San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

A property of the Solution of a Generalized
Helmholtz Equation of Any Dimension

The Helmholtz equation is usually expressed as

#### ∇²φ = −k²φ

with k being a constant. If k is a function of location it is a generalized Helmholtz equation. For now let k be a constant, but what is sought is a relationship that holds for constant k that will provide insights into the solution for nonconstant k.

The Laplacian operator is equal to the divergence of the gradient operator. Thus the Helmholtz equation is more properly expressed as

#### ∇·(∇φ) = −k²φ

This equation can be multiplied by the gradient of φ, ∇φ, to obtain

#### (1) ∇φ(∇·(∇φ)) = −k²φ ∇φ

The RHS is equivalent to (∇·(∇φ))∇φ = −½k²(∇(φ²))

The LHS is more complicated. There is the vector calculus identity.

#### ∇·(u·uT) = (∇·u)u + (u·∇)u

where (u·∇) is the directional derivative in the direction of u. Thus (u·∇)u is the directional derivative of u in the direction of u. It is thus something in the nature of a derivative of the vector u.

With u set equal to ∇φ this takes the form of

#### ∇·(∇φ·∇φT) = (∇·∇φ)∇φ + (∇φ·∇)∇φ

Therefore the LHS is given by

#### (∇·∇φ)∇φ = ∇·(∇φ·∇φT) − (∇φ·∇)∇φ

Thus Equation (1) can be expressed as

#### (2) ∇·(∇φ·∇φT) − (∇φ·∇)∇φ = −½k²(∇(φ²))

There is a generalized form of Gauss' Divergence Theorem, called the Generalized Stokes Theorem (GST), which takes the form of

#### ∫∂Vω = ∫Vdω

where ω is a differentiable form and dω is its exterior derivative. V denotes an orientable manifold and ∂V is its boundary surface. This theorem may be applied to all of the terms of the above equation.

The next step is to choose a volume over which Equation (2) will be inegrated. The one selected is a tubular one of infinitesimal cross section extending from a minimum of φ², namely zero, to a maximum of φ². The sides of this volume are parallel to ∇φ. The areas of the end plates of this volume are denoted as δS.

The GST applied to the RHS gives

#### ∫V[−½k²(∇(φ²))]dV = −½k²(φM²)δS + ½k²(φm²)δS

where M indicates at a point of Maximum φ² and m at a point of minimum φ², which happens to be be 0. Thus

#### ∫V[−½k²(∇(φ²))]dV = −½k²(φM²)δS

Since ∇φ=0 (a vector of zeroes) at a point of maximum φ² the integration of − (∇φ·∇)∇&phi yields

#### − (∇φ·∇φ)

The end surface is everywhere perpendicular to ∇φ. Therefore, by the GST,

#### ∫V∇·(∇φ·∇φT)dV = ∫∂V(∇φ·∇φT)dS

(To be continued.)

For two dimensions a relevant solution is of the form

#### φ²(r, θ) = A·cos²(nθ)Yn²(r)

where A is a constant, n is a positive integer, and Yn is a Bessel function of the first kind of index n.

If this function is depicted by showing its maxima the picture would look something like this.

(To be continued.)