Abstract. Let E be a Banach space. We consider a Cauchy problem of the type ( D k t u + P k−1

POLONICI MATHEMATICI LXV.1 (1996)

Partial differential equations in Banach spaces involving nilpotent linear operators by Antonia Chinn`i and Paolo Cubiotti (Messina)

Abstract. Let E be a Banach space. We consider a Cauchy problem of the type ( D ^k t u + P k−1

j=0

P

|α|≤m A j,α (D _t ^j D x ^α u) = f in R ⁿ⁺¹ ,

D ^j _t u(0, x) = ϕ j (x) in R ⁿ , j = 0, . . . , k − 1,

where each A j,α is a given continuous linear operator from E into itself. We prove that if the operators A j,α are nilpotent and pairwise commuting, then the problem is well-posed in the space of all functions u ∈ C ^∞ (R ⁿ⁺¹ , E) whose derivatives are equi-bounded on each bounded subset of R ⁿ⁺¹ .

Introduction. Let k, m, n ∈ N and let (E, k · k _E ) be a real or complex Banach space. Following [4], we denote by V (R ⁿ , E) the space of all functions u ∈ C ^∞ (R ⁿ , E) such that, for every non-empty bounded set Ω ⊆ R ⁿ , one has

kuk _Ω,E := sup

α∈N ⁿ ₀

sup

x∈Ω

kD ^α u(x)k _E < ∞,

where D ^α u = ∂ ^{α 1 +...+α n} u/∂x ^α ₁ ¹ . . . ∂x ^α _n ⁿ , α = (α 1 , . . . , α _n ) and N 0 = N ∪ {0}.

In the present paper, we are interested in the well-posedness in the space V (R ⁿ , E) of the Cauchy problem

(1)

( D _t ^k u + P k−1 j =0

P

|α|≤m A j,α (D ^j _t D _x ^α u) = f in R ⁿ⁺¹ ,

D _t ^j u(0, x) = ϕ j (x) in R ⁿ , j = 0, . . . , k − 1, where each A j,α is a given continuous linear operator from E into itself. We denote by L(E) the space of all continuous linear operators from E into

1991 Mathematics Subject Classification: Primary 35G10.

Key words and phrases: partial differential equations in Banach spaces, nilpotent operators.

[67]

itself, endowed with the usual norm kAk _L(E) = sup

kvk _E ≤1

kA(v)k _E .

Apparently, the only previous result on this subject is Theorem 1 of [4], where one assumes that

k−1 X

j=0

X

|α|≤m

kA _j,α k _L(E) < 1.

We wish here to prove another, independent result supposing that the oper- ators A j,α are nilpotent and pairwise commuting. However, a complete char- acterization of the well-posedness of the problem (1) in the space V (R ⁿ , E) remains still unknown.

We believe that such a characterization should be quite difficult. To support this, we now discuss a particularly simple case which shows the peculiarity of working in the space V (R ⁿ , E) rather than in the other spaces usually considered in the theory of linear partial differential equations.

Let σ be a non-negative real number. Denote by Γ ^(σ) (R ² ) the (real) Gevrey class in R ² of index σ. That is to say, Γ ^(σ) (R ² ) is the class of all real functions u ∈ C ^∞ (R ² ) such that, for every non-empty bounded set Ω ⊆ R ² , one has

λ>0 inf sup

(α,β)∈N ² ₀

sup

(t,x)∈Ω

 1

(α + β)!

 σ

λ ^α+β    

∂ ^α+β u(t, x)

∂t ^α ∂x ^β     < ∞.

Recall, in particular, that Γ ⁽¹⁾ (R ² ) coincides with the class of analytic func- tions in R ² . Also, observe that V (R ² , R) ⊆ Γ ⁽⁰⁾ (R ² ).

Given a real number a, consider now the differential operator P a : C ^∞ (R ² )

→ C ^∞ (R ² ) defined by

P _a (u) = ∂ ^k u

∂t ^k + a ∂ ^m u

∂x ^m .

Then, according to the classical work of Malgrange [3], we have P _a (C ^∞ (R ² )) = C ^∞ (R ² ) for every a ∈ R.

