Helmholtz-Weyl decomposition

By | April 5, 2021

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 u into of the form

    \[ u=u_1+u_2,\quad \text{where }\Div u_1=0\quad \text{and}\quad u_2=\nabla p  \]

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

    \[ u=(\Gamma *v)(x).\]

Then -\triangle u =v. Since

    \[  -\triangle u = \nabla \times (\nabla \times u)-\nabla (\Div u)   \]

it follows

    \[  v = \nabla \times (\nabla \times u)-\nabla (\Div u).  \]

Note that

    \[  \Div  \left[\nabla \times (\nabla \times u)\right]=0, \]

which implies the desired decomposition for smooth vector field v. Observe that the decomposition is not unique in general. Indeed, let \phi be a harmonic function in \mathbb{R}^3. Note that

    \[ u=(u_1+\nabla \phi)+(u_2-\nabla \phi)  \]

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 \Leb{q}(\Omega)^n into the direct sum of certain subspaces. Denote by \Leb{q}_\sigma(\Omega) the closure of C_{c,\sigma}^\infty(\Omega) under \Leb{q}-norm and G^q(\Omega) the set of all vector fields u_2 in \Leb{q} such that u_2=\nabla \varphi for some \varphi \in \Sob{1}{q}_{\loc}(\Omega). As an application of De Rham theorem, we can show that

(1)   \begin{equation*}  \Leb{q}_\sigma(\Omega)^n  =\left\{  u \in \Leb{q}(\Omega) : \int_\Omega u\cdot \nabla \psi \myd{x}=0\quad \text{for all } \psi \in D^{1,q'}(\Omega) \right\} \end{equation*}

In this short note, we will show that

    \[ \Leb{q}(\Omega)=\Leb{q}_{\sigma}(\Omega)\oplus G^q(\Omega).   \]

In other words, we wish to determine when an arbitrary vector u\in \Leb{q}(\Omega)^n can be uniquely expressed as the sum

    \[ u=w_1+w_2,\quad w_1\in \Leb{q}_\sigma(\Omega),\quad w_2\in G^q(\Omega). \]

The above decomposition holds for any domain \Omega in \mathbb{R}^n if q=2. The decomposition also holds for 1<q<\infty when \Omega=\mathbb{R}^n or bounded C^1-domain in \mathbb{R}^n.

One approach is to use the Leray projection:

    \[ \Proj(u)=u-\nabla \triangle^{-1}(\Div u)  \]

which is a Fourier multiplier operator whose symbol is

    \[ m(\xi)_{kj}=\delta_{kj}-\frac{\xi_k\xi_j}{|\xi|^2},\quad 1\leq k,j\leq n. \]

It can be shown that \Proj[\Proj(u)]=\Proj(u) for all u\in \mathcal{S}(\mathbb{R}^n)^n. Also, \Div \Proj(u)=0 for all u\in \mathcal{S}(\mathbb{R}^n)^n. Also, \Proj(u)=u for all u\in \mathcal{S}(\mathbb{R}^n)^n with \Div u=0. Finally, \Proj(\nabla \phi)=0 for all \phi \in \mathcal{S}(\mathbb{R}^n).

Another approach to obtain the Helmholtz decomposition is the following. We first show that the validity of Helmholtz decomposition in \Leb{q}(\Omega) is equivalent to the solvability for the following problems: find a unique p\in D^{1.q}(\Omega) such that

(2)   \begin{equation*} \int_{\Omega} (\nabla p-u)\cdot \nabla \varphi \myd{x}=0\quad \text{for all }\varphi \in D^{1,q'}(\Omega). \end{equation*}

Lemma 1. Let 1<q<\infty and let \Omega be a domain in \mathbb{R}^n. The following are equivalent:
(i) The Helmholtz-Weyl decomposition of \Leb{q}(\Omega)^n holds;
(ii) For any u\in \Leb{q}(\Omega)^n, there exists a unique function p\in D^{1,q}(\Omega) satisfying \eqref{eq:Neumann-problems}.

Proof. (ii) implies (i) : Let u\in \Leb{q}(\Omega)^n be given. By (ii), there exists a unique p\in D^{1,q}(\Omega) satisfying (2). Define w=u-\nabla p. Then w satisfies

    \[  \int_{\Omega} w \cdot \nabla \varphi \myd{x}=0 \]

for all \varphi \in D^{1,q'}(\Omega). By (1), we see that w\in \Leb{q}_\sigma(\Omega). It remains to show the uniqueness of the representation. Let u=u_1+\nabla p_1=u_2+\nabla p_2. Note that

    \[  \int_{\Omega} \nabla p_1 \cdot \nabla \varphi \myd{x}=\int_{\Omega} \nabla p_2 \cdot \nabla \varphi \myd{x}  \]

for all \varphi \in D^{1,q'}(\Omega). Since p_1-p_2 is a unique function satisfying

    \[ \int_{\Omega} \nabla (p_1-p_2)\cdot \nabla \varphi \myd{x}=0\quad \text{for all }\varphi \in D^{1,q'}(\Omega),  \]

it follows that u_1-u_2=\nabla p_1-\nabla p_2=0. This proves that u_1=u_2, which completes the proof of (i).

