An. Şt. Univ. Ovidius Constanţa
Vol. 11(1), 2003, 9–18
ON A SINGULARLY PERTURBED,
COUPLED ELLIPTIC-ELLIPTIC PROBLEM
Luminiţa Barbu and Emil Cosma
To Professor Silviu Sburlan, at his 60’ anniversary
Abstract
The behavior of the solution of the below problem (E² ), (BC² ),
(T C² ) is studied when the small parameter ² tends to 0.
1. Introduction.
We consider the following coupled boundary value problem of elliptic-elliptic
type, denoted by P² :
(
−²u00 (x) + α(x)u0 (x) + β(x)u(x) = f (x), x ∈ (a, b),
¡
¢0
− µ(x)v 0 (x) + α(x)v 0 (x) + β(x)v(x) = g(x), x ∈ (b, c),
(E² )
with homogeneous Dirichlet boundary conditions
u(a) = v(c) = 0
(BC² )
and transmission conditions at x = b
u(b) = v(b),
²u0 (b) = (µv 0 )(b).
(T C² )
The transmission conditions at x = b express the continuity of the solution
and of the flux.
The following assumptions will be required in the following:
(A1 ) a, b, c ∈ IR, a < b < c, ² > 0 is a small parameter;
Key Words: Coupled boundary value problem, transmission conditions, singularly perturbed, asymptotic expansion.
Mathematical Reviews subject classification: 34B05, 34E15.
9
10
L. Barbu and E. Cosma
(A2 ) α ∈ H 1 (a, c), β ∈ L∞ (a, c), µ ∈ H 1 (a, c);
0
(A3 ) α(x) ≤ α0 < 0 in [a, c], µ(x) ≥ µ0 > 0 in [b, c], β− α2 ≥ 0 a.e. in (a, c);
(A4 )f ∈ L2 (a, b), g ∈ L2 (b, c).
The aim of this paper is to investigate the problem P² for ² going to zero
from the view point of singular perturbation theory. This is a singularly
perturbed problem with respect to the norm of uniform convergence and the
boundary layer is the point x = a. To have an idea about this matter, let us
consider the particular case when α, β, µ are constant functions. If the solution
(u, v) of (P² ) converges in C[a, b] × C[b, c] to (U, V ), then it can easily be seen
that (U, V ) satisfies

αU 0 + βU = f, in (a, b),





 −µV 00 + αV 0 + βV = g, in (b, c),

U (a) = 0, U (b) = V (b),




 V 0 (b) = 0, V (c) = 0.
The condition U (a) = 0 is not satisfied in general (this exceeds the number
of conditions allowed). This fact is not acceptable from a physical point of
view. Actually we shall see that indeed, in the point x = a, the solution (u, v)
has a singular behaviour as ² goes to zero. The corresponding unperturbed
(reduced) problem, denoted by P0 , is the following [6]:
(
α(x)U 0 (x) + β(x)U (x) = f (x), x ∈ (a, b)
¡
¢0
− µ(x)V 0 (x) + α(x)V 0 (x) + β(x)V (x) = g(x),
(
V (c) = 0,
U (b) = V (b),
V 0 (b) = 0.
x ∈ (b, c),
(E0 )
(BC0 )
(T C0 )
This problem will be reobtained below by using the Vishik-Lusternik method
and it is the same as that derived in [6] by using a different way.
In [1], [2] we studied a similar transmission of type elliptic-elliptic but
we considered that α ≥ α0 > 0. Here we are assuming α < 0, hence the
asymptotic behavior is different from the case α > 0: actually we cannot have
a convergence of u(a) to U (a), in general. A comment is needed about the
conditions (T C0 ), (BC0 ). Note that the problem P0 has no conditions at all
for x = a and needs two conditions at b, in this case, when α < 0.
On a singularly perturbed, coupled elliptic-elliptic problem
11
In Section 2 we shall derive a formal zero-th order asymptotic expansion
for the solution (u, v) of the problem (P² ). The interface point x = a is a
boundary layer and the expansion of u contains a corresponding corrector
(boundary layer function).
In Section 3 we shall investigate the existence and uniqueness of the solutions to the problems (P² ) and P0 .
Finally, Section 4 is devoted to obtaining some estimates for the remainder
terms of the expansion established in Section 2 with respect to the uniform
convergence topology.
2. A formal asymptotic expansion for the solution of P²
The classical perturbation theory (see [7] for details) can be adapted to our
specific singular perturbation problem. Following this theory, we are going to
derive formally an expansion of the solution (u, v) of (P² ) of the form:
(
u(x) = U (x) + θ1 (ζ) + ρ1² (x), x ∈ [a, b],
(2.1)
v(x) = V (x) + θ2 (ζ) + ρ2² (x), x ∈ [b, c],
where ζ:=²−1 (x − a) is the fast variable; (U, V ) is the zero-th order term of
the regular series; θ1 , θ2 are boundary layer functions (correctors); ρ1² , ρ2²
denote the remainder terms of zero-th order.
We substitute formally in (E² ) (u, v) given by (2.1) and then we identify
the coefficients of ²k (k = −1, 0), separately those depending on x from those
depending on ζ. So, we get
(
α(x)U 0 (x) + β(x)U (x) = f (x), a < x < b,
(E0 )
0
−(µ(x)V 0 (x)) + α(x)V 0 (x) + β(x)V (x) = g(x), b < x < c.
For θ1 =θ1 (ζ) we derive the equation
θ100 (ζ) − α(a)θ10 (ζ) = 0.
(2.2)
Eq. (2.2) and the fact that θ1 is a boundary layer function (in particular,
θ1 (ζ) −→ 0, as ζ −→ ∞) implies that θ1 (ζ) = keα(a)ζ , where k is a constant
which will be determined from (BC² ). Also, we can deduce that θ2 = 0.
For the remainder terms we have the equations

