7.5 Two-dimensional modelings and simulations Ë 227
the formulation may be yielded as
T
X
0
X
ΩHTSC
휌|∇×h|^2 =
T
X
0
X
ΩHTSC
휌J^2 c(|h+He|). (7.38)
Correlating the above four functions (7.34), (7.36), (7.37), and (7.38), we obtain
휇 0
T
X
0
X
R^3
휕(h+He)
휕t
⋅ (휑−h)= −
T
X
0
X
ΩHTSC
휌∇×h⋅ ∇×(휑−h)
⩾
T
X
0
X
ΩHTSC
휌(J^2 c(|h+He|)−|∇×휑||∇×h|)⩾ 0. (7.39)
This proves thathis a solution of the problem
Find functionh∈K(h)such that휕(h+He)
휕t
,(휑−h)⩾ 0
for any휑∈K(h),h儨儨儨儨t= 0 =h 0 , (7.40)
where(u,w)=
T
X
0
X
R^3
u⋅wis the scalar product of two vector functions, withh 0 =
B 0 /휇 0 −He|t= 0 denoting the initial condition of the problem.
In the two-dimensional case, the magnetic field has only two components and it
can be expressed as
H=Hxx̂+Hyŷ. (7.41)
The current flowing in the HTSC has only thezcomponent, which is derived as
J=
( 0
0
8
x̂ ŷ ẑ
휕
휕x
휕
휕y
휕
휕z
Hx Hy Hz
) 1
1
9
=¤
휕Hy
휕x
−
휕Hx
휕y
¥ẑ. (7.42)
Since there is no imposed current in the HTSC, the net current induced by the variation
of the external magnetic field must satisfy
X
ΩHTSC
J=Itotal(t)= 0. (7.43)