Hardy type unique continuation properties for abstract Schrödinger equations and applications
Veli Shakhmurov
B1,21Department of Mechanical Engineering, Okan University, Akfirat, Tuzla 34959 Istanbul, Turkey
2Institute of Mathematics and Mechanics, Azerbaijan National Academy of Sciences, AZ1141, Baku, F. Agaev 9
Received 22 November 2019, appeared 21 December 2019 Communicated by Maria Alessandra Ragusa
Abstract. In this paper, Hardy’s uncertainty principle and unique continuation proper- ties of Schrödinger equations with operator potentials in Hilbert space-valuedL2classes are obtained. Since the Hilbert space Hand linear operators are arbitrary, by choosing the appropriate spaces and operators we obtain numerous classes of Schrödinger type equations and its finite and infinite many systems which occur in a wide variety of physical systems.
Keywords:Schrödinger equations, positive operators, groups of operators, unique con- tinuation, Hardy’s uncertainty principle
2010 Mathematics Subject Classification: 35Q41, 35K15, 47B25, 47Dxx, 46E40.
1 Introduction
Here, the unique continuation properties of the following abstract Schrödinger equation i∂tu+∆u+A(x)u+V(x,t)u=0, x∈Rn, t∈ [0,T], (1.1) are studied, where A = A(x)is a linear and V(x,t) is a given potential operator functions in a Hilbert space H; ∆ denotes the Laplace operator inRn and u = u(x,t) is the H-valued unknown function. This linear result was then applied to show that two regular solutionsu1, u2of non-linear abstract Schrödinger equations
i∂tu+∆u+A((x))u=F(u, ¯u), x∈Rn, t ∈[0,T] (1.2) for general non-linearities F must agree in Rn×[0,T], when u1−u2 and its gradient decay faster than any quadratic exponential at times 0 andT.
Hardy’s uncertainty principle and unique continuation properties for Schrödinger equa- tions studied e.g in [4–7] and the references therein. Abstract differential equations studied e.g. in [2,12–15,17–19,23,25]. However, there seems to be no such abstract setting for nonlin- ear Schrödinger equations except the local existence of weak solution (cf. [15]). In contrast to
BEmail: veli.sahmurov@okan.edu.tr
these results we will study the unique continuation properties of abstract Schrödinger equa- tions with the operator potentials. Since the Hilbert space H is arbitrary and A is a possible linear operator, by choosing H and A we can obtain numerous classes of Schrödinger type equations and its systems which occur in the different processes. Our main goal is to obtain sufficient conditions on a solutionu, the operator A, potentialV and the behavior of the so- lution at two different timest0andt1 which guarantee thatu(x,t)≡0 for x ∈ Rn,t ∈ [0,T]. If we choose Hto be a concrete Hilbert space, for example H = L2(Ω), A = L, where Ωis a domain inRm with sufficiently smooth boundary and Lis a regular elliptic operator then, we obtain the unique continuation properties of the anisotropic Schrödinger equation
∂tu=i(∆u+Lu) +V(x,t)u, x∈Rn, y∈ Ω, t ∈[0,T]. (1.3) Moreover, let we choose H= L2(0, 1)and Ato be differential operator with Wentzell–Robin boundary condition defined by
D(A) =u∈W2,2(0, 1), Au(j) =0, j=0, 1 , (1.4) A(x)u=a(x,y)u(2)+b(x,y)u(1),
where a, bare sufficiently smooth functions on Rn×(0, 1) andV(x,t) is a integral operator so that
V(x,t)u=
Z 1
0 K(x,y,t)u(x,y,t)dy,
where,K=K(x,τ,t)is a complex valued bounded function. From our general results we ob- tain the unique continuation properties of the Wentzell–Robin type boundary value problem (BVP) for the following Schrödinger equation
∂tu=i
∆u+a∂2u
∂y2 +b∂u
∂y
+
Z 1
0 K(x,y,t)u(x,y,t)dy, x∈Rn, y∈(0, 1), t∈[0,T],
(1.5)
a∂2yu(x,j,t) +b∂yu(x,j,t) =0, j=0, 1. (1.6) Note that, the regularity properties of Wentzell–Robin type BVP for elliptic equations were studied e.g. in[10, 11]and the references therein. Moreover, if putH=l2 and chooseAto be a infinite matrix
amj
,m,j=1, 2, . . . ,∞, then we derive the unique continuation properties of the following system of Schrödinger equation
∂tum =i
"
∆um+
∑
∞ j=1amj(x) +bmj(x,t)uj
#
, x∈Rn, t ∈(0,T), (1.7) whereamj are continuous andbmj are bounded functions.
