# BVP-Lesson-7

Boundary Value Problems

## Rectangular Domain ($R^2$)

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

## Disk Domain (Polar)

Disc of radius c

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

$u(c,\theta)=f(\theta)$ , 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:

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

Separate,

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

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

• $\displaystyle \lambda^2 > 0$

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

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

$\displaystyle \Theta(\theta) = A cos(\lambda \theta) + B sin( \lambda \theta )$
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 $\displaystyle A cos(\lambda \pi) + B sin( \lambda \pi )=A cos(\lambda -\pi) + B sin( \lambda -\pi )$
$\displaystyle B sin( \lambda \pi )=- B sin( \lambda \pi )$
$\displaystyle 2B sin( \lambda \pi )=0$
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 $\displaystyle \Theta_{\theta}(\pi)= \Theta_{\theta}( -\pi)$
$\displaystyle-A \lambda sin(\lambda \pi) + \lambda B cos( \lambda \pi )=-A \lambda sin(\lambda -\pi) + B \lambda cos( \lambda -\pi )$
$\displaystyle -A \lambda sin(\lambda \pi) =A \lambda sin(\lambda \pi)$
$\displaystyle 2A \lambda sin(\lambda \pi) =0$
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.

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