MHD Equations


The magnetohydrodynamics (MHD) equations are:

(1)\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho {\bf v}) = 0

(2)\rho\left(\frac{\partial {\bf v}}{\partial t} + ({\bf v} \cdot \nabla)
 {\bf v} \right) = -\nabla p +
 {1\over\mu}(\nabla\times{\bf B}) \times {\bf B} + \rho {\bf g}

(3){\partial {\bf B}\over\partial t}
        = \nabla\times({\bf v}\times{\bf B}) + \eta\nabla^2{\bf B}

(4)\nabla\cdot{\bf B} = 0

assuming \eta is constant. See the next section for a derivation. We can now apply the following identities (we use the fact that \nabla\cdot{\bf

\left[(\nabla\times{\bf B}) \times {\bf B}\right]_i =
    \varepsilon_{ijk}(\nabla\times{\bf B})_j B_k =
    \varepsilon_{ijk}\varepsilon_{jlm}(\partial_l B_m)B_k =
    (\delta_{kl}\delta_{im}-\delta_{km}\delta_{il})(\partial_l B_m)B_k =

=(\partial_k B_i)B_k - (\partial_i B_k)B_k
    =\left[({\bf B}\cdot\nabla){\bf B} -
    {1\over2}\nabla|{\bf B}|^2\right]_i

(\nabla\times{\bf B}) \times {\bf B} &=
    ({\bf B}\cdot\nabla){\bf B} - {1\over2}\nabla|{\bf B}|^2=
    ({\bf B}\cdot\nabla){\bf B} + {\bf B}(\nabla\cdot{\bf B})
        - {1\over2}\nabla|{\bf B}|^2
    =\nabla\cdot({\bf B}{\bf B}^T) - {1\over2}\nabla|{\bf B}|^2\\
\nabla\times({\bf v} \times {\bf B}) &=
    ({\bf B}\cdot\nabla){\bf v} - {\bf B}(\nabla\cdot{\bf v})
    +{\bf v}(\nabla\cdot {\bf B}) - ({\bf v}\cdot\nabla) {\bf B}
    \nabla\cdot({\bf B}{\bf v}^T - {\bf v}{\bf B}^T)\\
\nabla\cdot(\rho{\bf v}{\bf v}^T) &=
    \left(\nabla\cdot(\rho{\bf v})\right){\bf v}
    + \rho({\bf v}\cdot\nabla){\bf v}=
    -{\bf v}\frac{\partial \rho}{\partial t}
    + \rho({\bf v}\cdot\nabla){\bf v}

So the MHD equations can alternatively be written as:

(5)\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho {\bf v}) = 0

(6)\frac{\partial \rho{\bf v}}{\partial t} + \nabla\cdot(\rho{\bf v}{\bf v}^T)
    = -\nabla p +
    {1\over\mu}\left(\nabla\cdot({\bf B}{\bf B}^T)
        - {1\over2}\nabla|{\bf B}|^2\right) + \rho {\bf g}

(7){\partial {\bf B}\over\partial t}
        = \nabla\cdot({\bf B}{\bf v}^T - {\bf v}{\bf B}^T) + \eta\nabla^2{\bf B}

(8)\nabla\cdot{\bf B} = 0

One can also introduce a new variable p^* = p + {1\over2}\nabla|{\bf B}|^2, that simplifies (6) a bit.


The above equations can easily be derived. We have the continuity equation:

\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho {\bf v}) = 0

Navier-Stokes equations (momentum equation) with the Lorentz force on the right-hand side:

\rho\left(\frac{\partial {\bf v}}{\partial t} + ({\bf v} \cdot \nabla)
 {\bf v} \right) = -\nabla p + {\bf j} \times {\bf B} + \rho {\bf g}

where the current density {\bf j} is given by the Maxwell equation (we neglect the displacement current {\partial{\bf E}\over\partial t}):

{\bf j} = {1\over\mu}\nabla\times{\bf B}

and the Lorentz force:

{1\over\sigma}{\bf j} = {\bf E} + {\bf v}\times{\bf B}

from which we eliminate {\bf E}:

{\bf E} = - {\bf v}\times{\bf B} + {1\over\sigma}{\bf j} =
          - {\bf v}\times{\bf B} + {1\over\sigma\mu}\nabla\times{\bf B}

and put it into the Maxwell equation:

{\partial {\bf B}\over\partial t} = -\nabla\times{\bf E}

so we get:

{\partial {\bf B}\over\partial t} = \nabla\times({\bf v}\times{\bf B})
            - \nabla\times\left({1\over\sigma\mu}\nabla\times{\bf B}\right)

assuming the magnetic diffusivity \eta={1\over\sigma\mu} is constant, we get:

