San José State University

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A Solution to the Generalized
Helmholtz Equation of One Dimension

The Helmholtz equation arises in many contexts in the attempt to give a mathematical explanation of the physical world. These range from Alan Turing's explanation of animal coat patterns to Schrödinger's time-independent equation in quantum theory.

The Helmholtz equation per se is

#### ∇²φ = −k²φ

where k is a constant. The Generalized Helmholtz equation is that equation with k being a function of the independent variable(s).

## The One Dimensional Case

In one dimension the Helmholtz equation is

#### (d²φ/dx²) = −k²φ(x)

It just has the sinusoidal solution of φ(x) = A·sin(kx)+B·cos(kx). In one dimension the Generalized Helmholtz equation has a sinusoidal-like solution of varying amplitude and wavelength.

### Change of Variable

The sinusoidal solution being a function of kx suggests that the solution at the generalized equation may a function of

Then

#### (dφ/dx) = (dφ/dX)(dX/dx) = (dφ/dX)k(x) and (d²φ/dx²) = (d²φ/dX²)k²(x) + (dφ/dX)(dk/dX)k

Since (d²φ/dx²) is equal to −k²φ the above equation can be reduced to

## A Matric Equation

Let (dφ/dX) be denoted as ψ and (dk/dX)/k as γ. Then

In matric form

where

#### Φ = | φ | | ψ |   M = | 0     −1 |  | 1       γ |

Note that γ is a function of X and hence so is the matrix M.

For the analogous scalar differential equation the solution would go as follows:

#### (dy/dx) = −μ(x)y (1/y)(dy/dx) = −μ(x) Inegrating from 0 to x gives ln(y(x)) − ln(y(0)) = −∫0xμ(z)dz hence y(x) = exp(−∫0xμ(z)dz)y(0)

This suggests that the solution to the matrix equation is

#### Φ(X) = exp(−∫0XMdZ)Φ(0)

The integral of the matrix M is the following matrix

#### ∫0XM(Z)dZ = |   0             −X |  | X   ∫0Xγ(z)dz |

The solution is therefore

#### | φ(X) |    |   0             X         |       | φ(0) | || = exp{} | ψ(X) |    | −X   −∫0Xγ(z)dz     |     |ψ(0) |

(To be continued.)