Laplace equation with boundary condition

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SUMMARY

Anne, a PhD student researching the motions of damaged ships, is attempting to solve the Laplace equation ∇²φ=0 within a box under specific boundary conditions. The conditions require φ to equal 1 at x=-A and x=A, and to equal 0 at y=-B, y=B, z=Ztop, and z=Zbot. Despite trying a series solution involving cosine and hyperbolic functions, Anne struggles to satisfy both the boundary conditions and the Laplace equation simultaneously. A suggestion was made to consider a simpler solution of the form φ=x+c.

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ACG_PhD
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Good afternoon,

I am a PhD student in motions of damaged ships. I am trying to find a solution of Laplace equation inside a box with a set of boundary conditions such that:

2[itex]\phi[/itex]=0
[itex]\phi[/itex]x=1 when x=-A and x=A
[itex]\phi[/itex]y=0 when y=-B and y=B
[itex]\phi[/itex]z=0 when z=Ztop and z=Zbot


I have tried different solutions that look like
[itex]\phi[/itex]=x+Ʃ(cos(α(x+A))cos(β(y+B))cosh(γ(z-Ztop))

α=n*PI/A n=1...∞
β=m*PI/B m=1...∞
and α222

I can't find a solution that satisfy everything.
I either satisfy the boundary conditions or Laplace equation but not both.
The problem comes from the need to have a sinh(z)=0 for 2 values...
I can't figure out how to go around the problem.

Any advice would be really nice.

Thanks
Anne

PS: I have the three permutation to find [itex]\phi[/itex]x=1 then [itex]\phi[/itex]y=1 and [itex]\phi[/itex]z=1
I have the same problem in the three cases.
 
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ACG_PhD said:
Good afternoon,

I am a PhD student in motions of damaged ships. I am trying to find a solution of Laplace equation inside a box with a set of boundary conditions such that:

2[itex]\phi[/itex]=0
[itex]\phi[/itex]x=1 when x=-A and x=A
[itex]\phi[/itex]y=0 when y=-B and y=B
[itex]\phi[/itex]z=0 when z=Ztop and z=Zbot


I have tried different solutions that look like
[itex]\phi[/itex]=x+Ʃ(cos(α(x+A))cos(β(y+B))cosh(γ(z-Ztop))

α=n*PI/A n=1...∞

β=m*PI/B m=1...∞
and α222

I can't find a solution that satisfy everything.
I either satisfy the boundary conditions or Laplace equation but not both.
The problem comes from the need to have a sinh(z)=0 for 2 values...
I can't figure out how to go around the problem.

Any advice would be really nice.

Thanks
Anne

PS: I have the three permutation to find [itex]\phi[/itex]x=1 then [itex]\phi[/itex]y=1 and [itex]\phi[/itex]z=1
I have the same problem in the three cases.

Try x+c as the solution
 

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