Well-posedness and inverse Robin estimate for a multiscale elliptic/parabolic system

0.37 MB·

by  Martin Lind

WELL-POSEDNESS AND INVERSE ROBIN ESTIMATE FOR A
MULTISCALE ELLIPTIC/PARABOLIC SYSTEM
MARTIN LIND, ADRIAN MUNTEAN, AND OMAR RICHARDSON
Abstract. We establish the well-posedness of a coupled micro-macro parabolic-elliptic sys-
tem modeling the interplay between two pressures in a gas-liquid mixture close to equilibrium
that is filling a porous media with distributed microstructures. Additionally, we prove a local
7
stability estimate for the inverse micro-macro Robin problem, potentially useful in identifying
1
quantitatively a micro-macro interfacial Robin transfer coefficient given microscopic measure-
0
2 ments on accessible fixed interfaces. To tackle the solvability issue we use two-scale energy
estimates and two-scale regularity/compactness arguments cast in the Schauder’s fixed point
n
theorem. A number of auxiliary problems, regularity, and scaling arguments are used in ensur-
a
J ingthesuitableFr´echetdifferentiabilityofthesolutionandthestructureoftheinversestability

1 estimate.
3

]
P
A 1. Introduction
.
h We are interested in developing evolution equations able to describe multiscale spatial inter-
at actions in gas-liquid mixtures, targeting a rigorous mathematical justification of Richards-like
m equations - upscaled model equations generally chosen in a rather ad hoc manner by the engi-
[ neering communities to describe the motion of flow in unsaturated porous media. The main

1 issue is that one lacks a rigorous derivation of the Darcy’s law for such flow (see Hornung [2012]
v (chapter 1) for a derivation via periodic homogenization techniques of the Darcy law for the
2
saturated case).
2
1 If air-water interfaces can be assume to be stagnant for a reasonable time span, then aver-
9
aging techniques for materials with locally periodic microstructures (compare e.g. Chechkin
0
. and Piatnitski [1998]) lead in suitable scaling regimes to what we refer here as two-pressure
1
evolution systems. These are normally coupled parabolic-elliptic systems responsible for the
0
7 joint evolution in time t ∈ (0,T) (T < +∞) of a parameter-dependent microscopic pressure
1 Rρ(t,x,y) evolving with respect to y ∈ Y ⊂ Rd for any given macroscopic spatial position
:
v
x ∈ Ω and a macroscopic pressure π(t,x) with x ∈ Ω for any given t. Here R denotes the
i
X universal constant of gases. The two-scale geometry we have in mind is depicted in Figure 1
r below.
a
Figure 1. The macroscopic domain Ω and microscopic pore Y at x ∈ Ω
2010 Mathematics Subject Classification. 76S05, 35B27, 35R10, 35R30, 86A22.
Key words and phrases. Upscaled porous media, two-scale PDE, inverse micro-macro Robin problem.
1
2 MARTIN LIND, ADRIAN MUNTEAN, AND OMAR RICHARDSON
To cast the physical problem in mathematical terms as stated in (1), we need a number of
dimensional constant parameters (A (gas permeability), D (diffusion coefficient for the gaseous
species), p (atmospheric pressure), ρ (gas density)) and dimensional functions (k (Robin
F F
coefficient)andρ (initialliquiddensity)). ItisworthnotingthatexceptingtheRobincoefficient
I
k, all the model parameters and functions are either known or can be accessed directly via
measurements. Getting grip on a priori values of k is more intricate simply because this
coefficient is defined on the Robin part of the boundary of ∂Y, say Γ , where the micro-macro
R
information transfer takes actively place. The Neumann part of the boundary Γ := ∂Y −Γ
N R
is assumed to be accessible via measurements, while Γ is thought here as unaccessible.
R
Our aim is twofold:
(1) ensure the well-posedness in a suitable sense of our two-pressure system with k taken
to be known;
(2) prove stability estimates with respect to k for the inverse micro-macro Robin problem
(k is now unknown, but measured values of the microscopic pressure are available on
Γ ).
N
The main results reported here are Theorem 3.1 (the weak solvability of (1)) and Theorem 5.1
(the local stability for the inverse micro-macro Robin problem).
The choice of problem and approach is in line with other investigations running for two-scale
systems, or systems with distributed microstructures, like Lind and Muntean [2016], Meier
[2008], Peszynska and Showalter [2007]. As far as we are aware, this is for the first time that
an inverse Robin problem is treated in a two-scale setting. A remotely connected single-scale
inverse Robin problem is treated in Nakamura and Wang [2015].
2. Problem formulation
We shall consider the following parabolic-elliptic problem posed on two spatial scales x ∈ Ω
and y ∈ Y.

−Aρ ∆ π = f(π,ρ) in Ω
 F x

∂ ρ−D∆ ρ = 0 in Ω×Y
 t y


−D∇ ρ·n = k(π +p −Rρ) on Ω×Γ
y y F R
(1)
−D∇ ρ·n = 0 on Ω×Γ
 y y N

π = 0 at ∂Ω



ρ(t = 0) = ρ (x,y) in Ω×Y,
I
where the parameters, coefficients and the nonlinear function f satisfies the assumptions dis-
cussed below (see Section 2.1). The initial condition for p follows from the coupling between ρ
and p.
A prominent role in this paper is played by the micro-macro Robin transfer coefficient k,
which is selected from the following set
K := {k ∈ L2(Γ ) : 0 < k ≤ k(y) ≤ k¯ for y ∈ Γ }.
R R
2.1. Assumptions.
(A ) The domains Ω,Y have Lipschitz continuous boundaries.
1
(A ) The parameters satisfy A,D,ρ ,R ∈ (0,∞).
2 F
(A ) The initial value ρ ∈ H1(Ω×Y).
3 I
(A ) f(λu,µv) = λαµβf(u,v) where α+β = 1, α,β > 0.
4
WELL-POSEDNESS AND INVERSE ROBIN ESTIMATES 3
(A ) There is a structural constant C∗ > 0 such that
5
(cid:90)
|f(u ,v)−f(u ,v)|2dx ≤ C∗(cid:107)u −v (cid:107)2
1 2 1 2 L2(Ω)
Ω
uniformly in v ∈ L2(Ω;L2(Y)).
(A ) There is a constant C > 0 such that
6
(cid:90)
f(u,v)2dx ≤ C(cid:107)v(cid:107)2 .
L2(Ω;H1(Y))
Ω
(A ) The constant C∗ in (A ) satisfies
7 5
C∗c (Ω) < 1,
P
where c (Ω) is the Poincar´e constant of the domain Ω (see Proposition 2.1 below).
P
Remark 1. Assumptions (A )-(A ) have clear geometrical or physical meanings, while (A )-
1 3 4
(A ) are technical. The assumption (A ) is only used when deriving uniqueness of the weak
7 7
solution to (1). Note also that for some special classes of domains, the Poincar´e constant can
be quantitatively estimated, see e.g. Mikhlin [1981].
2.2. Auxiliary results. In this section, we state some auxiliary results that will be useful in
this context.
Proposition 2.1 (Poincar´e’s inequality). Let Ω ⊂ Rd be a fixed domain and denote by c (Ω)
P
the smallest constant such that
(cid:107)u(cid:107)2 ≤ c (Ω)(cid:107)∇ u(cid:107)2
L2(Ω) P x L2(Ω)
hold for all u ∈ H1(Ω). The constant c (Ω) is called the Poincar´e constant of the domain Ω.
0 P
Proposition 2.2 (Interpolation-trace inequality). Assume that Y ⊂ Rd is a Lipschitz domain
and u ∈ L2(Ω;H1(Y)). For any ρ > 0 we have
(cid:90) (cid:90) (cid:90) (cid:90) (cid:90) (cid:90)
u2dσ dx ≤ ρ |∇ u|2dydx+c |u|2dydx,
y y ρ
Ω ∂Y Ω Y Ω Y
where c ∼ 1/ρ. In particular,
ρ
(cid:107)u(cid:107) ≤ c(cid:107)u(cid:107) .
L2(Ω;L2(∂Y)) L2(Ω;H1(Y))
The next result provides a useful equivalent norm on H1.
Proposition 2.3. Let U ⊂ Rd be a domain and Γ ⊂ ∂Y where Γ has positive (d−1)-dimensional
surface measure. Then there are constants c ,c such that
1 2
(cid:90)
c (cid:107)u(cid:107)2 ≤ u2dσ +(cid:107)∇ u(cid:107)2 ≤ c (cid:107)u(cid:107)2 .
1 H1(U) x L2(Ω) 2 H1(U)
Γ
We shall also need the following two results, the first a Sobolev-type embedding and the
second a simple trace theorem.
Proposition 2.4. Assume that U ⊂ Rd, then Hd(U) ⊂ L∞(U) and
(cid:107)u(cid:107) ≤ C(cid:107)u(cid:107) .
L∞(U) Hd(U)
4 MARTIN LIND, ADRIAN MUNTEAN, AND OMAR RICHARDSON
Proposition 2.5. Assume that U ⊂ Rd and Γ ⊂ ∂U is Lipschitz continuous. Then
Hd+1(U) ⊂ Hd(Γ)
and
(cid:107)u(cid:107) ≤ c(cid:107)u(cid:107) .
Hd(Γ) Hd+1(U)
We have the following existence and regularity results.
Proposition 2.6 (see e.g. Evans [1998]). Consider the problem

∂ v −D∆ v = 0 on Ω×Y
 t y


−D∇ v ·n = k(g −Rv) on Ω×Γ
y y R
(2)
−D∇ v ·n = 0 on Ω×Γ
 y y N

v(t = 0) = v on Ω×Y.
I
If g ∈ L2(0,T;H1(Ω)) and v ∈ H1(Ω;Hm(Y)) (m ∈ N), then the problem (2) has a unique
I
weak solution v ∈ L2(0,T;H1(Ω;Hm+1(Y))).
Proposition 2.7 (see e.g Cazenave [2006]). Let v(t,x,y) be in L2(0,T;H1(Ω;Hm+1(Y))) (m ∈
N) and consider the problem
(cid:40)
−∆ u = f(u,v) on Ω×Y
x
(3)
u(t,x) = 0 on ∂Ω,t > 0.
where the nonlinear function satisfies (A )-(A ). Then the problem (3) has a unique weak
4 6
solution u ∈ L2(0,T;H1(Ω)).
Finally, we state the following two classical compactness results, see e.g. Zeidler [1986].
Theorem 2.8 (Aubin-Lions Theorem Aubin [1963]). Let B (cid:44)→ B (cid:44)→(cid:44)→ B . Suppose that B
0 1 0
is compactly embedded in B and that B is continuously embedded in B . Let
1
(cid:8) (cid:9)
W = u ∈ L2(0,T;B ) : ∂ u ∈ L2(0,T;B ) . (4)
0 t 1
Then the embedding of W into L2(0,T;B) is compact.
Theorem 2.9 (Schauder’s Fixed Point Theorem). Let B be a nonempty, closed, convex,
bounded set and T : B → B a compact operator. Then there exists at least one r ∈ B such that
T(r) = r.
3. Existence and uniquenes of the solution
3.1. Existence of weak solution. Themainresultofthissubsectionisthefollowingtheorem.
Theorem 3.1. Assume that (A )-(A ) hold. Then the problem (1) has at least a weak solution
2 6
solution (π,ρ) ∈ L2(0,T;H1(Ω))×L2(0,T;L2(Ω;H1(Y))).
0
Proof. Weshalldecoupletheproblem. Thefirstsub-problemisasfollows: givenπ ∈ L2(0,T;H1(Ω))
0
and ρ ∈ H1(Ω,H1(Y)), we let ξ be the weak solution to
I

∂ ξ −D∆ ξ = 0 on Ω×Y
 t y


−D∇ ξ ·n = k(π +p −Rξ) on Ω×Γ
y y F R
(5)
−D∇ ξ ·n = 0 on Ω×Γ
 y y N

ξ(t = 0) = ρ on Ω×Y.
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WELL-POSEDNESS AND INVERSE ROBIN ESTIMATES 5
The weak formulation of (5) is: find ξ such that for a.e. t ∈ [0,T] and every ψ ∈ L2(Ω,H1(Y))
there holds
(cid:90) (cid:90) (cid:90) (cid:90) (cid:90) (cid:90)
∂ ξψdydx+ D∇ ξ∇ ψdydx = k(π +p −Rξ)ψdσ dx, (6)
t y y F y
Ω Y Ω Y Ω ΓR
and ξ(t = 0) = ρ . Existence and regularity of ξ is provided by Proposition 2.6 (recall that
I
(A ) states that ρ ∈ H1(Ω×Y)).
3 I
The second sub-problem is: given data ξ, consider the problem
(cid:40)
−∆ π = f(π,ξ) on Ω
x
(7)
π = 0 on ∂Ω.
Let λ > 0 be a free parameter. By the scaling properties of f and uniqueness of weak solution,
we have that if π is the weak solution of (7) with data ξ, then π¯ = λπ is the weak solution to
(7) with data λξ. Hence, if π¯ is the weak solution to
(cid:40)
−∆ π¯ = λβf(π¯,ξ) on Ω
x
(8)
π¯ = 0 on ∂Ω,
then π¯ = λπ, again by the scaling properties of f. The weak form of (8) is as follows: find π¯
such that for a.e. t ∈ [0,T] and all ϕ ∈ H1(Ω), there holds
0
(cid:90) (cid:90)
∇ π ·∇ ϕdx = λβ f(π,ξ)ϕdx.
x x
Ω Ω
Existence and regularity of π¯ is guaranteed by Proposition 2.7.
We shall now use a fixed point argument a` la Schauder (see Theorem 2.9) to show that there
exists a λ > 0 for which the functions of the pair (π¯,ξ) are weak solutions to the sub-problems
(5) and (8). Then we recover (π,ρ), a weak solution to (1), by taking π = π¯/λ and ρ = ξ.
Define the operators
T : L2(0,T;L2(Ω)) → L2(0,T;L2(Ω;L2(Y)))
1
by T (π) = ξ (the weak solution of (5)) and
1
Tλ : L2(0,T;L2(Ω;H1(Y))) → L2(0,T;H1(Ω))
2
by Tλ(ξ) = π¯ (the weak solution of (8)). Finally, consider the operator Aλ on the space
2
L2(0,T;L2(Ω)) into itself defined by
Aλ(π) = Tλ(T (π)) = π¯. (9)
2 1
To obtain existence of solution, we shall prove that the operator Aλ has a fixed point. This
π will then give ξ. The idea of the proof is to first use the Schauder Fixed Point Theorem
(Theorem 2.9 above).
We shall prove that there exist a λ > 0 and a set B such that
(1) Aλ is a compact operator;
(2) B is convex, closed, bounded and satisfies Aλ(B) ⊂ B.
To obtain compactness of Aλ = Tλ ◦T , it is sufficient to demonstrate that T is compact and
2 1 1
that Tλ is continuous. Recall that we have
2
T : L2(0,T;H1(Ω)) → L2(0,T;L2(Ω;L2(Y))).
1
6 MARTIN LIND, ADRIAN MUNTEAN, AND OMAR RICHARDSON
However, since we assume that ξ ∈ H1(Ω,H1(Y)) we get that T (π) = ξ ∈ L2(0,T;H1(Ω×Y))
I 1
and ∂ ξ ∈ L2(0,T;L2(Ω×Y)). Whence,
t
T (L2(0,T;H1(Ω))) ⊂ V,
1
where
V = {u : u ∈ L2(0,T;H1(Ω×Y)), ∂ u ∈ L2(0,T;L2(Ω×Y))}
t
By Theorem 2.8,
V (cid:44)→(cid:44)→ L2(0,T;L2(Ω;L2(Y))).
Thus, for any bounded set M ⊂ L2(0,T;L2(Ω;L2(Y)))×L2(0,T;L2(Ω)), there holds T (M) ⊂
1
V and since V is compactly contained in L2(0,T;L2(Ω;L2(Y)) we have that T (M) is precom-
1
pact in L2(0,T;L2(Ω;L2(Y))). Hence, T is compact.
1
We continue to prove that Tλ is continuous. Assume we have two solutions π¯ = Tλ(ξ ) and
2 1 2 1
π¯ = Tλ(ξ ). Substituting these both in (9) and subtracting, we obtain
2 2 2
(cid:90) (cid:90)
∇ (π¯ −π¯ )·∇ ϕdx = λβ [f(π¯ ,ξ )−f(π¯ ,ξ )]ϕdx,
x 1 2 x 1 1 2 2
Ω Ω
and for ϕ = π¯ −π¯ , we get
1 2
(cid:90)
(cid:107)∇ (π¯ −π¯ )(cid:107)2 = λβ [f(π¯ ,ξ )−f(π¯ ,ξ )][π¯ −π¯ ]dx
x 1 2 L2(Ω) 1 1 2 2 1 2
Ω
(cid:90)
= λβ [f(π¯ ,ξ )−f(π¯ ,ξ )][π¯ −π¯ ]dx
1 1 2 1 1 2
Ω
(cid:90)
+λβ [f(π¯ ,ξ )−f(π¯ ,ξ )][π¯ −π¯ ]dx.
2 1 2 2 1 2
Ω
Using (A ) and (A ), we obtain that
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(cid:107)∇ (π¯ −π¯ )(cid:107)2 ≤ Cλβ(cid:107)π¯ −π¯ (cid:107) (cid:107)ξ −ξ (cid:107) .
x 1 2 L2(Ω) 1 2 L2(Ω) 2 1 L2(Ω;H1(Y))
By the Poincar´e’s inequality, we obtain
(cid:107)∇ (π¯ −π¯ )(cid:107) ≤ Cλβ(cid:107)ξ −ξ (cid:107)
x 1 2 L2(Ω) 1 2 L2(Ω;H1(Y))
and we conclude the mapping Tλ is continuous.
2
Let K > 0 be a fixed number that we specify later and let B be the collection of functions
K
u ∈ L2(0,T;H1(Ω)) such that
max{(cid:107)u(cid:107) ,(cid:107)∇ u(cid:107) } ≤ K.
L2(0,T;L2(Ω)) x L2(0,T;L2(Ω))
For each K > 0, the set
B ⊂ L2(0,T;H1(Ω))
K
is a convex, closed and bounded. We show that we may select K > 0 and λ > 0 such that
Aλ(B ) ⊂ B . (10)
K K
Note that T (B ) is a bounded subset of L2(0,T;L2(Ω;L2(Y))), with a bound depending only
1 K
on K. In other words,
(cid:107)ξ(cid:107) ≤ CK. (11)
L2(0,T;L2(Ω;L2(Y))))
Indeed, this follows from the fact that T is a compact operator.
1
We proceed by observing that we may choose λ > 0 such that if u ∈ B is arbitrary and
K
v = Tλ(T (u)), then
2 1
(cid:8) (cid:9)
max (cid:107)v(cid:107) ,(cid:107)v(cid:107) ≤ K. (12)
L2(0,T;L2(Ω)) L2(0,T;H1(Ω))
WELL-POSEDNESS AND INVERSE ROBIN ESTIMATES 7
Let ξ = T (u) so that v = Tλ(ξ). Testing the weak formulation of (8) with ϕ = u and using
1 2
Cauchy-Schwarz’ inequality and Poincar´e’s inequality, we get
(cid:107)∇ u(cid:107)2 ≤ Cλβ(cid:107)u(cid:107) (cid:107)ξ(cid:107) ≤ CKλβ(cid:107)ξ(cid:107) .
x L2(Ω) L2(Ω) L2(Ω;L2(Y)) L2(Ω;L2(Y))
Integrating over [0,T] and using (11), we obtain after using Poincar´e’s inequality
(cid:107)u(cid:107)2 ≤ C(cid:48)K2λβ.
L2(0,T;H1(Ω))
By taking λ small enough (depending on K), we obtain (12) whence (10) follows. (cid:3)
Remark 2. Instead of using scaling arguments and Schauder’s fixed point theorem, we could
have used alternatively the Schaefer/Leray-Schauder fixed point theorem.
3.2. Uniqueness of weak solutions. We proceed to prove the following uniqueness theorem.
Theorem 3.2. Assume that in addition to the assumptions of Theorem 3.1 the condition (A )
7
also holds. Then the weak solution to (1) is unique.
Proof. The weak formulation of the uncoupled problem is: find (π,ρ) ∈ L2(0,T;H1(Ω)) ×
0
L2(0,T;L2(Ω;H1(Y))) where ρ(t = 0) = ρ and for a.e. t ∈ [0,T] the equations
I
(cid:90) (cid:90)
Aρ ∇ π ·∇ ϕdx = f(π,ρ)ϕdx (13)
F x x
Ω Ω
and
(cid:90) (cid:90) (cid:90) (cid:90) (cid:90) (cid:90)
∂ ρψdydx+ D∇ ρ·∇ ψdydx = k(π +p −Rρ)ψdσ dx (14)
t y y F y
Ω Y Ω Y Ω ΓR
hold for all ϕ ∈ H1(Ω) and all ψ ∈ L2(Ω;H1(Y)).
0
Assumethattwopairsofsolutionsexist: (π ,ρ )and(π ,ρ ). Letq := π −π andz := ρ −ρ .
1 1 2 2 1 2 1 2
If we substitute the two solutions in (13) and (14) and subtract, we obtain that
(cid:90) (cid:90)
Aρ ∇ q ·∇ ϕdx = (f(π ,ρ )−f(π ,ρ ))ϕdx (15)
F x x 1 1 2 2
Ω Ω
and
(cid:90) (cid:90) (cid:90) (cid:90) (cid:90) (cid:90)
∂ zψdydx+ D∇ z ·∇ ψdydx = k(q −Rz)ψdσ dx (16)
t y y y
Ω Y Ω Y Ω ΓR
for all ϕ ∈ H1(Ω) and all ψ ∈ L2(Ω;H1(Y)).
0
Choosing specific test function ϕ = q, using Young’s inequality with parameter ε > 0 and
1
(A ), we obtain from (15) the first key estimate
5
Aρ (cid:107)∇ q(cid:107)2 ≤ (C∗ +ε )(cid:107)q(cid:107)2 +c (cid:107)z(cid:107)2 . (17)
F x L2(Ω) 1 L2(Ω) ε1 L2(Ω,L2(Y))
We focus on (16), which, using test function ψ = z, yields
(cid:90) (cid:90)
1 d
||z||2 +D||∇ z||2 = k(q −Rz)z. (18)
2dt L2(Ω;L2(Y)) y L2(Ω;L2(Y))
Ω ΓR
¯
Now, we estimate the right hand side of (18) by using trace inequality and the fact that k ≤ k
on Γ . We have
R
(cid:90) (cid:90) (cid:90) (cid:90) (cid:90) (cid:90)
|k(q −Rz)z|dσ dx ≤ k¯ |qz|dσ dx+Rk¯ z2dσ dx
y y y
Ω ΓR Ω ΓR Ω ΓR
¯ (cid:18) (cid:19) (cid:90) (cid:90)
k|Γ | 1
≤ R (cid:107)q(cid:107)2 + R+ k¯ z2dσ dx.
2 L2(Ω) 2 y
Ω ΓR
8 MARTIN LIND, ADRIAN MUNTEAN, AND OMAR RICHARDSON
The second term at the right-hand side of the previous inequality can be estimated by using
the trace inequality and Young’s inequality with parameter ε > 0:
(cid:90) (cid:90)
c
z2dσ dx ≤ ε(cid:107)∇ z(cid:107)2 + 0(cid:107)z(cid:107)2
y y L2(Ω;L2(Y)) ε L2(Ω;L2(Y))
Ω ΓR
for some absolute constant c > 0. Using the previous estimates and rearranging (18), we
0
obtain
1 d
||z||2 +(D−ε)||∇ z||2
2dt L2(Ω;L2(Y)) y L2(Ω;L2(Y))
¯ ¯
k|Γ | c k(R+1/2)
≤ R (cid:107)q(cid:107)2 + 0 (cid:107)z(cid:107)2 . (19)
2 L2(Ω) ε L2(Ω;L2(Y))
By Poincare’s inequality, we have
(cid:107)q(cid:107)2 ≤ c (Ω)(cid:107)∇ q(cid:107)2 ,
L2(Ω) P x L2(Ω)
where c (Ω) is the Poincar´e constant of the domain Ω. Using this in (17), we obtain
P
Aρ (cid:107)∇ q(cid:107)2 ≤ (C∗ +ε )c (Ω)(cid:107)∇ q(cid:107)2 +C(cid:107)z(cid:107)2 .
F x L2(Ω) 1 P x L2(Ω) L2(Ω,L2(Y))
By (A ), we may take ε > 0 small enough such that (C∗ +ε )c (Ω) < Aρ . Then we obtain
7 1 1 P F
Cc (Ω)
(cid:107)q(cid:107)2 ≤ c (Ω)(cid:107)∇ q(cid:107)2 ≤ P (cid:107)z(cid:107)2 . (20)
L2(Ω) P x L2(Ω) Aρ −(C∗ +ε )c (Ω) L2(Ω,L2(Y))
F 1 P
Whence, it follows from the previous estimate and (19) with ε = D/2 that
1 d D
||z||2 + ||∇ z||2 ≤ C(cid:107)z(cid:107)2
2dt L2(Ω;L2(Y)) 2 y L2(Ω;L2(Y)) L2(Ω;L2(Y))
By using Gro¨nwall’s inequality and the fact that z(0,x,y) = 0, it follows that z = 0. From
(20), we obtain q = 0 as well. This demonstrates the uniqueness. (cid:3)
4. Energy and stability estimates
We start this section by stating the following energy estimates for our problem.
Proposition 4.1. Assume (A )-(A ) and let (u,v) be a weak solution to
2 6

−∆ u = f(u,v) in Ω
 x

∂ v −D∆ v = 0 in Ω×Y
 t y


−D∇ v ·n +k(Rv −u) = g on Ω×Γ
y y R
(21)
−D∇ v ·n = 0 on Ω×Γ
 y y N

u = 0 at ∂Ω



v(t = 0) = v in Ω×Y.
I
Then the following energy estimate hold
(cid:107)u(cid:107)2 +(cid:107)v(cid:107)2
L2(0,T;H1(Ω)) L2(0,T;L2(Ω,H1(Y)))
(cid:16) (cid:17)
≤ C (cid:107)g(cid:107)2 +(cid:107)v (cid:107)2 . (22)
L2(0,T;L2(Ω;L2(ΓR))) I L2(Ω;L2(Y))
The proof of Proposition 4.1 follows by similar arguments as the proof of Theorem 4.2 below,
therefore we omit it.
Weproceedtostudythestabilityofsolutionswithrespecttosomeoftheparametersinvolved.
Some preliminary remarks:
WELL-POSEDNESS AND INVERSE ROBIN ESTIMATES 9
• We do not need to study the stability of the solution with respect to ρ ,p and R.
F F
Recall that R is an universal physical constant, while ρ ,p fix the type of fluid and
F F
gas we are considering.
• We could investigate the stability of (π,ρ) with respect to structural changes into the
non-linearity f(·,·). We omit to do so mainly because our main intent lies in under-
standing the role of the micro-macro Robin coefficient k.
• For this stability proof, we decide to use a direct method which relies essentially on
energy estimates; see e.g. Muntean [2009].
Fori ∈ {1,2},let(π ,ρ )betwoweaksolutionscorrespondingtothesetsofdata(ρ ,A ,D ,k ),
i i Ii i i i
where ρ ,A ,D ,k denote the initial data, diffusion coefficients and mass-transfer coefficients
Ii i i i
of the solution (π ,ρ ). Denote
i i
δu := u −u where u ∈ {π,ρ,ρ ,A,D,k}.
2 1 I
Theorem 4.2. Assume that for i = 1,2, (A ,D ) belongs to a fixed compact subset of R2,
i i
that k ∈ K and that (cid:107)ρ (cid:107) ≤ C. Let (π ,ρ ) (i = 1,2) be weak solutions to (1)
i Ii L2(Ω;L2(Y)) i i
corresponding to the choices of data above. Then the estimate
(cid:107)δπ(cid:107)2 +(cid:107)δρ(cid:107)2
L2(0,T;H1(Ω)) L2(0,T;L2(Ω;H1(Y)))
0
(cid:16) (cid:17)
≤ c (cid:107)δk(cid:107)2 +|δA|+|δD|+(cid:107)δρ (cid:107)2 , (23)
L2(ΓR) I L2(Ω;L2(Y))
holds
Proof. We have for i = 1,2
(cid:90) (cid:90)
A ρ ∇ π ·∇ ϕdx = f(π ,ρ )ϕdx
i F x i x i i
Ω Ω
and
(cid:90) (cid:90) (cid:90) (cid:90) (cid:90) (cid:90)
∂ ρ ψdydx+ D ∇ ρ ·∇ ψdydx = k (π +p −Rρ )ψdσ dx,
t i i y i y i i F i y
Ω Y Ω Y Ω ΓR
for all ϕ ∈ H1(Ω) and ψ ∈ L2(Ω;H1(Y)).
0
Subtracting the corresponding equations and then testing with ϕ := π −π and ψ := ρ −ρ
2 1 2 1
gives:
(cid:18) (cid:90) (cid:90) (cid:19) (cid:90)
ρ A ∇ π ·∇ ϕdx−A ∇ π ·∇ ϕdx = (f(π ,ρ )−f(π ,ρ ))ϕdx, (24)
F 2 x 2 x 1 x 1 x 2 2 1 1
Ω Ω Ω
and
(cid:90) (cid:90)
d
(cid:107)ψ(cid:107)2 + (D ∇ ρ −D ∇ ρ )·∇ ψdydx
