Vortex cusps

Vortex cusps

Volker Elling

Academia Sinica (Taipei)

Motivation Incompressible 2d Euler:

ω_t + v _{· ∇ω = 0 , ∇ ·} v = 0 , ω = ∇ × v Initial data

ω(t = 0, x) = |x|^−1/µ˚ω(∡x)

Thm [E., Comm. Math. Phys. 2016]: existence of selfsimilar solutions with ω algebraic spiral patterns, if

1. dominant sign: ^R₀^2π˚ω 6= 0

2. sufficiently high periodicity: ˚ω 2π/N -periodic with N ≫ 1 +

+ +

+ -

- -

-

t = 0 t > 0 t ≫ 0 t ≫ 0

What if neither sign dominates?

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Vorticity generation: Mach reflection From smooth initial data!

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Simple special case

N vortex-sheet pairs, equal angles between center lines

N = 4

t = 0

ω < 0 ω > 0

at infinity

t = 1

Angle

t ≫ 1

Limit: pair angle at infinity small Self-similar: µ similarity exponent,

dx ∼ t^µ

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Self-similarity (Euler):

0 = ω_t + v _{· ∇ω ,} 0 = ∇ · v _{, ω = ∇ ×} v Initial data ω(t = 0, x) = |x|^−1/µ˚ω(∡x). Ansatz:

ω(t, x) = t⁻¹ω(t^−µx) , v(t,x) = t^µ−1v(t^−µx) Selfsimilar incompressible Euler:

0 = (v _{− µ}x

| {z }

=q

) · ∇ω − ω = ∇ · ^(v _{− µ}x

| {z }

=q

)ω^ + (2µ − 1)ω q pseudo-velocity: along it ω smoother than transversally

0 = ∇ · v _{⇔ −2µ = ∇ ·} q

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Biot-Savart law Circulation: Γ(B) =

I

∂B v · dx =

Z

B ω dx dy

∂B Z^′

Z = x+iy, V = v^x−iv^y







∇ · v = 0

∇ × v = ω ⇒ V (Z) = 1 2πi

Z

dΓ(Z^′)

z }| {

ω(Z^′)dx dy Z − Z^′ Point vortex: V (Z) = Γ

2πi(Z − Z^′) ∼ 1 r

v = (+κ/2, 0) Vortex sheet

v = (−κ/2, 0) Straight uniform vortex sheet:

← infinitesimal point vortices on line, equal spacing dx, equal dΓ = κ dx

V (Z) =







−κ/2, y = Im Z > 0 +κ/2, Im Z < 0

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Cusp approximation: horizontal part

Γ(x) = ^R_C ω(x)dx, C ccw loop enclosing upper sheet from 0 to x V (Z) = 1

2πi

Z Γ_x(x^′)dx^′ Z − Z^′

ω(Z^′) =

v^x = 0 v_A^x = 0

graph(x 7→ ˆy(x))

graph(x 7→ −ˆy(x)) z

−Γx(x)δ(y + ˆy(x)) Γ_x(x)δ(y − ˆy(x))

v_S^x = ¹₂Γ_x(x) v_I^x = Γ_x(x)

Γ_x(x^′)δ(y^′ − ˆy(x^′))

−Γx(x^′)δ(y^′ + ˆy(x^′)) Z^′

Z^′

Distant blue points have nearby red point mirror image with equal ω of opposite-sign → strong cancellation

Nearby parts dominate V integral Horizontal velocity v^x = Re V :

v^x(x, y) ≈ v_I^x(x) = Γ_x(x) inside v^x(x, y) ≈ v_S^x(x) = 1

2Γ_x(x) on (upper) sheet

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Mass conservation

−2µ = ∇ · q _, q _· n = 0 at sheet (q = v _{− µ}x)

triangle mouth cusp

Integrate over any “cusp triangle”:

0 > −2µ · triangle area =

Z

upper,lower sides q _· n

| {z }

=0

ds +

Z

mouth q^x(x, y)dy Approximation has v^x(x, y) ≈ v^x(x) and hence

q^x(x, y) ≈ q^x(x) < 0 only consider solutions with

q_I^x < 0 , q_I^x → 0 as x ց 0

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Horizontal velocity modelling Selfsimilar vorticity equation:

S

y = 0 y = h A

I

(1 − 2µ)ω = ^(v^x − µx)ω^

x + ^(v^y − µy)ω^

y

(1−2µ)

Z _h

0 ωdy = ^{ Z h}

0 (v^x−µx)ωdy^

x+^h(v^y − µy)ω^ih

| {z 0}

=0

For small pair angle v^y, v_A^x, ˆy → 0, but v^x not. So:

ω = v_x^y − v_y^x ≈ −v_y^x ⇒ (1 − 2µ)

Z _h

0 v_y^xdy = ^{ Z h}

0 (v^x − µx)v_y^xdy^

x

= ^{ Z h}

0

(v^x)² 2



y − µ(xv^x)_ydy^

x

(1 − 2µ)[v^x]^h_y=0 = ^[1

2(v^x)² − µxv^x]^h_y=0^

x

v_A^x ≈ 0 ^{no h}⇒

only 0 (1 − 2µ)v_I^x = ^(v_I^x)²

2 − µxv_I^x

x

ODE for velocity v^x = v_I^x(x) between sheets!

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Horizontal velocity approximation

(1 − 2µ)v_I^x = ^(v_I^x)²

2 − µxv_I^x

x ⇒ (1 − µ)v_I^x = (v_I^x − µx)∂_xv_I^x v^x(x) = xu(x) ⇒ − ∂u

∂ log x = u(u − 1) u − µ

• Conservation: q_I^x ր 0 at cusp x ց 0, so u = v_I^x/x = o(¹_x).

