Green's function

This article covers Green's function as a tool for solving differential equations. For the Green function in stochastics see Green function (stochastics) .

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Green's functions are an important tool for solving inhomogeneous linear partial differential equations . They are named after the physicist and mathematician George Green . Using Green's formulas , he solved a special Dirichlet problem . A special solution to this partial boundary value problem that occurs in this procedure and with the help of which one can determine further solutions through the superposition principle , today bears the name Green's function. Until today, this solution method described by Green has been extended to a larger class of differential equations or of boundary value problems . Therefore, the concept of Green's function was placed in a much more general context. Laurent Schwartz transferred the Green function into the context of the distribution theory he developed . There it is understood as distribution itself and is often referred to as a fundamental solution. Other authors also refer to them as Green's function in the context of distributions.

In potential theory and gravity measurement it is u. a. used to solve the first boundary value problem . In theoretical physics , especially in high-energy and many-particle physics, an abundance of different functions is also defined, all of which are referred to as "Green's functions" and are related to the functions given here in one form or another, without this being related to the would be recognizable at first glance. These functions, especially the propagators of the relativistic quantum theories, are not meant in the following.

motivation

An inhomogeneous linear differential equation with constant complex coefficients has the form

${\displaystyle L\,y=f}$,

where is a linear differential operator . The aim is to find a particular solution to the inhomogeneity . One would now like to find something like a "reverse operator" , because then one could write the solution of the above equation as . But if has non-trivial solutions, it is not injective , so there can be no left inverse. But it is surjective if the equation has solutions for each of a suitable function space. Therefore one can look for a right inverse operator for which ${\displaystyle L}$ ${\displaystyle y_{p}}$ ${\displaystyle f}$ ${\displaystyle L^{-1}}$ ${\displaystyle y=L^{-1}f}$ ${\displaystyle Ly=0}$ ${\displaystyle L}$ ${\displaystyle L}$ ${\displaystyle f}$ ${\displaystyle G}$

${\displaystyle LG=1}$

applies. With you have found a particular solution of the initial equation, because it applies ${\displaystyle y=Gf}$

${\displaystyle Ly=L(Gf)=(LG)f=1f=f}$.

The general solution is obtained by adding the general solution of the homogeneous problem to the particular solution. Is chosen as inhomogeneity the delta function , then called the fundamental solution of . Depending on the author and the main topic, it is also referred to as Green's function. ${\displaystyle \delta }$ ${\displaystyle G}$ ${\displaystyle L}$ ${\displaystyle G}$

For any inhomogeneity , the question now arises as to how the fundamental solution can be used. The following then applies by means of the convolution ${\displaystyle f}$ ${\displaystyle y_{p}}$ ${\displaystyle G}$ ${\displaystyle *}$

${\displaystyle L\,y_{p}=f=\delta *f=(L\,G)*f=L\,(G*f)\quad \Rightarrow \quad y_{p}=G*f:=\int G(x-x')f(x')\,\mathrm {d} x'}$.

Physically this describes the superposition principle , mathematically one speaks of the linearity of . ${\displaystyle L}$

Explanation of the individual steps:

The first equal sign is the starting equation . For each function that is convolution with the delta distribution possible and again supplies the output function: . Use , that is, that solves the differential equation with -inhomogeneity. If the derivative of a convolution is formed, the derivative is simply drawn into it, that is to say . Finally , the particulate solution can be identified, namely as a convolution of the fundamental solution with the inhomogeneity . ${\displaystyle Ly_{p}=f}$ ${\displaystyle f}$ ${\displaystyle \delta }$ ${\displaystyle f=\delta *f}$ ${\displaystyle \delta =LG}$ ${\displaystyle G}$ ${\displaystyle \delta }$ ${\displaystyle L(f*g)=(Lf)*g=f*(Lg)}$ ${\displaystyle Ly_{p}=L(G*f)}$ ${\displaystyle f}$

