75
Unique Continuation and applications
Camille Laurent1 2
1CNRS, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France2UPMC Univ Paris 06, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France, email:
Contents
1 Introduction and generalities 31.1 Generalities about unique continuation . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Motivation to control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1.4 The general strategy of Carleman . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2 Operators depending on τ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.1 Dierential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.2 Pseudodierential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2 The classical case 182.1 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Carleman estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.1 Geometric interpretation of pseudoconvexity in the case of real symbol of order 2 252.3 Using Carleman estimates for unique continuation . . . . . . . . . . . . . . . . . . . 25
2.3.1 Convexication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3.2 Unique continuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.3.3 Quantitative estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.4 The Dirichlet case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.5 Other applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.5.1 Spectral estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.5.2 The heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.6 The general Theorem of Hörmander . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.7 A small bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.8 Further comments and problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3 The wave equation with coecients constant in time 453.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2 Proving unique continuation using the Carleman estimate . . . . . . . . . . . . . . . 46
3.2.1 Convexication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.2.2 Unique continuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.3 The Carleman estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.4 Global unique continuation and non characteristic hypersurfaces . . . . . . . . . . . . 57
3.4.1 Distance and metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.4.2 The global theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.5 Approximate controllability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.6 Further remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.6.1 The general theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
1
3.6.2 Quantitative estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
A Appendix 61A.1 Pseudodierential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61A.2 The Dirichlet problem for some second order elliptic operators . . . . . . . . . . . . . 61
A.2.1 Some notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61A.2.2 The Carleman estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
A.3 Link between bicharacteristic ow and geodesics . . . . . . . . . . . . . . . . . . . . . 66A.4 Useful computation for symbols of order 2 . . . . . . . . . . . . . . . . . . . . . . . . 66
B Correction of (some) exercices 69B.1 Feuille 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
B.1.1 Exercice 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69B.2 Feuille 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
B.2.1 Exercice 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71B.2.2 Exercice 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
2
Chapter 1
Introduction and generalities
We can nd good references about Carleman estimates and unique continuation. This coursewas very inuenced by the very good course of Nicolas Lerner that can be found on his websitehttp://webusers.imj-prg.fr/ nicolas.lerner/m2carl.pdf and the survey article by Jérôme Le Rousseauand Gilles Lebeau [9].
Other references about unique continuation are the book of Claude Zuily [14]. Chapter XXVIIIof Lars Hörmander [7] gives a more general framework for what is described in Chapter 2.1.
We also refer to the very complete notes of Daniel Tataru available athttps://math.berkeley.edu/ tataru/papers/ucpnotes.ps.
1.1 Generalities about unique continuation
1.1.1 The problem
The general problem of unique continuation can be set into the following form: given a dierentialoperator P on an open set Ω ⊂ Rn, and given a small subset U of Ω, do we have a bigger set U ⊂ Ustrictly bigger than U so that for u regular enough
Pu = 0 in Ω,u|U = 0
=⇒ u = 0 on U .(1.1)
More generally, in cases where (1.1) is known to hold, we are interested in proving a quantitativeversion of
Pu small in Ω,u small in U
=⇒ u small in U .
A more tractable problem than (1.1) is the so called local unique continuation across an hyper-surface problem: given x0 ∈ Rn and S = Φ = 0 an oriented local hypersurface containing x0, dowe have the following implication:
There is a neighborhood Ω of x0 so thatPu = 0 in Ω,u = 0 in Ω ∩ S+ =⇒ u = 0 in a neighborhood of x0.(1.2)
where S+ = Φ > 0.It turns out that proving (1.2) for a suitable class of hypersurface (with regards to the operator
P ) is in general a key step in the proof of properties of the type (1.1).They are some very easy already known cases
3
n = 1: if P has coecients regular enough and u is regular enough, Pu = 0 is only an ordinarydierential equation. The Cauchy-Lipschitz theorem gives directly the uniqueness since u = 0on x ≤ 0 implies dk
dxku(0) = 0 for all k ∈ N.
if n = 2, P = ∂ = 12 (∂x + i∂y). Pu = 0 implies that u is an analytic function. In particular, if
u = 0 on any open set, then u = 0.
Concerning Partial Dierential equations, it seems that there could be some geometrical condi-tions.
Theorem 1.1.1 (Finite speed of propagation for the wave equation). Let u be a C2(R1+n) (realvalued) solution of
∂2t u−∆u = 0
(u, ∂tu)|t=0 = (u0, u1)
Then, for any r0 > 0, we have for any t ∈ [0, r0]
Er0−t(t) ≤ Er0(0)
where Er(t) = 12
∫|x|≤r(∂tu(t, x))2 + |∇xu(t, x)|2 dx is the local energy in the ball of radius r at time
t.In particular, if u0(x) = u1(x) = 0 for |x| ≤ r0, then u = 0 in the cone
Cr0 =
(t, x) ∈ R1+n s.t. |x| ≤ r0 − t, t ∈ [0, r0].
Proof. Multiply the equation by ∂tu to obtain ∂2t t∂tu − ∆u∂tu = 0. First, we notice ∂2
t t∂tu =12(∂tu(t, x))2. Moreover, by the formula divx(fX) = fdiv(X) +∇f ·X for f C1 function and X C1
vector eld, we get ∆u∂tu = divx(∇u)∂tu = divx(∇xu∂tu)−∇xu · ∇x∂tu = divx(∇u∂tu)− ∂t |∇xu|2
2 .So, denoting e(t, x) = 1
2
((∂tu(t, x))2 + |∇xu(t, x)|2
)the density of energy, we have obtained.
∂te− div(∇u∂tu) = 0.
The main tool will be the Stokes theorem
Lemma 1.1.1 (Stokes formula). Let X = (X1, · · · , Xd) a C1? vector eld on a domain Ω ⊂ Rd with
boundary ∂Ω smooth by piece, with an outward pointing normal N(x) smooth by piece on ∂Ω. Then,we have the formula ∫
Ωdiv(X) dx =
∫∂ΩX ·N dσ
where dσ is the ??element of area??.
We want to apply this on the truncated cone, for t0 ≤ r0.
Cr0,t0 =
(t, x) ∈ R1+n s.t. |x| ≤ r0 − t, t ∈ [0, t0].
∂Cr0,t0 is the union of three pieces with the following normal.
the bottom part S0 =
(0, x) ∈ R1+n s.t. |x| ≤ r0
where the outward normal is N(x) =
(−1t, 0x)
the top part St0 =
(t0, x) ∈ R1+n s.t. |x| ≤ r0 − t0where the outward normal is N(x) =
(1t, 0x)
4
the lateral boundary M t00 =
(t, x) ∈ R1+n s.t. |x| = r0 − t, t ∈ [0, t0]
with the outward nor-
mal N(t, x) = (1, x/|x|)/|(1, x/|x|)| = (1, x/|x|)/√
2.
We will apply this to the vector elds
X1 = e(t, x)(1t, 0x) = e(t, x)∂t so that divt,xX1 = ∂te
X2 = (0t,∇xu∂tu) so that divt,xX2 = divx(∇xu∂tu)
So, we obtain
0 =
∫Cr0,t0
divt,xX1 + divt,xX2
=
∫S0
(X1 +X2) · (−1t, 0x) +
∫St0
(X1 +X2) · (1t, 0x) +1√2
∫Mt00
(X1 +X2) · (1, x/|x|) dσ
= −∫S0
e(0, x) +
∫St0
e(t0, x) +1√2
∫Mt00
(e(t, x) + ∂tu∇xu ·
x
|x|
)dσ
We prove that the integral on the lateral boundary is positive.∣∣∣∣∂tu∇xu · x|x|∣∣∣∣ ≤ |∂tu||∇xu| ≤ 1
2
(|∂tu|2 + |∇xu|2
)≤ e.
So it gives
−∫S0
e(0, x) +
∫St0
e(t0, x) ≤ 0.
This gives the rst inequality.Now, let us prove the unique continuation property. The assumption implies that Er0(0) = 0. So,
the inequality implies Er0−t(t) = 0 for t ∈ [0, r0] and in particular ∂tu = 0 and ∇xu = 0 in the coneCr0 . By connexity, this implies that u = cste in Cr0 . This constant needs to be zero since u(0, x) = 0for |x| ≤ r0.
We can infer two interesting consequences for the unique continuation:
unique continuation holds across the hypersurface t = 0 and actually, we have some nicelocal linear quantication of the unique continuation. This situation actually holds when P issaid to be hyperbolic with respect to P = 0. We refer to ♣ for more precisions.
unique continuation can not hold across an hypersurface tangent to the cone |x| = t+ r.
Consider for instance the wave equation P = ∂2t − ∂2
x (or more simply the transport operator P =∂t − ∂x) on Rt × Rx. All the regular solutions are of the form u = f(x + t) + g(x − t). Take forinstance g = 0 and f smooth with support exactly [0, 1]. The surface S = x+ t = 0 clearly doesnot satisfy the unique continuation
The rst general unique continuation result of the form (1.2) is the Holmgren Theorem, statingthat, for operators with analytic coecients, unique continuation holds across any noncharacteristichypersurface S.
5
Theorem 1.1.2 (Holmgren Theorem). Let P =∑|α|≤m aα(x)Dα
x be a linear dierential operatorwith analytic coecients on some open subset Ω ⊂ Rn. Denote pm its principal symbol.
Let x0 ∈ Ω and Φ ∈ C1(Ω). Assume that Φ is non characteristic for P at x0, that is
pm(x0,∇Φ(x0)) 6= 0.
Then, there exists a neighborhood V of x0 so that every u ∈ D′(Ω) satisfying Pu = 0 on Ω and u = 0in the set x ∈ Ω; Φ(x) > Φ(x0) is zero in V .
But we would like to avoid the assumption of analyticity of the coecients. This will requiresometimes some stronger assumption say of pseudoconvexity condition (see e.g. Denition 2.1.1below) and will be the object of Chapter 2. The following chapter 3 will deal with some intermediatecase where the analyticity is with respect to only one variable (we will actually treat the simpler casewhere it is independant on one variable).
1.1.2 Motivation to control
The unique continuation has been pretty much studied for problems of control because it is equivalentto approximate controllability. We give the example of the heat equation. The problem of control isthe following. Let Ω be a smooth domain of Rn and ω ⊂ Ω be an open subset. The Cauchy problem isfor instance well posed for initial data. Using semigroup theory (Hille Yosida) or the diagonalizationof the Laplacian ∆, we get
Theorem 1.1.3. For any u0 ∈ H10 (Ω), f ∈ L1([0, T ], H1
0 (Ω)), there exists a unique u ∈ C([0, T ], H10 (Ω))
solution of ∂tu−∆u = f on [0, T ]× Ω
u|t=0 = u0 on Ω
u = 0 on [0, T ]× ∂Ω(1.3)
Moreover, this ow map can be extended for u0 ∈ L2(Ω), f ∈ L1([0, T ], L2(Ω)) with solutions u ∈C([0, T ], L2(Ω))
The problem of controllability (to zero for instance) from ω at time T is the following.Given an initial data u0 ∈ L2(Ω), can we nd a source term f ∈ L1([0, T ], L2(ω)) supported in
ω so that the solution of (1.3) satises u(T ) = 0. If we only want approximate controllability, wewill ask that for any ε > 0, one can nd a control so that ‖u(T )‖L2(Ω ≤ ε.
We will precise this later, but it turns out that in many situations, the dual problem of the controlproblem is some observability problem for the free problem
∂tu−∆u = 0 on [0, T ]× Ωu = 0 on [0, T ]× ∂Ω
(1.4)
By observability, we mean the map u0 7→ u|[0,T ]×ω . That is the map of observing the free solutionon the set [0, T ] × ω. In many cases (we will precise this in each case of the heat or the wave), wecan expect
the exact controllability is equivalent to some observability estimate ‖u(T )‖2L2 ≤ C∫ T
0 ‖u(t)‖2L2(ω) dtfor solutions of the free equation (1.4)
the approximate controllability is equivalent to some unique continuation of the type: u = 0on [0, T ]× ω implies u = 0.
In particular, for evolution equation, we will be very interested in some unique continuation (even-tually quantitative) from some sets of the form [0, T ]× ω. It seems that these theorems will have toreect the way of propagation of the energy of solutions of heat or wave.
6
1.1.3 Notation
We consider complex valued functions dened on Rn.We will denote the duality in L2(Rn), denoted L2 when there is not ambiguity, by
(f, g)L2 =
∫Rnf(x)g(x)dx.
For any multiindex α = (α1, · · · , αn) ∈ Nn, we dene its length |α| = α1 + · · ·+ αn.If ζ = (ζ1, · · · , ζn) ∈ Cn, ζα is dened by ζα1
1 · · · ζαnn .
For 1 ≤ j ≤ n, we denote Dj =∂ji .
For α = (α1, · · · , αn) ∈ Nn, we denote
∂α = ∂α11 · · · ∂
αnn
Dα =∂α
i|α|.
This notation are made interesting by the following formula of Fourier transform
Dαu(ξ) = ξαu(ξ).
where the Fourier transform is dened with the convention
u(ξ) =
∫Rne−ix·ξu(x)dx.
With this denition, the Fourier inverse formula is
u(x) = F−1u =1
(2π)n
∫Rneix·ξu(ξ)dξ.
while Plancherel formula is
(u, v)L2 =1
(2π)n(u, v)L2(1.5)
‖u‖L2 =1
(2π)n/2‖u‖L2 .(1.6)
With this convention, we have
f ∗ g = f g(1.7)
fg =1
(2π)nf ∗ g(1.8)
And when we apply it to a = f , b = g and apply F−1
F−1(ab) = F−1(a) ∗ F−1(b)(1.9)
F−1(a ∗ b) = (2π)nF−1(a)F−1(b)(1.10)
We recall the Leibniz formula
∂α(fg) =∑
β+γ=α
(α
β
)(∂βf)(∂γg)
7
where(αβ
)=(α1
β1
)· · ·(αdβd
)with
(nk
)= Ckn = n!
k!(n−k)! .
We will sometime use the notation ‖·‖H1(K) for some subset K of Rn. This will be dened forf ∈ C∞0 (Rn) by
‖f‖2H1(K) =
∫K|∇f |2 +
∫K|f |2.
If K is complicated, the denition of H1(K) as a Banach space could lead to some diculties, butwe will not need that. We will only use the denition for smooth functions when the denition isclear.
Denition 1.1.1 (Classical dierential operators). Letm ∈ N. We denote Diffm the set of dierentialoperators of the form P =
∑|α|≤m pα(x)Dα with pα smooth and bounded as function of x ∈ Rn.
Its full symbol will be denoted p(x, ξ) =∑|α|≤m pα(x)ξα. It belongs to the set of polynomial of
degree m of the variable ξ, with coecients smooth functions of x ∈ Ω, that we denote Σm(Ω).Respectively, if p ∈ Σm, we will denote p(x,D) the operator with symbol p.We denote pm(x, ξ) =
∑|α|=m pα(x)ξα its principal symbol. It is homogeneous of degree m in ξ.
1.1.4 The general strategy of Carleman
When we want to prove unique continuation, across a surface S = Φ = 0, we could have a functionsmooth and at (all derivatives cancel) along. So, in some sense we need something to emphasizethe local behaviour close to S.
The general idea of Carleman to prove unique continuation is to consider some weighted estimatesof the form
τ∥∥eτΦu
∥∥2
H1 ≤ C∥∥eτΦPu
∥∥2
L2(1.11)
for u ∈ C∞0 (Rn) and uniformly for τ ≥ τ0.First, this inequality says directly that if u ∈ C∞0 (Rn) is solution of Pu = 0 on Φ ≥ 0, then the
right hand side will tend to zero as τ tends to innity. Therefore, the left hand side will converge tozero, which implies that u is supported in Φ ≤ 0.
But we want unique continuation not only for functions with compact support. Assume now theweaker assumption that supp(u) ∩ Φ > 0 is compact. Then, applying a cuto functions, we canarrive to the previous case. This conguration can actually happen if we replace the function Φ byanother function Ψ "a little bit more convex".
Now, what will we need to prove Carleman estimates? Actually, the exponential weight is not soconvenient to prove estimates and one might want to eliminate it by posing v = eτΦ. Then, we areleft to study the operator PΦ = eτΦPe−τΦ.
We easily check that eτΦDje−τΦ = Dj + iτ∂jΦ. Hence, we have changed Dj by another operator
with one derivative and one exponent of τ . So, we could expect that if P =∑
α pα(x)Dα, one is leftto study an operator operator PΦ with as many derivatives as exponents of τ . Yet, we want someestimates uniform in τ large. So, now we have to think τ as having the same weight as a derivative.We describe this calculus in the next section.
1.2 Operators depending on τ
In this section, we describe the setting that will be used in Chapter 2.1 and 3. The main thing is thepresence of a large parameter τ for which we want to make some estimates uniform for τ > 0 large.Morally, we want to think τ as having the same weight as a derivative.
8
Hsτ is the Hs norm with the following norm depending on τ
‖u‖Hsτ
=∥∥∥(|D|2 + τ2)
s2u∥∥∥L2(Rn)
= (2π)−n/2∥∥∥(|ξ|2 + τ2)
s2 u∥∥∥L2(Rn)
.
In the next subsection, we will describe some important properties of operator depending on alarge parameter τ . Since it is nally expected to be large, we will always assume τ ≥ 1 when dealingwith estimates uniform in τ .
Note also that when m is an integer, this expression can be written more simply. For instance, ifm = 1,
‖u‖H1τ≈ ‖u‖H1 + τ ‖u‖L2 .
1.2.1 Dierential operators
Denition 1.2.1 (Dierential operators depending on τ). Let m ∈ N. We denote Diffmτ the setof dierential operators of the form P =
∑|α|+β≤m pα,β(x)τβDα with pα,β smooth and bounded as
function of x ∈ Rn.Its full symbol will be denoted p(x, ξ, τ) =
∑|α|+β≤m pα,β(x)τβξα. It belongs to the set of polyno-
mial of degree m of the variable (ξ, τ), with coecients smooth functions of x ∈ Ω, that we denoteΣm(Ω).
Respectively, if p ∈ Σm, we will denote p(x,D, τ) the operator with symbol p.We denote pm(x, ξ, τ) =
∑|α|+β=m pα,β(x)τβξα its principal symbol. It is homogeneous of degree
m in ξ, τ .
Note that is it almost the same denition as Denition 1.1.1 except with the dependance on τwhich change the denition of the principal symbol.
Note that if p ∈ Σmτ , the inversion Fourier formula gives for u ∈ S(Rn)
p(x,D, τ) =1
(2π)n
∫Rneix·ξp(x, ξ, τ)u(ξ)dξ(1.12)
The following properties will be some examples of the general Heuristic (not true of course) thata pseudo dierential operator acts as if it was a multiplication by p(x, ξ, τ) modulo lower order terms.
The general philosophy is that we want to get some properties of some operators only from thereprincipal symbol.
If P , A, B of respective order m, m1 and m2, with respective principal symbol p, a, b, the roughsummary is the following
1. P act from Hsτ in Hs−m
τ
2. A B is of order m1 +m2 with principal symbol ab
3. [A,B] is of order m1 +m2 − 1 with principal symbol 1i a, b
4. P ∗ is of order m with principal symbol p
5. p ≥ C(ξ2 + τ2)m/2 implies Re(Pu, u)L2 ≥ C ′ ‖u‖Hm/2τ
for large τ
Proposition 1.2.1 (Action on Sobolev spaces). Let P ∈ Diffmτ (Rn).Then, for any s ∈ R, P is bounded from Hs
τ in Hs−mτ uniformly for τ ≥ 1.
9
Proof. We only prove it for s ∈ Z since we will only use it in this case. Yet, the other casescan be easily obtained by interpolation. By the triangular inequality, it is sucient to prove itfor pα,β(x)τβDα with |α| + β ≤ m. First, suppose s odd, s = 2j + 1 with j ∈ N. First, since(|ξ|2 + τ2)s ≤ C
(|ξ|2s + τ2s
), for u ∈ S(Rn), ‖u‖2Hs
τis equivalent to∫
Rn
(|ξ|2s + τ2s
)|u(ξ)|2dξ =
∫ξ||ξ|su(ξ)|2dξ +
∫Rnτ2s|u(ξ)|2dξ
=
∫Rn||ξ|2su(ξ)u(ξ)dξ + τ2s ‖u‖2L2
=
∫Rn
(−∆)2j+1u(x)u(x)dx+ τ2s ‖u‖2L2
=
∫Rn
(−∆)j∇u)(x) · ((−∆)j∇u(x))dx+ τ2s ‖u‖2L2
=∥∥(−∆)j∇u
∥∥2
L2 + τ2s ‖u‖2L2
This form easily gives that the multiplication by pα,β is bounded from Hsτ with s ∈ N odd. The even
case is actually easier. Moreover, we can extend this to any s ∈ Z by duality.So, we are left to prove that Dατβ applies Hs
τ into Hs−mτ for any s ∈ Z and for |α|+β ≤ m. This
is just a Fourier multiplier. We use the inequality |ξ| ≤(|ξ|2 + τ2
)1/2and the same for τ .∥∥∥Dατβ
∥∥∥2
Hs−mτ
=
∫Rn
(|ξ|2 + τ2)s−m|Dατβu(ξ)|2dξ =
∫Rn
(|ξ|2 + τ2)s−m|ξ|2ατ2β|u(ξ)|2dξ
≤∫Rn
(|ξ|2 + τ2)s−m(|ξ|2 + τ2
)α+β |u(ξ)|2dξ.
Moreover, since τ ≥ 1 and α + β ≤ m, we have(|ξ|2 + τ2
)m. This gives the expected estimate for
Schwarz functions. By density of the Schwarz functions, it gives the result.
Proposition 1.2.2 (Composition). Let A =∑|α|+β≤m1
aα,βτβDα, B =
∑|α|+β≤m2
bα,βτβDα be two
dierential operators with large parameter τ of respective order m1 and m2, with full symbol a(x, ξ, τ)and b(x, ξ, τ) and principal symbol am1(x, ξ, τ) and bm2(x, ξ, τ). Then, we have that A B is of orderm1 +m2 in (D, τ), that is in Diffm1+m2
τ (Ω). Moreover, it can be written
A B = (am1bm2)(x,D, τ) + r(x,D, τ)
with r(x,D, τ) ∈ Diffm1+m2−1τ (Ω).
In particular, the principal symbol of A B is am1bm2.
Proof. We prove it by iteration on m = m1 +m2.
m = 0: A = f(x) and B = g(x) so it is obvious.
m 7→ m+ 1: By linearity with respect to A and B, it is enough to prove it for A = f(x)τβDα
and B = g(x)τβ′Dα′ . Since m1 +m2 = m+ 1, at least one of the β, |α|, β′, |α′| is bigger than
1. If it is either β ≥ 1, β′ ≥ 1 or |α′| ≥ 1, the result is direct by iteration result for m (for the
composition by the right by Dα′
the result is easy). If it is |α|, take k such that αk ≥ 1. ThenA = ADk with A ∈ Diffm1−1
τ . Then,
ABu = ADk[g(x)τβ′Dα′ ]u = A(Dkg(x))τβ
′Dα′ + Ag(x)τβ
′Dα′Dku
The iteration result gives that the rst term is in Diffm−1τ and Ag(x)τβ
′Dα′ ∈ Diffm−1
τ withprincipal symbol ag(x)τβ
′ξα′. We obtain easily that Ag(x)τβ
′Dα′Dk ∈ Diffm−1
τ with principalsymbol ag(x)τβ
′ξα′ξk = ab.
10
See Section B.1.1 for the proof. For the commutator, we will need the following notation anddenition.
Denition 1.2.2 (Poisson bracket).
a, b :=∑j
[(∂ξja)(∂xjb)
](x, ξ, τ)−
∑j
[(∂xja)(∂ξjb)(x, ξ, τ)
]We check that we have the properties
a, b = −b, a(1.13)
a, bc = a, b c+ b a, c(1.14)
Proposition 1.2.3 (Commutation). [A,B] = AB − BA is of order m1 + m2 − 1 in (D, τ), that isin Diffm1+m2−1
τ . Moreover, it can be written
[A,B] =1
i
∑j
[(∂ξja)(∂xjb)
](x,D, τ)−
∑j
[(∂ξja)(∂xib)
](x,D, τ) + r(x,D, τ)
=1
ia, b (x,D, τ) + r(x,D, τ)
with r(x,D, τ) ∈ Diffm1+m2−2τ (Ω).
Proof. We rst prove by iteration on m the following property:For any A = f(x)Dk and B ∈ Diffmτ , the conclusion of the Proposition is true.
m = 0: B = g(x).
[A,B]u = [f(x)Dk, g(x)]u = f(x)Dk(g(x)u)− g(x)f(x)Dku = f(x)Dk(g(x))u =1
if(x)∂k(g(x))u.
Which is is Diff0τ with principal (and full) symbol 1
i f(x)∂xk(g(x)). And we check from thedenition of the Poisson bracket and for the principal symbols a1 = f(x)ξk, b0 = g(x), we havea1, b0 = a1, b = f∂xkg.
m = 1 (this is only needed as a partial result): By linearity (and the case m = 0), it is enoughto have the result forB = τ or B = g(x)Dl. The rst case is trivial so we just treat the second.
[A,B]u = [f(x)Dk, g(x)Dl]u = f(x)Dk [g(x)Dlu]− g(x)Dl [f(x)Dku]
= f(x)Dk(g(x))Dlu− g(x)Dl(f(x))Dku.
We recognize that it belongs to Diff1τ with principal (and full) symbol 1
i f(x)∂xk(g(x))ξl −g(x)∂xl(f(x))ξk which turns out to be equal to 1
i f(x)ξk, g(x)ξl.
m 7→ m+ 1: the main idea is to notice that we have, similarly to (1.13) and (1.14).
