Boundary Value Problems/Lesson 7

Boundary Value Problems

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 $$ 0 < r < c $$, 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.

$$\displaystyle - \frac {\Theta ''}{\Theta}= \lambda^2 $$ $$ \displaystyle \Theta ''+ \lambda^2 \Theta= 0 $$
 * $$\displaystyle \lambda^2 > 0 $$

$$\displaystyle \Theta(\theta) = A cos(\lambda \theta) + B sin( \lambda \theta ) $$ 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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