MAT237 - LEC5201 - 2019–2020 2020 Winter Term Notes

(1)

MAT237 - LEC5201 - 2019–2020 2020 Winter Term Notes

Jean-Baptiste Campesato April 1, 2020

Multivariable calculus

Change of variables: usual coordinate systems

Jean-Baptiste Campesato February 13^th, 2020

1 Polar coordinates

Φ ∶ (0, +∞) × (−𝜋, 𝜋) → ℝ²⧵ {(𝑥, 0) ∈ ℝ² ∶ 𝑥 ≤ 0}

(𝑟, 𝜃) ↦ (𝑟cos𝜃, 𝑟sin𝜃)

𝑥 𝑦

𝑟

𝑟cos𝜃

𝑟sin𝜃

𝜃

• Φis𝐶¹.

• Φis bijective.

• det𝐷Φ(𝑟, 𝜃) =det (

cos𝜃 −𝑟sin𝜃

sin𝜃 𝑟cos𝜃 )= 𝑟 (cos²𝜃 +sin²𝜃) = 𝑟 > 0.

• HenceΦis a𝐶¹-diffeomorphism.

• And|det𝐷Φ(𝑟, 𝜃)| = 𝑟.

Example 1. LetΔ = {(𝑥, 𝑦) ∈ ℝ² ∶ 1 ≤ 𝑥²+ 𝑦²≤ 9, 𝑥 ≥ 0}. We want to compute

∫

Δ

𝑒^𝑥²^+𝑦².

First, notice thatΔ = Φ ([1, 3] × [−𝜋/2, 𝜋/2]).

Hence,

∫

Δ

𝑒^𝑥²^+𝑦²𝑑𝑥𝑑𝑦 =

∫

[1,3]×[−𝜋/2,𝜋/2]

𝑒^𝑟²𝑟𝑑𝑟𝑑𝜃 by the CoV formula

=∫

𝜋/2

−𝜋/2∫

3 1

𝑟𝑒^𝑟²𝑑𝑟𝑑𝜃 by the iterated integrals theorem

=∫

𝜋/2

−𝜋/2

𝑒⁹− 𝑒 2 𝑑𝜃

=𝜋

2 (𝑒⁹− 𝑒)

(65)

2 Change of variables: usual coordinate systems

Example 2. We want to compute

∫

𝐵((1,1),1)

𝑥²+ 𝑦²− 2𝑦𝑑𝑥𝑑𝑦.

First notice thatΨ ∶ ℝ² → ℝ² defined byΨ(𝑥, 𝑦) = (𝑥 + 1, 𝑦 + 1)is a𝐶¹-diffeomorphism and that|det𝐷Ψ(𝑥, 𝑦)| = 1.

Moreover𝐵((1, 1), 1) = Ψ (𝐵(0, 1)). Hence

∫

𝐵((1,1),1)

𝑥²+ 𝑦²− 2𝑦𝑑𝑥𝑑𝑦 =

∫

𝐵(0,1)

(𝑥 + 1)²+ (𝑦 + 1)²− 2(𝑦 + 1)𝑑𝑥𝑑𝑦 by the CoV formula

= ∫

𝐵(0,1)

𝑥²+ 𝑦²− 2𝑥𝑑𝑥𝑑𝑦

Next, we have𝐵(0, 1) = Φ([0, 1] × [−𝜋, 𝜋]).

Notice that there is an issue for𝑟 = 0or𝜃 = ±𝜋(i.e.{(𝑥, 0) ∶ 𝑥 ∈ [−1, 0]}) but these sets have zero content.

Hence

∫

𝐵(0,1)

𝑥²+ 𝑦²− 2𝑥𝑑𝑥𝑑𝑦 =

∫

[0,1]×[−𝜋,𝜋]

(𝑟²− 2𝑟cos𝜃)𝑟𝑑𝑟𝑑𝜃 by the CoV formula

=∫

𝜋

−𝜋∫

1 0

𝑟³− 2𝑟²cos𝜃𝑑𝑟𝑑𝜃 by the iterated integrals theorem

=∫

𝜋

−𝜋

1 4−2

3cos𝜃𝑑𝜃

=𝜋 2

(66)

MAT237Y1 – LEC5201 – J.-B. Campesato 3

2 Cylindrical coordinates

Φ ∶ (0, +∞) × (−𝜋, 𝜋) × ℝ → ℝ³⧵ ((−∞, 0] × {0} × ℝ) (𝑟, 𝜃, 𝑧) ↦ (𝑟cos𝜃, 𝑟sin𝜃, 𝑧)

𝑥

𝑦 𝑧

𝜃 𝑟

(𝑟cos𝜃, 𝑟sin𝜃, 𝑧) 𝑧

𝑟cos𝜃 𝑟sin𝜃

• Φis𝐶¹.

