Decomposition of vector field into solenoidal and irrotational parts.

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Can a vector field $mathbfA: mathbbR^3 rightarrow mathbbR^3$ always be expressed as the sum of a solenoidal (divergenceless) part $mathbfA_S$, where $nabla cdot mathbfA_S = 0$, and an irrotational part $mathbfA_R$, where $nablatimesmathbfA_R = 0$?

$$
“,forallmathbfA;
existsmathbfA_S;
existsmathbfA_R :
mathbfA = mathbfA_S + mathbfA_R ;wedge;
nabla cdot mathbfA_S = 0 ;wedge;
nabla times mathbfA_R = 0
,”,?
$$

Since irrotational vector fields are gradients of some scalar field ($mathbfA_R = nabla R$), this is equivalent to asking:

$$
“,
forallmathbfA;
existsmathbfA_S;
exists R in left(mathbbR^3rightarrowmathbbRright) :
mathbfA = mathbfA_S + nabla R
;wedge;
nablacdotmathbfA_S = 0
,”,?
$$

What’s a proof of the result? If true, are $mathbfA_R$ and $mathbfA_S$ unique?
If it’s not always true, then what kind of fields can/cannot be decomposed in this way?

---

If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.

proof-writing vector-analysis

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asked Jul 20 at 11:20

Jollywatt

1447

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up vote
0
down vote

favorite

Can a vector field $mathbfA: mathbbR^3 rightarrow mathbbR^3$ always be expressed as the sum of a solenoidal (divergenceless) part $mathbfA_S$, where $nabla cdot mathbfA_S = 0$, and an irrotational part $mathbfA_R$, where $nablatimesmathbfA_R = 0$?

$$
“,forallmathbfA;
existsmathbfA_S;
existsmathbfA_R :
mathbfA = mathbfA_S + mathbfA_R ;wedge;
nabla cdot mathbfA_S = 0 ;wedge;
nabla times mathbfA_R = 0
,”,?
$$

Since irrotational vector fields are gradients of some scalar field ($mathbfA_R = nabla R$), this is equivalent to asking:

$$
“,
forallmathbfA;
existsmathbfA_S;
exists R in left(mathbbR^3rightarrowmathbbRright) :
mathbfA = mathbfA_S + nabla R
;wedge;
nablacdotmathbfA_S = 0
,”,?
$$

What’s a proof of the result? If true, are $mathbfA_R$ and $mathbfA_S$ unique?
If it’s not always true, then what kind of fields can/cannot be decomposed in this way?

---

If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.

proof-writing vector-analysis

share|cite|improve this question

asked Jul 20 at 11:20

Jollywatt

1447

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Can a vector field $mathbfA: mathbbR^3 rightarrow mathbbR^3$ always be expressed as the sum of a solenoidal (divergenceless) part $mathbfA_S$, where $nabla cdot mathbfA_S = 0$, and an irrotational part $mathbfA_R$, where $nablatimesmathbfA_R = 0$?

$$
“,forallmathbfA;
existsmathbfA_S;
existsmathbfA_R :
mathbfA = mathbfA_S + mathbfA_R ;wedge;
nabla cdot mathbfA_S = 0 ;wedge;
nabla times mathbfA_R = 0
,”,?
$$

Since irrotational vector fields are gradients of some scalar field ($mathbfA_R = nabla R$), this is equivalent to asking:

$$
“,
forallmathbfA;
existsmathbfA_S;
exists R in left(mathbbR^3rightarrowmathbbRright) :
mathbfA = mathbfA_S + nabla R
;wedge;
nablacdotmathbfA_S = 0
,”,?
$$

What’s a proof of the result? If true, are $mathbfA_R$ and $mathbfA_S$ unique?
If it’s not always true, then what kind of fields can/cannot be decomposed in this way?

---

If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.

proof-writing vector-analysis

share|cite|improve this question

asked Jul 20 at 11:20

Jollywatt

1447

Can a vector field $mathbfA: mathbbR^3 rightarrow mathbbR^3$ always be expressed as the sum of a solenoidal (divergenceless) part $mathbfA_S$, where $nabla cdot mathbfA_S = 0$, and an irrotational part $mathbfA_R$, where $nablatimesmathbfA_R = 0$?

$$
“,forallmathbfA;
existsmathbfA_S;
existsmathbfA_R :
mathbfA = mathbfA_S + mathbfA_R ;wedge;
nabla cdot mathbfA_S = 0 ;wedge;
nabla times mathbfA_R = 0
,”,?
$$

Since irrotational vector fields are gradients of some scalar field ($mathbfA_R = nabla R$), this is equivalent to asking:

$$
“,
forallmathbfA;
existsmathbfA_S;
exists R in left(mathbbR^3rightarrowmathbbRright) :
mathbfA = mathbfA_S + nabla R
;wedge;
nablacdotmathbfA_S = 0
,”,?
$$

What’s a proof of the result? If true, are $mathbfA_R$ and $mathbfA_S$ unique?
If it’s not always true, then what kind of fields can/cannot be decomposed in this way?

