DIRECT AND INVERSE DEGENERATE PARABOLIC DIFFERENTIAL

Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 198, pp. 1–22. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

DIRECT AND INVERSE DEGENERATE PARABOLIC DIFFERENTIAL EQUATIONS WITH MULTI-VALUED OPERATORS ANGELO FAVINI, ALFREDO LORENZI, HIROKI TANABE In memory of Alfredo Lorenzi

Abstract. Real interpolation spaces are used for solving some identification linear evolution problems in Banach spaces, under space regularity assumptions.

1. Introduction This note starts with the following direct problem in a Banach space X, d y(t) + Ay(t) 3 f (t), 0 ≤ t ≤ T, dt y(0) = y0 .

(1.1)

Here A is a possibly multivalued linear operator such that ρ(A) ⊃ Σα = {λ ∈ C; Re λ ≤ c(1 + | Im λ|)α },

(1.2)

and the following inequality holds for λ ∈ Σα k(λ − A)−1 kL(X) ≤ c(1 + |λ|)−β ,

(1.3)

where c, α and β are positive constants satisfying β ≤ α ≤ 1. It is shown in the book by Favini and Yagi [6] that −A generates a C ∞ -semigroup e−tA , 0 < t < ∞. We are interested in extending some of the results obtained in the paper by Favini, Lorenzi Tanabe [2], where A is assumed to be single-valued, to the case where A is multi˜ θ , f ∈ C([0, T ]; X)∩B([0, T ]; X eθ ) valued. In [2] supposing that y0 ∈ D(A), Ay0 ∈ X A A θ e or with (X, D(A))θ,∞ in place of XA , the existence and uniqueness of a solution to (1.1) with some regularity property is established, where B([0, T ]; Y ) stands for the set of all bounded functions defined in [0, T ] with values in a Banach space Y . In ˜θ this paper we show analogous results in case A is multivalued replacing Ay0 ∈ X A ˜ θ 6= ∅ or Ay0 ∩ (X, D(A))θ,∞ 6= ∅. or Ay0 ∈ (X, D(A))θ,∞ by Ay0 ∩ X A

2010 Mathematics Subject Classification. 35R30, 34G10, 35K20, 35K50, 45N05, 45Q05. Key words and phrases. Identification problems; first-order equations and systems in Banach spaces; linear parabolic integro-differential equations; existence and uniqueness. c

2015 Texas State University - San Marcos. Submitted June 11, 2015. Published July 30, 2015. 1

2

A. FAVINI, A. LORENZI, H. TANABE

EJDE-2015/198

θ e θ in case where A We begin by modifying the definitions of the spaces XA and X A −1 −tA is not necessarily single valued replacing A(t + A) and Ae by I − t(t + A)−1 −tA and de /dt respectively and prove some preliminary results on these spaces. The results proved for the direct problem are applied to the following identification problem: d y(t) + Ay(t) 3 f (t)z + h(t), t ∈ [0, T ], dt (1.4) y(0) = y0 ,

Φ[y(t)] = g(t),

t ∈ [0, T ],

where y ∈ C([0, T ]; X), f ∈ C([0, T ]; C) are unknown, z ∈ X, h ∈ C([0, T ]; X), g ∈ C([0, T ]; C) are given elements and Φ ∈ X ∗ . Under some regularity assumptions on z, h and g it is shown that a unique solution to problem (1.4) exists. An extension of this result to equations with several unknown scalar functions n X d y(t) + Ay(t) 3 fj (t)zj + h(t), t ∈ [0, T ], dt j=1 (1.5) y(0) = y0 , Φj [y(t)] = gj (t),

j = 1, . . . , n,

t ∈ [0, T ],

is also established. The above results are applied to the following problems d M u(t) + Lu(t) = f (t), t ∈ [0, T ], dt M u(0) = M u0 , and

(1.6)

n

X d M u(t) + Lu(t) = fj (t)zj + h(t), dt j=1

t ∈ [0, T ],

M u(0) = M u0 ,

(1.7)

Φj [M u(t)] = gj (t), j = 1, . . . , n, t ∈ [0, T ], where L and M are linear closed operators such that D(L) ⊂ D(M )

(1.8)

and for λ ∈ Σα a bounded inverse of λM + L exists and kM (λM + L)−1 kL(X) ≤ c(1 + |λ|)−β . −1

(1.9)

Then A = LM satisfies (1.2) and (1.3). A solution to (1.6) or (1.7) is easily obtained from that to (1.1) or (1.5) with y0 = M u0 . We refer to the monograph [5] for diverse problems concerning regular evolution equations via Mathematical Physics. These identification problems were discussed in the papers [1] and [5] by using at all different techniques. Our present approach allows us to weaken the assumptions and to improve the regularity of solutions. The plan of this paper is as follows. In section 2 preliminary results on intermediate spaces are collected. Section 3 is devoted to the existence and uniqueness of a solution to the direct problem (1.1). In Section 4 identification problem (1.4) is solved by transforming it to a simple Volterra integral equation for f . In Section 5 the general identification problem (1.4) is solved by applying the Banach fixed point theorem. This result is extended to equations (1.5) with several unknown scalar

