Preserving Taylor’s constraint in
magnetohydrodynamics
Jitse Niesen (Leeds)
in collaboration with: Glenn Ierley (UC San Diego)
Andrew Jackson (ETH Zürich)
Phil Livermore (Leeds)
SciCADE 2013, Valladolid
Outline
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Problem: The Earth’s magnetic field
Model
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Numerics
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Navier–Stokes & Maxwell equations
Setting Ekman number to zero
Stucture: Taylor’s constraint
Spatial discretization: spectral
Projection method
Midpoint rule
Conclusions
Structure of the Earth
1,2 Crust, solid silicate rock,
electrically insulating
3,4 Mantle, solid/plastic silicate
rock, insulating
5 Outer core, molten metal
(iron/nickel), conducting
6 Inner core, solid metal,
conducting
(Picture from Wikipedia)
Origin of Earth’s magnetic field
Magnetic field is (mostly) generated in outer core. Roughly:
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Electric currents generate magnetic field (Ampère’s law).
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Changing magn. field produces electric field (Faraday’s law).
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Electric and magnetic field exert force on fluid particles
(Lorentz force).
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Moving charged particles are an electric current.
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Inner core is hotter than mantle, causing fluid movements.
Origin of Earth’s magnetic field
Magnetic field is (mostly) generated in outer core. Roughly:
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Electric currents generate magnetic field (Ampère’s law).
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Changing magn. field produces electric field (Faraday’s law).
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Electric and magnetic field exert force on fluid particles
(Lorentz force).
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Moving charged particles are an electric current.
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Inner core is hotter than mantle, causing fluid movements.
However, details of process is unknown. There is no simple model.
Magnetic field is largely a dipole, approximately aligned to rotation
axis. But sometimes the dipole reverses; most recently 780,000
years ago. Mechanism for this is a mystery.
Equations of MHD
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electromagnetism (Maxwell’s equation)
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hydrodynamics (Navier–Stokes equation)
Equations of MHD
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electromagnetism (Maxwell’s equation)
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hydrodynamics (Navier–Stokes equation)
∂u
+ u · ∇u = −∇p+µ∇2 u + force
ρ
∂t
Incompressibility: ∇ · u = 0.
where
u
ρ
p
µ
=
=
=
=
fluid velocity
density
pressure
viscosity
Equations of MHD
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electromagnetism (Maxwell’s equation)
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hydrodynamics (Navier–Stokes equation)
∂u
+ u · ∇u = −∇p+µ∇2 u−2Ωẑ×u
ρ
∂t
Incompressibility: ∇ · u = 0.
where
u
ρ
p
µ
=
=
=
=
fluid velocity
density
pressure
viscosity
Ω = angular velocity
ẑ = rotation axis
Equations of MHD
I
electromagnetism (Maxwell’s equation)
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hydrodynamics (Navier–Stokes equation)
∂u
+ u · ∇u = −∇p+µ∇2 u−2Ωẑ×u+g ρr̂
ρ
∂t
Incompressibility: ∇ · u = 0.
where
u
ρ
p
µ
=
=
=
=
fluid velocity
density
pressure
viscosity
Ω
ẑ
g
r̂
=
=
=
=
angular velocity
rotation axis
gravitational constant
radial vector
Equations of MHD
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electromagnetism (induction equation)
∂B
= ∇ × (u × B) + σ −1 ∇2 B,
∂t
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∇ · B = 0.
hydrodynamics (Navier–Stokes equation)
∂u
+ u · ∇u = −∇p+µ∇2 u−2Ωẑ×u+g ρr̂
ρ
∂t
Incompressibility: ∇ · u = 0.
where
u
ρ
p
µ
B
=
=
=
=
=
fluid velocity
density
pressure
viscosity
magnetic field
Ω
ẑ
g
r̂
σ
=
=
=
=
=
angular velocity
rotation axis
gravitational constant
radial vector
conductivity
Equations of MHD
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electromagnetism (induction equation)
∂B
= ∇ × (u × B) + σ −1 ∇2 B,
∂t
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∇ · B = 0.
hydrodynamics (Navier–Stokes equation)
∂u
+ u · ∇u = −∇p+µ∇2 u−2Ωẑ×u+g ρr̂ +σ(∇×B)×B
ρ
∂t
Incompressibility: ∇ · u = 0.
where
u
ρ
p
µ
B
=
=
=
=
=
fluid velocity
density
pressure
viscosity
magnetic field
Ω
ẑ
g
r̂
σ
=
=
=
=
=
angular velocity
rotation axis
gravitational constant
radial vector
conductivity
Equations of MHD
Ro
∂u
+ (u · ∇)u
∂t
= −∇p + E ∇2 u − ẑ × u
+ Ra T (r )r̂ + (∇ × B) × B.