If, instead of a linear differential equation, one considers a linear differential equation with additional conditions such as boundary values or initial values , then the previously examined function is called Green's function. ${\displaystyle G}$

definition

Ordinary differential equations

Be

${\displaystyle \operatorname {L} :=L({\tfrac {\mathrm {d} }{\mathrm {d} t}})=\sum _{k=1}^{N}a_{k}(t){\tfrac {\mathrm {d} ^{k}}{\mathrm {d} t^{k}}}}$

a differential operator with its inhomogeneous differential equation . Then Green's function for this operator satisfies the fundamental equation: ${\displaystyle \operatorname {L} y=f}$ ${\displaystyle G}$

${\displaystyle \operatorname {L} G(t)=\delta (t)}$,

where is the delta distribution (ie it applies to an arbitrarily often differentiable function ). ${\displaystyle \delta (t)}$ ${\displaystyle \delta (\phi )=\phi (0)}$ ${\displaystyle \phi }$

Additional conditions may be added later, e.g. B. Retardation conditions (see below) or the equivalent "Sommerfeldian radiation condition" or an initial or boundary condition that makes it unambiguous. A special solution results from folding : ${\displaystyle G}$

${\displaystyle y(t)=(G*f)(t)\equiv \int _{-\infty }^{\infty }G(t-t')f(t')\mathrm {d} t'}$,

as can be seen as follows:

${\displaystyle \operatorname {L} y(t)=\int _{-\infty }^{\infty }\operatorname {L} G(t-t')f(t')\mathrm {d} t'=\int _{-\infty }^{\infty }\delta (t-t')f(t')\mathrm {d} t'=f(t).}$

For this corresponds to the stationary ("steady") response of the system, a damped harmonic oscillator, to a ballistic unit impact , i.e. H. on the special reduced driving force ${\displaystyle N=2}$

${\displaystyle f(t)=\delta (t)\equiv {\frac {1}{2\pi }}\,\int _{-\infty }^{\infty }e^{-\mathrm {i} \omega t}\mathrm {d} \omega }$.

Partial differential equations

The defining equation also applies to partial differential equations

${\displaystyle L\left(a_{ij}(x_{1},\ldots ,x_{n}){\frac {\partial ^{i}}{\partial x_{j}^{i}}}\right)G(x_{1},\ldots ,x_{n})=\delta (x_{1})\cdots \delta (x_{n})}$

and a special solution results from convolution:

${\displaystyle y=(2\pi )^{\frac {n}{2}}(G*f)(x_{1},\ldots ,x_{n})}$.

In this case, however, finding a Green function and calculating the multi-dimensional integrals are more problematic.

Green's function with boundary conditions

If you know a Green function for an operator , you can solve the inhomogeneous part of the differential equation without problems. For the general solution, however, one generally still has to meet boundary conditions . This can be done in many ways, but an elegant method is the addition of a solution to the homogeneous problem so that the boundary conditions are met. Clearly, when solving the Poisson equation , this corresponds to adding image charges and removing the edges, so that the previously specified values ​​are assumed where the edge was. As a simple example, think of a charged particle in front of a grounded plane. If you place an oppositely charged charge on the other side of the plane and mentally remove the plane, then where the plane was, the potential is zero, which fulfills the required boundary condition. ${\displaystyle L}$ ${\displaystyle LF=0}$

This method is often used to solve Poisson's equation (Gaussian units). Using the Gaussian integral theorem one finds ( ): ${\displaystyle \Delta \Phi =-4\pi \rho }$ ${\displaystyle G'=G+F}$

