﻿ The Solution to the Generalized Helmholtz Equation in Matrix Form
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

The Solution to the Generalized
Helmholtz Equation in Matrix Form

The equation under consideration is

#### (d²φ/dx²) = −k²(x)φ

If k is constant then the solution is φ(x)=A·sin(kx)+B·cos(kx). , where A and B are arbitrary constants.

The single second order equation may be replaced by a pair of first order equations by defining

Thus

#### (dφ/dx) = ψ (dψ/dx) = −k²(x)φ

In matrix form these are (dV/dx)=KV(x) where

#### | φ | V=|    | | ψ | |  0   1 | K =|          | | −k²0 |

If k is a constant the solution to the system of the two equations can be represented as

#### V(x) = exp(Kx)V(0)

More generally the solution is

#### V(x)=exp(L(x))V(0) where L(x) = ∫0xK(ζ)dζ

The matrix function L(x) takes the form

#### |     0     x | L(x) =|               | | −l(x)  0 |

where l(x) is equal to ∫0xk²(ζ)dζ.

## Proof that the above is indeed the solution to the System of First Order Differential Equations

Note that the exponential of an n×n matrix M is defined as

#### exp(M) = I/0! + M/1! + M²/2! + …

where I is the n×n identity matrix.

Consider now

Note that

Furthermore

Thus

#### V(x+Δx) − V(x) = (I+ΔL)V(x) − I·V(x) + higher order terms in ΔL hence V(x+Δx) − V(x) = ΔL·V(x) + higher order terms in ΔL

Ignoring the higher order terms ΔL takes the form

Therefore

#### |       0     1 | ΔL/Δx =|                | | −k²(x)  0 |

where x is a value between x and x+Δx. As Δx→0 the higher order terms divided by Δx go to zero and

#### |       0     1 | dL/dx =|                | | −k²(x)  0 |

The RHS of this is the matrix K(x). Thus

#### (dV/dx) = (dL/dx)V(x) = K(x)V(x)

which is what was to be proven.

(To be continued.)