{\partial {\bf B}\over\partial t} = \nabla\times({\bf v}\times{\bf B})
            - \eta\nabla\times\left(\nabla\times{\bf B}\right)
        = \nabla\times({\bf v}\times{\bf B})
            + \eta\left(\nabla^2{\bf B}-\nabla(\nabla\cdot{\bf B})\right)
        = \nabla\times({\bf v}\times{\bf B}) + \eta\nabla^2{\bf B}

where we used the Maxwell equation:

\nabla\cdot{\bf B} = 0

Finite Element Formulation

We solve the following ideal MHD equations (we use p^* = p + {1\over2}\nabla|{\bf B}|^2, but we drop the star):

(9)\frac{\partial {\bf u}}{\partial t} + ({\bf u} \cdot \nabla)
 {\bf u} - ({\bf B}\cdot\nabla){\bf B} + \nabla p = 0

(10){\partial {\bf B}\over\partial t} + ({\bf u}\cdot\nabla){\bf B}
    - ({\bf B}\cdot\nabla){\bf u} = 0

(11)\nabla\cdot{\bf u} = 0

(12)\nabla\cdot{\bf B} = 0

If the equation (12) is satisfied initially, then it is satisfied all the time, as can be easily proved by applying a divergence to the Maxwell equation {\partial {\bf B}\over\partial t} = -\nabla\times{\bf E} (or the equation (10), resp. (3)) and we get {\partial \over\partial t}(\nabla\cdot{\bf B}) = 0, so \nabla\cdot{\bf B} is constant, independent of time. As a consequence, we are essentially only solving equations (9), (10) and (11), which consist of 5 equations for 5 unknowns (components of {\bf u}, p and {\bf B}).

We discretize in time by introducing a small time step \tau and we also linearize the convective terms:

(13)\frac{{\bf u}^n-{\bf u}^{n-1}}{\tau} + ({\bf u}^{n-1} \cdot \nabla)
 {\bf u}^n - ({\bf B}^{n-1}\cdot\nabla){\bf B}^n + \nabla p = 0

(14){{\bf B}^n-{\bf B}^{n-1}\over\tau} + ({\bf u}^{n-1}\cdot\nabla){\bf B}^n
    - ({\bf B}^{n-1}\cdot\nabla){\bf u}^n = 0

(15)\nabla\cdot{\bf u}^n = 0

Testing (13) by the test functions (v_1, v_2), (14) by the functions (C_1, C_2) and (15) by the test function q, we obtain the following weak formulation:

(16)\int_\Omega {u_1 v_1\over\tau} + ({\bf u}^{n-1}\cdot\nabla)u_1 v_1
    - ({\bf B}^{n-1}\cdot\nabla)B_1 v_1
    -p {\partial v_1\over\partial x} \,{\rm d}{\bf x} =
    \int_\Omega {u_1^{n-1} v_1\over\tau}\,{\rm d}{\bf x}

\int_\Omega {u_2 v_2\over\tau} + ({\bf u}^{n-1}\cdot\nabla)u_2 v_2
    - ({\bf B}^{n-1}\cdot\nabla)B_2 v_2
    -p {\partial v_2\over\partial y} \,{\rm d}{\bf x} =
    \int_\Omega {u_2^{n-1} v_2\over\tau}\,{\rm d}{\bf x}

(17)\int_\Omega {B_1 C_1\over\tau} + ({\bf u}^{n-1}\cdot\nabla)B_1 C_1
    - ({\bf B}^{n-1}\cdot\nabla)u_1 C_1 \,{\rm d}{\bf x} =
    \int_\Omega {B_1^{n-1} C_1\over\tau}\,{\rm d}{\bf x}

\int_\Omega {B_2 C_2\over\tau} + ({\bf u}^{n-1}\cdot\nabla)B_2 C_2
    - ({\bf B}^{n-1}\cdot\nabla)u_2 C_2 \,{\rm d}{\bf x} =
    \int_\Omega {B_2^{n-1} C_2\over\tau}\,{\rm d}{\bf x}

(18)\int_\Omega {\partial u_1\over\partial x}q + {\partial u_2\over\partial y}q
    \,{\rm d}{\bf x} = 0

To better understand the structure of these equations, we write it using bilinear and linear forms, as well as take into account the symmetries of the forms. Then we get a particularly simple structure:

