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

$\nabla ^{2}\equiv \left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)$

is

$G(\mathbf {x} ,\mathbf {y} )=-{\frac {1}{4\pi }}{\frac {1}{|\mathbf {x} -\mathbf {y} |}}.$

As an important example of this Green function we mention that the formal solution of the Poisson equation of electrostatics, reading

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

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

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

Indeed,

$\nabla ^{2}\Phi (\mathbf {x} )={\frac {1}{4\pi \epsilon _{0}}}\iiint \;\rho (\mathbf {y} )\nabla ^{2}{\frac {1}{|\mathbf {x} -\mathbf {y} |}}\;\mathrm {d} \mathbf {y} =-{\frac {4\pi }{4\pi \epsilon _{0}}}\iiint \;\rho (\mathbf {y} )\delta (\mathbf {x} -\mathbf {y} )\;\mathrm {d} \mathbf {y} =-{\frac {\rho (\mathbf {x} )}{\epsilon _{0}}}.$

The integral form of the electrostatic field may be seen as a consequence of Coulomb's law. The field at the point x due to the charge dQ = ρ(y)dy is equal to

$d\Phi (\mathbf {x} )={\frac {dQ}{4\pi \epsilon _{0}|\mathbf {x} -\mathbf {y} |}}.$

The field is additive in the charges, so integration gives the total field at x.

Proof of Green function of ∇²

Integration regions: yellow domain is U_ε, large sphere is V_R, small sphere is V_ε

Without loss of generality we take x as the origin (0, 0, 0) and replace y by r = (x, y, z) in the above formulation. The length of r is indicated by r.

The proof uses Green's theorem:

$\iiint \limits _{V_{a}}{\Big (}\phi \nabla ^{2}{\frac {1}{r}}-{\frac {1}{r}}\nabla ^{2}\phi {\Big )}\,dV=\iint \limits _{S_{a}}{\big (}\phi {\boldsymbol {\nabla }}{\frac {1}{r}}{\big )}\cdot d\mathbf {S} -\iint \limits _{S_{a}}{\big (}{\frac {1}{r}}{\boldsymbol {\nabla }}\phi {\big )}\cdot d\mathbf {S} ,$

where V_a is a sphere with radius a and S_a is the surface of this sphere. The smooth test function φ and its gradient vanish for large r,

$\phi (x,y,z)=0,\quad {\boldsymbol {\nabla }}\phi (x,y,z)=\mathbf {0} \quad {\hbox{for}}\quad r\geq R\quad {\hbox{with}}\quad r\equiv {\sqrt {x^{2}+y^{2}+z^{2}}}.$

Further we notice that

$\nabla ^{2}{\frac {1}{r}}=0\quad {\hbox{for}}\quad r\in U_{\epsilon },$

because r ≠ 0 in that region (see the figure, where the region is indicated in yellow). This result is most easily proved if we recall that in spherical polar coordinates

$\nabla ^{2}{\frac {1}{r}}={\frac {1}{r}}{\frac {\partial ^{2}r}{\partial r^{2}}}{\frac {1}{r}}={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}1=0.$

First apply Green's theorem to the large sphere of radius R

$\iiint \limits _{V_{R}}\phi \nabla ^{2}{\frac {1}{r}}\,dV=\iiint \limits _{V_{R}}{\frac {1}{r}}\nabla ^{2}\phi \,dV+\iint \limits _{S_{R}}{\big (}\phi {\boldsymbol {\nabla }}{\frac {1}{r}}{\big )}\cdot d\mathbf {S} -\iint \limits _{S_{R}}{\big (}{\frac {1}{r}}{\boldsymbol {\nabla }}\phi {\big )}\cdot d\mathbf {S} =\iiint \limits _{V_{R}}{\frac {1}{r}}\nabla ^{2}\phi \,dV$

because by assumption φ and its gradient vanish on S_R.

We consider the integral on the right hand side and we will show that

$\iiint \limits _{V_{R}}{\frac {1}{r}}\nabla ^{2}\phi \,dV=-4\pi \phi (\mathbf {0} ),$

from which the result to be proved directly. The main trick is to write

$\iiint \limits _{V_{R}}{\frac {1}{r}}\nabla ^{2}\phi \,dV=\lim _{\epsilon \rightarrow 0}\iiint \limits _{U_{\epsilon }}{\frac {1}{r}}\nabla ^{2}\phi \,dV$

and to consider first the integral over U_ε (the yellow domain in the figure) for non-zero ε. After the integral has been evaluated, the limit for zero ε is taken.

