Directory UMM :Data Elmu:jurnal:A:Advances In Water Resources:Vol21.Issue8.1998:
Advances in Water Resources Vol. 21, No. 8, pp. 713–721, 1998
q 1998 Elsevier Science Ltd
Printed in Great Britain. All rights reserved
PII: S 0 3 0 9 - 1 7 0 8 ( 9 7 ) 0 0 0 2 6 - 2
0309-1708/98/$19.00 + 0.00
Explicit solution of a free boundary problem for
a nonlinear absorption model of mixed
saturated–unsaturated flow
Adriana C. Briozzo & Domingo A. Tarzia*
Departamento de Matema´tica, F.C.E., Universidad Austral, Paraguay 1950, (2000) Rosario, Argentina
(Received 11 November 1996; revised 23 March 1997; accepted 13 June 1997)
In wet soils, zones of saturation naturally develop in the vicinity of impermeable
strata, surface ponds and subterranean cavities. Hydrology must be then concerned
with transient flow through coexisting unsaturated and saturated zones. The models of
advancing saturated zones necessarily involve a nonlinear free boundary problem.
A closed-form analytic solution is presented for a nonlinear diffusion model under
conditions of ponding at the surface. The soil water diffusivity is restricted to the
special functional form D(v) ¼ a=(b ¹ v)2 , where v is the water content field to be
determined and a, b are positive constants. The explicit solution depends on a
parameter C (determined by the data of the problem), according to two cases:
1 , C , C1 or C $ C1 , where C1 is a constant which is obtained as the unique
solution of an equation. This result complements the study given in P. Broadbridge,
Water Resources Research, 1990, 26, 2435–2443, in order to established when the
explicit solution is available. The behavior of the bifurcation parameter C1 as a
function of the driving potential is studied with the corresponding limits for small and
large values. Moreover, the sorptivity is proven to be continuously differentiable
function of the variable C. q 1998 Elsevier Science Limited. All rights reserved
Key words: free boundary problem, mixed saturated–unsaturated flow, nonlinear
absorption model.
1 INTRODUCTION
In the saturated zone we have1
Following refs 1,6, we consider a homogeneous soil which
initially has some uniform volumetric water content vn . At
times t . 0, water is supplied at the surface x ¼ 0 under
pressure head W0 . Then, a mixed saturated–unsaturated
flow problem representing absorption of water by a soil
with a constant pond depth at the surface is presented. At
every time t the zone of saturation extends from x ¼ 0 to x ¼
s(t) (the free boundary), and the unsaturated zone extends
for x . s(t). By assuming the Darcy’s law and neglecting the
gravity, the water flux is given by
W(x, t) ¼ W0 ¹
W0 ¹ Ws
x; 0 , x , s(t);
s(t)
(2)
and, the following free boundary problem eqns (3)–(7)
arises for the unsaturated zone7:
v(s(t) þ , t) ¼ vs , t . 0,
(3)
]v
]
]v
¼
D(v)
, x . s(t), t . 0,
]t ]x
]x
(4)
]W
,
(1)
]x
where W is the soil water matric potential and K is the
hydraulic conductivity.
v(x, 0) ¼ v( þ `, t) ¼ vn , x . s(t), t . 0,
(6)
*Corresponding author.
s(0) ¼ 0
(7)
v ¼ ¹ K(W)
¹ DðvÞ
713
]vÿ þ
W ¹ Ws
, t . 0,
s(t) , t ¼ Ks 0
s(t)
]x
(5)
714
A. C. Briozzo, D. A. Tarzia
where
where
.
x
t
v
vn
vs
W
W0
Ws
K
Ks
sp (tp ) ¼
spatial coordinate
time
volumetric water content
initial volumetric water content
volumetric water content at saturation
soil water matric potential
pond depth
soil water potential at x ¼ s(t), Ws , W , W0
hydraulic conductivity
hydraulic conductivityat saturation
dW
soil water diffusivity D ¼ K
dv
D
a
(b ¹ v)2
yp ¼ xp ¹ sp (tp ) . 0, tp ¼ tp . 0;
(8)
where a and b are positive constants. With this form of
diffusivity, the nonlinear diffusion eqn (4) may be transformed in a linear one. Following ref. 2, we normalize the
water content variable as follows
Q¼
v ¹ vn
vs ¹ vn
and we consider
8
b ¹ vn
>
>
> C ¼ v ¹ v . 1 parameter;
>
>
s
n
>
>
>
>
a
>
>
length scale;
ls ¼
>
>
(vs ¹ vn )C(C ¹ 1)Ks
>
>
>
<
a
time scale;
ts ¼
C(C
¹
1)Ks2
>
>
>
>
>
x
>
>
dimensionless length;
xp ¼
>
>
>
l
s
>
>
>
>
>
t
>
: tp ¼ dimensionless time:
ts
(9)
(10)
sp (0) ¼ 0,
(12)
Q(xp , 0) ¼ Q( þ `, tp ) ¼ 0, xp . sp (tp ), tp . 0,
(13)
Q(sp (tp ) þ , tp ) ¼ 1, tp . 0,
(14)
W ¹ Wsp
C(C ¹ 1) ]Q ÿ
, tp . 0,
sp (tp ) þ , tp ¼ 0p
sp (tp )
(C ¹ Q)2 ]xp
(15)
(17)
hence, we have the dimensionless free boundary problem
eqns (18)–(22)
]Q
] C(C ¹ 1) ]Q
ds ]Q
¼
, y . 0, tp . 0,
þ p
2
dtp ]yp p
]tp ]yp (C ¹ Q) ]yp
(18)
sp (0) ¼ 0,
(19)
Q(yp , 0) ¼ Q( þ `, tp ) ¼ 0, yp . 0, tp . 0,
(20)
Q(0, tp ) ¼ 1, tp . 0,
(21)
¹
where
C(C ¹ 1) ]Q ÿ þ Wop ¹ Wsp
, tp . 0,
0 , tp ¼
sp (tp )
(C ¹ Q)2 ]yp
W0p ¼
Wsp ¼
Then, problem eqns (3)–(7) is transformed into problem
eqns (11)–(15)
]Q
] C(C ¹ 1) ]Q
¼
(11)
, xp . sp (tp ), tp . 0,
]tp ]xp (C ¹ Q)2 ]xp
¹
(16)
is the position of the free boundary.
Now we define a dimensionless depth coordinate moving
with the saturated–unsaturated interface
From now on we consider the free boundary problem
eqns (3)–(7), where the position s(t) of the free boundary
and the water field v(x,t) must be determined. We restrict
our attention to the special functional form of the soil water
diffusivity expressed by
D(v) ¼
s(t) s(ts tp )
¼
ls
ls
(22)
W0
dimensionless pond depth
ls
Ws
dimensionless soil water potential at the
ls
moving saturated ¹ unsaturated interface:
The goal of the paper is to solve the dimensionless free
boundary problem eqns (18)–(22). We will show an explicit to this problem which depends on a parameter C,
according to two cases: 1 , C , C1 or C $ C1 , where C1
is a constant (the bifurcation parameter) obtained as the
unique solution of the following equation:
p
d
2
(23)
C ¹ 1 ¼ , C . 1,
Q
2
C
where Q is a real function defined by
p
Q(x) ¼ pxexp(x2 )erfc(x), x . 0,
(24)
and d . 0 is a parameter defined in eqn (43).
2 CLOSED-FORM ANALYTIC SOLUTION OF THE
FREE BOUNDARY EQNS (18)–(22).
In order that the two boundary conditions eqns (3) and (5)
are compatible, s(t) must be of the form
p
s(t) ¼ m t,
(25)
m being an unknown constant. By eqns (2) and (5), the
unknown m is related to the unknown sorptivity S by the
Solution of free boundary problem
following expression
ÿ
2Ks W0 ¹ Ws
,
(26)
m¼
S
p
and v(s(t), t) ¼ S=2 t is the infiltration rate, where v is
related to W through the Darcy eqn (1). The sorptivity S
is a basic hydraulic property relating cumulative intake I(t)
(expressed as a length) to the square root of time for a onedimensional
p sorption into a soil without gravity, i.e.
I(t) ¼ S t (Ref. 8). It has been shown in4,5 that the dominant parameter governing the dynamics of infiltration at
small times is the sorptivity S. Since S ia measure of the capillary uptake or removal of water, is essentially a property of the
medium with some resemblance to permeability.
When v is in
1
cm s ¹1 and t in s, the unit of S is cm s ¹ 2 (Ref. 4).