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DIRECT AND INVERSE DEGENERATE PROBLEMS

3

functions in Section 6. Finally, in Section 7 these results are applied to problems (1.6) and (1.7). 2. Preliminaries Let A be a possibly multivalued linear operator in the complex Banach space X satisfying (1.2) and (1.3). Then it is shown in Chapter III of Favini and Yagi [6] that −A generates a semigroup e−tA , 0 < t < ∞, satisfying di −tA e kL(X) ≤ C0 t(β−i−1)/α , i = 0, 1, 2, dti The set D(A) makes a Banach space with norm k

kukD(A) = inf kφkX φ∈Au

C0 > 0.

for u ∈ D(A).

It is known that d d A−1 e−tA = e−tA A−1 = −e−tA , lim e−tA u = u for u ∈ D(A). t→0 dt dt If u ∈ D(A), then in view of (2.3) and (2.1) one has for φ ∈ Au

(2.1)

(2.2)

(2.3)

d d −tA e ukX = k e−tA A−1 φkX = ke−tA φkX ≤ C0 t(β−1)/α kφkX . dt dt This implies, by (2.2), k

d −tA e ukX ≤ C0 t(β−1)/α kukD(A) . dt This inequality and the one obtained with the aid of an analogous argument imply k

k

d −tA e kL(D(A),X) ≤ C0 t(β−1)/α , dt

k

d2 −tA e kL(D(A),X) ≤ C0 t(β−2)/α . dt2

Definition. For 0 < θ < 1, θ XA = {u ∈ X; |u|XAθ = sup tθ ku − t(t + A)−1 ukX < ∞}, 0 2 we may apply Lemma 2.5 and (3.34) to obtain that (2α+2β−4+θ)/α y20 (t) − f (t) ∈ XA and |y20 (t) − f (t)|X (2α+2β−4+θ)/α ≤ Γ((2α + 2β − 4 + θ)/α)|y20 (t) − f (t)|Xe α+β+θ−2 A

A

C1 Γ((2α + 2β − 4 + θ)/α) ≤ kf kB([0,T ];Xe θ ) . A (4 − α − 2β + θ)/α From this inequality, (3.13) and (3.30) one concludes that (2α+2β−4+θ)/α

y20 − f ∈ B([0, T ]; XA

).

(2α+2β−4+θ)/α

This and (3.28) imply y 0 − f ∈ B([0, T ]; XA ). The proof that y defined by (3.2) satisfies (1.1) is the same as that in the proof of Theorem 3.1. Remark 3.3. Suppose that the assumptions of Theorem 3.1 are satisfied, and let k ∈ R. Consider the problem d y(t) + Ay(t) + ky(t) 3 f (t), dt y(0) = y0 .

0 ≤ t ≤ T,

(3.35)

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13

The operator −(A + k) generates a differentiable semigroup e−t(A+k) = e−kt e−tA . Then define C00 = C0 emax{k,0}T , so that for 0 < t ≤ T one gets di −t(A+k) e kL(X) ≤ C00 t(β−i−1)/α , i = 0, 1, 2. dti The unique solution to (3.35) is given by Z t y(t) = e−t(A+k) y0 + e−(t−s)(A+k) f (s)ds. k

0

Since ekt y(t) is a solution to (1.1) with f (t) replaced by ekt f (t), y satisfies y ∈ C 1 ([0, T ]; X) and (2α+β−3+θ)/α

y 0 + ky − f ∈ C (2α+β−3+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

).

Furthermore, the following results hold just as (3.13) and (3.11): Z Z t d t −(t−s)(A+k) ∂ −(t−s)(A+k) e f (s)ds, e f (s)ds = f (t) + dt 0 ∂t 0 Z t C00 t(α+β−2+θ)/α ∂ −(t−s)(A+k) k e f (s)dskX ≤ kf kB([0,T ];(X,D(A))θ,∞ ) . (α + β − 2 + θ)/α 0 ∂t An analogous remark holds also for Theorem 3.2. 4. An identification problem for the differential equation In this section we consider problem (1.4): d y(t) + Ay(t) 3 f (t)z + h(t), dt y(0) = y0 , Φ[y(t)] = g(t),

t ∈ [0, T ],

(4.1) (4.2)

t ∈ [0, T ],

(4.3)

Theorem 4.1. Suppose that 2α + β + θ > 3, y0 ∈ D(A), Ay0 ∩ (X, D(A))θ,∞ 6= ∅, z ∈ (X, D(A))θ,∞ , h ∈ C([0, T ]; X) ∩ B([0, T ]; (X, D(A))θ,∞ ), g ∈ C 1 ([0, T ]; C), Φ ∈ X ∗ , Φ[y0 ] = g(0), Φ[z] 6= 0. Then problem (4.1)-(4.3) admits a unique solution (y, f ) such that y ∈ C 1 ([0, T ]; X),

f ∈ C([0, T ]; C), (2α+β−3+θ)/α

y 0 − f (·)z − h ∈ C (2α+β−3+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

Proof. Supposing that f ∈ C([0, T ]; C) is known, define a function y by Z t Z t −tA −(t−s)A y(t) = e y0 + f (s)e zds + e−(t−s)A h(s)ds. 0

(4.4) ).

