One-phase Solidification Problem:
FEM approach via Parabolic Variational
Inequalities
Ali Etaati
May 14th 2008
Content:
• One-phase Stefan problem-solidification example
• Derivation of a complementarity system
• Parabolic variational inequality
• Finite element method
One-Phase Stefan problem
 2
 (t )
1
 (t )
 (t )
0
u ( x, t )  0

 u ( x, t )  0
u ( x, t )  0

x   0 (t )
x  (t )
x    (t )
l ( x)  0,
x    (0)  (0),

l ( x)  0, l  0
x   0 (0)
  (t )  {x : l ( x)  t},
 0
 (t )  {x : l ( x)  t},
(t )  {x : l ( x)  t},

Solidification:
 2
 (t )

 (t )
 0 (t )
0  t1  t2
  (t1 )    (t2 )
1
T    (0, T )
(T: final time)
Solidification (cont’d):
Heat conduction equation (in the solid)

u
 u ,
t
x  T
Freezing temperature
u ( x, l ( x))  0
Energy balance (Stefan condition) at  (t )
solid
[u( x, l ( x))]liquid
.l  L
(L is the latent heat)
Solidification (cont’d):
Initial and Boundary conditions (temperature distribution):
u0 ( x),
u ( x,0)  
 0,
u   g ( x, t )  0,
x   0 (0)
u  0,
x   2
x    (0)
x  1
Initial enthalpy:
u0 ( x),
H 0 ( x)  
 L,
x   (0)
x  0 (0)
Freezing index:
t
  u ( x ,  ) d ,
l ( x )

U ( x, t )  
0
,
 t
  u ( x ,  ) d ,
 0
t  l ( x),
x   0 (0)
t  l ( x),
x   0 (0)
x    (0)
Then:
 U
 t  U  H 0 ( x),


U  0,
U  0, U  0,



x  T
x  T0
x  (t )
U ( x,0)  0,
t

~
U    g ( x, ) d ,

0
U  0,

x 
x  1
x   2
Linear complementarity system:
U

 U  H 0 ( x)  0,
t
U  0,
(
a.e. in T
U
 U  H 0 ( x))U  0,
t
Parabolic Variational Inequalities:
Let
k (t )  {v  H 01 () : v  G( x, t )
a.e.
K  {v  L2 (0, T , H 01 ()) : v  G
where
G  L (0, T , H ()),
2
and
G0
2
a.e.
on
on
a.e.
}
on
G
 L2 (T )
t
  (0, T )
2
Suppose: f  L (T )
Define:
(w, v)   wvdx,

a(w, v)   w.vdx

T }
Parabolic Variational Inequalities (cont’d)
Problem 1:
Find w  L2 (0, T ; H 01 ()) with w / t  L2 (0, T ; L2 ())
such that w( x,0)  w0  k (0)
and for almost all t  (0, T ), w(t )  k (t ) and is such that:
w
( , v  w)  a( w, v  w)  ( f , v  w)
t
2
1
for all v  L (0, T ; H 0 ()) with v(t )  k (t ) a.e. in (0,T).
Parabolic Variational Inequalities (cont’d)
Problem 2:
Find w K with w / t  L2 (0, T ; L2 ()) such that
w( x,0)  w0 ( x)  k (0)
and
w
0 {( t , v  w)  a(w, v  w)  ( f , v  w)}dt  0
T
for all v  K.
Parabolic Variational Inequalities (cont’d)
Equivalence:
Consider a solution of Problem 2, for any 
 w,
v  ~
v  K ,
 0 and   (0, T ),
t  (   ,   )
t  (   ,   )
then
 
w ~
~  w)  ( f , v~  w)}dt  0
{
(
,
v

w
)

a
(
w
,
v

  t
We obtain the solution for Problem 1.
• Clearly a solution of Problem 1 solves Problem 2.
Parabolic Variational Inequalities (cont’d)
Theorem
Solution to Problem 2 satisfies the linear complementarity
system:
w
 w  f  0,
t
w  G,
w
(
 w  f )( w  G )  0.
t
a.e. in T
Parabolic Variational Inequalities (cont’d)
Proof
For any non-negative   C0 (T ),
Problem 2:
v  w    K and so, from
w
w
[(


w

f
)