Then, in terms of dimensionless variables we have
p
(27)
sp (tp ) ¼ mp tp
where
r
ÿ
2Ks W0 ¹ Ws (vs ¹ vn ) C(C ¹ 1) m p
t:
¼
mp ¼
a
ls s
S
(28)
3
To linearize the diffusion eqn (4) we define the variables
8
C(C ¹ 1)
>
>
>m¼ C¹Q ,
>
>
<
Zyp ÿ
1
(29)
p
C ¹ Q(n, tp ) dn,
x
¼
>
>
0
>
C(C
¹
1)
>
>
:
t ¼ tp ;
and we obtain the problem eqns (30)–(33)
]m ]2 m
g ]m
¼
þ p , x . 0, t . 0,
]t ]x2 2 t ]x
m(0 , t) ¼ C, t . 0,
þ
p
¹ C(C ¹ 1) ]m þ
S
(0 , t) ¼ pp , t . 0,
m
]x
2 t
m(x, 0) ¼ C ¹ 1 ¼ m( þ `, t), x . 0, t . 0,
(30)
r
C(C ¹ 1)
,
a
g(0 þ ) ¼ C:
(31)
(32)
(33)
(34)
Now we assume a similarity solution
x
m ¼ g(f), f ¼ p:
t
(35)
1
g9(f)(f þ g) þ g0(f) ¼ 0, f . 0,
2
(36)
g( þ `) ¼ C ¹ 1,
(37)
Then the problem eqns (30)–(33) reduces to the problem
eqns (36)–(39)
(38)
(39)
The solution to the conditions eqns (36)–(38) is given by
r 2
CS p
g
fþg
erfc
, f.0
exp
g(f) ¼ C ¹ 1 þ
4
2
a
2
(40)
where the coefficient g is unknown.
The extra boundary condition eqn (39) is consistent with
this solution provided that
r 2
g
1 S p
g
¼
exp
:
(41)
erfc
C 2 a
2
4
Since S and g verify the following relation (another method
is given in Remark 3 and Appendix A)
p
q
a
g 6 g2 ¹ g20 (C) , g $ g0 (C), C . 1 (42)
S¼
2
where
8
8ÿ
>
2
2
>
>
< g0 (C) ¼ C W0p ¹ Wsp ¼ d (C ¹ 1),
r
ÿ
>
8Ks W0 ¹ Ws (vs ¹ vn )
>
>
:
d¼
a
(43)
we have that the above eqn (41) in variable g ¼ g(C) is
given by
0
1
s
2
1 1@
g0 (C) A g
¼
1 6 1¹
, g $ g0 (C),
Q
C 2
g
2
C . 1:
where
r
p
2 aÿ
S
C(C ¹ 1)
W0p ¹ Wsp , Sp ¼ S
:
g ¼ p þ
a
CS
a
p
CSp CS
¹ C(C ¹ 1)g9(0 þ ) ¼
¼
2
2
715
ð44Þ
In studying eqn (44), we shall consider two cases respectively corresponding to choose the sign (þ) or (¹).
Case 1: (sign þ in the expression of S as a function of g)
The eqn (44) may be written as
g
1
¼ H1 (g, C)Q
(45)
, g $ g0 (C), C . 1:
C
2
where H1 is defined by
0
1
s
1@
g0 (C) 2 A
1þ 1¹
, g $ g0 (C),
H1 (g, C) ¼
2
g
C . 1:
The function H1 satisfies the following properties
8
1
>
>
H1 (g0 (C), C) ¼ , C . 1,
> (i)
>
2
>
<
(ii) H1 ( þ `, C) ¼ 1, C . 1,
>
>
>
]H1
>
>
: (iii)
(g, C) . 0, g . g0 (C), C . 1
]g
ð46Þ
(47)
716
A. C. Briozzo, D. A. Tarzia
Now we define the real function
1
, g $ g0 (C), C . 1:
F1 (g, C) ¼
CH1 (g, C)
which satisfies the following properties
8
2
>
>
(i)
F1 (g0 (C), C) ¼ , C . 1,
>
>
C
>
>
>
<
1
(ii) F1 ( þ `, C) ¼ , C . 1,
>
C
>
>
>
>
]F1
>
>
: (iii)
(g, C) , 0, g . g0 (C), C . 1
]g
Then, we have that the eqn (45) is equivalent to
g
F1 (g, C) ¼ Q
, g $ g0 (C), C . 1:
2
Since Q satisfies the properties
8
(i)
Q(0) ¼ 0,
>
>
>
>
>
< (ii) Q( þ `) ¼ 1,
>
(iii)
>
>
>
>
:
(iv)
Q9(0) ¼ p, Q9(x) . 0, x . 0,
(48)
C . 1:
(49)
(50)
(51)
Q0(x) , 0, x . 0
we conclude that eqn (50) admits a unique solution in the
variable g if and only if
2
g0 (C)
⇔ M(C) # 2
F1 (g0 (C), C) ¼ $ Q
C
2
where the real function M is defined by
p
g0 (C)
d
¼ CQ
C ¹ 1 , C . 1:
M(C) ¼ CQ
2
2
The function M satisfies the following properties
8
(i)
M9(C) . 0, C . 1,
>
>
>
>
>
>
(ii)
M(1) ¼ Q(0) ¼ 0,
>
>
>
<
(iii)
M( þ `) ¼ þ `,
>
>
> (iv) M(C) , C, C . 1,
>
>
>
>
>
M(C)
>
: (v)
¼ 1:
lim
C→ þ `
C
(52)
which satisfies the following properties
8
1
>
>
(i)
H2 (g0 (C), C) ¼ , C . 1,
>
>
2
>
<
(ii) H2 ( þ `, C) ¼ 0, C . 1,
>
>
>
]H2
>
>
: (iii)
(g, C) , 0, g . g0 (C), C . 1
]g
Now we define the real function
1
F2 (g, C) ¼
, g $ g0 (C), C . 1:
CH2 (g, C)
which satisfies the following properties
8
2
>
>
F2 (g0 (C), C) ¼ , C . 1,
> (i)
>
C
>
<
(ii) F2 ( þ `, C) ¼ þ `, C . 1,
>
>
>
]F2
>
>
: (iii)
(g, C) . 0, g . g0 (C), C . 1
]g
ð57Þ
(58)
(59)
(60)
Hence, the eqn (56) is equivalent to
g
, g $ g0 (C), C . 1:
(61)
F2 (g, C) ¼ Q
2
Taking into account the properties of functions Q and F2 ,
we deduce that eqn (59) admits a unique solution in the
variable g if and only if
2
g (C)
⇔
F2 (g0 (C), C) ¼ # Q 0
C
2
M(C) $ 2, C . 1 ⇔ C $ C1 :
Then, we have obtained the following result:
(53)
Therefore, there exist a unique constant C1 . 1 such that
d p
C1 ¹ 1 ¼ 2
(54)
M(C1 ) ¼ C1 Q
2
and
M(C) # 2 ⇔ KC # C1 :
Moreover, by using eqn (53)iv we deduce
C1 . 2
where H2 is defined by
0
1
s
1@
g0 (C) 2 A
1¹ 1¹
H2 (g, C) ¼
, g $ g0 (C),
2
g
(55)
Case 2: (sign ¹ in the expression of S as a function of g)
The eqn (44) may be written as
g
1
¼ H2 (g, C)Q
, g $ g0 (C), C . 1:
(56)
C
2
Theorem 1. Assume that C ¼ ðb ¹ vn Þ=ðvs ¹ vn Þ . 1.
