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Helmholtz Equation in Matrix Form |
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The equation under consideration is
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
In matrix form these are (dV/dx)=KV(x) where
| φ | | ||
V | = | | | |
| ψ | |
| 0 | 1 | | ||
K | = | | | | |
| −k² | 0 | |
If k is a constant the solution to the system of the two equations can be represented as
More generally the solution is
The matrix function L(x) takes the form
| 0 | x | | ||
L(x) | = | | | | |
| −l(x) | 0 | |
where l(x) is equal to ∫_{0}^{x}k²(ζ)dζ.
Note that the exponential of an n×n matrix M is defined as
where I is the n×n identity matrix.
Consider now
Note that
Furthermore
Thus
Ignoring the higher order terms ΔL takes the form
| 0 | Δx | | ||
ΔL | = | | | | |
| −∫_{x}^{x+Δx}k²(ζ)dζ | 0 | |
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
which is what was to be proven.
(To be continued.)
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