]
dxdt

{
(
 t
0 t , v  w)  a(w, v  w)  ( f , v  w)}dt  0
T
T
Which implies that
w
 w  f  0, a.e. in
t
T
Parabolic Variational Inequalities (cont’d)
Proof (cont’d)

Now let T  {( x, t )  T : w( x, t )  G( x, t )}. Then for any
  C0 (T ), v  w    K for sufficiently small  so that
T
0   {(
0
w
w
, v  w)  a( w, v  w)  ( f , v  w)}dt   [(  w  f ) ]dxdt.
t
t

T
Hence
w
 w  f  0, a.e. in T
t
Conversely, by noting that if w K satisfies the complementarity
system, then, for v  K
w
(  w  f )(v  w)  0,
a.e. in
T ,
t
It is then clear that w solves Problem 2.
Finite Element approximation (FEM)
General Discretisation by FEM for Problem 1:
find
w n 1  k hn 1 such that
(( wn 1  wn ) / k , v  wn 1 ) h  a( wn  , v  wn 1 )  ( f n  , v  wn 1 ) h
For all
v  khn1  {v   vi bi ( x) V h : vi  G( xi , (n  1)k )}
iI
w
n 
w
n 1
(for all interior points)
 (1   ) w ,
n
  [0,1]
V h : Space of continuous functions which are linear on each
element and which vanish on the boundaries.
bi (x) : a piecewise linear basis function.
Finite Element approximation (cont’d)
If f is continuous:
Otherwise:
w0  ( w0 ) I ,
f n  { f (nk )}I
(f
if
a( w0 , v)  a( w0 , v)
n 
1
, v) h 
k
( n 1) k
 ( f (t ), v)dt
h
(for all v  V )
nk
w0  H 2 () ,
h
(for all v  V )
In any case:
f h,k (t )  f
in
L2 (T ) ,
w0  w0
in
H 01 ()
Time marching of the discrete system
z
n 1
 0,
where
z
w
n 1
n 1
G
 M (w
n 1
n 1
M ij  (bi , b j ) h ,
 0,
(z
n 1 T
) (w
n 1
G
n 1
)0
(*)
n
n 
n 
 w ) / k  Aˆ w  f ,
Aˆ ij  a(bi , b j )
n
n
n
{w , f , G } are nodal vectors.
M  kAˆ : is a symmetric positive definite matrix which causes
the problem (*) to have unique solution.
FEM; Stability
Let
v  v~,
2
T
~
~
 u, v  u Mv ,
| v |  v, v ,
|| v ||2  v~ 2 Aˆ v~,
 k v n  (v n 1  v n ) / k
and let’s assume:
k (t * )  k (t )
for
t*  t
and
0  k (0)
Or equivalently that
G( x, t )  G( x, t * )
for
t*  t
and
G( x,0)  0.
The complementarity problems are equivalent to
~ n 1 T ~ ~ n 1
~
( Z ) (v  w )  0 for all v~  G n1
~w
~ n in (**), then
We may take v
(**)
FEM; Stability (cont’d)
Stability theorem:
Providing the stability condition
(1  2 )k[ S (h)]2  2
holds when   1/ 2 , there is a constant C, independent of spaceand time-steps such that:
n
n
j 2
n 1 2
0 2
k
|

w
|

||
w
||

C
{||
w
||  k | f
 k
j 0
j 0
j  2
|}
FEM; Stability (cont’d)
Lemma:
The explicit method
~ n ~ n 1 
~ n 1
n 1
n
~
~
w  G  {( I  kA) w  kf  G }
is stable under the following conditions
1  kAii  0;
Aij  0, i  j ;
A
ij
0
j
f ( x, t )  f ( x)  0,
G ' (t )  0,
G '' (t )  0
G( x, t )  G(t )  0
and
w( x,0)  G(0)  0
n
for which {w } is bounded by
~ n1 ~ n ~ n1 ~ n ~
G G  w  w  0
Reference
• Weak and variational methods for moving boundary problems,
C M Elliott & J R Ockendon.