how to satisfy Laplace's equation ?


by hotel
Tags: equation, laplace, satisfy
hotel
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#1
Jun2-05, 11:10 AM
P: 13
Hi

I am not quit sure I have understand the laplace equation correctly. I hope some one can help me with it.

As far as I understand if we are able to differentiate any function twice, then the function is harmonic.

so we assume [tex]V(x,y)[/tex] is harmonic because of the above.

Does it means that [tex]\nabla ^2V[/tex] is consequently equal to zero ?

How would V behave if

[tex]\nabla ^2V>0[/tex]
and
[tex]\nabla ^2V<0[/tex]
?

thanku
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dextercioby
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#2
Jun2-05, 11:14 AM
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Nope,a function is harmonic on a domain if it is a solution of Laplace equation in that domain.

Daniel.
hotel
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#3
Jun2-05, 11:48 AM
P: 13
so for which conditions are the Laplace equation not satisfied ?

Is it only at discontinuities in some region of a function and discontinuities at the boundaries of a function ?

dextercioby
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#4
Jun2-05, 12:07 PM
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how to satisfy Laplace's equation ?


One requirement is that the function be [itex] C^{2} [/itex] class on that domain.If that domain is open,you don't have any boundary conditions.

Daniel.
hotel
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#5
Jun2-05, 12:30 PM
P: 13
what if the domain is closed ? how are the boundary conditions in this case ?

Or the equation can only be satisfied under open domains?
dextercioby
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#6
Jun2-05, 01:13 PM
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It doesn't matter whether the domain is open or closed,the equation is the same,but the requirements on the function may differ from one case to another.

Daniel.
mathwonk
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#7
Jun2-05, 11:52 PM
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take any complex differentiable function, i.e. any analytic function such as a complex polynomial. then the real part is harmonic, and vice versa.

the property of harmonicity has a physical meaning with respect to perhaps the distribution of temperature in a disc.

certainly not all C^2 functions are harmonic. for one thing harmonic functions have the famous maximum principle satisfied by holomorphic i.e. analytic functions, they never take their maximum on any open set. (i think.)


e.g. let us consider z^2 = (x+iy)^2 = x^2 -y^2 + 2ixy. then x^2 -y^2 and also xy are harmonic, but just x^2 is not harmonic because the second derivative wrt x is 2, while the second deriv wrt y is 0. so they are not negatives of each other.

in the classical theory of holomorphic functions, proving they exist (with certain boundary properties) was done first by proving harmonic functions exist.

this is the famous Dirichlet principle, assumed by riemann and proved by Neumann and Hilbert, and others.

there is a beautiful treatment in the wonderful trilogy of books by c.l.siegel, "topics in complex analysis", where he essentially goes through riemann's thesis and part of his work on abelian fucntions and provides almost all rigorous details for riiemann's claims, especially in siegel's chapters 2 and 4.


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