Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
8.4 Many measurements: methods of boundary control
8.4 Many measurements: methods of boundary control
We consider the initial value problem (8.0.1)–(8.0.3), where (8.4.1)
Au= −a,a0=1,a∈C∞() does not depend ont, anda>0 on. We assume zero initial conditionsu0=0,u1=0, which is natural in many appli- cations (say, in geophysics). We define the Riemannian distance
da(x,y)=inf
γ(x,y)
a−1/2dγ
where inf is over all smooth regular curvesγinwith endpointsx,y. Letγ1=∂ and−γ01,T be the local lateral Neumann-to-Dirichlet map corresponding to a part γ0⊂∂: suppg1⊂γ0×(0,T) andg0is given only onγ0×(0,T).
Theorem 8.4.1. The lateral Neumann-to-Dirichlet map−γ01,luniquely determines a in the domainγ0,T = {y∈:da(y, γ0)<T/2}.
This result belongs to Belishev [Be1], [Be2], [Be3]. We will reproduce a mod- ified scheme of his proof. It is more convenient to assume that we are given the local lateral Dirichlet-to-Neumann mapγ0,T :g0→g1.
Outline of proof.Let us assume that
(8.4.2) T =.
This is certainly true for smallT. By reconstructinga for smallT and then in- creasingT, we will cover the general case. For smallT one can introduce inT
the geodesic coordinatesx(η, ξ), η∈∂, ξ =dista(x, ∂).
LetWγ0,Tg0=u(,T;g0) whereu(;g0) be the solution to the hyperbolic problem (8.0.1), (8.0.2), u=g0 on∂×(0,T),suppg0⊂γ0×(0,T) on× {T}. We introduce the weighted spaceL2,a() by using the scalar product
(u,v)2,a()=
a−1uv
inL2(). In fact, it is the same space L2() with a different scalar product and (equivalent) norm. As observed in section 8.0, the operator Wγ0,T is linear and continuous from L2(∂×(0,T)) into L2,a(), so we can define the operator Cγ0,T =Wγ∗0,TWγ0,T ( where W∗ is the adjoint of W) from L2(γ0×(0,T) into itself. We have
(8.4.3) (Cγ0,Tg01,g02)2(∂×(0,T))=(Wγ0,Tg01,Wγ0,Tg02)2,a().
ByCg0we denote the restriction ofContog0supported inγ0.
Lemma 8.4.2. The operator Cγ0,T is a symmetric, positive linear operator in L2(γ0×(0,T)). It has zero kernel and is uniquely determined byγ0,2T.
PROOF. Assume thatCg0=0. Then (W g0,W g0)2,a()=0 and W g0=0. We will extend the solutionu=u(;g0) onto×Ras follows. Let us defineu(x,t)=
−u(x,2T−t) whenT ≤t <2T, andu =0 when 2T ≤t. Then the limits of u(x,t) ast →Tfromt<T andT <tare 0, and the limits of∂tu(x,t) are equal.
By using the definition of a generalized solution, we conclude thatu solves the same hyperbolic equation in×R. By assumption (8.4.2), the set\T is not empty. Apparently,u =0 on this set for allt ∈R. The Laplace transformU(x,s) ofu with respect tot satisfies the elliptic equations2U−aU =0 in×R, and it is zero on\T. By uniqueness of the continuation for elliptic equations (Section 3.3) we haveU=0 on. By inverting the Laplace transform we obtain u=0, and so its Dirichlet datag0are zero as well.
To show thatγ0,2T uniquely determinesCγ0,T, observe that (∂t2−∂s2)(u(,t;g01),u(,s;g02))2,a()
=
((u(,t;g01))u(,s;g02)−u(,t;g01)u(,s;g02))
=
γ0
((∂νu(,t;g01))u(s;g02))−u(,t;g01)∂νu(,;g02))
=
γ0
(g01(t)γ0,2TT g02(s)−γ0,2Tg01(t)g02(s))
is given fors<2T,t<2T when we are giveng01,g02. In addition,u(t;g01)=0 whent<0. By solving the Cauchy problem for the one-dimensional wave operator (∂t2−∂s2) we uniquely determine (u(,t;g01),u(,s;g02))2,a() whent <2T,s+ t <2T. Lettings=y=T, we obtain (Wγ0,Tg01,Wγ0,Tg02)2,a(), which uniquely determineCγ0,T by (8.4.3).