[A,B] = −[B,A](1.15)
[A,BC] = [A,B]C +B[A,C].(1.16)
More precisely, by linearity, it is enough to consider B = τg(x)B or B = g(x)DlB withB ∈ Diffmτ . In the rst case, we have
[A,B] = τ [A, g(x)B].
11
So the iteration step m gives the result. In the second case
[A,B] = [A, g(x)DlB] = [A, g(x)Dl]B + g(x)Dl[A, B].
The case m = 1 and the iteration assumption then gives (after using Proposition 1.2.2) that[A,B] ∈ Diffmτ with principal symbol
1
i
(a, g(x)ξl bm + g(x)ξl
a, bm
)=
1
i
a, g(x)ξlbm
=
1
i
a, g(x)ξlbm
=
1
ia, b
where we have used (1.14).
The result is now proved for any A = f(x)Dk and B ∈ Diffmτ (or A ∈ Diffmτ B = g(x)Dl byantisymmetry). The nal result can then be proved easily by iteration on m = m1 + m2 using thesame strategy.
m = 0: the result is obvious since then A = f(x) and B = g(x).
m 7→ m + 1 By linearity with respect to both variable, it is enough to get the result forA = f(x)τβDα and B = g(x)τβ
′Dα′ . Since m1 +m2 = m+ 1, at least one of the β, |α|, β′, |α′|
is bigger than 1. By symmetry of the role, we can assume that it is either β ≥ 1 or |α| ≥ 1. Inthe rst case, we apply directly the iteration result for m. If it is |α|, take k such that αk ≥ 1.Then A = f(x)DkA where A is of order m1 − 1. We have again
[A,B] = [f(x)DkA, B] = f(x)Dk[A, B] + [f(x)Dk, B]A
and we conclude similarly by the iteration step m for the rst term and for the second term bythe previous result proved in the specic case A = f(x)Dk.
Denition 1.2.3. Let P an operator, we will write P ∗ (in some cases where it is well dened) itsformal dual, that is that satises
(Pu, u)L2 = (u, P ∗u)L2(1.17)
for all u ∈ C∞0 (Rn)
Proposition 1.2.4 (Adjoint). There exists one unique operator P (x,D, τ)∗ ∈ Diffmτ satisfying (1.17).Moreover, P (x,D, τ)∗ − P (x,D, τ) ∈ Diffm−1
τ where P (x,D, τ) is the operator with symbol p.
Proof. By linarity, it is enough to prove it for P = a(x)τβDα. By looking on the Fourier side using(1.5), we have
(Dαu, v)L2 =
∫RnDαuv =
∫Rnξαuv =
∫Rnuξαv =
∫RnuDαv = (u,Dαv)L2
So that (a(x)τβDαu, v
)L2
=(τβDαu, p(x)v
)L2
=(u,Dαa(x)v
)L2.
So P ∗ = Dαa(x) and we know by Theorem 1.2.2 that P ∗ ∈ Diffmτ with principal symbol a(x)τβξα =pm.
12
We will also write a weak form of Gårding estimates for operator of order 2. The general case (thatwe shall not need for Carleman estimates will be stated in the next subsection about pseudodierentialoperators with large parameters.
Lemma 1.2.1 (An easy local Gårding inequality in some particular case). Let P be an operator oforder 2 of the form
P = A+k∑i=1
Bi 1
−∆ + τ2Bi(1.18)
with A,Bi ∈ Diff2τ with real principal symbol a2(x, ξ, τ), b2,i(x, ξ, τ) satisfying
p2(0, ξ, τ) = a2(0, ξ, τ) +k∑i=1
b22,i(0, ξ, τ)
|ξ|2 + τ2≥ C(|ξ|2 + τ2).(1.19)
for all ξ ∈ Rn and τ ≥ 0. Then, there exist r > 0 and C1, C2 > 0, so that we have
Re (P (x,D, τ)u, u)L2 ≥ C1 ‖u‖2H1τ− C2 ‖u‖2L2
for any u ∈ C∞0 (B(0, r)). In particular, we have, for τ large enough
Re (P (x,D, τ)u, u)L2 ≥ C1 ‖u‖2H1τ.
Proof. Denote a(x, ξ, τ), bi(x, ξ, τ) the full symbol of A, Bi.The idea of the proof is to "freeze" the coecients. The theorem is an easy consequence of the
following two Lemma.
Lemma 1.2.2. Under assumption (1.19), we have (with the same constant C)
Re (P2(0, D, τ)u, u)L2 ≥ C ‖u‖2H1τ
Lemma 1.2.3. If P is of the form (1.18) with the above assumptions, for any ε > 0, there exitsC > 0 and r > 0 so that
|(P (x,D, τ)u, u)L2 − (P2(0, D, τ)u, u)L2 | ≤ ε ‖u‖2H1τ
+ C ‖u‖2L2
for any u ∈ C∞0 (B(0, r)), τ ≥ 1.
Proof of Lemma 1.2.2. Since the coecients of P2(0, D, τ) are constant, we have P2(0, D, τ)u(ξ) =p(0, ξ, τ)u(ξ). So, using Parseval formula, we get
(P2(0, D, τ)u, u)L2 =1
(2π)n
∫ξp2(0, ξ, τ)u(ξ)u(ξ)dξ =
1
(2π)n
∫ξp2(0, ξ, τ)|u(ξ)|2dξ ≥ C
∫ξ(|ξ|2 + τ2)|u(ξ)|2dξ
≥ C
∫ξ|√|ξ|2 + τ2u(ξ)|2dξ = C ‖u‖2H1
τ.
Note that the previous calculation actually gives that (P2(0, D, τ)u, u)L2 is actually real.
13
Proof of Lemma 1.2.3. Using the estimate ‖u‖H1τ‖u‖L2 ≤ ε ‖u‖2H1
τ+ Cε ‖u‖2L2 , any bound of the
remainder of the form ‖u‖H1τ‖u‖L2 or ε ‖u‖2H1
τwill be sucient.
(Pu, u)L2 = (Au, u)L2 +
(B 1
−∆ + τ2Bu, u
)L2
A is of order 2 with principal symbol. We denote r = a− a2 of order 1.
(A(x,D, τ)u, u)L2 = (A2(x,D, τ)u, u)L2 + (R(x,D, τ)u, u)L2
with
|(R(x,D, τ)u, u)L2 | ≤ C ‖u‖H1τ‖u‖L2 .
Moreover, a2 is a quadratic form in the variables ζ = (ξ, τ), that can be decomposed according tothe exponent in τ
a2(x, ξ, τ) =
n∑i,j=1
a2(x)ξiξj +
n∑i=1
ai,0(x)τξi + a0,0(x)τ2
(a2,i,j(x)DiDju, u)L2 = (a2,i,j(x)Dju,Diu)L2 − ((Dia2,i,j)u,Dju)L2
where the second term can be bounded by C ‖u‖H1τ‖u‖L2 .
(a2,i,j(x)Dju,Diu)L2 = (a2,i,j(0)Dju,Diu)L2 + ((a2,i,j(x)− a2,i,j(0))Dju,Diu)L2
But again, the second term is bounded by ‖a2,i,j(x)− a2,i,j(0)‖L∞(Supp(u)) ‖u‖2H1τthat can be bounded
by ε ‖u‖2H1τif r is chosen small enough. So, at that stage, we have proved
(A(x,D, τ)u, u)L2 = (A2(0, D, τ)u, u)L2 +R
with |R| ≤ Cε ‖u‖2H1τ
+ Cε ‖u‖2L2 .Let us now see the other term with B.(
B 1
−∆ + τ2Bu, u
)L2
=
(1
−∆ + τ2Bu,B∗u
)L2
=
(1
−∆ + τ2Bu,B2(0, D, τ)u
)L2
+
(1
−∆ + τ2Bu, (B∗(x,D, τ)−B2(0, D, τ)u
)L2
=
(1
−∆ + τ2Bu,B2(0, D, τ)u
)L2
+R1
Using that B sends H1τ into H−1
τ , we get∥∥∥∥ 1
−∆ + τ2Bu
∥∥∥∥H1τ
≤ C ‖Bu‖H−1τ≤ C ‖u‖H1
τ.
So, we are left to estimate ‖(B∗(x,D, τ)−B2(0, D, τ))u‖H−1τ. Since the dierence is of order 1, we
can replace B(x,D, τ)∗ by B2(x,D, τ)∗ up to a bound by C ‖u‖L2 .So, we need to estimate ‖(B∗2(x,D, τ)−B2(0, D, τ))u‖H−1
τ. By duality and the locality of the
considered dierential operators, it leads to estimate
sup‖f‖
H1τ
=1((B∗2(x,D, τ)−B2(0, D, τ))u, f)L2 = sup
‖f‖H1τ
=1,supp(f)⊂B(0,2r)(u, (B2(x,D, τ)−B2(0, D, τ)∗)f)L2
14
Since B2(0, D, τ) has real valued and constant coecients B2(0, D, τ)∗ = B2(0, D, τ).So, we are left to estimate, for ‖f‖H1
τ= 1, supp(f) ⊂ B(0, 2r)
(u, (B2(x,D, τ)−B2(0, D, τ))f)L2
Similarly as before, for instance for terms of the form b2,i,j(x)DiDju
(u, (b2,i,j(x)− b2,i,j(0))DiDjf)L2 = (Diu, (b2,i,j(x)− b2,i,j(0))Djf)L2 − ((Dib2,i,j)(x)u,Djf)L2
The rst term is bounded by ε ‖u‖H1τ‖f‖H1
τif |b2,i,j(x)− b2,i,j(0)| ≤ ε on B(0, 2r). The second term
is bounded by C ‖u‖L2 ‖f‖H1τ.
The other terms of the form∑n
i=1 bi,0(x)τξi + b0,0(x)τ2 are treated similarly.
So, at this stage, it remains to replace(
1−∆+τ2 Bu,B2(0, D, τ)u
)L2
by the same with the rst
B replaced by B2(0, D, τ). That means to estimate(1
−∆ + τ2 (B(x,D, τ)−B2(0, D, τ))u,B2(0, D, τ)u
)L2
≤ ‖(B(x,D, τ)−B2(0, D, τ))u‖H−1τ‖u‖H1
τ
This last term has already been estimated with B∗ instead of B, but it works the same.
Proposition 1.2.5 (Semiglobal Gårding inequality). Under the same notations as Lemma 1.2.1, butreplacing the local assumption (1.19) by a semiglobal one
p2(x, ξ, τ) ≥ C(|ξ|2 + τ2) ∀(x, ξ, τ) ∈ K × Rn × R+(1.20)
for one compact K.Then, the same estimate holds for any u ∈ C∞0 (Rn) supported in K.
Proof. Apply Lemma 1.2.1 for any x0 ∈ K. Recover K by a nite number of such balls:K ⊂∪i∈IB(xi, ri). Dene a smooth partition of unity χi ∈ C∞(B(xi, ri)) so that∑
i
χ2i = 1 on K.
We have u =∑
i χ2iu.
(Pu, u)L2 =∑i
(χ2iPu, u
)L2 =
∑i
(χiPu, χiu)L2
=∑i
(Pχiu, χiu)L2 +∑i
([χi, P ]u, χiu)L2
[χi, P ] is of order 1 for the part of P coming from A. The part coming from the Bi is also of order1 (see the next subsection) but is not a dierential operator. So, using the general fact [T, S−1] =TS−1 − S−1T = S−1[S, T ]S−1 for S the dierential operator (∆ + τ2). This allows to estimate thecommutator. Moreover, applying the local Lemma, we get∑
i
(Pχiu, χiu)L2 ≥∑i
‖χiu‖2H1τ− C ‖χiu‖2L2 .
And with similar arguments,∑
i ‖χiu‖2H1τ≥ C ‖u‖2H1
τ− C ‖u‖2L2 .
As an immediate consequence of we get the following result (we only proved it for m = 2 but theresult is true more generally)
15
Proposition 1.2.6 (Elliptic estimates). Let P ∈ Diffmτ elliptic, in the sense that pm(x, ξ, τ) 6= 0 forall (x, ξ, τ) with (ξ, τ) 6= 0 for one compact K.
Then, we have, for τ large enough and for u ∈ C∞0 (Rn) supported in K,
‖u‖Hmτ≤ C ‖Pu‖L2 .
Proof.
‖Pu‖2L2 = Re (Pu, Pu)L2(Rn) = Re (P ∗Pu, u)L2(Rn) = Re (P ∗Pu, u)L2(Rn) .
But the principal symbol of P ∗P is |pm|2 thanks to Proposition 1.2.4. Therefore, since |pm|2 ishomogeneous of order 2m with pm(x, ξ, τ) 6= 0 for all (x, ξ, τ) with (ξ, τ) 6= 0, then using thecompactness of
(x, ξ, τ)
∣∣x ∈ K, |ξ|2 + τ2 = 1we can nd C > 0 so that |pm|2 ≥ C(ξ2 + τ2)m for
x ∈ K. Proposition 1.2.5 allows to conclude.
1.2.2 Pseudodierential operators
We would like to extend formula (1.12) to some symbol p(x, ξ, τ) not necessarily polynomial. Forinstance, the operator −∆ + τ2 is well dened for τ 6= 0, and we would like to say that it is of order−2 with symbol 1
|ξ|2+τ2 . One good class is the Smτ . Note that this theory will not be strictly necessary
for obtaining Carleman estimates where the previous dierential operators will be enough. Yet, webelieve that it is good to know that the previous results can be set into a more general framework.
We say that p(x, ξ, τ) belongs to Smτ , if it is smooth in (x, ξ) and for any (α, β), there exists Cα,βso that ∣∣∣∂αx ∂βξ p(x, ξ, τ)
∣∣∣ ≤ Cα,β (1 + |(ξ, τ)|)m−|β| ∀(x, ξ, τ) ∈ Rn × Rn × R+(1.21)
Note that if p ∈ Σm, then p ∈ Smτ , so this denition is an extension of the dierential opearators.By mimicking formula (1.12), for p(x, ξ, τ) ∈ Smτ , we dene the associated pseudodierential
operator by the formula
P (x,D, τ) =1
(2π)d
∫Rneix·ξp(x, ξ, τ)u(ξ)dξ.
that can (rst) be seen to be well dened as a continuous function if u ∈ S(Rn).But it actually has much better properties similar to the dierential case.The set of such operators will be denoted Ψm
τ .
Proposition 1.2.7 (Action on Sobolev spaces). Let P ∈ Ψmτ (Rn).
Then, for any s ∈ R, P is bounded from Hsτ in Hs−m
τ uniformly in τ .
Since the notion of principal symbol is more complicated, we will state the results a bit dierently.
Proposition 1.2.8 (Composition). Let A ∈ Ψm1τ , B ∈ Ψm2
τ with full symbol a(x, ξ, τ) and b(x, ξ, τ).Then, we have the following AB is of order m1 +m2 in (D, τ), that is in Diffm1+m2
τ (Ω). Moreover,it can be written
A B = (ab)(x,D, τ) +1
i
∑j
[(∂ξja)(∂xjb)
](x,D, τ) + r(x,D, τ)
with r(x,D, τ) ∈ Ψm1+m2−2τ (Ω).
16
Proposition 1.2.9 (Commutation). [A,B] ∈ Ψm1+m2−1τ . Moreover, it can be written
[A,B] =1
ia, b (x,D, τ) + r(x,D, τ)
with r(x,D, τ) ∈ Ψm1+m2−2τ (Ω).
Proposition 1.2.10 (Adjoint). P (x,D, τ)∗ ∈ Ψmτ and P (x,D, τ)∗ − P (x,D, τ) ∈ Diffm−1
τ .
Lemma 1.2.4 (Gårding). Assume p ∈ S2τ real valued so that
Re p(x, ξ, τ) ≥ C(ξ2 + τ2)
then, we have
Re (Pu, u)L2 ≥ C ‖u‖H1τ− C2 ‖u‖L2
for all u ∈ C∞0 (Rn).
Link with the semiclassical operators of the course [11] Let p ∈ S0τ . Denote h = τ−1 and
rename psc(x, ξ, h) = p(x, ξ, τ−1). Then, property (1.21) can be rewritten∣∣∣∂αx ∂βξ psc(x, ξ, h)∣∣∣ ≤ Cα,β (1 + |(ξ, h−1)|
)−|β| ≤ Cα,βhβ|for all (x, ξ, h) ∈ Rn × Rn × (0, 1]. In particular, psc(x, ξ, h) ∈ S0
scl as dened in Section 4.5 of [11].In the case of operator of order 0,
Proposition 1.2.7 in L2 is a consequence of Theorem 4.5.1 of [11].
Proposition 1.2.8 1.2.9 and 1.2.10 are consequence of Corollary 4.5.6 of [11].
The other cases can be obtained by changing p by multiplying by some power of (∆ + τ2)s with theappropriate s, and sometimes change u by (∆ + τ2)su.
For example, here is how to obtain Lemma 1.2.4 from Theorem 4.5.11 of [11]. Take p ∈ S2τ . r =
Re p(x, ξ, τ)−C(ξ2 + τ2) ≥ 0. a = (∆ + τ2)−1/2r(∆ + τ2)−1/2 is of order 0. Dene asc = a(x, ξ, h−1).So, it gives
Re (asc(x,D, h)u, u)L2 ≥ −C1h ‖u‖L2
Re(
(∆ + τ2)−1/2p(∆ + τ2)−1/2u, u)L2− C ‖u‖L2 ≥ −C1τ
−1 ‖u‖L2
If we apply it to (∆ + τ2)1/2v, it gives
Re (pv, v)L2 − C∥∥∥(∆ + τ2)1/2v
∥∥∥L2≥ −C1τ
−1∥∥∥(∆ + τ2)1/2v
∥∥∥L2
which gives the result.
17
Chapter 2
The classical case
2.1 Main results
In this chapter, we will prove a very general result of unique continuation for operators, with realprincipal symbol
Before stating the general result, it is interesting to state it in the cases that will be the moreuseful for us, namely when they are of order 2.
Theorem 2.1.1 (Elliptic real operator of order 2). Let Ω an open set of Rn. Let P =∑n
i,j=1 ai,j(x)∂i∂j+∑k bk(x)∂k+c(x) be a dierential operator of order 2 with ai,j ∈ C∞(Ω) real valued, bk, c ∈ L∞(Rn).
Assume also that P is elliptic, that is there exists C > 0 so that
n∑i,j=1
ai,j(x)ξiξj ≥ C|ξ|2, for all (x, ξ) ∈ Ω× Rn.
Let Φ ∈ C2(Ω) so that ∇Φ 6= 0 on Ω.Let x0 ∈ Ω. Then, there exists V one neighborhood of x0 so that for any u ∈ C∞(Ω),
Pu = 0 in Ω,u = 0 in Ω ∩ Φ > Φ(x0) =⇒ u = 0 on V.(2.1)
The previous theorem actually says that for elliptic operator of order 2, there is no geometriccondition for unique continuation across hypersurface.
The previous theorem is a particular case of a more general theorem that we state below. Forthat, we will need to dene some notion of convexity with respect to an operator.
Denition 2.1.1 (Strict pseudo-convexity for surfaces dened by a function). Let P ∈ Diff2 be a(classical) dierential operator with real valued principal symbol p ∈ Σ2 and Φ ∈ C∞ real valued.
We say that the function Φ is pseudoconvex in the sense of surface with respect to P in Ω if itsatises
p(x, ξ) = ∇ξp(x, ξ) · ∇Φ(x) = 0 =⇒ p, p,Φ > 0 ∀(x, ξ) ∈ Ω× Rn \ 0(2.2)
We can check that Denition 2.1.1 is invariant if we change the function Φ by another one g(Φ)which has the same level sets (with dierent value of course) and denes the same level sets. Thatis why this property only depends on the surface. It is a geometric property of the surface. SeeSubsection 2.2.1 for an interpretation as convexity with respect to the bicharacteristics curves.
18
Theorem 2.1.2 (Real operator of order 2). Let Ω an open set of Rn. Let P =∑n
i,j=1 ai,j(x)∂i∂j +∑k bk(x)∂k + c(x) be a dierential operator of order 2 with ai,j ∈ C∞(Ω) real valued, bk, c ∈
L∞(Rn). Let Φ ∈ C2(Ω) so that ∇Φ 6= 0 on Ω. Assume also that the principal symbol of P ,p(x, ξ) = −
∑ni,j=1 ai,j(x)ξiξj satises
p(x, ξ) = ∇ξp(x, ξ) · ∇Φ(x) = 0 =⇒ p, p,Φ > 0 ∀(x, ξ) ∈ Ω× Rn \ 0(2.3)
Let x0 ∈ Ω. Then, there exists V one neighborhood of x0 so that for any u ∈ C∞(Ω),Pu = 0 in Ω,u = 0 in Ω ∩ Φ > Φ(x0) =⇒ u = 0 on V.(2.4)
Remark 2.1.1 (Elliptic case). The previous Theorem 2.1.2 actually includes Theorem 2.1.1. Indeed,the condition p(x, ξ) = 0 is never fullled.
Remark 2.1.2 (The example of the wave equation). In the case of the wave equation, P = ∂2t −∆,
p = −ξ2t + |ξx|2, we compute (using that Φ does not depend on ξ)
p,Φ = ∇ξp · ∇(t,x)Φ = −2ξt∂tΦ + 2ξx · ∇xΦ
p, p,Φ = ∇ξp · ∇(t,x)p,Φ − ∇(t,x)p · ∇ξp,Φ= ∇ξp · ∇(t,x)p,Φ= −2ξt∂t [−2ξt∂tΦ + 2ξx · ∇xΦ] + 2ξx · ∇x [−2ξt∂tΦ + 2ξx · ∇xΦ]
= 4[ξ2t ∂
2t Φ− 2ξtξx · ∇x∂tΦ +Hessx(Φ)(ξx, ξx)
]If for instance, we choose Φ(t, x) of the form −f(t) + ϕ(x), we can write (2.3) (specialized in thepoint (t, x) = (0, 0)) as
ξ2t = |ξx|2 and f ′(0)ξt = ξx · ∇xϕ(0) =⇒ −f ′′(0)|ξx|2 +Hessx(ϕ)(0)(ξx, ξx) > 0 ∀ξ ∈ Rn \ 0.
We select several possibilities
|f ′(0)|2 > |∇xϕ(0)|2, that is |∂tΦ|2 > |∇xΦ|2: ∇t,xΦ is timelike. The rst two conditions imply|f ′(0)ξt|2 = |ξx · ∇xϕ(0)|2 ≤ |ξx|2|ϕ(0)|2 = |ξt|2|∇xϕ(0)|2. This implies ξt = 0 and so ξx = 0.That means that the condition is empty.
Any timelike surface satises the unique continuation. This is very natural. Actually, theCauchy problem is hyperbolic and indeed locally wellposed for any timelike hypersurface (likefor instance the wave equation posed with initial data at t = 0).
|f ′(0)|2 > |∇xϕ(0)|2, the typical example is unique continuation across an hypersurface of theform Φ(t, x) = ϕ(x), the condition is then
ξ2t = |ξx|2 and ξx · ∇xϕ(0) = 0 =⇒ Hessx(ϕ)(0)(ξx, ξx) > 0 ∀ξ ∈ Rn \ 0.
Typically, if Φ(t, x) = |x|2 − 1, the condition holds when we want to prove the unique contin-uation from the exterior of the ball to the interior and not in the other direction. There areactually counterexamples if we allow a potential smooth in t and x (see Alinhac-Baouendi [1])
Note also that for the 1D wave equation, the constraint ξx · ∇xϕ(0) is much more demandingand implies ξx = 0 and ξ = 0 if ξ2
t = |ξx|2. This is natural since we can exchange the time andspace variable. So the nite speed of propagation (or a more rened version of it) implies easilythe unique continuation across any non caracteristic hypersurface (|f ′(0)|2 6= |∂xϕ(0)|2 in oursetting).
The previous Theorem will be nally proved in section 2.3.2. The next sections are some pre-liminary, especially Carleman estimates that will be used for the proof of the unique continuation.
19
2.2 Carleman estimates
As we have seen in the Introduction in subsection 1.1.4, our goal is to get some estimates of the typeof (1.11). Our main result in this section is Theorem 2.2.1. Yet, its assumptions may seem not sonatural at rst sight. So, before stating we describe some properties of the conjugated operator.
Hence, we will need to study the conjugated operator PΦ := eτΦPe−τΦ that we will see as anoperator depending on τ as dened in the previous section. We can actually compute its principalsymbol.
Lemma 2.2.1 (The conjugated operator). Let P =∑
α pα(x)Dα ∈ Diffm be a (classical) dierentialoperator with principal symbol pm ∈ Σm and Φ ∈ C∞ real valued.
Then, PΦv = eτΦPe−τΦv ∈ Diffmτ . Hence, it is a dierential operator depending on τ of order min (ξ, τ). Moreover, its principal symbol (as an operator depending on τ) is
pφ,m = pm(x, ξ + iτ∇Φ) =∑α
pα(x)(ξ + iτ∇Φ)α
We will denote it pφ for simplicity in the sequel.
Roughly speaking, the previous Lemma says that pΦ is obtained by replacing ξ by ξ + iτ∇Φ inpm.
Note that it implies for instance that pΦ may have a complex symbol even if p is real valued.
Proof. We easily check that eτΦDje−τΦu = Dju+ iτ(∂jΦ)u. In particular, the conjugated operator
eτΦDje−τΦ ∈ Diff1
τ is a dierential operator of order 1 is (D, τ) with principal symbol ξj + iτ∂jΦ.Moreover,
eτΦDαjj e−τΦ = eτΦDje
−τΦeτΦDj · · · eτΦDje−τΦ
= (eτΦDje−τΦ)(eτΦDje
−τΦ) · · · (eτΦDje−τΦ).
Therefore, using Proposition 1.2.2 |αj | times, we get that it is a dierential operator depending on τof order |αj | with principal symbol (ξj + iτ∂jΦ)|αi| (note that the full symbol is more complicated).
So, since Dα = Dα11 · · ·D
αjj · · ·Dαn
n , we obtain similarly that eτΦDαe−τΦ has principal symbol
n∏i=1
(ξj + iτ∂jΦ)|αi| = (ξ + iτ∇Φ)α
using the notation dened in section 1.1.3.Since P =
∑α pα(x)Dα and pα commutes with eτΦ, it gives the result by summing up.
Example 1. Take P = −∆ with symbol |ξ|2. We compute
eτΦ(−∆)e−τΦu = −eτΦ[∆(e−τΦ)u+ e−τΦ∆u+ 2∇u · ∇(e−τΦ)
]= −eτΦ
[−τ(∆Φ)e−τΦ + τ2|∇Φ|2e−τΦu+ e−τΦ∆u− 2τ∇u · ∇Φe−τΦ
]= τ(∆Φ)u− τ2|∇Φ|2u−∆u+ 2τ∇u · ∇Φ
where we have used
∇(e−τΦ) = −τ∇Φe−τΦ
∆(e−τΦ) = div(∇(e−τΦ)) = −τdiv(∇Φe−τΦ) = −τ(∆Φ)e−τΦ − τ∇Φ · ∇(e−τΦ)
= −τ(∆Φ)e−τΦ − τ2|∇Φ|2e−τΦ.
20
τ2|∇Φ|2u, ∆u and τ∇u · ∇Φ are of order 2 with respective symbol τ2|∇Φ|2, −|ξ|2 and iτξ · ∇Φ(remember that ∇u = (∂1u, · · · , ∂nu) = i(D1u, · · · , Dnu) has complex symbol denoted for short iD ).
So, denoting pΦ the full symbol of P and pΦ,2 its principal symbol, we have
pΦ = |ξ|2 − τ2|∇Φ|2 + 2iτξ · ∇Φ + τ(∆Φ)u
pΦ,2 = |ξ|2 − τ2|∇Φ|2 + 2iτξ · ∇Φ
Note that we have pΦ,2 = |ξ + iτ∇Φ|2 = p2(x, ξ + τ∇Φ)
The right condition for obtaining Carleman estimates is the condition of pseudoconvexity.
Denition 2.2.1 (Strict pseudo-convexity for functions). Let P ∈ Diff2 be a (classical) dierentialoperator with real valued principal symbol p2 ∈ Σ2 and Φ ∈ C∞ real valued.
We say that the function Φ is pseudoconvex with respect to P at x0 if it satises
p2, p2,Φ (x0, ξ) > 0, if p2(x0, ξ) = 0 and ξ 6= 0;(2.5)
1
iτpΦ, pΦ(x0, ξ) > 0, if pΦ(x0, ξ) = 0 and τ > 0,(2.6)
where pΦ(x, ξ) = p2(x, ξ + iτ∇Φ).
Note that the previous denition is clearly stronger than Denition 2.1.1. Moreover, it is not onlydependent on the level set of the functions. It also depends on the "convexity with respect to thelevel sets". This is not only a geometric quantity, contrary to Denition 2.2.1 of pseudoconvexity forsurfaces. Yet, we will see later that for a surface satisfying Denition 2.1.1, there is a good choice offunction satisfying the assumption of Denition 2.2.1 and leading to a unique continuation theorem.
Note also, that in some sense, we could say that for real operators, the rst line is the limit ofthe second line as τ tends to 0, see Lemma 2.2.4 below.
Theorem 2.2.1 (The Carleman estimate). Let P ∈ Diff2 be a (classical) dierential operator withreal valued principal symbol p ∈ Σ2 and Φ ∈ C∞ real valued, strictly pseudoconvex with respect to Pat x0, as in Denition 2.2.1.
Then, there exists a neighborhood V of x0, C > 0 and τ0 > 0 so that we have the followingestimate
τ∥∥eτΦu
∥∥2
H1τ≤ C
∥∥eτΦPu∥∥2
L2 ,(2.7)
for any u ∈ C∞0 (V ) and τ ≥ τ0.
Proof. Before going further, let us notice that Lemma 2.2.1 only depends on the leading order of theoperator P . More precisely, if it is true for one P , it is true for P + R with R ∈ Diff1. Indeed,eτΦ (P +R) e−τΦ = Pφ +RΦ with RΦ ∈ Diff1
τ . In particular, if (2.8) is true for P , we can write
τ ‖v‖2H1τ≤ C ‖PΦv‖2L2 ≤ 2C ‖(PΦ +RΦ)v‖2L2 + 2C ‖RΦv‖2L2 ≤ 2C ‖(PΦ +RΦ)v‖2L2 +D ‖v‖2H1
τ
where we have used RΦ ∈ Diff1τ and Proposition 1.2.1. But for τ large enough, we have τ −D ≥ τ/2,
which gives (2.8) with P replaced by P +R and with dierent constants C and τ0.P can be written P =
∑k,l ak,lDkDl +R1 with ak,l = al,k real valued and R1 ∈ Diff1. Moreover,
since ak,lDkDl = Dkak,lDl − (Dkak,l)Dl where the second term is of order 1. So P can be writtenP =
∑k,lDkak,lDl + R2 with 2R2 ∈ Diff1. We can assume now that P is a sum of terms Dkak,lDl
with ak,l = al,k real valued. The previous simplication is only technical and allows to impose Pformally selfadjoint.
21
As already described in the introduction, if we make the change of unknown v = eτΦu, theexpected inequality (2.7) is equivalent to
τ ‖v‖2H1τ≤ C ‖PΦv‖2L2 ,(2.8)
where PΦv = eτΦPe−τΦv.Lemma 2.2.1 says that PΦ ∈ Diff2
τ with principal symbol p(x, ξ + iτ∇Φ). Yet, this symbol cancancel and we cannot apply Proposition 1.2.6. In some sense, we have to go to the "next order".
Denote
QR =PΦ + P ∗Φ
2; QI =
PΦ − P ∗Φ2i
.
So that
PΦ = QR + iQI .(2.9)
This is more or less the decomposition of PΦ according to its real and imaginary part (in the senseof operators). This can be seen for instance at the level of principal symbols since, by Proposition1.2.4, QR and QI are in Diff2
τ with principal symbols
qR =pΦ + pΦ
2= Re pΦ; qR =
pΦ − pΦ
2i= Im pΦ.
Moreover, we easily check that they are formally selfadjoint: Q∗R = QR and Q∗I = QI . Using, thisproperty and (2.9), we compute for v ∈ C∞0 (Rn)
‖PΦv‖2L2 = (PΦv, PΦv)L2 = ((QR + iQI)v, (QR + iQI)v)
= (QRv,QRv) + (iQIv, iQIv) + (QRv, iQIv) + (iQIv,QRv)
= ‖QRv‖2L2 + ‖QRv‖2L2 − i (QI QRv, v) + i (QR QIv, v)
= ‖QRv‖2L2 + ‖QRv‖2L2 + (i[QR, QI ]v, v) .
Now, we have 2 kinds of terms
the one with ‖QRv‖2L2 (and resp. ‖QIv‖2L2) that corresponds to(Q2Rv, v
)where Q2
R is of order4 with principal symbol (Re pΦ)2 (resp. (Im pΦ)2)
the one with i[QR, QI ] which is of order 2 + 2− 1 = 3 and principal symbol Re pΦ, Im pΦ byProposition 1.2.3
The rst kind has stronger order but they can cancel and are therefore not sucient to get the"coercivity" estimate. So, the idea is to consider that in the case when qR or qI cancel, we use thenext term of commutator. But to compare them, we will need to bring them to the same order andin some sense make a "sacrice" of this main order 4. More precisely, let C > 0 large to be xedlater. For τ large enough we have 1
τ1/2 ≥ C(|ξ|2+τ2)1/2 for any ξ ∈ Rn. So, this gives after Fourier
transform,
1
τ‖QRv‖2L2 =
∥∥∥∥QRvτ1/2
∥∥∥∥2
L2
≥∥∥∥∥ C
(−∆ + τ2)1/2QRv
∥∥∥∥2
L2
≥(
C2
(−∆ + τ2)1/2QRv,
1
(−∆ + τ2)1/2QRv
)≥
(QR
C2
(−∆ + τ2)QRv, v
).
22
Note that we have applied Q∗R = QR for some function (−∆+τ2)−1QRv which does not have compactsupport, but it can be easily justied for Schwarz functions also. The same applies to QI .
So, at that step, we have proved
1
τ‖PΦv‖2L2 ≥ (Lv, v)(2.10)
with L = QRC2
(−∆+τ2)QR +QI
C2
(−∆+τ2)QI + i
τ [QR, QI ].
We would like to apply Lemma 1.2.1 to L. First, we need to remark that iτ [QR, QI ] ∈ Diff2
τ . Theorder are correct since [QR, QI ] is of order 3 while 1
τ is "morally" of order −1. We just need to showthat we can "factorize by τ".
Let us prove that QI (of order 2 in (ξ, τ)) can be written τQI with QI of order 1.Indeed, in the beginning of the proof, we proved that it was enough to consider p(x, ξ) a quadratic
form. Hence, p(x,D) can be decomposed as a sum of terms of the form Dkak,l(x)Dl with ak,l(x) realvalued and ak,l = al,k. For this kind of terms, we have the following conjugate operator
e−τΦDkak,l(x)Dle−τΦ = (Dk + iτ∂kΦ)ak,l(x)(Dl + iτ∂lΦ)
=[Dkak,l(x)Dl − τ2ak,l(x)(∂kΦ)(∂lΦ)
+iτ (Dkak,l(x)∂lΦ + (∂kΦ)ak,l(x)Dl)]
We directly recognize it can be written PΦ = P + τM with M ∈ Diff1τ . Since by choice, P = P ∗, we
have the decomposition QR = P + τM+M∗
2 and QI = τM−M∗
2i . So, we can take QI = M−M∗2i .
So, we have in that case
L = QRC2
(−∆ + τ2)QR +QI
C2
(−∆ + τ2)QI + i[QR, QI ]
with QI of order 1. So, i[QR, QI ] is of order 2 + 1 − 1 = 2 with principal symbol 1τ Re pΦ, Im pΦ.
We have the following Lemma which allows to conclude thanks to (2.10) and Lemma 1.2.1.
Lemma 2.2.2. Under the assumptions of pseudoconvexity for functions, that is Denition 2.2.1,there exists C1 and C2 > 0 so that
C1
|ξ|2 + τ2
[(Re pΦ)2 + (Im pΦ)2
]+
1
τRe pΦ, Im pΦ ≥ C2
(|ξ|2 + τ2
)(2.11)
taken at the point (x0, ξ, τ), for all ξ ∈ Rn, τ > 0.
Remark 2.2.1. This kind of inequality is often called subelliptic estimates. Indeed, if the operator PΦ
was elliptic in the (ξ, τ) variable, we would have some estimates by below with the norm H2τ instead
of H1τ as in Proposition 1.2.6. But here the principal symbol of PΦ, p2(x, ξ + iτ∇Φ) can have some
zero (with (ξ, τ) 6= 0) even if p2 is elliptic.Take for instance the Laplace operator described in Example 1. The principal symbol of pΦ is
|ξ|2 − τ2|∇Φ|2 + 2iτξ · ∇Φ. It cancels if we take ξ ⊥ ∇Φ and τ2 = |ξ|2/|∇Φ|2, which is alwayspossible.
Note that it can seem surprising since for xed τ , the operator PΦ is elliptic in the ξ variable. Wecould expect to have some inequalities of the form
‖u‖H2 ≤ Cτ ‖PΦu‖L2 .
It is indeed possible if P is elliptic, but the constant Cτ will then blow-up as τ1/2.
23
Proof of Lemma 2.2.2. Note rst that since f, f = 0 and f, g = −g, f for any f and g, wehave
1
iτpΦ, pΦ =
1
iτRe pΦ − i Im pΦ,Re pΦ + i Im pΦ
=1
τRe pΦ, Im pΦ −
1
τIm pΦ,Re pΦ
=2
τRe pΦ, Im pΦ.
Moreover, by the previous computation, we can write Im pΦ = τ qI (note that it could be seen as aconsequence of the fact that p is real and pΦ = p on the set τ = 0, so we can factorize Im pΦ byTaylor expansion).
We notice that all the terms in (2.11) are homogeneous in (ξ, τ) of order 2 and continuous thanks tothe previous remark. Therefore, it is enough to prove (2.11) on the setK =
(ξ, τ)
∣∣|ξ|2 + τ2 = 1; τ ≥ 0.
It is a consequence of the following Lemma with f = (Re pΦ)2 + (Im pΦ)2 and g = 2Re pΦ, qIandh = 0 (the function h will be usefull for another application later.
The Lemma 2.2.4 after, proves that actually, the rst assumption in Denition 2.2.1 is the limit ofthe second one on the set τ = 0. So that the hypothesis hold, up to the set τ = 0∩|ξ|2+τ2 = 1.
Lemma 2.2.3. Let K be a compact set and f, g, h three continuous real valued functions on K.Assume that f ≥ 0 on K, and g > 0 on f = 0. Then, there exists A0, C > 0 such that for allA ≥ A0, we have g +Af − 1
Ah ≥ C on K.
Lemma 2.2.4. Let p be real valued, then limτ→0
1iτ pΦ, pΦ = 2 p p,Φ.
Proof of Lemma 2.2.3. The set N = f = 0 ∩K is compact. Since g is continuous, its minimumon the set N is reached. So, we have g ≥ C1 > 0 on N . Since g is continuous and N compact,there exists an open neighborhood V of N so that g ≥ C1/2 on V . Now, K \ V is closed in Kand indeed compact. So, f reach its minimum on K \ V . But since (K \ V ) ∩ N = ∅, we havef 6= 0 on K \ V and indeed f > 0. So the minimum C2 is actually strictly positive. Dene nowC3 = min g(x), x ∈ K \ V and C4 = max |h(x)|, x ∈ K.
We are in the following situation, for some A still to be chosen:
on V , we have g +Af − 1Ah ≥
C12 −
1AC4.
on K \ V , we have g +Af − 1Ah ≥ C3 +AC2 − 1
AC4.
So, we need to choose A so that it leads to positive lower bound. If we want the nal estimate withC = C1/4, for instance, we need
A ≥ 4C4C1
A2C2 +A(C3 − C1
4
)− C4 > 0
Since C2 > 0, the last case is fullled if A is large enough since the polynomial of order 2 convergesto +∞ as A goes to ∞.
Proof of Lemma 2.2.4. We rst notice that for τ = 0, pΦ, pΦ = p, p so since p is real, pΦ, pΦ = 0for τ = 0. So, we have by Taylor expansion in τ , lim
τ→0
1τ pΦ, pΦ = ∂
∂τ pΦ, pΦ∣∣τ=0
. Also, we easily
verify ∂τ (pΦ, pΦ) = ∂τpΦ, pΦ+ pΦ, ∂τpΦ.
24
But since p is real, pΦ = p(x, ξ − iτ∇Φ), so that
∂τpΦ = i∇Φ · ∇ξp(x, x, ξ + iτ∇Φ) = ipΦ,Φ∂τpΦ = −i∇Φ · ∇ξp(x, x, ξ − iτ∇Φ) = −ipΦ,Φ.
So, we get ∂τ (pΦ, pΦ) = −ipΦ,Φ, pΦ+ ipΦ, pΦ,Φ. When specied for τ = 0, we get
∂
∂τpΦ, pΦ
∣∣∣∣τ=0
= −ip,Φ, p+ ip, p,Φ = 2ip, p,Φ.
This gives the result.
The proof of Lemma 2.2.3 is left to the reader.
Remark 2.2.2. In the elliptic case, the "trick" of factorisation by τ of the Imaginary part of QI canbe avoided. Indeed, close to τ = 0 the symbol pΦ is actually close to p and is therefore non zero.
Note that a more general assumption for treating the behavior of pΦ, pΦ close to τ = 0 is touse the principal normality assumption
|p, p| ≤ C|p||ξ|m−1.
The inequality is obviously fullled in the elliptic case where |p||ξ|m−1 ≥ C|ξ|2m−1 and for real valuedreal principal symbol where p, p = 0.
In the more general case of principal normality, for the behavior of 1iτ pΦ, pΦ, close to τ = 0,
the inequality allows to take advantage of the term |pφ|2for proving some inequality related to (2.11).In that case, a variant of Lemma 2.2.4 remains true, but only close to the set pΦ = 0.
We refer to Section 2.6 for the statemenent of the Theorem and to Hörmander [7] Chapter XXVIIIfor the proof.
2.2.1 Geometric interpretation of pseudoconvexity in the case of real symbol of
order 2
2.3 Using Carleman estimates for unique continuation
2.3.1 Convexication
Up to now, we have proved some Carleman estimates under some complicated conditions of Denition2.2.1. The main purpose of this section is to prove that for some function satisfying Denition 2.1.1,we can nd some other function approriate for the Carleman estimate, that is satisfying Denition2.2.1. We will rst give a dierent formulation of Denition 2.1.1.
Proposition 2.3.1 (Usual pseudoconvexity for surfaces). Let Φ satisfying the pseudoconvexity as-sumptions for surfaces of Denition 2.1.1 for an operator P ∈ Diff2 with real valued principal symbol.Then, it satises the stronger property for any x0 ∈ Ω
p2, p2,Φ (x0, ξ) > 0, if p2(x0, ξ) = p2,Φ(x0, ξ) = 0 and ξ 6= 0;(2.12)
1
iτpΦ, pΦ(x0, ξ) > 0, if pΦ(x0, ξ) = pΦ,Φ(x0, ξ) = 0 and τ > 0,(2.13)
where pΦ(x, ξ) = p2(x, ξ + iτ∇Φ).
25
Note that we have added a second property that makes it look very similar to the strict pseudo-convexity for functions of Denition 2.2.1. It is just slightly weaker because the inequality is askedin some more constrained cases because we have asked the additional conditions p2,Φ = 0 andpΦ,Φ = 0.
The conclusion of the previous Proposition 2.3.1 are actually the usual conditions of pseudocon-vexity for surfaces that are the assumptions for unique continuation for operators of higher order.
Proof. The rst property is exactly the same, so , we just need to prove that the second (2.13) isautomatically fullled in that case.
For xed ξ ∈ Rn, denote f(s) = p2(x0, ξ+ s∇Φ(x0)). Since p is a polynomial of order 2 (on Rn),f is a polynomial in one variable s of order at most 2, with real valued coecients. Moreover, wehave f ′(s) = ∇Φ(x0) · ∇ξp2(x0, ξ + s∇Φ(x0)) = p2,Φ(x0, ξ + s∇Φ(x0)). This is actually pΦ,Φtaken at the point x0, ξ and τ = s/i. Therefore, the assumptions of (2.13) are equivalent to the factthat there exists one s = iτ with τ ∈ R so that f(s) = 0 and f ′(s) = 0. That means that s = iτ is aroot of order 2 and s is a double root of f .
Note that since p2 is homogeneous of order 2, Lemma A.4.1 gives
pΦ(x0, ξ, τ) = p2(x0, ξ + iτ∇Φ) = p2(x0, ξ)− τ2p2(x0,∇Φ) + iτp,Φ(x0, ξ).(2.14)
We will distinguish 2 cases
f is of order exactly 2, that is the case if p2(x0,∇Φ) 6= 0, we say that the surface Φ(x) = Φ(x0)non characteristic for P at x0
f is of order at most 1, p2(x0,∇Φ) = 0, we say that the surface Φ(x) = Φ(x0) is characteristicfor P at x0
Assume now that f is of order exactly 2. Since f is real valued, its root are either both reals or twocomplex conjugates. So, the double root can only be s = 0, which is forbidden in the assumption(actually this case refers to the assumption (2.12), see Lemma 2.2.4).
Assume now that f is of order at most 1. The fact that f has a double root implies that actually,f is the zero polynomial. That is p2(x0, ξ) = p2(x0,∇Φ) = p,Φ(x0, ξ) = 0. Note that thanks to(A.22), it also implies pΦ,Φ(x0, ξ, τ) = 0 for every τ ∈ R.
Since p2 (noted p) is homogeneous of order 2, real valued , we obtain after some computations(see Lemma A.4.2) that
1
iτpΦ, pΦ = 2 p, p,Φ+ 2τ2 p, p,Φ (x,∇Φ).
Yet, we are in some case where ξ satises p2(x0, ξ) = p,Φ(x0, ξ) = 0, so the pseudoconvexity forsurface implies either p, p,Φ > 0 or ξ = 0.
Moreover, we are in some case where p2(x0,∇Φ) = 0 and, by (A.22), pΦ,Φ(x0,∇Φ, τ) =2R(∇Φ,∇Φ)+2iτp(x,∇Φ) = 2p(x,∇Φ)+2iτp(x,∇Φ) = 0. So the pseudoconvexity condition takenat the point ∇Φ implies p, p,Φ (x,∇Φ) > 0.
In particular, 1iτ pΦ, pΦ (x0, ξ, τ) > 0 when τ > 0.
A simpler proof, could have been made if we assume that we are in some coordinates so thatΦ = x1. Actually, this is not a loss of generality since we could prove (but we did not do it yet) thatthe assumptions and conclusions of Proposition 2.3.1 are invariant by change of coordinates and ofdening function for the surface. In that case, we can check that actually f can never be identicallyzero. Indeed, if it happens, we have 0 = f(s) = pΦ = p2(x0, ξ + se1) and ∂ξ1p = 0. It gives
p, p,Φ = p, p, x1 = p, ∂ξ1p = ∇ξp · ∇x∂ξ1p−∇xp · ∇ξ∂ξ1p = 0.
26
This is impossible since we have p(x, ξ) = p,Φ = 0 in the considered points. It contradict the rstassumption.
Note that the cancellation of p, p,Φ under these assumptions is specic to the chosen coordi-nates.
Now, we will present a systematic way to produce some functions satisfying the stronger assump-tion for some function satisfying the weaker pseudoconvexity for surfaces.
Proposition 2.3.2 (Analytic convexication). Let Φ pseudoconvex in the sense of surfaces, withΦ(x0) = 0, that is satisfying the assumptions of Denition 2.1.1 (and therefore, the one of Proposition2.3.1)
Then there exists λ large enough so that the functions Ψ = eλΦ satises the pseudoconvexity forfunctions of Denition 2.2.1.
Note that the value of the level sets of Ψ are dierent, but the level set of Φ and Ψ are actuallythe same: the geometry of the level sets did not change. That means for a function that satisfy theunique conditions of the Theorem of Unique Continuation, we produce some new function admissiblefor the Carleman estimates, whose level sets have exactly the same geometric properties.
Proof. pΨ = p(x0, ξ + iτλΨ∇Φ). We have generally
1
iτpΨ, pΨ(x0, ξ, τ) =
1
iτ[∇ξp(x0, ξ − iτ∇Ψ) · (∇xp(x0, ξ + iτ∇Ψ))]
+Hess(Ψ) [∇ξp(x0, ξ − iτ∇Ψ);∇ξp(x0, ξ + iτ∇Ψ)]
− 1
iτ[∇xp(x0, ξ − iτ∇Ψ) · (∇ξp(x0, ξ + iτ∇Ψ))]
+Hess(Ψ) [∇ξp(x0, ξ − iτ∇Ψ);∇ξp(x0, ξ + iτ∇Ψ)]
=2
τIm [∇ξp(x0, ξ − iτ∇Ψ) · (∇xp(x0, ξ + iτ∇Ψ))]
+2Hess(Ψ) [∇ξp(x0, ξ − iτ∇Ψ);∇ξp(x0, ξ + iτ∇Ψ)] .
In order to simplify the notations, once x0 is xed, we will denote cΨ(ξ, τ) = 1iτ pΨ, pΨ(x0, ξ, τ) and
the same thing for cΦ. Moreover, we will use the convention of replacing cΦ(ξ, τ) = 1iτ pΦ, pΦ(x0, ξ, τ)
by its limit 2p, p,Φ if τ = 0. And this extension is still continuous.We compute
∂jΨ = λ∂jΦeλΦ
∂j,kΨ = λ∂j,kΦeλΦ + λ2(∂jΦ)(∂kΦ)eλΦ
that we can write in a shorter way
∇Ψ = λ∇ΦeλΦ
Hess(Ψ)(ξ, ξ) = λHess(Φ)(ξ; ξ)eλΦ + λ2(ξ · ∇Φ)(ξ · ∇Φ)eλΦ
And taken at the point x0, this gives, noticing several times that eλΦ(x0) = 1
∇Ψ(x0) = λ∇Φ
Hess(Ψ)(x0)(ξ, ξ) = λHess(Φ)(x0)(ξ; ξ)eλΦ + λ2(ξ · ∇Φ(x0)(ξ · ∇Φ(x0))
27
Using the previous computations for Ψ, we get, (we drop the fact that Φ and the dierent derivativesof Φ are taken at x0)
cΨ(ξ, τ) =1
iτpΨ, pΨ(x0, ξ, τ)
=2
τIm [∇ξp(x0, ξ − iτλ∇Φ) · (∇xp(x0, ξ + iτλ∇Φ))]
+2λHess(Φ) [∇ξp(x0, ξ − iτλ∇Φ);∇ξp(x0, ξ + iτλ∇Φ)]
+2λ2 (∇ξp(x0, ξ − iτλ∇Φ) · ∇Φ) (∇ξp(x0, ξ + iτλ∇Φ) · ∇Φ)
= λcΦ(ξ, λτ) + 2λ2 |p,Φ(x0, ξ + iτλ∇Φ)|2
= λcΦ(ξ, λτ) + 2λ2 |pΦ,Φ(x0, ξ, λτ)|2 .
That is this additional term that comes from the convexication that will allow to get some morepositivity when pΦ,Φ 6= 0. The positivity when pΦ,Φ 6= 0 being ensured by the assumptions onΦ.
Using Lemma 2.2.3 (combined with Lemma 2.2.4 for the limit when τ = 0, see the proof ofCarleman estimates), Property 2.13 gives some constants C1, C2 so that
cΦ(ξ, τ) + C1 |pΦ,Φ(x0, ξ, τ)|2 + C1|pΦ(x0, ξ, τ)|2
|ξ|2 + τ2≥ C2(|ξ|2 + τ2).
for any τ ≥ 0, |ξ|2 + τ2 = 1, with the convention of replacing cΦ(ξ, τ) = 1iτ pΦ, pΦ(x0, ξ, τ) by its
limit 2p, p,Φ if τ = 0. Replacing τ by λτ for λ ≥ 1 and by homogeneity, it can be reformulated
cΦ(ξ, λτ) + C1 |pΦ,Φ(x0, ξ, λτ)|2 + C1|pΦ(x0, ξ, λτ)|2
|ξ|2 + λ2τ2≥ C2(|ξ|2 + λ2τ2).
for any (ξ, τ) 6= (0, 0) with τ ≥ 0.
Moreover, we notice that |pΦ(x0,ξ,λτ)|2|ξ|2+λ2τ2 = |pΨ(x0,ξ,τ)|2
|ξ|2+λ2τ2 .
Taking 2λ ≥ C1, it gives
1
iτpΨ, pΨ(x0, ξ, τ) + C1λ
|pΨ(x0, ξ, τ)|2
|ξ|2 + τ2
= cΨ(ξ, τ) + C1|pΨ(x0, ξ, τ)|2
|ξ|2 + τ2
= λcΦ(ξ, λτ) + 2λ2 |pΦ,Φ(x0, ξ, λτ)|2 + C1λ|pΨ(x0, ξ, τ)|2
|ξ|2 + τ2
= λ[cΦ(ξ, λτ) + 2λ |pΦ,Φ(x0, ξ, λτ)|2
]+ C1
|pΨ(x0, ξ, τ)|2
|ξ|2 + τ2
≥ λ
[cΦ(ξ, λτ) + C1 |pΦ,Φ(x0, ξ, λτ)|2 + C1
|pΨ(x0, ξ, τ)|2
|ξ|2 + λ2τ2
]≥ C2λ(|ξ|2 + λ2τ2) ≥ C2λ(|ξ|2 + τ2).
This implies the strong convexity for functions (still with the same convention when τ = 0).
Proposition 2.3.3 (Stability and Geometric convexication). Let Ψ pseudoconvex in the sense offunctions, that is satisfying the assumptions of Denition 2.2.1.
Then there exists ε0 small enough so that for any 0 < ε < ε0, the functions Ψε = Ψ− ε|x− x0|2still satises the pseudoconvexity for functions.
28
Note that this step has slightly changed the level sets of Ψ. The level sets Ψε = 0 are nowslightly bended (except at x0) into the set Ψ > 0 where u will be assumed to be zero. This slightchange will be crucial for the unique continuation.♣ faire un dessin
Proof. First, we notice that we can prove as in the previous Proposition before that Denition 2.2.1(still using Lemma 2.2.3 combined with Lemma 2.2.4 for the limit when τ = 0) implies (and isactually equivalent to) the existence of an inequality of the form
1
iτpΨ, pΨ(x0, ξ, τ) + C1
|pΨ(x0, ξ, τ)|2
|ξ|2 + τ2≥ C2(|ξ|2 + τ2).
uniformly for (ξ, τ) with |ξ|2 + τ2 = 1, τ ≥ 0, still with the same convention as before when τ = 0.It can be easily checked that this estimates only involves some derivatives of Ψ up to order 2 at
the point x0. It is therefore stable by the addition of a function small for the C2 norm around x0.
2.3.2 Unique continuation
In this section, we give the nal proof of Theorem 2.1.2.
Proof of Theorem 2.1.2. Let u solution of Pu = 0 in Ω so that u = 0 on Ω ∩ Φ > 0. Φ is pseu-doconvex for surfaces at x0. Proposition 2.3.2 allows to produce a new function Ψ pseudoconvex forfunctions so that (up to changing Ψ by Ψ− 1) Ψ ≥ 0 = Φ ≥ 0 and Ψ < 0 = Φ < 0.
Proposition 2.3.3 proves that for some small ε > 0, Ψε = Ψ− ε|x− x0|2 satises the pseudocon-vexity for functions. It therefore satises the following properties
1. there exists R > 0, C > 0 and τ0 > 0 so that we have the following estimate
τ∥∥eτΨεw
∥∥2
H1τ≤ C
∥∥eτΨεPw∥∥2
L2 ,(2.15)
for any w ∈ C∞(B(x0, R)) and τ ≥ τ0.
2. there exists η > 0 so that Ψε(x) ≤ −η for x ∈ Φ ≤ 0 ∩ |x− x0| ≥ R/2,
3. there exists a small neighborhood V ⊂ B(x0, R/2) of x0 so that Ψε(x) ≥ −η/2 for x ∈ V .
The property 1 is a consequence of Theorem 2.2.1.Property 2 is true thanks to the parameter ε in the geometric convexication. Indeed, for |x−x0| ≥
R/2, we have Ψε(x) ≤ Ψ(x)− εR2/4. Moreover, Φ(x) ≤ 0 implies Ψ ≤ 0, so that Ψε(x) ≤ −εR2/4.So we can take η = −εR2/4.
Property 3 is only a continuity argument since Ψε(x0) = 0.Pick χ ∈ C∞0 (B(x0, R)) so that χ = 1 on B(x0, R/2). We want to apply the Carleman estimate
to w = χu solution of Pw = χPu + [P, χ]u = [P, χ]u. Notice that [P, χ] is a classical dierentialoperator of order 1 with coecients supported in the set Φ ≤ 0 ∩ R2 ≤ |x − x0| ≤ R where wehave Ψε(x) ≤ −η. In particular, we have
∥∥eτΨεPw∥∥L2 ≤ Ce−τη ‖u‖H1 .
Moreover, since Ψε(x) ≥ −η/2 on V , we have e−τη/2 ‖u‖L2(V ) ≤∥∥e−τη/2u∥∥
L2(V )≤∥∥eτΨεu
∥∥L2(V )
≤∥∥eτΨεχu∥∥L2(V )
≤ τ1/2∥∥eτΨεw
∥∥H1τ. So the Carleman estimate implies
e−τη/2 ‖u‖L2(V ) ≤ τ1/2∥∥eτΨεw
∥∥H1τ≤ C
∥∥eτΨεPw∥∥L2 ≤ Ce−τη ‖u‖H1 .
This gives ‖u‖L2(V ) ≤ Ce−τη/2 ‖u‖H1 and u = 0 on V by letting τ tend to innity.
29
Note that when the operator P is replaced by P +∑
k bk(x)∂k + c(x) with bk, c ∈ L∞(Rn), themodied P does not exactly belong to the class Diff2 because the coecients are not smooth. Forinstance, the commutator could involve some derivative in the coecients of the lower order terms.Yet, we proceed slighltly dierently. We apply the same reasoning to P where P only contains theterms of order 2 that are smooth. The only dierence is that we don't have any more Pχu = [P, χu]but Pχu = [P, χu]−
∑k bk(x)∂k(χu)− c(x)χu.
For instance, for the 1 order terms, these terms can be estimated by∥∥eτΨεbk(x)∂kw∥∥L2 ≤ ‖bk‖L∞
∥∥eτΨε∂kw∥∥L2 ≤ ‖bk‖L∞
[∥∥∇ [eτΨεw]∥∥L2 + τ
∥∥(∇Ψε)eτΨεw
∥∥L2
]≤ C
∥∥eτΨεw∥∥H1τ.
where C depends on ‖bk‖L∞ and ‖Ψε‖H1(B(0,R)). This term can be made smaller than 12τ
1/2∥∥eτΨεw
∥∥H1τ
for τ large enough and therefore, (2.15) remains true with the constant replaced by a bigger one. Wecan then continue the proof similarly.
The unique continuation result that we obtain is only local. Yet, we could expect to iterate thisresult with a well chosen sequence of hypersurfaces. It turns out that it is not easy to do in thegeneral case. Take for instance the wave equation. Suppose that you are given an open set ω andT > 0 so that a solution u satises
∂2t −∆u = 0 on [0, T ]× Ω
u = 0 on [0, T ]× ω
The question of dening what is the "domain of dependence" for the unique continuation that can beobtained with our unique continuation using iterated pseudoconvex surfaces (as described in Remark2.1.2) is not clear.
An easier case is the elliptic case where the pseudoconvexity condition for the surface is almostempty. This allows to get the following global result.
Theorem 2.3.1 (Global result in the elliptic case). Let Ω a connected open set. Let P satisfying thesame assumptions as Theorem 2.1.1. Let u smooth solution of Pu = 0 on Ω, that satises u = 0 onan arbitrary open set ω ⊂ Ω.
Then, u = 0 on Ω.
Proof. The support of u, denoted by F is a closed set. Let us prove that it is also open. Let x ∈ F . Wedistinguish two cases: If x ∈ Int(F ), it by denition that there exists a neighborhood of x includedin F . In the other case, x ∈ Fr(F ) = F \ Int(F ), the boundary in Ω. Dene R so that B(x,R) ⊂ Ω.Take x1 ∈ Ω \ F with dist(x, x1) < R/2 (it exists since x ∈ Fr(F )). So, we have u(y) = 0 in aneighborhood of x1 by denition of the support. Dene r1 = sup r ∈ [0, R/2];u(x) = 0 in B(x1, r).We know that r1 > 0.
Assume r1 < R/2. So, we have u = 0 in B(x1, r). Moreover, since dist(x, x1) < R/2 andB(x,R) ⊂ Ω, B(x1, R/2) ⊂ Ω. So, we can apply Theorem 2.1.2 to any point x0 ∈ S(x1, r1)the sphere of radius r1 and of center x1 to get that for any x0 ∈ S(x1, r1), there exists rx0 so thatu(y) = 0 in B(x0, rx0). Covering S(x1, r1) by a nite number of such balls and using the compactnessof S(x1, r1) we get one ε so that u(y) = 0 on B(x1, r1 + ε) contradicting the denition of r1. So, wehave r1 = R/2.
But since dist(x, x1) < R/2, there exists a neighborhood of x included in B(x1, R/2). In partic-ular, u = 0 in this neighborhood. This contradicts the fact that x ∈ Fr(F ). So, Fr(F ) = ∅ and Fis open. So F is open and close and not Ω. So, F = ∅.
♣ faire un dessin
30
2.3.3 Quantitative estimates
In this section, we want to give some estimates that quantify the unique continuation, that is someinequality proving in some sense the implication
Pu small in Ω,u small in U
=⇒ u small in U .
described in the introduction. More precisely, we would like to have some estimates of the kind‖u‖
U≤ ϕ(‖u‖U + ‖Pu‖ω , ‖u‖Ω), with ϕ(a, b)→ 0 as a→ 0 when b is bounded. The ideal situation
would be some linear estimate in a, independent on b. This is the case when the Cauchy problem iswellposed. For instance the wave operator across the surface t = 0. Yet, in those cases, Carlemanestimates are generally not the best way to get uniqueness and good estimates. We will be interestedin some case where the Cauchy problem is ill-posed and linear estimates are not expected to occur.
In our situation, the estimates that we can expect are more of Hölder type, that is ϕ(a, b) = aθb1−θ.We can obtain this kind of estimates locally, with the same generality for the operators as we provedthe unique continuation.
In what follows, we will remain in the elliptic framework where "almost any reasonable" functionis pseudoconvex functions for surfaces at some point x0. It allows to simplify 2 important facts:
we can pick some functions pseudoconvex for surfaces whose level sets are compact. The easierexample being |x− x0|. One advantage is that we can skip the geometric convexication.
the globalization is much easier, as we saw in Theorem 2.3.1 for the unique continuation.
Yet, in that generality, it will be quite complicated to get global one. So, we will keep in the ellipticframework for simplicity.
We rst state the local result
Theorem 2.3.2 (Local quantitative estimates for real elliptic operator of order 2). Let Ω, P be asin Theorem 2.1.2. x0 ∈ Ω. Let r > 0 so that B(x0, 3r) ⊂ Ω.
Then, there exists C > 0, 0 < δ < 1 so that
‖u‖H1(B(x0,2r))≤ C
[‖u‖H1(B(x0,r)
+ ‖Pu‖L2(B(x0,3r))
]δ‖u‖1−δH1(B(x0,3r))
for any u ∈ C∞(Ω).
Note that in the context of complex analysis (for the usual Laplacian), this inequality is usuallycalled 3 sphere inequality and can be proved quite dierently, but only for analytic coecients.
Proof. Denote Φ = −|x−x0|. Using compactness arguments and Proposition 2.3.2, we get that for λlarge enough, Ψ = eλΦ is pseudoconvex for functions, uniformly for every x0 ∈ B(x0, 3r)\B(x0, r/2).The "uniformly" meaning that some estimates like (2.11) are true with one xed C > 0, τ0 > 0uniformly for x0 ∈ B(x0, 3r) \B(x0, r/2). Then, we get exactly by the same reasoning as before butusing the "semiglobal" Gårding inequality of Proposition 1.2.5, we get the Carleman estimate
τ1/2∥∥eτΨv
∥∥H1τ≤ C
∥∥eτΨPv∥∥L2 ,(2.16)
for any v ∈ C∞0 (B(x0, 3r) \B(x0, r/2)), τ ≥ τ0.Take χ ∈ C∞0 (B(x0, 3r)\B(x0, r/2)) so that χ = 1 on B(x0, 5r/2)\B(x0, r). Apply the estimate
to v = χu.We have Pv = χPu+ [P, χ]u where [P, χ] is of order 1 supported in two parts of Rn
31
|x− x0| ∈ [r/2, 1], where Ψ ≤ e−λr/2 := ρ3. The corresponding term is bounded by∥∥eτΦ[P, χ]u∥∥L2(|x−x0|∈[r/2,3r/4])
≤ Ceτρ3 ‖u‖H1(B(x0,r))
|x− x0| ∈ [5r/2, 3r], where Ψ ≤ e−λ5r/2 := ρ1. The corresponding term is bounded by∥∥eτΦ[P, χ]u∥∥L2(|x−x0|∈[5r/2,3r])
≤ Ceτρ1 ‖u‖H1(B(x0,3r))
The term corresponding to χPu is bounded by Ceτρ3 ‖Pu‖L2(B(0,3r)) since Ψ ≤ ρ3 on Supp(χ).
The rst term must be estimated by below. We rst write the H1τ norm slightly dierently∥∥eτΨ∇v
∥∥L2(Ω)
+ τ∥∥eτΨv
∥∥L2(Ω)
≤ C∥∥∇(eτΨv)
∥∥L2(Ω)
+ τ∥∥∇ΦeτΨv
∥∥L2 + τ
∥∥eτΨv∥∥L2(Ω)
≤ C∥∥eτΨv
∥∥H1τ.
where the constant C may depend on ‖∇Φ‖L∞(B(x0,3r)).
But since χ = 1 on |x− x0| ∈ [r, 2r], we have, for a dierent constant C∥∥eτΨv∥∥H1τ≥ C
∥∥eτΨ∇v∥∥L2(Ω)
+ C∥∥eτΨv
∥∥L2(Ω)
≥ C∥∥eτΨ∇v
∥∥L2(|x−x0|∈[r,2r])
+ C∥∥eτΨv
∥∥L2(|x−x0|∈[r,2r])
≥ C∥∥eτΨ∇u
∥∥L2(|x−x0|∈[r,2r])
+ C∥∥eτΨu
∥∥L2(|x−x0|∈[r,2r])
≥ Ceτρ2
[‖∇u‖L2(|x−x0|∈[r,2r]) + ‖u‖L2(|x−x0|∈[r,2r])
]where ρ2 := e−λr is chosen so that eτΨ ≥ eρ2 on the set |x− x0| ∈ [r, 2r].
We nally get
eτρ2 ‖u‖H1(|x−x0|∈[r,2r]) ≤ Ceτρ1 ‖u‖H1(B(x0,3r))
+ Ceτρ3
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(B(0,3r))
].
with ρ1 < ρ2 < ρ3. This gives
‖u‖H1(|x−x0|∈[r,2r]) ≤ Ce−C1τ ‖u‖H1(B(x0,3r))
+ CeC2τ[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(B(0,3r))
].
with C1 = ρ2 − ρ1 > 0 and C2 = ρ3 − ρ2 > 0. Next, we apply the following Lemma of interpolationtype, for which we postpone the proof.
Lemma 2.3.1. Let C1 , C2, C3 > 0 and τ0 > 0. Then, there exists some constants δ ∈]0, 1[ andC > 0 so that:
For any a, b, c real positive numbers that satises the following estimate, uniformly for τ ≥ τ0
a ≤ e−C1τ b+ eC2τ c
a ≤ C3b.
we also have the estimate
a ≤ Cbδc1−δ.
Applying this Lemma with a = ‖u‖H1(|x−x0|∈[r,2r]) /C, b = ‖u‖H1(B(x0,3r)), c =
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(B(0,3r))
]and noticing that a ≤ c/C, we get, with a dierent constant C > 0,
‖u‖H1(|x−x0|∈[r,2r]) ≤ C ‖u‖δH1(B(x0,3r))
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(B(0,3r))
]1−δ.
Moreover, we have obvisously, if C ≥ 1,
‖u‖H1(B(x0,r))≤ C ‖u‖δH1(B(x0,3r))
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(B(0,3r))
]1−δ.
This gives the expected result by summing up.
32
Proof of the Lemma 2.3.1. We minimize in τ . The minimum is reached for τ =ln(bC1cC2
)C1+C2
. To simplify
(actually, it is just changing b by bC1), we apply the formula for τ1 =ln( bc)C1+C2
. It gives, if τ1 ≥ τ0,
a ≤ e− C1C1+C2
ln( bc)b+ eC2
C1+C2ln( bc)c
≤(b
c
)δ−1
b+
(b
c
)δc = 2bδc1−δ.
where we have denoted δ = C2C1+C2
.
In the case τ1 ≤ τ0, this means bc ≤ e
τ0(C1+C2), so b ≤ C(τ0, C1, C2)c. So, the assumption a ≤ C3bgives a ≤ C3b
δb1−δ ≤ Cbδc1−δ with a new constant depending on τ0, C1, C2, C3.This gives the expected estimate in both cases with an appropriate constant C > 0.
Now, we want to globalize this estimates
Theorem 2.3.3 (Global quantitative estimates for real elliptic operator of order 2). Let Ω connected,and P be as in Theorem 2.1.2. x0 ∈ Ω and r0 > 0. Let K be a compact subset of Ω.
Then, there exists C > 0, 0 < δ < 1 so that
‖u‖H1(K) ≤ C[‖u‖H1(B(x0,r0)) + ‖Pu‖L2(Ω)
]δ‖u‖1−δH1(Ω)
for any u ∈ C∞(Ω).
Proof. By compactness, it is enough to prove the following similar result for any x ∈ K, there exists0 < rx < r0, δx so that B(x1, rx)) ⊂ Ω and
‖u‖H1(B(x1,rx)) ≤ C[‖u‖H1(B(x0,r0)) + ‖Pu‖L2(Ω)
]δx‖u‖1−δxH1(Ω) .(2.17)
Indeed, we recover K by a nite number of such balls K ⊂ ∪i∈IB(xi, ri). Take δ = mini∈I δi.Indeed, the inequality (2.17) is still true with δx replaced by δ. Indeed, since δ ≤ δx, we can
decompose ‖u‖H1(B(x1,rx)) = ‖u‖δδx
H1(B(x1,rx))‖u‖
1− δδx
H1(B(x1,rx))where both exponent are positive. Using
(2.17) for the rst term and ‖u‖H1(B(x1,rx)) ≤ ‖u‖H1(Ω) for the second, we get, with eventuallydierent constants,
‖u‖H1(B(x1,rx)) ≤ C[‖u‖H1(B(x0,r0)) + ‖Pu‖L2(Ω)
]δ‖u‖
δδx−δ
H1(Ω)‖u‖
1− δδx
H1(Ω)
≤ C[‖u‖H1(B(x0,r0)) + ‖Pu‖L2(Ω)
]δ‖u‖1−δH1(Ω)
By summing up and using the recovering property, we would get
‖u‖H1(K) ≤ C[‖u‖H1(B(x0,r0)) + ‖Pu‖L2(Ω)
]δ‖u‖1−δH1(Ω) .
So, we are left to prove (2.17) for any x ∈ K.We also assume ‖Pu‖L2(Ω) ≤ ‖u‖H1(Ω, otherwise, the result is trivial because the right hand side
is bigger than ‖u‖H1(Ω).We will need the following geometric Lemma that we prove later.
Lemma 2.3.2. Under the previous assumptions. Let x0 and x1 ∈ Ω and r0 > 0. Then, there exists0 < r < r0, N ∈ N and a sequence of points yk, k = 0, · · · , N so that
33
y0 = x0, yN = x1.
B(yk+1, r) ⊂ B(yk, 2r).
B(yk, 3r) ⊂ Ω.
Assuming this Lemma, we prove recursively the following property.There exists Ck and δk ∈]0, 1[ so that
‖u‖H1(B(yk,r))≤ Ck
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(Ω)
]δk‖u‖1−δk
H1(Ω).(2.18)
The property is true for k = 0 for C = 1 and any δk ∈ [0, 1] since ‖u‖H1(B(x0,r))≤ ‖u‖H1(Ω).
Assume the property true for k < N . Theorem 2.3.2 applied at the point yk (which can beapplied since B(yk, 3r) ⊂ Ω) gives C > 0, 0 < δ < 1 so that
‖u‖H1(B(yk,2r))≤ C
[‖u‖H1(B(yk,r)
+ ‖Pu‖L2(B(yk,3r))
]δ‖u‖1−δH1(B(yk,3r))
.
Since B(yk+1, r) ⊂ B(yk, 2r) and B(yk, 3r) ⊂ Ω, it gives
‖u‖H1(B(yk+1,r))≤ C
[‖u‖H1(B(yk,r))
+ ‖Pu‖L2(Ω)
]δ‖u‖1−δH1(Ω) .
The assumption at step k and the fact that δ > 0 gives
‖u‖H1(B(yk+1,r))≤ C
[Ck
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(Ω)
]δk‖u‖1−δk
H1(Ω)+ ‖Pu‖L2(Ω)
]δ‖u‖1−δH1(Ω) .
Since we have assumed ‖Pu‖L2(Ω) ≤ ‖u‖H1(Ω), we have ‖Pu‖L2(Ω) ≤[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(Ω)
]δk‖u‖1−δk
H1(Ω).
So, we are left with some dierent constant Ck+1
‖u‖H1(B(yk+1,r))≤ Ck+1
[[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(Ω)
]δk‖u‖1−δk
H1(Ω)
]δ‖u‖1−δH1(Ω)
≤ Ck+1
[‖u‖H1(B(x0,r))
+ ‖Pu‖L2(Ω)
]δkδ‖u‖1−δkδ
H1(Ω).
So, it gives the result with δk+1 = δk+1δ.
Proof of Lemma 2.3.2. Since Ω is a connected set of Rn, it is connected by arc and we can nd γ(t)be a continuous path in Ω so that γ(0) = x0, γ(1) = x1.
[0, 1] is a compact set. Denote d = maxt∈[0,1](dist(γ(t),Ωc). We x r = d/3. By compactness, γ isalso uniformly continuous on [0, 1]. So, there exists ε > 0 so that |t−t′| ≤ ε implies |γ(t)−γ(t′)| ≤ r/2.We take N =
⌊1ε
⌋+ 1 and dene
yk = γ(kε) for k = 0, · · ·N − 1
yN = x1 = γ(1).
This fullls the expected criterium. For instance, B(yk+1, r) ⊂ B(yk, 2r) is fullled if |yk+1−yk| < r.This works since for k ≤ N − 2, |yk+1 − yk| = |γ((k + 1)ε)− γ(kε)| ≤ r/2 by the uniform continuityassumption. For the last step, k = N − 1, the same argument applies since yN = γ(1) and yN−1 =γ(b1/εc ε). We observe that |1−
⌊1ε
⌋ε| ≤ ε because
∣∣1ε −
⌊1ε
⌋∣∣ ≤ 1 by denition.
34
2.4 The Dirichlet case
Theorem 2.4.1 (Global quantitative estimates for real elliptic operator of order 2). Let Ω connectedwith smooth boundary, and P be as in Theorem 2.1.2. Let Γ ⊂ ∂Ω a non empty open subset of theboundary. Let K be a compact subset of Ω.
Then, there exists C > 0, 0 < δ < 1 so that
‖u‖H1(K) ≤ C[‖∂νu‖L2(Γ) + ‖Pu‖L2(Ω)
]δ‖u‖1−δH1(Ω)
for any u ∈ C∞(Ω) with u = 0 on ∂Ω.
Obtaining the previous estimates follows a similar path as previously, except that we need to provesome Carleman estimates until the boundary. By some change of variables, it is always possible (seeLemma ??) to get to the following situation.
We decompose x ∈ Rn with x = (x′, xn) x′ ∈ Rn−1, xn ∈ R. The boundary ∂Ω becomes the setxn = 0 and P is of the form D2
xn + r(x,Dx′) where r(x,Dx′) is a family of operator depending onx = (x′, xn), but with derivatives only in x′.
We denote Kr0 = Rn+ ∩ B(x0, r0) and C∞0 (Kr0) is the set of functions in C∞(Rn+) supported inB(x0, r0). The index + in the norms means that it is taken on Rn+.
Theorem 2.4.2 (Local Carleman estimate). Let r0 > 0 and P = D2xn + r(x,Dx′) be a dierential
operator of order two on a neighborhood of Kr0, with real principal part, where r(x,Dx′) is a smoothxn family of second order operators in the (tangential) variable x′.
Let ψ be quadratic polynomial such that ψ′xn 6= 0 on Kr0 and
p, p, ψ (x, ξ) > 0, if p(x, ξ) = 0, x ∈ Kr0 , ξ 6= 0;(2.19)
1
iτpψ, pψ(x, ξ) > 0, if pψ(x, ξ) = 0, x ∈ Kr0 , τ > 0,(2.20)
where pψ(x, ξ) = p(x, ξ + iτ∇ψ).Then, there exist C > 0, τ0 > 0 such that for any τ > τ0, we have for all u ∈ C∞0 (Kr0/4)
τ‖eτψu‖21,+,τ ≤ C(∥∥eτΨPu
∥∥2
0,++ τ3|(eτΨu)|xn=0|20
+τ |(D(eτΨu)
)|xn=0
|20).(2.21)
If moreover ∂xnψ > 0 for (x′, xn = 0) ∈ Kr0, then we have for all u ∈ C∞0 (Kr0/4) such thatu|xn=0 = 0,
τ‖eτΨu‖21,+,τ ≤ C∥∥eτΨPu
∥∥2
0,+.(2.22)
Note that the Theorem applies to real elliptic operators, but also to wave type operators withthe associated pseudoconvexity condition.
We give a proof of this theorem in the appendix. The general idea is the following.We would like to apply the same reasoning as before. Yet, we have to be more careful about
the Gårding inequality in the case of boundary. One possibility is to use symbolic calculus only inthe tangential variable x′ where integration by parts are allowed without boundary terms. But theintegration by parts for D2
n and its conjugated operator produce some boundary terms that we needto take into account.
How to deal with boundary terms?
35
In the variable xn, the operator is D2n and the conjugated operator is explicit (Dn + iτ(∂nψ))2 =
D2n− τ2(∂nψ)2 + 2iτ(∂nψ)Dn− (∂2
nψ). The integration by parts can be explicitely computed. Whatis not a priori obvious is that there is no term of order 2 and 3.
What will save us is that fact that the real part and the imaginary part don't have the samenumber of derivative in xn. Decomposing Pψ = Qr+iQi as before, we check that Qr has 2 derivativesin xn (D2
n) while Qi only has one (2iτ(∂nψ)Dn).
Let us look at each term, using the integration by part formula (f,Dng) = (Dnf, g) + i(f, g)xn=0
integrating by part of (Qru,Qiu) = (QiQru, u)+boundaryterms: the worst terms should comefrom the integration by part of (D2
nu, (∂nψ)Dnu) and we expect the boundary term to be ofthe form i(D2
nu, (∂nψ)u)xn=0. For instance, the boundary term corresponding to the to thisshould be of the form
integrating by part of (Qiu,Qru) = (QrQiu, u)+boundaryterms: the worst terms should comefrom the integration by part of
((∂nψ)Dnu,D2nu) = (Dn [(∂nψ)Dnu] , Dnu) + i((∂nψ)Dnu,Dnu)xn=0
= (D2n [(∂nψ)u] , Dnu) + i(Dn [(∂nψ)Dnu] , u)xn=0 + i((∂nψ)Dnu,Dnu)xn=0
This gives the boundary terms
i((∂nψ)D2nu, u)xn=0 + i((Dn∂nψ)Dnu, u)xn=0 + i((∂nψ)Dnu,Dnu)xn=0.
The two terms of order 2 cancel. So we are left with some terms of order 1 that come into theboundary terms that can be handled by the Carleman method. The true computation contains somemore terms, but with less derivative in x1.
How to deal with interior terms?The interior terms are more or less the same as in the boundaryless case. So, we could expect
that their symbol satisfy the same positivity condition. Yet, we would like to use only a tangentialGårding inequality, that is only in the derivatives in the variable x′ (with symbol only depending onthe cotangent variable ξ′.)
The idea is to perform a kind of euclidian division of the commutator i[Qr, Qi] by Dn. Indeed,we can factorize i[Qr, Qi] = τ
[C0D
2n + C1Dn + C2
]where Ci are tangential operators (we have also
used that Qi can be written τQi). Moreover, since Qr contains some derivative in xn with maincoecient D2
n while the main derivative of Qi in xn is 2iτ(∂nψ)Dn where (∂nψ) 6= 0. This allows toperform a similar "euclidian division" with Qr, Qi which allows to write
i[Qr, Qi] = τD0Qr +D1Q
i + τD2.
Since the terms τD0Qr are in some sense weaker than ‖Qr‖L2 (and the same for Qi), we are left with
some tangential operator. D2 is not always positive, but the nal task is to transfer the informationwe have on pψ to this tangential operator.
2.5 Other applications
2.5.1 Spectral estimates
From the quantitative estimate of Theorem 2.3.3, we can already get some applications about spectralestimates of eigenfunctions of second order elliptic operators. We rst describe the context.
36
We will denote Tn = Rn/Zn the n-dimensional torus. This can be seen as [0, 1]n with thenecessary identication of points. Functions on Tn can be seen as functions on Rn with periodicboundary conditions. Let A = (ai,j)
ni,j=1 a symmetric matrix with ai,j ∈ C∞(Tn) real valued. We
dene the operator Pu = −div(A∇u) = −∑
i,j ∂i (ai,j∂ju). Assume also that P is elliptic, that isthere exists C so that
n∑i,j=1
ai,j(x)ξiξj ≥ C|ξ|2, for all (x, ξ) ∈ Ω× Rn.
P is also symmetric, that is (Pu, v)L2(Tn) = (u, Pv)L2(Tn) for u, v ∈ C∞(Tn).We can check that its can be extended to a positive self-adjoint operator with domain H2(Tn).
Therefore, since the embedding ofH2(Tn) into L2(Tn) is compact. Therefore, the resolvent (P+Id)−1
is well dened and compact on L2.All this allows to dene an orthonormal basis of L2(Tn). There exists some functions Φi ∈
C∞(Tn), λi ∈ R (actually λi ≥ 0 since P is positive) so that
(Φi)i∈N is an orthonormal basis of the Hilbert space L2(Tn)
PΦi = λiΦi.
We refer to Brézis [2] Chapter VI and IX for more details about this construction.
Remark 2.5.1. The same construction holds for a general compact Riemannian manifold M . Themetric g induces some natural volume form ωg, denition of ∇g, divergence and nally the Laplace-Beltrami operator dened by ∆gu = divg(∇gu) that can be written in local coordinates as
∆gu =1√det(g)
∑i,j
∂i
(√det(g)gi,j∂ju
)where gi,j is the inverse of the matrix g = (gi,j)
ni,j=1.
We have the following Theorem which states some localization property of eigenfunctions of P
Theorem 2.5.1. Under the previous assumptions on P and the related Φi, λi. Let ω be an opensubset of Tn. There exists C and c > 0 so that we have the estimate uniform in i ∈ N
1 = ‖Φi‖2L2(Tn) ≤ Cec√λj ‖Φi‖2L2(ω) .
This is a kind of observability estimate for eigenfunctions: observing one part of the eigenfunctions
gives a proportion of the energy of at least Ce−c√λj .
Proof. Let Q = −∂2s + P the operator dened on Rs × Tnx. We can easily verify that Q is elliptic
with real coecients.Dene fj = e
√λjsΦj(x). We verify that Qfj = −λjfj + e
√λjsPΦj = 0. We want to apply
Theorem 2.3.3 to Q and fj on an open set Ω =]0, 3[×Tn and K = [1, 2]× Tn. Note that we are notexactly in the conguration of the Theorem since Tn is not an open set of Rn. But it can be easilychecked that some localization in Rn using the periodicity gives the expected result.
If (t0, x0), r > 0 are so that B((t0, x0), r) ⊂]1, 2[×ω, the Theorem gives C > 0, 0 < δ < 1 so that
‖fj‖H1([1,2]×Tn) ≤ C[‖fj‖H1(]1,2[×ω)
]δ‖fj‖1−δH1(]0,3[×Tn) .
37
We make it even worse by
‖fj‖L2([1,2]×Tn) ≤ C[‖fj‖H2(]1,2[×ω)
]δ‖fj‖1−δH2(]0,3[×Tn) .
But on [1, 2], we have e√λjs ≥ e
√λj . So ‖fj‖L2([1,2]×Tn) ≥ e
√λj ‖Φj‖L2(×Tn) = e
√λj .
Similarly, using that P is elliptic of order 2
‖fj‖H2(]1,2[×ω) ≤ ‖fj‖L2(]1,2[×ω) +∥∥∂2
sfj∥∥L2(]1,2[×ω)
+ ‖Pfj‖L2(]1,2[×ω)
≤ ‖fj‖L2(]1,2[×ω) + ‖λjfj‖L2(]1,2[×ω) + ‖λjfj‖L2(]1,2[×ω)
≤ e2√λj (1 + 2λj) ‖Φj‖L2(ω) ≤ Ce
3√λj ‖Φj‖L2(ω) .
where we have used (1 + 2λj) ≤ Ce√λj for some well chosen C. Similarly
‖fj‖H2(]0,3[×Tn) ≤ ‖fj‖L2(]0,3[×Tn) +∥∥∂2
sfj∥∥L2(]0,3[×Tn)
+ ‖Pfj‖L2(]0,3[×Tn)
≤ Ce3√λj (1 + 2λj) ‖Φj‖L2(Tn) ≤ Ce
4√λj .
We nally obtain, with some dierent constant C
e√λj ≤ Ce3δ
√λj ‖Φj‖δL2(ω) e
4(1−δ)√λj .
Which gives the expected result.
Remark 2.5.2. The rate ec√λj is not always optimal, but there are some geometric situation where
it is optimal. The typical example is when there is a stable geodesic. On the sphere S2 parameterizedby
(x, y, z);x2 + y2 + z2 = 1, for instance we have some exact eigenvalues of the Laplace-Beltrami
operator of the form Φn = cn Re(x + iy)n with eigenvalue λn = n(n), with some appropriate nor-malization constants cn. Since |Φn|2(x, y, z) ≤ cn(x2 + y2)n ≤ (1 − z2)n, we easily observe thatΦn(x)/cn is exponentially small outside the set S2 ∩ z = 0. Yet, we can check that cn . n1/2 since∫ 1/2−1/2(1− z2)ndz =
∫ 1/2−1/2 e
n ln(1−z2)dz ≥∫ 1/2−1/2 e
−cnz2dz ≥ n−1/2
∫ √n/2−√n/2
e−cs2ds.
There are some geometries where we can replace c(λj) = ec√λj by a constant or sometimes a
power of λj or log(λj). The general question of making the link between the geometric properties ofω with respect to Ω and determining the appropriate c(λ) is a big open problem in spectral theory.
Using the boundary estimates of Theorem 2.4.1, it is possible to get a more precise result. Actually,the previous result remain true not only for eigenfunctions, but also for nite sum of eigenfunction.Since now, the stability estimate is still true for an open set with boundary, we state the result foran elliptic operator P with the Dirichlet boundary conditions. The framework will be quite similarto the previous one
Let Ω be a smooth compact open set with boundary. Let A = (ai,j)ni,j=1 a symmetric matrix
with ai,j ∈ C∞(Ω) real valued. We dene the operator Pu = −div(A∇u) = −∑
i,j ∂i (ai,j∂ju).We consider (and we will still denote it P the selfadjoint extension of P associated to the Drichletboundary condition, that is u = 0 on ∂Ω. We use the same notation Φj and λj the eigenfunctionsand eigenvalues.
Theorem 2.5.2. Under the previous assumptions on P and the related Φi, λi. Let ω be an opensubset of Ω. There exists C and c > 0 so that we have the estimate uniform in λ
‖u‖2L2(Ω) ≤ Cec√λ ‖u‖2L2(ω) .
for any u =∑
λj≤λ ujΦj.
38
Proof. As before, we consider the elliptic operator Q = −∂2s + P the operator dened on R+
s × Ω.
Dene f(s, x) =∑
i,λj≤λ ujsinh(√λjs)√
λjΦj(x) and we easily verify that it satises Qf = 0 and f = 0
on R × ∂Ω and 0 × Ω. Theorem 2.4.1 gives, if we take Γ = 0 × ω. Note that we can assumewithout loss of generality that ω is far from ∂Ω, in order to avoid problems with "corners" at thepoints 0 × ∂Ω. A weak form of the inequality is then
‖f‖L2([0,1]×Ω) ≤ C ‖∂sf‖δL2(0×ω) ‖f‖
1−δH1([0,2[×Ω)
We have ∂sf(0, x) = u(x), so ‖∂sf‖L2(0×ω) = ‖u‖L2(ω).As before, using Parseval identity
‖f‖2H1([0,2[×Ω) ≤ ‖f‖2L2([0,2[×Ω) + ‖∂sf‖2L2([0,2[×Ω) +
∫ 2
0(−∆f, f)L2(Ω)
≤ C
∫ 2
0
∑λj≤λ
|uj |2(
cosh(√λjs)
2 + sinh(√λjs)
2)ds
≤ Cec√λ∑λj≤λ
|uj |2 ≤ Cec√λ ‖u‖L2(Ω) .
And similarly
‖f‖2L2([0,1[×Ω) ≥ ‖f‖2L2([0,1[×Ω)
≥ C
∫ 1
0
∑λj≤λ
|uj |2sinh(
√λjs)
λj
2
ds
≥ C∑λj≤λ
|uj |2∫ √λj
0
sinh(y)2
λ3/2j
dy ≥ C ′ ‖u‖2L2(Ω) .
So, we obtain, with some dierent constants C, c
‖u‖L2(Ω) ≤ Cec√λ ‖u‖δL2(ω) ‖u‖
1−δL2(Ω) .
This gives the result.
This type growth of the type e√λ is optimal whatever the geometry if ω 6= Ω. See [9].
2.5.2 The heat equation
Our previous Theorem gives immediatly the following corollary for solutions of the heat equation atlow frequency
Corollary 2.5.1. Under the previous assumptions on P and the related Φi, λi. Let ω be an opensubset of Ω. There exists C and c > 0 so that we have the estimate uniform in λ ≥ 0 and T > 0
‖u(T )‖2L2(Ω) ≤ Cec√λ 1
T
∫ T
0‖u(t)‖2L2(ω) dt.
for any f =∑
λj≤λ fjΦj with u solution of the heat equation∂tu−∆u = 0 on [0, T ]× Ω
u = 0 on [0, T ]× ∂Ωu(0, x) = f(x) on Ω
39
Proof. The decay of the energy gives ‖u(T )‖2L2(Ω) ≤ ‖u(t)‖2L2(Ω) for any 0 ≤ t ≤ T . Moreover, for
any t ∈ [0, T ], the spectral estimates can be written ‖u(t)‖2L2(ω) ≤ Cec√λ ‖u‖2L2(ω). Integrating in
time gives the previous estimates and using the
T ‖u(T )‖2L2(Ω) ≤∫ T
0‖u(t)‖2L2(ω) ≤ Ce
c√λ
∫ T
0‖u‖2L2(ω) .
We will prove some observability estimate for the heat equation on a bounded domain. It isknown that some observability estimates are equivalent to some result of control.
We have chosen to prove the observability and then to deduce the related result of control.In the original paper, the idea was the following:
Use the "observability inequality" of Theorem 2.5.2 to get some result about control of the low
frequency up to λ in time T/4 with a cost ≈ 4T e√λ. This allows to control the λ rst frequency
to zero.
Use the decay of the heat equation on the remaining eigenvalues to get decay of e−λT/4 in timeT/4.
Iterate the result on dyadic times and tending to T .
The important fact is that the power 1/2 in λ in ec√λ is strictly smaller than the exponential decay
e−λT/4. We will work directly on the observability estimate, but still using the decay provided bythe heat equation.
Theorem 2.5.3 (Observability for the heat equation). Let ω ⊂ Ω a non empty open set and T > 0.Then, there exists C > 0 so that we have the estimate
‖u(T )‖2L2 ≤ C∫ T
0‖u‖2L2(ω)
for any u solution of ∂tu−∆u = 0 on [0, T ]× Ω
u = 0 on [0, T ]× ∂Ωu(0, x) = u0(x) on Ω
with u0 ∈ L2(Ω).
Proof. The idea is that our spectral estimate gives good estimates only when there are few highfrequencies, that is after the decay of the wave operator have operated, that is close to times T .
We will divide the interval [0, T ] as the union of the intervals [Tk+1, Tk] with T0 = T , Tk+1 =Tk − T2−k. We check that Tk converges to T −
∑k∈N∗ T2−k = 0. To simplify the notations, we
denote Lk = T2−k the length of the interval.For each interval [Tk+1, Tk], we will select a frequency cuto µk and decompose
u = uk,L + uk,H =∑λj≤µk
+∑λj>µk
40
We will cut [Tk+1, Tk] in two pieces, [Tk+1, Tk+1 +Lk/2] where we only use the damping and [Tk+1 +Lk/2, Tk] where we observe (using that the high frequency have been damped). We apply Corollary2.5.1 on [Tk+1 + Lk/2, Tk]
‖uk,L(Tk)‖2L2(Ω) ≤ Cec√µk
2
Lk
∫ Tk
Tk+1+Lk/2‖uk,L(t)‖2L2(ω) dt.(2.23)
So, by triangular inequality, noticing that the error we do from the cut o in frequency is small. For‖uk,L(t)‖L2(ω) ≤ ‖u(t)‖L2(ω) + ‖uk,H(t)‖L2(ω)
‖uk,H(t)‖L2(ω) ≤ ‖uk,H(t)‖L2(Ω) ≤ ‖uk,H(Tk+1 + Lk/2)‖L2(Ω) ≤ e−µkLk/2 ‖u(Tk+1)‖L2(Ω)(2.24)
where we have used the damping of high frequency.Moreover, we have similarly
‖uk,H(Tk)‖L2(ω) ≤ e−µkLk/2 ‖u(Tk+1)‖L2(Ω) .(2.25)
So, putting together (2.23), (2.24) and (2.25), we get
‖u(Tk)‖2L2(Ω) ≤ Cec√µk
2
Lk
∫ Tk
Tk+1+Lk/2‖u(t)‖2L2(ω) dt+ Cec
√µke−µkLk/2 ‖u(Tk+1)‖L2(Ω)
≤ Cec√µk
2
Lk
∫ T
0‖u(t)‖2L2(ω) dt+ Cec
√µke−µkLk ‖u(Tk+1)‖2L2(Ω) .
Now, we can choose µk. Recall that Lk = T2−k converge to zero. Pick for instance, µk =√C1L
−2k
with C1 large. We have µkLk =√C1√µk and µk+1 = 2µk. If C1 is large enough, we have
Cec√µke−µkLk ≤ e−3c
õk+1 .
Indeed,
−c√µk + µkLk − 3c√µk+1 =
õk
(√C1 − c(1 + 3
√2))
= C12k(C1 − c(1 + 3
√2)).
This can be made arbitrary arbitrary large uniformly for k ∈ N. Once C1 and µk are xed, we haveone constant C so that
ecõk
2
Lk≤ Ce2c
õk
So, we obtain
‖u(Tk)‖2L2(Ω) ≤ Ce2c√µk
∫ T
0‖u(t)‖2L2(ω) dt+ e−3c
√µk+1 ‖u(Tk+1)‖2L2(Ω) .
e−3c√µk ‖u(Tk)‖2L2(Ω) ≤ Ce−c
õk
∫ T
0‖u(t)‖2L2(ω) dt+ e−3c
√µk+1 ‖u(Tk+1)‖2L2(Ω) .
Denoting zn = e−3c√µk ‖u(Tk)‖2L2(Ω), we get
zn − zn+1 ≤ Ce−c√µk
∫ T
0‖u(t)‖2L2(ω) dt.
41
We recognize a telescopic series and that e−c√µk is summable. So, by summing up, we get with a
new constant, uniform in k ∈ N,
e−3c√µ0 ‖u(T )‖2L2(Ω) − e
−3c√µk ‖u(Tk)‖2L2(Ω) ≤ C
∫ T
0‖u(t)‖2L2(ω) dt
Since ‖u(Tk)‖L2(Ω) is bounded, e−3c√µk ‖u(Tk)‖2L2(Ω) converges to zero, which gives the result.
Theorem 2.5.4 (Control to zero of the heat equation). Let ω ⊂ Ω a non empty open set and T > 0.Let u0 ∈ L2(Ω). Then, there exists g ∈ L2([0, T ], L2(ω)) so that the solution of
∂tu−∆u = g on [0, T ]× Ωu = 0 on [0, T ]× ∂Ω
u(0, x) = u0(x) on Ω
satises u(T ) = 0.
Proof. We consider the dual to the heat equation, v is a solution of−∂tv −∆v = 0 on [0, T ]× Ω
v = 0 on [0, T ]× ∂Ωv(T, x) = vT (x) on Ω
(2.26)
This is actually exactly the backward heat equation. It is made interesting, because at least forsmooth solutions u and v with Dirichlet boundary conditions, we have the formula that can be easilyobtained by multiplying the equation (2.26) by v (consider v real valued for simplicity), integratingover [0, T ]× Ω and integrating by parts∫
Ωu(T )v(T )−
∫Ωu(0)v(0) =
∫ T
0
∫Ωgv.
The formula can also be extended to the case where u0 ∈ L2(Ω), g ∈ L2([0, T ];L2(Ω)) , vT ∈ L2(Ω)by a density argument.
Our hope if u(T ) = 0 would be to get∫ T
0 gv =∫
Ω u0v(0). Reciprocally, we can check that if∫ T0 gv =
∫Ω u0v(0) for any solution of (2.26) with vT ∈ L2(Ω), then u(T ) = 0.
Now, consider the quadratic form
a(vT , vT ) =
∫ T
0
∫ωvvdx dt.
where v, v are the associated solutions to (2.26). a is well dened for vT , vT ∈ L2(Ω) and denes apositive quadratic form. Our observability estimates says that it is a scalar product. Yet, it is weakerthan the L2(Ω) norm. We dene the completion H of L2(Ω) with respect to this norm.
Dene the linear form
l(vT ) =
∫Ωu0v(0).
Our observability estimates can be written
‖v(0)‖2L2(Ω) ≤ C∫ T
0
∫ω|v|2dx dt.
42
This says exactly that l is linear continuous in H, since u0 ∈ L2(Ω). By the Riesz representation (orLax-Milgram), there exists vu0
T ∈ H so that
a(vT , vu0T ) = l(vT )
for all vT ∈ H.The application θ : L2(Ω) 7→ L2([0, T ] × ω) dened by θ(vT ) = v|[0,T ]×ω where v is solution of
(2.26) is well dened in L2(Ω), but also bounded for the norm a on H. Therefore, it can be extendedto H.
Take g = θ(vu0T ) ∈ L2([0, T ]× ω). By choice, we have∫ T
0
∫Ωgθ(vT ) = l(vT )
for any vT ∈ H. If we take in particular vT ∈ L2(Ω), this gives∫ T
0
∫Ωgθ(vT ) =
∫Ωu0v(0)
for v solution of (2.26). This gives the expected result.
2.6 The general Theorem of Hörmander
Theorem 2.6.1 (The Theorem of Hörmander). Let Ω an open set of Rn and x0 ∈ Ω. Let P bea dierential operator of order m with some eventually complex coecients with C∞(Ω) principalsymbol and all coecients in L∞loc(Ω). Assume that P is principally normal, that is the principalsymbol p of P satises: for any compact K of Ω, there is C > 0
p, p ≤ C|p||ξ|m−1.
for all (x, ξ) ∈ K × Rn.Let Φ ∈ C2(Ω) real valued so that ∇Φ(x0) 6= 0. Assume that it satises
Re p, p,Φ (x0, ξ) > 0, if p(x0, ξ) = p,Φ(x0, ξ) = 0 and ξ 6= 0;
1
iτpΦ, pΦ(x0, ξ) > 0, if pΦ(x0, ξ) = pΦ,Φ(x0, ξ) = 0 and τ > 0,
where pΦ(x, ξ) = p(x, ξ + iτ∇Φ) and p is the principal symbol of P .Then, there exists V one neighborhood of x0 in Ω so that for any u ∈ Hm−1
loc (V ),Pu = 0 in V,u = 0 in V ∩ Φ > Φ(x0) =⇒ u = 0 on V.
This is Theorem 28.3.4 of [7].
2.7 A small bibliography
The result of unique continuation have a long history going back to Carleman [4] who rst, had theidea to conjugate the operator with an exponential weight to get unique continuation. He proved theresult in the case of elliptic operators of order 2 in dimension 2. Calderón [3] extended the result tosome operators with simple characteristics. Namely, that was in situations where pφ = pφ, φ never
43
happens. The general version was given by Hörmander [6] for real operators and [7]. Note that otherworks consider the limit case where there is some higher order of cancelation. We refer to Zuily [14]for more details.
The boundary Carleman estimates where proved by Lebeau-Robbiano [10] in order to give thesame application that we give in Section 2.5.2, that is the controllability of the heat equation. Theyalso proved the spectral estimates of Section 2.5.1.
Note that there is also another proof (independently) by Fursikov-Immanuvilov [5] of the control-lability of the heat equation using directly some Carleman estimates for the heat equation. Detailsabout this and some link with the elliptic Carleman estimate are given in [9].
2.8 Further comments and problems
Many things have not been written in an optimal way in the previous theorems and can be improved:
the fact that the ai,j are real valued is not necessary♣ .
the regularity of u can be much lowered. Note also, that if the coecients are regular enough,the regularity of u can often be recovered using classical elliptic regularity results, see Brézis[2] for instance.
the regularity of the coecients is not optimal. The main coecients should actually be C1
while the lower order terms can be in some Lp spaces.
the fact to be an exact solution of Pu = 0 can be replaced by some weaker assumption like|Pu|(x) ≤ C (|u(x)|+ |D(x)|) for almost every x ∈ Ω.
Counterexemples of AlinhacRough coecients, nonlinear problemsboundary conditions, interfacesSystemsGlobal result (cf Einstein??)
44
Chapter 3
The wave equation with coecientsconstant in time
3.1 The problem
In this section, we will specialize to some more specic operators of wave type ∂2t −Q(x,Dx) where
Q is elliptic negative. In Remark 2.1.2, we have seen that, in the framework of regular coecient,the unique continuation function from cylinder, that is for set of the form ϕ(x) ≤ 0 requires someconvexity assumptions of the surface. Yet, in the Holmgren Theorem 1.1.2, the condition is muchweaker. For the at wave equation, ∂2
t −∆, it only requires |∂tϕ|2 6= |∇xϕ|2. That means, we shouldnot be tangent to the cone of light. More or less, this is the weaker condition that one could expectbut would not contradict the nite speed of propagation. But we would like to relax the analyticityassumption. The counterexample of Alinhac-Baouendi [1] actually tell us that we can not relax thiscompletely.
It turns out that some analyticity with respect to part of the variables can be sucient, forinstance the time in our example of the wave equation.
In the following, the variable will be z = (t, x) ∈ R1+n with dual variable ξ = (ξt, ξx) ∈ R1+n.To keep the notation coherent with the elliptic case, we will denote ξt = ξ0 and ξx will be writtenξx = (ξ1, · · · , ξn).
The main theorem of this chapter will be the following.
Theorem 3.1.1 (Wave type operator with coecients constant in time). Let T > 0 and Ωx an openset of Rn. Denote Ω =]− T, T [×Ωx.
Let Q =∑n
i,j=1 ai,j(x)∂i∂j +∑
k bk(x)∂k + c(x) be a dierential operator of order 2 with ai,j ∈C∞(Ωx) real valued, bk, c ∈ L∞(Ωx). Assume also that Q is negative elliptic, that is there existsC > 0 so that
n∑i,j=1
ai,j(x)ξiξj ≥ C|ξx|2, for all (x, ξx) ∈ Ωx × Rn.
Dene P = ∂2t −Q dened on Ω of principal symbol p.
Let z0 = (t0, x0) ∈ Ω and Φ ∈ C2(Ω) so that p(z0,∇t,xΦ) 6= 0, or more precisely
(∂tΦ(z0))2 6=∑i,j
ai,j(x0)(∂iΦ)(∂jΦ).
45
Then, there exists V one neighborhood of z0 so that for any u ∈ C∞(Ω),Pu = 0 in Ω,u = 0 in Ω ∩ Φ > Φ(z0) =⇒ u = 0 on V.(3.1)
The main tool will be an inequality of Carleman type, but with an additional weight in the Fouriervariable.
Namely, for a smooth real valued function ψ (later on, we will assume that it is polynomial oforder 2), we dene the operator
Qε,τu = e−ε|Dt|
2
2τ eτψ
where e−|Dt|
2
2τ is the Fourier multiplier dened for u ∈ S(Rn) by(
e−ε|Dt|2
2τ u
)(ξ) = e−ε
|ξt|2
2τ u(ξ) where
ξt is the Fourier variable corresponding to the variable t.
Denition 3.1.1 (Pseudoconvexity for functions in |ξt| = 0). With the above assumptions for P , letψ smooth real valued. We say that ψ is satises the pseudoconvexity for functions in |ξt| = 0 at z0 if
p, p, ψ (z0, ξ) > 0, if p(z0, ξ) = 0, ξt = 0, ξ 6= 0;(3.2)
1
iτpψ, pψ(z0, ξ) > 0, if pψ(z0, ξ) = 0, ξt = 0, τ > 0,(3.3)
where pψ(x, ξ) = p(x, ξ + iτ∇ψ).
Theorem 3.1.2 (Carleman estimate for wave type operators with coecients constant in time).With the above assumptions for P , let ψ be quadratic real valued polynomial such that Ψ satises thepseudoconvexity for functions in ξt = 0 at z0 of Denition 3.1.1.
Then, there exists r > 0, ε > 0, d > 0, C > 0, τ0 > 0 such that for all τ > τ0 and u ∈C∞0 (B(x0, r))
τ‖Qψε,τu‖2H1τ≤ C
∥∥∥Qψε,τPu∥∥∥2
L2+ Ce−dτ
∥∥∥eτψu∥∥∥2
H1τ
(3.4)
Note that if we set ε = 0, this would be a classical Carleman estimate. Yet, the role of the Fouriermultiplier will be to kill the high frequency in the variable t. So, we will just need to look at the verysmall frequency in ξt. That is why the pseudoconvexity assumption is only made in ξt = 0.
3.2 Proving unique continuation using the Carleman estimate
In this section, we assume that Theorem 3.1.2 is proved and we will prove Theorem 3.1.1. Some partwill be similar to the classical case, that means constructing an approriate function ψ, pseudoconvexfor functions in ξt = 0 from the function φ pseudoconvex for surfaces.
The main dierences are the following:
the pseudoconvexity is only on ξt = 0, so it requires a small adaptation of the convexicationprocedure. Moreover, we want ψ quadratic.
the Carleman estimates implies an exponential weight in Fourier that change the proof of uniquecontinuation.
46
3.2.1 Convexication
Quite similarly to the classical case, we will rst give a dierent formulation of fact that Φ is noncharateristic. It implies a property of pseudoconvexity quite similar to the one of Proposition 2.3.1,but on the set ξt = 0.
Proposition 3.2.1 (Usual pseudoconvexity for surfaces in ξt = 0). Let P and Φ satisfying theassumptions of Theorem 3.1.1. Then, Φ satises the stronger property
p, p,Φ (z0, ξ) > 0, if p(z0, ξ) = p,Φ(z0, ξ) = ξt = 0 and ξ 6= 0;(3.5)
1
iτpΦ, pΦ(z0, ξ) > 0, if pΦ(z0, ξ) = pΦ,Φ(z0, ξ) = ξt = 0 and τ > 0,(3.6)
where pΦ(x, ξ) = p(x, ξ + iτ∇Φ) and p is the principal symbol of P .
Proof. The principal symbol of P is p(t, x, ξt, ξx) = −ξ2t − q where q(x, ξx) = −
∑i,j ai,j(x)ξiξj .
So, we notice that for ξt = 0, we have p(t, x, 0, ξx) = −q(x, ξx). Since q is assumed to be elliptic,the assumption p(z0, ξ) = ξt = 0 implies ξ = 0 and therefore, 3.5 is empty.
We will use the computations of Lemma A.4.1 (note that we have actually already proved that3.6 is empty if p(x,∇Φ) 6= 0 in Proposition 2.3.1). Noting Rz0 the symmetric quadratic form so thatRz0(ξ, ξ) = p(z0, ξ), we get
pΦ,Φ(z0, ξ, τ) = 2Rz0(ξ,∇Φ) + 2iτp(z0,∇Φ)
In particular, since by assumption, p(z0,∇Φ) 6= 0, this gives that pΦ,Φ never cancel and (3.6) isalso empty.
Next, we will follow the same previous steps of convexication as Section 2.3.1.
Proposition 3.2.2 (Analytic convexication). Let P and Φ satisfying the assumptions of Theorem3.1.1 with Φ(z0) = 0, that is satisfying the assumptions of Denition 2.1.1 (and therefore, the one ofProposition 3.2.1)
Then there exists λ large enough so that the functions Ψ = eλΦ satises the pseudoconvexity forfunctions in ξt = 0 of Denition 3.1.1.
Proof. The proof is very similar to Proposition 2.3.2.We denote again for τ > 0
cΨ(ξ, τ) =1
iτpΨ, pΨ(z0, ξ, τ).
Since p is real Lemma 2.2.4 still applies and we can extend cΨ(ξ, τ) by continuity to τ = 0 by2p, p,Φ. Then, using Lemma 2.2.3, the consequences of Proposition 3.2.1 can be reformulated bythe existence of some constants C1, C2 > 0 so that
cΦ(ξ, τ) + C1
[|pΦ,Φ(z0, ξ, τ)|2 +
|pΦ(x0, ξ, τ)|2
|ξ|2 + τ2+ |ξt|2
]≥ C2(|ξ|2 + τ2).
The same computation lead to
cΨ(ξ, τ) = λcΦ(ξ, λτ) + 2λ2 |pΦ,Φ(x0, ξ, λτ)|2 .
The same arguments then lead to
cΨ(ξ, τ) + C1
[|pΨ(x0, ξ, τ)|2
|ξ|2 + τ2+ |ξt|2
]≥ C2(|ξ|2 + τ2).
for λ large enough. This implies the result.
47
It remains to perform the Geometric convexication and to ensure that we can take ψ quadratic.
Proposition 3.2.3 (Geometric convexication). Let Ψ pseudoconvex in the sense of functions inξt = 0, that is satisfying the assumptions of Denition 3.1.1. Set also Ψ(x0) = 0.
Then there exists ϕ quadratic that still satises the pseudoconvexity for functions in ξt = 0 at z0
and fullls the following Geometric requirements for some R0 > 0, for any 0 < R < R0, there existsr > 0, η > 0 so that
1. there exists η > 0 so that ϕ(x) ≤ −η for z ∈ Ψ ≤ 0 ∩ R/2 ≤ |z − z0| ≤ R,
2. ψ(0) = 0.
Proof. For δ > 0, we take
ϕ(z) = ΨT (x)− δ|z − z0|2.
where
ΨT (z) =∑|α|≤2
1
α!(∂αΨ)(0)(z − z0)α
that is ΨT is the Taylor expansion of Ψ at order 2. Indeed, this is almost the same construction asin the classical case, except that we have replaced Ψ by its Taylor expansion at order 2.
First, we notice that since the pseudoconvexity only involves the derivative up to order 2 at z0,ΨT is also pseudoconvexity for functions in ξt = 0 at z0. Moreover, the same stability argument asin Proposition 2.3.3 applies. So, for δ small enough, ϕ(x) is pseudoconvexity for functions in ξt = 0at z0. We x δ > 0 suciently small. It remains to prove the geometric properties.
Since ΨT is the Taylor expansion of Ψ at order 2, there exists R0 small enough so that |ΨT −Ψ| ≤|z − z0|2δ/2 for |z − z0| ≤ R0. Now, take R ≤ R0.
Let z ∈ Ψ ≤ 0 ∩ R/2 ≤ |z − z0| ≤ R. Since Ψ(z) ≤ 0, we have ΨT (z) ≤ |z − z0|2δ/2.Therefore,
ϕ(z) ≤ −δ|z − z0|2/2.
So, in particular since |z − z0|2 ≥ R2/4, we get ϕ(x) ≤ −δR2/8 and we can take η = δR2/8.
3.2.2 Unique continuation
Proof of Theorem 3.1.1. Let u solution of Pu = 0 in Ω so that u = 0 on Ω ∩ Φ > 0. Φ is pseudo-convex for surfaces at x0. Proposition 3.2.2 and 3.2.3 allows to produce some quadratic function ψthat satises the pseudoconvexity for functions in ξt = 0 at z0 . It therefore satises the followingproperties
1. there exists R > 0, C > 0 d, ε > 0 and τ0 > 0 so that we have the following estimate
τ‖Qψε,τw‖2H1τ≤ C
∥∥∥Qψε,τPw∥∥∥2
L2+ Ce−dτ
∥∥∥eτψw∥∥∥2
H1τ
(3.7)
for any w ∈ C∞(B(x0, R)) and τ ≥ τ0.
2. there exists η > 0 so that ψ(z) ≤ −η for z ∈ Φ ≤ 0 ∩ |x− x0| ≥ R/2,
3. ψ(z0) = 0.
48
4. ψ(z) ≤ d/4 in B(z0, R).
All the properties were already obtained. We only added 4 which can easily be obtained up toreducing R.
Pick χ ∈ C∞(B(x0, R)) so that χ = 1 on B(x0, R/2). As before, we want to apply the Carlemanestimate to w = χu solution of Pw = χPu+ [P, χ]u = [P, χ]u. Again, [P, χ] is a classical dierentialoperator of order 1 with coecients supported in the set Φ ≤ 0 ∩ R2 ≤ |x − x0| ≤ R where we
have Ψε(x) ≤ −η. In particular, we have∥∥∥Qψε,τPw∥∥∥
L2≤∥∥eτψPw∥∥
L2 ≤ Ce−τη ‖u‖H1 .
For the second term in the right hand side, we use Property 4 to get
e−dτ∥∥∥eτψw∥∥∥2
H1τ
≤ e−dτedτ/2 ‖w‖2H1τ≤ e−dτ/2τ2 ‖w‖2H1 ≤ e−dτ/4 ‖u‖2H1
for τ large enough.So, we have obtained that there exists C > 0, δ > 0 so that for all τ ≥ τ0
‖Qψε,τw‖L2 ≤ Ce−δτ .
We will use the following Lemma that we prove below
Lemma 3.2.1. Let ψ ∈ C∞(Ω) a real valued function. Let v ∈ C∞0 (Ω) so that
‖Qψε,τv‖L2 ≤ C ∀τ ≥ τ0.
Then, v is supported in ψ ≤ 0.
The Lemma gives that w is supported in the set ψ ≤ −δ. Yet, we have ψ(0) = 0. Then thereexists a neighborhood V included in the set χ = 1 of z0 so that ψ(z) ≥ −δ/2 in V . Therefore,w = 0 and u = 0 in V .
We need to prove Lemma 3.2.1. Note that if we had ε = 0, the proof would be easy. The idea isthe following.
1. We make a kind of foliation along the level sets of ψ: if we want to measure u, we rather denethe distribution hf = ψ∗(fv) by 〈hf , w〉E ′(R),C∞(R) = 〈fv, w(ψ)〉E ′(Rn),C∞(Rn) and estimate itfor any test function f . Heuristically, hf (s) is the integral of fu on the level set ψ(x) = s.
2. We notice that the Fourier transform of hf is hf (ζ) = 〈fv, e−iζψ〉 and can be extended to the
complex domain if v is compactly supported. In particular, on the imaginary axis, hf (iτ) =〈f, veτψ〉. Since the estimate gives information on the norm of eτψv for τ large, this can be
translated in some information on hf on the upper imaginary axis. . A Phragmén-Lindelöftype argument allows to transfer this estimate to the (almost) whole upper plan.
3. This will give some exponential bound on the upper plan. A Paley-Wiener allows to conclude.
More precisely:
Proof of Lemma 3.2.1. We will work by duality. Let f ∈ S(Rn+1) with Fourier transform f com-pactly supported in B(0, R) for R large. We dene the distribution hf ∈ E ′(R) by
〈hf , w〉E ′(R),C∞(R) = 〈fv, w(ψ)〉E ′(Rn+1),C∞(Rn+1).
hf is indeed compactly supported because Supp(hf ) ⊂ ψ(z); z ∈ Supp(v) which is compact.
49
Since hf ∈ E ′(R), the Fourier transform of hf is an analytic function on C that can be computedwith the formula
hf (ξs) =⟨hf , e
−isξs⟩E ′(Rs),C∞(Rs)
= 〈fv, eiξsψ〉E ′(Rn+1),C∞(Rn+1)
On ξs ∈ R, we have the general bound
|hf (ξs)| ≤ ‖f‖L∞(Supp(v)) ‖v‖L∞(Supp(v)) ≤ Cf,v.
But our estimates gives some bound on the imaginary axis ξs = iτ , for τ ≥ τ0,
|hf (iτ)| =∣∣∣〈fv, eτψ〉∣∣∣
E ′(Rn+1),C∞(Rn+1)
=∣∣∣〈f, veτψ〉S′(Rn+1),S(Rn+1)
∣∣∣=
∣∣∣∣〈eε |Dt|22τ f, e−ε|Dt|
2
2τ veτψ〉S′(Rn+1),S(Rn+1)
∣∣∣∣≤
∥∥∥∥eε |Dt|22τ f
∥∥∥∥L2(Rn+1)
∥∥∥∥e−ε |Dt|22τ veτψ∥∥∥∥L2(Rn+1)
≤∥∥∥∥eε |ξt|22τ
∥∥∥∥L∞(Supp(f))
‖f‖L2(Rn+1)
∥∥∥Qψε,τv∥∥∥L2(Rn+1)
≤ CeεR2
2τ ‖f‖L2(Rn+1)
≤ Cf,τ0C.
Note that at that point, the term eε|Dt|
2
2τ was harmless because the Fourier transform of f is compactly
supported. Otherwise eε|Dt|
2
2τ f does not have meaning, even for f ∈ S(Rn+1). That is why we hadto work by duality.
Moreover, for τ ∈ [0, τ0], the estimate
|hf (iτ)| ≤ C
follows easily with some appropriate constant C independant on τ . Moreover, the compact supportof hf ensure |hf (z)| ≤ Cec|z| on C.
Now, we have some nice estimates on R ∪ iR+, we will transfer them to the upper plane by thePhragmén-Lindelöf Theorem.
Lemma 3.2.2 (Phragmén-Lindelöf Theorem). Let φ be a holomorphic function in Q1 = x+ iy;x ≥ 0, y ≥ 0,continuous in Q1. Assume that there exist c > 0 and C > 0 such that
|φ(z)| ≤ Cec|z|, z ∈ Q1,
|φ(z)| ≤ 1 , z ∈ ∂Q1 = R+ ∪ iR+.
Assume moreover that there exist C, c > 0 so that |φ(z)| ≤ Cec|z|.Then |φ(z)| ≤ 1 for all z ∈ Q1.
This gives us (the result is also true for the up left quarter plane with the same method)
|hf (ξs)| ≤ C ∀ξs ∈ C, Im(ξs) ≥ 0
The Paley-Wiener Theorem gives Supp(hf ) ⊂] − ∞,+]. Therefore, we have proved that for χ ∈C∞0 (R+),
0 = 〈fv, χ(ψ)〉 = 〈f, χ(ψ)v〉
Since this is true for a subset of function f dense in S, this means that v = 0 for ψ > 0.
50
Proof of Lemma 3.2.2. First note that the sector Q1 can be rotated, say to quadrant
Q = z ∈ C, arg(z) ∈ [−π4,π
4].
Let
uδ(z) = φ(z)e−zδ 3
2 ,
(with the principal determination of the logarithm that is if z = reiθ with −π < θ < π, z3/2 =r3/2e3iθ/2) which is harmonic in Q.
Also, we have
|e−zδ 3
2 | = e−δr3/2 cos(3θ/2)
On Q, we have |θ| ≤ π/4 and therefore |3θ/2| ≤ 3π/8 < π/2 and cos(3θ/2) ≥ η > 0.So, the assumption on Φ gives lim supz∈Q,|z|→∞ uδ(z) = 0. As a consequence, there exists R > 0
such that |uδ(z)| < 1/2 on |z| ≥ R ∩Q. Now, on the bounded set QR = Q ∩ |z| ≤ R, we applythe maximum principle to the function uδ, satisfying |uδ| ≤ 1 on ∂QR. This yields |uδ| ≤ 1 on QR
and hence |uδ| ≤ 1 on Q. Finally letting δ tend to zero, we obtain the sought result.Note that the limit power could be 2 − ε with ε > 0: the result is false for ε = 0, as showed by
the holomorphic function z 7→ ez2on the quarter plane Q.
Theorem 3.2.1 (Paley-Wiener). Suppose f ∈ S(R). Then f(s) = 0 for all s > 0 if and onlyif f can be extended to a continuous and bounded function in the closed upper half-plane C+ =z = x+ iy : y ≥ 0 with f holomorphic in the interior.
Proof. One implication is simpler. Assume f(s) = 0 for s ≥ 0, then f(ξ) =∫s≤0 e
−isξf(s)ds. Then,
is can easily be continuously extended on C+ with the estimate |f(x + iy)| ≤∫s≤0 e
ys|f(s)|ds ≤∫s≤0 |f(s)|ds < +∞.
Let us now prove the converse. We denote fε,δ(s) = 12π
∫ξ∈R e
isξgε,δ(ξ) dξ the bounded continuous
function with Fourier transform gε,δ(z) = f(z+iδ)(1−iεz)2 well dened and holomorphic in a neighborhood
of C+, bounded by C(1+εy)2+(εx)2 . But we have for s > 0
fε,δ(s) = limN+∞
1
2π
∫ N
−Neisξgε,δ(ξ) dξ = lim
N+∞
1
2π
∫γN
eiszgε,δ(z)
where we have used a change of rectangular contour γN composed of three segment
[−N,−N + iN ] where we have the estimate |eiszgε,δ(z)| ≤ C1+(εN)2 , using that Im(z) ≥ 0 and
s ≥ 0.
[−N + iN,N + iN ] where we have the estimate |eiszgε,δ(z)| ≤ Ce−sN
(1+εN)2 , using that Im(z) = N
and s ≥ 0.
[N + iN,N ] with the same estimates.
Making N converge to +∞ and taking into account that the length of the path are of the order of N ,we get that fε,δ(s) = 0 for s ≥ 0. By dominated convergence, we obtain rst fε,0(s) = 0 for s ≥ 0,
and then f = 0 for s ≥ 0 by dominated convergence again since f is Schwarz on R.
51
3.3 The Carleman estimate
As in the classical case, we need to check the eect of the conjugated operator. Yet, we have to be a
little careful, rst because eε|Dt|
2
2τ is not well dened on any Sobolev space and even not on S.As before, we make the change of variable v = eτψu and for getting (3.4), we are left to prove
τ‖e−ε|Dt|
2
2τ v‖2H1τ≤ C
∥∥∥∥e−ε |Dt|22τ Pψv
∥∥∥∥2
L2
+ Ce−dτ ‖v‖2H1τ
Our operator P commutes with e−ε|Dt|
2
2τ since its coecients are independent on t. Yet, the operatorPψ = eτψPe−τψ may depend on t because ψ depends on t. We will take advantage of the fact thatsince ψ is quadratic, the principal symbol of Pψ only involve some derivative of ψ of order at least 1and is therefore linear. We rst prove the following simple Lemma.
Lemma 3.3.1. Let u ∈ S(Rn+1), then
e−ε|Dt|
2
2τ (tu) =
(t+ iε
Dt
τ
)e−ε
|Dt|2
2τ u.
Proof. We have the formula
tv(ξ) =
∫Rn+1
e−itξte−ix·ξxtv(t, x)dxdt =
∫Rn+1
i∂ξte−itξte−ix·ξxtv(t, x)dxdt = i∂ξt u(ξ).
(e−ε
|Dt|22τ (tu)
)(ξ) = e−ε
|ξt|2
2τ (tu)(ξ) = e−ε|ξt|
2
2τ i∂ξt u(ξ) = i∂ξt
[e−ε
|ξt|2
2τ u(ξ)
]+ i
εξtτe−ε
|ξt|2
2τ u(ξ)
=[
te−ε|Dt|2
2τ u
](ξ) + i
[εDt
τe−ε
|Dt|22τ u
](ξ)
That is
e−ε|Dt|
2
2τ (tu) = te−ε|Dt|
2
2τ u+ iεDt
τe−ε
|Dt|2
2τ u
=
(t+ iε
Dt
τ
)e−ε
|Dt|2
2τ u.
Remark 3.3.1. Lemma 3.3.1 could easily be iterated to get the formula
e−ε|Dt|
2
2τ (tku) =
(t+ iε
Dt
τ
)ke−ε
|Dt|2
2τ u.
where the exponent k is meant in the sense of composition. For f polynomial in t, we would get
e−ε|Dt|
2
2τ (f(t)u) = f
(t+ iε
Dt
τ
)e−ε
|Dt|2
2τ u.
This means that the "formal" conjugated operator of e−ε|Dt|
2
2τ with f(t) is a dierential of the sameorder as f .
52
For a general (even smooth) function f(t), it seems therefore very hard to give a precise meaningto f
(t+ iεDtτ
).
Even in the analytic case, f(t+ iεDtτ
)would be an innite sum of dierential operators, that
means an operator of "innite order". This is not clear how to dene this in an exact way. Yet,some authors managed to give some meaning of an approximation of this formula. Namely, the idea isto replace t+ iεDtτ by some approximate operator χ
(tκ
)t+ iχ
(εDtκτ
)εDtτ for κ small. These operators
have the advantage to be bounded and we can consider some innite series. We refer to Hörmander[8]. Similarly, it is possible to replace the holomorphic function f with a cuto near small x and ξt,see Tataru [12, 13].
We compute, if P =∑|α|≤2 pα(x)Dα
z
Pψ =∑|α|≤2
pα(x)(Dz + iτ∇zψ)α.
Lemma 3.3.2 (The "conjugated operator"). Let P =∑
α pα(z)Dαz ∈ Diffm be a (classical) dif-
ferential operator on Rn+1 = Rt × Rx, with principal symbol pm ∈ Σm. Assume also that all itscoecients are independent on t, that is pα(z) = pα(x). Let ψ real valued and quadratic in t.
Then, for any ε > 0, there exists a unique Pψ,εv ∈ Diffmτ so that we have
Qψε,τPu = Pψ,εQψε,τu.
for any u ∈ S(Rn+1).Moreover, the principal symbol of Pψ,ε is
pψ,m,ε = pm(z, ξ + iτ∇ψ − εψ′′t,zξt) =∑|α|=m
pα(x)(ξ + iτ∇Φ− εψ′′t,zξt)α,
where we use the notation ψ′′t,zξt = Hess(ψ)((ξt, 0, · · · , 0); ·) = ξtV with V the constant vector withcoecients Vk = (∂t∂kψ). We will denote it pψ,ε for simplicity in the sequel.
Additionally, Pψ,ε can be decomposed Pψ,ε = PR,ε+ iτ PI,ε with (PR,ε)∗ = PR,ε and (PI,ε)
∗ = PI,ε.
Remark 3.3.2. The expression "conjugated operator" is a bit abusive since eε|Dt|
2
2τ is not well denedas an operator. Yet, we would like to write formally
Pψ,εv = Qψε,τP(Qψε,τ
)−1v = e−ε
|Dt|2
2τ eτψPe−τψeε|Dt|
2
2τ v = e−ε|Dt|
2
2τ Pψeε|Dt|
2
2τ v.
Remark 3.3.3. The previous expression is actually a consequence of the fact that for a quadratichomogeneous function f in t, we have iτ∇f
(t+ iεDtτ
)= iτ∇f(t)− ε∇f(Dt)
Proof. Since ψ is quadratic, for any k = 0, ·, n, ∂kψ is polynomial of order 1 and can be written
∂kψ = f1(x) + tf0(x).
where f1(x) (resp. f0) is polynomial in x of order 1 (resp. 0). In particular, Lemma 3.3.1 gives
e−ε|Dt|
2
2τ [(Dk + iτ∂kψ)u] = e−ε|Dt|
2
2τ [(Dk + iτ(f1(x) + tf0(x)))u]
=
[Dk + iτ
(f1(x) +
(t+ iε
Dt
τ
)f0(x)
)]e−ε
|Dt|2
2τ u
= [(Dk + iτ∂kψ − εf0(x)Dt] e−ε |Dt|
2
2τ u.
53
To get an intrinsic expression, we notice that f0(x) = ∂t∂kψ, so f0(x)Dt can be written ∂t∂kψDt.So, the full (and principal symbol) of Dk + iτ∂kψ − εf0(x)Dt is
ξk + iτ∂kψ − ε(∂t∂kψ)ξt = ξk + iτ∂kψ(x)− ε(∂t∂kψ)ξt.
So, since Dα = Dα11 · · ·D
αjj · · ·Dαn
n , as in Lemma 2.2.1, we obtain similarly that the "conjugatedoperator" for Dα has principal symbol
n∏k=1
(ξk + iτ∂kΦ− ε(∂t∂kψ)ξt)|αk| = (ξ + iτ∇Φ− εψ′′t,zξt)α,
where we use the notation ψ′′t,zξt = Hess(ψ)((ξt, 0, · · · , 0); ·).Since all the functions pα(x) do not depend on t, they commute with Qψε,τ . So, we get the
conclusion of the rst two statements of the Lemma.We prove the last one for the conjugated operator of Dα by iteration on |α|. We prove the more
precise statement that it can be written A + τB with A∗ = A and A with constant coecients oforder |α|. If |α| = 0, it is obvious. Otherwise, assume Mα = Dα and Mα
ψ,ε = A + τB with A, B asbefore. Let l = 0 1, ... or n.
e−ε|Dt|
2
2τ eτψMαDlu = Mαψ,εe
−ε |Dt|2
2τ eτψDlu
= (A+ τB) [Dl + iτ∂lψ(x)− ε(∂t∂lψ)Dt] e−ε |Dt|
2
2τ eτψu
But since ψ is quadratic, ∂t∂lψ is a constant and we have the expression
(A+ τB) [Dl + iτ∂lψ(x)− ε(∂t∂lψ)Dt] = ADl − εA(∂t∂lψ)Dt + τC.
for an appropriate C. This gives the result. The nal result can also be obtained using that pα doesnot depend on t.
The important fact of the previous formula is that the principal symbol of Pψ,ε is actually closeto the principal symbol of Pψ if ε is small. So, we can expect that it satises the same subellipticestimates.
We rst write the following Lemma on pψ, that we have actually already used and proved inProposition 3.2.2, using Lemma 2.2.3 and homogeneity, so we skip the proof.
Lemma 3.3.3. Assume that ψ satises the pseudoconvexity for functions in ξt = 0 at z0 of Denition3.1.1, then there exist C1, C2 > 0 so that we have the estimate taken at the point z0 and for anyτ ≥ 0, ξ ∈ Rn,
1
iτpψ, pψ+ C1
[|pψ(z0, ξ, τ)|2
|ξ|2 + τ2+ |ξt|2
]≥ C2(|ξ|2 + τ2).
where we have extended 1iτ pψ, pψ by continuity at τ = 0 with the value 2p, p, ψ.
By perturbation, we can get a similar conclusion for the perturbated operator.
Lemma 3.3.4. Assume that ψ satises the pseudoconvexity for functions in ξt = 0 at z0 of Denition3.1.1. Then, there exists ε0 > 0 so that for any 0 ≤ ε < ε0, there exist C1, C2 > 0 so that we havethe estimate for any τ ≥ 0, ξ ∈ Rn,
1
iτpψ,ε, pψ,ε+ C1
[|pψ,ε(z0, ξ, τ)|2
|ξ|2 + τ2+ |ξt|2
]≥ C2(|ξ|2 + τ2).
where we have extended 1iτ pψ,ε, pψ,ε by continuity at τ = 0 with the appropriate value.
54
Proof. The Lemma mainly follows by saying that pψ,ε is a perturbation of pψ and using Lemma 3.3.3.Yet, we have to be a little careful because of the factor 1
τ . Noticing as before that 1iτ pψ,ε, pψ,ε =
2τ Re pψ,ε, Im pψ,ε. Then, using the last part of Lemma 3.3.2, we can write Im pψ,ε = τ pψ,ε
i. More-
over, pψ,εi and all its derivative are all continuous in ε. Hence, we can write 1
iτ pψ,ε, pψ,ε =
2Re pψ,ε, pψ,εi. It can therefore be extended by continuity to τ = 0 and the result follows by
a perturbation of Lemma 3.3.3.
We are now ready to prove a rst subelliptic estimate that will be crucial for the nal proof ofTheorem 3.1.2
Proposition 3.3.1. Under the previous assumptions for P , ψ, z0, there exist ε > 0, a neighborhoodV of z0, C > 0 and τ0 > 0 so that we have the following estimate
τ ‖v‖2H1τ≤ C ‖Pψ,εv‖2L2 + Cτ ‖Dtv‖2L2 ,(3.8)
for any v ∈ C∞0 (V ) and τ ≥ τ0.
Proof. Using the same computations as in Theorem 2.2.1 with the decomposition Pψ,ε = PR,ε+PI,ε =
PR,ε + iτ PI,ε, we get
‖Pψ,εv‖2L2 = ‖PR,εv‖2L2 + ‖PI,εv‖2L2 + (i[PR,ε, PI,ε]v, v)
= ‖PR,εv‖2L2 + ‖PI,εv‖2L2 + τ(i[PR,ε, PI,ε]v, v
).
The same computations lead to
1
τ‖Pψ,εv‖2L2 ≥ (Lv, v)
with
L = PR,εC2
(−∆ + τ2)PR,ε + PI,ε
C2
(−∆ + τ2)PI,ε +
i
τ[PR,ε, PI,ε]
= PR,εC2
(−∆ + τ2)PR,ε + PI,ε
C2
(−∆ + τ2)PI,ε + i[PR,ε, PI,ε].
for τ large enough and C to be chosen. So, we have
1
τ‖Pψ,εv‖2L2 + C ‖Dtv‖2L2 ≥
((L+ C2D2
t )v, v)
(3.9)
The principal symbol of L+ C2D2t is
1
2iτpψ,ε, pψ,ε+ C2
[|pψ,ε(z0, ξ, τ)|2
|ξ|2 + τ2+ |ξt|2
].
We conclude as before using Lemma 3.3.4 and Gårding inequality.
Proof of Theorem 3.1.2. Let ε > 0 xed and r0 so that B(z0, r0) ⊂ V where V is given by theprevious theorem. In the proof, we consider functions u ∈ C∞0 (B(z0, r0/4)). Let χ ∈ C∞0 (]− r0, r0[)such that χ = 1 on ]− r0/2, r0/2[.
Setting v = Qψε,τu = e−ε
2τ|Da|2(eτψu), we need to prove
τ ‖v‖2H1τ≤ C ‖Pψ,εv‖2L2 + Ce−dτ
∥∥∥eτψu∥∥∥2
H1τ
.
55
Yet, v is not compactly supported in the variable t. So, we set f = χ(t)v(x) and we have
supp(f) ⊂ B(z0, r0) so that we may apply Proposition 3.8 to f . We have v − f = (1 − χ)Qψε,τu =
(1−χ)e−ε
2τ|Da|2(χeτψu) for some χ ∈ C∞c (BRt(0, r0/3)) with χ = 1 in a neighborhood of BRt(0, r0/4)
so that χu = u.
Lemma 3.3.5. Let χ1 ∈ C∞(Rn+1), χ2 ∈ C∞(Rn+1) with all derivatives bounded such that dist(supp(χ1), supp(χ2)) >0. Then there exist C, c > 0 such that for all u ∈ S(Rn) and all λ ≥ 0, we have∥∥∥∥χ1e
− |Dt|2
λ (χ2u)
∥∥∥∥H1
≤ Ce−cλ ‖u‖H1 .
As a consequence of Lemma 3.3.5, we have, for τ ≥ τ0
‖v‖1,τ ≤ ‖f‖1,τ + Ce−Cτε ‖eτψu‖H1
τ(3.10)
The subelliptic estimate (3.8) applied to f gives
τ ‖f‖2H1τ≤ C ‖Pψ,εf‖2L2 + Cτ ‖Dtf‖2L2 ,
We need to estimate the terms on the RHS in terms of v.Second, we estimate ‖Pψ,εf‖L2 = ‖Pψ,εχv‖L2 = ‖χPψ,εv‖L2+‖[Pψ,ε, χ]v‖L2 . For the commutator,
we write [Pψ,ε, χ]v = [Pψ,ε, χ]e−ε
2τ|Da|2χeτψu. We notice that [Pψ,ε, χ] is a dierential operator of
order 1 in (D, τ) with some coecients supported on supp(χ′t) that is, away from supp(χ). Inparticular, Lemma 3.3.5 implies ‖[Pψ,ε, χ]v‖L2 ≤ Ce−c
τε
∥∥eτψu∥∥H1τ. This yields
‖Pψ,εf‖L2 ≤ ‖Pψ,εv‖L2 + Ce−cτε
∥∥∥eτψu∥∥∥H1τ
Now, it remains to treat the term ‖Dtf‖L2 . Similarly, we obtain
‖Dtf‖L2 = ‖Dt(χv)‖L2 ≤ ‖χDtv‖L2 + ‖χ′(t)e−ε
2τ|Dt|2χeτψu‖L2 ≤ ‖Dtv‖L2 + Ce−c
τε
∥∥∥eτψu∥∥∥L2
where we have used again Lemma 3.3.5.Let ς a small constant to be xed later on. We distinguish between frequencies of size smaller
and bigger than ςτ . We get for τ ≥ 1ς2ε
large enough (so that the function s 7→ se−ε
2τs2 is decreasing
on s ≥√
τε )
‖Dtv‖L2 = ‖Dte− ε
2τ|Dt|2eτψu‖L2 ≤ ‖Dt1|Dt|≤ςτv‖L2 + ‖Dt1|Dt|≥ςτe
− ε2τ|Dt|2eτψu‖0
≤ ςτ‖v‖L2 + ςτe−τς2ε
2 ‖eτψu‖L2
So, at that point, we have proved that there are some constants c, C > 0 so that for any ς > 0, wehave for τ large enough
τ ‖v‖2H1τ≤ C ‖Pψ,εv‖2L2 + Cς2τ3‖v‖2L2 + C
(e−cτ + ς2τ3e−τς
2ε)∥∥∥eτψu∥∥∥2
L2.
This gives the result if ς is chosen small enough so that the term Cς2τ3‖v‖2L2 ≤ Cς2τ‖v‖2H1τcan be
absorbed.
56
Proof of Lemma 3.3.5. Using the Fourier transform of the Gaussian (see exercice), we have
(e−|Dt|
2
λ f)(s) =
(λ
4π
) 12∫Rse−
λ4|s−t|2f(s) ds.
We have
χ1e− |Dt|
2
λ (χ2u)(t, x) =
(λ
4π
) 12∫Rsχ1(t, x)e−
λ4|s−t|2(χ2u)(s, x) ds
=
(λ
4π
) 12
χ1(t, x)
∫s,|t−s|≥d
e−λ4|s−t|2(χ2u)(s, x) ds
where we have used the properties of support for the second equality. so that
|χ1e− |Dt|
2
λ (χ2u)|(t, x) ≤ ‖χ1‖L∞(λ
4π
) 12∫s,|t−s|≥d
e−λ4|s−t|2 |χ2u|(s, x) ds
≤ ‖χ1‖L∞(λ
4π
) 12 (
1|·|≥de−λ
4|·|2 ∗Rs |χ2u|(·, x)
)(t).
As a consequence, using the Young inequality, we have
‖χ1e− |Dt|
2
λ (χ2u)‖L2 ≤ ‖χ1‖L∞(λ
4π
) 12∥∥∥1|·|≥de
−λ4|·|2∥∥∥L1(R)
‖χ2u‖L2(Rn+1),
and, using∥∥∥1|·|≥de
−λ4|·|2∥∥∥L1(R)
≤ Ce−cλ for some appropriate c > 0, we obtain
‖χ1e− |Dt|
2
λ (χ2u)‖L2 ≤ ‖χ1‖L∞ Ce−cλ‖u‖L2(Rn+1),
which implies the result.
Note that to estimate∥∥∥1|·|≥de
−λ4|·|2∥∥∥L1(R)
, we could use∫ +∞r e−s
2ds ≤ e−
r2
2
∫ +∞r e−
s2
2 ds ≤ Ce−r2
2
so that∫ +∞r e−λs
2ds =
∫ +∞√λre−y
2dy ≤ Ce−λr2/2.
3.4 Global unique continuation and non characteristic hypersurfaces
3.4.1 Distance and metric
Let Ωx, P be as in Theorem 3.1.1. Assume Ωx connected. We are going to dene the Riemanniandistance related to the operator Q.
We can assume that ai,j(x) is symmetric without changing the operator P . The ellipticity andpositivity assumption shows that for any x ∈ Ωx, we can dene the matrix (gi,j) = (ai,j)
−1 which isstill positive.
For any x ∈ Ωx and ξ ∈ Rn, we dene ‖ξ‖g(x) =√∑n
i,j=1 gi,j(x)ξiξj . Moreover, if γ : [0, 1] 7→ Ωx
is a smooth path, we dene
length(γ) =
∫ 1
0‖γ(t)‖g(γ(t)) dt.
This allows to dene the Riemannian distance
dist(x1, x2) = infγ(0)=x1;γ(1)=x2
length(γ).
57
3.4.2 The global theorem
Theorem 3.4.1 (Global unique continuation). Let Ωx, P be as in Theorem 3.1.1. Let x0, x1 ∈ Ωx.Let ω0 be neighborhood of x0 in Ωx. Then, for any T > dist(x0, x1), there exist ε > 0 and Vx1 oneneighborhood of x1 so that for any u ∈ C∞(Ω),
Pu = 0 in ]− T, T [×Ωx,u = 0 in ]− T, T [×ω0
=⇒ u = 0 in ]− ε, ε[×Vx1 .(3.11)
Proof. According to Lemma 3.4.1 below, we can nd local coordinates (w, xn) near γ in which thepath γ by γ(s) = (0, s`0) and the metric is given by the matrix m(w, xn) ∈Mn(R) with
m(w, xn) =
(m′(xn) 0
0 1
)+OMn(R)(|w|), for w ∈ BRn−1(0, δ), δ > 0,(3.12)
withm′(xn) ∈Mn−1(R) (uniformly) denite symmetric. With these coordinates in the space variable,and still using the straight time variable, the symbol of the wave operator is given by
p(t, w, xn, τ, ξw, ξn) = p(w, xn, τ, ξw, ξn) = −τ2 + 〈m(w, xn)ξ, ξ〉, ξ = (ξw, ξn),(3.13)
where we have used τ for the dual of the time variable and ξw, ξn for the dual to w ∈ BRn−1(0, δ)and xn ∈ [0, `0].
We now aim to apply Theorem 3.1.1 and we need to construct appropriate non characteristichypersurfaces.
Pick again t0 with `0 < t0 < T . For b < δ small, to be xed later on, we dene
xn = l, x′ = (t, w), D =
(t, w)
∣∣∣∣(wb )2+( tt0
)2≤ 1
G(t, w, ε) = ε`0ψ
(√(wb
)2+( tt0
)2), φε(t, w, xn) := G(t, w, ε)− xn, ε ∈ [0, 1]
where ψ is such that
ψ even, ψ(±1) = 0, ψ(0) = 1,
ψ(s) ≥ 0, |ψ′(s)| ≤ α, for s ∈ [−1, 1],
with 1 < α < t0`0. This is possible since t0
`0> 1.
Note also that the fact that ψ is even gives that G(t, w, ε) is actually smooth.Note also that the point (t = 0, w = 0, xn = `0) corresponding in the local coordinates to x1
belongs to φ1 = 0. We have
dφε(t, w, xn) = ε`0
((wb
)2+( tt0
)2)−1/2
ψ′(√(w
b
)2+( tt0
)2)(
tdt
t20+wdw
b2
)− dxn.
Given the form of the principal symbol of the wave operator in these coordinates (see (3.12)-(3.13)),we obtain
p(w, xn, dφε(t, w, xn)) = −ε2`20t2
t40
((wb
)2+( tt0
)2)−1
|ψ′|2
+`20ε2
b4〈m′(xn)w,w〉
((wb
)2+( tt0
)2)−1
|ψ′|2 + 1
+O(|w|2)
(1 +
ε2`20b4|w|2
((wb
)2+( tt0
)2)−1
|ψ′|2),
58
where |ψ′|2 is taken at the point
(√(wb
)2+(tt0
)2). Now, since α < t0
`0and m′(xn) is uniformly
(for xn ∈ [0, `0]) denite positive, there is η > 0 so that for |w| ≤ b small enough, we have
1 +O(|w|2) ≥ α2 `20
t20η
〈m′(xn)w,w〉+O(|w|2)|w|2 ≥ 1
2〈m′(xn)w,w〉 ≥ 0.
Hence, there is a suciently small neighborhood (taking again b small enough) of the path (i.e. ofw = 0), in which we have (for any ε ∈ [0, 1]), and any (t, w, xn) ∈ D × [0, `0]
p(w, xn, dφε(t, w, xn)) ≥ −ε2
t20`20
( tt0
)2((w
b
)2+( tt0
)2)−1
|ψ′|2 + α2 `20
t20+ η
≥ −`20
t20|ψ′|2 + α2 `
20
t20+ η ≥ η.
So, the surface φε = 0 is noncharacteristic for any ε ∈ [0, 1] and, therefore, strictly pseudoconvexwith respect to the wave operator.
Now, dene Kε = xn ≤ G(t, w, ε) ∩ xn ≥ 0.Consider ε0 = sup ε;u = 0 on Kε. A continuity argument yields that that u = 0 on Kε0 . A
compactness argument on the compact set (taking into account the "corners") and the successiveapplication of Theorem 3.1.1 gives the result. ♣ un peu rapide...
Lemma 3.4.1. Let γ : [0, 1] → Ωx be a smooth path without self intersection of length `0 so thatγ(0) = x0 and γ(1) = x1.
Then, there are some coordinates (w, l) ∈ BRn−1(0, ε) × [0, `0] in an open neighborhood U nearγ([0, 1]) so that
γ([0, 1]) = w = 0 × [0, `0],
the metric g is of the form m(l, w) =
(1 00 m′(l)
)+OMn(R)(|w|),
Proof. The path γ is of length `0 so, we can reparametrize it by γ : [0, `0] → Ωx such that γ isunitary (that is ‖γ(s)‖γ(s) = 1). Moreover, since γ does not have self intersection, there exist U aneighborhood in Ωx of γ and a dieomorphism ψ such that
ψ(U) ⊂ (x, y) ∈ Rn |x ∈ [−ε, `0 + ε], |y| ≤ ε,
ψ(γ(s)) = (s, 0),
ψ(U) = (x, y) ∈ Rn, f1(y) ≤ x ≤ f2(y) |x ∈ [−ε, `0 + ε], |y| ≤ ε for some smooth functions filocally dened
Then, we make some change of variable to diagonalize the metric on γ. By unitarity of thecoordinates, the metric on γ has the form
m(x, 0) =
(1 l(x)
tl(x) g(x)
),
where l is a line vector and g is a positive denite matrix. We perform the change of variable
Φ : (x, y) 7→ (x, y) = (x− ax · y, y). In y = 0, we have DΦ(x, 0) =
(1 −ax0 Id
)with tDΦ(x, 0) =
59
(1 0−tax Id
)(in particular, the change of variable is valid for small y) andDΦ(x, 0)−1 =
(1 ax0 Id
)with tDΦ(x, 0)−1 =
(1 0tax Id
). Moreover, in the new coordinates, the set in y = 0 and the
metric there is given by
tDΦ(x, 0)−1m(x, 0)DΦ(x, 0)−1 =
(1 l(x) + a(x)
tl(x) +t a(x) ∗
)So, we choose a(x) = −l(x) so that in this new coordinates m(x, 0) is of the form
m(x, 0) =
(1 00 ∗
).(3.14)
The expected property of m is then obtained by the mean value theorem using the diagonal form(3.14) on γ.
3.5 Approximate controllability
3.6 Further remarks
3.6.1 The general theorem
of Tataru (Robbiano-Zuily, Hörmander)
3.6.2 Quantitative estimates
boundary Carleman estimates
60
Appendix A
Appendix
A.1 Pseudodierential operators
A.2 The Dirichlet problem for some second order elliptic operators
In this section, we shall consider a particular class of operators as described in Remark ??, thatis, with symbols the form p2(x, ξ) = Qx(ξ) where Qx is a smooth family of real quadratic forms.Assuming that the variables xa are tangent to the boundary, and that the functions satisfy Dirichletboundary conditions, we prove a counterpart of the local estimate of Theorem ?? for this boundaryvalue problem. For this, the main goal to achieve is to prove a Carleman estimate adapted to thisboundary value problem. All local, semiglobal and global results shall then follow.
This situation is of particular interest for the wave equation for which xa is the time variable,which is always tangent to the boundary of cylindrical domains.
For the sake of simplicity, we shall further assume that the operator principal symbol of P isindependent of the xa variable (we would otherwise need to assume the coecients of P to beanalytic with respect to xa). This allows to avoid some additional technicalities in the (alreadyrather technical) proofs.
A.2.1 Some notation
Here, we shall always assume that the analytic variables are tangential to the boundary, that is
x = (xa, xb) ∈ Rna × Rnb+ , with Rnb+ = Rnb−1 × R+, and xb = (x′b, xnb ).
When the distinction between analytic and non-analytic variables is not essential, we shall split thevariables according to
x = (x′, xn) ∈ Rn+ = Rn−1 × R+, with x′ = (xa, x′b) ∈ Rna+nb−1, and xn = xnb ∈ R+.
We also denote by ξ′ ∈ Rn−1 the cotangential variables and ξn the conormal variable, by D′ = 1i (∂x′)
the associated tangential derivations and Dn = 1i ∂xn the normal derivation.
For any r0 > 0, we dene
Kr0 =x ∈ Rn+; |x| ≤ r0
= BRn(0, r0) ∩ xn ≥ 0.(A.1)
We denote by C∞0 (Rn+) the space of restrictions to Rn+ of functions in C∞0 (Rn), and by C∞0 (Kr0)the space of functions C∞0 (Rn+) supported in Kr0 . the trace of a function f ∈ C∞0 (Rn+) at xn = 0 isdenoted by f|xn=0.
61
We denote by (f, g) =∫Rn+fg, ‖f‖20,+ = (f, f) the L2(Rn+) inner product and norm. For k ∈ N,
the norm ‖·‖k,+ will denote the classical Sobolev norm on Rn+ and ‖·‖k,+,τ the associated weightednorms, that is,
‖f‖2k,+,τ =∑
j+|α|≤k
τ2j ‖∂αf‖20,+ , τ ≥ 1.
We also dene the tangential Sobolev norms, given by
|f |2k,τ =∥∥∥(|D′|+ τ)kf
∥∥∥2
0,+∼
∑j+|α|≤k
τ2j ‖∂αx′f‖20,+ , τ ≥ 1.
We shall also use, for f, g ∈ C∞0 (Rn+), the notation (f, g)0 =∫Rn−1 f|xn=0(x′)g|xn=0(x′)dx′.
Finally, for j ∈ N, we denote by Dkτ , the space of tangential dierential operators, i.e. operatorsof the form
P (x,D′, τ) =∑
j+|α|≤k
aj,α(x)τ jD′α,
and by
σ(P ) = p(x, ξ′, τ) =∑
j+|α|=k
aj,α(x)τ jξ′α
their principal symbol.
A.2.2 The Carleman estimate
In this section, we state and prove the Carleman estimate of Theorem (2.4.2) asociated to the Dirichletproblem . Note that it applies also to elliptic operator, but also to wave type operators.
To prove Theorem 2.4.2, we dene the conjugated operator Pψ = eτψPe−τψ = P (x,D + iτψ′).When proving the theorem, we shall drop the index + in the norms to lighten the notation; of
course, all inner norms and integrals are meant on Rn+. We rst need the following proposition.
Theorem A.2.1. Under the assumptions of Theorem 2.4.2, there exist C > 0, τ0 > 0 such that forany τ > τ0 and f ∈ C∞0 (Kr0), we have
τ‖f‖21,τ ≤ C ‖Pψf‖20 + τ3|f|xn=0|20 + τ |Df|xn=0|20.(A.2)
If moreover ∂xnψ > 0 for (x′, xn = 0) ∈ Kr0, then
τ‖f‖21,τ ≤ C ‖Pψf‖20 , for all f ∈ C∞0 (Kr0) such that f|xn=0 = 0.(A.3)
Proof. Dening Q2 = 12(Pψ + P ∗ψ) and Q1 = 1
2iτ (Pψ − P ∗ψ), we have
Pψ = Q2 + iτQ1,
and denote by qj the principal symbol of Qj , j = 1, 2. We haveQ2 = D2
n +Q2
Q1 = Dnψ′xn + ψ′xnDn + 2Q1,
(A.4)
where Q2 ∈ D2τ and Q1 ∈ D1
τ with principal symbols
q2 = −τ2(ψ′xn)2 + r(x, ξ′)− τ2r(x, ψ′x′)
q1 = r(xb, ξ′, ψ′x′),
62
where r is the bilinear form associated with the quadratic form r. Note that, even if it does notappear in the notation, all these operators depend upon the parameter τ .
With this notation, we hence have pψ = q02 + iτ q0
1, so that 1iτ pψ, pψ = 2q0
2, q01. Assump-
tions (??) and (??) then translate respectively into
q02, q
01(x, ξ) > 0, if p(x, ξ) = 0, x ∈ Kr0 , τ = 0;(A.5)
q02, q
01(x, ξ) > 0, if pψ(x, ξ) = 0, x ∈ Kr0 , τ > 0,(A.6)
where the second assertion is a direct consequence of (??), and the rst one follows from (??) togetherwith the fact that, using that p is real, we have
limτ→0+
1
iτpψ, pψ =
∂
∂τ
1
ipψ, pψ
∣∣∣∣τ=0
= 2 p, p, ψ .
Next, we have the integration by parts formulæ:(g, Q2f) = (Q2g, f)− i [(g,Dnf)0 + (Dng, f)0)0] ,
(g, Q1f) = (Q1g, f)− 2i(ψ′xng, f
)0.
(A.7)
So, we have for f ∈ C∞0 (Kr0)
‖Pψf‖0 =∥∥∥Q2f
∥∥∥2
0+ τ2
∥∥∥Q1f∥∥∥2
0+ iτ
[(Q1f, Q2f
)−(Q2f, Q1f
)].(A.8)
So, we get, using the integration by parts formulæ (A.7)
‖Pψf‖0 =∥∥∥Q2f
∥∥∥2
0+ τ2
∥∥∥Q1f∥∥∥2
0+ iτ
([Q2, Q1]f, f
)+ τB(f),(A.9)
with the boundary term
B(f) =[(Q1f,Dnf)0 + (DnQ1f, f)0
]− 2
(ψ′xnQ2f, f
)0
= 2(ψ′xnDnf,Dnf)0 + (M1f,Dnf)0 + (M ′1Dnf, f)0 + (M2f, f)0,(A.10)
for some tangential operator M1 of order 1 (in ξ′, τ) (note that terms of order two in Dn cancel).Now that we have made the exact computations, we will make some estimates on the symbols
of the interior part of the commutator. The idea is to tranfer the positivity assumption of the fullsymbol to some positivity of a tangential symbol, which will then allow to apply the tangentialGårding.
The rst step is to perform a factorisation of [Q2, Q1] with respect to Q1 and Q2 to have atangential reminder. Since [Q2, Q1] is of order 2, it can be written i[Q2, Q1] = C2 + C1Dn + C0D
2n
where Ci ∈ Diτ . But using (A.4), and ψ′xn 6= 0 on Kr0 , we can replace Dn = 12ψ′xn
Q1 + D1τ and
D2n = Q2 −Q2. So, in particular, we can write
i[Q2, Q1] = B0Q2 +B1Q1 +B2.(A.11)
where Bi ∈ Diτ with real symbol bi. Now, we need to
use the assumption to get some positivity of the symbol pψ, pψ, this is Lemma A.2.1;
transfer this information to a tangential information on the symbol, this is Lemma A.2.2.
63
Lemma A.2.1. There exist C1, C2 > 0 such that for all (x, ξ) ∈ Kr0 × Rn and τ > 0, we have
(|ξ|2 + τ2) ≤ C1q02, q
01(x, ξ) + C2
[|pψ(x, ξ)|2
|ξ|2 + τ2
].
Proof. All the terms are homogeneous of order 2 in (ξ, τ) and continuous on the compact (x, ξ, τ) ∈Kr0 ×(ξ, τ) ∈ Rn×R+, |ξ|2 + τ2 = 1. Thus, on this set, the result is a consequence of (A.5), (A.6)
and Lemma 2.2.3 applied to f =|pψ(x,ξ)|2|ξ|2+τ2 ≥ 0, g = q0
2, q01 and h = 0. The result on the whole
Kr0 × Rn × R+ follows by homogeneity.
Now, we set
µ(x, ξ′) = (q1)2 + (ψ′xn)2q2.
The symbol µ(x, ξ′) satises the property that µ(x, ξ′) = 0 if and only if there exists ξn real suchthat pψ(x, ξ′, ξn) = 0. This is easily seen by noticing that the zero of q1 can only be with ξn = − q1
ψ′xn.
Notice also that µ(x, ξ′) is a tangential symbol of order 2.
Lemma A.2.2. There exist C1, C2 > 0 such that for all (x, ξ′) ∈ Kr0 × Rn−1 and τ > 0, we have
(|ξ′|2 + τ2) ≤ C1b2 + C2
[[µ(x, ξ′)]2
|ξ′|2 + τ2
].(A.12)
Proof. Note rst that for any (x, ξ′, ξn) with ξn = − q1(x,ξ′)ψ′xn
, we have q1(x, ξ′, ξn) = 0 and
pψ(x, ξ′, ξn) = q2(x, ξ′, ξn) = (ψ′xn)−2µ(x, ξ′).
Now, assume µ(x, ξ′) = 0 and ξa = 0. Setting ξn = − q1(x,ξ′)ψ′xn
, we have pψ(x, ξ′, ξn) = 0. Using
Lemma ??, we have q2, q1(x, ξ′, ξn) > 0. According to the denition of B2 in (A.11), we haveb2(x, ξ′, ξn) > 0. As a consequence, we have
µ(x, ξ′) = 0 =⇒ b2(x, ξ′, ξn) > 0.
Moreover, all terms in (A.12) are homogeneous of order 2 in the variables (ξ′, τ) and continuous on(ξ′, τ) 6= (0, 0). Hence, applying Lemma 2.2.3 below on the compact set Kr0 × (ξ′, τ) ∈ Rn−1 ×R+, |ξ′|2 + τ2 = 1 yields (A.12) on that set. The conclusion follows by homogeneity.
Taking the real part of (A.9) and using (A.11), we obtain
‖Pψf‖0 − τ Re (B(f)) =∥∥∥Q2f
∥∥∥2
0+ τ2
∥∥∥Q1f∥∥∥2
0+ τ Re (B2f, f) + τ Re
((B0Q2 +B1Q1)f, f
).(A.13)
Concerning the remainder term, we have
τ |Re(
(B0Q2 +B1Q1)f, f)| ≤ τ‖f‖0‖Q2f‖0 + τ |f |1‖Q1f‖0
≤ τ−1/2(τ |f |21,τ + ‖Q2f‖20 + τ2‖Q1f‖20
).(A.14)
Dening nowΣ = (Q1)2 + (ψ′xn)2Q2,
64
with principal symbol µ, and for an operator G with principal symbol µε(x,ξ′)|ξ′|2+τ2 , the tangential Gårding
inequality (that means with some derivatives only in the variable x′), in which symbols are allowedto depend smoothly upon the variable xn yields, for τ suciently large,
|f |21,τ ≤ C Re (B2f, f) + Re (Σf,Gf) .(A.15)
Writing ψ′xnDn = 12(Q1 − [Dn, ψ
′xn ]) − Q1 (where ψ′xn does not vanish), this allows to estimate the
full norm ‖f‖1,τ according to
‖f‖1,τ ≤ C(‖Q1f‖0 + |f |1,τ ).(A.16)
Recalling the denitions of Qi in (A.4), we also have
Σ =
(1
2(Q1 − [Dn, ψ
′xn ])− ψ′xnDn
)2
+(ψ′xn)2(Q2 −D2
n
)=
(1
2(Q1 − [Dn, ψ
′xn ])− ψ′xnDn
)1
2(Q1 − [Dn, ψ
′xn ])
+(ψ′xn)2(Q2
),(A.17)
and hence
Σ ∈ (ψ′xn)2Q2 −1
2ψ′xnDnQ1 +D1
τ Q1 +D1τ +D0
τDn.
We now want to estimate the term Re (Σf,Gf) in (A.15). For this, integrating by parts in thetangential direction xa, we have∣∣(ψ′′xn,xa ((ψ′xn)2Dn +Q1ψ
′xn ;Da
)f,Gf
)∣∣ ≤ C‖ 〈Da〉 f‖‖f‖1,τ .
This yields
| (Σf,Gf) | ≤ C‖Q2f‖0‖f‖0 +
∣∣∣∣( 1
2iψ′xnQ1f,Gf
)0
∣∣∣∣+‖Q1f‖0‖f‖1,τ + ‖f‖0‖f‖1,τ + C‖ 〈Da〉 f‖‖f‖1,τ
≤∣∣∣∣( 1
2iψ′xnQ1f,Gf
)0
∣∣∣∣+ C‖f‖1,τ(τ−1‖Q2f‖0 + +‖Q1f‖0 + τ−1‖f‖1,τ + ‖Daf‖0
)(A.18)
According to (A.15) and (A.16) and (A.18), this now implies
‖f‖21,τ . Re (B2f, f) + ‖Q1f‖20 +
∣∣∣∣( 1
2iψ′xnQ1f,Gf
)0
∣∣∣∣+ τ−2‖Q2f‖20 + ‖Daf‖20.
Coming back to (A.13), we obtain, for τ large enough,
τ‖f‖21,τ . ‖Pψf‖20 − τ Re (B(f))−∥∥∥Q2f
∥∥∥2
0− τ2
∥∥∥Q1f∥∥∥2
0+ τ
∣∣∣∣( 1
2iψ′xnQ1f,Gf
)0
∣∣∣∣. ‖Pψf‖20 − τ Re (B(f)) + τ
∣∣∣∣( 1
2iψ′xnQ1f,Gf
)0
∣∣∣∣ .Recalling te denition of Q1, we have ψ
′xnQ1 = Dn +G1, where G1 ∈ D1
τ is a dierential operator oforder 1 (in (τ,D′)), we nally have
τ‖f‖21,τ . ‖Pψf‖20 − τ Re (B(f)) + τ |(Dnf +G1f,Gf)0| ,(A.19)
65
where G a tangential pseudodierential operator of order zero, Recalling the form of B(f) in (A.10)gives the bound |B(f)| ≤ τ2|f|xn=0|20 + |Df|xn=0|20, which concludes the proof of (A.2).
Now if f|xn=0 = 0, all tangential derivatives vanish. With (A.19) and the form of B(f) in (A.10),this yields
τ‖f‖21,τ . ‖Pψf‖20 − 2τ(ψ′xnDnf,Dnf)0,
which proves (A.3) since ψ′xn > 0 for (x′, xn = 0) ∈ K. This concludes the proof of Proposition A.2.1.
A.3 Link between bicharacteristic ow and geodesics
A.4 Useful computation for symbols of order 2
Many computation can be simplied for symbols homogeneous of order 2 using the language ofquadratic forms. Indeed, we have the following properties.
Lemma A.4.1. Let p(x, ξ) =∑
k,l ak,l(x)ξiξj be a symbol of order 2. For any x, denote Rx the
associated quadratic form Rx(ξ, ξ) =∑
k,lak,l(x)+al,k(x)
2 ξkξl.Then, we have
p,Φ = 2Rx(ξ,∇Φ)(A.20)
pΦ(x, ξ, τ) = p(x, ξ + iτ∇Φ) = p(x, ξ)− τ2p(x,∇Φ) + iτp,Φ(x, ξ)(A.21)
pΦ,Φ = 2R(ξ,∇Φ) + 2iτp(x,∇Φ).(A.22)
Moreover, assume f only depends on x, then
f, p,Φ = −p, f(x,∇Φ)(A.23)
Proof. Note that we check that p(x, ξ) = Rx(ξ, ξ).(A.20), (A.21) and (A.22) are linear in ak,l, so it is enough to prove it for p(x, ξ) = a(x)ξkξl and
Rx(ξ, ξ) = a(x)2
[ξkξl + ξlξk
].
p,Φ = ∇ξ(a(x)ξkξl) · ∇Φ = a(x) [ξk(∂lΦ) + ξl(∂kΦ)] = 2Rx(ξ,∇Φ)
This gives (A.20).
pΦ(x, ξ, τ) = p(x, ξ + iτ∇Φ) = a(x)(ξk + iτ∂kΦ)(ξl + iτ∂lΦ)
= a(x)[ξkξl − τ2(∂kΦ)(∂lΦ) + iτ(∂kΦ)ξl + iτ(∂lΦ)ξk
]= p(x, ξ)− τ2p(x,∇Φ) + 2iτRx(ξ,∇Φ).
This gives (A.21). For (A.22), using (A.20), we get
pΦ,Φ = (∇ξp)(x, ξ + iτ∇Φ) · ∇Φ = p,Φ(x, ξ + iτ∇Φ)
= 2Rx(ξ + iτ∇Φ,∇Φ) = 2Rx(ξ,∇Φ) + 2iτRx(∇Φ,∇Φ) = 2R(ξ,∇Φ) + 2iτp(x,∇Φ).
Finally,
f, p,Φ = f, a(x) [ξk(∂lΦ) + ξl(∂lΦ)] = −a(x) [(∂kf)(∂lΦ) + (∂lf)(∂lΦ)] = −2Rx(∇f,∇Φ) = −p, f(x,∇Φ)
66
Lemma A.4.2. If p is homogeneous of order 2, real valued
1
iτpΦ, pΦ = 2 p, p,Φ+ 2τ2 p, p,Φ (x,∇Φ)
Proof. Using (A.22) and the easy formula a− b, a+ b = 2 a, b, we have
1
iτpΦ, pΦ = 2
p− τ2p(x,∇Φ), p,Φ
= 2 p, p,Φ − 2τ2 p(x,∇Φ), p,Φ
Next, using Lemma A.4.3 below,
1
iτpΦ, pΦ = 2 p, p,Φ+ 2τ2 p, p,Φ (x,∇Φ)
Lemma A.4.3. If p is of homogeneous order 2
p(x,∇Φ), p,Φ = −p, p,Φ (x,∇Φ).
Proof. Note rst that we have the general formula (true for any order), see Lemma below
p(x,∇Φ), q(x,∇Φ) + p, q(x,∇Φ)(x,∇Φ) = p, q(x,∇Φ.
With q = p,Φ, it gives
p(x,∇Φ), p,Φ+ p, p,Φ(x,∇Φ) (x,∇Φ) = p, p,Φ (x,∇Φ).(A.24)
But for quadratic operators, (A.20) gives p,Φ(x,∇Φ) = 2Rx(∇Φ,∇Φ) = 2p(x,∇Φ), so (A.23)gives
p, p,Φ(x,∇Φ) (x,∇Φ) = 2p, p(x,∇Φ)(x,∇Φ)
= −2p(x,∇Φ), p,Φ
So, when we insert it into (A.24), we obtain
p(x,∇Φ), p,Φ − 2p(x,∇Φ), p,Φ = p, p,Φ (x,∇Φ).
Lemma A.4.4. Let p, q some symbols, Φ(x) function. Then
p(x,∇Φ), q(x,∇Φ) + p, q(x,∇Φ)(x,∇Φ) = p, q(x,∇Φ(A.25)
Proof.
p, q(x,∇Φ) = (∇ξp)(x,∇Φ) · (∇xq)(x,∇Φ)− (∇xp)(x,∇Φ) · (∇ξq)(x,∇Φ)
while
p(x,∇Φ), q(x, ξ) = −(∇xp)(x,∇Φ) · (∇ξq)(x, ξ)−HessΦ [(∇ξp)(x,∇Φ); (∇ξq)(x, ξ)]
So applied with ξ = ∇Φ
p(x,∇Φ), q(x,∇Φ) = −(∇xp)(x,∇Φ) · (∇ξq)(x,∇Φ)−HessΦ [(∇ξp)(x,∇Φ); (∇ξq)(x,∇Φ)]
67
By symmetry
p, q(x,∇Φ)(x,∇Φ) = (∇xq)(x,∇Φ) · (∇ξp)(x,∇Φ) +HessΦ [(∇ξp)(x,∇Φ); (∇ξq)(x,∇Φ)]
That is
p(x,∇Φ), q(x,∇Φ) + p, q(x,∇Φ)(x,∇Φ) = −(∇xp)(x,∇Φ) · (∇ξq)(x,∇Φ) + (∇xq)(x,∇Φ) · (∇ξp)(x,∇Φ)
= p, q(x,∇Φ
68
Appendix B
Correction of (some) exercices
B.1 Feuille 1
B.1.1 Exercice 1
Let A = aα,β(x)Dατβ . B = bα′,β′(x)Dα′τβ′of respective order m1 and m2 and full symbol a and b.
A Bu = aα,β(x)Dατβ[bα′,β′(x)Dα′τβ
′u]
= aα,β(x)τβ+β′Dα[bα′,β′(x)Dα′u
]Using Leibniz formula
∂α(fg) =∑
γ+δ=α
(α
γ
)(∂γf)(∂δg),
we get
Dα[bα′,β′(x)Dα′u
]=
1
i|α|
∑γ+δ=α
(α
γ
)(∂γbα′,β′)(∂δD
α′u)
So, we get
A Bu =1
i|α|
∑γ+δ=α
aα,β(x)τβ+β′(α
γ
)(∂γbα′,β′)(∂δD
α′u)
Each term in the sum is a dierential operator of order β+β′+|δ|+|α′| ≤ β+β′+|α|+|α′| = m1+m2.This maximum is reached only for the term δ = α, γ = 0,
(αγ
)= 1 where we have the term
1
i|α|aα,β(x)τβ+β′bα′,β′(x)(∂αD
α′u) = aα,β(x)τβ+β′bα′,β′(x)(DαDα′u) = (ab)(x,D, τ).
Let us now see the terms of order m1 +m2 − 1. They are so that β + β′ + |δ|+ |α′| = m1 +m2 − 1,that is |δ| = m1 − 1 and |γ| = 1. Moreover, γ = (1, 0, 0 · · · , 0) or γ = (0, 1, 0 · · · , 0), etc... Wedenote these vectors ej . The sum is amongst terms so that αj ≥ 1. In each of these cases,
(αej
)=(
α1
0
)· · ·(αj
1
)· · ·(αn0
)= αj .
1
i|α|
n∑j=1,αj≥1
aα,β(x)τβ+β′αj(∂jbα′,β′)(∂α−ejDα′u)
=1
i
n∑j=1,αj≥1
αjaα,β(x)τβ+β′(∂jbα′,β′)(Dα−ejDα′u).
69
Its symbol is
1
i
n∑j=1,αj≥1
aα,β(x)τβ+β′(∂jbα′,β′)αjξα−ejξα
′.
We ony recognize αjξα−ej = ∂ξjξ
α. And the formula is still true and equal to zero if αj = 0. So, theterm of order m1 +m2 − 1 is therefore
1
i
n∑j=1
aα,β(x)τβ+β′(∂jbα′,β′)(∂ξjξα)ξα
′=
1
i
n∑j=1
(∂ξja)(∂xjb)
.Now, take A =
∑|α|+β≤m1
aα,β(x)Dατβ and B = bα′,β′(x)Dα′τβ′. Decompose A = am1(x,D, τ)+
am1−1(x,D, τ) + r(x,D, τ) with am1(x,D, τ) homogeneous of order m1, am1−1(x,D, τ) homogeneousof order m1 − 1 and r(x,D, τ) of order at most m1 − 2.
The previous calculation shows
A B = am1(x,D, τ) B + am1−1(x,D, τ) B + r(x,D, τ) B
= (am1b)(x,D, τ) +1
i
n∑j=1
[(∂ξjam1)(∂xjb)
](x,D, τ) + (am1−1b)(x,D, τ) + r(x,D, τ) B
= (ab)(x,D, τ) +1
i
n∑j=1
[(∂ξjam1)(∂xjb)
](x,D, τ)− (rb)(x,D, τ) + r(x,D, τ) B
where a and b are the full symbol of A and B (actually the coecients greater than m1 − 1 andm1 − 1 are enough). So, we can write the formula in this case
A B = (ab)(x,D, τ) +1
i
n∑j=1
[(∂ξjam1)(∂xjb)
](x,D, τ) + C(x,D, τ)(B.1)
where C is of order at most m1 +m2 − 2.Let us now nally get to the general case, take B = bm2(x,D, τ)+bm2−1(x,D, τ)+s(x,D, τ) with
bm1(x,D, τ) homogeneous of order m2, bm2−1(x,D, τ) homogeneous of order m2 − 1 and s(x,D, τ)of order at most m2 − 2. Applying Formula (B.1) to B equal to bm2(x,D, τ) and bm2−1(x,D, τ), weget
A B = (abm2)(x,D, τ) +1
i
n∑j=1
[(∂ξjam1)(∂xjbm2)
](x,D, τ) + C1(x,D, τ)(B.2)
+(abm2−1)(x,D, τ) +1
i
n∑j=1
[(∂ξjam1)(∂xjbm2−1)
](x,D, τ) + C2(x,D, τ)(B.3)
where C1 is of order at most m2 − 2 and C2 m2 − 3.In particular, since (abm2)(x,D, τ) + (abm2−1)(x,D, τ) = (ab)(x,D, τ) + C3(x,D, τ) where C3 is
of order at most m1 +m2 − 2 and (∂ξjam1)(∂xjbm2−1) is of order at most m1 +m2 − 2, we have theequivalent of Formula (B.1) in the general case. This also proves Proposition 1.2.2.
Note that it means that
the symbol of order m1 +m2 is am1bm2 .
70
the symbol of order m1 +m2 − 1 is
am1bm2−1 + am1−1bm2−1 +1
i
n∑j=1
[(∂ξjam1)(∂xjbm2−1)
]This directly gives that [A,B] is of order at most m1 + m2 − 1 with principal symbol of orderm1 +m2 − 1
1
i
n∑j=1
[(∂ξjam1)(∂xjbm2)
]− 1
i
n∑j=1
[(∂xjam1)(∂ξjbm2)
]=
1
iam1 , bm2 .
B.2 Feuille 4
B.2.1 Exercice 1
1. We denote f(s) = e−s2dierentiate
df(ξ)
dξ=
d
dξ
∫Rde−isξe−s
2ds = −i
∫Rdse−isξe−s
2ds =
i
2
∫Rde−isξ∂s(e
−s2)ds = −ξ2
∫Rde−isξe−s
2ds = −ξ
2f(ξ).
So, this equation can be explicitely solved f(ξ) = f(0)e−ξ2
4 . It is still a Gaussian, but we haveto nd the normalization constant.
To compute f(0) =∫R e−s2ds, notice that
(∫R e−s2ds
)2=∫R2 e
−(x2+y2)dxdy =∫r>0 e
−r2(2πr)dr =
−π∫r>0 ∂r(e
−r2)dr = π.
So,∫R e−s2ds =
√π and f(ξ) =
√πe−
ξ2
4 .
2. In higher dimension, we compute∫Rne−ix·ξe−|x|
2=
∫Rx1
· · ·∫Rxn
e−ix1ξ1 · · · e−ixdξde−x21 · · · e−x2
n = f(ξ1) · · · f(ξn) = πn/2e−|ξ|2
4 .
3. Now, we want to compute gλ(ξ) where gλ = e−|x|2λ . By scaling, we have
gλ(ξ) =
∫Rne−ix·ξe−
|x|2λ dx = λn/2
∫Rne−i√λy·ξe−|y|
2dy = λd/2g1(
√λξ) = (πλ)n/2e−λ
|ξ|24 .
That is
e−|x|2λ (ξ) = (πλ)n/2e−λ
|ξ|24(B.4)
4. Now, we want to give a convolution formulation for Sλ. We have for u ∈ S(Rn), Sλu(ξ) =
e−|ξ|2λ u(ξ). So, using (1.9)
Sλu = F−1
(e−|ξ|2λ u
)= F−1
(e−|ξ|2λ
)∗ u = fλ ∗ u
with fλ = F−1
(e−|ξ|2λ
)= 1
(2π)n
e−|·|2λ = (πλ)n/2
(2π)n e−λ|x|2
4 =(λ4π
)n/2e−λ
|x|24 .
71
5. Assume u ∈ S(Rn) and v is a smooth solution of∂tv −∆v = 0v(0, x) = u(x).
with v(t) ∈ S(Rn) for t ≥ 0. Then, applying the Fourier transform in x gives∂tv(t, ξ) + |ξ|2v(t, ξ) = 0
v(0, ξ) = u(ξ).
Leading to v(t, ξ) = e−t|ξ|2u(ξ) and v(t) = e−t|Dx|
2u = S1/tu. Note that we recover the heat
Kernel v(t) = Kt ∗ u with Kt(x) =(
14πt
)n/2e−|x|24t
B.2.2 Exercice 2
Formally, we want to replace t by it in the previous formula and use analytic continuation. Moreprecisely
Take ϕ ∈ S. The function z 7→ fϕ(z) =
⟨e−|x|2z , ϕ
⟩S′,S
denes an holomorphic function on the
half plane C+ = z = a+ ib; a > 0, b ∈ R (use derivation under the integral). By formula B.4, it is
equal to
⟨(πt)d/2e−t
|·|24 , ϕ
⟩S′,S
on R+. But the function gϕ(z) =
⟨(πz)d/2e−z
|·|24 , ϕ
⟩S′,S
also dene
an analytic function (use the branch of the square root dened on C+). So, since fϕ and gϕ are twoholomorphic functions on C+ that are equal on R+, we have fϕ = gϕ on C+.
Moreover, gϕ can be extended by continuity to the set z = a+ ib; a ≥ 0, b ∈ R∗, simply by thesame formula. It is also clear that fϕ can also be extended by continuity. So, both functions areequal on that set, leading to the nal formula on the imaginary set iR∗.
e−|x|2it (ξ) = (πit)d/2e−it
|ξ|24 .
72
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