Then, there exists a bifurcation parameter C1 ¼ C1 (d) ¼
C1 (a, Ks , W0 ¹ Ws , vs ¹ vn ) . 1, which is the unique solution
of the eqn (23). We have:
I) If 1 , C # C1 : There exist a unique g1 (C) $ g0 (C)
such that
0
1
s
2
1 1@
g0 (C) A
g (C)
¼
1þ 1¹
;
(62)
Q 1
C 2
g1 (C)
2
and, the solution of the problem eqns (36)–(39) is given by
r 2
S (C)C p
g (C)
exp 1
g1 (f) ¼ C ¹ 1 þ 1
4
2
a
f þ g1 (C)
, f . 0,
ð63Þ
3 erfc
2
where
p
q
a
g1 (C) þ g21 (C) ¹ g20 (C) :
S1 (C) ¼
2
(64)
Solution of free boundary problem
II) If C $ C1 : There exist a unique g2 (C) $ g0 (C) such that
0
1
s
1 1@
g0 (C) 2 A
g2 (C)
¼
1¹ 1¹
;
(65)
Q
C 2
g2 (C)
2
and, the solution of the problem eqns (36)–(39) is given by
r 2
S2 (C)C p
g (C)
exp 2
g2 (f) ¼ C ¹ 1 þ
4
2
a
f þ g2 (C)
, f . 0,
ð66Þ
3 erfc
2
where
S2 (C) ¼
p
q
a
g2 (C) ¹ g22 (C) ¹ g20 (C) :
2
(67)
Remark 1. For the case C ¼ C1 , we have that M(C1 ) ¼ 2,
that is
1
1
g0 (C1 )
¼ Q
:
(68)
C1 2
2
Then, g0 (C1 ) satisfies the two equations eqns (45) and (55)
because
1
H1 (g0 (C), C) ¼ H2 (g0 (C), C) ¼ , C . 1:
2
Remark 2. For C ¼ C1 both solutions g1 and g2 coincide
because
p
(69)
g1 (C1 ) ¼ g2 (C1 ) ¼ g0 (C1 ) ¼ d C1 ¹ 1,
and
p
a
d p
a(C1 ¹ 1)
g0 (C1 ) ¼
S(C1 ) ¼ S1 (C1 ) ¼ S2 (C1 ) ¼
2
2
p
ð70Þ
¼ 2(W0 ¹ Ws )(vs ¹ vn )(C1 ¹ 1):
Therefore, the solution of the problem eqns (36)–(39) is
given by
r 2
S(C1 )C1 p
g0 (C1 )
exp
g(f) ¼ C1 ¹ 1 þ
2
a
4
f þ g0 (C1 )
, f . 0:
ð71Þ
3 erfc
2
The sorptivity S as a function of the variable C is given by
8
p
p
a
>
>
>
g
(C)
¼
(C)
þ
g21 (C) ¹ g20 (C) , 1 , C1
S
>
1
1
>
2
>
>
>
p
<
a
S(C)¼
C ¼ C1
g (C ),
>
2 0 1
>
>
>
p
>
>
p
>
>
: S2 (C) ¼ a g2 (C) ¹ g2 (C) ¹ g2 (C) , C . C1
2
0
2
where g0 (C) is defined in eqn (43), and g1 (C) and g2 (C) are
defined by eqns (62) and (65) respectively.
The function S ¼ S(C) is continuously differentiable.
717
Moreover, we have
S(1 þ ) ¼ þ `, S( þ `) ¼ 0:
(72)
Proof. We get S1 (C1¹ ) ¼ S2 (C1þ ) because of eqn (70). By
elementary but tedious computations we obtain
]S1 ¹
]S2 þ
d p
(C1 ) ¼
(C1 ) ¼
a(C1 ¹ 1)
]C
]C
2C1
2
d
C1 (C1 ¹ 2) ¹ 1
3
8
On the other hand, by elementary computations we get eqn
(72).
Remark 3. An alternative method to prove Theorem 1
was suggested by an anonymous referee and it is shown in
the Appendix.
Finally, we invert the relations eqns (35), (29), (10) and
(9) to obtain the parametric solution to the problem eqns
(3)–(7), which depends on C.
Corollary 2. There exists a bifurcation parameter
C1 ¼ C1 (d) ¼ C1 (a, Ks , W0 ¹ Ws , vs ¹ vn ) . 1, for the solution of problem (3)–(7) which is given by:
(I) Case 1 , C # C1 . We have
0
1
B
C
B
(C ¹ 1) C
C þ vn , x . 0,
!
1
¹
v1 (x, t) ¼ (vs ¹ vn )C B
B
C
@
x A
g1 p
t
t . 0,
x ¼ ls y1p (x, t) þ m1
ð73Þ
p
ts t, x . 0, t . 0,
t ¼ ts t, x . 0, t . 0,
s1 (x, t) ¼ m1
(74)
(75)
p
ts t, x . 0, t . 0, (the free boundary)
(76)
with
r
ÿ
ÿ
2Ks W0 ¹ Ws vs ¹ vn
C(C ¹ 1) ls
p
m1 ¼
a
S1 (C)
ts
ÿ
2Ks W0 ¹ Ws
,
ð77Þ
¼
S1 (C)
!
Zx
1
n
y1p (x, t) ¼ p g1 p dn
t
C(C ¹ 1) 0
(
p
r
t
x
p
¼ p ðC ¹ 1Þ p þ S1 (C)C
a
t
C(C ¹ 1)
"
!
2
g (C)
x
g (C)
p þ 1
3 exp 1
:
4
2
2 t
718
A. C. Briozzo, D. A. Tarzia
x
g (C)
3 erfc p þ 1
2
2 t
!
!2 !
1
x
g1 (C)
p þ
¹ pexp ¹
2
p
2 t
g (C)
g (C)
¹ 1 erfc 1
2
2
#)
1
g21 (C)
, x . 0, t . 0:
þ pexp ¹
4
p
ð78Þ
(II) Case C $ C1 : We have
0
t . 0,
ð79Þ
p
ts t, x . 0, t . 0,
t ¼ ts t, x . 0, t . 0,
s2 (x, t) ¼ m2
(80)
(81)
p
ts t, x . 0, t . 0, (the free boundary)
(82)
with
ÿ
ÿ
r
2Ks W0 ¹ Ws vs ¹ vn
C(C ¹ 1) ls
p
m2 ¼
a
S2 (C)
ts
ÿ
2Ks W0 ¹ Ws
,
ð83Þ
¼
S2 (C)
!
Zx
1
n
y2p (x, t) ¼ p g2 p dn
t
C(C ¹ 1) 0
(
p
r
t
x
p
¼ p ðC ¹ 1Þ p þ S2 (C)C
a
t
C(C ¹ 1)
"
!
2
g (C)
x
g (C)
p þ 2
:
3 exp 2
4
2
2 t
!
x
g (C)
1
¹ p
3 erfc p þ 2
2
p
2 t
!2 !
x
g (C)
g (C)
p þ 2
¹ 2
3 exp ¹
2
2
2 t
#)
g2 (C)
1
g22 (C)
þ pexp ¹
,
3 erfc
4
2
p
x . 0, t . 0:
B
C
B
(C1 ¹ 1) C
B
C
!
(x,
t)
¼
v
(x,
t)
¼
(v
¹
v
)C
1
¹
v1
2
s
n
1B
C
@
x A
g p
t
ð85Þ
x ¼ ls yp (x, t) þ m
(86)
þ vn , x . 0, t . 0,
ð84Þ
p
ts t, x . 0, t . 0,
t ¼ ts t, x . 0, t . 0,
s(x, t) ¼ m
1
B
C
B
(C ¹ 1) C
C þ vn , x . 0,
!
1
¹
v2 (x, t) ¼ (vs ¹ vn )CB
B
C
@
x A
g2 p
t
x ¼ ls y2p (x, t) þ m2
Remark 4. For the case C ¼ C1 , the two parametric solutions coincide one each other, that is
0
1
p
ts t, x . 0, t . 0, (the free boundary)
(87)
(88)
with
ÿ
ÿ
2Ks W0 ¹ Ws vs ¹ vn
m ¼ m1 ¼ m2 ¼
S(C1 )
r
ÿ
2Ks W0 ¹ Ws
C1 (C1 ¹ 1) ls
p ¼
3
a
S(C1 )
ts
s
ÿ
2Ks W0 ¹ Ws
ÿ
,
¼
vs ¹ vn (C1 ¹ 1)
ð89Þ
!
Zx
1
n
yp (x, t) ¼ p g p dn
t
C1 (C1 ¹ 1) 0
(
p
r
ÿ
x
t
p
¼ p C1 ¹ 1 p þ S(C1 )C1
a
t
C1 (C1 ¹ 1)
"
!
2
g (C )
x
g (C )
p þ 0 1
3 exp 0 1 :
4
2
2 t
!
x
g (C )
3 erfc p þ 0 1
2
2 t
!2 !
1
x
g0 (C1 )
p þ
¹ pexp ¹
2
p
2 t
g (C )
g (C )
¹ 0 1 erfc 0 1
2
2
#)
1
g20 (C1 )
þ pexp ¹
, x . 0, t . 0:
4
p
ð90Þ
3 BEHAVIOR OF THE BIFURCATION
PARAMETER C 1 UPON THE DATA
We shall study the bifurcation parameter C1 , the unique
Solution of free boundary problem
719
p!
