1&2. Helmholtz decomposition theorem is wrong(Chapter 1 and 2)

Helmholtz decomposition theorem is wrong

　　　　　　　　　　　

There is a theorem called “Helmholtz decomposition” in Meteorology.

Helmholtz decomposition says that any flow(vector)F can be perfectly decomposed into two flows(vectors). One of them would be an irrotational flow(vector)Ve, and another would be a nondivergent flow(vector)Vr.

that is, F=Vr+Ve

This theorem may be believed in Meteorological Society in the World.

But I can prove this theorem is mathematically wrong.

１．The proof of Helmholtz decomposition

Helmholtz decomposition states that,

　　　

any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved
　　　into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free)
　　　 vector field;

If you want to prove that any vector F can be resolved into a irrotational vector Ve and a solenoidal vector Vr, you need to find some identity which can be described by like F=Ve + Vr.

As far as I know, the person who has most correctly tried to prove this theorem is James R.Holton.

Holton has tried to prove this theorem in the Appendix of the book "An Introduction to Dynamic Meteorology(first edition)" as follows.

In this appendix, Holton used V instead F as any flow.
If you could derive (C.1), you might say that Helmholtz decomposition theorem has been certainly proven.

Holton uses “The vector triple product identity” to derive (C.3). It is described as follow.
（Ａ×Ｂ）×Ｃ＝−Ａ（Ｂ・Ｃ）＋Ｂ（Ａ・Ｃ）

For example, “The vector triple product identity” is posted in the next homepage.
(reference:http://en.wikipedia.org/wiki/Triple_product#Proof)

If you replace Ａ and Ｂ with ▽, and Ｃ with Ｗ, you may get

( ▽×▽）×Ｗ=-▽(▽・Ｗ)+▽( ▽・Ｗ)

then
you can get (C.3) in Holton’s text. So, you might say that Helmholtz decomposition is perfectly proved.

２．Collapse of Helmholtz decomposition

Though Holton might have thought there was no problem in the appendix of his book, but (C.2) actually contains the big problem.

Any flow F certainly exist. But Ｗ which is given as (C.2) is not guaranteed to exist. You need to make sure that there exists Ｗ for any flow F.

I mean that there are many flows which do not include W.

I will show you that.

The only expression that guarantees Helmholtz decomposition theorem is the vector triple product identity (C.3).

You have got following expression from the vector triple product identity.

It is given on terms and conditions as required by

That is, χ and A are deriven from the common function W.
If χ is decided from W, then A should be decided at the same instance.

And, the flow given by ▽× A is supposed to exist independently from other flows.

It is called solenoidal flow. That means a flow in the tube. According to Helmholtz Decomposition there exists such a flow.

I do not think that such a solenoidal flow exists in the real world except mathematical world.
If there were such solenoidal flows, I would be able to show the collapses of Helmholtz decomposition.

Assuming that Helmholtz decomposition theorem is correct. you can consider two flows as following.

χ1 and A1 are functions which are derived from W1. and, χ2, and A2 are derived from W2.

Here, because an arbitrary flow (vector function) must be possible , you can consider the flow F3 which includes divergent component of (∇・χ1) and rotational component of (∇×A2).
where 　χ1＝▽・Ｗ１, 　 A2＝▽×Ｗ２

I must say that again F3 should have (∇・χ1) as divergent component, and have (∇×A2) as rorational component.

But according to Helmholtz decomposition, F3 can be decomposed into two flows only by the vector triple product identity (C.3).

Then
　

where F3 has (∇・χ3) as divergent component, and have (∇×A3) as rorational component.
But, because F1≠F3, and F2≠F3

So,

Here, you must say that F3 which has combined with divergent component of (∇・χ1) and rotational component of (∇×A2) can not decomposed into divergent component of (∇・χ1) and rotational component of (∇×A2).

So, you have to say that Helmholtz Decomposition have collapsed.

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