San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

A Transformation of a
Generalized Helmholtz Equation

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). This suggests that the solution to the general equation may be a function of ∫0xk(ζ)dζ.

Consider a change of the independent variable such that

dz = k(x)dx
This means that
(dz/dx) = k(x)
z = ∫0xk(ζ)dζ

The first derivative of φ with respect to x is given by

(dφ/dx) = (dφ/dz)(dz/dx) = (dφ/dz)k

And the second derivative is

(d²φ/dx²) = [d(dφ/dx)/dz]k = [d((dφ/dz)k))/dz]k = [(d²φ/dz²)k + ((dφ/dz)(dk/dz)]k

When this expression is substituted into the Helmholtz equation under consideration the result is

[(d²φ/dz²)k + ((dφ/dz)(dk/dz)]k = −k²φ
which reduces to
[(d²φ/dz²)k + ((dφ/dz)(dk/dz)] = −kφ
and further to
(d²φ/dz²) + ((dφ/dz)(dk/dz)/k = −φ

Note that

(dk/dx) = (dk/dz)k
and thus
(dk/dz) = (1/k)(dk/dx)
and hence
(dk/dz)/k = (1/k²)(dk/dx) = −d(1/k)/dx

Let d(1/k)/dx be denoted as g(z). The transformed equation is then

(d²φ/dz²) − (dφ/dz)g(z) = −φ

HOME PAGE OF applet-magic
HOME PAGE OF Thayer Watkins