The proof is complete.
Exercise 8.4.3. Prove that for any harmonic functionv in,v∈C1(),v=0 on∂\γ0we have
(u(,T;g),v)2,a()=(g, •v)2(0×(0,T)) where•v=−1∗θv−θ∂νv, θ(t)=(T −t).
{Hint: integrate by parts in 0=
×(0,T)
(a−1∂t2u−u)θv}
This result is important for the proof of Theorem 8.4.1 because it enables to
“replace” unknown functionu(,T;g0) by known functionv.
Letγ be an open smooth subsurface ofγ0andWγ,Tthe operatorWconsidered only on functionsg0that are zero outsideγ ×(0,T).
Lemma 8.4.4. The range of Wγ,T is dense in L2,a(γ,2T).
8.4. Many measurements: methods of boundary control 245 PROOF. Let us assume the opposite. Then by the Hahn-Banach theorem, there is a functionu1∈L2() that is L2,a()-orthogonal to all solutionsu(,T;g0) with g0=0 outsideγ ×(0,T). Since all these solutions are zero outsideγ,2T, we can letu1be zero onto\γ,2T.
Let us consider the solutionvto the backward initial boundary value problem (∂t2−a)v =0 onQ,
v =0, ∂tv=u1on× {T}, v =0 on∂×(0,T). (8.4.4)
Whenu1∈ H˚(1)() andg0 ∈H˚(1)(×(0,T)), we havev,u ∈H(2)(Q). Multi- plying the equation∂t2−av=0 bya−1u and using integration by parts with respect totand Green’s formula with respect tox, we obtain
0=
Q
a−1(∂t2−a)vu
=
×{T}a−1u1u−
∂×(0,T)
(∂νvu−∂νuv)
=
×{T}a−1u1u−
γ×(0,T)
g0∂νv.
(8.4.5)
Some integrals over the boundary are zero due to the zero initial and boundary value conditions (8.0.2), (8.4.4) foruandv. So for all such regular solutions the last expression is zero. Approximatingu1 by ˚H(1)-functions in L2(), we can obtain this equality in the less regular case under consideration. By the choice of u1, we have
γ×(0,T)
g0∂νv=
×{T}a−1u1u =0
when u=u(,T;g0),g0∈ H˚(1)(∂×(0,T)),g0=0 outside γ×(0,T). Since the space of such gis dense in L2(γ ×(0,T)), we conclude thatv=0 onγ × (0,T). Therefore,v has zero Cauchy data onγ×(0,T). As above, we extend v onto×(T,2T) by lettingv(x,t)= −v(x,2T−t). Then using mollifying (described in the book of H¨ormander [H¨o2]) with respect tot and Theorem 8.1, we can assume thatv∈H(2)(×(0,T)).
Now we can apply Corollary 3.4.6 to conclude thatv=0,∂tv=0 onγ,2T × {T}, because any point of this surface can be reached by noncharacteristic defor- mations ofγ ×(0,2T). Sou1=∂tv=0, which contradicts the choice ofu1.
The proof is complete.
In the proof we will use the following known (see, e.g. [Be3]) formula for propagation of jump singularities of solutionsvto the problem (8.4.4)
(8.4.6) lim
τ→T−ξ∂νv(η, τ)(PT −Pξ)φ=α(η, ξ)φ(x(η, ξ),T), η∈γ0
providedφ∈C∞(),0< ξ <T, andαis a positive function determined only by a. HerePξ is the projector defined as the operator of multiplication by the function χ(γ0,2χ). On the other hand, from (8.4.5) we have
(Wγ0,Tg0, φ)2,a()=(g0, ∂νv)2(γ0×(0,T))
for allg0∈L2(γ0×(0,T)), so∂νv=Wγ,∗Tφ. Hence we have the relation (8.4.7) ∂νv(; (PT −Pξ)φ)=Wγ∗0,T(PT−Pξ)φ
A crucial observation is thatγ0,2Tuniquely determines the right side of (8.4.7).