(i) implies (ii) : Let u\in \Leb{q}(\Omega)^n. Then there exists u=w+\nabla p, where w\in \Leb{q}_\sigma(\Omega). By (1), we have

    \[  \int_{\Omega} (\nabla p-u)\cdot \nabla \varphi \myd{x}=0  \]

for all \varphi \in D^{1,q'}(\Omega). Hence by the uniqueness of the representation, such a p is unique (up to a constant). This completes the proof of (ii).

Now we are ready to prove the following Helmholtz-Weyl decomposition on \Leb{q}(\Omega)^n:

Theorem 1 (Helmholtz-Weyl decomposition). Let 1<q<\infty and let \Omega be either \mathbb{R}^n, \mathbb{R}^n_+, or bounded C^1-domain in \mathbb{R}^n. Then we have

    \[   \Leb{q}(\Omega)^n=\Leb{q}_\sigma(\Omega)\oplus G^q(\Omega). \]

Proof. By Lemma 1, it suffices to solve problem (2). We mainly focus on the case \Omega=\mathbb{R}^n. Let u\in C_c^\infty(\mathbb{R}^n). For simplicity, we assume that n\geq 3. Define

    \[    p(x)=-\int_{\mathbb{R}^n}  \Gamma(x-y) \Div u(y)\myd{y}. \]

Then we have p\in C^\infty(\mathbb{R}^n). By Calder\’on-Zygmund estimate, we have

    \[  \norm{\nabla p}{\Leb{q}}\leq C \norm{u}{\Leb{q}} \]

for some constant C=C(n,q)>0. Since u vanishes outside \supp u, it follows that for sufficiently large R, we have

    \[ \int_{\ball{R}} (u-\nabla p )\cdot \nabla  \varphi \myd{x}\myd{x}=-\int_{\partial \ball{R}} (\nabla p\cdot \nu) \varphi \myd{\sigma}  \]

for all \varphi\in C^\infty(\mathbb{R}^n) with \nabla \varphi \in \Leb{q'}(\mathbb{R}^n). A change of variable shows that

    \[ \int_{\partial \ball{R}} (\nabla p\cdot \nu)\varphi \myd{\sigma} =R^{n-1}\int_{\partial \ball{1}} \nabla p(R\omega) \cdot \omega \varphi(R\omega) \myd{\omega}.   \]

Since |\nabla p(x)|\leq C|x|^{-n} for sufficiently large x, it follows that for sufficiently large R\gg 1, we have

    \[  \left|\int_{\partial \ball{R}} |\nabla p||\varphi|\myd{\sigma} \right|\leq \frac{1}{R} \int_{\partial \ball{1}} |\varphi(Ry)|\myd{\sigma(y)}. \]

We claim that

    \[  \lim_{R\rightarrow \infty}\int_{\partial \ball{1}} \frac{1}{R}|\varphi(Ry)|\myd{\sigma}(y)=0.\]

By Fundamental theorem of calculus, we have

    \[ \varphi(R\omega)-\varphi(\omega)=\int_1^R \nabla \varphi(r\omega)\cdot \omega \myd{r}.  \]

A change of variable into polar coordinate shows that

    \begin{align*}  \frac{1}{R}\int_{\partial \ball{1}} |\varphi(R\omega )|\myd{\sigma}(\omega)&\leq \frac{1}{R}\int_{\partial \ball{1}}|\varphi(\omega)|\myd{\sigma(\omega)}+\frac{1}{R}\int_{\partial\ball{1}}\int_{1}^R |\nabla \varphi(r\omega)|\myd{\sigma(\omega)} \\  &= \frac{1}{R}\int_{\partial \ball{1}}|\varphi(\omega)|\myd{\sigma(\omega)}+\frac{1}{R}\int_{\ball{R}\setminus \ball{1}} \frac{|\nabla \varphi(x)|}{|x|^{n-1}}\myd{x}.\end{align*}

By H\”older’s inequality, we have

    \[ \frac{1}{R}\int_{\ball{R}\setminus\ball{1}}\frac{|\nabla \varphi(x)|}{|x|^{n-1}}\myd{x}\leq \frac{1}{R}\norm{\nabla \varphi}{\Leb{q'}}\left(\int_{\ball{R}\setminus \ball{1}} \frac{1}{|x|^{(n-1)q}}\myd{x} \right)^{1/q}  \]