C1 (d) ¹ 1
2
3
p!
p!
d C1 (d) ¹ 1
d C1 (d) ¹ 1
C1 (d)d
þ pQ9
Q
2
2
4 C1 (d) ¹ 1
p
d
¹ C1 (d) C1 (d) ¹ 1Q9
, 0, d . 0:
Fig. 1. The bifurcation parameter C1 versus variable which are
related by the eqn (23).
solution of the eqn (23), as a function of the variable d
defined by eqn (43). See Fig. 1 and Table 1.
Table 1. Values for C 1 as a function of d
(iii) If the limit of C1 ðdÞ is finite when d → 0 þ we have
a contradiction with eqn (23) because its left hand side
goes to 0 and its right hand side goes to a positive
number. Therefore eqn (91)iii holds.
(iv) If limd→ þ ` C1 ðdÞ ¼ þ ` then we have a contradiction with eqn (23) because its left hand side goes to 1
when d goes to þ ` (because of eqn (51)ii) while its
right hand side goes to 0. Then, the limit of C1 ðdÞ is
finite ( $ 2) when d goes to þ `. Then, by eqn (51)ii,
we get eqn (91)iv.
Analogously, we can study the bifurcation parameter C1
as a function of the driving potential e defined by
e ¼ W0 ¹ Ws
(92)
Theorem 4. The function C1 ¼ C1 (e) satisfies the following properties:
8
]C1
>
< (i)
(ii)
, 0, ;e . 0;
C1 . 2;
]e
>
: (iii) lim C1 (e) ¼ þ `; (iv)
lim C1 (e) ¼ 2:
þ
e→0
e→ þ `
(93)
Lemma 3. We have that C1 ¼ C1 (d) satisfies the following
properties:
8
]C1
>
< (i)
(ii)
, 0, ;d . 0;
C1 . 2;
]d
>
: (iii) lim C1 (d) ¼ þ `; (iv)
lim C1 (d) ¼ 2:
þ
d→0
d→ þ `
(91)
Moreover, we have that the inverse function d ¼ d(C1 ) is
given explicitly by
2
2
¹1
, C1 . 2,
d ¼ pQ
C
C1 ¹ 1
1
Proof. By taking into account the Lemma 4, and the parameters d and e are related by the following expression
ÿ
p
8Ks vs ¹ vn
(94)
d ¼ me, with m ¼
a
the results (i)–(iv) hold.
From Theorem 5 we obtain the following conclusions:
(i) It is clear that only the ‘ þ ’ branch, in relation eqn
(42), occurs when the driving potential e ¼ W0 ¹ Ws
goes to zero because of eqn (93)iii). Therefore, the
‘¹’ branch has not physical meaning.
(ii) For the other cases (the driving potential
e ¼ W0 ¹ Ws is positive) the two branches (‘þ’ and
‘¹’) in relation (42) have a physical meaning.
where Q ¹ 1 is the inverse function of Q.
ACKNOWLEDGEMENTS
Proof.
(i) is the condition eqn (55).
(ii) By using eqn (51) we have
dC1
1
(d) ¼
2
dd
This paper has been sponsored by the Project #4798 Free
Boundary Problems for the Heat-Diffusion Equation from
CONICET, Rosario-Argentina. The authors appreciate the
valuable suggestions by two anonymous referees which
improved the paper.
720
A. C. Briozzo, D. A. Tarzia
APPENDIX A
We shall show a new proof of Theorem 1 by studying a
single trascendental equation for the sorptivity S. If we substitute eqn (34) in eqn (41), we obtain for the unknown S the
following equation
F(S, C) ¼ Qp (S, C), S . 0 (C . 1 : parameter)
(A1)
p
a
¼
1 ad2 (C ¹ 1)
þ
, S . 0, C . 1
C
4S2 C
(A2)
and
g(S, C)
, S . 0, C . 1
Q (S, C) ¼ Q
2
p
(A3)
with
S
g(S, C) ¼ p þ
a
p 2
ad (C ¹ 1)
, S . 0, C . 1
4S
M(C) , 2 ⇔ 1 , C , C1 :
2
g0 (C)
⇔
(b) F(S , C) , Q (S , C) ⇔ , Q
C
2
p
where
g(S, C)
F(S, C) ¼
CS
Proof. Functions F and Qp satisfy the following relations:
2
g0 (C)
p
p p
⇔
(a) F(S , C) . Q (S , C) ⇔ . Q
C
2
(A4)
By elementary computations we deduce that the functions
Qp and F satisfy the following properties.
Lemma 5. We have:
Therefore, for a fixed C, we have that if 1 , C , C1 the
abscisa Sp1 of the intersection point of the graphs of the
functions F and Qp is to the right of the minimum point
Sp (Sp1 . Sp ), in other case this point Sp2 is to the left of the
minimum point (Sp2 , Sp ).
Now, we can relate the solutions Sp1 and Sp2 of the eqn (A1)
according to the two cases 1 , C , C1 and C . C1 respectively, which are given by the above Theorem 8, with the
expressions eqns (64) and (67) obtained in Theorem 1.
Theorem 8. We have
p
q
a
p
(i) S1 ¼ S1 (C) ¼
g1 (C) þ g21 (C) ¹ g20 (C) ,
2
1 , C , C1
Qp (0, C) ¼ Qp ( þ `, C) ¼ 1, C . 1
(ii)
;
(ii)
Sp2
p
q
a
g2 (C) ¹ g22 (C) ¹ g20 (C) ,
¼ S2 (C) ¼
2
C . C1
C.1
(iii)
C.1
C.1
where
p
p
a
d a(C ¹ 1)
p
p
g (C) ¼
(A5)
S ¼ S (C) ¼
2 0
2
is the minimum point of function Qp with respect to S, for
all C . 1.
g0 (C)
p p
, C . 1.
(iii) Q (S , C) ¼ Q
2
Lemma 6. We have:
(i)F(0, C) ¼ þ `, C . 1; (ii)F( þ `, C) ¼
1
, C . 1;
C
Sp1
p
a
d p
a(C1 ¹ 1)
g0 (C1 ) ¼
¼ ¼ S(C1 ) ¼
2
2
p
¼ 2(W0 ¹ Ws )(vs ¹ vn )(C1 ¹ 1), C ¼ C1 :
Sp2
Proof. Sp1 and Sp2 must satisfy the expression eqn (42). On
the
otherpa hand,
for
1 , C , C1
we
have
p
p
S1 . S ¼ 2 g0 (C). Then Sp1 is given by the sign ‘ þ ’ in
eqn (42) (analogously for C . C1 we have Sp2 is given by
the sign ‘¹’ in eqn (42)) because the functions
q
q
g3 (x) ¼ x þ x2 ¹ g20 (C), g4 (x) ¼ x ¹ x2 ¹ g20 (C),
x . g0 (C)
satisfy the following properties:
(A6)
(iii)
p
M(C) . 2 ⇔ C . C1 :
(i)
8
, 0 if S , Sp ,
>
>
<
p
]Q
(S, C) ¼
0 if S ¼ Sp ,
>
]S
>
:
. 0 if S . Sp ,
p
]F
2
(S, C) , 0, S . 0, C . 1; (iv)F(Sp , C) ¼ ,
]S
C
C . 1:
Theorem 7. The eqn (A1) for the sorptivity S with a
parameter C . 1, admits a unique solution Sp1 . Sp if
1 , C , C1 or, Sp2 , Sp if C . C1 , where C1 is the
unique solution of the eqn (23).
g3 (g0 (C)) ¼ g4 (g0 (C)) ¼ g0 (C),
g3 ( þ `) ¼ þ `, g4 ( þ `) ¼ 0,
g3 9(x) . 0, g4 9(x) , 0:
REFERENCES
1. Broadbridge, P. Solution of a nonlinear absorption model
Solution of free boundary problem
2.
3.
4.
5.
of mixed saturated-unsaturated flow. Water Resources
Research, 1990, 26, 2435–2443.
Broadbridge, P. & White, I. Constant rate rainfall infiltration:
A versatile nonlinear model. 1. Analytic solutions. Water
Resources Research, 1988, 24, 145–154.
Knight, J. H. & Philip, J. R. Exact solutions in nonlinear
difussion. J. Eng. Math., 1974, 8, 219–227.
Philip, J. R. The theory of infiltration. 4. Sorptivity and algebraic infiltration equations. Soil. Sci., 1957, 84, 257–264.