(4.5)

0

Since α + β + θ − 2 ≥ 2α + β + θ − 3 > 0, by Lemma 2.3 and (2.5) the following statements hold: d (4.6) e−tA z → z as t → 0, k e−tA zkX ≤ C0 t(β−2+θ)/α kzk(X,D(A))θ,∞ . dt Rt Hence 0 f (s)e−(t−s)A zds is differentiable and Z Z t d t ∂ −(t−s)A f (s)e zds = f (t)z + f (s) e−(t−s)A zds. (4.7) dt 0 ∂t 0

14


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Rt According to the proof of Theorem 3.1 (cf. (3.13) and (3.11)) 0 e−(t−s)A h(s)ds is differentiable and Z t Z ∂ −(t−s)A d t −(t−s)A e h(s)ds = h(t) + e h(s)ds, (4.8) dt 0 ∂t 0 Z t ∂ −(t−s)A t(α+β−2+θ)/α e h(s)dskX ≤ C0 khkB([0,T ];(X,D(A))θ,∞ ) . (4.9) k (α + β − 2 + θ)/α 0 ∂t Hence y is differentiable and d d y(t) = e−tA y0 +f (t)z + dt dt

Z 0

t

∂ d f (s) e−(t−s)A zds+ ∂t dt

Z

t

e−(t−s)A h(s)ds. (4.10)

0

Assuming that y(t) satisfies (4.3) one deduces from (4.10) the following identity Z t hd i h∂ i g 0 (t) = Φ e−tA y0 + f (t)Φ[z] + e−(t−s)A z ds f (s)Φ dt ∂t 0 (4.11) i hd Z t e−(t−s)A h(s)ds . +Φ dt 0 Rewriting (4.11) one obtains the following integral equation to be satisfied by f : Z t i h∂ f (t) + χ e−(t−s)A z ds f (s)Φ ∂t 0 (4.12) i hd Z t i hd −tA −(t−s)A 0 e y0 − χΦ e h(s)ds , = χg (t) − χΦ dt dt 0 where χ = Φ[z]−1 . Set hd i e−tA z , dt hd i hd Z t i −tA 0 e y0 − χΦ ψ(t) = χg (t) − χΦ e−(t−s)A h(s)ds . dt dt 0 κ(t) = χΦ

(4.13)

Then (4.12) is rewritten as Z

t

κ(t − s)f (s)ds = ψ(t),

f (t) + 0

or briefly f + κ ∗ f = ψ.

(4.14)

By (4.6) one has |κ(t)| ≤ C0 |χ|kΦkt(β−2+θ)/α kzk(X,D(A))θ,∞ . In view of (3.5), (4.8) and (4.9) one observes that ψ ∈ C([0, T ]; C). Let r be the solution to the integral equation κ + r + r ∗ κ = 0. This equation is solved by successive approximations, and the solution r satisfies κ + r + κ ∗ r = 0,

|r(t)| ≤ C2 t(β−2+θ)/α ,

C2 > 0.

The integral equation (4.14) admits a unique solution f ∈ C([0, T ]; C) given by f = ψ + r ∗ ψ, or Z t f (t) = ψ(t) + r(t − s)ψ(s)ds. (4.15) 0

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15

It is easy to verify that if we define y by (4.5) with f given by (4.15), the pair (y, f ) satisfies (4.1) and (4.2). From (4.10) and (4.11) it follows that hd i d g 0 (t) = Φ y(t) = Φ[y(t)]. dt dt From this equality and the compatibility condition Φ[y0 ] = g(0) (4.3) follows. Since f (·)z + h ∈ C([0, T ]; X) ∩ B([0, T ]; (X, D(A))θ,∞ ), the assertion (4.4) follows from Theorem 3.1. Theorem 4.2. Suppose that α + β > 3/2, 2(2 − α − β) < θ < 1, y0 ∈ D(A), e θ 6= ∅, z ∈ X e θ , limt→0 e−tA z = z, g ∈ C 1 ([0, T ]; C), Φ ∈ X ∗ , Φ[y0 ] = Ay0 ∩ X A A e θ ), limτ →0 e−τ A h(t) = h(t) for every g(0), Φ[z] 6= 0, h ∈ C([0, T ]; X) ∩ B([0, T ]; X A t ∈ [0, T ]. Then problem (4.1)-(4.3) admits a unique solution y such that y ∈ C 1 ([0, T ]; X) and (2α+2β−4+θ)/α

y 0 − f ∈ C (2α+2β−4+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

).