Indeed, letg0j(;ξ) be a complete system of functions inL2(γ0×(T −ξ,T)) which is orthonormal in the following sense: (Cγ0,2Tg0j,g0k)2(γ0×(T −ξ,T))=δj k
(Kronecker delta). Such a system can be obtained from anyL2(γ0×(T −ξ,T))- complete system by using the Gram-Schmidt process. Lemma 8.4.2 guarantees that (Cγ0,T g01,g02)2 is a scalar product. Lemma 8.4.2 and Lemma 8.4.4 also imply thatWγ0,T g0j(;ξ) form an orthonormal basis inL2,a((γ0, χ)). We have
Wγ∗0,T(PT −Pξ)φ
=Wγ∗0,T ∞
j=1
(φ,Wγ0,Tg0j(;T))2,aWγ0,Tg0j(;T)
−(φ,Wγ0,Tg0j(;ξ))2,aWγ0,Tg0j(;ξ)
=∞
j=1
((φ,Wγ0,Tg0j(;T))2,aCγ0,Tg0j(;T)
−(φ,WTg0j(;ξ))2,aCγ0,Tg0j(;ξ))
Hence, due to Lemma 8.4.2 and Exercise 8.4.3, for any harmonic functionφ∈ C∞(),φ=0 on∂\γ0, the operatorγ0,2T uniquely determines WT∗(PT − Pξ)φ. Combining this observation with the relations (8.4.6), (8.4.7) we conclude thatγ0,2T uniquely determinesα(η, ξ)φ(x(η, ξ),T) for any suchφ.
We will complete the reconstruction in the normal neighborhoodN(T) ofγ0
by using appropriateφ. Here the normal neighborhood is defined as{x(η, ξ) :η∈ γ0,0< ξ <T/2}. Let K be any compact in ∪γ0. From the Runge property for harmonic functions (see the proof of Lemma 5.7.2) and interior Schauder estimates of Theorem 4.1 it follows that we can approximate the harmonic functions 1,x1, . . . ,xninC1(K) by harmonicC∞()-functionsφ0, . . . , φnwhich are zero on∂\γ0. Therefore we can chooseφ0, . . . , φn in such a way thatφ0> 12 and the mapx→(φ1(x), . . . , φn(x)) is one-to-one onK. We already found thatγ0,T
uniquely determinesφj/φ0(x(η, ξ)),j =1, . . . ,nwhenη∈γ0, ξ <T/2. Due to the choice ofφjit uniquely determinesx(η, ξ) for theseη, ξ. Since the mapx(η, ξ) is one-to-one we can uniquely find ξ =ξ(x),x∈N(T). Then by the eikonal equation
a−1(x)= |∇xξ|2,x∈N(T).
8.4. Many measurements: methods of boundary control 247 To show uniqueness in the remaining part ofγ0,T and for largerT we can use a smaller domain2⊂providedais known in\2.
Let2be a subdomain ofsuch thatγ2=∂2\∂is in,∂2∈C∞, and any geodesicx(η, ξ), η∈γ0, 0< ξ <T, intersects\2over a connected set. LetQ0 be∪γ0,T−|2t−T|× {t}overt ∈(0,T) and20=γ2×(0,T)∩Q0. We claim that ifais given on\2then the Dirichlet-to-Neumann operator2:g02 →∂νu2on γ20 is uniquely determined by the original Neumann-to-Dirichlet operator. Here u2is the solution to the following hyperbolic problem
∂t2u2−au2=0 in2×(0,T) u2=∂tu2=0 on2× {0}
u2=g02on∂2×(0,T),suppg02⊂γ20
Indeed, letube the solution of the above problem with2,g02, γ20replaced by ,g0, γ0×(0,T). The original Dirichlet-to-Neumann map uniquely determines
∂νu onγ0×(0,T) when g0 is given. Due to our assumptions about2 the set Q0can be covered by differentiable family of noncharacteristic surfaces with the boundaries onγ0×(0,T). Since the hyperbolic equation is known in\2× (0,T) Corollary 3.4.6 guarantees thatu is uniquely determined inQ0. Hence for any g02=u on∂2×(0,T) we are given∂νu onγ20. Now our claim follows from completeness of suchuinL2(γ20) guaranteed by Lemma 8.4.5.
Lemma 8.4.5. Any function g02∈ L2(γ2×(0,T)) withsuppg02⊂20 can be L2-approximated on20by functions u solving the equations(8.4.1)on×(0,T) with zero initial conditions on× {0}and with u=0on(∂\γ0)×(0,T).