Philip, J. R. The theory of infiltration. 5. The influence of the
initial moisture content. Soil. Sci., 1958, 85, 329–339.
721
6. Philip, J. R. The theory of infiltration. 6. Effect of water
depth over soil. Soil. Sci., 1958, 85, 278–286.
7. Tarzia, D. A., A bibliography on moving-free boundary problems for the heat-difussion equation. The Stefan problem.
Progetto Nazionale M. P. I., Equazione di evoluzione e applicazioni fisico-matematiche, Firenze, 1988 (with 2528 references). An updated one will give at the end of 1997 with
more than 5000 on the subject.
8. Warrick, A. W. & Broadbridge, P. Sorptivity and macroscopic capillary length relationship. Water Resources
Research, 1992, 28, 427–431.
q 1998 Elsevier Science Ltd
Printed in Great Britain. All rights reserved
PII: S 0 3 0 9 - 1 7 0 8 ( 9 7 ) 0 0 0 2 6 - 2
0309-1708/98/$19.00 + 0.00
Explicit solution of a free boundary problem for
a nonlinear absorption model of mixed
saturated–unsaturated flow
Adriana C. Briozzo & Domingo A. Tarzia*
Departamento de Matema´tica, F.C.E., Universidad Austral, Paraguay 1950, (2000) Rosario, Argentina
(Received 11 November 1996; revised 23 March 1997; accepted 13 June 1997)
In wet soils, zones of saturation naturally develop in the vicinity of impermeable
strata, surface ponds and subterranean cavities. Hydrology must be then concerned
with transient flow through coexisting unsaturated and saturated zones. The models of
advancing saturated zones necessarily involve a nonlinear free boundary problem.
A closed-form analytic solution is presented for a nonlinear diffusion model under
conditions of ponding at the surface. The soil water diffusivity is restricted to the
special functional form D(v) ¼ a=(b ¹ v)2 , where v is the water content field to be
determined and a, b are positive constants. The explicit solution depends on a
parameter C (determined by the data of the problem), according to two cases:
1 , C , C1 or C $ C1 , where C1 is a constant which is obtained as the unique
solution of an equation. This result complements the study given in P. Broadbridge,
Water Resources Research, 1990, 26, 2435–2443, in order to established when the
explicit solution is available. The behavior of the bifurcation parameter C1 as a
function of the driving potential is studied with the corresponding limits for small and
large values. Moreover, the sorptivity is proven to be continuously differentiable
function of the variable C. q 1998 Elsevier Science Limited. All rights reserved
Key words: free boundary problem, mixed saturated–unsaturated flow, nonlinear
absorption model.
1 INTRODUCTION
In the saturated zone we have1
Following refs 1,6, we consider a homogeneous soil which
initially has some uniform volumetric water content vn . At
times t . 0, water is supplied at the surface x ¼ 0 under
pressure head W0 . Then, a mixed saturated–unsaturated
flow problem representing absorption of water by a soil
with a constant pond depth at the surface is presented. At
every time t the zone of saturation extends from x ¼ 0 to x ¼
s(t) (the free boundary), and the unsaturated zone extends
for x . s(t). By assuming the Darcy’s law and neglecting the
gravity, the water flux is given by
W(x, t) ¼ W0 ¹
W0 ¹ Ws
x; 0 , x , s(t);
s(t)
(2)
and, the following free boundary problem eqns (3)–(7)
arises for the unsaturated zone7:
v(s(t) þ , t) ¼ vs , t . 0,
(3)
]v
]
]v
¼
D(v)
, x . s(t), t . 0,
]t ]x
]x
(4)
]W
,
(1)
]x
where W is the soil water matric potential and K is the
hydraulic conductivity.
v(x, 0) ¼ v( þ `, t) ¼ vn , x . s(t), t . 0,
(6)
*Corresponding author.
s(0) ¼ 0
(7)
v ¼ ¹ K(W)
¹ DðvÞ
713
]vÿ þ
W ¹ Ws
, t . 0,
s(t) , t ¼ Ks 0
s(t)
]x
(5)
714
A. C. Briozzo, D. A. Tarzia
where
where
.
x
t
v
vn
vs
W
W0
Ws
K
Ks
sp (tp ) ¼
spatial coordinate
time
volumetric water content
initial volumetric water content
volumetric water content at saturation
soil water matric potential
pond depth
soil water potential at x ¼ s(t), Ws , W , W0
hydraulic conductivity
hydraulic conductivityat saturation
dW
soil water diffusivity D ¼ K
dv
D
a
(b ¹ v)2
yp ¼ xp ¹ sp (tp ) . 0, tp ¼ tp . 0;
(8)
where a and b are positive constants. With this form of
diffusivity, the nonlinear diffusion eqn (4) may be transformed in a linear one. Following ref. 2, we normalize the
water content variable as follows
Q¼
v ¹ vn
vs ¹ vn
and we consider
8
b ¹ vn
>
>
> C ¼ v ¹ v . 1 parameter;
>
>
s
n
>
>
>
>
a
>
>
length scale;
ls ¼
>
>
(vs ¹ vn )C(C ¹ 1)Ks
>
>
>
<
a
time scale;
ts ¼
C(C
¹
1)Ks2
>
>
>
>
>
x
>
>
dimensionless length;
xp ¼
>
>
>
l
s
>
>
>
>
>
t
>
: tp ¼ dimensionless time:
ts
(9)
(10)
sp (0) ¼ 0,
(12)
Q(xp , 0) ¼ Q( þ `, tp ) ¼ 0, xp . sp (tp ), tp . 0,
(13)
Q(sp (tp ) þ , tp ) ¼ 1, tp . 0,
(14)
W ¹ Wsp
C(C ¹ 1) ]Q ÿ
, tp . 0,
sp (tp ) þ , tp ¼ 0p
sp (tp )
(C ¹ Q)2 ]xp
(15)
(17)
hence, we have the dimensionless free boundary problem
eqns (18)–(22)
]Q
] C(C ¹ 1) ]Q
ds ]Q
¼
, y . 0, tp . 0,
þ p
2
dtp ]yp p
]tp ]yp (C ¹ Q) ]yp
(18)
sp (0) ¼ 0,
(19)
Q(yp , 0) ¼ Q( þ `, tp ) ¼ 0, yp . 0, tp . 0,
(20)
Q(0, tp ) ¼ 1, tp . 0,
(21)
¹
where
C(C ¹ 1) ]Q ÿ þ Wop ¹ Wsp
, tp . 0,
0 , tp ¼
sp (tp )
(C ¹ Q)2 ]yp
W0p ¼
Wsp ¼
Then, problem eqns (3)–(7) is transformed into problem
eqns (11)–(15)
]Q
] C(C ¹ 1) ]Q
¼
(11)
, xp . sp (tp ), tp . 0,
]tp ]xp (C ¹ Q)2 ]xp
¹
(16)
is the position of the free boundary.
Now we define a dimensionless depth coordinate moving
with the saturated–unsaturated interface
From now on we consider the free boundary problem
eqns (3)–(7), where the position s(t) of the free boundary
and the water field v(x,t) must be determined. We restrict
our attention to the special functional form of the soil water
diffusivity expressed by
D(v) ¼
s(t) s(ts tp )
¼
ls
ls
(22)
W0
dimensionless pond depth
ls
Ws
dimensionless soil water potential at the
ls
moving saturated ¹ unsaturated interface:
The goal of the paper is to solve the dimensionless free
boundary problem eqns (18)–(22). We will show an explicit to this problem which depends on a parameter C,
according to two cases: 1 , C , C1 or C $ C1 , where C1
is a constant (the bifurcation parameter) obtained as the
unique solution of the following equation:
p
d
2
(23)
C ¹ 1 ¼ , C . 1,
Q
2
C
where Q is a real function defined by
p
Q(x) ¼ pxexp(x2 )erfc(x), x . 0,
(24)
and d . 0 is a parameter defined in eqn (43).
2 CLOSED-FORM ANALYTIC SOLUTION OF THE
FREE BOUNDARY EQNS (18)–(22).
In order that the two boundary conditions eqns (3) and (5)
are compatible, s(t) must be of the form
p
s(t) ¼ m t,
(25)
m being an unknown constant. By eqns (2) and (5), the
unknown m is related to the unknown sorptivity S by the
Solution of free boundary problem
following expression
ÿ
2Ks W0 ¹ Ws
,
(26)
m¼
S
p
and v(s(t), t) ¼ S=2 t is the infiltration rate, where v is
related to W through the Darcy eqn (1). The sorptivity S
is a basic hydraulic property relating cumulative intake I(t)
(expressed as a length) to the square root of time for a onedimensional
p sorption into a soil without gravity, i.e.