Let E be a Banach space. Lp(Ω;E) denotes the space of strongly measurable E-valued functions that are defined on the measurable subsetΩ⊂Rnwith the norm
kfkLp =kfkLp(Ω;E) = Z
Ωkf(x)kpEdx 1p
, 1≤ p<∞. Let Hbe a Hilbert space and
kuk= kukH = (u,u)
1 2
H = (u,u)12 foru∈ H.
For p=2 andE= H, Lp(Ω;E)becomes a H-valued function space with inner product:
(f,g)L2(Ω;H) =
Z
Ω(f(x),g(x))Hdx, f,g ∈L2(Ω;H).
Here,Ws,2(Rn;H), −∞< s <∞ denotes the H-valued Sobolev space of orders which is defined as:
Ws,2=Ws,2(Rn;H) = (I−∆)−2s L2(Rn;H) with the norm
kukWs,2 =(I−∆)2s u
L2(Rn;H)< ∞.
It clear that W0,2(Rn;E) = L2(Rn;H). Let H0 and H be two Hilbert spaces and H0 is con- tinuously and densely embedded into H. LetWs,2(Rn;H0,H)denote the Sobolev–Lions type space, i.e.,
Ws,2(Rn;H0,H) =nu∈Ws,2(Rn;H)∩L2(Rn;H0),
kukWs,2(Rn;H0,H)=kukL2(Rn;H0)+kukWs,2(Rn;H) <∞o. LetC(Ω;E)denote the space ofE-valued uniformly bounded continuous functions onΩ with norm
kukC(Ω;E) =sup
x∈Ω
ku(x)kE.
Cm(Ω;E)will denote the spaces ofE-valued uniformly bounded strongly continuous and m-times continuously differentiable functions onΩwith norm
kukCm(Ω;E)= max
0≤|α|≤msup
x∈Ω
kDαu(x)kE.
Here,Or = {x∈Rn, |x|< r} forr > 0. Let N denote the set of all natural numbers, C denote the set of all complex numbers. Let E1 and E2 be two Banach spaces. B(E1,E2) will denote the space of all bounded linear operators from E1 to E2. For E1 = E2 = E it will be denoted by B(E). By (E1,E2)θ,p, 0 < θ < 1, 1 ≤ p ≤ ∞ we will denote the interpolation spaces obtained from {E1,E2}by the K-method [24, §1.3.2]. Here, S = S(Rn;E) denotes the E-valued Schwartz class, i.e. the space of E-valued rapidly decreasing smooth functions on Rn, equipped with its usual topology generated by seminorms. S(Rn;C)will be denoted by justS. LetS0(Rn;E)denote the space of all continuous linear operators,L :S →E, equipped with topology of bounded convergence.
LetA= A(x),x∈Rnbe closed linear operator inEwith independent onx∈Rndomain D(A)that is dense onE. The Fourier transformation of A(x), i.e. ˆA= FA = Aˆ(ξ)is a linear operator defined as
Aˆ(ξ)u(ϕ) =A(x)u(ϕˆ) foru∈S0(Rn;E), ϕ∈ S(Rn). (For details see e.g. [1, Section 3]).
For linear operators AandB,[A,B]-denotes a commutator operator, i.e.
[A,B] =AB−BA.
These kind of operators are of fundamental importance in real analysis, potential theory and in the study of elliptic and parabolic differential equations (see e.g. [9,16,22] and refer- ences therein).
Sometimes we use one and the same symbol C without distinction in order to denote positive constants which may differ from each other even in a single context. When we want to specify the dependence of such a constant on a parameter, sayα, we write Cα.
2 Main results
Let A = A(x), x ∈ Rn be closed linear operator in a Hilbert space H with independent on x∈RndomainD(A)that is dense on H. Let
H(A) =nu∈D(A), kukH(A) =kAukH+kukH <∞o, X= L2(Rn;H), X(A) = L2(Rn;H(A)), X(A) =L2(Rn;H(A)) X∞(A) =L∞(Rn;H(A)), Ys=Ws,2(Rn;H), Ys(A) =Ws,2(Rn;H(A)),
B= L∞(Rn;B(H)) and µ(t) =αt+β(1−t).