• Φis bijective.

• det𝐷Φ(𝑟, 𝜃, 𝑧) =det

⎛⎜

⎜⎝

cos𝜃 −𝑟sin𝜃 0 sin𝜃 𝑟cos𝜃 0

0 0 1

⎞⎟

⎟⎠

=det (

cos𝜃 −𝑟sin𝜃

sin𝜃 𝑟cos𝜃 )= 𝑟 > 0.

• And|det𝐷Φ(𝑟, 𝜃, 𝑧)| = 𝑟.

(67)

Example 3. We want to compute

∫_Δ𝑧𝑑𝑥𝑑𝑦𝑑𝑥

whereΔ = {(𝑥, 𝑦, 𝑧) ∈ ℝ³ ∶ 𝑥²+ 𝑦²≤_𝐻^𝑅²₂(𝐻 − 𝑧)², 0 ≤ 𝑧 ≤ 𝐻}.

𝑅 𝐻 Δ

𝜃

𝑟 𝑧

−𝜋 𝐻

Γ

𝜋 𝑅

Notice thatΔ = Φ(Γ)whereΓ = {(𝑟, 𝜃, 𝑧) ∶ 0 ≤ 𝑟 ≤ 𝐻^𝑅(𝐻 − 𝑧), 𝜃 ∈ [−𝜋, 𝜋], 𝑧 ∈ [0, 𝐻]}. Again,Γgoes outside the domain ofΦbut the involved sets have zero content.

Hence

∫_Δ𝑧𝑑𝑥𝑑𝑦𝑑𝑥 =

∫_Γ𝑧𝑟𝑑𝑟𝑑𝜃𝑑𝑧 by the CoV formula

=∫

𝐻

0 ∫

𝜋

−𝜋∫

𝑅 𝐻(𝐻−𝑧) 0

𝑟𝑧𝑑𝑟𝑑𝜃𝑑𝑧

=∫

𝐻

0 ∫

𝜋

−𝜋

𝑅²

2𝐻²(𝐻 − 𝑧)²𝑧𝑑𝜃𝑑𝑧

= 𝜋𝑅² 𝐻² ∫

𝐻 0

(𝐻 − 𝑧)²𝑧𝑑𝑧

= 𝜋𝑅²𝐻² 12

(68)

MAT237Y1 – LEC5201 – J.-B. Campesato 5

3 Spherical coordinates

Φ ∶ (0, +∞) × (0, 2𝜋) × (0, 𝜋) → ℝ³⧵ ([0, +∞) × {0} × ℝ) (𝑟, 𝜃, 𝜑) ↦ (𝑟cos𝜃sin𝜑, 𝑟sin𝜃sin𝜑, 𝑟cos𝜑) In this course, we use the following convention ^⋆:

(𝑟, 𝜃, 𝜑) = (radius/distance to the origin,longitude,colatitude)

𝑥

𝑦 𝑧

𝑟sin 𝜑

𝑟cos𝜑

𝜃 𝜑

𝑟

𝑥

𝑦 𝑧

𝑟cos𝜃sin𝜑 𝑟sin𝜃sin𝜑 𝜃

𝜑 𝑟

• Φis𝐶¹.

• Φis bijective.