${\displaystyle {\begin{aligned}\Phi (\mathbf {r} )&=\int _{\Omega }\delta (\mathbf {r} -\mathbf {r} ')\Phi (\mathbf {r} ')\mathrm {d} ^{3}r'=\int _{\Omega }(\nabla \cdot \nabla G'(\mathbf {r} -\mathbf {r} '))\Phi (\mathbf {r} ')\mathrm {d} ^{3}r'\\&=\int _{\partial \Omega }\!\!\!\Phi (\mathbf {r} ')\nabla G'(\mathbf {r} -\mathbf {r} ')\cdot \mathrm {d} \mathbf {f} \,-\int _{\partial \Omega }\!\!\!G'(\mathbf {r} -\mathbf {r} ')\nabla \Phi (\mathbf {r} ')\cdot \mathrm {d} \mathbf {f} \,+\int _{\Omega }\!G'(\mathbf {r} -\mathbf {r} ')(-4\pi \rho (\mathbf {r} '))\mathrm {d} ^{3}r'\end{aligned}}}$

Depending on whether you have specified the potential or its derivative on the edge, you now choose the function that is to be added to in such a way that in the first case applies and is usually called Dirichlet's Green function . In the second case one does not choose - as would be obvious - such that it disappears, since this would violate Gaussian theorem. Instead, one chooses so that ${\displaystyle F}$ ${\displaystyle G}$ ${\displaystyle G'|_{\partial \Omega }=0}$ ${\displaystyle G'}$ ${\displaystyle G_{D}}$ ${\displaystyle F}$ ${\displaystyle \nabla G'\cdot \mathbf {n} |_{\partial \Omega }}$ ${\displaystyle F}$

${\displaystyle \nabla G'\cdot \mathbf {n} |_{\partial \Omega }={\frac {1}{|\partial \Omega |}}}$

applies (which in the above integral only produces the mean value of the potential across the surface, a constant around which the solution is undetermined anyway) and usually calls Neumann's Green function . The Green functions to be determined are often found in symmetrical problems from geometrical considerations. Alternatively, one can develop according to an orthonormal system of the operator. If a solution has been found, it is clearly determined, as follows directly from the maximum principle for elliptic differential equations. ${\displaystyle G'}$ ${\displaystyle G_{N}}$ ${\displaystyle F}$

Examples

Poisson problem

Often one understands by Green's function the integral kernel of the Laplace operator taking into account certain boundary values , that is to say for applies ${\displaystyle G}$ ${\displaystyle \Delta }$ ${\displaystyle G}$

${\displaystyle u(x)=\int G(x,y)\Delta u(y)\mathrm {d} y}$.

George Green used this function with the boundary value problems that follow from potential theory to determine Green's formulas . However, the importance of this result was only realized after his death.

In this section, Green's function of Dirichlet's problem is the Poisson equation

${\displaystyle {\begin{aligned}\Delta u=&f\quad {\mbox{in}}\;D\\u=&g\quad {\mbox{auf}}\;\partial D\end{aligned}}}$

where is the Laplace operator and an open bounded region with a smooth edge . The fundamental solution of the Laplace operator is ${\displaystyle \Delta }$ ${\displaystyle D\subset \mathbb {R} ^{n}}$ ${\displaystyle \partial D}$

${\displaystyle \Phi (x)={\begin{cases}{\frac {1}{2\pi }}\ln |x|&{\mbox{für}}\;\;n=2\\-{\frac {1}{n(n-2)\alpha (n)}}{\frac {1}{|x|^{n-2}}}&{\mbox{für}}\;\;n\geq 3\,,\end{cases}}}$

where the volume of the unit ball is in. Now fix and choose a ball around with radius , so very located. Define . On this set, the fundamental solution is smooth. It then follows from Green's formula ${\displaystyle \alpha (n)}$ ${\displaystyle \mathbb {R} ^{n}}$ ${\displaystyle x\in D}$ ${\displaystyle B(x,\epsilon )}$ ${\displaystyle x}$ ${\displaystyle \epsilon >0}$ ${\displaystyle B(x,\epsilon )}$ ${\displaystyle D}$ ${\displaystyle V_{\epsilon }:=D\setminus B(x,\epsilon )}$ ${\displaystyle y\mapsto \Phi (y-x)}$

