Laplace's equation (2D, cartesian)

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SUMMARY

The discussion centers on Laplace's equation in a rectangular pipe configuration, specifically addressing boundary conditions that prevent a non-trivial solution. The sides at x=0 and x=a are held at V=0, while the sides at y=0 and y=b are maintained at V=V0. The general solution form is given as (A*eky + B*e-ky)*(C*sin(kx)+D*cos(kx)), where D is determined to be zero due to the boundary condition at x=0. The lack of symmetry in the problem compared to Griffiths' example leads to the conclusion that insufficient conditions exist to solve for constants, thus eliminating the possibility of a non-trivial solution.

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JJfortherear
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Homework Statement



Rectangular pipe, infinite in the z direction. The sides in the y-z plane (at x=0 and x=a) are held at V=0, while the sides in the x-z plane (at y=0 and y=b) are held at V=V0

Explain why there cannot be a non-trivial solution to this configuration.


Homework Equations



General form of solution is (A*eky + B*e-ky)*(C*sin(kx)+D*cos(kx))

D is zero from b.c. (V=0 at x=0)

The Attempt at a Solution



So Griffiths has this problem (and a non-trivial solution), except the pipe is symmetric about the x axis, so V(y)=V(-y), and from that, A=B. My version lacks this symmetry (and thus lacks enough conditions to solve for the constants), but I can't imagine the solutions fly out the window just from a shift of coordinates. Any help?
 
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I agree with you (and Griffiths), a non-trivial solution does exist.
 

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