---

If true, this would be a very nice result: any vector field could then be thought of in terms of its unique solenoidal and irrotational components.

proof-writing vector-analysis

share|cite|improve this question

asked Jul 20 at 11:20

Jollywatt

1447

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 11:20

Jollywatt

1447

asked Jul 20 at 11:20

Jollywatt

1447

asked Jul 20 at 11:20

Jollywatt

1447

1447

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $mathbfF$ on $mathbbR^3$ can be written as
$$
mathbfF =
underbrace-nablaPhi_textirrotational +
underbracenablatimesmathbfA_textsolenoidal
$$
provided 1) that $mathbfF$ is twice continuously differentiable and 2) that $mathbfF$ vanishes faster than $1/r$ as $r rightarrow infty$.
If $mathbfF$ is on a bounded domain $V subset mathbbR^3$, then the condition 2) that $mathbfF$ vanishes can be relaxed.

Helmholtz’s theorem even gives $Phi$ and $mathbfA$ in terms of $mathbfF$. More details can be found here:
https://en.wikipedia.org/wiki/Helmholtz_decomposition

share|cite|improve this answer

edited Jul 20 at 23:30

Jollywatt

1447

answered Jul 20 at 12:39

Matrefeytontias

572110

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $mathbfF$ on $mathbbR^3$ can be written as
$$
mathbfF =
underbrace-nablaPhi_textirrotational +
underbracenablatimesmathbfA_textsolenoidal
$$
provided 1) that $mathbfF$ is twice continuously differentiable and 2) that $mathbfF$ vanishes faster than $1/r$ as $r rightarrow infty$.
If $mathbfF$ is on a bounded domain $V subset mathbbR^3$, then the condition 2) that $mathbfF$ vanishes can be relaxed.

Helmholtz’s theorem even gives $Phi$ and $mathbfA$ in terms of $mathbfF$. More details can be found here:
https://en.wikipedia.org/wiki/Helmholtz_decomposition

share|cite|improve this answer

edited Jul 20 at 23:30

Jollywatt

1447

answered Jul 20 at 12:39

Matrefeytontias

572110

add a comment | 

up vote
1
down vote



accepted

This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $mathbfF$ on $mathbbR^3$ can be written as
$$
mathbfF =
underbrace-nablaPhi_textirrotational +
underbracenablatimesmathbfA_textsolenoidal
$$
provided 1) that $mathbfF$ is twice continuously differentiable and 2) that $mathbfF$ vanishes faster than $1/r$ as $r rightarrow infty$.
If $mathbfF$ is on a bounded domain $V subset mathbbR^3$, then the condition 2) that $mathbfF$ vanishes can be relaxed.

Helmholtz’s theorem even gives $Phi$ and $mathbfA$ in terms of $mathbfF$. More details can be found here:
https://en.wikipedia.org/wiki/Helmholtz_decomposition

share|cite|improve this answer

edited Jul 20 at 23:30

Jollywatt

1447

answered Jul 20 at 12:39

Matrefeytontias

572110

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $mathbfF$ on $mathbbR^3$ can be written as
$$
mathbfF =
underbrace-nablaPhi_textirrotational +
underbracenablatimesmathbfA_textsolenoidal
$$
provided 1) that $mathbfF$ is twice continuously differentiable and 2) that $mathbfF$ vanishes faster than $1/r$ as $r rightarrow infty$.
If $mathbfF$ is on a bounded domain $V subset mathbbR^3$, then the condition 2) that $mathbfF$ vanishes can be relaxed.

Helmholtz’s theorem even gives $Phi$ and $mathbfA$ in terms of $mathbfF$. More details can be found here:
https://en.wikipedia.org/wiki/Helmholtz_decomposition

share|cite|improve this answer

edited Jul 20 at 23:30

Jollywatt

1447

answered Jul 20 at 12:39

Matrefeytontias

572110

This is called Helmholtz decomposition, a.k.a., the fundamental theorem of vector calculus. Helmholtz’s theorem states that any vector field $mathbfF$ on $mathbbR^3$ can be written as
$$
mathbfF =
underbrace-nablaPhi_textirrotational +
underbracenablatimesmathbfA_textsolenoidal
$$
provided 1) that $mathbfF$ is twice continuously differentiable and 2) that $mathbfF$ vanishes faster than $1/r$ as $r rightarrow infty$.
If $mathbfF$ is on a bounded domain $V subset mathbbR^3$, then the condition 2) that $mathbfF$ vanishes can be relaxed.

Helmholtz’s theorem even gives $Phi$ and $mathbfA$ in terms of $mathbfF$. More details can be found here:
https://en.wikipedia.org/wiki/Helmholtz_decomposition

share|cite|improve this answer

edited Jul 20 at 23:30

Jollywatt

1447

answered Jul 20 at 12:39

Matrefeytontias

572110

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 23:30

Jollywatt

1447

edited Jul 20 at 23:30

Jollywatt

1447

edited Jul 20 at 23:30

Jollywatt

1447

1447

answered Jul 20 at 12:39

Matrefeytontias

572110

answered Jul 20 at 12:39

Matrefeytontias

572110

answered Jul 20 at 12:39

Matrefeytontias

572110

572110

add a comment | 

add a comment | 

 

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