Math 634 Lecture #131.8.5: Polar Coordinates

The “polar coordinate” system plays an especially important role in the analysis of planarODEs x = f(x), x ∈ R2.

We start with the angular wrapping function P : R → T defined by

P (θ) = (cos θ, sin θ).

This map is smooth and surjective, but not injective, and so P is not a diffeomorphism.

The angular wrapping function is an example of a covering map from Algebraic Topology:each point of T is contained in a open set of T on which a local continuous inverse of Pexists.

In this case, the local continuous inverse of P is actually a diffeomorphism.

Thus, each open set together with its local inverse of P is a coordinate system (recall Defi-nition 1.87 on p.47) that we all an angular coordinate system.

[We have added an additional condition to this coordinate system, namely that the diffeo-morphism from an open set of T to R be a local inverse of P .]

The image of a point of T under an angular coordinate system is called its angular coor-dinate, or simply its angle, relative to the angular coordinate system.

Example. The pair (V, Ψ) where V = {x2 + y2 = 1 : x > 0} ⊂ T and Ψ : V → (−π/2, π/2)given by Ψ(x, y) = arctan(y/x) is an angular coordinate system.

The number θ = Ψ(x, y) is the angle assigned to the point (x, y) ∈ V by this angularcoordinate system.

The point (1/√

2, 1/√

2) ∈ V is assigned the angle of θ = π/4 = arctan(√

2/√

2).

Another angular coordinate system is the pair (V, Ψ) where Ψ(x, y) = 2nπ +arctan(y/x) forn ∈ N, for which the angle assigned to the point (1/

√2, 1/

√2) ∈ V is θ = 2nπ + π/4.

[The pair (V, ϕ) where ϕ(x, y) = π + arctan(y/x) is technically a coordinate system, but isnot an angular coordinate system. Why?]

This illustrates that there are infinitely many angles assigned to each point of T.

But all of the angular coordinate systems are compatible in the sense each one is a localinverse of the angular wrapping function.

Now we move to coordinates on R2 that are compatible with the polar wrapping functionP : R2 → R2 defined by

P (r, θ) = (r cos θ, r sin θ).

This function is smooth and surjective, but is not injective, so that P is not a diffeomorphism.

This function is not a covering map because it has no local inverse at the origin of its image:P (0, θ) = (0, 0) for all θ ∈ R.

But near every point of the “punctured plane” R2 \ {(0, 0)} in the codomain, the map Pdoes have a local inverse.

Definition 1.130. A polar coordinate system on the punctured plane is a coordinatesystem (V, Ψ) where V ⊂ R2 \ {(0, 0)}, the range W = Ψ(V ) ⊂ R2, and Ψ : V → W isthe inverse of the polar wrapping function P restricted to W . The collection of all polarcoordinate systems is called polar coordinates.

Example. Consider the sets

V = {(x, y) ∈ R2 : y > 0}, W = {(r, θ) : r > 0, 0 < θ < π},

and the map Ψ : V → W defined by

Ψ(x, y) =(√

x2 + y2,π

2− arctan(x/y)

).

We check that the pair (V, Ψ) is a polar coordinate system, that is, that Ψ is a local inverseof the polar wrapping function P using the appropriate trig identities:

P (Ψ(x, y)) =

(√x2 + y2 cos

(π

2− arctan

(x

y

)),√

x2 + y2 sin

(π

2− arctan

(x

y

)))=

(√x2 + y2 sin

(arctan

(x

y

)),√

x2 + y2 cos

(arctan

(x

y

)))=

(√x2 + y2

x√x2 + y2

,√

x2 + y2y√

x2 + y2

)= (x, y).

Now we look at the push-forward of a planar ODE u = f(u), u ∈ R2, u = (x, y), to polarcoordinates.

For a polar coordinate system (V, Ψ) on the punctured plane with W = Ψ(V ) and (r, θ) =Ψ(x, y), the push-forward of the vector field f is

y = DΨ(Ψ−1(y))f(Ψ−1(y)), y = (r, θ) ∈ W.

Since Ψ is a local inverse of the polar wrapping function P , the (principal part of the)push-forward is

F (r, θ) = DΨ(P (r, θ))f(P (r, θ)).