I(t) ¼ S t (Ref. 8). It has been shown in4,5 that the dominant parameter governing the dynamics of infiltration at
small times is the sorptivity S. Since S ia measure of the capillary uptake or removal of water, is essentially a property of the
medium with some resemblance to permeability.
When v is in
1
cm s ¹1 and t in s, the unit of S is cm s ¹ 2 (Ref. 4).
Then, in terms of dimensionless variables we have
p
(27)
sp (tp ) ¼ mp tp
where
r
ÿ
2Ks W0 ¹ Ws (vs ¹ vn ) C(C ¹ 1) m p
t:
¼
mp ¼
a
ls s
S
(28)
3
To linearize the diffusion eqn (4) we define the variables
8
C(C ¹ 1)
>
>
>m¼ C¹Q ,
>
>
<
Zyp ÿ
1
(29)
p
C ¹ Q(n, tp ) dn,
x
¼
>
>
0
>
C(C
¹
1)
>
>
:
t ¼ tp ;
and we obtain the problem eqns (30)–(33)
]m ]2 m
g ]m
¼
þ p , x . 0, t . 0,
]t ]x2 2 t ]x
m(0 , t) ¼ C, t . 0,
þ
p
¹ C(C ¹ 1) ]m þ
S
(0 , t) ¼ pp , t . 0,
m
]x
2 t
m(x, 0) ¼ C ¹ 1 ¼ m( þ `, t), x . 0, t . 0,
(30)
r
C(C ¹ 1)
,
a
g(0 þ ) ¼ C:
(31)
(32)
(33)
(34)
Now we assume a similarity solution
x
m ¼ g(f), f ¼ p:
t
(35)
1
g9(f)(f þ g) þ g0(f) ¼ 0, f . 0,
2
(36)
g( þ `) ¼ C ¹ 1,
(37)
Then the problem eqns (30)–(33) reduces to the problem
eqns (36)–(39)
(38)
(39)
The solution to the conditions eqns (36)–(38) is given by
r 2
CS p
g
fþg
erfc
, f.0
exp
g(f) ¼ C ¹ 1 þ
4
2
a
2
(40)
where the coefficient g is unknown.
The extra boundary condition eqn (39) is consistent with
this solution provided that
r 2
g
1 S p
g
¼
exp
:
(41)
erfc
C 2 a
2
4
Since S and g verify the following relation (another method
is given in Remark 3 and Appendix A)
p
q
a
g 6 g2 ¹ g20 (C) , g $ g0 (C), C . 1 (42)
S¼
2
where
8
8ÿ
>
2
2
>
>
< g0 (C) ¼ C W0p ¹ Wsp ¼ d (C ¹ 1),
r
ÿ
>
8Ks W0 ¹ Ws (vs ¹ vn )
>
>
:
d¼
a
(43)
we have that the above eqn (41) in variable g ¼ g(C) is
given by
0
1
s
2
1 1@
g0 (C) A g
¼
1 6 1¹
, g $ g0 (C),
Q
C 2
g
2
C . 1:
where
r
p
2 aÿ
S
C(C ¹ 1)
W0p ¹ Wsp , Sp ¼ S
:
g ¼ p þ
a
CS
a
p
CSp CS
¹ C(C ¹ 1)g9(0 þ ) ¼
¼
2
2
715
ð44Þ
In studying eqn (44), we shall consider two cases respectively corresponding to choose the sign (þ) or (¹).
Case 1: (sign þ in the expression of S as a function of g)
The eqn (44) may be written as
g
1
¼ H1 (g, C)Q
(45)
, g $ g0 (C), C . 1:
C
2
where H1 is defined by
0
1
s
1@
g0 (C) 2 A
1þ 1¹
, g $ g0 (C),
H1 (g, C) ¼
2
g
C . 1:
The function H1 satisfies the following properties
8
1
>
>
H1 (g0 (C), C) ¼ , C . 1,
> (i)
>
2
>
<
(ii) H1 ( þ `, C) ¼ 1, C . 1,
>
>
>
]H1
>
>
: (iii)
(g, C) . 0, g . g0 (C), C . 1
]g
ð46Þ
(47)
716
A. C. Briozzo, D. A. Tarzia
Now we define the real function
1
, g $ g0 (C), C . 1:
F1 (g, C) ¼
CH1 (g, C)
which satisfies the following properties
8
2
>
>
(i)
F1 (g0 (C), C) ¼ , C . 1,
>
>
C
>
>
>
<
1
(ii) F1 ( þ `, C) ¼ , C . 1,
>
C
>
>
>
>
]F1
>
>
: (iii)
(g, C) , 0, g . g0 (C), C . 1
]g
Then, we have that the eqn (45) is equivalent to
g
F1 (g, C) ¼ Q
, g $ g0 (C), C . 1:
2
Since Q satisfies the properties
8
(i)
Q(0) ¼ 0,
>
>
>
>
>
< (ii) Q( þ `) ¼ 1,
>
(iii)
>
>
>
>
:
(iv)
Q9(0) ¼ p, Q9(x) . 0, x . 0,
(48)
C . 1:
(49)
(50)
(51)
Q0(x) , 0, x . 0
we conclude that eqn (50) admits a unique solution in the
variable g if and only if
2
g0 (C)
⇔ M(C) # 2
F1 (g0 (C), C) ¼ $ Q
C
2
where the real function M is defined by
p
g0 (C)
d
¼ CQ
C ¹ 1 , C . 1:
M(C) ¼ CQ
2
2
The function M satisfies the following properties
8
(i)
M9(C) . 0, C . 1,
>
>
>
>
>
>
(ii)
M(1) ¼ Q(0) ¼ 0,
>
>
>
<
(iii)
M( þ `) ¼ þ `,
>
>
> (iv) M(C) , C, C . 1,
>
>
>
>
>
M(C)
>
: (v)
¼ 1:
lim
C→ þ `
C
(52)
which satisfies the following properties
8
1
>
>
(i)
H2 (g0 (C), C) ¼ , C . 1,
>
>
2
>
<
(ii) H2 ( þ `, C) ¼ 0, C . 1,
>
>
>
]H2
>
>
: (iii)
(g, C) , 0, g . g0 (C), C . 1
]g
Now we define the real function
1
F2 (g, C) ¼
, g $ g0 (C), C . 1:
CH2 (g, C)
which satisfies the following properties
8
2
>
>
F2 (g0 (C), C) ¼ , C . 1,
> (i)
>
C
>
<
(ii) F2 ( þ `, C) ¼ þ `, C . 1,
>
>
>
]F2
>
>
: (iii)
(g, C) . 0, g . g0 (C), C . 1
]g
ð57Þ
(58)
(59)
(60)
Hence, the eqn (56) is equivalent to
g
, g $ g0 (C), C . 1:
(61)
F2 (g, C) ¼ Q
2
Taking into account the properties of functions Q and F2 ,
we deduce that eqn (59) admits a unique solution in the
variable g if and only if
2
g (C)
⇔
F2 (g0 (C), C) ¼ # Q 0
C
2
M(C) $ 2, C . 1 ⇔ C $ C1 :
Then, we have obtained the following result:
(53)
Therefore, there exist a unique constant C1 . 1 such that
d p
C1 ¹ 1 ¼ 2
(54)
M(C1 ) ¼ C1 Q
2
and
M(C) # 2 ⇔ KC # C1 :
Moreover, by using eqn (53)iv we deduce
C1 . 2
where H2 is defined by
0
1
s
1@
g0 (C) 2 A
1¹ 1¹
H2 (g, C) ¼
, g $ g0 (C),
2
g
(55)
Case 2: (sign ¹ in the expression of S as a function of g)
The eqn (44) may be written as
g
1
¼ H2 (g, C)Q
, g $ g0 (C), C . 1:
(56)
C
2
Theorem 1. Assume that C ¼ ðb ¹ vn Þ=ðvs ¹ vn Þ . 1.
