﻿ A Fourier Analysis of the Generalized Helmholtz Equation
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A Fourier Analysis of the
Generalized Helmholtz Equation

The Generalized Helmholtz Equation is

#### ∇²φ(z) = −H(z)φ(z)

where z repesents location and H is positive. In one dimension this is

#### (d²φ/dx²) = −H(x)φ(x)

One approach to analyzing this equation is to take its Fourier transforms F(ω); i.e., the Fourier transform of f(x) defined over the interval [−∞, +∞] is

#### Ff(ω) = ∫−∞+∞f(x)exp(−2πωx)dx

Suppose H(x) is a constant H0 and assume that

#### φ(x) = A·exp(βx)

where A is an arbitrary constant whose value is established by initial conditions. Then

#### dφ(x)/dx = βA·exp(βx) and d²φ(x)/dx² = β²A·exp(βx)

Therefore for the equation to be satisfied it must be that

Thus

#### φ(x)=A·exp( i(H0)½x) = C·cos(ω0x) + D·sin(ω0x)

where =ω0=(H0)½. Thus for a constant coefficient the spectrum of the solution equals a single value of frequency.

When the Fourier transform is applied to the generalized Helmholtz equation the result is

#### F[(∂²φ/∂x²)] = −F[H(x)φ(x)] which reduces to ω² Fφ(ω) = −∫−∞+∞FH(ω−ν) Fφ(ν)dν

.

The integral on the RHS of the above is known as the convolution of FH and Fφ.

The Fourier transform can be interpreted as the decomposition of a function into constant frequency sinusoidal components. That decomposition is called the spectrum of the function. What is desired is an analysis that demonsttrates the relationship of the spectrum of the solution φ(x) to the coefficient function H(x).

In the above it was found that for the case of a constant H the spectrum of the solution is a single spike at (H0)½.

More generally the convolution is in the nature of a weighted sum. The weighted sum can be repesented as the weight total times a weighted average. Thus

#### ω²Fφ(ω) = −W∫−∞+∞ w(ω−ν) Fφ(ν)dν

where W= ∫−∞+∞FH(ν)dν and

#### w(ν) = FH(ν)/W

.

The differential generalized Helmholtz equation is related to an integral equation in the frequency domain for the spectrum of its solution. The equation has the value of Fφ(ω) tied to a weighted average of nearby values this means that ω² is tied to W and hence ω to W½.

The integral equation above is like a system of linear equations and a system of linear equations can be represented in terms of matrices. Suppose ω can take on only integral values positive, negative and zero. Then Fφ(ω) is an infinite order vector denoted as Fφ. Let Ω be an infinite order diagonal square matrix with ω; on the diagonal. The LHS of the matrix equation is then Ω²Fφ.

The construction of the RHS is more involved. Let Γ be the weights of the weighted sum expressed as a row vector and let Λ be the infinite square matrix created by placing Γ on each row centered on the principal diagonal; i.e.,

#### |.......................| | .......Γ...............| | ...........Γ ........,| | ...............Γ.....| |.......................|

The matrix equation for the integral equation is

#### Ω²Fφ = ΛFφor, equivalently [Ω² − Λ]Fφ = 0

A solution to this equation would have to be normalized.

(To be continued.)