Since U_ε has two surfaces, Green's theorem cannot be applied directly, and therefore we write (see the figure),

$\iiint \limits _{U_{\epsilon }}=\iiint \limits _{V_{R}}-\iiint \limits _{V_{\epsilon }}$

and apply Green's theorem to the two terms. Recalling that we saw already the first term, we get

${\begin{aligned}\iiint \limits _{U_{\epsilon }}{\frac {1}{r}}\nabla ^{2}\phi \,dV&=\left[\iiint \limits _{V_{R}}\phi \nabla ^{2}{\frac {1}{r}}\,dV-\iiint \limits _{V_{\epsilon }}\phi \nabla ^{2}{\frac {1}{r}}\,dV\right]+\iint \limits _{S_{\epsilon }}{\big (}\phi {\boldsymbol {\nabla }}{\frac {1}{r}}{\big )}\cdot d\mathbf {S} -\iint \limits _{S_{\epsilon }}{\big (}{\frac {1}{r}}{\boldsymbol {\nabla }}\phi {\big )}\cdot d\mathbf {S} \\&=\left[\iiint \limits _{U_{\epsilon }}\phi \nabla ^{2}{\frac {1}{r}}\,dV\right]+\iint \limits _{S_{\epsilon }}{\big (}\phi {\boldsymbol {\nabla }}{\frac {1}{r}}{\big )}\cdot d\mathbf {S} -\iint \limits _{S_{\epsilon }}{\big (}{\frac {1}{r}}{\boldsymbol {\nabla }}\phi {\big )}\cdot d\mathbf {S} \\\end{aligned}}$

The integral between square brackets is zero because ∇²(1/r) is zero on U_ε. The last integral can be shown to vanish for small ε. Because φ and its gradient are smooth and finite, and r is constant (equal to ε) on the surface, we may write for small ε

$\iint \limits _{S_{\epsilon }}{\big (}{\frac {1}{r}}{\boldsymbol {\nabla }}\phi {\big )}\cdot d\mathbf {S} \approx {\frac {\langle \mathbf {n} \cdot {\boldsymbol {\nabla }}\phi (\epsilon )\rangle }{\epsilon }}\iint \limits _{S_{\epsilon }}dS=4\pi \epsilon \langle \mathbf {n} \cdot {\boldsymbol {\nabla }}\phi (\epsilon )\rangle \rightarrow 0,$

where we assumed that

${\boldsymbol {\nabla }}\phi (\mathbf {r} )\cdot d\mathbf {S} ={\boldsymbol {\nabla }}\phi (\mathbf {r} )\cdot \mathbf {n} \;dS\approx \langle {\boldsymbol {\nabla }}\phi (\epsilon )\cdot \mathbf {n} \rangle \;dS$

and that the value of the gradient averaged over the surface may be taken out of the integral. The remaining surface integral is equal to 4πε².

In order to evaluate the final integral we use

${\boldsymbol {\nabla }}{\frac {1}{r}}=-{\frac {\mathbf {r} }{r^{3}}}\equiv -{\frac {\mathbf {e} _{r}}{r^{2}}}\qquad {\hbox{and}}\qquad d\mathbf {S} =\mathbf {e} _{r}\;r^{2}\sin \theta \,d\phi d\theta$

so that

$\iint \limits _{S_{\epsilon }}{\big (}\phi {\boldsymbol {\nabla }}{\frac {1}{r}}{\big )}\cdot d\mathbf {S} =-\iint \limits _{S_{\epsilon }}\phi (\mathbf {r} )\sin \theta \,d\phi d\theta =-\phi (\epsilon )\;4\pi .$

The limit ε → 0 gives the desired result.

References

• P. Roman, Advanced Quantum Theory, Addison-Wesley, Reading, Mass. (1965) Appendix 4.
• I. M. Gel'fand and G. E. Shilov, Generalized Functions, Vol. 1, Academic Press, New York (1964)