Definition 2.1. A functionu∈ L∞(0,T;X(A))is called a local weak solution to (1.1) on(0,T) if u satisfies (1.1). In particular, if (0,T) coincides with R, then u is called a global weak solution to (1.1). If the solution of (1.1) belongs to C [0,T];X(A)∩Y2
, then it is called a strong solution.
Our main result in this paper is the following.
Theorem 2.2. Assume that the following condition are satisfied:
(1) A = A(x) and ∂x∂A
k are symmetric operators in a Hilbert space H with independent on x ∈ Rn domain D ∂x∂A
k
= D(A) that is dense on H. Moreover, (A(x)u,u) ≥ 0 and (A(x)u,u) ∈ L2(Rn) for u∈D(A);
(2) n
k
∑
=1xk
A∂f
∂xk − ∂A
∂xk f
,f
X
≥0 for f ∈ L∞
0,T;Y1(A)
;
(3) A(x)A−1(x0)∈ L1(Rn;B(H))for some x0∈Rnand V(x,t)∈B(H)for(x,t)∈Rn×[0, 1]; (4) either, V(x,t) =V1(x) +V2(x,t), where V1(x)∈ B(H)for x ∈Rn and
M1= sup
x∈Rn
kV1(x)kB(H) <∞, sup
t∈[0,1]
e|x|2µ−2(t)V2(·,t) B <∞ or
rlim→∞kVkL1(0,1;L∞(Rn/Or);B(H))=0;
(5) u∈C([0, 1];X(A))is a solution of the equation(1.1)and
eβ−2|x|2u(·, 0)
X< ∞, eα−2|x|2u(·, 1)
X <∞.
where
α, β>0, αβ <2.
Then u(x,t)≡0.
As a result of Theorem2.2 we get the following Hardy’s uncertainty principle result for the non linear equation (1.2).
Theorem 2.3. Suppose that the assumptions (1)–(2) of Theorem 2.2 are satisfied. Let u1,u2 ∈ C [0, 1];Yk(A), k∈Z+be strong solutions of the equation(1.2)with k> n2. Moreover, assume:
(1) F∈ Ck C2,C
and F(0) =∂uF(0) =∂u¯F(0) =0.There areα,β>0withαβ<2such that e−β−2|x|2(u1(·, 0)−u2(·, 0))∈X, e−α−2|x|2(u1(·, 1)−u2(·, 1))∈ X
(2) there exists a constant B0>0such that
kF(u, ¯u)kH ≤ B0kuk(H(A),H)
1p,p
for all u∈ (H(A),H)1 p,p. Then u1≡ u2.
One of the results we get is the following one.
Theorem 2.4. Assume that the all conditions of Theorem 2.2 are satisfied. Suppose ∆+A+V1 generates a bounded continuous group. Let u ∈ C([0, 1];X(A)) be a solution of (1.1). Then
e|x|2µ−2(t)u(·,t)
1 µ(t)
X is logarithmically convex in[0, 1]and there is N= N(α,β)such that
e|x|2µ−2(t)u(·,t)
1 µ(t)
X ≤eN(M1+M2+M21+M22)
eβ−2|x|2u(·, 0)
β(1−t)µ(t) X
eα−2|x|2u(·, 1)
αtµ(t)
X ,
when
M2 =e2B(V2) sup
t∈[0,1]
e|x|2µ−2(t)V2(·,t)
B, B(V2) = sup
t∈[0,1]
kReV2(·,t)kB. Moreover,
q
t(1−t)
e|x|2µ−2(t)∇u
L2(Rn×[0,1];H)
≤eN(M1+M2+M21+M22)h
eβ−2|x|2u(·, 0)
X+eα−2|x|2u(·, 1)
X
i .
Consider the Cauchy problem for abstract parabolic equations with variable operator co- efficients
∂tu=∆u+A(x)u+V(x,t)u, (2.1) u(x, 0) = f(x), x ∈Rn, t∈[0, 1],
where A(x)is a linear andV(x,t)is the given potential operator functions in H. By employ- ing Theorem2.2we obtain the following result for the abstract parabolic equation (2.1):
Theorem 2.5. Assume the assumptions (1)–(3) of Theorem 2.2 are satisfied. Suppose that u ∈ L∞(0, 1;X(A))∩L2 0, 1;Y1
is a solution of (2.1)and kfkX <∞,
eδ−2|x|2u(·, 1) X <∞ for some δ<1. Then, f(x)≡0for x∈Rn.
