Example of Potential equation on semi-annulus.

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