Boundary Value Problems/Lesson 7

Boundary Value Problems

Rectangular Domain (R²)

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

Disk Domain (Polar)

File:Bvpdisc.jpg

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.

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

$r^2{R}^{″}\mathrm{\Theta }+r{R}^{\prime }\mathrm{\Theta }+R{\mathrm{\Theta }}^{″}=0$

Separate,

$\frac{r^2{R}^{″}+r{R}^{\prime }}{R}+\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}=0$
$\frac{r^2{R}^{″}+r{R}^{\prime }}{R}=-\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}=\text{Constant}$

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

• ${\lambda }^2>0$
  

$-\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}={\lambda }^2$

${\mathrm{\Theta }}^{″}+{\lambda }^2\mathrm{\Theta }=0$

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

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

Example of Potential equation on semi-annulus.

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