−ε ρ001² (x) + α(x)ρ01² (x) + β(x)ρ1² (x) =



¡
¢ ¡
¢
¡
¢
= ²U 00 (x) + ²−1 α(x) − α(a) θ10 ζ(x) − β(x)θ1 ζ(x) , a < x < b,


¢0
 ¡
− (µρ02² )(x) + α(x)ρ02² (x) + β(x)ρ2² (x) = 0, b < x < c,
(ER)
12
L. Barbu and E. Cosma
where ζ(x) = (x − a)/².
From (BC² ), we can derive
k = −U (a), hence θ1 (ξ) = −U (a)eα(a)ζ
V (c) = 0,
(
ρ1² (a) = 0,
ρ2² (c) = 0.
(2.3)
(BC0 )
(BCR)
By replacing (2.1) into (T C)² , we can see that U and V satisfy the transmission conditions
(
U (b) = V (b),
(T C0 )
V 0 (b) = 0,
and for the remainder terms we have
(
ρ1² (b) = ρ2² (b) + U (a)eα(a)θ1 (ζ(b)) ,
¢
¡
²ρ01² (b) = −²U 0 (b) + U (a)α(a)eα(a)θ1 (ζ(b)) + µρ02² (b).
(T CR)
Summarizing, the reduced problem, P0 , is (E0 ) − (BC0 ) − (T C0 ), while the
problem satisfied by the remainder terms is (ER) − (BCR) − (T CR). As we
shall see later on, the last problem is satisfied by (ρ1² , ρ2² ) in a generalized
sense.
3. Existence and regularity for the problems (P² ) and P0 .
For the problem (P² ) we have the following result, whose proof is essentially
known (see [1]):
Proposition 3.1.
Assume that (A1 )−(A4 ) are satisfied. Then, the
problem P² admits a unique solution (u, v) ∈ H 2 (a, b)×H 2 (b, c) satisfying (E² )
a.e. in (a, b) and in (b, c), respectively, as well as (BC² ) and (T C² ).
In the following, we are going to investigate the reduced problem (P0 ).
In fact, we can split it in two separate problems, with the unknowns U and V ,
respectively. Clearly, V is a solution of (E0 )2 , with the boundary conditions
V 0 (b) = 0, V (c) = 0.
(3.1)
By the Lax-Milgram lemma, there exists a unique solution V ∈H 2 (b, c)
of this problem. Obviously, Eq. (E0 )1 , with U (b) = V (b), has a unique
solution U ∈ H 1 (a, b). Therefore, we have the following result
13
On a singularly perturbed, coupled elliptic-elliptic problem
Proposition 3.2.
Assume that (A1 ) − (A4 ) are satisfied. Then,
the problem P0 has a unique solution U ∈H 1 (a, b), V ∈H 2 (a, c), which satisfies
(E0 ) a.e. in (a, b) and (b, c), respectively, as well as (BC0 ) and (T C0 ).
If we denote
ρ̃1² (x) := ρ1² (x) + A² x + B² , A² := (b − a)χ² , B² := −aA² ,
(3.2)
where χ² = −U (a)eα(a)θ1 (ζ(b) then we have ρ̃1² (a)=0, ρ̃1² (b) = ρ2² (b) and
taking into account (P² ), (P0 ), we can see that
(
ρ̃1² in [a, b]
ρ² :=
,
ρ2² in (b, c]
satisfies ρ² ∈H01 (a, c) and
Z b
Z
0
0
ρ̃1² ϕ dx +
²
a
Z
b
= −²
0
Z
c
U 0 ϕ0 dx −
a
µρ2² ϕ dx +
a
a
c
+
a
Z
αv 0 ϕdx +
b
c
αρ2² 0 ϕdx
b
a
Z
0
c
0
b
Z
βV ϕdx− µV
b
b
βU ϕdx =
c
0
gϕdx, ∀ ϕ ∈ H01 (a, c),
Z
b
a
αV ϕdx+
b
c
f ϕdx +
Z
(µv 0 )ϕ0 dx+
b
Z
a
αU 0 ϕdx +
c
βuϕdx +
b
βvϕdx =
b
Z
b
a
Z
c
b
Z
(µV )ϕ dx+
b
+
¡
¢
¡
¢ ¡
¢

