Green's function for Helmholtz Equation

In summary: Arfken and Weber 9.7.3In summary, the Arfken and Weber problem 9.7.2 has two appropriate criteria that the student needs to satisfy in order to be able to solve the equation. The criteria states that the function satisfies the two appropriate criteria and is therefore a Green's function for the Helmholtz equation. Symmetricity of Green's functions is also assumed. The attempt at a solution is given by the hankel function of first kind which only takes into account the outgoing wave. Physical cases only have one outgoing wave so that is the solution that is used.
  • #1
Demon117
165
1

Homework Statement


Arfken & Weber 9.7.2 - Show that

[itex]\frac{exp(ik|r_{1}-r_{2}|)}{4\pi |r_{1}-r_{2}|}[/itex]

satisfies the two appropriate criteria and therefore is a Green's function for the Helmholtz Equation.


Homework Equations


The Helmholtz operator is given by

[itex]\nabla ^{2}A+k^{2}A[/itex]

Symmetricity of Green's functions.


The Attempt at a Solution


Right off the bat I am not sure what is mean't by "the two appropriate criteria" phrase. What exactly are the two appropriate criteria that they ask for in Arfken & Weber problem 9.7.2? Where can I find this criteria so that I know how to answer this question?
 
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  • #2
matumich26 said:

Homework Statement


Arfken & Weber 9.7.2 - Show that

[itex]\frac{exp(ik|r_{1}-r_{2}|)}{4\pi |r_{1}-r_{2}|}[/itex]

satisfies the two appropriate criteria and therefore is a Green's function for the Helmholtz Equation.


Homework Equations


The Helmholtz operator is given by

[itex]\nabla ^{2}A+k^{2}A[/itex]

Symmetricity of Green's functions.


The Attempt at a Solution


Right off the bat I am not sure what is mean't by "the two appropriate criteria" phrase. What exactly are the two appropriate criteria that they ask for in Arfken & Weber problem 9.7.2? Where can I find this criteria so that I know how to answer this question?

I answered my own question. I read the material 6 hours ago, looked at the assignment and did it but kept coming up with the solution that [itex](\nabla ^{2}+k^2)G=0[/itex], but I kept claiming that was incorrect. Well, looking back for the 100th time I realized this has to be true based on page 598 of Arfken & Weber, not only because it says so but also because the Helmholtz equation indicates a Green's function corresponding to an outgoing wave, which means that G(r1,r2) must satisfy a homogenous differential equation. Wow, that was a lot but I think it makes sense.
 
  • #3
afkern and weber problem 9.7.3

Hey I am still struggling with the solution of the problem and trying to figure out your explanation. Can you explain it more elaborately.
 
  • #4
The solutions are given by hankel functions of first kind and second kind.A time dependence of exp(-iwt) is assumed.In physical cases only outgoing wave is present so only one is chosen and not the other.
 
  • #5


As a scientist, it is important to carefully read and understand the problem before attempting to solve it. In this case, the problem is asking for the two criteria that need to be satisfied in order for the given function to be considered a Green's function for the Helmholtz Equation. These criteria can be found in the textbook, Arfken & Weber, in section 9.7.2.

The first criterion is that the given function must satisfy the Helmholtz Equation, which is given by:

\nabla ^{2}G+k^{2}G= -\delta (r_{1}-r_{2})

where G is the Green's function and k is the wavenumber. In this case, the given function satisfies this equation, as can be seen by substituting it into the equation and simplifying.

The second criterion is that the given function must satisfy the symmetricity property, which means that the function must be equal to its mirror image with respect to the point r1 and r2. In other words, if we swap the positions of r1 and r2, the function should remain the same. This property can also be seen by substituting the given function into the Helmholtz Equation and using the fact that the delta function is symmetric.

Therefore, since the given function satisfies both of these criteria, it can be considered a Green's function for the Helmholtz Equation. This function can then be used to solve problems in which the Helmholtz Equation is involved, making it a useful tool for scientists studying this equation.
 

1. What is the Helmholtz Equation and why is it important?

The Helmholtz Equation is a partial differential equation that describes the propagation of waves in physical systems. It is important in many fields of science, including physics, engineering, and mathematics, as it allows us to understand and predict the behavior of waves in various systems.

2. What is a Green's function for the Helmholtz Equation?

A Green's function for the Helmholtz Equation is a mathematical tool used to solve the Helmholtz Equation. It represents the response of a system to a point source excitation and allows us to find a solution to the Helmholtz Equation for any arbitrary source distribution.

3. How is a Green's function for the Helmholtz Equation used in practical applications?

Green's functions for the Helmholtz Equation have a wide range of practical applications, including in electromagnetics, acoustics, and seismology. They are used to model and analyze the behavior of waves in complex systems, such as antennas, acoustic chambers, and earthquake zones.

4. What are the properties of a Green's function for the Helmholtz Equation?

A Green's function for the Helmholtz Equation has several important properties, including being symmetric, satisfying the Helmholtz Equation itself, and having a singularity at the source point. These properties make it a powerful tool for solving the Helmholtz Equation in various scenarios.

5. How is a Green's function for the Helmholtz Equation derived?

The derivation of a Green's function for the Helmholtz Equation involves using the method of images and the principle of superposition. It is a complex mathematical process, but the end result is a function that satisfies the properties needed to solve the Helmholtz Equation for any source distribution.

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