Green's function

In physics and mathematics, Green's functions are auxiliary functions in the solution of linear partial differential equations. Green's function is named for the British mathematician George Green (1793 – 1841).

Definition

Let L_x be a given linear differential operator in n variables x = (x₁, x₂, ..., x_n), then the Green function of L_x is the function G(x,y) defined by

$L_{\mathbf {x} }G(\mathbf {x} ,\mathbf {y} )=\delta (\mathbf {x} -\mathbf {y} ),$

where δ(x-y) is the Dirac delta function. Once G(x,y) is known, any differential equation involving L_x is formally solved. Suppose we want to solve,

$L_{\mathbf {x} }\,\phi (\mathbf {x} )=\rho (\mathbf {x} )$

for a known right hand side ρ(x). The formal solution is

$\phi (\mathbf {x} )=\int \;G(\mathbf {x} ,\mathbf {y} )\;\rho (\mathbf {y} )\;\mathrm {d} \mathbf {y} .$

The proof is by verification,

$L_{\mathbf {x} }\,\phi (\mathbf {x} )=\int \;L_{\mathbf {x} }\;G(\mathbf {x} ,\mathbf {y} )\;\rho (\mathbf {y} )\;\mathrm {d} \mathbf {y} =\int \;\delta (\mathbf {x} -\mathbf {y} )\;\rho (\mathbf {y} )\mathrm {d} \mathbf {y} =\rho (\mathbf {x} )$

where in the last step the defining property of the Dirac delta function is used.

The integral operator that has the Green function as kernel may be seen as the inverse of a linear operator,

$L_{\mathbf {x} }\;\phi (\mathbf {x} )=\rho (\mathbf {x} )\quad \Longrightarrow \quad \phi (\mathbf {x} )=L_{\mathbf {x} }^{-1}\;\rho (\mathbf {x} )=\int G(\mathbf {x} ,\mathbf {y} )\rho (\mathbf {y} )\;\mathrm {d} \mathbf {y} .$

It is illuminating to make the analogy with matrix equations. Let $\mathbb {L}$ and $\mathbb {G}$ be n×n matrices connected by

$\mathbb {L} \mathbb {G} =\mathbb {E} \quad \Longleftrightarrow \quad \left(\mathbb {L} \mathbb {G} \right)_{ij}=\delta _{ij},\quad {\hbox{i.e.,}}\quad \mathbb {G} =\mathbb {L} ^{-1},$

then the solution of a matrix-vector equation is

$\mathbb {L} {\boldsymbol {\phi }}={\boldsymbol {\rho }}\quad \Longrightarrow \quad \phi _{i}=\sum _{j}\mathbb {G} _{ij}\rho _{j}.$

Make the correspondence i ↔ x, j ↔ y, and compare the sum over j with the integral over y, and the correspondence is evident.

Example

We consider a case of three variables, n = 3 with x = (x, y, z).

The Green function of

$-{\frac {1}{4\pi }}{\boldsymbol {\nabla }}^{2}\equiv -{\frac {1}{4\pi }}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)$

is

$G(\mathbf {x} ,\mathbf {y} )={\frac {1}{|\mathbf {x} -\mathbf {y} |}}.$

That is, the formal solution of the Poisson equation of electrostatics

$\nabla ^{2}\Phi (\mathbf {x} )=-{\frac {1}{\epsilon _{0}}}\rho (\mathbf {x} ),$

where ε₀ is the electric constant and ρ is a charge distribution, is given by

$\Phi (\mathbf {x} )={\frac {1}{4\pi \epsilon _{0}}}\iiint \;{\frac {\rho (\mathbf {y} )}{|\mathbf {x} -\mathbf {y} |}}\;\mathrm {d} \mathbf {y}$

Proof

(To be continued)

Reference

P. Roman, Advanced Quantum Theory, Addison-Wesley, Reading, Mass. (1965) Appendix 4.