+A(u_1, v_1) &&  && -X(p, v_1) && -B(B_1, v_1) &&  &=& l_1(v_1)\\
  && +A(u_2, v_2) && -Y(p, v_2) &&  && -B(B_2, v_2) &=& l_2(v_2)\\
+X(q, u_1) && +Y(q, u_2) &&  &&  &&  &=& 0\\
-B(u_1, C_1) &&  &&  && +A(B_1, C_1) &&  &=& l_4(C_1)\\
 && -B(u_2, C_2) &&  &&  && +A(B_2, C_2) &=& l_5(C_2)


A(u, v) &= \int_\Omega {u v\over\tau} +
    ({\bf u}^{n-1}\cdot\nabla)u v\,{\rm d}{\bf x}\\
B(u, v) &= \int_\Omega ({\bf B}^{n-1}\cdot\nabla)uv\,{\rm d}{\bf x}\\
X(u, v) &= \int_\Omega u {\partial v\over\partial x} \,{\rm d}{\bf x}\\
Y(u, v) &= \int_\Omega u {\partial v\over\partial y} \,{\rm d}{\bf x}\\
l_1(v) &= \int_\Omega {u_1^{n-1} v\over\tau} \,{\rm d}{\bf x}\\
l_2(v) &= \int_\Omega {u_2^{n-1} v\over\tau} \,{\rm d}{\bf x}\\
l_4(v) &= \int_\Omega {B_1^{n-1} v\over\tau} \,{\rm d}{\bf x}\\
l_5(v) &= \int_\Omega {B_2^{n-1} v\over\tau} \,{\rm d}{\bf x}

E.g. there are only 4 distinct bilinear forms. Schematically we can visualize the structure by:

A   -X -B  
  A -Y   -B
X Y      
-B     A  
  -B     A

In order to solve it with Hermes, we first need to write it in the block form:

a_{11}(u_1, v_1) &+& a_{12}(u_2, v_1) &+& a_{13}(p, v_1) &+&
    a_{14}(B_1, v_1) &+& a_{15}(B_2, v_1) &=& l_1(v_1)\\
a_{21}(u_1, v_2) &+& a_{22}(u_2, v_2) &+& a_{23}(p, v_2) &+&
    a_{24}(B_1, v_2) &+& a_{25}(B_2, v_2) &=& l_2(v_2)\\
a_{31}(u_1, q) &+& a_{32}(u_2, q) &+& a_{33}(p, q) &+&
    a_{34}(B_1, q) &+& a_{35}(B_2, q) &=& l_3(q)\\
a_{41}(u_1, C_1) &+& a_{42}(u_2, C_1) &+& a_{43}(p, C_1) &+&
    a_{44}(B_1, C_1) &+& a_{45}(B_2, C_1) &=& l_4(C_1)\\
a_{51}(u_1, C_2) &+& a_{52}(u_2, C_2) &+& a_{53}(p, C_2) &+&
    a_{54}(B_1, C_2) &+& a_{55}(B_2, C_2) &=& l_5(C_2)

comparing to the above, we get the following nonzero forms:

a_{11}(u_1, v_1) &+& 0 &+& a_{13}(p, v_1) &+&
    a_{14}(B_1, v_1) &+& 0 &=& l_1(v_1)\\
0 &+& a_{22}(u_2, v_2) &+& a_{23}(p, v_2) &+&
    0 &+& a_{25}(B_2, v_2) &=& l_2(v_2)\\
a_{31}(u_1, q) &+& a_{32}(u_2, q) &+& 0 &+&
    0 &+& 0 &=& 0\\
a_{41}(u_1, C_1) &+& 0 &+& 0 &+&
    a_{44}(B_1, C_1) &+& 0 &=& l_4(C_1)\\
0 &+& a_{52}(u_2, C_2) &+& 0 &+&
    0 &+& a_{55}(B_2, C_2) &=& l_5(C_2)


a_{11}(u_1, v_1) &= A(u_1, v_1)\\
a_{22}(u_2, v_2) &= A(u_2, v_2)\\
a_{44}(B_1, C_1) &= A(B_1, C_1)\\
a_{55}(B_2, C_1) &= A(B_2, C_2)\\
a_{13}(p, v_1) &= -X(p, v_1)\\
a_{31}(u_1, q) &= X(q, u_1)\\
a_{23}(p, v_2) &= -Y(p, v_2)\\
a_{32}(u_2, q) &= Y(q, u_2)\\
a_{14}(B_1, v_1) &= -B(B_1, v_1)\\
a_{41}(u_1, C_1) &= -B(u_1, C_1)\\
a_{25}(B_2, v_2) &= -B(B_2, v_2)\\
a_{52}(u_2, C_2) &= -B(u_2, C_2)

and l1, ..., l5 are the same as above.

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