${\displaystyle \;\int \limits _{V_{\epsilon }}\left[\Phi (y-x)\Delta u(y)-u(y)\Delta \Phi (y-x)\right]\mathrm {d} y=\int \limits _{\partial V_{\epsilon }}\left[\Phi (y-x){\frac {\partial u}{\partial n}}(y)-u(y){\frac {\partial \Phi }{\partial n}}(y-x)\right]\mathrm {d} S(y)}$,

where is the partial derivative with respect to the outer unit normal vector . There and is, it arises ${\displaystyle {\tfrac {\partial }{\partial n}}}$ ${\displaystyle \Delta \Phi =0}$ ${\displaystyle \partial V_{\epsilon }=\partial D\cup \partial B(x,\epsilon )}$

${\displaystyle \;\int \limits _{V_{\epsilon }}\Phi (y-x)\Delta u(y)\mathrm {d} y=\int \limits _{\partial D}\Phi (y-x)\left[{\frac {\partial u}{\partial n}}(y)-u(y){\frac {\partial \Phi }{\partial n}}(y-x)\right]\mathrm {d} S(y)+\int \limits _{\partial B(x,\epsilon )}\left[\Phi (y-x){\frac {\partial u}{\partial n}}(y)-u(y){\frac {\partial \Phi }{\partial n}}(y-x)\right]\mathrm {d} S(y)}$.

for apply ${\displaystyle \epsilon \to 0}$

${\displaystyle \left|\;\int \limits _{\partial B(x,\epsilon )}\Phi (y-x){\frac {\partial u}{\partial n}}(y)\mathrm {d} S(y)\right|\leq C\epsilon |\ln(\epsilon )|\to 0}$

and

${\displaystyle \;\int \limits _{\partial B(x,\epsilon )}u(y){\frac {\partial \Phi }{\partial n}}(y-x)\mathrm {d} S(y)={\frac {1}{2\pi \epsilon }}\int \limits _{\partial B(x,\epsilon )}u(y)\mathrm {d} S(y)\to u(x)}$,

from what

${\displaystyle u(x)=\int \limits _{\partial D}\left[\Phi (y-x){\frac {\partial u}{\partial n}}(y)-u(y){\frac {\partial \Phi }{\partial n}}(y-x)\right]\mathrm {d} S(y)-\int \limits _{D}\Phi (y-x)\Delta u(y)\mathrm {d} y}$

follows. This is one way of presenting the solution to the Poisson problem. However, in this context the normal derivation of is unknown. For this reason a correction function is introduced to solve the boundary value problem ${\displaystyle {\tfrac {\partial u}{\partial n}}}$ ${\displaystyle u}$ ${\displaystyle \phi _{x}}$

${\displaystyle {\begin{aligned}\Delta \phi _{x}=&0\;\qquad \qquad {\mbox{in}}\ D\\\phi _{x}=&\Phi (y-x)\quad {\mbox{auf}}\ \partial D\end{aligned}}}$

solves. Using the same reasoning as before, it follows from Green's formula

${\displaystyle -\int \limits _{D}\phi _{x}(y)\Delta u(y)\mathrm {d} y=\int \limits _{\partial D}u(y){\frac {\partial \phi _{x}}{\partial n}}(y)-\Phi (y-x){\frac {\partial u}{\partial n}}(y)\mathrm {d} S(y)}$.

