In physics, the Green's function (or fundamental solution) for the Laplacian (or Laplace operator) in three variables is used to describe the response of a particular type of physical system to a point source. In particular, this Green's function arises in systems that can be described by Poisson's equation, a partial differential equation (PDE) of the form
∇ 2 u ( x ) = f ( x ) {\displaystyle \nabla ^{2}u(\mathbf {x} )=f(\mathbf {x} )}
where ∇ 2 {\displaystyle \nabla ^{2}} is the Laplace operator in R 3 {\displaystyle \mathbb {R} ^{3}} , f ( x ) {\displaystyle f(\mathbf {x} )} is the source term of the system, and u ( x ) {\displaystyle u(\mathbf {x} )} is the solution to the equation. Because ∇ 2 {\displaystyle \nabla ^{2}} is a linear differential operator, the solution u ( x ) {\displaystyle u(\mathbf {x} )} to a general system of this type can be written as an integral over a distribution of source given by f ( x ) {\displaystyle f(\mathbf {x} )} :
u ( x ) = ∫ G ( x , x ′ ) f ( x ′ ) d x ′ {\displaystyle u(\mathbf {x} )=\int G(\mathbf {x} ,\mathbf {x'} )f(\mathbf {x'} )d\mathbf {x} '}
where the Green's function for Laplacian in three variables G ( x , x ′ ) {\displaystyle G(\mathbf {x} ,\mathbf {x'} )} describes the response of the system at the point x {\displaystyle \mathbf {x} } to a point source located at x ′ {\displaystyle \mathbf {x'} } :
∇ 2 G ( x , x ′ ) = δ ( x − x ′ ) {\displaystyle \nabla ^{2}G(\mathbf {x} ,\mathbf {x'} )=\delta (\mathbf {x} -\mathbf {x'} )}
and the point source is given by δ ( x − x ′ ) {\displaystyle \delta (\mathbf {x} -\mathbf {x'} )} , the Dirac delta function.
Motivation One physical system of this type is a charge distribution in electrostatics. In such a system, the electric field is expressed as the negative gradient of the electric potential, and Gauss's law in differential form applies:
E = − ∇ ϕ ( x ) ∇ ⋅ E = ρ ( x ) ε 0 {\displaystyle {\begin{aligned}\mathbf {E} &=-\mathbf {\nabla } \phi (\mathbf {x} )\\[1ex]{\boldsymbol {\nabla }}\cdot \mathbf {E} &={\frac {\rho (\mathbf {x} )}{\varepsilon _{0}}}\end{aligned}}}
Combining these expressions gives us Poisson's equation:
− ∇ 2 ϕ ( x ) = ρ ( x ) ε 0 {\displaystyle -\mathbf {\nabla } ^{2}\phi (\mathbf {x} )={\frac {\rho (\mathbf {x} )}{\varepsilon _{0}}}}
We can find the solution ϕ ( x ) {\displaystyle \phi (\mathbf {x} )} to this equation for an arbitrary charge distribution by first considering the distribution created by a point charge q {\displaystyle q} located at x ′ {\displaystyle \mathbf {x'} } :
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