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to a Generalized Helmholtz Equation |
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The equation under consideration is
where k(x) is a positive and symmetric about x=0 but is a declining function of |x|. This equation can be termed a generalized Helmholtz equation.
Now consider the points of relative maxima and minima of φ², labeled as x_{i} and y_{i}, respectively. These are points such that dφ²/dx=2φ(dφ/dx) equals zero; (dφ/dx)=0 for the maxima and φ=0 for the minima. Note that because of the equation, when φ=0 then d(dφ/dx)/dx=0 and hence (dφ/dx) is at a maximum or minimum.
Let the labeling be such that x_{i} > y_{i}.
Multiply the equation by (dφ/dx) to get
Now integrate the above equation from x=y_{i} to x=x_{i}. The result may be represented as
At x_{i} (dφ/dx) is equal to 0. By the Generalized Mean Value Theorem the RHS of the above equation may be expressed as
where z_{i} is some point between y_{i} and x_{i}.
But φ² is equal to zero at y_{i} and (dφ²/dx) is equal to zero at x_{i}.
Therefore the previous equation reduces to
As formulated φ²(x_{i}) is a local relative maximum of φ² and (dφ/dx)²_{yi} is the next lower local relative maximum of (dφ/dx)². The same relationship would prevail between a local relative maximum of φ² and the next lower local relative maximum of (dφ/dx)². Or, between a local relative maximum of (dφ/dx)² and the next lower and next higher local relative maximum of φ².
This gives
This means that
This provides a chain of ratios that links φ²(x_{i}) to some φ²(x_{0}). If z_{i+1}≅w_{i} the formula would reduce to
Instead
The factor Γ_{i} is the product of i terms each of which is less than one.
If k(x)² were constant over some interval of x about x_{0} then over that interval
where A is a constant. The wavelength L is given by
The difference between y_{i} and x_{i} is then ½L=π/k, and
A more convnient representation is
where L_{j}=2π/k(y_{j}).
The simplified formula previously derived for φ²(x_{i}) can be expressed as
where
For some cases φ² is at a local maximum at x=0; for other cases it is at a local minimum there.
The average value of φ²(x) in the interval about x_{i} is equal to ½φ²(x_{i}).
As E → ∞ so k(x)→∞ for all x. Thus L_{j} → 0 and hence Γ_{j} → 1 for all j. Likewise (x_{i}+½L_{i})→x_{i}.
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