Proof. Supposing that f ∈ C([0, T ]; X) is known, define the function y by (4.5). e θ . Then in view of Lemma 2.5, Let φ ∈ Ay0 ∩ X A d d −tA e y0 = e−tA A−1 φ = −e−tA φ dt dt converges as t → 0. Hence d the function t 7→ e−tA y0 belongs to C([0, T ]; X). dt e θ one has By the definition of X

(4.16)

A

k

d −tA e zkX ≤ t(β−2+θ)/α |z|Xe θ . A dt

(4.17) Rt

Hence using the assumption limt→0 e−tA z = z one observes that 0 f (s)e−(t−s)A zds is differentiable and (4.7) holds. According to the proof of Theorem 3.2 (cf. (3.30)) R t −(t−s)A e h(s)ds is differentiable, and equality (4.8) and the inequality 0 Z t t(α+β−2+θ)/α ∂ −(t−s)A e h(s)dskX ≤ khkB(0,T ;Xe θ ) (4.18) k A (α + β − 2 + θ)/α 0 ∂t holds. Hence y is differentiable and (4.10) holds. Assuming that y(t) satisfies (4.3) one deduces (4.14) from (4.10) as in the proof of Theorem 4.1, where κ and ψ are functions defined by (4.13). By virtue of (4.17) one has |κ(t)| ≤ |χ|kΦkX ∗ t(β−2+θ)/α |z|Xe θ . A

In view of (4.16), (4.8) and (4.18) one observes ψ ∈ C([0, T ]; C). The remaining part of the proof is the same as that of Theorem 4.1. 5. Equations with several unknown scalar functions In this section we consider the problem consisting of recovering several unknown scalar functions f1 , . . . , fn and a vector function y such that n X d y(t) + Ay(t) 3 fj (t)zj + h(t), t ∈ [0, T ], (5.1) dt j=1 y(0) = y0 ,

(5.2)

16


Φj [y(t)] = gj (t),

j = 1, . . . , n,

EJDE-2015/198

t ∈ [0, T ].

(5.3)

Theorem 5.1. Suppose 2α + β + θ > 3, y0 ∈ D(A), Ay0 ∩ (X, D(A))θ,∞ 6= ∅, zj ∈ (X, D(A))θ,∞ , j = 1, . . . , n, h ∈ C([0, T ]; X) ∩ B([0, T ]; (X, D(A))θ,∞ ), gj ∈ C 1 ([0, T ]; C), Φj ∈ X ∗ , Φj [y0 ] = gj (0), j = 1, . . . , n, and   Φ1 [z1 ] . . . Φ1 [zn ] . . .  6= 0. (5.4) det  . . . Φn [z1 ] . . . Φn [zn ] Then, problem (5.1)–(5.3) admits a unique solution (y, f1 , . . . , fn ) such that y ∈ C 1 ([0, T ]; X), y0 −

n X

f1 , . . . , fn ∈ C([0, T ]; C), (2α+β−3+θ)/α

fj (·)zj − h ∈ C (2α+β−3+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

).

(5.5)

j=1

The proof is performed in parallel to the proof of Theorem 4.1. If f1 , . . . , fn ∈ C([0, T ]; C) are known, y is given by Z t Z tX n −(t−s)A −tA fj (s)e zj ds + e−(t−s)A h(s)ds. y(t) = e y0 + 0

0 j=1

Just as the proof of (4.10) one deduces from this equality y 0 (t) =

n X d −tA fj (t)zj e y0 + dt j=1 Z tX Z n d t −(t−s)A −(t−s)A + fj (s)Dt e zj ds + e h(s)ds. dt 0 0 j=1

(5.6)

It follows from (5.3) and (5.6) that gi0 (t) = Φi [y 0 (t)] n X = Φi Dt e−tA y0 + fj (t)Φi [zj ] + j=1

h

Z

+ Φi Dt

t

Z tX n

fj (s)Φi [Dt e−(t−s)A zj ]ds

0 j=1

i e−(t−s)A h(s)ds .

0

This is rewritten as    Φ1 [z1 ] . . . Φ1 [zn ] f1 (t)  ... ...  ...  Φn [z1 ] . . . Φn [zn ] fn (t) Rt   0 g1 (t) − Φ1 [Dt e−tA y0 ] − Φ1 Dt 0 e−(t−s)A h(s)ds  ... = R t −(t−s)A 0 −tA gn (t) − Φn [Dt e y0 ] − Φn Dt 0 e h(s)ds    −(t−s)A −(t−s)A Z t Φ1 [Dt e z1 ] . . . Φ1 [Dt e zn ] f1 (s)    . . .  ds. ... ... −(t−s)A −(t−s)A 0 fn (s) Φn [Dt e z1 ] . . . Φn [Dt e zn ]

(5.7)

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Set



Φ1 [z1 ] . . . A =  ... Φn [z1 ] . . .