2dt L2(Ω;L2(Y)) 2 y 2 1 y 1 y
Ω Y (25)
(cid:90) (cid:90)
= (k (π +p −Rρ )−k (π +p −Rρ ))ψdσ dx.
2 2 F 2 1 1 F 1 y
Ω ΓR
Regarding (24), note that
(cid:90) (cid:90)
A ∇ π ·∇ ϕdx−A ∇ π ·∇ ϕdx
2 x 2 x 1 x 1 x
Ω Ω
(cid:90)
= A (cid:107)∇ ϕ(cid:107)2 +(A −A ) ∇ π ·∇ ϕdx
2 x L2(Ω) 2 1 x 1 x
10 MARTIN LIND, ADRIAN MUNTEAN, AND OMAR RICHARDSON
Using (A ) and (A ), we may estimate the right-hand side of (24) and obtain
5 6
(cid:18) (cid:90) (cid:19)
A ρ (cid:107)∇ ϕ(cid:107)2 ≤ C∗(cid:107)ϕ(cid:107)2 +c (cid:107)ψ(cid:107)2 +|δA| |∇ π ||∇ ϕ|dx .
2 F x L2(Ω) L2(Ω) L2(Ω;L2(Y)) x 1 x
Ω
Using Poincar´e’s inequality, assumptions on f and Young’s inequality with parameter ε > 0,
we get
(cid:18) (cid:90) (cid:19)
(cid:107)∇ ϕ(cid:107)2 ≤ c (cid:107)ψ(cid:107)2 +|δA| |∇ π ||∇ ϕ|dx
x L2(Ω) L2(Ω;L2(Y)) x 1 x
Ω
(cid:16) (cid:17)
≤ cε(cid:107)∇ ϕ(cid:107)2 +c (cid:107)ψ(cid:107)2 +|δA|(cid:107)∇ π (cid:107)2 .
x L2(Ω) L2(Ω;L2(Y)) x 1 L2(Ω)
Choosing ε = 1/(2c), rearranging and using energy estimates for π , we obtain
1
(cid:16) (cid:17)
(cid:107)∇ ϕ(cid:107)2 ≤ c (cid:107)ψ(cid:107)2 +|δA| . (26)
x L2(Ω) L2(Ω;L2(Y))
We proceed to estimate (cid:107)ψ(cid:107)2 , using (25). Note that
L2(Ω;L2(Y))
(cid:90) (cid:90)
(D ∇ ρ −D ∇ ρ )·∇ ψdydx
2 y 2 1 y 1 y
Ω Y
(cid:90) (cid:90)
= (D −D ) ∇ ρ ·∇ ψ +D (cid:107)∇ ψ(cid:107)2 .
2 1 y 2 y 1 y L2(Ω;L2(Y))
Ω Y
Hence, it follows that
(cid:90) (cid:90)
d
(cid:107)ψ(cid:107)2 +D (cid:107)∇ ψ(cid:107)2 ≤ |δD| |∇ ξ ||∇ ψ(cid:107)dydx
2dt L2(Ω;L2(Y)) 1 y L2(Ω;L2(Y)) y 1 y
Ω Y
(cid:90) (cid:90)
+ |(k −k )ψ|+|(k π −k π )ψ|+|R(k ρ −k ρ )ψ|dσ dx. (27)
2 1 2 2 1 1 2 2 1 1 y
Ω ΓR
We have
(cid:90) (cid:90) (cid:90) (cid:90)
|k −k ||ψ|dσ dx ≤ ε ψ2dσ dx+c |Ω|(cid:107)k −k (cid:107)2
2 1 y y ε 2 1 L2(ΓR)
Ω ΓR Ω ΓR
(cid:16) (cid:17)
≤ cε (cid:107)ψ(cid:107)2 +(cid:107)∇ ψ(cid:107)2 +c(cid:107)k −k (cid:107)2 .
L2(Ω;L2(Y)) y L2(Ω;L2(Y)) 2 1 L2(ΓR)
Further,
(cid:90) (cid:90) (cid:90) (cid:90)
|(k π −k π )ψ|dσ dx ≤ c(cid:107)k −k (cid:107)2 + |k ||π −π ||ψ|dσ dx
2 2 1 1 y 2 1 L2(ΓR) 1 2 1 y
Ω ΓR Ω ΓR
(cid:90) (cid:90)
≤ c(cid:107)k −k (cid:107)2 +εk¯|Γ |(cid:107)π −π (cid:107)2 +c ψ2dσ dx
2 1 L2(ΓR) R 2 1 L2(Ω) ε y
Ω ΓR
(cid:16) (cid:17)
≤ c(cid:107)k −k (cid:107)2 +ε (cid:107)∇ ϕ(cid:107)2 +(cid:107)∇ ψ(cid:107)2 +c (cid:107)ψ(cid:107)2 .
2 1 L2(ΓR) x L2(Ω) y L2(Ω;L2(Y)) ε L2(Ω;L2(Y))
Finally,
(cid:90) (cid:90) (cid:90) (cid:90)
R |(k ρ −k ρ )ψ|dσ dx ≤ R k2ψ2 +(k −k )2ρ2dσ dx.
2 2 1 1 y 2 2 1 1 y
Ω ΓR Ω ΓR
We assume that for all y ∈ Γ , we have
R
(cid:90)
ρ2dx ≤ K,
1
Ω