# Boundary Value Problems/Lesson 7

## Rectangular Domain (R2)

$\displaystyle u_{xx}+u_{yy}=0$ ## Disk Domain (Polar)

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

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

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

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

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

$\displaystyle {\frac {r^{2}R''+rR'}{R}}+{\frac {\Theta ''}{\Theta }}=0$ $\displaystyle {\frac {r^{2}R''+rR'}{R}}=-{\frac {\Theta ''}{\Theta }}={\mbox{Constant}}$ The constant may be greater than , equal to or less than zero.

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

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

%%(php) <?php echo "Hello, World!"; ?> %%