β(x)θ1 (ζ(x) + α(a) α(x) − α(a) θ1 ζ(x) +


 h
i
h² (x) :=
+
A
in (a, b),
² α(x) + β(x)(A² x + B² ) ,



0, in (b, c).
Z
c
Z
αρ̃01² ϕdx
h² ϕdx + ²A² ϕ(b) + α(a)ϕ(b)χ² , ∀ ϕ ∈ H01 (a, c), (3.3)
Indeed, by (E² ) and (E0 ), we obtain
Z b
Z b
Z
²u0 ϕ0 dx +
αu0 ϕdx +
Z
b
0
b
a
where
Z
c
f ϕdx,
a
0
c
ϕ|b
Z
=
b
c
ϕgdx, ∀ ϕ ∈ H01 (a, c).
Now, subtracting the last two equalities from the first one, we obtain that
Z b ³
Z b ³
´
´
d
d
0
0
0
θ1 (ζ(x)) + %1² ϕ dx +
α
θ1 (ζ(x)) + ρ01² ϕdx+
² U +
dx
dx
a
a
14
L. Barbu and E. Cosma
Z
b
+
Z
c
+
b
a
c
Z
(µρ02² )ϕ0 dx +
³
´
β θ1 (ζ(x)) + ρ1² ϕdx+
b
Z
αρ02² ϕdx +
c
βρ2² ϕdx − (µV 0 )(b)ϕ(b) = 0.
b
From
Z
b
²
a
d
d
b
θ1 (ζ(x))dx = ²ϕ(x) θ1 (ζ(x))|a − ²
ϕ
dx
dx
Z
0
Z
= ϕ(b)χ² α(a) − ²−1 α(a)
b
ϕ
a
d2
θ1 (ζ)dx =
dx2
b
2
ϕθ1 (ζ(x))dx,
a
we can see that,
Z
b
¡
² ρ̃1² 0 − A² ϕ0 )dx +
a
Z
b
¡
¢
α ρ̃1² 0 − A² − ²−1 α(a)θ1 (ζ) ϕdx+
a
Z
b
+
¡
¢
β θ1 (ξ) + ρ̃1² − A² x − B² ) ϕdx+
a
Z
c
+
b
−²−1 α(a)
Z
µρ02² ϕ0 dx +
Z
b
c
Z
αρ02² ϕdx +
b
2
Z
ϕθ1 (ζ)dx = −²
a
b
c
βρ2² ϕdx−
b
U 0 ϕ0 dx + α(a)ϕ(b)χ² .
a
In conclusion, we obtain (3.3).
An elementary computation shows that, if ρ1² , ρ2² are smooth functions,
then they satisfy problem (ER) − (BCR) − (T CR) in a classical sense.
On a singularly perturbed, coupled elliptic-elliptic problem
15
4. Estimates for the remainder terms
The main result of this section is
Theorem 4.1. Assume that (A1 )−(A4 ) are satisfied and α is Lipschitzian in [a, b] (i.e., ∃ L>0, such that |α(x)−α(y) |≤L|x−y|, ∀ x, y∈[a, b]).
Then, for every ²>0, the problem (P² ) has a unique solution
(u, v) ∈H 2 (a, b)×H 2 (b, c) of the form (2.1), where col(U, V )∈H 1 (a, b)×H 2 (b, c)
is the solution of (P0 ), θ1 is given by (2.3), θ2 = 0 and (ρ1² , ρ2² ) ∈H 1 (a, b)×H 2 (b, c).
√
√
In addition, we have the estimates kρ1² kC[a,b] =O( ²), kρ2² kC[b,c] =O( ²).
Proof. By Propositions 3.1 and 3.2, (ρ1² , ρ2² )∈H 1 (a, b)×H 2 (b, c). For
the sake of simplicity, we assume that β−α0 /2≥γ0 >0 a.e. in (a, c). If we choose
in (3.3) ϕ=ρ² ∈H01 (a, c),, we can see that
Z
²
a
Z
+
b
c
b
Z
2
(ρ̃0 1² ) dx +
b
c
Z
2
