Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
5.3 Completeness of products of solutions of PDE
The property of completeness of products for harmonic function inRn,n ≥2, was observed by Calderon [C]. Apparently, it is not valid inR1(linear combinations of linear functions are not dense). Since it plays a fundamental role in the thoery of inverse problems, we discuss its connection to potential theory (uniqueness theorems for Riesz potentials).
Let be a bounded domain in Rn, n≥2. Let us consider functions u1,u2 that are harmonic near. We will show that span{u1u2}is dense in L1(), for n≥3. Assume that it is not so. Then by Hahn-Banach Theorem there is a nonzero measureàsupported insuch that
u1(y)u2(y)dà(y)=0 for all suchu1,u2. The functionsu1(y)= |x−y|2−n =u2(y) are harmonic nearwhenxis outside, so the Riesz potential
|x−y|αdà(y)=0 (α= −2) whenxis outside. It was proven using asymptotic behavior of this potential at infinity by M. Riesz [Ri] or using the Fourier transformation by Isakov [Is4], p. 79, that whenα=2k, α+n = 2k+2, the exterior Riesz potential determines a measureàin a unique way, so we haveà=0, which is a contradiction.
Now we will demonstrate the Calderon’s approach, which turned out to be fruitful for some equations with variable coefficients.
First we give some auxiliary results.
Lemma 5.3.1. Letξ(0)∈Rn\{0}, n≥2. Then there areζ(1), ζ(2)∈Cnsuch that ζ(1)ãζ(1)=0=ζ(2)ãζ(2), ζ(1)+ζ(2)=ξ(0).
(5.3.1)
If n≥3, then for any R >0there are suchζ(1),ζ(2)with the additional property R+ |ξ(0)|/2≤ |ζ(j)|, |Iζ(j)| ≤ |ξ(0)|/2+R.
(5.3.2)
PROOF. Let ζ(1)=ξ +iη. Then ζ(2)=ξ(0)−ξ−iη, where ξ, η are vectors inRn to be found. The relations (5.3.1) are equivalent to the equalities ξãξ = ηãη, ξãη=0,|ξ(0)|2=2ξãξ(0), ξ(0)ãη=0.
5.3. Completeness of products of solutions of PDE 139 Ifn=2, letξ(0)=te1, ande1,e2 be an orthonormal basis inR2. When the ξj are coordinates ofξ in this basis then all solutions to the above system of the equations forξ, ηareξ1=t/2, ξ2=0, η1=0, η2=t/2 or−t/2.
Ifn≥3, then we similarly have the solutions
ξ1 =t/2, ξ2=0, ξ3 =R, η1=0, η2 =(t2/4+R2)1/2, η3=0.
Observing that|Iζ(j)| = |η| =(|ξ(0)|2/4+R2)1/2and using the inequalitya2+ b2≤(a+b)2for nonnegativea,b, we complete the proof.
Now we are ready to demonstrate Calderon’s proof of completeness of products of harmonic functions.
Since linear combinations of exponential functions exp(iξ(0)ãx), ξ(0)∈ Rn\{0}are dense (in L2()), it suffices to find two harmonic functions whose product is this exponential function. We let u1(x)=exp(iζ(1)ãx), u2(x)= exp(iζ(2)ãx), where the ζ(j) satisfy conditions (5.3.1), which guarantee that u1,u2are harmonic. Suchζ(j) exist by Lemma 5.3.1. By multiplying the expo- nential functions and using condition (5.3.1) again, we obtain exp(iξ(0)ãx).
This approach was transferred by Sylvester and Uhlmann [SyU2] onto solutions of the multidimensional Schr¨odinger equation (5.2.1). Motivated by the Calderon’s idea and by geometrical optics, they suggested using almost exponential solutions to this equation,
uj(x)=exp(iζ(j)ãx)(1+wj(x)), (5.3.3)
with the property thatwj goes to zero (in L2()) when R(from Lemma 5.3.1) goes to infinity. We will extend their method to the more general equations
(Pj(−i∂)+cj)uj =0 (5.3.4)
(which include parabolic and hyperbolic equations with constant leading coeffi- cients) by using the following auxiliary result.
Lemma 5.3.2. Let P be a linear partial differential operator of order m inRn with constant coefficients. Then there is a bounded linear operator E from L2() into itself such that
P(−i∂)E f = f for all f ∈L2() (5.3.5)
and
Q E f2()≤Csup( ˜Q(ξ)/P˜(ξ))f2(), (5.3.6)
where C depends only on m, n,diam, andsupis overξ ∈Rn. Here ˜P(ξ)=
|α|≤m|∂ξαP(ξ)|21/2 .
Actually, this result was obtained by Ehrenpreis and Malgrange in the 1950s.
For complete proofs we refer to the book of H¨ormander [H¨o2] and to the paper of Isakov [Is7].
Theorem 5.3.3. Let0 be an open nonvoid subset ofRn. Suppose that for any ξ(0)∈0and for any number R there are solutionsζ(j)to the algebraic equations
ζ(1)+ζ(2)=ξ(0), Pj(ζ(j))=0with|ζ(j)|>R.
(5.3.7)
and there is a positive number C such that for theseζ(j)we have
|ζ(j)| ≤CP˜j(ξ+ζ(j))for allξ ∈Rn. (5.3.8)
If f ∈L∞()and
f u1u2=0 (5.3.9)
for all L2-solutions ujto the equations(5.3.4)near, then f =0.
This theorem claims that linear combinations of products of solutions of these equations are dense inL1(). It easily follows from the following construction of almost exponential solutions to the equations (5.3.4).
Theorem 5.3.4. Suppose that conditions(5.3.7),(5.3.8)are satisfied.
Then for anyξ(0)∈0there are solutions to equations(5.3.4)nearof the form(5.3.3), wherewj2()≤C/|ζ(j)|and C depends only oncj∞()and ondiam.
PROOF. From Leibniz’s formula we have P(−i∂)(uv)=
α
P(α)(−i∂)u(−i∂)αv/α!.
Letting u =exp(iζãx), v=1+wj and using the relations P(α)(−i∂)u = P(α)(ζ) exp(iζ ãx) as well as the equality Pj(ζ(j))=0, we conclude thatuj in (5.3.3) solves equation (5.3.4) if and only if
Pj(−i∂+ζ(j))wj =
α
P(α)j (ζ(j))(−i∂)αwj/α!= −cj(1+wj).
(5.3.10)
LetEbe the operator of Lemma 5.3.2 forP(−i∂)=Pj(−i∂+ζ(j)). Then any solutionwjto the equation
wj = −E(cj(1+wj)) inL2() (5.3.11)
is a solution to (5.3.4). From condition (5.3.7) and from the estimate (5.3.6) with Q(∂)= |ζ(j)| we have E f2()≤C|ζ(j)|−1f2(). So using condi- tions (5.3.7), (5.3.8) we can choose|ζ(j)|so large that the operator from the right side of equation (5.3.11) is a contraction of the ball {w2()≤2Ccj∞() (measn)1/2/|ζ(j)|}inL2() into itself. By the Banach contraction theorem there is a solutionwjto the equation (5.3.11) in this ball. This proves Theorem 5.3.4.
PROOF OFTHEOREM5.3.3. Letξ(0)∈0and letu1,u2be solutions constructed in Theorem 5.3.4. Thenζ(1)+ζ(2)=ξ(0), and functionsu1u2(x)=exp(iξ(0)ã x)(1+w1+w2+w1w2) converge to exp(iξ(0)ãx) inL1() as the|ζ(j)|go to
5.3. Completeness of products of solutions of PDE 141 infinity. So the Fourier transform of f (extended as zero outside) is zero on0. Since f is compactly supported, its Fourier transform is analytic onRn. Therefore it is zero everywhere, and so is f. This completes the proof.
Corollary 5.3.5. Let n≥3. If(5.3.9)is valid for all solutions ujto the equations
−uj+cjuj =0, j =1,2, near, then f =0.
PROOF. To prove this result we check the conditions of Theorem 5.3.3 forPj(ζ)= ζ12+ ã ã ã +ζn2. Letξ(0)∈Rn. Due to rotational invariancy we may assumeξ(0)= (ξ1(0),0, . . . ,0). The vectors
ζ(1)=(ξ1(0)/2,i(ξ12(0)/4+R2)1/2,R,0, . . . ,0), ζ(2)=(ξ1(0)/2,−i(ξ12(0)/4+R2)1/2,−R,0, . . . ,0)
are solutions to the equationζ ãζ =0 with the absolute values greater thanR, so condition (5.3.7) is satisfied. To check condition (5.3.8) we observe that
P˜2(ξ+ζ)≥ |2ζ1+2ξ1|2+ ã ã ã + |2ζn+2ξn|2+12
≥4
|Iζ1|2+ ã ã ã + |Iζn|2 ≥ |ζ|2,
provided thatζãζ =0, because then|Rζ| = |Iζ|. This completes the proof.