Then, there exists a bifurcation parameter C1 ¼ C1 (d) ¼
C1 (a, Ks , W0 ¹ Ws , vs ¹ vn ) . 1, which is the unique solution
of the eqn (23). We have:
I) If 1 , C # C1 : There exist a unique g1 (C) $ g0 (C)
such that
0
1
s
2
1 1@
g0 (C) A
g (C)
¼
1þ 1¹
;
(62)
Q 1
C 2
g1 (C)
2
and, the solution of the problem eqns (36)–(39) is given by
r 2
S (C)C p
g (C)
exp 1
g1 (f) ¼ C ¹ 1 þ 1
4
2
a
f þ g1 (C)
, f . 0,
ð63Þ
3 erfc
2
where
p
q
a
g1 (C) þ g21 (C) ¹ g20 (C) :
S1 (C) ¼
2
(64)
Solution of free boundary problem
II) If C $ C1 : There exist a unique g2 (C) $ g0 (C) such that
0
1
s
1 1@
g0 (C) 2 A
g2 (C)
¼
1¹ 1¹
;
(65)
Q
C 2
g2 (C)
2
and, the solution of the problem eqns (36)–(39) is given by
r 2
S2 (C)C p
g (C)
exp 2
g2 (f) ¼ C ¹ 1 þ
4
2
a
f þ g2 (C)
, f . 0,
ð66Þ
3 erfc
2
where
S2 (C) ¼
p
q
a
g2 (C) ¹ g22 (C) ¹ g20 (C) :
2
(67)
Remark 1. For the case C ¼ C1 , we have that M(C1 ) ¼ 2,
that is
1
1
g0 (C1 )
¼ Q
:
(68)
C1 2
2
Then, g0 (C1 ) satisfies the two equations eqns (45) and (55)
because
1
H1 (g0 (C), C) ¼ H2 (g0 (C), C) ¼ , C . 1:
2
Remark 2. For C ¼ C1 both solutions g1 and g2 coincide
because
p
(69)
g1 (C1 ) ¼ g2 (C1 ) ¼ g0 (C1 ) ¼ d C1 ¹ 1,
and
p
a
d p
a(C1 ¹ 1)
g0 (C1 ) ¼
S(C1 ) ¼ S1 (C1 ) ¼ S2 (C1 ) ¼
2
2
p
ð70Þ
¼ 2(W0 ¹ Ws )(vs ¹ vn )(C1 ¹ 1):
Therefore, the solution of the problem eqns (36)–(39) is
given by
r 2
S(C1 )C1 p
g0 (C1 )
exp
g(f) ¼ C1 ¹ 1 þ
2
a
4
f þ g0 (C1 )
, f . 0:
ð71Þ
3 erfc
2
The sorptivity S as a function of the variable C is given by
8
p
p
a
>
>
>
g
(C)
¼
(C)
þ
g21 (C) ¹ g20 (C) , 1 , C1
S
>
1
1
>
2
>
>
>
p
<
a
S(C)¼
C ¼ C1
g (C ),
>
2 0 1
>
>
>
p
>
>
p
>
>
: S2 (C) ¼ a g2 (C) ¹ g2 (C) ¹ g2 (C) , C . C1
2
0
2
where g0 (C) is defined in eqn (43), and g1 (C) and g2 (C) are
defined by eqns (62) and (65) respectively.
The function S ¼ S(C) is continuously differentiable.
717
Moreover, we have
S(1 þ ) ¼ þ `, S( þ `) ¼ 0:
(72)
Proof. We get S1 (C1¹ ) ¼ S2 (C1þ ) because of eqn (70). By
elementary but tedious computations we obtain
]S1 ¹
]S2 þ
d p
(C1 ) ¼
(C1 ) ¼
a(C1 ¹ 1)
]C
]C
2C1
2
d
C1 (C1 ¹ 2) ¹ 1
3
8
On the other hand, by elementary computations we get eqn
(72).
Remark 3. An alternative method to prove Theorem 1
was suggested by an anonymous referee and it is shown in
the Appendix.
Finally, we invert the relations eqns (35), (29), (10) and
(9) to obtain the parametric solution to the problem eqns
(3)–(7), which depends on C.
Corollary 2. There exists a bifurcation parameter
C1 ¼ C1 (d) ¼ C1 (a, Ks , W0 ¹ Ws , vs ¹ vn ) . 1, for the solution of problem (3)–(7) which is given by:
(I) Case 1 , C # C1 . We have
0
1
B
C
B
(C ¹ 1) C
C þ vn , x . 0,
!
1
¹
v1 (x, t) ¼ (vs ¹ vn )C B
B
C
@
x A
g1 p
t
t . 0,
x ¼ ls y1p (x, t) þ m1
ð73Þ
p
ts t, x . 0, t . 0,
t ¼ ts t, x . 0, t . 0,
s1 (x, t) ¼ m1
(74)
(75)
p
ts t, x . 0, t . 0, (the free boundary)
(76)
with
r
ÿ
ÿ
2Ks W0 ¹ Ws vs ¹ vn
C(C ¹ 1) ls
p
m1 ¼
a
S1 (C)
ts
ÿ
2Ks W0 ¹ Ws
,
ð77Þ
¼
S1 (C)
!
Zx
1
n
y1p (x, t) ¼ p g1 p dn
t
C(C ¹ 1) 0
(
p
r
t
x
p
¼ p ðC ¹ 1Þ p þ S1 (C)C
a
t
C(C ¹ 1)
"
!
2
g (C)
x
g (C)
p þ 1
3 exp 1
:
4
2
2 t
718
A. C. Briozzo, D. A. Tarzia
x
g (C)
3 erfc p þ 1
2
2 t
!
!2 !
1
x
g1 (C)
p þ
¹ pexp ¹
2
p
2 t
g (C)
g (C)
¹ 1 erfc 1
2
2
#)
1
g21 (C)
, x . 0, t . 0:
þ pexp ¹
4
p
ð78Þ
(II) Case C $ C1 : We have
0
t . 0,
ð79Þ
p
ts t, x . 0, t . 0,
t ¼ ts t, x . 0, t . 0,
s2 (x, t) ¼ m2
(80)
(81)
p
ts t, x . 0, t . 0, (the free boundary)
(82)
with
ÿ
ÿ
r
2Ks W0 ¹ Ws vs ¹ vn
C(C ¹ 1) ls
p
m2 ¼
a
S2 (C)
ts
ÿ
2Ks W0 ¹ Ws
,
ð83Þ
¼
S2 (C)
!
Zx
1
n
y2p (x, t) ¼ p g2 p dn
t
C(C ¹ 1) 0
(
p
r
t
x
p
¼ p ðC ¹ 1Þ p þ S2 (C)C
a
t
C(C ¹ 1)
"
!
2
g (C)
x
g (C)
p þ 2
:
3 exp 2
4
2
2 t
!
x
g (C)
1
¹ p
3 erfc p þ 2
2
p
2 t
!2 !
x
g (C)
g (C)
p þ 2
¹ 2
3 exp ¹
2
2
2 t
#)
g2 (C)
1
g22 (C)
þ pexp ¹
,
3 erfc
4
2
p
x . 0, t . 0:
B
C
B
(C1 ¹ 1) C
B
C
!
(x,
t)
¼
v
(x,
t)
¼
(v
¹
v
)C
1
¹
v1
2
s
n
1B
C
@
x A
g p
t
ð85Þ
x ¼ ls yp (x, t) þ m
(86)
þ vn , x . 0, t . 0,
ð84Þ
p
ts t, x . 0, t . 0,
t ¼ ts t, x . 0, t . 0,
s(x, t) ¼ m
1
B
C
B
(C ¹ 1) C
C þ vn , x . 0,
!
1
¹
v2 (x, t) ¼ (vs ¹ vn )CB
B
C
@
x A
g2 p
t
x ¼ ls y2p (x, t) þ m2
Remark 4. For the case C ¼ C1 , the two parametric solutions coincide one each other, that is
0
1
p
ts t, x . 0, t . 0, (the free boundary)
(87)
(88)
with
ÿ
ÿ
2Ks W0 ¹ Ws vs ¹ vn
m ¼ m1 ¼ m2 ¼
S(C1 )
r
ÿ
2Ks W0 ¹ Ws
C1 (C1 ¹ 1) ls
p ¼
3
a
S(C1 )
ts
s
ÿ
2Ks W0 ¹ Ws
ÿ
,
¼
vs ¹ vn (C1 ¹ 1)
ð89Þ
!
Zx
1
n
yp (x, t) ¼ p g p dn
t
C1 (C1 ¹ 1) 0
(
p
r
ÿ
x
t
p
¼ p C1 ¹ 1 p þ S(C1 )C1
a
t
C1 (C1 ¹ 1)
"
!
