Green's formulas

In mathematics , especially vector analysis , the two Green formulas (sometimes also Green identities , Green theorems or theorems ) are special applications of the Gaussian integral theorem . They are named after the mathematician George Green . They are used, among other things, in electrostatics when calculating potentials. The formulas are not to be confused with Green's theorem , which deals with plane integrals.

In the following, let it be compact with a smooth edge in sections and let two functions be on , where it is continuously differentiable single and double . is the Nabla operator . ${\displaystyle U\subset \mathbb {R} ^{n}}$ ${\displaystyle \phi }$ ${\displaystyle \psi }$ ${\displaystyle U}$ ${\displaystyle \phi }$ ${\displaystyle \psi }$ ${\displaystyle \nabla }$

First Green identity

${\displaystyle \int \limits _{U}(\phi \nabla ^{2}\psi +\nabla \phi \cdot \nabla \psi )\,\mathrm {d} U=\int \limits _{\partial U}\phi {\frac {\partial \psi }{\partial n}}\mathrm {d} S}$,

where denotes the normal derivative of , i.e. the normal component of the gradient of on the edge . ${\displaystyle {\tfrac {\partial \psi }{\partial n}}=\nabla \psi \cdot {\vec {n}}}$ ${\displaystyle \psi }$ ${\displaystyle \psi }$ ${\displaystyle \partial U}$

This identity can be proven as follows:

${\displaystyle \;\int \limits _{\partial U}\phi {\frac {\partial \psi }{\partial n}}\mathrm {d} S=\int \limits _{\partial U}(\phi \,\nabla \psi )\cdot {\vec {n}}\,\mathrm {d} S=\int \limits _{U}\nabla \cdot (\phi \,\nabla \psi )\,\mathrm {d} U=\int \limits _{U}(\phi \nabla ^{2}\psi +\nabla \phi \cdot \nabla \psi )\,\mathrm {d} U}$,

where in the second step the Gaussian integral theorem in the form

${\displaystyle \;\int \limits _{\partial U}{\vec {F}}\cdot {\vec {n}}\;\mathrm {d} S=\int \limits _{U}\nabla \cdot {\vec {F}}\;\mathrm {d} U}$

was used.

Second Green identity

${\displaystyle \int \limits _{U}(\phi \nabla ^{2}\psi -\psi \nabla ^{2}\phi )\,\mathrm {d} U=\int \limits _{\partial U}\left(\phi {\frac {\partial \psi }{\partial n}}-\psi {\frac {\partial \phi }{\partial n}}\right)\,\mathrm {d} S}$

The second Greensian identity follows from the first Greensian identity, whereby it is now assumed that it is also continuously differentiable twice: ${\displaystyle \phi }$

${\displaystyle \int \limits _{U}(\phi \nabla ^{2}\psi +\nabla \phi \cdot \nabla \psi )\,\mathrm {d} U=\int \limits _{\partial U}\phi {\frac {\partial \psi }{\partial n}}\,\mathrm {d} S}$,

${\displaystyle \int \limits _{U}(\psi \nabla ^{2}\phi +\nabla \psi \cdot \nabla \phi )\,\mathrm {d} U=\int \limits _{\partial U}\psi {\frac {\partial \phi }{\partial n}}\,\mathrm {d} S}$

If you now subtract the second equation from the first equation, the second Green identity results.

Applications in electrostatics (3D)

Uniqueness set

For an electrostatic potential , the Poisson's equation applies where the electrical charge density is ( Gaussian system of units ). If the charge density is given in a volume , and if the values ​​of are also given on the edge ( Dirichlet boundary condition ), then: ${\displaystyle \phi \!\,}$ ${\displaystyle \nabla ^{2}\phi =-4\pi \rho }$ ${\displaystyle \rho \!\,}$ ${\displaystyle U\!\,}$ ${\displaystyle \partial U}$ ${\displaystyle \phi \!\,}$

Within is clearly defined. ${\displaystyle U\!\,}$ ${\displaystyle \phi ({\vec {r}})}$

Proof: Let and two potentials meet the same requirements for charge density and boundary values. The following then applies to the difference function ${\displaystyle \phi _{1}({\vec {r}})}$ ${\displaystyle \phi _{2}({\vec {r}})}$

${\displaystyle \chi ({\vec {r}}):=\phi _{1}({\vec {r}})-\phi _{2}({\vec {r}})\qquad {\begin{array}{rl}\nabla ^{2}\chi ({\vec {r}})=0&{\vec {r}}\in U\\\chi ({\vec {r}})=0&{\vec {r}}\in \partial U\end{array}}}$