17

 Φ1 [zn ] ... . Φn [zn ]

Then by assumption (5.4), A−1 exists. Set Rt   0 g1 (t) − Φ1 [Dt e−tA y0 ] − Φ1 Dt 0 e−(t−s)A h(s)ds , ... Φ(t) = A−1  R t −(t−s)A 0 −tA gn (t) − Φn [Dt e y0 ] − Φn Dt 0 e h(s)ds     −tA −tA Φ1 [Dt e z1 ] . . . Φ1 [Dt e zn ] f1 (t)  , F (t) =  . . .  . ... ... K(t) = A−1  fn (t) Φn [Dt e−tA z1 ] . . . Φn [Dt e−tA zn ] Tt follows from (5.7) that Z F (t) = Φ(t) −

t

K(t − s)F (s) ds. 0

Since kK(t)kL(Cn ) ≤ Ct(β−2+θ)/α , the remaining part of the proof is the same as that of Theorem 4.1. Analogously the following theorem is established. e θ 6= ∅, Theorem 5.2. Suppose 2(2 − α − β) < θ < 1, y0 ∈ D(A), Ay0 ∩ X A θ −tA e e θ ), zj ∈ XA , limt→0 e zj = zj , j = 1, . . . , n, h ∈ C([0, T ]; X) ∩ B([0, T ]; X A −τ A 1 ∗ limτ →0 e h(t) = h(t) for every t ∈ [0, T ], gj ∈ C ([0, T ]; C), Φj ∈ X , Φj [y0 ] = gj (0), j = 1, . . . , n, and   Φ1 [z1 ] . . . Φ1 [zn ] . . .  6= 0. det  . . . Φn [z1 ] . . . Φn [zn ] Then, problem (5.1)-(5.3) admits a unique solution (y, f1 , . . . , fn ) such that y ∈ C 1 ([0, T ]; X), y0 −

n X

f1 , . . . , fn ∈ C([0, T ]; C), (2α+2β−4+θ)/α

fj (·)zj − h ∈ C (2α+2β−4+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

).

j=1

6. Problems (1.6) and (1.7) Let L and M be two linear closed operators satisfying (1.8) and (1.9). Set A = LM −1 . Namely D(A) = M D(L) = {M u : u ∈ D(L)}, Ay = {Lu : y = M u, u ∈ D(L)} for y ∈ D(A).

(6.1)

It is shown in Favini and Yagi [6] that A satisfies (1.2) and (1.3). The graph-norm of D(A) is defined by kykD(A) = inf{kLukX : y = M u, u ∈ D(L)}

for y ∈ D(A).

Consider the problem d M u(t) + Lu(t) = f (t), dt

t ∈ [0, T ],

(6.2)

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M u(0) = M u0 .

(6.3)

Theorem 6.1. Suppose that 2α + β + θ > 3, u0 ∈ D(L), Lu0 ∈ (X, D(A))θ,∞ , f ∈ C([0, T ]; X) ∩ B([0, T ]; (X, D(A))θ,∞ ). Then problem (6.2)–(6.3) admits a unique solution u such that M u ∈ C 1 ([0, T ]; X), (2α+β−3+θ)/α

Lu ∈ C (2α+β−3+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

(6.4) ).

Proof. Set y0 = M u0 . Then y0 ∈ D(A), and Ay0 ∩ (X, D(A))θ,∞ is not empty, since it contains Lu0 . In view of Theorem 3.1 there exists a solution y to (1.1), A being defined by (6.1). Set u(t) = L−1 (f (t) − y 0 (t)). Then Lu(t) = f (t) − y 0 (t) ∈ Ay(t) = LM −1 y(t).

(6.5)

−1

Since L is bijective, it follows from (6.5) that u(t) ∈ M y(t), or y(t) = M u(t). From the first equation of (6.5) equation (6.2) follows. It is obvious that u satisfies (6.3). Analogously, using Theorem 3.2 instead of Theorem 3.1, the following theorem is obtained. eθ , f ∈ Theorem 6.2. Suppose that θ > 2(2 − α − β), u0 ∈ D(L), Lu0 ∈ X A −τ A θ e ), limτ →0 e f (t) = f (t) for every t ∈ [0, T ]. Then C([0, T ]; X) ∩ B([0, T ]; X A problem (6.2)–(6.3) admits a unique solution u such that M u ∈ C 1 ([0, T ]; X), (2α+2β−4+θ)/α

Lu ∈ C (2α+2β−4+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

(6.6) ).