First of all, we generalize the result of G. H. Hardy (see e.g. [20, p. 131]) about uncertainty principle for Fourier transform:
Lemma 2.6. Let f (x)be H-valued function for x∈Rnand kf(x)k=O
e−
|x|2 β2
,
fˆ(ξ) =O
e−4
|ξ|2 α2
, x,ξ ∈Rn forαβ<4.
Then f(x)≡0. Also, ifαβ=4, thenkf(x)kis a constant multiple of e−
|x|2 β2
.
Proof. Indeed, by employing Phragmén–Lindelöf theorem to the classes of Hilbert-valued an- alytic functions and by reasoning as in[8]we obtain the assertion.
Consider the Cauchy problem for free abstract Schrödinger equation
i∂tu+∆u+Au=0, x∈Rn, t∈[0, 1], (2.2) u(x, 0) = f(x),
where A = A(x) is a linear operator in a Hilbert space H with independent on x domain D(A).
The above result can be rewritten for solution of the (2.2) onRn×(0,∞). Indeed, assume ku(x, 0)k=O
e−
|x|2 β2
, ku(x,T)k=O
e−|x|
2 α2
forαβ<4T.
Then u(x,t) ≡ 0. Also, if αβ = 4T, then u has as a initial data a constant multiple of e−
1
β2+4Ti |x|2
.
Lemma 2.7. Assume that A is a symmetric operator in H with independent on x ∈ Rn domain D(A)that is dense on H. Moreover, A(x)A−1(x0) ∈ L1(Rn;B(H))for some x0 ∈ Rn. Then for
f ∈Ws,2(Rn;H),s≥0there is a generalized solution of (2.2)expressing as u(x,t) = F−1h
eiAˆξtfˆ(ξ)i, Aˆξ = Aˆ(ξ)− |ξ|2, (2.3) where F−1is the inverse Fourier transform and Aˆ(ξ)denotes the Fourier transform of A(x).
Proof. By applying the Fourier transform to the problem (2.2) we get
i∂tuˆ(ξ,t) +Aˆξuˆ(ξ,t) =0, x ∈Rn, t∈[0, 1], (2.4)
ˆ
u(ξ, 0) = fˆ(ξ), ξ ∈ Rn. It is clear to see that the solution of (2.4) can be exspressed as
ˆ
u(ξ,t) =eiAˆξtfˆ(ξ). Hence, we obtain (2.3)
3 Estimates for solutions
We need the following lemmas for proving the main results. Consider the abstract Schrödinger equation
∂tu= (a+ib) [∆u+Au+Vu+F(x,t)], x∈Rn, t∈ [0, 1], (3.1) where a, bare real numbers, A = A(x)is a linear operator, V = V(x,t)is a given potential operator function in Hand F(x,t)is a given H-valued function.
Condition 3.1. Assume that:
(1) A = A(x) is a symmetric operator in Hilbert space H with independent on x ∈ Rn domain D(A) that is dense on H; moreover, (A(x)u,u) ≥ 0 and (A(x)u,u) ∈ L2(Rn) foru∈ D(A);
(2) ∂A∂x
k are symmetric operators in H with independent onx ∈ Rn domain D ∂x∂A
k
= D(A). Moreover,
∑
n k=1xk
A∂f
∂xk − ∂A
∂xk f
,f
X
≥0, for f ∈ L∞
0,T;Y1(A); (3.2)
(3) a>0,b∈R;V =V(x,t)∈B(H). Let
|∇υ|2H =
∑
n k=1
∂υ
∂xk
2
forυ∈W1,2(Rn;H).
Lemma 3.2. Assume that the Condition 3.1 holds. Then the solution u of (3.1) belonging to L∞(0, 1;X(A))∩L2 0, 1;Y1
satisfies the following estimate eMT
eφ(·,T)u(·,T)
X≤eγ|x|2u(·, 0)
X+κ eφ(t)F
L1(0,T;X)+ak(Au,u)kX, (3.3) where
φ(x,t) = γa|x|2
a+4γ(a2+b2)t, MT = kaReV−bImVkB, κ =pa2+b2, γ≥0.
Proof. Letυ=eϕu, whereϕis a real-valued function to be chosen later. The functionυverifies
∂tυ=Sυ+Kυ+ (a+ib)eϕF, (x,t)∈Rn×[0, 1],
whereS,Kare symmetric and skew-symmetric operators respectively given by S= aA1−ibB1+ϕt+aReV−bImV, K=ibA1−aB1+i(bReυ+aImυ), here
A1 =∆+A(x) +|∇ϕ|2, B1 =2∇ϕ.∇+∆ϕ.
