Jackson p244,Green function for wave equation

In summary, G=\frac{e^{ikR}}{R} is a solution for a homogenous equation that is not well behaved for R=0, and we can change the integration over the ball K(0,r) to integrate over the sphere to get G=-4\pi \delta(\vec{R}) .
  • #1
NinjaSlayer
2
0
Jackson electrodynamics 3rd. p244

I understood that
G=\frac{e^{ikR}}{R}
is a spetial solution for
( \nabla ^2 + k^2 )G =0 (R>0) .

but,why G=\frac{e^{ikR}}/{R} satisfy
( \nabla ^2 + k^2 )G =-4\pi \delta (\mathbf{R}) ?

How to normalize the Green function?
( \nabla ^2 + k^2 )\frac{e^{ikR}}{R}=...calculate...=0.

I can't understand.Please help me...
 
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  • #2
NinjaSlayer said:
Jackson electrodynamics 3rd. p244

I understood that
G=\frac{e^{ikR}}{R}
is a spetial solution for
( \nabla ^2 + k^2 )G =0 (R>0) .
Firstly
[tex]
G=\frac{e^{ikR}}{R}
[/tex]
is not strictly a solution of homogenous equation (i.e. equation without delta function
on RHS) because G is not well behaved for R=0 and because of it one has to consider all
derivatives near the point R=0 in distributional sense. It is true that for all points
[tex]R \neq 0 [/tex] we have [tex]( \nabla ^2 + k^2 )G =0 [/tex] but it's not true for
R=0 as G blows up there.

NinjaSlayer said:
but,why G=\frac{e^{ikR}}/{R} satisfy
( \nabla ^2 + k^2 )G =-4\pi \delta (\mathbf{R}) ?

There is a nice trick which shows why it is the case. Let's integrate the expression
[tex] ( \nabla ^2 + k^2 )G [/tex] over small ball of radious r in the limit r goes to 0.
Notice that:
[tex]
\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2 + k^2 )G =
\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2)G
[/tex]
Gauss theorem tells
us that we can change integration over ball K(0,r) for integration over sphere:
[tex]
\int_{K(0,r)} d^3 x \nabla ^2 G = \int_{K(0,r)} d^3 x ~ \textrm{div} ~\textrm{grad} G
= \int_{S(0,r)} dS \vec{n} ~\textrm{grad} G
[/tex]
So we have:
[tex]
\lim_{r \rightarrow 0} \int_{K(0,r)} d^3 x ( \nabla ^2 + k^2 )G = \lim_{r \rightarrow 0}
\int_{S(0,r)} dS \left( \frac{(ik R -1)\exp(ikR)}{R^2} \right) = -4\pi
[/tex]
Because integration of [tex] ( \nabla ^2 + k^2 )G [/tex] gives [tex]-4\pi[/tex] and
[tex] ( \nabla ^2 + k^2 )G [/tex] is zero everywher beside R=0 it has
to be equal to [tex] -4\pi \delta(\vec{R}) [/tex].

How to normalize the Green function?
( \nabla ^2 + k^2 )\frac{e^{ikR}}{R}=...calculate...=0.
Because [tex]( \nabla ^2 + k^2 )G =-4\pi \delta (\vec{R}) [/tex] is not a homogenous
equation if you multiply G by a constant it won't be a solution to this eqaution.
 
  • #3
Oh...It's a very nice trick.
Thank you for your help!
 

1. What is the Green function for the wave equation?

The Green function for the wave equation is a mathematical function that represents the solution to the wave equation for a given set of initial conditions. It can be used to find the solution for any point in space and time.

2. How is the Green function for the wave equation derived?

The Green function for the wave equation is derived using the method of separation of variables, where the solution is expressed as a product of two functions - one depending only on space and the other only on time. These functions are then solved using boundary conditions to obtain the Green function.

3. What are the properties of the Green function for the wave equation?

The Green function for the wave equation has several important properties, including linearity, causality, and time-reversibility. It also satisfies the wave equation itself and has a singularity at the point of excitation.

4. How is the Green function for the wave equation used in practice?

The Green function for the wave equation is used in various fields of science and engineering to solve problems involving wave propagation. It can be used to calculate the response of a system to a given excitation, or to determine the propagation of waves in a medium with varying properties.

5. Are there any limitations to using the Green function for the wave equation?

While the Green function for the wave equation is a powerful tool, it does have some limitations. It assumes a linear medium and does not take into account any nonlinear effects. It also requires knowledge of the initial conditions and boundary conditions, which may not always be easily obtainable in practical situations.

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