Next, consider the problem n

X d M u(t) + Lu(t) = fj (t)zj + h(t), dt j=1

t ∈ [0, T ],

(6.7)

M u(0) = M u0 , Φj [M u(t)] = gj (t),

j = 1, . . . , n,

(6.8) t ∈ [0, T ].

(6.9)

Theorem 6.3. Suppose 2α + β + θ > 3, u0 ∈ D(L), Lu0 ∈ (X, D(A))θ,∞ , zj ∈ (X, D(A))θ,∞ , j = 1, . . . , n, h ∈ C([0, T ]; X) ∩ B([0, T ]; (X, D(A))θ,∞ ), Φj ∈ X ∗ , gj ∈ C 1 ([0, T ; C), Φj [M u0 ] = gj (0), j = 1, . . . , n, and (5.4) holds. Then problem (6.7)–(6.9) admits a unique solution (u, f1 , . . . , fn ) such that M u ∈ C 1 ([0, T ]; X),

f1 , . . . , fn ∈ C([0, T ]; C), (2α+β−3+θ)/α

Lu ∈ C (2α+β−3+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA

(6.10) ).

Proof. Let (y, f1 , . . . , fn ) be a solution to problem (5.1)-(5.3) with A defined by (6.1) and define a function u by n hX i u(t) = L−1 fj (t)zj + h(t) − y 0 (t) , t ∈ [0, T ]. j=1

Then in view of (5.1) Lu(t) =

n X j=1

fj (t)zj + h(t) − y 0 (t) ∈ Ay(t) = LM −1 y(t).

(6.11)

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Since L is injective, one gets u(t) ∈ M −1 y(t). This implies M u(t) = y(t).

(6.12)

The first equation in (6.11) and (6.12) imply (6.7). It is obvious that (6.8) and (6.9) hold. From the first equation in (6.11) and the second equation in (5.5) it follows that n X (2α+β−3+θ)/α Lu = fj (·)zj + h − y 0 ∈ C (2α+β−3+θ)/α ([0, T ]; X) ∩ B([0, T ]; XA ). j=1

Thus the regularity property (6.10) is proved.

The following theorem is analogously established. e θ , zj ∈ X eθ , Theorem 6.4. Suppose θ > 2(2 − α − β), u0 ∈ D(L), Lu0 ∈ X A A e θ ), limτ →0 e−τ A h(t) limt→0 e−tA zj = zj , j = 1, . . . , n, h ∈ C([0, T ]; X) ∩ B([0, T ]; X A = h(t) for every t ∈ [0, T ], gj ∈ C 1 ([0, T ]; C), Φj ∈ X ∗ , Φj [M u0 ] = gj (0), j = 1, . . . , n, and (5.4) holds. Then, problem (6.7)–(6.9) admits a unique solution (u, f1 , . . . , fn ) such that M u ∈ C 1 ([0, T ]; X), Lu ∈ C

(2α+2β−4+θ)/α

f1 , . . . , fn ∈ C([0, T ]; C), (2α+2β−4+θ)/α

([0, T ]; X) ∩ B([0, T ]; XA

).

Remark 6.5. When L is the realization in L2 (Ω) of a second order strongly elliptic linear differential operator L with the Dirichlet boundary condition in a bounded domain Ω and M is the multiplication operator by a function belonging to L∞ (Ω) one has α = 1, β = 1/2. Hence the assumption 2α + β + θ > 3 of Theorems 6.1 and 6.3 is not satisfied for θ = 1/2, and the assumption θ > 2(2 − α − β) of Theorems 6.2 and 6.4 is not satisfied for θ ∈ (0, 1). A treatment of this case is e ⊂ L2 (Ω)1/2 ⊂ given in [7]. Furthermore, owing to the inclusion relations D(A) A e (L2 (Ω), D(A))1/2,∞ the assumptions are described by using a clearer space D(A) 1/2 e 2 (Ω) , (L2 (Ω), D(A))1/2,∞ in [7], where A e = LM e −1 and L e is the realization than L A

of L in H −1 (Ω) = H01 (Ω)∗ . 7. Problems for systems Let us consider the following inverse problem: Recover yi , fij , i = 1, . . . , n, j = 1, . . . , N , such that y10 = A1 y1 + B11 y1 + · · · + B1n yn + f11 (t)z1 + · · · + f1N (t)zN + h1 (t), ... yn0

(7.1)

= An yn + Bn1 y1 + · · · + Bnn yn + fn1 (t)z1 + · · · + fnN (t)zN + hn (t), y1 (0) = y10 , . . . , yn (0) = yn0 , Φj [yi (t)] = gji (t),

i = 1, . . . , n; j = 1, . . . , N.

(7.2) (7.3)

Assume 0 < β ≤ α ≤ 1 and 2α+β +θ > 3. It is also assumed that for i, j = 1, . . . , n, Ai and Bij satisfy k(λ − Ai )−1 kL(X) ≤

C (1 + |λ|)β

for λ ∈ Σα = {λinC; Re λ ≥ −c(1 + | Im λ|)α },

θ Bij ∈ L(D(Aj ), XA ). i

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Also assume that yi0 ∈ D(Ai ), Ai yi0 ∈ (X, D(Ai ))θ,∞ , zj ∈ ∩nk=1 (X, D(Ak ))θ,∞ , hi ∈ C([0, T ]; X) ∩ B([0, T ]; (X, D(Ai ))θ,∞ , gji ∈ C 1 ([0, T ]; C), Φj ∈ X ∗ , Φj [yi0 ] = gji (0), i = 1, . . . , n, j = 1, . . . , N . Set       y1 A1 0 . . . 0 B11 . . . B1n . . . , B =  . . . ... . y = . . . , A = . . . yn 0 0 . . . An Bn1 . . . Bnn Then A and A + B generate infinitely differentiable semigroups in X n . The system (7.1) is written as       zN z2 z1       0 0 0    y 0 = (A + B)y + f11 (t)  . . . + f12 (t) . . . + · · · + f1N (t) . . . 0 0 0       0 0 0  z1   z2  zN       + f21 (t)  . . . + f22 (t) . . . + · · · + f2N (t) . . . + . . . 0 0 0         h1 (t) 0 0 0  0   h2 (t)  0 0        + fn1 (t)  . . . + fn2 (t) . . . + · · · + fnN (t) . . . +  . . .  . hn (t) zN z2 z1 Theorem 6.3 applies provided that  Φ1 [z1 ] . . . det  . . . ΦN [z1 ] . . .

 Φ1 [zN ] . . .  6= 0. ΦN [zN ]

Indeed, the further information reduces to the N linear systems in the unknowns fi1 , . . . , fiN , i = 1, . . . , n, whose determinant is just the indicated above. Identification problem (7.1)–(7.3) admits a unique solution y = (y1 , . . . , yn )t , fij , i = 1, . . . , n, j = 1, . . . , N such that y ∈ C 1 ([0, T ]; X n ), (A + B)y ∈ C

fij ∈ C 1 ([0, T ]; C),

(2α+β−3+θ)/α

n

i = 1, . . . , n, j = 1, . . . , N, (2α+β−3+θ)/α

([0, T ]; X ) ∩ B([0, T ]; XA+B ) ⊂ C (2α+β−3+θ)/α ([0, T ]; X n ) ∩ B([0, T ]; (X, D(A1 ))(2α+β−3+θ)/α,∞ ) × · · · × B([0, T ]; (X, D(An ))(2α+β−3+θ)/α,∞ ) .

Example. Let us consider the inverse problem Z t d (A + 1)y + Ay = k(t − s)Ay(s)ds + f (t)z + h(t), dt 0 (A + 1)y(0) = (A + 1)y0 , Φ[(A + 1)y(t)] = Φ[g(t)],

0 ≤ t ≤ T.

0 ≤ t ≤ T,

(7.4) (7.5) (7.6)

We suppose that −1 is a simple pole for the resolvent of A, i.e. (A + 1 + λ)−1 exists for 0 < |λ| ≤ ε and C . k(A + 1 + λ)−1 kL(X) ≤ |λ|

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A change of variable y(t) = eκt x(t) transforms equation (7.4) into d (A + 1)x(t) + κ(A + 1)x(t) + Ax(t) dt Z t = k(t − s)e−κ(t−s) Ax(s)ds + f1 (t)z + e−κt h(t), 0

where f1 (t) = e

−κt

f (t). Now

λ(A + 1) + κ(A + 1) + A = (λ + κ + 1)A + λ + κ 1 λ+κ = (λ + κ + 1) A + 1 − . = (λ + κ + 1) A + λ+κ+1 λ+κ+1 Hence if 0 < |λ + κ + 1|−1 < ε, i.e. |λ + κ + 1| > ε−1 , λ(A + 1) + κ(A + 1) + A has a 1 bounded inverse. Take κ so large that κ+1 > ε−1 . Then (A+1− λ+κ+1 )−1 ∈ L(X) exists for λ ∈ C \ S(−1 − κ, ε−1 ) and (A + 1)(λ(A + 1) + κ(A + 1) + A)−1 −1 1 λ+κ+1 −1 1 1 −1 A+1− , = (λ + κ + 1) 1+ λ+κ+1 λ+κ+1 = (A + 1)(λ + κ + 1)−1 A + 1 −

so that k(A + 1)(λ(A + 1) + κ(A + 1) + A)−1 kL(X) ≤ C|λ + κ + 1|−1 for |λ + κ + 1| > ε−1 . Hence the previous results (See also Favini and Tanabe [5]) apply for α = β = 1. However, this pole case allows a better treatment. First of all the change of variable y = e−t x transforms the given problem (7.4)–(7.6) into Z t d (A + 1)x(t) − x(t) = k1 (t − s)Ax(s)ds + f1 (t)z + h1 (t), (7.7) dt 0 (A + 1)x(0) = (A + 1)y0 , (7.8) Φ[(A + 1)x(t)] = g1 (t).

(7.9)

where k1 (t) = et k(t), f1 (t) = et f (t), h1 (t) = et h(t), g1 (t) = et g(t). If −1 is a simple pole for (A − λ)−1 , so that X = N (A + 1) ⊕ R(A + 1), and P denotes the projection onto N (A + 1), problem (7.7)–(7.9) reduces to d (A + 1)(1 − P )x(t) − (1 − P )x(t) dt Z t (7.10) = k1 (t − s)A(1 − P )x(s)ds + f1 (t)(1 − P )z + (1 − P )h1 (t), 0

(A + 1)(1 − P )x(0) = (A + 1)(1 − P )y0 ,

(7.11)

Φ[(A + 1)(1 − P )x(t)] = g1 (t),

(7.12)

t

Z −P x(t) =

k1 (t − s)P (A + 1 − 1)x(s)ds + f1 (t)P z + P h1 (t) 0

Z =−

(7.13)

t

k1 (t − s)P x(s)ds + f1 (t)P z + P h1 (t). 0

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Since the restriction of A + 1 to R(A + 1) is boundedly invertible, the change of variable (A + 1)(1 − P )x(t) = ξ(t) transforms (7.10)–(7.12) into Z t d k1 (t − s)[1 − R]ξ(s)ds + f1 (t)(1 − P )z + (1 − P )h1 (t), ξ(t) − Rξ(t) = dt 0 (7.14) ξ(0) = (A + 1)(1 − P )y0 ,

(7.15)

Φ[ξ(t)] = g1 (t),

(7.16)

where R indicates the inverse of the restriction of A + 1 to R(A + 1). Therefore, if k is continuous in [0, T ], Φ[(1 − P )z] 6= 0, h ∈ C([0, T ]; X), (1 − P )y0 ∈ D(A), g ∈ C 1 ([0, T ]; C), problem (7.14)–(7.16) admits a unique strict solution (ξ, f1 ). Hence, we have a unique strict solution ((1 − P )x, f1 ) to (7.10)–(7.12). Since f1 is now known, we only remain to solve integral equation (7.13), that is uniquely solvable. Notice that this improves the preceding result, since condition Φ[z] 6= 0 is replaced by the weaker condition Φ[(1 − P )z] 6= 0. References [1] A. Favini, A. Lorenzi, G. Marinoschi, H. Tanabe; Perturbation methods and identification problems for degenerate evolution systems, contribution to the Seventh Congress of Romanian Mathematicians, Brasov, 2011, Eds. L. Beznea, V. Brinzanescu, M. Iosifescu, G. Marinoschi, R. Purice, D. Timotin, Publishing House of the Romanian Academy of Science, 88-96, 2013. [2] A. Favini, A. Lorenzi, H. Tanabe: Direct and inverse problems for systems of singular differential boundary value problems, Electronic Journal of Diffential Equations, 2012 (2012), 1-34. [3] A. Favini, A. Lorenzi, H. Tanabe; Degenerate integrodifferential equations of parabolic type with Robin boundary conditions: Lp -theory, preprint. [4] A. Favini, A. Lorenzi, H. Tanabe, A. Yagi; An Lp -approach to singular linear parabolic equations with lower order terms, Discrete and Continuous Dynamical Systems, 22 No.4, 989-1008. [5] A. Favini, H. Tanabe; Degenerate differential equations of parabolic type and inverse problems, Proceedings of Seminar on Partial Differential Equations in Osaka 2012, Osaka University, August 20-24, 2012, 89-100. [6] A. Favini and A. Yagi; Degenerate Differential Equations in Banach spaces, Marcel Dekker, Inc., New York-Barsel-Hong Kong, 1999. [7] H. Tanabe; Identification problem for degenerate parabolic equations, Proceedings of Seminar on Partial Differential Equations in Osaka 2012, Osaka University, August 20-24, 2012, 83-88. [8] A. Prilepko, D. G. Orlovsky, I. Vasin; Methods for Solving Inverse Problems in Mathematical Physics, Marcel Dekker, Inc., New York-Barsel-Hong Kong, 2000. Angelo Favini Dipartimento di Matematica, University of Bologna, Piazza di Porta San Donato 5, 40126, Bologna, Italy E-mail address: [email protected] Alfredo Lorenzi ` degli Studi di Milano via Saldini Dipartimento di Matematica “F. Enriques”, Universita 50, 20133 Milano, Italy E-mail address: [email protected] Hiroki Tanabe Takarazuka, Hirai Sanso 12-13, 665-0817, Japan E-mail address: [email protected]