µ (ρ02² ) dx +
Z
(β − α0 /2)ρ22² dx = −²
a
b
a
b
(β − α0 /2) ρ̃1² 2 dx+
Z
U 0 ρ̃1² 0 dx −
b
h² ρ̃1² dx + γ² ,
(4.1)
a
where γ² = α(a)ϕ(b) ρ̃1² (b) + ²A² ρ̃1² (b). In the case β−α0 /2≥0 a.e. in (a, c),
we choose in (3.3)
( −x
e r̃1² (x)
in [a, b]
ϕ(x) :=
e−b r2² (x)
in (b, c]
and we can use a slight modification of our reasoning below. Denote by k · k1 ,
k · k2 the norms of L2 (a, b), L2 (b, c), respectively. As β− α0 /2≥γ0 >0 a.e. in
(a, c) and µ≥µ0 >0 in [b, c], it follows from (4.1) that
¡
2
2
2
2¢
²k ρ̃1² 0 k1 + µ0 kρ02² k2 + γ0 kρ̃1² k1 + kρ2² k2 ≤
h
i
2
2
2
2
(4.2)
≤ (1/2) ²kU 0 k1 + ²kρ̃01² k1 + γ0−1 kh² k1 + γ0 kρ̃1² k1 + | γ² |.
√
In the following, we can show that kh² k1 =O( ²) and γ(²)= O(²k ), ∀k≥0.
From equations (E² ), we obtain
¡
2
2
2
2¢
²ku0 k1 + µ0 kv 0 k2 + γ0 kuk1 + kvk2 ≤
h
i
¡
¡
2
2¢
2
2¢
≤ (1/2) γ0−1 kf k1 + kgk2 + γ0 kuk1 + kvk2 .
(4.3)
16
L. Barbu and E. Cosma
2
This implies that kuk1 =O(1), kvk2 =O(1) and kv 0 k2 = =O(1), ² ku0 k2 =O(1).
Since v(c)=0 and H 1 (b, c)⊂C[b, c], with a compact injection, we get kvkC[b,c] =O(1).
For γ² , we have
lim ²−k γ² = 0, for every k ≥ 0,
²→0
k
and therefore γ² = O(² ). Now, as β∈L∞ (a, c), | A² |, | B² |= O(²k ), k ≥ 1, α
2
is Lipschitzian in [a, b], one gets by an easy computation that kh² k1 = O(²).
Now, from (4.2), it follows
√
√
√
kρ̃01² k1 = O(1), kρ̃1² k1 = O( ²), kρ02² k2 = O( ²), kρ2² k2 = O( ²),
therefore
√
kρ1² k1 ≤ kρ̃1² k1 + k(A² x + B² )ω(²)k1 = O( ²).
From
Z
ρ22² (x)
c
= (−1/2)
x
one gets
(4.4)
ρ2² 0 (s)ρ2² (s)ds ≤ 2 kρ2² k2 kρ2² k2 ,
√
kρ2² kC[b,c] = O( ²).
(4.5)
√
In that which follows, we shall prove that kρ1² kC[b,c] =O( ²). To do that, we
integrate (E² )1 on [y, b], y∈[a, b]:
¡
¢
² u0 (y) − u0 (b) +
Z
b
Z
α(s)u0 (s)ds +
y
By replacing
Z
b
β(s)u(s)ds =
y
b
f (s)ds.
(4.6)
y
¡
¢
u( x) = U (x) + θ1 ζ(x) + ρ1² (x),
in (4.6), one obtains
Z
² ρ01² (y) +
y
b
Z
α(s)ρ01² (s)ds +
b
β(s)ρ1² (s)ds =
y
h
¢i
d ¡
= ² U 0 (y) +
θ1 ζ(y) −
dy
Z
−
y
b
³d ¡
³ ¡
h
¢´
¢´i
α(s)
θ1 ζ(s) + β(s) θ1 ζ(s)
ds + ²u0 (b)a.e. in (a, x).
ds
(4.7)
Now we multiply the above inequality by ρ01² (y) and then integrate on [a, x].
17
On a singularly perturbed, coupled elliptic-elliptic problem
Finally, we obtain that
¡
¢
α(x)/2 ρ21² (x) + b̃(x)ρ1² (x) + c̃(x) = 0,
(4.8)
where
Z
b
b̃(x) := −
³
´
α0 (y) − β(y) (ρ1² (y)dy − ²u0 (b) − α(b)θ1 (ζ(b) + θ1 (ζ(y)),
x
Z
x
c̃(x) := −
Z
−
a
Z
x
−
a
a
x
¡ 0
¢
α /2 − β (y)ρ̃21² (y)dy−
h
¡
¢ ¡
¢i
ρ01² (y) ²U 0 (y) − α(y) − α(a) θ1 ζ(y) dy−
Z
h³
´ ¡
¢
0
β(y) − α (y) θ1 ζ(y) ρ1² (y)dy − ²
x
a
2
(ρ01² ) (y)dy.
By (4.4), one
Integrating (4.6) on [a, b], one obtains
√ obtains that c̃(x)=O(²).
√
²u0 (b)=O( ²), hence b̃(x)=O( ²).
Finally, from
√
(1/2)α(x)ρ21² (x) + O( ²)ρ1² (x) + O(²) = 0 a.e. in (a, b),
α(x)/2 ≤ α0 /2, < 0 in [a, b],
√
we can deduce that | ρ1² (x) | ≤ C ², x ∈ [a, b], so we have kr1² kC[a,b] =
√
O( ²).
References
[1] L. Barbu and Gh. Morosanu, On a class of singularly perturbed, coupled boundary
value problems, Math. Sci. Res. Hot-Line 4(6) (2000), 25-37.
[2] L. Barbu and Gh. Moroşanu, Asymptotic Analysis of Some Boundary Value Problems
with Singular Perturbations,Editura Academiei, Bucharest, 2000 (in Romanian).
[3] C.A. Coclici, Gh. Moroşanu and W.L. Wendland, On the viscous-viscous and the
viscous-inviscid interactions in Computational Fluid Dynamics, Comput. Visualization
Sci. 2 (1999), 95-105.
[4] C.A. Coclici, Gh. Moroşanu and W.L. Wendland, One-dimensional singularly perturbed coupled boundary value problems, Math. Sci. Res. Hot-Line 3(10) (1999),
1-21.
[5] C.A. Coclici, Gh. Moroşanu and W.L. Wendland, The coupling of hyperbolic and
elliptic boundary value problems with variable coefficients, Math. Meth. Appl. Sci. 23
(2000), 401-440.
18
L. Barbu and E. Cosma
[6] F. Gastaldi and A. Quarteroni, On the coupling of hyperbolic and parabolic systems:
analytical and numerical approach, Appl. Numer. Math. 6 (1990), 3-31.
[7] A.B. Vasilieva, V.F. Butuzov and I.V. Kalashev, The Boundary Function Method for
Singular Perturbation Problems, SIAM, Philadelphia, 1995.
”Ovidius” University of Constantza,
Faculty of Mathematics and Informatics,
Mamaia Bd., 124,
8700 Constantza,
Romania