- **Differential Equations**
(*http://www.physicsforums.com/forumdisplay.php?f=74*)

- - **Solve laplace's equation on semi infinite strip**
(*http://www.physicsforums.com/showthread.php?t=577619*)

Solve laplace's equation on semi infinite strip[itex]\nabla[/itex]^2(Z)=0
Z= 0 for x=0, y=0 Z= x(1-x) for y=0 Z=0 for y=infinity Range 0<x<1 and y>0 (suppose strictly speaking should be x=1 and x=0 too) So all I want to do is solve this Use separation of variables: X''/X = a^2 = -Y''/Y Gives X = Aexp(ax) + Bexp(-ax) and Y=Ccos(ay) + Dsin(ay) Or completely free to swap these around to give Y=Aexp(ay)+ Bexp(-ay) and X= Ccos(ax) +Dsin(ax) which I shall do as get further with boundary conditions Know that Z=XY As at y=infinity, Z=0 ==> A=0 At x=0, Z=0 ==>C = 0 At x=1, Z=0 ==> a=n*pi where n is an integer so have Z=Esin(n*pi*x)exp(-n*pi*y) where E is a new constant but how on earth do I make this compatible with the remaining boundary condition for y=0 ==> Z=x(1-x) = Esin(n*pi*x) ?????? Clearly must have gone wrong somewhere??? |

Re: Solve laplace's equation on semi infinite stripThe next step is very standard in solving PDEs by separation of variables, any book/article on the topic will explain it. You have to expand [itex]x(x-1)[/itex] as a series using [itex] \sin(\pi n x)[/itex]. You should find that
[tex] x(x-1) = \sum_{n=1}^\infty \frac{4 \left(-1+(-1)^n\right)}{n^3 \pi ^3}\sin(\pi n x)[/tex] |

All times are GMT -5. The time now is 06:44 AM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums