AND THE NAVIER-STOKES EQUATIONS IN THE PLANE
RU-YU LAI, GUNTHER UHLMANN, AND JENN-NAN WANG
Abstract. In this paper, we prove in two dimensions global identifiability of the viscosity in an incompressible fluid by making boundary measurements. The main contribution of this work is to use more natural boundary measurements, the Cauchy forces, than the Dirichlet-to-Neumann map previously considered in [7] to prove the uniqueness of the viscosity for the Stokes equations and for the Navier-Stokes equations.
Contents
1. Introduction 1
2. Equivalence of boundary data for the plate and for the 2D Stokes
equations 3
3. Global uniqueness for the Stokes equations 6
3.1. (∂2
z, ∂z2) system 7
3.2. ∂z system 10
3.3. Proof of the uniqueness result 13
4. Global uniqueness for the stationary Navier-Stokes equations 15
References 17
1. Introduction
Let Ω be a simply connected bounded domain in R2 with smooth boundary.
Assume that Ω is filled with an incompressible fluid. Let u = (u1, u2)T be the
velocity vector field satisfying the Stokes equations div σ(u, p) = 0 in Ω,
div u = 0 in Ω, (1.1)
where σ(u, p) = 2µε − pI2 is the stress tensor and ε = ((∇u) + (∇u)T)/2, µ is the
viscosity and p is the pressure. Here the notation I2 is the 2 × 2 identity matrix.
Physically, zero viscosity is observed only in superfluids the have the ability to self-propel and travel in a way that defies the forces of gravity and surface tension. Otherwise all fluids have positive viscosities. Thus, we can assume that µ > 0 in Ω. The second equation of (1.1) is the incompressibility condition. Because of the
The first and second author were supported in part by the National Science Foundation. The second author was also supported by a Simons Fellowship.
The third author was supported in part by the National Science Council of Taiwan. 1
conservation of mass, the incompressibility condition is equivalent to the material derivative of the density function ρ to be zero, that is,
Dρ Dt =:
∂ρ
∂t + u · ∇ρ = 0. (1.2)
When ρ is constant, (1.2) is satisfied. The above equation also holds for nonconstant density functions. We can conclude that a nonconstant viscosity µ is possible since the viscosity function is a function of density. A fluid with nonconstant viscosity is called a non-Newtonian fluid which is relatively common, such as blood, shampoo and custard.
Let g ∈ H3/2(∂Ω) satisfy the compatibility condition
Z
∂Ω
g · ndS = 0, (1.3)
where n is the unit outer normal to ∂Ω. This condition leads to the uniqueness of (1.1), that is, there exists a unique solution (u, p) ∈ H2(Ω) × H1(Ω) (p is unique up to a constant) of (1.1) and u|∂Ω = g. We could define the Cauchy data for the
Stokes equations (1.1) by
Cµ= {(u, σ(u, p)n) |∂Ω: (u, p) satisfies (1.1)} .
The inverse problems we consider in this paper is to determine µ from the knowledge of the Cauchy data Cµ. Recently, Imanuvilov and Yamamoto [7] studied the same
inverse problem with the Dirichlet-to-Neumann (DN) map defined by Λµ(g) = ∂u ∂ν, p ∂Ω .
They showed that the knowledge of the DN map uniquely determines the viscosity µ of the Navier-Stokes equations. Unlike their boundary measurements, we use Cauchy data (u, σ(u, p)n) |∂Ω to deduce the uniqueness of µ. The physical sense of
σ(u, p)n is the stress acting on ∂Ω and is called the Cauchy force. Mathematically, the pressure function p plays the role of the Lagrange multiplier corresponding to the incompressibility condition. The information of p on ∂Ω is coupled in the Cauchy force. Given the measurement of p alone on ∂Ω is unnatural.
The main result of this paper is the following global uniqueness result. Note that the following theorem also holds for the Navier-Stokes equations.
Theorem 1.1. Let Ω be a simply connected bounded domain in R2 with smooth boundary. Suppose that µ1 and µ2 are two viscosity functions for the Stokes
equa-tions. Assume that µj ∈ C3(Ω) and µj > 0 with
∂αµ1|∂Ω= ∂αµ2|∂Ω for all |α| ≤ 1.
Let Cµ1 and Cµ2 be the Cauchy data associated with µ1 and µ2, respectively. If
Cµ1 = Cµ2, then µ1 = µ2 in Ω.
In higher dimensions, the global uniqueness of identifying the viscosity using the Cauchy data has been well studied. For the Stokes equations, the uniqueness for the inverse boundary problem was established by Heck, Li and Wang [5] in dimension
three. In [11], Li and Wang proved the unique determination of µ for the Navier-Stokes equations in dimension three. To study the Navier-Navier-Stokes equations they applied the linearization technique due to Isakov [8]. The idea is to reduce the semilinear inverse boundary value problem to the corresponding linear one. When applying the linearization method to the Navier-Stokes equation, the difficulty is to show the existence of particular solutions to the Navier-Stokes equations with certain controlled asymptotic properties. The idea used in [11] is independent of the spatial dimension. It works for the two-dimensional case as well. We will briefly describe the result in Section 4.
Our first strategy for proving Theorem 1.1 is to show that the inverse boundary value problem for the 2D Stokes equations and that for the thin plate-like are equiv-alent. The equivalence is known to hold for the 2D isotropic elastic equation and the thin plate equation. Recently, Kang, Milton and Wang [9] gave explicit formulas showing that the Cauchy data of the elasticity system determines the Cauchy data of the thin plate equations, and vice versa (see also [6]). Since the Stokes equa-tions can be viewed as an elasticity system with incompressibility, we can prove a similar equivalence by using the similar arguments in [9]. Having established the equivalence of two inverse boundary value problems, we then transform the thin plate equations into a first order system. Albin, Guillarmou, Tzou and Uhlmann [1] showed that the Cauchy data of the first order system D + V uniquely determine V if V is diagonal, where D is an operator with ∂ or ¯∂ at its diagonal. When V is not diagonal, they reduced it to the diagonal case so that the similar result holds for the non-diagonal one. For the Stokes equation, the potential V contains the function µ up to the second order derivative, we apply their result and the assumption on the boundary of µ to deduce the global uniqueness.
The paper is organized as follows. We show the equivalence of the inverse bound-ary problems for the thin plate-like and for the Stokes equations in Section 2. In Section 3, we derive a first order system from the thin plate-like equations. We then show that the Cauchy data of the first order system uniquely determines the viscosity µ. In Section 4, we study the same inverse problem for the Navier-Stokes equations.
2. Equivalence of boundary data for the plate and for the 2D Stokes equations
In this section we would like to connect the inverse boundary value problem for the thin plate equations to that for the Stokes equations. We define the 4-th order tensor R by
RM = RT⊥M R⊥
for any 2 × 2 matrix M , where
R⊥= 0 1 −1 0 .
Hereafter, for any function u, the notation u,j means the derivative of u with respect
equivalent to
σ11,1+ σ12,2= 0, σ21,1+ σ22,2 = 0.
It follows that there exist potentials ψ1 and ψ2 such that
σ11 = ψ1,2, σ12 = −ψ1,1, σ21 = ψ2,2, σ22 = −ψ2,1.
(2.1)
Since σ12= σ21, we have
ψ1,1+ ψ2,2 = 0.
Thus there exists a potential φ such that
ψ1 = φ,2, ψ2 = −φ,1.
(2.2)
The potential φ is called the Airy stress function. Substituting (2.2) into (2.1), we see that σ = R∇2φ = φ,22 −φ,12 −φ,21 φ,11 , (2.3)
where ∇2φ denotes the Hessian of φ, i.e.,
∇2φ = φ,11 φ,12 φ,21 φ,22
. In light of σ = 2µε − pI2 and (2.3), we get
ε = 1 2µ R∇
2
φ + pI2 .
(2.4)
The divergence-free condition div u = 0 implies that 0 = u1,1+ u2,2= Trace(ε) = 1 2µ(∆φ + 2p) , thus p = −∆φ 2 . (2.5)
Note that the physical significance of the pressure p is that −p is the mean of the two normal stresses at a point, that is,
p = −1
2(σ11+ σ22). From (2.4) and (2.5), it follows that
0 = div div(Rε) = div div 1
2µR R∇ 2φ + pI 2 = div div 1 2µ ∇2φ − ∆φ 2 I2 = div div 1 2µ 1 2(φ,11− φ,22) φ,12 φ,12 12(φ,22− φ,11) . (2.6)
Conversely, if div div(Rε) = 0, then
where ε = (εij). If Ω is simply connected, then there exists a function u such that
ε = (∇u) + (∇u)T /2. (For the proof of the existence of such function u, we refer
to [4], page 99-103). Based on (2.6), the function u also satisfies div u = 0. Let p = −∆φ/2, then (u, p) satisfies the Stokes equations (1.1). Thus we have proved that the two systems (1.1) and (2.6) are equivalent if Ω is simply connected.
Next we would like to discuss the equivalence of the Cauchy data. We define the operator Pµ(φ) by Pµ(φ) := div div 1 2µ 1 2(φ,11− φ,22) φ,12 φ,12 12(φ,22− φ,11)
and denote u,n = ∇u · n and u,t = ∇u · t, where n = (n1, n2) is the unit normal
and t = (−n2, n1) = R⊥Tn is the unit tangent vector field along ∂Ω in the positive
orientation. The Dirichlet data associated with (2.6) is described by the pair {φ, φ,n}
and the Neumann data by the pair Mn:= n · 1 2µ ∇2φ − ∆φ 2 I2 n, (Mt),t:= div 1 2µ ∇2φ − ∆φ 2 I2 · n + t · 1 2µ ∇2φ − ∆φ 2 I2 n ,t .
We define the Cauchy data for (2.6) by
Cµ∗ =(φ, φ,n, Mn, (Mt),t) |∂Ω: φ ∈ H4(Ω), Pµ(φ) = 0 .
We now adopt the arguments used in [9] to show that σn|∂Ωdetermines {φ, φ,n} on
∂Ω and u|∂Ω determines {Mn, (Mt),t} on ∂Ω, and vice versa. Therefore, the Cauchy
data Cµ for the Stokes equations and the Cauchy data Cµ∗ for (2.6) are equivalent.
Assume for the moment that u ∈ C2+α(Ω) for some α ∈ (0, 1). It follows from (2.3) that RT⊥σn = (∇2φ)t = ∇φ,1· t ∇φ,2· t . (2.7)
For j = 1, 2, we integrate ∇φ,j· t along ∂Ω from some point x0 ∈ ∂Ω, we recover ∇φ
(up to a constant) on ∂Ω. Hence φ,n and φ,t are recovered. We integrate φ,t along
∂Ω, then φ on ∂Ω is known (also up to a constant). The appearance of integrating constants is evident from (2.6). In other words, the traction σn uniquely determines the Dirichlet data φ and φ,n. On the other hand, if φ and φ,n are given, then ∇φ is
known. Hence, the boundary traction σn is recovered via (2.7).
To show that Mnand (Mt),tcan be recovered from u. Since ε = ((∇u)+(∇u)T)/2,
we get that Rε = RT⊥εR⊥ = u2,2 −12(u2,1+ u1,2) −1 2(u2,1+ u1,2) u1,1
and thus div (Rε) = u2,21−12(u2,12+ u1,22) −1 2(u2,11+ u1,21) + u1,12 = 1 2 u2,12− u1,22 u1,12− u2,11 = 1 2R T ⊥∇ (u1,2− u2,1) . Consequently, we obtain div (Rε) · n = 1 2(R⊥n) · ∇ (u1,2− u2,1) = − 1 2t · ∇ (u1,2− u2,1) = 1 2(u2,1− u1,2),t.
Recall that (Mt),t = div(Rε) · n + (t · (Rε)n),t and therefore
(Mt),t =
1
2(u2,1− u1,2),t+ (t · (Rε)n),t.
(2.8)
Integrating (2.8) along ∂Ω from some point x0 ∈ ∂Ω and choosing an appropriate
(u2,1− u1,2)(x0), we obtain Mt= 1 2(u2,1− u1,2) + t · (Rε)n. (2.9) We observe that 1 2(u2,1− u1,2) = RT⊥ 0 12(u1,2− u2,1) 1 2(u2,1− u1,2) 0 R⊥n · t. The second term on the right side of (2.9) can be written as
t · (Rε)n = RT⊥ u1,1 12(u1,2+ u2,1) 1 2(u1,2+ u2,1) u2,2 R⊥n · t. Thus we have Mt= R⊥T u1,1 u1,2 u2,1 u2,2 R⊥n · t = −RT⊥(∇u)t · t = −n · (∇u)t. (2.10)
Moreover, using the definition of Mn, we get
Mn= n · Rεn = n · RT⊥εR⊥n = t · (∇u)t.
(2.11)
From (2.10) and (2.11), we deduce that
u,t = −Mtn + Mnt,
(2.12)
which implies the Neumann data Mn and Mt can be recovered from u,t. On the
other hand, we use the formula (2.12) and integrate −Mtn + Mnt along ∂Ω. Thus,
the velocity field u is determined.
By a density argument the above discussion holds for the slightly relaxed regu-larity assumption on the boundary data g ∈ H3/2(∂Ω). Hence, we can remove the assumption that u ∈ C2+α(Ω). We therefore conclude that knowing the Cauchy data
of the Stokes equations is equivalent to knowing that of the thin plate-like equations (2.6).
3. Global uniqueness for the Stokes equations
From the previous section, we have concluded that to study the inverse boundary value for the Stokes equations (1.1), it suffices to consider the same question for the plate-like equation (2.6). Our strategy now is to deduce a first order system DU + V U = 0 from (2.6). The most nontrivial property that we will show is that Cµ∗ determines the Cauchy data of the first order system DU + V U = 0. Having obtained this result, the global identifiability of µ for the Stokes equations is reduced to the uniqueness problem for this first order system. The global uniqueness of the inverse boundary value problem for such a first order system was recently studied by Albin, Guillarmou, Tzou and Uhlmann in [1]. Consequently, the proof of the uniqueness question for the Stokes equations follows from their result.
3.1. (∂2
z, ∂z2) system. As usual, we define z = x + iy,
∂z = 1 2 ∂ ∂x − i ∂ ∂y , ∂z = 1 2 ∂ ∂x + i ∂ ∂y . The complex version of Gauss integral formulas are given by
Z Ω ∂zw(z)dxdy = 1 2i Z ∂Ω w(z)dz, Z Ω ∂zw(z)dxdy = − 1 2i Z ∂Ω w(z)dz (3.1)
for w ∈ C1(Ω) ∩ C(Ω) lead to the Cauchy Pompeiu representations
w(z) = 1 2πi Z ∂Ω w(ζ) dζ ζ − z − 1 π Z Ω ∂ζw(ζ)dξdη ζ − z, z ∈ Ω, (3.2) w(z) = − 1 2πi Z ∂Ω w(ζ) dζ ζ − z − 1 π Z Ω ∂ζw(ζ) dξdη ζ − z, z ∈ Ω, (3.3)
where ζ = ξ + iη. Iterations of these formulas give the following higher order representations w(z) = 1 2πi Z ∂Ω w(ζ) dζ ζ − z − 1 2πi Z ∂Ω ∂ζw(ζ)ζ − z ζ − zdζ + 1 π Z Ω ∂ζ2w(ζ)ζ − z ζ − zdξdη (3.4) and w(z) = − 1 2πi Z ∂Ω w(ζ) dζ ζ − z + 1 2πi Z ∂Ω ∂ζw(ζ) ζ − z ζ − zdζ + 1 π Z Ω ∂ζ2w(ζ)ζ − z ζ − zdξdη. (3.5)
for w ∈ C2(Ω) ∩ C1(Ω) (see [2, Page 272]). In the sequel, we need a technical lemma.
Lemma 3.1. Let Ω be an open bounded domain in C and f ∈ Ck(Ω) for k ≥ 2. Define u(z) = 1 π Z Ω f (ζ)z − ζ z − ζdξdη
Then u(z) is in Ck(Ω) and satisfies (3.6) ∂z2u(z) = f (z) in Ω. Likewise, if we define u(z) = 1 π Z Ω f (ζ)z − ζ z − ζdξdη, then u(z) is in Ck(Ω) and satisfies
∂z2u(z) = f (z) in Ω.
Proof. We adopt the proof of a similar result in [3, Theorem 2.1.2] to our case here. We only prove the first part of the lemma, the other part is treated similarly. We first consider f ∈ Ck
0(C). Changing variable ζ
0 = z − ζ in u and differentiation under
the integral sign implies that u ∈ Ck(C). To verify (3.6), we apply Gauss integral
formula twice and (3.4) (note that f is compactly supported). We get ∂z2u(z) = 1 π Z C f (ζ)∂ζ2z − ζ z − ζdξdη = 1 π Z C ∂ζ2f (ζ)z − ζ z − ζdξdη = f (z).
For the general situation, let z0 ∈ Ω and χ ∈ C0∞(C), 0 ≤ χ ≤ 1, χ = 1 in some
neighborhood V of z0 and suppχ ⊂ Ω. Thus,
u(z) = 1 π Z Ω f (ζ)z − ζ z − ζdξdη = 1 π Z Ω χf (ζ)z − ζ z − ζdξdη + 1 π Z Ω (1 − χ(ζ))f (ζ)z − ζ z − ζdξdη =: u1(z) + u2(z). Since ∂2
zu2 = 0 in V , from the previous argument for Ω = C, we have
∂z2u(z) = ∂z2u1(z) + ∂z2u2(z) = χ(z)f (z) = f (z)
for z ∈ V .
Lemma 3.2. Let Ω be an open bounded domain with smooth boundary ∂Ω. Suppose that f, g ∈ C2(Ω) ∩ C1(Ω). Suppose that the compatibility condition
(3.7) ∂z2f = ∂z2g in Ω
is satisfied. Then there exists a function w ∈ C2(Ω) satisfies
∂2
zw = f in Ω,
∂z2w = g in Ω. (3.8)
Proof. Let us make an ansatz w(z) = 1 π Z Ω g(ζ)z − ζ z − ζdξdη + 1 π Z Ω f (ζ)z − ζ z − ζdξdη − 1 π2 Z Ω Z Ω ∂λ2f (λ)ζ − λ ζ − λdsdt z − ζ z − ζdξdη + φ1(z) + φ2(z), where φ1(z) = − 1 (2πi)2 Z ∂Ω Z ∂Ω ∂λf (λ)ζ − λ ζ − λdλ (z − ζ) log(z − ζ)dζ − 1 (2πi)2 Z ∂Ω Z ∂Ω f (λ) 1 ζ − λdλ (z − ζ) log(z − ζ)dζ − 1 (2πi)2 Z ∂Ω Z ∂Ω ∂λf (λ) 1 ζ − λdλ |z − ζ|2log(z − ζ)dζ and φ2(z) = − 1 2πi Z ∂Ω ∂ζg(ζ)|z − ζ|2log(z − ζ)dζ − 1 2πi Z ∂Ω g(ζ)(z − ζ) log(z − ζ)dζ.
Here we take the principal value for the log. Since z − ζ does not vanish for all z ∈ Ω and ζ ∈ ∂Ω, h(z, ζ) = log(z − ζ) is well-defined on Ω × D where D = {ζ ∈ ∂Ω, 0 < arg(z − ζ) < 2π}. Moreover, for fixed ζ ∈ ∂Ω, the function h(z, ζ) is holomorphic in Ω. We can interchange the differentiation and the integral sign see Chapter 8 in [10] and get
∂z2φ1(z) = 0, ∂z2φ2(z) = 0 in Ω.
On the other hand, we can compute that
∂z2φ1(z) = 1 (2πi)2 Z ∂Ω Z ∂Ω ∂λf (λ)ζ − λ ζ − λdλ ∂ζ z − ζ z − ζ dζ + 1 (2πi)2 Z ∂Ω Z ∂Ω f (λ) 1 ζ − λdλ ∂ζ z − ζ z − ζ dζ − 1 (2πi)2 Z ∂Ω Z ∂Ω ∂λf (λ) 1 ζ − λdλ z − ζ z − ζdζ, and ∂z2φ2(z) = 1 2πi Z ∂Ω g(ζ)∂ζ z − ζ z − ζ dζ − 1 2πi Z ∂Ω ∂ζg(ζ) z − ζ z − ζdζ.
Using the compatibility condition (3.7), Lemma 3.1, and Gauss’s formula (3.1) twice, we can see that
∂z2 1 π2 Z Ω Z Ω ∂λ2f (λ)ζ − λ ζ − λdsdt z − ζ z − ζdξdη = ∂z2 1 π2 Z Ω Z Ω ∂λ2g(λ)ζ − λ ζ − λdsdt z − ζ z − ζdξdη = 1 π Z Ω ∂ζ2g(ζ)z − ζ z − ζdξdη = 1 π Z Ω g(ζ)∂ζ2 z − ζ z − ζ dξdη + 1 2πi Z ∂Ω g(ζ)∂ζ z − ζ z − ζ dζ − 1 2πi Z ∂Ω ∂ζg(ζ) z − ζ z − ζdζ.
By the above relation and the ansatz, we then deduce ∂z2w(z) = 1 π Z Ω g(ζ)∂z2 z − ζ z − ζ dξdη + f (z) − ∂2 z 1 π2 Z Ω Z Ω ∂λ2g(λ)ζ − λ ζ − λdsdt z − ζ z − ζdξdη + ∂z2φ1(z) + ∂z2φ2(z) = f (z).
On the other hand, from (3.4), we have that Z Ω ∂λ2f (λ)ζ − λ ζ − λdsdt = f (ζ)π + 1 2i Z ∂Ω f (λ) 1 ζ − λ dλ + 1 2i Z ∂Ω ∂λf (λ)ζ − λ ζ − λdλ, which implies that
h(z) := ∂z2 1 π2 Z Ω Z Ω ∂λ2f (λ)ζ − λ ζ − λdsdt z − ζ z − ζdξdη = 1 π Z Ω f (ζ)∂z2 z − ζ z − ζ dξdη + 1 2π2i Z Ω Z ∂Ω f (λ) 1 ζ − λ dλ + Z ∂Ω ∂λf (λ)ζ − λ ζ − λdλ ∂2z z − ζ z − ζ dξdη. Applying (3.1) twice yields
h(z) = 1 π Z Ω f (ζ)∂z2 z − ζ z − ζ dξdη + 1 (2πi)2 Z ∂Ω Z ∂Ω f (λ) 1 ζ − λ dλ ∂ζ z − ζ z − ζ dζ + 1 (2πi)2 Z ∂Ω Z ∂Ω ∂λf (λ)ζ − λ ζ − λdλ ∂ζ z − ζ z − ζ dζ − 1 (2πi)2 Z ∂Ω Z ∂Ω ∂λf (λ) 1 ζ − λdλ z − ζ z − ζdζ.
In view of Lemma 3.1 and h, we conclude that ∂z2w(z) = g(z) + Z Ω f (ζ)∂z2 z − ζ z − ζ dξdη − ∂2 z 1 π2 Z Ω Z Ω ∂η2f (η)ζ − η ζ − ηdsdt z − ζ z − ζdξdη + ∂z2φ1(z) + ∂z2φ2(z) = g(z). Note that the above lemma also holds when f, g ∈ H2(Ω) since we can
approxi-mate a H2 function by a sequence in C∞(Ω) in the H2(Ω) space.
3.2. ∂z system. Let A and B be two 2×2 matrices. We define A·B = Trace(ABT).
We write equation (2.6) in nondivergence form 0 = div div 1 2µ ∇2φ − ∆φ 2 I2 = 1 4µ∆ 2φ +1 2∇ 1 µ · ∇(∆φ) + 1 2∇ 2 1 µ · ∇2φ − ∆φ 2 I2 . Since µ > 0, the equation above is equivalent to
∆2φ + 2µ∇ 1 µ · ∇ (∆φ) + 2µ∇2 1 µ · ∇2φ − ∆φ 2 I2 = 0, (3.9)
which implies that
(3.10) ∂z2∂z2φ + α∂z2∂zφ + β∂z2φ + α∂z∂z2φ + β∂ 2 zφ = 0, where α = µ∂z 1 µ , β = µ 2∂ 2 z 1 µ . (3.11)
With equation (3.10) in mind, we define a first order system D + V acting on functions with values in C4 as follows
D + V = ∂z 0 0 0 0 ∂z 0 0 0 0 ∂z 0 0 0 0 ∂z + α β α β −1 0 0 0 α β α β 0 0 −1 0 . (3.12)
The corresponding Cauchy data of D + V is
CD+V =U |∂Ω: U ∈ H1(Ω, C4), U is a solution of (D+V)U=0 .
The next key step is to show that the Cauchy data Cµ∗ for (2.6) determine CD+V.
To do so, we begin the following lemma saying that Cµ∗ determines all derivatives of the solution on the boundary up to third order under suitable assumption.
Lemma 3.3. Assume that ∂κµ
1|∂Ω= ∂κµ2|∂Ω for all |κ| ≤ 1. If Cµ∗1 = C
∗ µ2, i.e.,
{φ1, φ1,n, M1,n, (Mt)1,t} = {φ2, φ2,n, M2,n, (Mt)2,t},
where φj is the solution to the equation Pµj(φj) = 0, j = 1, 2, then
∂κφ1 = ∂κφ2 on ∂Ω for |κ| ≤ 3.
Proof. The equalities φ1 = φ2 and φ1,n= φ2,n gives ∇φ1 = ∇φ2 on ∂Ω, i.e.,
φ1,1 = φ2,1, φ1,2 = φ2,2 on ∂Ω
and thus
∇φ1,k · t = ∇φ2,k · t, k = 1, 2 on ∂Ω.
(3.13)
Moreover, since M1,n = M2,n, by the definition of Mn and the hypothesis µ1|∂Ω =
µ2|∂Ω, we obtain (n21− n2 2)(φ1,11− φ2,11) − (n21 − n 2 2)(φ1,22− φ2,22) + 4n1n2(φ1,12− φ2,12) = 0. (3.14)
From (3.13) and (3.14), we have AU := −n2 n1 0 0 −n2 n1 n2 1− n22 4n1n2 n22− n21 φ1,11− φ2,11 φ1,12− φ2,12 φ1,22− φ2,22 = 0 on ∂Ω. (3.15)
Since the matrix A is invertible, we get that φ1,ij = φ2,ij on ∂Ω for 1 ≤ i, j ≤ 2.
With φ1,ij = φ2,ij on ∂Ω, we can deduce
∇φ1,ij · t = ∇φ2,ij · t,
that is,
−n2φ1,1ij+ n1φ1,2ij = −n2φ2,1ij + n1φ2,2ij.
(3.16)
Using the condition (Mt)1,t = (Mt)2,t and φ1,ij = φ2,ij on ∂Ω for 1 ≤ i, j ≤ 2, it
follows that div 1 2µ1 (∇2φ1− ∆φ1 2 I2) · n = div 1 2µ2 (∇2φ2− ∆φ2 2 I2) · n. (3.17)
Putting (3.16), (3.17) together and using the boundary assumption of µ, we obtain that −n2 n1 0 0 0 −n2 n1 0 0 0 −n2 n1 n1 n2 n1 n2 φ1,111− φ2,111 φ1,112− φ2,112 φ1,122− φ2,122 φ1,222− φ2,222 = 0.
Since the matrix above is invertible, we deduce that φ1,ijk= φ2,ijk for 1 ≤ i, j, k ≤ 2.
We are now ready to prove the crucial step.
Lemma 3.4. Assume that µ ∈ C3(Ω). Suppose that ∂κµ
1|∂Ω= ∂κµ2|∂Ω, ∀ |κ| ≤ 1.
Proof. Assume that Cµ∗1 = Cµ∗2 with two parameters µ1and µ2. Let U1 = (u1, u2, u3, u4)T be a solution of (D + V1)U1 = 0, then (D + V1)U1 = ∂z 0 0 0 0 ∂z 0 0 0 0 ∂z 0 0 0 0 ∂z + α1 β1 α1 β1 −1 0 0 0 α1 β1 α1 β1 0 0 −1 0 u1 u2 u3 u4 = 0,
where αj, βj are defined in (3.11) with respect to µj, j = 1, 2, respectively. The 2nd
and 4th equations of the system (D + V1)U1 = 0 gives
∂zu2 = u1, ∂zu4 = u3.
(3.18)
Likewise, the 1st and 3rd equations of (D + V1))U1 = 0 implies
∂zu1 = ∂zu3
(3.19)
It immediately follows from (3.18) and (3.19) that ∂z2u2 = ∂z2u4.
(3.20)
In view of (3.20) and Lemma 3.2, there exists a function Φ1 satisfying
∂z2Φ1 = u2, ∂z2Φ1 = u4.
(3.21)
Substituting (3.21) into (3.18) gives
u1 = ∂zu2 = ∂z2∂zΦ1, u3 = ∂zu4 = ∂z∂z2Φ1.
(3.22)
The 1st equation of (D + V1)U1 = 0, i.e.,
∂zu1+ αu1+ βu2+ αu3+ βu4 = 0
with u1, · · · , u4 replaced by (3.21) and (3.22) above, is equivalent to
Pµ1(Φ1) = 0 in Ω
(cf. (3.10)). Similarly, for V2 and U2 satisfying (D + V2)U2 = 0 in Ω associated
with µ2, we obtain a Φ2 solving Pµ2(Φ2) = 0 in Ω where the components of U2 and
Φ2 satisfy corresponding equations like (3.21), (3.22). The assumption Cµ∗1 = C
∗ µ2
implies
{Φ1, Φ1,n, M1,n, (Mt)1,t} = {Φ2, Φ2,n, M2,n, (Mt)2,t}
and Lemma 3.3 gives
∂κ(Φ1− Φ2)|∂Ω = 0 for |κ| ≤ 3.
Since U2 = (∂2z∂zΦ2, ∂z2Φ2, ∂z∂z2Φ2, ∂z2Φ2)T, we obtain (U1 − U2)|∂Ω = 0 and thus
CD+V1 = CD+V2.
3.3. Proof of the uniqueness result. We denote A = α β −1 0 and Q = α β 0 0 , then the system D + V can be represented as
D + V = ∂zI2 0 0 ∂zI2 + A Q Q A .
In the following lemma, we show that µ is uniquely determined by the Cauchy data CD+V.
Lemma 3.5. Let (αj, βj), j = 1, 2 be in C1(Ω). Assume that ∂κµ1|∂Ω= ∂κµ2|∂Ω for
all |κ| ≤ 1. If CD+V1 = CD+V2, then µ1 = µ2 in Ω.
Proof. Using that CD+V1 = CD+V2, we apply Theorem 4.1 in [1] to obtain that there
exist invertible matrices Fj ∈ C1(Ω, C2⊕ C2) such that F1 = F2 on ∂Ω. Moreover,
(3.23) ∂zFj = FjAj and Q1 = F Q2F−1,
where F := F1−1F2 is an invertible matrix.
Let us denote the two rows of the matrix Fj−1 by aj and bj, then the first relation
of (3.23) implies ∂zFj−1 = −AjFj−1 and hence
Fj−1 = ∂zbj bj
(3.24)
with the help of the form of Aj. We now write
F−1 = h v m r . Using the condition Q1 = F Q2F−1, we have that
mα1 = mβ1 = 0
and
(3.25) hα1 = hα2+ mβ2, hβ1 = vα2+ rβ2.
Then m = 0 in Ω0, where Ω0 = {x ∈ Ω : α1(x) 6= 0 or β1(x) 6= 0}. Note that if
x is in the complement of Ω0, then (α2(x), β2(x)) must be zero by (3.25) since F
is invertible. Thus α1 = α2 = 0 in the complement of Ω0. If Ω0 is empty, then
α1 = 0 = α2 in Ω. By the boundary condition ∂κµ1 = ∂κµ2 for |κ| ≤ 1, we conclude
that µ1 = µ2. Actually, in this case, we obtain that µ1 = µ2 = constant.
Now we suppose that Ω0 is a nonempty open set. Since m = 0 in Ω0, F−1 can be rewritten as
F−1 = h v 0 r
in Ω0. Using F = F1−1F2 and (3.24), we can deduce that
which implies F−1 = r ∂zr 0 r in Ω0. (3.26)
In deriving (3.26), we used the fact that ∂zb1 and b1 are linearly independent due
to the invertibility of F1−1. Note that since F is invertible, r never vanishes at any point in Ω0.
We observe that
∂zF−1 = ∂z(F2−1F1) = ∂zF2−1F1+ F2−1∂zF1
= −A2F−1+ F−1A1,
then it follows that
2∂zr = (α1 − α2)r (3.27) and ∂z2r = (β1− β2)r − α2∂zr. (3.28) From (3.27), we have 2∂z2r = r∂z(α1− α2) + (α1− α2)∂zr. (3.29)
Substituting (∂zr = (α1− α2)r/2) into (3.28) and (3.29) gives
2∂z2r = (2β1− 2β2− α2(α1− α2)) r = ∂z(α1− α2) + (α1 − α2)2/2 r,
which implies
2β1− 2β2− α2(α1− α2) = ∂z(α1− α2) + (α1− α2)2/2.
(3.30)
Note that r does not vanish in Ω0. By direct computation and the definition of αj
and βj in (3.11), it follows that
2βj = ∂zαj+ α2j.
(3.31)
Then we obtain
α12 = α22 in Ω0 (3.32)
by substituting (3.31) into (3.30). Combining (3.32) and the previously derived fact α1 = α2 = 0 in Ω \ Ω0,
we have that
α21 = α22 in Ω, which is equivalent to
(∇ log µ1)2 = (∇ log µ2)2 in Ω.
Since µ1|∂Ω= µ2|∂Ω and by the continuity of µj and ∇µj, j = 1, 2, we obtain
∇ log µ1 = ∇ log µ2 in Ω.
Using the boundary condition µ1|∂Ω= µ2|∂Ω again, we finally conclude that µ1 = µ2
in Ω.
Proof of theorem 1.1. From Section 2 we have known that the Cauchy data for the Stokes equations and that for the equation Pµ(φ) = 0 are equivalent, that is,
Cµ1 = Cµ2 is equivalent to C
∗ µ1 = C
∗
µ2. Therefore, Theorem 1.1 follows from Lemma
3.4 and Lemma 3.5.
4. Global uniqueness for the stationary Navier-Stokes equations In this section we consider the unique determination of the viscosity in an in-compressible fluid described by the stationary Navier-Stokes equations. In higher dimensions, this problem has been solved by Li and Wang in [11] using the lineariza-tion technique. Since their methods are independent of spatial dimensions, we could apply their ideas to show the uniqueness result of µ for the Navier-Stokes equations in the two dimensional case.
Let u = (u1, u2)T be the velocity vector field satisfying the stationary
Navier-Stokes equations
div σ(u, p) − (u · ∇)u = 0 in Ω, div u = 0 in Ω, (4.1)
and the corresponding Cauchy data is denoted by ˜
Cµ= {(u, σ(u, p)n) |∂Ω: (u, p) satisfies (4.1)} .
Let u|∂Ω = φ ∈ H3/2(∂Ω) satisfy (1.3). We choose φ = εψ with ψ ∈ H3/2(∂Ω)
and let (uε, pε) = (εvε, εqε) satisfy (4.1). The problem (4.1) is reduced to
div σ(vε, qε) − ε(vε· ∇)vε = 0 in Ω, div vε = 0 in Ω, vε= ψ on ∂Ω. (4.2)
We are looking for a solution of (4.2) with the form vε = v0+ εv and qε = q0 + εq,
where (v0, q0) satisfies the Stokes equations
div σ(v0, q0) = 0 in Ω, div v0 = 0 in Ω, v0 = ψ on ∂Ω, (4.3) and (v, q) satisfies − div σ(v, q) + ε(v0· ∇)v + ε(v · ∇)v0+ ε2(v · ∇)v = f in Ω, div v = 0 in Ω, v = 0 on ∂Ω, (4.4) with f = −(v0· ∇)v0.
In [11], it is shown that for any ψ ∈ H3/2(∂Ω), let (v
0, q0) ∈ H2(Ω) × H1(Ω) be
the unique solution (q0 is unique up to a constant) of the Stokes equations (4.3).
There exists a solution (uε, pε) of (4.1) of the form
with the boundary data uε|∂Ω= εψ for all |ε| ≤ ε0, where ε0depends on kψkH3/2(∂Ω).
Here (v, p) is a solution of (4.4) and satisfies the regularity result kvkH2(Ω)+ kqkH1(Ω)/R ≤ C
16
X
j=2
kψkjH3/2(∂Ω)
where C is independent of ε and kqkH1(Ω)/R := infc∈Rkq + ckH1(Ω). Hence, we have
kε−1uε− v0kH2(Ω) = kεvkH2(Ω) → 0, kε−1pε− q0kH1(Ω)/R = kεqkH1(Ω)/R→ 0, as ε → 0, which imply kε−1uε|∂Ω− v0|∂ΩkH3/2(∂Ω) → 0, (4.5) and kε−1σ(uε, pε)n|∂Ω− σ(v0, q0)n|∂ΩkH1/2(∂Ω) → 0, (4.6) provided Z Ω pεdx = Z Ω q0dx = 0.
From (4.5) and (4.6), we can deduce that the Cauchy data ˜Cµ of the Navier-Stokes
equations uniquely determines the Cauchy data Cµof the Stokes equations. In other
words, ˜Cµ1 = ˜Cµ2 implies Cµ1 = Cµ2. Therefore, the uniqueness of the viscosity
for the Navier-Stokes equations follows from Theorem 1.1. We have the following theorem.
Theorem 4.1. Let Ω be a simply connected bounded domain in R2 with smooth
boundary. Suppose that µ1 and µ2 are two viscosity functions for the Navier-Stokes
equations. Assume that µj ∈ C3(Ω) and µj > 0 with
∂κµ1|∂Ω= ∂κµ2|∂Ω for all |κ| ≤ 1.
Let ˜Cµ1 and ˜Cµ2 be the Cauchy data associated with µ1 and µ2, respectively. If
˜
Cµ1 = ˜Cµ2, then µ1 = µ2 in Ω.
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Department of Mathematics, University of Washington, Seattle, WA 98195-4350, USA
E-mail address: rylai@math.washington.edu
Department of Mathematics, University of Washington, Seattle, WA 98195-4350, USA
E-mail address: gunther@math.washington.edu
Institute of Applied Mathematics, NCTS (Tapei), National Taiwan University, Taipei 106, Taiwan
