An example

In this section we present an example of the discussions above. We will work in normalized units, where the magnetic field has been normalized by some typical magnetic field strength B0, and the pressures

(a) (b) (c)

Figure 4.7: Two distinct branches join together as they pass through a critical point, Ac: (a) A < Ac,

(b)A=Ac and (c) A > Ac.

pressure function of the form

Pk=eA B

B+ 1. (4.7)

If we choose the free function to beF =A2_{(an arbitrary choice for illustrative purposes), the shear field}

follows from (2.36) and is found to be

By=

A2_{(B+ 1)}2

B

B3_{+ 2B}2_{+B−exp (A)}. (4.8)

One finds that this shear field profile has a contradictory region which varies with the value ofA. The points which define this region where the shear field becomes larger than the total magnetic field strength are given by finding the positive solutions to|By|=B, which in this case is

(B+ 1)2 A2±B=±eA. (4.9)

Since the above equation is cubic, it is possible to solve in generality to find the B(A) and thus the contradictory region, however the general form is extremely lengthy and does not grant much insight into the problem. Suffice to say that (4.9) has either one or two positive solutions forB depending on the value of A. If there is only one solution, then the contradictory region spans from B = 0 to that solution. If there are two, then the contradictory regions spans between those two values. It is possible to find the value of A at which the change from one to two contradictory regions occurs by taking the limit of (4.9) asB →0. This will be a critical point as described in Section 4.3.2, since at this point a new branch will appear due to the existence of a second valid region in which we can have values for the

shear field. Thus, by taking the limit we find that the the critical point,A=Ac must satisfy

Ac2=eAc. (4.10)

The positive sign is taken in the above equation, as it is necessary for a real solution. Solving (4.10) for the critical value givesAc≈ −0.7.

There remains one point we have not considered, and that is when F= 0. This is a common critical point, since it forces the shear field into a structure reminiscent of a pitchfork bifurcation. To see this more clearly, consider the shear field constraint (2.36). When F = 0, the right hand side vanishes, and we are left with (in normalized form)

By 1− 1 B ∂Pk ∂B = 0. (4.11)

This has two possible solutions. The first is the trivial By = 0 which, in this analogy, generates the

handle and middle prong of the pitchfork. The other solution to (4.11) occurs when the bracketed quantity becomes zero, suppose this happens atB=Bc. The crucial idea here is that there is a constant

value of the total magnetic field strength that satisfies (4.9). If the total magnetic field strength is constant atB=Bc then, from the analysis of the grid lines in Section 4.3.1, we must have semi-circles

in theBy-Bpplane of radiusBc. These are the outer prongs of the pitchfork, if you will. In our example

we have 1− 1 B ∂Pk ∂B = B3+ 2B2+B−exp (A) (B+ 1)2B . (4.12)

Recall that we must be at the point F = 0. In general this will happen for some value of A which we will callA0, and in our specific example A0= 0. Thus we substitute this into (4.12) and set the result

equal to zero to find the radius of the semicircle, i.e. we wish to solve

B3 c+ 2Bc2+Bc−1 (Bc+ 1) 2 Bc = 0. (4.13)

This can be solved analytically, but again for simplicity we will give the result approximately to be

Bc ≈0.466.

Now consider the quantity S (see (4.6)), which will determine the points at which the shear field becomes multi-valued. By substituting in the form of the shear field (4.8),S becomes

S=−A

2_{(B+ 1) exp (A)(3B+ 1)}

(B3_{+ 2B}2_{+B−exp (A))}2 −

B3+ 2B2+B−exp (A)

A2_{(B+ 1)}2 . (4.14)

critical points, and checkSfor changes in sign in order to characterise the shear field for all values of A

andBp.

For A < Ac the contradictory region spans fromB = 0 to some positive valueB =Bc, after which

the shear field is valid for all further values of B. Thus we should expect only a single branch of the shear field forA < Ac. We then check S forA < Ac and find that it does not change sign, hence the

branch is only single-valued. Indeed that is seen when we graph the shear field in Figure 4.8a.

For A > Ac, butA 6=A0, the contradictory region spans between two positive values of B. Thus,

for these values ofA we should expect precisely two branches. Again, we also check for sign changes in

S. One finds that for these values ofAprior to the contradictory region there is a sign change inS, but after that region there are no sign changes inS. Therefore we deduce that, forA > Ac, butA6=A0, one

of the branches bends back on itself, and the other branch does not. We see this behaviour in Figures 4.8b and 4.8c.

For the singular case whereA=Ac, there is no contradictory region, and the shear field takes on the

pitchfork like structure noted above. This is shown in Figure 4.8d. We also show a surface plot of the shear field in Figure 4.9 to give another impression of the shear field structure as it varies withA and

Bp. For clarity it has been split into two parts, showing the values of By for A >0 and A <0 on the

left and right respectively.