• The Jacobian determinant is

det𝐷Φ(𝑟, 𝜃, 𝜑) =det

⎛⎜

⎜⎝

cos𝜃sin𝜑 −𝑟sin𝜃sin𝜑 𝑟cos𝜃cos𝜑 sin𝜃sin𝜑 𝑟cos𝜃sin𝜑 𝑟sin𝜃cos𝜑

cos𝜑 0 −𝑟sin𝜑

⎞⎟

⎟⎠

=cos𝜑det (

−𝑟sin𝜃sin𝜑 𝑟cos𝜃cos𝜑

𝑟cos𝜃sin𝜑 𝑟sin𝜃cos𝜑)− 𝑟sin𝜑det (

cos𝜃sin𝜑 −𝑟sin𝜃sin𝜑 sin𝜃sin𝜑 𝑟cos𝜃sin𝜑 )

= 𝑟²cos²𝜑sin𝜑det (

−sin𝜃 cos𝜃

cos𝜃 sin𝜃 )− 𝑟²sin³𝜑det (

cos𝜃 −sin𝜃 sin𝜃 cos𝜃 )

= −𝑟²cos²𝜑sin𝜑 − 𝑟²sin³𝜑

= −𝑟²sin𝜑 < 0since𝜑 ∈ (0, 𝜋)

• And|det𝐷Φ(𝑟, 𝜃, 𝜑)| = 𝑟²sin𝜑.

⋆This convention may differ from the one used in other courses in math or in physics (the meaning of𝜃and𝜑may be swapped, some people use the latitude and not the colatitude…).

I believe that the usual convention in physics is(𝑟, 𝜃, 𝜑) = (radius,colatitude,longitude)as in ISO 80000-2, i.e. the meaning of𝜃and𝜑are swapped from our convention in MAT237.

(69)

Example 4. We want to compute∫_Δ𝑧𝑑𝑥𝑑𝑦𝑑𝑧whereΔ = {(𝑥, 𝑦, 𝑧) ∈ ℝ³ ∶ 𝑥²+ 𝑦²+ 𝑧²≤ 1, 𝑧 ≥ 0}. Notice thatΔ = Φ ([0, 1] × [0, 2𝜋] × [0, 𝜋/2]).

Again, there is an issue with the domain ofΦbut the involved sets have zero content.

Hence

∫_Δ𝑧𝑑𝑥𝑑𝑦𝑑𝑧 =

∫

[0,1]×[0,2𝜋]×[0,𝜋/2]

𝑟³cos𝜑sin𝜑𝑑𝑟𝑑𝜃𝑑𝜑 by the CoV formula

=∫

𝜋/2

0 ∫

2𝜋

0 ∫

1 0

𝑟³sin(2𝜑)

2 𝑑𝑟𝑑𝜃𝑑𝜑 by the iterated integrals theorem

= 2𝜋1 4 (

cos0 4 −cos𝜋

4 )

=𝜋 4

(70)

(71)

(72)

(73)

(74)

(75)

(76)

(77)

(78)

(79)

(80)

(81)

(82)

(83)

(84)

(85)

(86)

(87)

(88)

(89)

(90)

(91)

(92)

(93)

(94)

(95)

(96)

(97)

(98)

(99)

(100)

(101)

(102)

(103)

(104)

(105)

(106)

(107)

(108)

(109)

(110)

(111)

(112)

(113)

(114)

(115)

(116)

(117)

(118)

(119)

(120)

(121)

(122)

(123)

(124)

(125)

(126)

(127)

3

(128)

(129)

(130)

(131)

(132)

(133)

(134)

Assume that S is "closed"

In this corollary, I am not very formal.

It is possible to define properly what do I mean by "closed", but I think that's not necessary for our purpose.

You can believe your intuition: "closed" means that there is no "boundary" in the above sense, so the line integral is 0.

By the way, (*) doesn't depend on the coordinate system!

(Image from Wikipedia) By "counterclockwise", I mean that the force pushes

the particle in the direction of the orientation of the boundary, as in the "right hand rule" on the right:

� n

(135)

(136)

(137)

(138)

(139)

(140)

(141)

(142)

(143)

- G eitc ~ c_~ ^{~~ ~} ^M ^(VIL ^{~ )}

~ \)) ~ ~ ~ ~ 9~CDiL ~ ;