Adding this equation to the representation of so found above gives the representation ${\displaystyle u}$

${\displaystyle u(x)=-\int \limits _{\partial D}u(y){\frac {\partial }{\partial n}}\left(\Phi (y-x)-\phi _{x}(y)\right)\mathrm {d} S(y)-\int \limits _{D}(\Phi (y-x))-\phi _{x}(y))\Delta u(y)\mathrm {d} y}$

without the term . The function is called Green's function of the Laplace operator for the area . It can also be shown that the function depends symmetrically on its arguments, i.e. it is true . ${\displaystyle {\tfrac {\partial u}{\partial n}}}$ ${\displaystyle G(x,y):=\Phi (y-x)-\phi _{x}(y)}$ ${\displaystyle D}$ ${\displaystyle G(x,y)=G(y,x)}$

Determination of the static electric field

According to Maxwell's equations , the source strength of the time-invariable electric field in a homogeneous, linear and isotropic material applies

${\displaystyle \mathbf {\nabla } \cdot \mathbf {E} (\mathbf {r} )={\frac {1}{\varepsilon _{0}}}\rho (\mathbf {r} )}$,

where is the electric field strength and the electric charge density. Since the electrostatic case is a conservative system , the following applies ${\displaystyle \mathbf {E} }$ ${\displaystyle \rho }$

${\displaystyle \mathbf {E} =-\mathbf {\nabla } \,\varphi }$,

where is the electrical potential. Insertion yields the Poisson equation ${\displaystyle \varphi }$

${\displaystyle \Delta \varphi =-{\frac {1}{\varepsilon _{0}}}\rho (\mathbf {r} )}$,

thus an inhomogeneous linear partial differential equation of the 2nd order. If one knows a Green function of the Laplace operator , then reads a particular solution ${\displaystyle G_{\Delta }}$ ${\displaystyle \Delta =\mathbf {\nabla } \cdot \mathbf {\nabla } }$

${\displaystyle \varphi _{p}(\mathbf {r} )=\int _{\Omega }G_{\Delta }(\mathbf {r} -\mathbf {r} ')\left(-{\frac {1}{\varepsilon _{0}}}\rho (\mathbf {r} ')\right)\mathrm {d} ^{n}r'}$.

A (not clearly determined) Green function of the Laplace operator in 3 dimensions is

${\displaystyle G_{\Delta }(\mathbf {r} )=-{\frac {1}{4\pi }}{\frac {1}{|\mathbf {r} |}}}$,

with which after insertion

${\displaystyle \varphi _{p}(\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int _{\Omega }{\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\mathrm {d} ^{3}r'\equiv \int _{Q}{\frac {-1}{4\pi |\mathbf {r} -\mathbf {r} '(q)|}}\,\mathrm {d} \left(-{\frac {q}{\varepsilon _{0}}}\right)}$

results. The last equation is intended to clarify the physical interpretation of Green's function. The Green function together with the differential represent a "potential impulse", the total potential is then obtained by superposing all "potential impulses", ie by executing the integral.

Inhomogeneous wave equation

This case is a little more difficult and of a different nature, because you are dealing with a hyperbolic differential equation rather than an elliptical one. This is where the complications indicated above arise.

Green's function by Fourier analysis

The inhomogeneous wave equation has the form

${\displaystyle \Box u(\mathbf {r} ,t)\equiv \left({\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\Delta \right)u(\mathbf {r} ,t)=f(\mathbf {r} ,t)}$.

By means of Fourier decomposition one finds after executing the operator for the Fourier transform

${\displaystyle \left(k^{2}-{\frac {\omega ^{2}}{c^{2}}}\right){\tilde {u}}(\mathbf {k} ,\omega )={\tilde {f}}(\mathbf {k} ,\omega )}$.

According to the convolution theorem we have:

${\displaystyle {\tilde {G}}(\mathbf {k} ,\omega )={\frac {1}{4\pi ^{2}}}\,{\frac {1}{k^{2}-{\frac {\omega ^{2}}{c^{2}}}}}}$.

