Dresner, Stability of superconductors.2002
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210 |
Appendix B |
S(c)ct = ∂/∂z(∂L/∂cz) – ∂L/∂c |
(B.4.1) |
where
(B.4.2)
The action integral of this Lagrangian
(B.4.3)
is a nonincreasing function of time (Dresner, 1982):
(B.4.4)
Here we have integrated the first term by parts; the integrated term vanishes since cz(±∞) = 0. The stationary values of A correspond to the steady states. For all
nonsteady states, A decreases monotonically with time.
When there are just two steady states, the one with the larger action is always unstable. Let the two steady states be c1(z) and c2(z) and let A1 > A2. Consider the one-parameter family of initial conditions
c(z,0) = ac1 + (1 – a)c 2, 0
≥
a
≥
1
(B.4.5)
Sketched in Fig. B.2 is the initial action as a function of a, which is shown as a continuous curve. This curve intersects the horizontal line corresponding to the intermediate value A* of the action in point P. As time goes on, all points on this curve move down except the fixed end points A1 and A2 that correspond, respectively, to the steady states c1 and c2. The point P can therefore only move to the
right. Since P is bounded on the right by the line a = 1, it must approach a limiting value a* ≥ 1.
Stability of the MPZ |
211 |
Figure B.2. A sketch of the initial action as a function of a when there are two steady states c1 and c2.
If a* < 1, the action of the solution corresponding to the initial condition B.4.5 having a = a* approaches A* as t → ∞ . But then this solution approaches a steady state that is neither c1 nor c2, contrary to the hypothesis that c1 and c2 are the only steady states. Therefore, a* = 1, which means that eventually the point P lies in every neighborhood of a = 1. Thus there are initial conditions arbitrarily close to c1 whose solutions do not approach c1. Thus c1 is unstable, as was to be proved.
B.5. ACTION INTEGRAL OF THE MPZ
It remains to prove that the action of the steady state c = 0 is less than that of the MPZ. Now, if we examine Fig. 4.8 we see that Q(c) must have the general form shown in Fig. B.3. Using the same argument that led to Eq. (4.6.2) we find that
(B.5.1)
Thus when c < cmax,
(B.5.2)
so that LMPZ > 0. Thus AMPZ > 0. On the other hand, A(c = 0) = 0.
212 |
Appendix B |
Figure B.3. The general form of Q(c).
B.6. STABILITY OF THE STEADY STATES OF AN UNCOOLED SEGMENT OF A SUPERCONDUCTOR
The problem of Sections 10.1–10.2 can also be recast into the form (B.1.2), now with the boundary conditions c(±a,t) = 0 and the stipulation Q(c) > 0. The Lagrangian is again given by Eq. (B.4.2), but the action is now
(B.6.1)
The proof that dA/dt ≥ 0 goes through as before since ct(±a,t) = 0. In Section 10.2, it was found that there are two steady states (when steady states exist at all), and to study their stability we must determine the value of the action A of each. In a manner similar to that of Eq. (B.4.4) we prove that dA/da < 0 for steady states:
= –cz2 (a)/2 < 0 |
(B.6.2) |
since |
|
czz + Q(c) = 0 |
(B.6.3) |
for a steady state and ca(a) = –cz(a) (cf. Fig. B.4). |
|
Stability of the MPZ |
213 |
Figure B.4. Geometric relations near the foot of the solution showing that ca(a) = –cz(a).
If we multiply Eq. (B.6.3) by cz and integrate from z = 0 to z = a, we find
(B.6.4)
Thus A decreases with increasing a and does so faster on the upper branch (larger cmax [or larger Tmax in the parlance of Sections 10.1–10.2]) than on the lower branch. If we go backwards from the single state of largest a (cf. Fig. 10.1), the action A increases on both branches but it increases faster on the upper branch than the lower. Thus for a given a, A is larger on the upper branch than on the lower. From the results at the end of Section B .4, we see that the upper state is unstable and the lower state stable.
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