~ ^, ^Ql e.. 9? ~ ' f · . l\l - '9i ³ c~-

~ '\\ f"" ~Gr ~ ^G ^- ^l\,l_.,,~ ^C ² ~ d.>-~F '° o

(I) ~ ~ ^{-ù ~} ~ ^Il>-. VI>.\°"- -- ~ - -

~ ^{(}l \h} ~~-~ ostl ~t '.:: ⁰ ~

~ ~'vs G~ ftl:~~ C ~\- . t-=: ^~&

m. ~ M ^b ^Ch ^0- 1- -cl~I\. ~ J ~ F

b. eù ~ r ⁼ ~ l ^~ ^~ ^t ^c;, ^î ^~

⁰

® ~ \l<'l<L. {l},_ in

~ ( I N - ~ ,

3 ?o t- C1,L .,, \. • V<\ t ^it.L ^-\k.

~ ~ r ^? ^o ^{~ ~} ^\A ^V0 ^AL

~ ~ ~ \ ~J-u:)<.

te'-- '\ ~ ^\?.,~I ^l ⁾ ^f_ ^ll.,l ^1\,6.Q_ là- '

^,7

G('\) ⁼⁼ ^\ r ^l~ ^~ ^--\; ² ^{?,, '"} ^{\:~) )'-} lt ^l ^{C\ -} ^?o)) ^J.~

-Ct\. ^Jp ~ ~

~ , ~ ~"'- te- ^2. ~~ "~ ~ ~ ~~

^I

""'- 1tf\::: \:k1- G 1.1 C o.J ~ ^c,,,.Q ^& ^{c=: \} ^{w,.Q. \_} ~ l M--

~ -lk t.wl ^0M.

^W ^,^{ ^·^\--^.

^?\.,~ ^, t ~ ^~ ^/M._

ti ~-b ^~ ~ - ,~

¹

^~

^}/.\^:

(144)

(\'.)

Ft"-~,,i;, ~ F \ \.\-\.) ?., ·" \-\) - t ~ ~d"••,i..1.) - -

::: F \\~-\.)

^~01'X.

~\-,c; ^·· l~-\--')'r'o,~ - .ç-~j / t~---\-">- ro.~ ~ - t,; - ~ - ) --

~

_: f ^(.Q ^b ^-\- \-('t'll.,"\1't"')) _,. __ - - - - - - -· -- ---- - --- --

\ c~, ^,11-::;.. ^_ ^_ ^__ ^;::: ^--

^'X._^- ^- ^_^{- -} ^-

^--- ^{-=-· ---} --- ---- ---- --

":I C\ ~ C ^_ \>,,.... % - ^- ^P.- ^~ ^r1:--- ^~,è--> ^- ^- ^{- -} ^- ^. ^-

~

^\).)Q..__

^U,\L ^- ^~ ^- _~~ ^.- ^- ^- ^{- - -} ^L~-- ^ç1- ^- ^_ ^- ~v ---- ~ - ^~ ---

\'.tJ; .

l _ ^Cr) ^_ ^_ J_G __ ^j_E ^__ ^~N_ ^G:) ^_E ^__ ^----

- - . -~(f ^~1) ^{__ 6:_} ^~ ^~F)~---

l ^~ _ ^~ , ^~~- ~ - ^- ^ ^~~ ^~ _ --

~ ^{~ \} ~ ^Ù>- ^~ ^{- - - -} ---== ----

tt,(Q. _( F ,c;_ ~ _ r)) ^-=- Ctr ~ - ^yiJ- ^F~ ⁺ ^-- ^fd,ï.,iI~ ^r ^\l F - ^cFJ)~r)- ^J.t.,cfK ^_

-- - - u) - -- - - -Cf) ___ - ~ ^(g

cb(__lc _: VJ - ~ "' - ^'r~ ^:: ^~< ^~ ^î ^::\ ^'è -~,~~'!> - '!_ _ l>s-\>o,~}j i::- ~

. ~ -

-

- - - - - - - -

. J .