To compute DΨ(P (r, θ)), we differentiate Ψ◦P (r, θ) = (r, θ) to get DΨ(P (r, θ))DP (r, θ) = I,or

DΨ(P (r, θ)) = [DP (r, θ)]−1 =

[cos θ −r sin θsin θ r cos θ

]−1

=1

r

[r cos θ r sin θ− sin θ cos θ

].

Thus, if f(x, y) = (f1(x, y), f2(x, y)), then

F (r, θ) =1

r

[r cos θ r sin θ− sin θ cos θ

] [f1(r cos θ, r sin θ)f2(r cos θ, r sin θ)

].

Does this depend on the local coordinate system (V, Ψ)? No, it only depends on P .

Generally, there is a “singularity” for F (r, θ) when r = 0, in that F is not defined on theline {(r, θ) : r = 0} in the polar coordinate plane.

In practice, the best way to change to polar coordinates is to obtain the components ofF (r, θ) by differentiating the identities

r2 = x2 + y2, θ = arctan(y/x)

which give

rr = xx + yy = xf1(x, y) + yf2(x, y),

r2θ = r2

(1

1 + (y/x)2

xy − yx

x2

)= xy − yx = xf2(x, y)− yf1(x, y),

and then substituting x = r cos θ and y = r sin θ on right-hand sides.

Example (Exercise 1.131(2)). Convert x = 1− y2, y = x in to polar coordinates.

Following the practical approach above, we have that

rr = x(1− y2) = r cos θ(1− r2 sin2 θ),

r2θ = x(x)− y(1− y2) = r2 cos2 θ − r sin θ(1− r2 sin2 θ).

Simplifying we have

r = cos θ(1− r2 sin2 θ),

θ =r cos2 θ − sin θ(1− r2 sin2 θ)

r.

Notice that original vector field does not have an equilibrium at the origin, and the push-forward vector field has a singularity, i.e., is not defined, when r = 0.

There is one situation where the push-forward vector field is defined at the origin.

Proposition 1.135. If u = f(u), u = (x, y) ∈ R2, satisfies f(0) = 0, then the correspondingODE in polar coordinates has a removable singularity. Also, when f is Ck, the desingularizedvector field in polar coordinates is Ck−1.

See the appendix for a sketch of the proof of this.

When the push-forward vector field F (r, θ) is defined on the whole polar coordinate plane, itis technically an “extension” of the push-forward of the original vector field to the singularline r = 0.

Example. Converting the system

x = y, y = −(x2 + y2 − 1)y − x

into polar coordinates gives

r = −r(r2 − 1) sin2 θ, θ = −1− (r2 − 1) sin θ cos θ.

The equilibrium at the origin (x, y) = (0, 0) corresponds to r = 0. What does r = 1correspond to?

Appendix. Sketch of the Proof of Proposition 1.135. By Taylor’s Theorem we can write

f1(x, y) = 0 + a1x + b1y + c1x2 + d1xy + e1y

2 + · · ·,f2(x, y) = 0 + a2x + b2y + c2x

2 + d2xy + e2y2 + · · ·.

Substituting x = r cos θ and y = r sin θ gives

f1 = r(a1 cos θ + b1 sin θ) + r2(c1 cos2 θ + d1 cos θ sin θ + e1 sin2 θ) + · · ·,f2 = r(a2 cos θ + b2 sin θ) + r2(c2 cos2 θ + d2 cos θ sin θ + e2 sin2 θ) + · · ·.

Then we have that

rr = xf1 + yf2 = r2(a1 cos2 θ + (a2 + b1) cos θ sin θ + b2 sin2 θ) + O(r3),

r2θ = xf2 − yf1 = r2(a2 cos2 θ + (b2 − a1) cos θ sin θ − b1 sin2 θ) + O(r3).

The common factors of r now cancel to leave

r = r(a1 cos2 θ + (a2 + b1) cos θ sin θ) + b2 sin2 θ) + O(r2),

θ = a2 cos2 θ + (b2 − a1) cos θ sin θ − b1 sin2 θ + O(r),

which are defined when r = 0.

Homework. Exercises 1.128, 1.129 on p.66, Exercises 1.131 (1,3), 1.132 (1,3), 1.133 onp.68, and Exercises 1.136, 1.137 on p.70.