- #1
netheril96
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I read a lot of books on the uniqueness theorem of Poisson equation,but all of them are confined to a bounded domain [tex]\Omega[/tex] ,i.e.
"Dirichlet boundary condition: [tex]\varphi[/tex] is well defined at all of the boundary surfaces.
Neumann boundary condition: [tex]\nabla\varphi[/tex]is well defined at all of the boundary surfaces.
Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.
"
However,in the method of mirror image,the domain is usually unbounded.For instance,consider the electric field induced by a point charge with a infinely large grounded conductor plate.In all of the textbooks,it is stated that "because of the uniqueness theorem...",but NO book has ever proved it in such a domain!
Some may say that we can regard the infinity as a special surface,but we CAN'T since this "surface" has a infinite area.I tried to prove it using the same way as that in a bounded domain,i.e. with the electric potential known in a bounded surface and [tex]\varphi \to 0{\rm{ }}(r \to \infty )[/tex],
I ended up with[tex]\int_S {\phi \frac{{\partial \phi }}{{\partial n}}} dS = {\int_V {\left( {\nabla \phi } \right)} ^2}dV[/tex]in which [tex]\phi[/tex] is the difference between two possible solution of the electric potential.
Let S be the surface of an infinite sphere,we have
[tex]4\pi \int_{r \to \infty } {{r^2}\phi \frac{{\partial \phi }}{{\partial r}}} dr = {\int_V {\left( {\nabla \phi } \right)} ^2}dV[/tex]
We have[tex]\phi \to 0(r \to \infty )[/tex],but it doesn't indicate [tex]{r^2}\phi \frac{{\partial \phi }}{{\partial r}} \to 0(r \to \infty )[/tex] So we can't conclude that [tex]\nabla \phi \equiv 0[/tex] so that the uniqueness theorem doesn't hold (or we cannot prove it with the same way proving uniqueness theorem in a bounded domain)
Or can anybody here prove that [tex]{r^2}\frac{{\partial \phi }}{{\partial r}}[/tex] is bounded?
"Dirichlet boundary condition: [tex]\varphi[/tex] is well defined at all of the boundary surfaces.
Neumann boundary condition: [tex]\nabla\varphi[/tex]is well defined at all of the boundary surfaces.
Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann boundary conditions): the uniqueness theorem will still hold.
"
However,in the method of mirror image,the domain is usually unbounded.For instance,consider the electric field induced by a point charge with a infinely large grounded conductor plate.In all of the textbooks,it is stated that "because of the uniqueness theorem...",but NO book has ever proved it in such a domain!
Some may say that we can regard the infinity as a special surface,but we CAN'T since this "surface" has a infinite area.I tried to prove it using the same way as that in a bounded domain,i.e. with the electric potential known in a bounded surface and [tex]\varphi \to 0{\rm{ }}(r \to \infty )[/tex],
I ended up with[tex]\int_S {\phi \frac{{\partial \phi }}{{\partial n}}} dS = {\int_V {\left( {\nabla \phi } \right)} ^2}dV[/tex]in which [tex]\phi[/tex] is the difference between two possible solution of the electric potential.
Let S be the surface of an infinite sphere,we have
[tex]4\pi \int_{r \to \infty } {{r^2}\phi \frac{{\partial \phi }}{{\partial r}}} dr = {\int_V {\left( {\nabla \phi } \right)} ^2}dV[/tex]
We have[tex]\phi \to 0(r \to \infty )[/tex],but it doesn't indicate [tex]{r^2}\phi \frac{{\partial \phi }}{{\partial r}} \to 0(r \to \infty )[/tex] So we can't conclude that [tex]\nabla \phi \equiv 0[/tex] so that the uniqueness theorem doesn't hold (or we cannot prove it with the same way proving uniqueness theorem in a bounded domain)
Or can anybody here prove that [tex]{r^2}\frac{{\partial \phi }}{{\partial r}}[/tex] is bounded?