:: t ^- ~~f-l~!'.'. ~~- ~\-!_ «\') - - - -

® - ^d,i..r l\ ^r J f - ~ :ft ^'f ^C\,1:.,~ ⁹ ^; ^~ ^-\'\} ^{_ _} --=- ^~~

~ \_f , ,v)_l \-r')_ ~ _ \- f l\.1~\--J îé)_.\:_\ ~~ --- ---~ _

® c}..t~ lf) _ _ ⁼ + ^{2fit-) -} ^~ ^-\- ^- ^~\- ^, ^-~ ^-~ ^-]E~.1-~ ^..== ^.. ^1. ^~fl~ ^~ ^\r ^l ^P ^o ^_ ^~ ^l~_} ^~

?,(, . '7)~ - 0~

_ ~ U L e»s ^Q_ G-_l °' \ ^~ - -ç;) ^~ ^f ^w ^{-\-) '?o} ^~ ^t ^9.) ^~ ^t ^~%li ~V-)J>,, _:1 hù1JC

_- ^~ s: . - ^ï-Lf ^f ^\U~ ^\- ^} ^\'~~:\ ^)J ^d'r ^_ ^__ ^-_---~ ^- ^-~

::: [ t~f"( ~=t'Wo- ~\-~JJ~ - ~ • _:-__ -=--~ _ ~-

EJ

(145)

l~~ ~ ~ ~i r ^o._~Tu- ^{~ ~}

G ( "--~,t0 ~ (' ⁾ f ^{(l ~} ^- ^\--) <i>o ~t ^{\i,4,~ ')} J ~ (.\-(C~,~.1:) ^-ï>

⁰⁾

Î ^dJ-

o

~ ?

₀

^t,llk ,J . 'f\ ^Èill ~ L ~_,_,__\- ~ ^Po ^1-

⁰

₁

V\ ~ ~ \ . l,_, M._

~ü.-_¼..~ ~ ( Vv\ r~ct_, ^l\,l ^V\ ^~t<.J ^J- ^ê) ^io ^~

fo :=:O ~

l ~ ~~~

-

~

^-

~

..

5),~ ~ ... - f l"-,.,,i) -o 1

• v'I. t'>l ,'11~1 ~ ail"~~\){<;~

y~ . q.. 'l,,Ç-tt"Î

~ ~V~ ^o ^~\- ~ ^V) ~ (y - . ttl~~~

C., 'L ~J" · f ~ tv,\.Q_ G

1~J \ . ^~ ^WLri k \\f . -; ^~

~

^,x,?~'1^'l~'l; {.;:::; \

· 1

eot~

(L\- ~~ f. - ~ ~ 4Ïl4 ⁰

(146)

University of Toronto – MAT237Y1 – LEC5201

Multivariable calculus

Poincaré lemma: summary of the last two sections

Jean-Baptiste Campesato April 2^nd, 2020

1 Star-shaped sets

Definition 1. We say that𝑆 ⊂ ℝ^𝑛isstar-shapedif there exists𝑝₀ ∈ 𝑆such that for any𝑞 ∈ 𝑆the line segment from𝑝₀to𝑞is included in𝑆.

Proposition 2. A non-empty convex set is star-shaped.

Beware:the empty set is convex but not star-shaped (because “∀𝑝 ∈ ∅, 𝑃 (𝑝)” is vacuously true whereas

“∃𝑝 ∈ ∅, 𝑃 (𝑝)” is always false).

2 Conservative vector fields

Theorem 3. Let𝑈 ⊂ ℝ^𝑛be open (𝑛 ≥ 2), andF∶ 𝑈 → ℝ^𝑛be a continuous vector field.

The following are equivalent:

1. There exists𝑓 ∶ 𝑈 → ℝ 𝐶¹such thatF= ∇𝑓.

2. For any piecewise smooth closed oriented curve𝐶in𝑈, we have

∫_𝐶F⋅dx= 0.

3. Fsatisfies thepath-independence property: for any two piecewise smooth oriented curves𝐶₁and𝐶₂with same startpoint and same endpoint, we have

∫_𝐶

1

F⋅dx=

∫_𝐶

2

F⋅dx.

Then we say thatFisconservativeand that𝑓is apotentialofF(beware: in physics we usually say that−𝑓 is a potential ofF).

(147)

2 Poincaré lemma: summary of the last two sections 2.1 𝑛 = 2

Proposition 4. Let𝑈 ⊂ ℝ²be open andF∶ 𝑈 → ℝ²be a𝐶¹vector field.

