Today, one of my students, who is a cadet in Republic of Korea Air Force Academy, asked a proof of Helmholtz-Weyl decomposition for general vector field, which was considered in the electromagnetic class.

In the mathematical fluid dynamics, we need to decompose a vector field into of the form

for some scalar function . This is known as the *Helmholtz-Weyl decomposition*. If is a sufficiently smooth vector field, define

Then . Since

it follows

Note that

which implies the desired decomposition for smooth vector field . Observe that the decomposition is not unique in general. Indeed, let be a harmonic function in . Note that

is another decomposition.

A systematic study of such decomposition in terms of function spaces was initiated by Weyl in 1940 and developed by several authors. Motivated by the above decomposition, we decompose into the direct sum of certain subspaces. Denote by the closure of under -norm and the set of all vector fields in such that for some . As an application of De Rham theorem, we can show that

(1)

In this short note, we will show that

In other words, we wish to determine when an arbitrary vector can be uniquely expressed as the sum

The above decomposition holds for any domain in if . The decomposition also holds for when or bounded -domain in .

One approach is to use the Leray projection:

which is a Fourier multiplier operator whose symbol is

It can be shown that for all . Also, for all . Also, for all with . Finally, for all .

Another approach to obtain the Helmholtz decomposition is the following. We first show that the validity of Helmholtz decomposition in is equivalent to the solvability for the following problems: find a unique such that

(2)

Lemma 1.Let and let be a domain in . The following are equivalent:

(i) The Helmholtz-Weyl decomposition of holds;

(ii) For any , there exists a unique function satisfying \eqref{eq:Neumann-problems}.

Proof. (ii) implies (i) : Let be given. By (ii), there exists a unique satisfying (2). Define . Then satisfies

for all . By (1), we see that . It remains to show the uniqueness of the representation. Let . Note that

for all . Since is a unique function satisfying

it follows that . This proves that , which completes the proof of (i).

(i) implies (ii) : Let . Then there exists , where . By (1), we have

for all . Hence by the uniqueness of the representation, such a is unique (up to a constant). This completes the proof of (ii).

Now we are ready to prove the following Helmholtz-Weyl decomposition on :

Theorem 1 (Helmholtz-Weyl decomposition).Let and let be either , , or bounded -domain in . Then we have

Proof. By Lemma 1, it suffices to solve problem (2). We mainly focus on the case . Let . For simplicity, we assume that . Define

Then we have . By Calder\’on-Zygmund estimate, we have

for some constant . Since vanishes outside , it follows that for sufficiently large , we have

for all with . A change of variable shows that

Since for sufficiently large , it follows that for sufficiently large , we have

We claim that

By Fundamental theorem of calculus, we have

A change of variable into polar coordinate shows that

By H\”older’s inequality, we have

Hence it remains to show that

Suppose first that . Since , it follows that

This implies the desired result when . Suppose next that . A change of variable shows that

Finally, suppose that . If we set , then . A change of variable shows that

since . This proves the desired claim. Hence it follows that satisfies

for all and

Note also that if and satisfy

for all , then

for all . So is harmonic in . Since are harmonic in , so is . Hence by Lemma ??, . This implies that for some constant . This implies the theorem when .

Now for any , there exists such that in . Set . Then satisfies

for all and . Hence forms a Cauhcy sequence in . Denote the limit by . Then

for all . By de Rham theorem, we can show that for some . This proves the existence of a solution of the problem (2). To show the uniqueness part, suppose that and satisfy

for all . This implies that is harmonic in , and so is . Hence . This implies that for some constant . This completes the proof of Theorem 1.