- #1
nathangrand
- 40
- 0
[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?
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?