IfFis conservative then 𝜕𝐹₂

𝜕𝑥 =𝜕𝐹₁

𝜕𝑦 on𝑈. When𝑈is star-shaped, the converse is true:

Theorem 5(Poincaré lemma). Let𝑈 ⊂ ℝ²be astar-shapedopen set andF∶ 𝑈 → ℝ²be a𝐶¹vector field.

If𝜕𝐹₂

𝜕𝑥 =𝜕𝐹₁

𝜕𝑦 on𝑈then there exists𝑓 ∶ 𝑈 → ℝ 𝐶²such thatF= ∇𝑓.

Beware:the above theorem is false with no assumption on the domain.

Indeed, defineF(𝑥, 𝑦) = (𝑥²+𝑦^𝑦 ²,_𝑥₂^−𝑥_+𝑦₂)on𝑈 = ℝ²⧵ {0}. Then𝜕𝐹₂

𝜕𝑥 =𝜕𝐹₁

𝜕𝑦 but

∫_𝐶F⋅dx= −2𝜋 ≠ 0where 𝐶is the unit circle with the counterclockwise orientation, soFis not conservative.

Alexis Clairaut proved a first version of this Poincaré lemma in 1739 but there was a flaw since he didn’t real- ize that he needed some assumptions on the domain. Then Jean Le Rond d’Alembert gave this counter-example in 1768. Notice that the notation

∫_𝐶𝑃 (𝑥, 𝑦)d𝑥 + 𝑄(𝑥, 𝑦)d𝑦was introduced by A. Clairaut in the above cited paper.

Proposition 6. Let𝑈 ⊂ ℝ²be an open rectangle andF∶ 𝑈 → ℝ²be a𝐶¹conservative vector field.

Let𝑝 = (𝑎, 𝑏) ∈ 𝑈and define𝑓 ∶ 𝑈 → ℝby𝑓 (𝑥, 𝑦) =

∫

𝑥 𝑎

𝐹₁(𝑡, 𝑏)d𝑡 +

∫

𝑦 𝑏

𝐹₂(𝑥, 𝑡)d𝑡.

Then𝑓is𝐶²andF= ∇𝑓.

2.2 𝑛 = 3

Proposition 7. Let𝑈 ⊂ ℝ³be open andF∶ 𝑈 → ℝ³be a𝐶¹vector field.

IfFis conservative thencurlF=0on𝑈. When𝑈is star-shaped, the converse is true:

Theorem 8(Poincaré lemma). Let𝑈 ⊂ ℝ³be astar-shapedopen set andF∶ 𝑈 → ℝ³be a𝐶¹vector field.

IfcurlF=0on𝑈then there exists𝑓 ∶ 𝑈 → ℝ 𝐶²such thatF= ∇𝑓.

Proposition 9. Let𝑈 ⊂ ℝ³be an open rectangle andF∶ 𝑈 → ℝ³be a𝐶¹conservative vector field.

Let𝑝 = (𝑎, 𝑏, 𝑐) ∈ 𝑈and define𝑓 ∶ 𝑈 → ℝby𝑓 (𝑥, 𝑦, 𝑧) =

∫

𝑥 𝑎

𝐹₁(𝑡, 𝑏, 𝑐)d𝑡 +

∫

𝑦 𝑏

𝐹₂(𝑥, 𝑡, 𝑐)d𝑡 +

∫

𝑧 𝑐

𝐹₂(𝑥, 𝑦, 𝑡)d𝑡.

2.3 General case

This subsection is not part of MAT237.

The proof of the above Poincaré Lemma for𝑛 = 2relied on the Gradient Theorem, and the proof of the above Poincaré Lemma for𝑛 = 3relied on Kelvin–Stokes theorem. They admit a generalization to any𝑛 from which we may derive the following general results.

Proposition 10. Let𝑈 ⊂ ℝ^𝑛be open andF∶ 𝑈 → ℝ^𝑛be a𝐶¹vector field.

IfFis conservative then∀𝑖, 𝑗 = 1, … , 𝑛, 𝜕𝐹_𝑖

𝜕𝑥_𝑗 =𝜕𝐹_𝑗

𝜕𝑥_𝑖 on𝑈. When𝑈is star-shaped, the converse is true:

Theorem 11(Poincaré lemma). Let𝑈 ⊂ ℝ^𝑛be astar-shapedopen set andF∶ 𝑈 → ℝ^𝑛be a𝐶¹vector field.