PROOF. Let us assume the opposite. Then there is a non-zero functionψ∈L2(γ2× (0,T)),suppψ⊂20such that
(8.4.8)
γ2×(0,T)
ψu=0
for allu described in Lemma 8.4.5. Using completeness of smooth functions in L2we can assume that (extended)ψ∈C∞(Rn+1).
Letu∗be the solution of the following transmission problem
∂t2u−−au−=0 in−2 ×(0,T), ∂t2u+−au+=0 in+2 ×(0,T) u−=∂tu−=0 on−2 × {T},u+=∂tu+=0 on∂+2 × {T} u−=u+, ∂νu+−∂νu−=ψon∂+2 ×(0,T)
u−=0 on∂×(0,T).
(8.4.9)
Here+2 is a smooth subdomain of2with the closure insuch that suppψ⊂
∂+2,−2 =\+2 andu∗=u− on−2 ×(0,T),u∗=u+ on+2 ×(0,T). By subtractingψ fromu−2 and using Theorem 8.1 we obtain existence of a (weak) solution to the transmission problem with∂tku∗ ∈L2(2×(0,T)). By subtracting (1+a)∂t2u−,(1+a)∂t2u+from the both sides of the differential equations (8.4.9) we can consider these equations and the relations on∂+2 as an elliptic transmission
problem. The solution of the elliptic transmission problem is the sum of volume potentials over+2 ×(0,T), −2 ×(0,T) and of the single layer potentialUwith densityψover∂+2 ×(0,T). The known properties of potentials (for the Laplace equation) imply that for the sequence of domains+2k whose boundaries are im- ages of∂+2 under the maps(x;k)=x−k−1ν(x) (interior normal deformation) we have|∇U((;k),)|<g∈L2(∂+2 ×(0,T)) and the functions∇U((;k), ) have limits almost everywhere on∂+2 ×(0,T) whenk→ ∞. The similar prop- erty holds for outside limits. From theory of elliptic boundary value problems it follows thatu+2 ∈H(2)(+3 ×(0,T)),u−2 ∈ H(2)(−3 ×(0,T)) where+3, −3 are subdomains of+2, −2 such that their closures are inside+2, −2 ∪∂. From the differential equation (8.4.9) we have
0=
+3×(0,T)
(a−1∂t2u+−u+)u=
∂+3×(0,T)
(∂νuu+−u∂νu+) where we did integrate by parts two times, have used zero initial and final conditions (8.4.9) foru andu+ and the differential equation foru. Letting+3 =+2k and passing to the limit ask→ ∞we obtain that
(8.4.10)
∂+2×(0,T)
(∂νuu+−u∂νu+)=0 We can similarly consideru−to obtain
(8.4.11) −
γ0×(0,T)
u∂νu−−
∂−2×(0,T)
(∂νuu−−u∂νu−)=0
where we have used that∂−=∂∪∂+, thatu−=0 on∂and thatu=0 on∂\0. Adding the equalities (8.4.10) and (8.4.11) and using the transmission condition (8.4.9) we conclude that
−
γ0×(0,T)
u∂νu−−
∂+2×(0,T)
ψu =0
From (8.4.8) we conclude that the first integral is zero. Sinceu on γ0×(0,T) can be any smooth function we have∂νu−=0 on γ0×(0,T). Using also the boundary condition (8.4.9) we obtain thatu−has zero Cauchy data onγ0×(0,T).
As above from uniqueness in the lateral Cauchy problem we derive thatu−=0 onQ0∩(∂+2 ×(0,T)).
Now we will complete the proof by showing thatψ=0. Our first claim is that u−=0 on ∂+2 ×(T/2,T). Indeed, we have this equality on20. Due to the definition20∩ {T/2<t <T}is the intersection of∂+2 ×(T/2,T) and of the influence domain for our hyperbolic equation. Since ψ is supported in 20 we conclude that a solutionu−to the backward hyperbolic problem (8.4.9) is zero on
∂+2 ×(T/2,T)\20. Finally,u+=u− =0 on∂+2 ×(T/2,T) and therefore a solutionu+to the hyperbolic equation (8.4.9) with zero final conditions is zero on +2 ×(T/2,T). Therefore from the transmission condition (8.4.9) we derive that ψ=0 on∂+2 ×(T/2,T). To show thatψis zero when 0<t<T/2 we consider the hyperbolic equation (8.4.9) foru+2 in the intersection Q+ of +2 ×(0,T/2)