2
g (C )
x
g (C )
p þ 0 1
3 exp 0 1 :
4
2
2 t
!
x
g (C )
3 erfc p þ 0 1
2
2 t
!2 !
1
x
g0 (C1 )
p þ
¹ pexp ¹
2
p
2 t
g (C )
g (C )
¹ 0 1 erfc 0 1
2
2
#)
1
g20 (C1 )
þ pexp ¹
, x . 0, t . 0:
4
p
ð90Þ
3 BEHAVIOR OF THE BIFURCATION
PARAMETER C 1 UPON THE DATA
We shall study the bifurcation parameter C1 , the unique
Solution of free boundary problem
719
p!
C1 (d) ¹ 1
2
3
p!
p!
d C1 (d) ¹ 1
d C1 (d) ¹ 1
C1 (d)d
þ pQ9
Q
2
2
4 C1 (d) ¹ 1
p
d
¹ C1 (d) C1 (d) ¹ 1Q9
, 0, d . 0:
Fig. 1. The bifurcation parameter C1 versus variable which are
related by the eqn (23).
solution of the eqn (23), as a function of the variable d
defined by eqn (43). See Fig. 1 and Table 1.
Table 1. Values for C 1 as a function of d
(iii) If the limit of C1 ðdÞ is finite when d → 0 þ we have
a contradiction with eqn (23) because its left hand side
goes to 0 and its right hand side goes to a positive
number. Therefore eqn (91)iii holds.
(iv) If limd→ þ ` C1 ðdÞ ¼ þ ` then we have a contradiction with eqn (23) because its left hand side goes to 1
when d goes to þ ` (because of eqn (51)ii) while its
right hand side goes to 0. Then, the limit of C1 ðdÞ is
finite ( $ 2) when d goes to þ `. Then, by eqn (51)ii,
we get eqn (91)iv.
Analogously, we can study the bifurcation parameter C1
as a function of the driving potential e defined by
e ¼ W0 ¹ Ws
(92)
Theorem 4. The function C1 ¼ C1 (e) satisfies the following properties:
8
]C1
>
< (i)
(ii)
, 0, ;e . 0;
C1 . 2;
]e
>
: (iii) lim C1 (e) ¼ þ `; (iv)
lim C1 (e) ¼ 2:
þ
e→0
e→ þ `
(93)
Lemma 3. We have that C1 ¼ C1 (d) satisfies the following
properties:
8
]C1
>
< (i)
(ii)
, 0, ;d . 0;
C1 . 2;
]d
>
: (iii) lim C1 (d) ¼ þ `; (iv)
lim C1 (d) ¼ 2:
þ
d→0
d→ þ `
(91)
Moreover, we have that the inverse function d ¼ d(C1 ) is
given explicitly by
2
2
¹1
, C1 . 2,
d ¼ pQ
C
C1 ¹ 1
1
Proof. By taking into account the Lemma 4, and the parameters d and e are related by the following expression
ÿ
p
8Ks vs ¹ vn
(94)
d ¼ me, with m ¼
a
the results (i)–(iv) hold.
From Theorem 5 we obtain the following conclusions:
(i) It is clear that only the ‘ þ ’ branch, in relation eqn
(42), occurs when the driving potential e ¼ W0 ¹ Ws
goes to zero because of eqn (93)iii). Therefore, the
‘¹’ branch has not physical meaning.
(ii) For the other cases (the driving potential
e ¼ W0 ¹ Ws is positive) the two branches (‘þ’ and
‘¹’) in relation (42) have a physical meaning.
where Q ¹ 1 is the inverse function of Q.
ACKNOWLEDGEMENTS
Proof.
(i) is the condition eqn (55).
(ii) By using eqn (51) we have
dC1
1
(d) ¼
2
dd
This paper has been sponsored by the Project #4798 Free
Boundary Problems for the Heat-Diffusion Equation from
CONICET, Rosario-Argentina. The authors appreciate the
valuable suggestions by two anonymous referees which
improved the paper.
720
A. C. Briozzo, D. A. Tarzia
APPENDIX A
We shall show a new proof of Theorem 1 by studying a
single trascendental equation for the sorptivity S. If we substitute eqn (34) in eqn (41), we obtain for the unknown S the
following equation
F(S, C) ¼ Qp (S, C), S . 0 (C . 1 : parameter)
(A1)
p
a
¼
1 ad2 (C ¹ 1)
þ
, S . 0, C . 1
C
4S2 C
(A2)
and
g(S, C)
, S . 0, C . 1
Q (S, C) ¼ Q
2
p
(A3)
with
S
g(S, C) ¼ p þ
a
p 2
ad (C ¹ 1)
, S . 0, C . 1
4S
M(C) , 2 ⇔ 1 , C , C1 :
2
g0 (C)
⇔
(b) F(S , C) , Q (S , C) ⇔ , Q
C
2
p
where
g(S, C)
F(S, C) ¼
CS
Proof. Functions F and Qp satisfy the following relations:
2
g0 (C)
p
p p
⇔
(a) F(S , C) . Q (S , C) ⇔ . Q
C
2
(A4)
By elementary computations we deduce that the functions
Qp and F satisfy the following properties.
Lemma 5. We have:
Therefore, for a fixed C, we have that if 1 , C , C1 the
abscisa Sp1 of the intersection point of the graphs of the
functions F and Qp is to the right of the minimum point
Sp (Sp1 . Sp ), in other case this point Sp2 is to the left of the
minimum point (Sp2 , Sp ).
Now, we can relate the solutions Sp1 and Sp2 of the eqn (A1)
according to the two cases 1 , C , C1 and C . C1 respectively, which are given by the above Theorem 8, with the
expressions eqns (64) and (67) obtained in Theorem 1.
Theorem 8. We have
p
q
a
p
(i) S1 ¼ S1 (C) ¼
g1 (C) þ g21 (C) ¹ g20 (C) ,
2
1 , C , C1
Qp (0, C) ¼ Qp ( þ `, C) ¼ 1, C . 1
(ii)
;
(ii)
Sp2
p
q
a
g2 (C) ¹ g22 (C) ¹ g20 (C) ,
¼ S2 (C) ¼
2
C . C1
C.1
(iii)
C.1
C.1
where
p
p
a
d a(C ¹ 1)
p
p
g (C) ¼
(A5)
S ¼ S (C) ¼
2 0
2
is the minimum point of function Qp with respect to S, for
all C . 1.
g0 (C)
p p
, C . 1.
(iii) Q (S , C) ¼ Q
2
Lemma 6. We have:
(i)F(0, C) ¼ þ `, C . 1; (ii)F( þ `, C) ¼
1
, C . 1;
C
Sp1
p
a
d p
a(C1 ¹ 1)
g0 (C1 ) ¼
¼ ¼ S(C1 ) ¼
2
2
p
¼ 2(W0 ¹ Ws )(vs ¹ vn )(C1 ¹ 1), C ¼ C1 :
Sp2
Proof. Sp1 and Sp2 must satisfy the expression eqn (42). On
the
otherpa hand,
for
1 , C , C1
we
have
p
p
S1 . S ¼ 2 g0 (C). Then Sp1 is given by the sign ‘ þ ’ in
eqn (42) (analogously for C . C1 we have Sp2 is given by
the sign ‘¹’ in eqn (42)) because the functions
q
q
g3 (x) ¼ x þ x2 ¹ g20 (C), g4 (x) ¼ x ¹ x2 ¹ g20 (C),
x . g0 (C)
satisfy the following properties:
(A6)
(iii)
p
M(C) . 2 ⇔ C . C1 :
(i)
8
, 0 if S , Sp ,
>
>
<
p
]Q
(S, C) ¼
0 if S ¼ Sp ,
>
]S
>
:
. 0 if S . Sp ,
p
]F
2
(S, C) , 0, S . 0, C . 1; (iv)F(Sp , C) ¼ ,
]S
C
C . 1:
Theorem 7. The eqn (A1) for the sorptivity S with a
parameter C . 1, admits a unique solution Sp1 . Sp if
1 , C , C1 or, Sp2 , Sp if C . C1 , where C1 is the
unique solution of the eqn (23).
g3 (g0 (C)) ¼ g4 (g0 (C)) ¼ g0 (C),
g3 ( þ `) ¼ þ `, g4 ( þ `) ¼ 0,
g3 9(x) . 0, g4 9(x) , 0:
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