If∀𝑖, 𝑗 = 1, … , 𝑛, 𝜕𝐹_𝑖

𝜕𝑥_𝑗 =𝜕𝐹_𝑗

𝜕𝑥_𝑖 on𝑈then there exists𝑓 ∶ 𝑈 → ℝ 𝐶²such thatF=∇𝑓.

(148)

MAT237Y1 – LEC5201 – J.-B. Campesato 3 Proposition 12. Let𝑈 ⊂ ℝ^𝑛be an open rectangle andF∶ 𝑈 → ℝ^𝑛be a𝐶¹conservative vector field.

Let𝑝 = (𝑎₁, … , 𝑎_𝑛) ∈ 𝑈and define𝑓 ∶ 𝑈 → ℝby𝑓 (𝑥₁, … , 𝑥_𝑛) =

𝑛

∑𝑖=1∫

𝑥𝑖 𝑎𝑖

𝐹_𝑖(𝑥₁, … , 𝑥_𝑖−1, 𝑡, 𝑎_𝑖+1, … , 𝑎_𝑛)d𝑡.

3 Vector potentials

Proposition 13. Let𝑈 ⊂ ℝ³be open andF∶ 𝑈 → ℝ³be a𝐶¹vector field.

IfF=curlGforG∶ 𝑈 → ℝ³𝐶²thendivF= 0on𝑈.

Then we say thatGis avector potentialofF.

When𝑈is star-shaped, the converse is true:

Theorem 14(Poincaré lemma). Let𝑈 ⊂ ℝ³be astar-shapedopen set andF∶ 𝑈 → ℝ³be a𝐶¹vector field.

IfdivF= 0on𝑈then there existsG∶ 𝑈 → ℝ³𝐶²such thatF=curlG.

Beware:the above theorem is false with no assumption on the domain.

Indeed, defineF=_‖r‖^r₃ onℝ³⧵ {0}wherer(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦, 𝑧).

Then divF= 0but there is noG∶ ℝ³⧵ {0} → ℝ³such thatF=curlG.

Proposition 15. Let𝑈 ⊂ ℝ³be astar-shapedopen set andF∶ 𝑈 → ℝ³be a𝐶¹vector field such thatdivF= 0.

Let𝑝₀= (𝑥₀, 𝑦₀, 𝑧₀) ∈ 𝑈such that for any𝑞 ∈ 𝑈 the line segment from𝑝₀to𝑞is included in𝑈. DefineG∶ 𝑈 → ℝ³by

G(𝑥, 𝑦, 𝑧) =

∫

1

0

F

⎛⎜

⎜⎝

𝑥₀+ 𝑡(𝑥 − 𝑥₀) 𝑦₀+ 𝑡(𝑦 − 𝑦₀) 𝑧₀+ 𝑡(𝑧 − 𝑧₀)

⎞⎟

⎟⎠

×

⎛⎜

⎜⎝ 𝑥 − 𝑥₀ 𝑦 − 𝑦₀ 𝑧 − 𝑧₀

⎞⎟

⎟⎠ 𝑡d𝑡

ThenGis𝐶²andF=curlG.

Quite often (but not always),𝑈iscenteredat0so that we can take𝑝₀ =0, then the above formula may rewritten

G=

∫

1 0

F(𝑡r) × (𝑡r)d𝑡 wherer(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦, 𝑧).

4 The general Poincaré lemma

In all the above results namedPoincaré Lemma, we may relax the assumption on the domain by assuming that𝑈 is open and contractible (which is weaker than open and star-shaped since a star-shaped set is contractible) but this notion is not part of MAT237.

Then, they are all special cases of the following very general result:

Theorem 16(Poincaré Lemma).

If𝑀is a contractible manifold then𝐻_dR^𝑛 (𝑀) = {

ℝ if𝑛 = 0 0 otherwise , i.e. the closed differential forms on𝑀are exact.

You should come back here after you learn differential forms and de Rham cohomology.

MAT237 - LEC5201 - 2019–2020 2020 Winter Term Notes