Hence it remains to show that

    \[  \lim_{R\rightarrow \infty} \frac{1}{R}\left(\int_{\ball{R}\setminus \ball{1}} \frac{1}{|x|^{(n-1)q}}\myd{x} \right)^{1/q}=0. \]

Suppose first that n'<q<\infty. Since (n-1)q>n, it follows that

    \[ \lim_{R\rightarrow \infty} \left(\int_{\ball{R}\setminus \ball{1}}\frac{1}{|x|^{(n-1)q}}\myd{x} \right)<\infty.  \]

This implies the desired result when n'<q<\infty. Suppose next that q=n'. A change of variable shows that

    \begin{align*}\lim_{R\rightarrow\infty} \left(\int_{\ball{R}\setminus \ball{1}}\frac{1}{|x|^{(n-1)q}}\myd{x} \right)&=\lim_{R\rightarrow \infty}\frac{c}{R}\left(\int_{1}^R \frac{1}{\rho}\myd{\rho}   \right)^{1/q}\\&= c\lim_{R\rightarrow \infty} \frac{(\ln R)^{1/q}}{R}=0. \end{align*}

Finally, suppose that 1<q<n'. If we set \alpha=(n-1)(q-1), then 0<\alpha<1. A change of variable shows that

    \begin{align*}  \lim_{R\rightarrow \infty}\left(\int_{\ball{R}\setminus \ball{1}}\frac{1}{|x|^{(n-1)q}}\myd{x} \right)&=\lim_{R\rightarrow \infty} \frac{c}{R}\left(\frac{1}{1-\alpha}(R^{-\alpha+1}-1) \right)^{1/q}\\  &=c\lim_{t\rightarrow \infty}\frac{t^{1/q}}{(1+t)^{1/(1-\alpha)}}=0\end{align*}

since 1/(1-\alpha)>1/q. This proves the desired claim. Hence it follows that p satisfies

    \[ \int_{\mathbb{R}^n} (\nabla p-u)\cdot \nabla \varphi \myd{x}=0  \]

for all \varphi \in C^\infty(\mathbb{R}^n)\cap   D^{1,q'}(\mathbb{R}^n) and

    \[ \norm{\nabla p}{\Leb{q}}\leq C \norm{u}{\Leb{q}}. \]

Note also that if p_1 and p_2 satisfy

    \[ \int_{\mathbb{R}^n} \nabla p \cdot \nabla \varphi \myd{x}=\int_{\mathbb{R}^n} u\cdot \nabla \varphi \myd{x}  \]

for all \varphi \in C^\infty(\mathbb{R}^n)\cap   D^{1,q'}(\mathbb{R}^n), then

    \[ \int_{\mathbb{R}^n} \nabla (p_1-p_2)\cdot \nabla \varphi \myd{x}=0 \]

for all \varphi \in C^\infty_c(\mathbb{R}^n). So p_1-p_2 is harmonic in \mathbb{R}^n. Since p_1-p_2 are harmonic in \mathbb{R}^n, so is \nabla (p_1-p_2). Hence by Lemma ??, \nabla (p_1-p_2)=0. This implies that p_1-p_2=c for some constant c. This implies the theorem when u\in C_c^\infty(\mathbb{R}^n)^n.

Now for any u\in \Leb{q}(\mathbb{R}^n)^n, there exists \{u_k\}\subset C_c^\infty(\mathbb{R}^n)^n such that u_k\rightarrow u in \Leb{q}. Set p_k = (\Gamma *\Div u_k). Then p_k satisfies

    \[ \norm{\nabla p_k-\nabla p_l}{\Leb{q}}\leq C \norm{u_k-u_l}{\Leb{q}}  \]

for all k and l. Hence \{\nabla p_k \} forms a Cauhcy sequence in \Leb{q}. Denote the limit by \Phi =\lim_{k\rightarrow \infty } \nabla p_k. Then

    \[  \int_{\mathbb{R}^n} (\Phi-u)\cdot \nabla \varphi \myd{x}=\lim_{k\rightarrow \infty}\int_{\mathbb{R}^n} (\nabla p_k-u_k)\cdot \nabla \varphi  \myd{x}=0 \]

for all \varphi \in D^{1,q'}(\mathbb{R}^n). By de Rham theorem, we can show that \Phi=\nabla p for some p. This proves the existence of a solution of the problem (2). To show the uniqueness part, suppose that p_1 and p_2 satisfy

    \[ \int_{\mathbb{R}^n} \nabla p_1 \cdot \nabla \varphi \myd{x}=\int_{\mathbb{R}^n} \nabla p_2 \cdot \nabla \varphi \myd{x}  \]

for all \varphi \in D^{1,q'}(\mathbb{R}^n). This implies that p_1-p_2 is harmonic in \mathbb{R}^n, and so is \nabla (p_1-p_2). Hence \nabla (p_1-p_2)=0. This implies that p_1-p_2=c for some constant c. This completes the proof of Theorem 1.

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