By differentiating the inner product in X, we get
∂tkυk2X =2 Re(Sυ,υ)X+2 Re(Kυ,υ)X
+2 Re((a+ib)eυF,υ)X+2 Re(a+ib) (Vυ,υ)X, t ≥0. (3.4)
A formal integration by parts gives that Re(Sυ,υ)X=−a
Z
Rn|∇υ|2Hdx+
Z
Rn
a|∇ϕ|2+ϕt
kυk2dx+a Z
Rn(Aυ,υ)dx, Re(Kυ,υ)X =−aγ
Z
Rn
h
(2∇ϕ.∇υ,υ) +∆ϕkυk2idx. (3.5) It is clear that
Re(a+ib) (Vυ,υ)X= a Z
Rn
(ReVυ,υ)dx−b Z
Rn
(ImVυ,υ)dx,
Re((a+ib)eϕF,υ)X =aRe Z
Rn
(eϕF,υ)dx−bIm Z
Rn
(eϕF,υ)dx
=aeϕRe(F,υ)X−beϕIm(F,υ)X.
Then by using the Cauchy–Schwarz inequality, by assumptions (2), (3), in view of (3.4) and (3.5) we obtain
∂tkυk2X ≤2kaReV−bImVkBkυk2X+2κkeϕF(t,·)kXkυkX+ak(Aυ,υ)kX, wherea,band ϕare such that
a+ b
2
a
|∇ϕ|2+ϕt ≤0 inRn++1. (3.6) The remaining part of the proof is obtained by reasoning as in [7, Lemma 1] .
When ϕ(x,t) =q(t)ψ(x), it suffices that
a+b
2
a
q2(t)|∇ψ|2+q0(t)ψ(x)≤0. (3.7) If we putψ(x) =|x|2then (3.6) holds, when
q0(t) =−4
a+ b
2
a
q2(t), q(0) =γ, γ≥0. (3.8) Let
ψr(x) =
(|x|2, |x|<r,
∞, |x|>r.
Regularizeψrwith a radial mollifierθρ and set
ϕρ,r(x,t) =q(t)θρ∗ψr(x), υρ,r(x,t) =eϕρ,ru, whereq(t) =γa
a+4γ a2+b2 t−1
is the solution to (3.7). Because the right hand side of (3.5) only involves the first derivatives of ϕ, ψris Lipschitz and bounded at infinity,
θρ∗ψr(x)≤θρ∗ |x|2 =C(n)ρ2
and (3.6) holds uniformly in ρ and r, when ϕ is replaced by ϕρ,r. Hence, it follows that the estimate
eMT
eφ(T)u(T)
X≤ MT
eγ|x|2u(0)
X+pa2+b2keϕρ,rFkL1(0,T;X)
holds uniformly in ρ and r. The assertion is obtained after letting ρ tend to zero and r to infinity.
Remark 3.3. It should be noted that if H=C,A=0 andV(x,t)is a complex valued function, then the abstract condition (3.3) can be replaced by
MT =kaReV−bImVkL1(0,T;L∞(Rn)) <∞.
Let
Q(t) = (f,f)X, D(t) = (S f,f)X, N(t) =D(t)Q−1(t), ∂tS=St. In a similar way as in [7, Lemma 2] we have the following result.
Lemma 3.4. Assume that S = S(t)is a symmetric, K = K(t)is a skew-symmetric operators in H, G(x,t)is a positive and f(x,t)is an X-valued reasonable function. Then,
Q00(t) =2∂tRe(∂tf−S f −K f, f)X+2(Stf + [S,K]f, f)X
+k∂tf−S f +K fk2X− k∂tf −S f −K fk2X (3.9) and
∂tN(t)≥ Q−1(t)
(Stf + [S,K]f,f)X− 1
2k∂tf−S f −K fk2X
.
Moreover, if
k∂tf−S f −K fkH ≤M1kfkH+G(x,t), St+ [S,K]≥ −M0 for x ∈Rn, t∈[0, 1]and
M2= sup
t∈[0,1]
kG(.,t)kL2(Rn)(kf(·,t)kX)−1 <∞.
Then Q(t)is logarithmically convex in[0, 1]and there is a constant M such that Q(t)≤ eM(M0+M1+M2+M21+M22)Q1−t(0)Qt(1), 0≤t ≤1.
Lemma 3.5. Assume that the Condition3.1holds. Moreover, suppose sup
t∈[0,1]
kV(·,t)kB ≤ M1,
eγ|x|2u(·, 0)
X<∞,
eγ|x|2u(·, 1)
X< ∞, M2= sup
t∈[0,1]
eγ|x|2F(·,t) X
kukX <∞.
Then, for solution u ∈ L∞(0, 1;X(A))∩ L2 0, 1;Y1
of (3.1), eγ|x|2u(·,t) is logarithmically convex in[0, 1]and there is a constant N such that
eγ|x|2u(·,t)
X ≤eN M(a,b)
eγ|x|2u(·, 0)
1−t X
eγ|x|2u(·, 1)
t
X, (3.10)
where
M(a,b) =κ2 γM12+M22
+κ(M1+M2) when0≤t ≤1.
Proof. Let f =eγϕu, whereϕis a real-valued function to be chosen. The function f(x)verifies
∂tf = S f +K f + (a+ib) (V f+eγϕF) inRn×[0, 1], (3.11) whereS,Kare symmetric and skew-symmetric operator, respectively given by
S= aA1−ibγB1+ϕt+aReV−bImV, K=ibA1−aγB1+i(bReυ+aImυ), where
A1=∆+A(x) +γ2|∇ϕ|2, B1=2∇ϕ.∇+∆ϕ.
A calculation shows that,
St+ [S,K] =γ∂2tϕ+2γ2a∇ϕ.∇ϕt−2ibγ(2∇ϕt.∇+∆ϕt)
−γκ24∇. D2ϕ∇−4γ2D2ϕ∇ϕ+∆2ϕ
+2[A(x)∇ϕ.∇ − ∇ϕ.∇A]. (3.12) If we put ϕ= |x|2, then (3.12) reduces to the following
St+ [S,K] =−γκ2
h8∆−32γ2|x|2i+ d
dtA+2[A(x)∇ϕ.∇ − ∇ϕ.∇A].
Moreover by assumption (2), (Stf+ [S,K]f,f) =γκ
Z
Rn
8|∇f|2H+32γ2|x|2kfk2dx +2
Z
Rn([A(x)∇ϕ.∇f− ∇ϕ.∇A f],f)dx≥0. (3.13) This identity, the condition onV and (3.13) imply that
k∂tf−S f −K fkX≤κ2(M1kfkX+eγϕkFkX). (3.14) If we knew that the quantities and calculations involved in the proof of Lemma 3.4 (similar as in [7, Lemma 2]) were finite and correct, when f = eγ|x|2u we would have the logarithmic convexity ofQ(t) =eγ|x|2u(·,t)
X and the estimate (3.10) from Lemma3.4. But this fact is verifying by reasoning as in [7, Lemma 3].
Let
σ= q
t(1−t)eγ|x|2, Y= L2([0, 1]×Rn;H). Lemma 3.6. Assume that a, b, u, A and V are as in Lemma3.5andγ>0. Then,
kσ∇ukY+kσ|x|ukY ≤N[(1+M1)]
"
sup
t∈[0,1]
eγ|x|2u(.,t)
X+ sup
t∈[0,1]
eγ|x|2F(.,t)
Y
# ,
where N is bounded number, whenγandκ are bounded below.
Proof. The integration by parts shows that Z
Rn
|∇f|2H+4γ2|x|2kfkdx=
Z
Rn
h
e2γ|x|2
|∇u|2H−2nγ
kuk2idx,
when f = eγ|x|2u, while integration by parts, the Cauchy–Schwarz inequality and the identity, n=∇·x, give that
Z
Rn
|∇f|2H+4γ2|x|2kfkdx≥2γnkfk2X. The sum of the last two formulae gives the inequality
2 Z
Rn
|∇f|2H+4γ2|x|2kfkdx≥
Z
Rneγ|x|2|∇f|2Hdx. (3.15) Integration over [0, 1]of t(1−t)times the formula (3.6) for Q00(t) and integration by parts, shows that
2 Z 1
0 t(1−t) (Stf+ [S,K]f,f)Xdt+
Z 1
0 Q(t)dt
≤Q(1) +Q(0) +2 Z 1
0
(1−2t)Re(∂tf−S f −K f,f)Xdt +
Z 1
0 t(1−t)k∂tf−S f −K fk2Xdt. (3.16) Assuming again that the last two calculations are justified for f =eγ|x|2. Then (3.13)–(3.16) imply the assertion.
4 Appell transformation in abstract function spaces
Let
ρ(t) =α(1−t) +βt, ϕ(x,t) = (α−β)x2 4(a+ib)ρ(t), ν(s) =
γαβρ2(s) + (α−β)a 4(a2+b2)ρ(s)
.
Lemma 4.1. Assume A and V are as in Lemma3.5and u =u(x,s)is a solution of the equation
∂su= (a+ib) [∆u+Au+V(y,s)u+F(y,s)], y ∈Rn, s∈ [0, 1]. Let a+ib6=0,γ∈Randα,β∈R+. Set
˜
u(x,t) =pαβρ−1(t)
n
2 up
αβxρ−1(t),βtρ−1(t)eϕ. (4.1)
Then,u˜(x,t)verifies the equation
∂tu˜ = (a+ib)∆u˜+Au˜+V˜ (x,t)u+F˜(x,t), x∈Rn, t∈[0, 1] with
V˜ (x,t) =αβρ−2(t)Vp
αβxρ−1(t),βtρ−1(t),
F˜(x,t) =pαβρ−1(t)
n
2+2p
αβxρ−1(t),βtρ−1(t). Moreover,
eγ|x|2F˜(·,t)
X =αβρ−2(t)eν|y|2kF(s)kX ,
eγ|x|2u˜(·,t)
X =eν|y|2ku(s)kX, when s=µ(t)andγ∈R.
Proof. Ifuis a solution of the equation
∂su= (a+ib) [∆u+Au+Q(y,s)], y∈Rn, s∈[0, 1] (4.2) then, the functionu1(x,t) =u √
rx,rt+τ
verifies
∂tu1 = (a+ib)∆u1+Au1+rQ √
rx,rt+τ
, y∈ Rn, s ∈[0, 1] andu2(x,t) =t−n2u xt,1t
e
|x|2
4(a+ib)t is a solution to
∂tu2 =−(a+ib)
∆u2+Au+t−(2+n2)Qx t,1
t
e
|x|2 4(a+ib)t
, y ∈Rn, s ∈[0, 1]. These two facts and the sequel of changes of variables below verify the lemma, whenα> β, i.e.
u s
αβ
α−βx, αβ
α−βt− β α−β
!
is a solution to the same non-homogeneous equation but with right-hand side αβ
α−βQ s
αβ
α−βx, αβ
α−βt− β α−β
! . The function,
1 (α−t)n2u
pαβx pα−β(α−t),
αβ
(α−β) (α−t)− β α−β
! e
|x|2 4(a+ib)(α−t)
verifies (4.2) with right-hand side αβ
(α−β) (α−t)n2+2Q
pαβx p
α−β(α−t),
αβ
(α−β) (α−t)− β α−β
! e
|x|2 4(a+ib)(α−t). Replacing(x,t)by p
α−βx,(α−β)t
we get that ρ−
n 2 (t)u
p
αβρ−1(t)x,αβρ−1(t)−β α−β
e(α−β)
|x|2ρ(t)
4(a+ib) (4.3) is a solution of (4.2) but with right-hand
ρ−(n2+2) (t)Q
p
αβρ−1(t)x,αβρ−1(t)−β α−β
e(α−β)
|x|2ρ(t)
4(a+ib). (4.4) Finally, observe that
s =βtρ(t) = αβρ
−1(t)−β α−β
and multiply (4.3) and (4.4) we obtain the assertion for α > β. The case β > α follows by reversing by changes of variables,s0 =1−s andt0 =1−t.
5 Variable coefficients. Proof of Theorem 2.4
We are ready to prove Theorem2.4. Let
B= L1(0, 1;L∞(Rn;B(H))).
Proof of Theorem2.4. We may assume that α 6= β. The case α = β follows from the latter by replacing β by β+δ, δ > 0, and letting δ tend to zero. We may also assume that α < β.
Otherwise, replace u by ¯u(1−t). Assume a > 0. Set W = ∆+A+V1. By Lemma2.7 the problem
∂tu= (a+ib) [∆u+A(x)u+V1(x)u], x∈Rn, t∈ [0, 1], (5.1) u(x, 0) =u0(x).
has a solution u=U(t)u0= et(a+ib)Wu0 ∈C([0, 1];X(A)), where U(t) =F−1h
e−Q(ξ)i
, Q(ξ) = (a+ib)h− |ξ|2+Aˆ(ξ) +Vˆ1(ξ)i,
here,F−1is the inverse Fourier transform, ˆA(ξ), ˆV1(ξ)respectively denote the Fourier trans- forms of A(x),V1(x). By reasoning as the Duhamel principle we get that the problem
∂tu= (a+ib) [∆u+A(x)u+V(x,t)]u, x∈Rn, t∈[0, 1], u(x, 0) =u0(x)
has a solution expressing as u(x,t) =H(t)u0+i
Z t
0 H(t−s)V2(x,s)u(x,s)ds for x∈Rn, s∈[0, 1], (5.2) where
eitW = H(t) =H(t,x) = F−1h eiQ(ξ)i
. For 0≤ε≤1 set
Fε(x,t) = i
ε+ieεtWV2(x,t)u(x,t) (5.3) and
uε(x,t) =e(ε+i)tWu0+ (ε+i)
Z t
0
e(ε+i)(t−s)WFε(x,s)u(x,s)ds. (5.4) Then,uε(x,t)∈ L∞(0, 1;X(A))∩L2 Rn;Y1
and satisfies
∂tuε = (ε+i) (Wu+Fε) inRn×[0, 1], uε(·, 0) =u0(·).
The identities
e(z1+z2)W =ez1Wez2W, when Rez1, Rez2 ≥0, and (5.2)–(5.4) show that
uε(x,t) =eεtWu(x,t), fort∈ [0, 1]. (5.5)
In particular, the equalityuε(x, 1) =eεWu(x, 1), Lemma3.2with a+ib=ε,γ= 1
β, F≡0 and the fact thatuε(0) =u(0)imply that
e
|x|2
β2+4εuε(·, 1) X
≤eεkV1kB e
|x|2 β2 u(·, 1)
X
,
e
|x|2 α2
uε(·, 0) X
= e
|x|2 α2
u(·, 0) X
.
A second application of Lemma3.2 witha+ib= ε,F ≡ 0, the value ofγ = µ−2(t)and (5.2) show that
ε
|x|2
µ2(t)+4εtFε(·,t) X
≤eεkV1kB ε
|x|2
µ2(t)V2(·,t) B
ku(·,t)kX
fort ∈[0, 1]. Setting,αε =α+2εandβε = β+2ε, the last three inequalities give that
e
|x|2 β2
ε uε(·, 1) X
≤eεkV1kB e
|x|2
β2 u(·, 1) X
, (5.6)
e
|x|2 α2
ε uε(·, 0) X
≤eεkV1kB e
|x|2 α2
u(·, 0) X
,
ε|x|
2µ−2(t)Fε(·,t)
X≤ eεkV1kB ε|x|
2µ−2(t)V2(·,t)
Bku(·,t)kX, t∈[0, 1]. (5.7) A third application of Lemma3.2witha+ib=b,F≡0,γ=0, and (5.2), (5.5) implies that
kFε(·,t)kX ≤eεkV1kBkV2(·,t)kBku(·,t)kX, (5.8) kuε(·,t)kX ≤eεkV1kBku(·,t)kX, t ∈[0, 1].
Setγε = α1
εβε and let
˜
uε(x,t) = p
αεβερ−ε 1(t) n2
up
αεβεxρ−ε 1(t),βεtρ−ε 1(t)
eϕε
be the function associated to uε in Lemma 4.1, where a+ib = ε+i and α, β are replaced respectively byαε,βε when
ρε(t) =αε(1−t) +βεt, ϕε = ϕε(x,t) = (αε−βε)|x|2 4(a+ib)ρε(t). Becauseα< β, ˜uε ∈ L∞(0, 1;X)∩L2 0, 1;Y1
and satisfies the equation
∂tu˜ε = (ε+i) ∆u˜ε+A(x)u˜ε+V˜1ε(x,t)u˜ε+F˜ε(x,t) inRn×[0, 1], where
V˜1ε(x,t) =αεβερ−ε 2(t)V1p
αεβερ−ε 1(t)x
, sup
t∈[0,1]
V˜1ε(·,t)
B ≤ β αM1, F˜ε(x,t) =hpαεβερ−ε 1(t)i
n 2+2
Fεp
αεβερ−ε 1(t)x,βεtρ−ε 1(t)eϕε, (5.9)