Proof That If a Harmonic Function Zero on a Surface is Zero Interior to the Surface

Theorem
If  
\[\phi\]
  is a harmonic function on a volume  
\[V\]
  with surface  
\[S\]
  on which  
\[\phi =0\]
  then  
\[\phi =0\]
  on  
\[V\]
.
Proof
By Green's First Theorem for Harmonic Functions, if  
\[\psi = \phi\]
  we have
\[\int \int_V (\mathbf{\nabla \phi} ) \cdot (\mathbf{\nabla \phi} ) dV = \int \int_S \phi \frac{\partial \phi}{\partial n} dS \]

By the statement in the theorem  
\[\phi =0\]
  on  
\[S\]
.
\[\int \int_V (\mathbf{\nabla \phi} ) \cdot (\mathbf{\nabla \phi} ) dV = 0\]

Hence  
\[\phi = CONSTANT\]
  on  
\[S\]
.
But  
\[\phi\]
  is continuous and twice differentiable in  
\[V\]
, so  
\[\phi =0\]
  throughout  
\[V\]
.

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