• Need q_I^x = v_I^x − µx < 0 everywhere, so u < µ.

• At x → ∞ need v^x = o(x) (← “uniqueness”), so u = o(1).

• Analysis: get O(1/x) blowup as x ց 0, or u → µ and hence q^x → 0 at finite x > 0 (no use), unless u → 0 or u → 1.

• u = 0 unstable for any µ > ¹₂.

• u = 1: for µ < 1: unstable.

For µ = 1: u = 1 absent (cancels).

For µ > 1: u = 1 asy. stable nontrivial v^x = x + o(x) solutions.

Conclusion: only for µ > 1 have useful solution:

v_I^x(x) = x + ... , v_S^x(x) = 1

2x + ... , q_S^x(x) = (1

2 − µ)x + ...

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Vertical velocity approximation 1: just zero

Assume v^y inside and outside is tiny, can be neglected:

v_S^y ≈ 0

ODE: q_S tangential to sheet graph ˆy, so yˆ_x = q_S^y

q_S^x = v_S^y − µˆy

v_S^x − µx ≈ 0 − µˆy

(¹₂ − µ)x ⇒ (1

2 − µ)xˆy_x ≈ −µˆy Solution:

ˆy = Cx^µ/(µ−12) for x ≈ 0

Cusp exponent:

µ µ − ¹₂

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v^y approximation 2: non-horizontal sheet

Sheet tangent: (1, ˆy_x)/^q1 + ˆy_x². To first order in ˆy: tangent (1, ˆy_x) (v^x, v^y) = −¹₂v_I^x(1, ˆyx)

(v^x, v^y) = ¹₂v_I^x(1, ˆyx) (v^x, v^y) = ¹₂v_I^x(1, −ˆy_x)

(v^x, v^y) = −¹₂v_I^x(1, −ˆyx)

v_A^y ≈ −v_I^xˆy_x , v_S^y ≈ −1

2v_I^xyˆ_x , v_I^y ≈ 0 ODE:

ˆy_x = v_S^y − µˆy

v_S^x − µx ≈ −¹₂xˆy_x − µˆy

(¹₂ − µ)x ⇒ (1 − µ)xˆy_x ≈ −µˆy Cusp exponent:

µ µ − 1

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v^y approximation 3: mass conservation

Inside v_I^x = x + ..., i.e. v_x^x = 1 + ..., so by v_x^x+ v_y^y = 0 get v_y^y = −1 + ..., hence using v^y = 0 at y = 0 integration to y = ˆy yields

v^y ≈ −ˆy on lower side of sheet Since v ^jump to upper side is tangential,

v^y ≈ −ˆy − v_I^xˆy_x ≈ −ˆy − xˆy_x on upper side and averaging yields

v_S^y ≈ −ˆy − 1

2xˆy_x on sheet (p.v.) ODE

yˆ_x = v_S^y − µˆy

v_S^x − µx ≈ −ˆy − 1

2xˆy_x − µˆy(1

2 − µ)x (1 − µ)xˆy_x = −(1 + µ)ˆy

Cusp exponent

µ + 1 µ − 1

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Numerics Use Birkhoff-Rott equation (⇔ Euler for smooth sheets):

Z = Z(Γ) , (1 − 2µ)ΓZ_Γ = V ^∗ − µZ , V =

Z dΓ^′

Z(Γ) − Z(Γ^′) Derivative: by finite differences

Approximate graph ˆy by points, interpolate by polynomial segments

→ can evaluate Biot-Savart integrals

• Linear interpolation: Convergence to cusp with µ/(µ − 1) exponent (2nd approx). Not very stable especially as µ ց 1.

• Quadratic/cubic/... interpolation: convergence to (µ + 1)/(µ − 1) exponent cusp (3rd approx). More robust if oscillation avoided.

• Linear/higher mixed:

V (Z) = 1 2πi

Z ω(Z^′)dx dy Z − Z^′

Enough to use quadratic/higher for same z segment and mirror image, for rest can use linear

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Velocity field (zoom in):

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µ = 1.3, φ_∞ = 10^◦ numerics:

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Pseudo-velocity q:

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Increase φ_∞ to 25^◦:

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Increase φ_∞ to 40^◦:

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Increase to 80^◦:

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Maximum cusp angle at infinity φ_∞ Numerics:

µ φ∞

Spirals

Cusp

2.5 3 1.5 2

0.5 1 90 80 70 60 50 40 30 20 10 0

• for µ ր ∞, max angle ր 90^◦

• for µ ց 1, max angle ց 0

• Cusp if µ > 1 and angle below max, otherwise: spirals

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µ = 1.3 with φ_∞ = 80^◦ solution:

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Rotate: solution for µ = 1.3 with φ_∞ = 10^◦, opposite circulation

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Concluding remarks

• “Most” vortex cusps appear to have asymptotics v^x = x + ... , y = xˆ ^α + ... , α = µ + 1

µ − 1 as x ց 0

• No cusps unless µ > 1 and sufficiently small angles-at-infinity and positive (ccw) circulation on the upper (second in ccw direction) sheet! (= inner v points away from cusp.)

Single Mach reflection does produce such circulation.

• For larger angles, µ ≤ 1, or negative circulation: numerics sug- gest non-cusp flows with algebraic spiral ends.

• µ = 1 key (compressible) but borderline: may need to superim- pose harmonic strain flow to modify exponent.

• Low-order numerics not necessarily stable, yield wrong exponent.

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