The inverse transformation can be calculated and found with the help of the residual calculus

${\displaystyle G(\mathbf {r} ,t)={\frac {1}{(2\pi )^{4}}}\int \mathrm {d} ^{3}k\int \mathrm {d} \omega {\frac {\mathrm {e} ^{-\mathrm {i} (\mathbf {k} \cdot \mathbf {r} +\omega t)}}{\left(k^{2}-{\frac {\omega ^{2}}{c^{2}}}\right)}}={\frac {1}{4\pi r}}\left(\delta \left(t-{\frac {r}{c}}\right)+\delta \left(t+{\frac {r}{c}}\right)\right)}$,

which naturally gives rise to two parts (“retarded” or “advanced” part) of Green's function. The argument in the first delta function,, means that a "cause" generated at the point in time at , due to the finite speed of propagation of the wave, only causes its "effect" at the point in time . For the second delta function, the result is that the field leads the inhomogeneity by the corresponding time interval. For reasons of causality, that would be unphysical if one viewed the inhomogeneity as a cause and the field as an effect; but it is entirely physical when the inhomogeneity acts as an absorber (receiver) of the wave. ${\displaystyle t-{\tfrac {r}{c}}}$ ${\displaystyle t=0}$ ${\displaystyle \mathbf {r} =0}$ ${\displaystyle {\tfrac {r}{c}}}$ ${\displaystyle \mathbf {r} }$

The retarded Green function , in which the inhomogeneity causally corresponds to a "transmission process" of expiring spherical waves, is thus

${\displaystyle G_{\mathrm {ret} }({\boldsymbol {r}},t)={\frac {\delta \left(t-{\frac {r}{c}}\right)}{4\pi r}}}$

The retarded solution of the wave equation then results from convolution:

${\displaystyle {\begin{aligned}u(\mathbf {r} ,t)&=(2\pi )^{\frac {4}{2}}(G_{\text{ret}}*f)(\mathbf {r} ,t)=\int \mathrm {d} ^{3}r'\int _{-\infty }^{t}\mathrm {d} t'\,G_{\text{ret}}(\mathbf {r} -\mathbf {r} ',t-t')f(\mathbf {r} ',t')\\&={\frac {1}{4\pi }}\int \mathrm {d} ^{3}r'{\frac {f(\mathbf {r} ',t-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}})}{|\mathbf {r} -\mathbf {r} '|}}\end{aligned}}}$

So there is a superposition principle with retardation : The solution is a superposition of escaping spherical waves ( Huygens principle , Sommerfeld radiation condition ), the formation of which takes place similar to electrostatics.

The advanced Green function , in which the inhomogeneity corresponds causally to a “receiving process” of incoming spherical waves, is

${\displaystyle G_{\mathrm {av} }({\boldsymbol {r}},t)={\frac {\delta \left(t+{\frac {r}{c}}\right)}{4\pi r}}}$.

Alternative derivation

If one assumes that the Green function of the Laplace operator is known (see main article Laplace operator and Poisson equation ), the retarded Green function of the wave equation can be obtained without Fourier transformation. First of all applies to any “smooth” function ${\displaystyle f}$

${\displaystyle \left({\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\Delta \right){\frac {f\left(t-{\frac {r}{c}}\right)}{4\pi r}}=\delta ({\boldsymbol {r}})f(t)}$,

where is the three-dimensional delta function. To see that the left side in the area is always zero, write the Laplace operator in spherical coordinates with the radial part in the shape . In the immediate vicinity of , the smooth function can be viewed as spatially constant . Applying the Laplace operator to the factor then creates the three-dimensional delta function. ${\displaystyle \delta ({\boldsymbol {r}})}$ ${\displaystyle r>0}$ ${\displaystyle \Delta f=r^{-1}\partial ^{2}(rf)/\partial r^{2}}$ ${\displaystyle r=0}$ ${\displaystyle f(t)}$ ${\displaystyle 1/4\pi r}$

The argument can be made more precise by expanding from to powers of , whereby the leading power must be treated separately when using the Laplace operator. ${\displaystyle f(t-r/c)}$ ${\displaystyle r}$