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Equations of MHD
Ro
∂u
+ (u · ∇)u
∂t
= −∇p + E ∇2 u − ẑ × u
+ Ra T (r )r̂ + (∇ × B) × B.
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Dimensionless quantities:
Ro Rossby number — inertia versus rotation
Ra Modified Rayleigh number — buoyancy versus rotation
E Ekman number — E = Ω/νL2 , viscosity versus rotation
Ekman number, E
In Earth, Ekman number is extremely small (10−12 – 10−15 ).
This leads to extremely thin boundary layers.
Numerical experiments cannot handle such small Ekman numbers;
typically, people use big computers and take E around 10−6 .1
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Kono & Roberts (2002), Rev. Geophys. 40:1–53 (2002).
Ekman number, E
In Earth, Ekman number is extremely small (10−12 – 10−15 ).
This leads to extremely thin boundary layers.
Numerical experiments cannot handle such small Ekman numbers;
typically, people use big computers and take E around 10−6 .1
Our approach is to take E = 0.
1
Kono & Roberts (2002), Rev. Geophys. 40:1–53 (2002).
Equations of MHD
Ro
∂u
+ (u · ∇)u
∂t
= −∇p + E ∇2 u − ẑ × u
+ Ra T (r )r̂ + (∇ × B) × B.
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Taking E = 0 changes character of equation (second-order term
disappears). Do not impose no-slip condition, so no boundary layer.
Equations of MHD
Ro
∂u
+ (u · ∇)u
∂t
= −∇p + E ∇2 u − ẑ × u
+ Ra T (r )r̂ + (∇ × B) × B.
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Taking E = 0 changes character of equation (second-order term
disappears). Do not impose no-slip condition, so no boundary layer.
We also take Ro = 0 (in reality, Ro ≈ 10−9 ), killing the ∂u
∂t term.
This is called the magnetostrophic balance, and predicts long-term
evolution (and hopefully, reversals).
It is difficult to find stable numerical methods for this case.
J. B. Taylor’s constraint
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B,
∂B
= ∇ × (u × B) + ∇2 B,
∂t
2
J. B. Taylor (1963), Proc. Roy. Soc. A 9:274–283.
∇ · u = 0,
∇ · B = 0.
J. B. Taylor’s constraint
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B,
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Let C (s) be cylinder aligned with rotation axis with radius s.
Integrate azimuthal component of first equation over C (s).
2
J. B. Taylor (1963), Proc. Roy. Soc. A 9:274–283.
J. B. Taylor’s constraint
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B,
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Let C (s) be cylinder aligned with rotation axis with radius s.
Integrate azimuthal component of first equation over C (s).
Z
Z
Z
(ẑ × u)ϕ dS = u · dS = ∇ · u dx = 0,
Z
Z
(∇p)ϕ dS = (∇ × ∇p)ϕ dx = 0,
Z
Ra T (r )r̂ ϕ dS = 0,
2
J. B. Taylor (1963), Proc. Roy. Soc. A 9:274–283.
J. B. Taylor’s constraint
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B,
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
Let C (s) be cylinder aligned with rotation axis with radius s.
Integrate azimuthal component of first equation over C (s).
Z
Z
Z
(ẑ × u)ϕ dS = u · dS = ∇ · u dx = 0,
Z
Z
(∇p)ϕ dS = (∇ × ∇p)ϕ dx = 0,
Z
Ra T (r )r̂ ϕ dS = 0,
R
so (∇ × B) × B ϕ dS = 0. This is Taylor’s constraint.
It is an infinite family (depends on s) of quadratic constraints.2
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J. B. Taylor (1963), Proc. Roy. Soc. A 9:274–283.
J. B. Taylor’s constraint II
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B,
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
The equation
a solution if Taylor’s constraint is
R for u has only
satisfied:
(∇ × B) × B ϕ dS = 0.2
2
J. B. Taylor (1963), Proc. Roy. Soc. A 9:274–283.
J. B. Taylor’s constraint II
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B,
∂B
= ∇ × (u × B) + ∇2 B,
∂t
∇ · u = 0,
∇ · B = 0.
The equation
a solution if Taylor’s constraint is
R for u has only
satisfied:
(∇ × B) × B ϕ dS = 0.2
The equation for u determines flow up to terms of the form ug (s)ϕ̂
(geostrophic components). We can find ug by differentiating
Taylor’s constraint.