In the paper [SyU2] the authors used the more standard fundamental solution E(the Faddeev Green’s function) of the operator+2iζ ã ∇, and they obtained estimates (5.3.6) in the Sobolev spaces Hδm(Rn), which are constructed from the weightedL2;δ(Rn)-spaces with the normf(x)(1+ |x|2)δ/22(Rn). They showed also thatEis invertible in these weighted spaces. H¨ahner [Ha] observed that in the Sylvester-Uhlmann scheme it suffices to use periodic solutions of the Schr¨odinger equation, because c is compactly supported. To explain his idea we let R = (−R,R)ì ã ã ã ì(−R,R) be the cube inRnand we consider functions f(x) which are 2R-periodic with respect to all variablesx1, . . . ,xn. H¨ahner proved that there is a periodic fundamental solution Epof the operator+2iζ ã ∇satisfying the bound
Epf2(R)≤ R/(π|ζ|)f2(R).
The bound on the norm of this operator is explicit and one can use it instead ofE in the proof of Theorem 5.3.4 obtaining explicit bounds on constantsC. A proof of this estimate is straightforward if one uses Fourier series representations for f. Exercise 5.3.6. Let P = −−2ζ ã ∇,ζ ãζ =0, and E be its regular funda- mental solution. Prove the estimateEw(1)()≤Cw(0)() for all functions w ∈L2() that are zero outside withC depending only on diamand the dimensionnof the space. Using this result prove that there are solutionsujto the Schr¨odinger equation (−+cj)uj =0 f the form (5.3.3) with
|ζ(j)|wj2()+ wj(1)()≤C.
{Hint: make use of Lemma 5.3.2 and of the proof of Corollary 5.3.5.}
Exercise 5.3.7. Let P= −−2ζã ∇,ζ ãζ =λ2, λ∈R+, andE be its regular fundamental solution. Prove the estimateEw(0)()≤C/λw(0)() for allw from Exercise 5.3.6 withCdepending on the same parameters as in that exercise.
Now we will demonstrate an approach of Bukhgeim and Uhlmann [BuU] based on Carleman estimates which leads to uniqueness of the coefficientc(and hence of scalar principal coefficient a) from some partial boundary data. We will use notation+, −from section 5.2.
Lemma 5.3.8. Let e be a unit vector inRnandϕ(x)= −eãx.
Then there is a constant C depending only on,c∞()such that τ2/C
|u|2e2τϕ+2τ
+
eãν|∂ν|2e2τϕ
≤
| −u+cu|2e2τϕ−2τ
−
eãν|∂νu|2e2τϕ for all functions u∈H(2)()with u=0on∂.
A proof can be obtained following advices to solve Exercise 3.4.5. Lemma 5.3.8 and the standard scheme from functional analysis (similar to proofs of Lax- Milgram theorem) lead to the following
Corollary 5.3.9. There is an operator Efrom L2()into H(2)()and a constant C depending only onandc∞()such that v =Ef solves the boundary value problem
−v+cv= f in, v=0on−
and
|v|2e2τϕ≤C/τ2
|f|2e2τϕ.
Corollary 5.3.9 is an analogue of Lemma 5.3.2 for a bounded domain and E can be viewed as a special fundamental solution satisfying a partial boundary condition. Using this corollary in the proof of Theorem 5.3.3 instead of Lemma 5.3.2 one can construct almost exponential solutions (5.3.3) to the Schr¨odinger equation inwhich satisfy partial boundary conditions.
Now we give an outline of a proof of Theorem 5.2.2. Letξ ∈Rnbe orthogonal toe,|ξ|< τ, andη∈Rnbe orthogonal to botheandξwith|η|2=τ2− |ξ|2. Let us defineζ(1)= −τe−i(ξ +η), ζ(2)=τe−i(ξ−η). By Theorem 5.3.4 there is an almost exponential solutionu∗1(x)=eζ(1)ãx(1+w1∗(x)) withw∗12()≤C/τ. Using Corollary 5.3.9 instead of theorem 5.3.3 and repeating the proof of Theorem 5.3.4 one can construct a solutionu2(x)=eζ(2)ãx(1+w2(x)) withw2=0 on− andw22()≤C/τ withw22()≤C/τ. Letu1solve the Dirichlet problem