In particular, a Gaussian function can be chosen for. Since the delta distribution can be represented as a limit of Gaussian functions, the defining equation for the Green function of the wave equation is obtained in the limit . ${\displaystyle f}$ ${\displaystyle f\to \delta }$

Further examples

In the following table the Green functions are given for some operators.

comment	Differential operator ${\displaystyle L}$	Green's function ${\displaystyle G}$
	${\displaystyle \partial _{t}}$	${\displaystyle \Theta (t)}$
	${\displaystyle \partial _{t}+\gamma }$	${\displaystyle \Theta (t)\mathrm {e} ^{-\gamma t}}$
	${\displaystyle (\partial _{t}+\gamma )^{2}}$	${\displaystyle \Theta (t)t\mathrm {e} ^{-\gamma t}}$
one-dimensional harmonic oscillator	${\displaystyle \partial _{t}^{2}+2\gamma \partial _{t}+\omega _{0}^{2}}$	${\displaystyle \Theta (t)\mathrm {e} ^{-\gamma t}{\frac {\sin(t{\sqrt {\omega _{0}^{2}-\gamma ^{2}}})}{\sqrt {\omega _{0}^{2}-\gamma ^{2}}}}}$
two-dimensional Laplace operator	${\displaystyle \Delta _{2}=\partial _{x}^{2}+\partial _{y}^{2}}$	${\displaystyle {\frac {\ln(\varrho )}{2\pi }}}$
three-dimensional Laplace operator	${\displaystyle \Delta =\partial _{x}^{2}+\partial _{y}^{2}+\partial _{z}^{2}}$	${\displaystyle -{\frac {1}{4\pi \mathrm {r} }}}$
Helmholtz equation	${\displaystyle \Delta +k^{2}}$	${\displaystyle -{\frac {\mathrm {e} ^{-\mathrm {i} kr}}{4\pi r}}}$
Diffusion equation	${\displaystyle \partial _{t}-D\Delta }$	${\displaystyle \Theta (t)\left({\frac {1}{4\pi Dt}}\right)^{\frac {3}{2}}\mathrm {e} ^{-{\frac {r^{2}}{4Dt}}}}$
D'Alembert operator	${\displaystyle \Box ={\frac {1}{c^{2}}}\partial _{t}^{2}-\Delta }$	${\displaystyle {\frac {\delta (t-{\frac {\mathrm {r} }{c}})}{4\pi \mathrm {r} }}}$

See also

• distribution

Web links

• Eric W. Weisstein : Greens Function . In: MathWorld (English).

Individual evidence

1. ↑ Hans Niels Jahnke (Ed.): A History of Analysis . AMS, 2003, ISBN 0-8218-2623-9 , pp. 204 .
2. ↑ Eric W. Weisstein : Greens Function . In: MathWorld (English).
3. ^ Lars Hörmander : The Analysis of Linear Partial Differential Operators. Volume 1: Distribution Theory and Fourier Analysis. Second edition. Springer-Verlag, Berlin et al. 1990, ISBN 3-540-52345-6 ( Basic Teachings of Mathematical Sciences 256), p. 80.
4. ↑ Y. Chover, J. Rubinstein: An Introduction to partial differential equations. Pp. 208-214.
5. ↑ Y. Chover, J. Rubinstein: An Introduction to partial differential equations. P. 208.
6. ↑ Lawrence C. Evans : Partial Differential Equations. Reprinted with corrections. American Mathematical Society, Providence RI 2008, ISBN 978-0-8218-0772-9 ( Graduate studies in mathematics 19), pp. 33-35.
7. ↑ Y. Chover, J. Rubinstein: An Introduction to partial differential equations. Pp. 208-215.
8. ^ RP Feynman , Lectures on Physics , Volume 2: Electrodynamics, Oldenbourg-Verlag 2001, Section 21-2.
9. ↑ In part taken from H. Schulz , Physik mit Pencil , Verlag Harri Deutsch 2009, page 155.