2
J. B. Taylor (1963), Proc. Roy. Soc. A 9:274–283.
Spatial discretization — spectral method
Decompose B in toroidal and poloidal parts:
B = ∇ × T (x)x + ∇ × ∇ × P(x)x .
This decomposition is unique and sum B is divergence-free.
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Livermore, Ierley & Jackson (2008), Proc. Roy. Soc. A 464:3149–3174.
Spatial discretization — spectral method
Decompose B in toroidal and poloidal parts:
B = ∇ × T (x)x + ∇ × ∇ × P(x)x .
This decomposition is unique and sum B is divergence-free.
Expand T and P in spherical harmonics Y`m (so that BCs are easy
(α,β)
to apply) and Jacobi polynomials Pn
in radial direction:
X
(−1/2,`+3/2)
P(x) =
a`mn r ` Pn
(2r 2 − 1)Y`m (θ, ϕ).
`,m,n
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Livermore, Ierley & Jackson (2008), Proc. Roy. Soc. A 464:3149–3174.
Spatial discretization — spectral method
Decompose B in toroidal and poloidal parts:
B = ∇ × T (x)x + ∇ × ∇ × P(x)x .
This decomposition is unique and sum B is divergence-free.
Expand T and P in spherical harmonics Y`m (so that BCs are easy
(α,β)
to apply) and Jacobi polynomials Pn
in radial direction:
X
(−1/2,`+3/2)
P(x) =
a`mn r ` Pn
(2r 2 − 1)Y`m (θ, ϕ).
`,m,n
Truncate
to a certain degree.
R the expansion up
Then C (s) (∇ × B) × B ϕ dS (lhs of constraint) becomes a
polynomial in s. So infinite family reduces to a finite one.3
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Livermore, Ierley & Jackson (2008), Proc. Roy. Soc. A 464:3149–3174.
Projection method
0
1. Evolve magnetic field: Bn 7→ Bn+1
0
2. Linearize constraints around Bn and project: Bn+1
7→ Bn+1
3. Compute velocity implied by field: un+1
Step 3: Split u(x) = um (x) + ug (s)ϕ̂.
Determine um exactly from “Navier–Stokes equation”
ẑ × u = −∇p + Ra T (r )r̂ + (∇ × B) × B
(but degree of um twice as big as B).
Compute ug by solving a two-point BVP which follows by
differentiating Taylor’s constraint. This is hard: need to
convert between spherical and cylindrical coordinates;
endpoints are regular singular points.
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Livermore, Ierley & Jackson (2012), unpublished.
Projection method
0
1. Evolve magnetic field: Bn 7→ Bn+1
0
2. Linearize constraints around Bn and project: Bn+1
7→ Bn+1
3. Compute velocity implied by field: un+1
Linearized projection method preserves constraint up 10−8 .4
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Livermore, Ierley & Jackson (2012), unpublished.
Implicit midpoint rule
Consider u as a function of B, computed as in projection method.
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Solve induction equation ∂B
∂t = ∇ × (u × B) + ∇ B using the
implicit midpoint rule:
Bn+1 = Bn + 12 h ∇×(un ×Bn )+∇2 Bn ∇×(un+1 ×Bn+1 )+∇2 Bn+1 .
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Taylor’s constraint C (s) (∇ × B) × B ϕ dS = 0 is quadratic and
thus satisfied exactly by midpoint rule.
Implicit midpoint rule
Consider u as a function of B, computed as in projection method.
2
Solve induction equation ∂B
∂t = ∇ × (u × B) + ∇ B using the
implicit midpoint rule:
Bn+1 = Bn + 12 h ∇×(un ×Bn )+∇2 Bn ∇×(un+1 ×Bn+1 )+∇2 Bn+1 .
R
Taylor’s constraint C (s) (∇ × B) × B ϕ dS = 0 is quadratic and
thus satisfied exactly by midpoint rule.
Implicit methods are expensive.
Alternative is to use Störmer–Verlet: if we as a splitting
B
Taylor’s constraint becomes
R = B1 + B2 such that
(∇
×
B
)
×
B
1
2 ϕ dS = 0, then it is still automatically
C (s)
satisfied (work in progress).
Conclusions / further work
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Mathematically, solving MHD equations with E = 0 is an
interesting problem.
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“SciCADE community” has something to contribute, but
should not underestimate MHD community.
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DAE theory is probably useful. Use derivative of equation
for u instead of derivative of Taylor’s constraint?
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If we can do E = 0, we can think about small-E expansions.
(I don’t have background to say whether E = 0 by itself is
scientifically interesting).
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Current implementation is very much a prototype.
(Based on Maple, can only handle low degree).
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Very inconvenient that Earth is round.