Grad-Shafranov Equation for 2D MHS Equilibria

Coronal arcades and prominences are examples of two dimensional equilibria with variations in the $y$ direction ignored, $\partial /\partial y = 0$. Expressing the magnetic field in terms of the vector magnetic potential we have,

$${\bf B} = \nabla \times {\bf A},$$ (3.48)

where ${\bf A}$ is written as

$${\bf A} = \left ( A_{x}(x,z), A(x,z), A_{z}(x,z) \right ).$$

Thus, the magnetic field is expressed in general as

$${\bf B} = \left (-{\partial A \over \partial z}, B_{y}, {\partial A \over \partial x}\right ),$$ (3.49)

and

$$B_{y}(x,z) = {\partial A_{x}\over \partial z} - {\partial A_{z}\over \partial x}.$$ (3.50)

Notice that $\nabla \cdot {\bf B} = 0$ is satisfied immediately. The current is

$${\bf j} = {1\over \mu}\nabla \times {\bf B} = {1\over \mu} ... ...z}, - \nabla^{2}A, {\partial B_{y}\over \partial x}\right ),$$ (3.51)

where

$$\nabla^{2}A = {\partial^{2}A \over \partial x^{2}} + {\partial^{2}A \over \partial z^{2}}.$$

If the magnetic field is force-free then $\nabla \times {\bf B} = \alpha {\bf B}$ and so we have the three equations

$$-{\partial B_{y}\over \partial z} = - \alpha {\partial A \over \partial z},$$ (3.52)

$$-\nabla^{2} A = \alpha B_{y},$$ (3.53)

$${\partial B_{y}\over \partial x} = \alpha {\partial A \over \partial x}.$$ (3.54)

Combining (3.53) and (3.55) to eliminate $\alpha$ we obtain

$${\partial A\over \partial z}{\partial B_{y}\over \partial x... ...\partial x}{\partial B_{y}\over \partial z} = J(B_{y}, A) = 0,$$ (3.55)

where $J$ is the Jacobian of $B_{y}$ and $A$. When the Jacobian vanishes identically the most general solution is

$$B_{y} = B_{y}(A),$$ (3.56)

so that $B_{y}$ is an arbitrary function of $A$. Thus,

$${\partial B_{y}\over \partial x} = {dB_{y}\over dA}{\partial A \over \partial x},$$

and so

$$\alpha = {dB_{y}\over dA}.$$ (3.57)

Thus, (3.54) becomes a nonlinear equation for $A$, namely

$$- \nabla^{2}A = B_{y}{dB_{y}\over dA} = {d\over dA}\left ({1\over 2}B_{y}^{2}(A) \right )$$ (3.58)

The Grad-Shafranov equation for two dimensional magnetohydrostatic equilibria is then

$$\nabla^{2}A + {d\over dA}\left ({1\over 2}B_{y}^{2}(A) \right ) = 0,$$ (3.59)

and

$${\bf B} = \left ( -{\partial A\over \partial z}, B_{y}(A), {\partial A \over \partial x} \right ).$$ (3.60)

Example 3.7.1   Consider the particular form of $B_{y}(A)$ given by

$$B_{y} = \lambda e^{-A}.$$

Thus, the Grad-Shafranov equation becomes

$$\nabla^{2}A = \lambda^{2}e^{-2A},$$

which is a non-linear, elliptic, partial differential equation for $A(x,z)$. Assume the field lines are circular so that $A= A(r)$ and $r^{2} = x^{2} + (z - z_{0})^{2}$, $x = r\cos \theta$, $z - z_{0} = r \sin \theta$. Changing to the polar coordinates we find

$$\nabla^{2}A = {1\over r}{d\over dr}\left ( r{dA\over dr}\right ),$$

since we are assuming that there is no $\theta$ dependence in $A$. Thus, the equation we need to solve is

$${1\over r}{d\over dr}\left ( r{dA\over dr}\right ) = \lambda^{2}e^{-2A}.$$

Now we consider the case where the vertical component of the magnetic field is specified, in theory from photospheric magnetograms, as

$$B_{z} = {2x\over 1+x^{2}},$$

which translates into a boundary condition for $A$ by integrating with respect to $x$ to get

$$A(x,0) = \log (1 + x^{2}).$$

Therefore, we try a solution of the form

$$A(x,z) = \log \left ( b^{2} + x^{2} + (z - z_{0})^{2}\right ).$$

Substituting into the Grad-Shafranov equation we find that we can have a solution provided

$$b^{2} = {\lambda^{2}\over 4}.$$

Then $A(x,0)$ implies that

$$b^{2} + z_{0}^{2} = 1, \qquad \Rightarrow z_{0}^{2} = 1 - {\lambda^{2}\over 4}.$$

Hence,

$$z_{0} = \pm \sqrt{1 - {\lambda^{2}\over 4}}$$

From this result we see that there are two possible solution to the Grad-Shafranov equation that have the same boundary condition, provided $\lambda < 2$. There are no possible solutions for $\lambda > 2$. This is a feature of non-linear equations