If you insert for and also for in the first Green formula , it follows ${\displaystyle \chi \!\,}$ ${\displaystyle \phi \!\,}$ ${\displaystyle \psi \!\,}$

${\displaystyle \int _{U}\nabla \chi \cdot \nabla \chi \,\mathrm {d} U=0}$

So the gradient must vanish everywhere in , thus be constant, and because of its zero boundary value even be constantly equal to zero. So within . ${\displaystyle \nabla \chi }$ ${\displaystyle U\!\,}$ ${\displaystyle \chi \,}$ ${\displaystyle \phi _{1}({\vec {r}})=\!\,\phi _{2}({\vec {r}})}$ ${\displaystyle U\!\,}$

NB In the proof the Poisson equation and thus the charge density is only within of uses. ${\displaystyle U\!\,}$

Shielding by closed conductor surface

${\displaystyle \partial U}$is a closed conductor surface, so that the electrostatic potential to a constant value has ( equipotential surface ). For example, it can be physically implemented by grounding the conductor surface. According to the uniqueness theorem, the potential curve within is already determined by the charge distribution in and by the boundary value. As a result, electrical charges in the exterior have no influence on the potential curve in the interior. ${\displaystyle \phi \!\,}$ ${\displaystyle \partial U}$ ${\displaystyle \phi _{0}\!\,}$ ${\displaystyle \phi _{0}\!\,=0}$ ${\displaystyle U\!\,}$ ${\displaystyle U\!\,}$

If the closed conductor surface is not earthed, then the boundary values of are still constant, but with an unknown value. This value can depend on which charges are present outside of . The proof of the uniqueness theorem can be generalized to the effect that the difference function is still constant in , but no longer equal to zero. The constant does not play a role for the electric field strength, which is obtained by deriving from the potential; the electric field strength is therefore shielded even without grounding. ${\displaystyle \phi \!\,}$ ${\displaystyle U\!\,}$ ${\displaystyle \chi \!\,}$ ${\displaystyle U\!\,}$

Symmetry of the Green function

The Green function with Dirichlet boundary condition and with a vector parameter is defined by ${\displaystyle {\vec {r}}_{1}\in U}$

${\displaystyle {\begin{array}{cl}\nabla ^{2}G_{D}({\vec {r}},{\vec {r}}_{1})=\delta ({\vec {r}}-{\vec {r}}_{1})&{\vec {r}}\in U\\G_{D}({\vec {r}},{\vec {r}}_{1})=0&{\vec {r}}\in \partial U\end{array}}}$

Except for one factor, this corresponds to the Poisson's equation for a potential that is generated by a point charge at the location and that has the boundary value 0 on the grounded surface . The existence of such a function is physically clear, and because of the uniqueness theorem it is uniquely determined. Although the roles of (measuring point) and (position of the charge) are physically different, there is a mathematical symmetry: ${\displaystyle -4\pi }$ ${\displaystyle \phi ({\vec {r}})}$ ${\displaystyle {\vec {r}}_{1}}$ ${\displaystyle \partial U}$ ${\displaystyle {\vec {r}}}$ ${\displaystyle {\vec {r}}_{1}}$

${\displaystyle G_{D}({\vec {r}}_{2},{\vec {r}}_{1})=G_{D}({\vec {r}}_{1},{\vec {r}}_{2})}$

Proof: Put in the second Green formula

${\displaystyle \phi ({\vec {r}})=G_{D}({\vec {r}},{\vec {r}}_{1})\qquad \psi ({\vec {r}})=G_{D}({\vec {r}},{\vec {r}}_{2})}$

so on the left side one obtains integrals with delta functions, which result in. On the right hand side the integrands vanish because of the boundary values ​​of . ${\displaystyle G_{D}({\vec {r}}_{2},{\vec {r}}_{1})-G_{D}({\vec {r}}_{1},{\vec {r}}_{2})}$ ${\displaystyle G_{D}}$

Potential expressed by charge density and boundary values

If one uses the second Green formula as an integration variable and one lets the electrostatic potential be, one obtains with and with the help of the symmetry of the explicit expression ${\displaystyle {\vec {r}}_{1}}$ ${\displaystyle \phi \!\,}$ ${\displaystyle \psi ({\vec {r}}_{1})=G_{D}({\vec {r}},{\vec {r}}_{1})}$ ${\displaystyle G_{D}}$

