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Hardy type unique continuation properties for abstract Schrödinger equations and applications

Veli Shakhmurov

B1,2

1Department 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

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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(), 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+a2u

∂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=1

amj(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

kf(x)kpEdx 1p

, 1≤ p<. Let Hbe a Hilbert space and

kuk= kukH = (u,u)

1 2

H = (u,u)12 foru∈ H.

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For p=2 andE= H, Lp(Ω;E)becomes a H-valued function space with inner product:

(f,g)L2(Ω;H) =

Z

(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

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

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).

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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

=1

xk

Af

∂xk∂A

∂xk f

,f

X

0 for f ∈ L

0,T;Y1(A)

;

(3) A(x)A1(x0)∈ L1(Rn;B(H))for some x0Rnand 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

xRn

kV1(x)kB(H) <∞, sup

t∈[0,1]

e|x|2µ2(t)V2(·,t) B < or

rlimkVkL1(0,1;L(Rn/Or);B(H))=0;

(5) u∈C([0, 1];X(A))is a solution of the equation(1.1)and

eβ2|x|2u(·, 0)

X< ∞, eα2|x|2u(·, 1)

X <∞.

where

α, β>0, αβ <2.

Then u(x,t)≡0.

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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 eβ2|x|2(u1(·, 0)−u2(·, 0))∈X, eα2|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|2µ2(t)u(·,t)

1 µ(t)

X is logarithmically convex in[0, 1]and there is N= N(α,β)such that

e|x|2µ2(t)u(·,t)

1 µ(t)

X ≤eN(M1+M2+M21+M22)

eβ2|x|2u(·, 0)

β(1t)µ(t) X

eα2|x|2u(·, 1)

αtµ(t)

X ,

when

M2 =e2B(V2) sup

t∈[0,1]

e|x|2µ2(t)V2,t)

B, B(V2) = sup

t∈[0,1]

kReV2,t)kB. Moreover,

q

t(1−t)

e|x|2µ2(t)∇u

L2(Rn×[0,1];H)

≤eN(M1+M2+M21+M22)h

eβ2|x|2u(·, 0)

X+eα2|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 <∞,

eδ2|x|2u(·, 1) X < for some δ<1. Then, f(x)≡0for x∈Rn.

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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

e4

|ξ|2 α2

, x,ξRn 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)A1(x0) ∈ L1(Rn;B(H))for some x0Rn. Then for

f ∈Ws,2(Rn;H),s≥0there is a generalized solution of (2.2)expressing as u(x,t) = F1h

eiAˆξtfˆ(ξ)i, Aˆξ = Aˆ(ξ)− |ξ|2, (2.3) where F1is 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), ξRn. 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)

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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=1

xk

Af

∂xkA

∂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)

Xeγ|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+(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)

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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

tkυk2X2kaReV−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)θρψr(x), υρ,r(x,t) =eϕρ,ru, whereq(t) =γa

a+4γ a2+b2 t1

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,

θρψr(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.

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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)Q1(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

+ktf−S f +K fk2X− ktf −S f −K fk2X (3.9) and

tN(t)≥ Q1(t)

(Stf + [S,K]f,f)X1

2ktf−S f −K fk2X

.

Moreover, if

ktf−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)Q1t(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)

1t X

eγ|x|2u(·, 1)

t

X, (3.10)

where

M(a,b) =κ2 γM12+M22

+κ(M1+M2) when0≤t ≤1.

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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−4γ2D2ϕϕ+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

ktf−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.

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Proof. The integration by parts shows that Z

Rn

|∇f|2H+4γ2|x|2kfkdx=

Z

Rn

h

e|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+2|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)ktf−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) = (αβ)x2 4(a+ib)ρ(t), ν(s) =

γαβρ2(s) + (αβ)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,γRandα,βR+. Set

˜

u(x,t) =pαβρ1(t)

n

2 up

αβxρ1(t),βtρ1(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) =αβρ2(t)Vp

αβxρ1(t),βtρ1(t),

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F˜(x,t) =pαβρ1(t)

n

2+2p

αβxρ1(t),βtρ1(t). Moreover,

eγ|x|2F˜(·,t)

X =αβρ2(t)eν|y|2kF(s)kX ,

eγ|x|2u˜(·,t)

X =eν|y|2ku(s)kX, when s=µ(t)andγR.

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) =tn2u 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

αβx, αβ

αβt− β αβ

!

is a solution to the same non-homogeneous equation but with right-hand side αβ

αβQ s

αβ

αβx, αβ

αβt− β αβ

! . The function,

1 (α−t)n2u

pαβx pαβ(α−t),

αβ

(αβ) (α−t)− β αβ

! e

|x|2 4(a+ib)(αt)

verifies (4.2) with right-hand side αβ

(αβ) (α−t)n2+2Q

pαβx p

αβ(α−t),

αβ

(αβ) (α−t)− β αβ

! e

|x|2 4(a+ib)(αt). Replacing(x,t)by p

αβx,(αβ)t

we get that ρ

n 2 (t)u

p

αβρ1(t)x,αβρ1(t)−β αβ

e(αβ)

|x|2ρ(t)

4(a+ib) (4.3) is a solution of (4.2) but with right-hand

ρ(n2+2) (t)Q

p

αβρ1(t)x,αβρ1(t)−β αβ

e(αβ)

|x|2ρ(t)

4(a+ib). (4.4) Finally, observe that

s =βtρ(t) = αβρ

1(t)−β αβ

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.

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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) =F1h

eQ(ξ)i

, Q(ξ) = (a+ib)h− |ξ|2+Aˆ(ξ) +Vˆ1(ξ)i,

here,F1is 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) = F1h 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)(ts)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)

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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+uε(·, 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 = µ2(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|

2µ2(t)Fε(·,t)

X≤ eεkV1kB ε|x|

2µ2(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

αεβερε 1(t) n2

up

αεβεε 1(t),βεε 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) = (αεβε)|x|2 4(a+ib)ρε(t). Becauseα< β, ˜uε ∈ L(0, 1;X)∩L2 0, 1;Y1

and satisfies the equation

tε = (ε+i) ε+A(x)u˜ε+V˜1ε(x,t)u˜ε+F˜ε(x,t) inRn×[0, 1], where

1ε(x,t) =αεβερε 2(t)V1p

αεβερε 1(t)x

, sup

t∈[0,1]

1ε(·,t)

Bβ αM1, F˜ε(x,t) =hpαεβερε 1(t)i

n 2+2

Fεp

αεβερε 1(t)x,βεε 1(t)eϕε, (5.9)

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