${\displaystyle \phi ({\vec {r}})=-4\pi \int _{U}G_{D}({\vec {r}},{\vec {r}}_{1})\rho ({\vec {r}}_{1})dU_{1}+\int _{\partial U}\phi ({\vec {r}}_{1}){\frac {\partial G_{D}({\vec {r}},{\vec {r}}_{1})}{\partial n_{1}}}dS_{1}}$

Integral equation for the potential

Using the Green formula shown above, expressions for the electrostatic potential of a charge distribution can be derived. It should be the charge density at the site . The local potential is denoted by. The function is wanted . ${\displaystyle \rho ({\vec {r}}\,')}$ ${\displaystyle {\vec {r}}\,'}$ ${\displaystyle \phi ({\vec {r}})}$ ${\displaystyle {\vec {r}}}$ ${\displaystyle \phi }$

We bet for . The following then applies: ${\displaystyle \psi ({\vec {r}}\,')={\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}}$

1. ${\displaystyle \Delta '\psi ({\vec {r}}\,')=\Delta '{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}=-4\pi \delta ({\vec {r}}-{\vec {r}}\,')}$,

   • where is the Laplace operator , ${\displaystyle \Delta =\nabla ^{2}}$
   • the dash indicates that this operator affects the deleted variable
   • and the delta distribution is. ${\displaystyle \delta }$

   This identity is to be understood in the sense of distributional derivations .

2. ${\displaystyle \Delta '\phi ({\vec {r}}\,')=-4\pi \rho ({\vec {r}}\,')}$with the load distribution on site . ${\displaystyle \rho }$ ${\displaystyle {\vec {r}}\,'}$

If we put both in the second Green identity, we get on the left:

${\displaystyle \int \limits _{V}(\phi ({\vec {r}}\,')(-4\pi \delta ({\vec {r}}-{\vec {r}}\,'))-{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}(-4\pi \rho ({\vec {r}}\,')))\,\mathrm {d} V'=-4\pi \int \limits _{V}\phi ({\vec {r}}\,')\delta ({\vec {r}}-{\vec {r}}\,')\,\mathrm {d} V'+4\pi \int \limits _{V}{\frac {\rho ({\vec {r}}\,')}{|{\vec {r}}-{\vec {r}}\,'|}}\,\mathrm {d} V'}$.

The right side of the identity is:

${\displaystyle \int \limits _{F}\left(\phi ({\vec {r}}\,'){\frac {\partial }{\partial n'}}{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}-{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}{\frac {\partial }{\partial n'}}\phi ({\vec {r}}\,')\right)\mathrm {d} F'}$.

Written as identity:

${\displaystyle -4\pi \int \limits _{V}\phi ({\vec {r}}\,')\delta ({\vec {r}}-{\vec {r}}\,')\,\mathrm {d} V'+4\pi \int \limits _{V}{\frac {\rho ({\vec {r}}\,')}{|{\vec {r}}-{\vec {r}}\,'|}}\,\mathrm {d} V'=\int \limits _{F}\left(\phi ({\vec {r}}\,'){\frac {\partial }{\partial n'}}{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}-{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}{\frac {\partial }{\partial n'}}\phi ({\vec {r}}\,')\right)\mathrm {d} F'}$.

Within the volume, the following applies because of the function ${\displaystyle {\vec {r}}}$ ${\displaystyle \delta }$

${\displaystyle -4\pi \int \limits _{V}\phi ({\vec {r}}\,')\delta ({\vec {r}}-{\vec {r}}\,')\mathrm {d} V'=-4\pi \phi ({\vec {r}})}$

With this we can finally resolve and obtain the above identity according to the potential:

${\displaystyle \phi ({\vec {r}})=\int \limits _{V}{\frac {\rho ({\vec {r}}\,')}{|{\vec {r}}-{\vec {r}}\,'|}}\,\mathrm {d} V'-{\frac {1}{4\pi }}\int \limits _{F}\left(\phi ({\vec {r}}\,'){\frac {\partial }{\partial n'}}{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}-{\frac {1}{|{\vec {r}}-{\vec {r}}\,'|}}{\frac {\partial }{\partial n'}}\phi ({\vec {r}}\,')\right)\mathrm {d} F'}$.

literature

• John David Jackson: Classical Electrodynamics . Walter de Gruyter, Berlin 2006, ISBN 3-11-018970-4
• Walter Greiner: Theoretical Physics Volume 3 - Classical Electrodynamics . Verlag Harri Deutsch, Frankfurt am Main, Thun ISBN 3-8171-1184-3
• Otto Forster : Analysis 3. Integral calculus in R ⁿ with applications. 3rd edition. Vieweg-Verlag, 1996. ISBN 3-528-27252-X