Analogously, we have

P _a (Γ ^(σ) (R ² )) = Γ ^(σ) (R ² ) for every a ∈ R and σ ∈ [0, 1[ ∪ ([1, ∞[∩Q).

Precisely, this follows from Theorems 9.4 and 9.6 of [5] (see also p. 408 and pp. 467–468) when σ ∈ [0, 1[, and from Theorem 4.1 of [1] when σ ∈ [1, ∞[∩Q (in the case σ = 1 the result was previously proved in [2]).

Now, we come to the space V (R ² , R) (for short V (R ² )). On the basis of Theorem 4 of [4], we have

P a (V (R ² )) = V (R ² ) if and only if a 6= ±1.

1. The result. Our result is the following.

Theorem 1. Let k, n, m ∈ N, and let {A j,α } _j =0,...,k−1, α∈N ⁿ ₀ , |α|≤m be a family of pairwise commuting elements of L(E) such that for some q ∈ N one has

kA ^q _j,α ^¯ k L(E) = 0 for each j = 0, 1, . . . , k − 1 and α ∈ N ⁿ ₀ with |α| ≤ m.

Then for each f ∈ V (R ⁿ⁺¹ , E) and each ϕ ₀ , ϕ ₁ , . . . , ϕ _k−1 ∈ V (R ⁿ , E) there exists a unique function u ∈ V (R ⁿ⁺¹ , E) such that for each t ∈ R and each x ∈ R ⁿ one has

(2)

( D _t ^k u(t, x) + P k−1 j=0

P

|α|≤m A _j,α (D _t ^j D ^α _x u(t, x)) = f (t, x), D _t ^j u(0, x) = ϕ j (x), j = 0, . . . , k − 1.

Moreover , if p is the cardinality of the set {A _j,α : A _j,α 6= 0, j = 0, . . . , k − 1, α ∈ N ⁿ ₀ , |α| ≤ m} and s := k ² pq, then for each bounded set Ω ⊆ R ⁿ , and each r ≥ 0 and λ > 0, if one puts

σ := max n λ,

k−1 X

j =0

X

|α|≤m

λ ^j−k+1 kA j,α k L(E)

o , one has the following inequality:

0≤i≤k−1 max λ ⁻ⁱ kD _t ⁱ uk _{[−r,r]×Ω,E}

≤ min n ( max

0≤j≤¯ s−1 σ ^j ) max

0≤i≤k−1 (λ ⁻ⁱ kϕ _i k _Ω,E )e ^rσ + 

( max

0≤j≤¯ s−1 σ ^j )re ^rσ +

¯

X s−1 j=0

σ ^j 

λ ^1−k kf k _{[−r,r]×Ω,E} ,

e ^r  ( max

0≤j≤¯ s−1 σ ^j )( max

0≤i≤k−1 λ ⁻ⁱ kϕ i k Ω,E ) +  ^s−1 X ^¯

j=0

σ ^j 

λ ^1−k kf k _{0}×Ω,E o .

Before giving the proof of Theorem 1, we need some preliminary results.

Proposition 2. Let A ∈ L(E), B ∈ V (R, E) and v ∈ C ¹ (R, E) be such that P ∞

n=1 kA ⁿ k _L(E) < ∞ and

v ^′ (t) = A(v(t)) + B(t) in R.

Then v ∈ V (R, E) and for each r ≥ 0 the following inequality holds:

(3) kvk [−r,r],E ≤ min n ( sup

n∈N 0

kA ⁿ k L(E) )kv(0)k E e ^{rkAk L (E)}

+  ( sup

n∈N 0

kA ⁿ k _L(E) )re ^{rkAk L (E)} + X ∞ n=0

kA ⁿ k _L(E) 

kBk _[−r,r],E ,

e ^r  ( sup

n∈N 0

kA ⁿ k _L(E) )kv(0)k _E +  X ^∞

n=0

kA ⁿ k _L(E) 

kBk _0,E o , where we put kA ⁰ k _L(E) = 1.

P r o o f. First, observe that sup _n∈N kA ⁿ k L(E) ≤ P ∞

n=1 kA ⁿ k L(E) < ∞.

Since B ∈ C ^∞ (R, E) we get v ∈ C ^∞ (R, E) and, arguing by induction, we have

v ^(m) (t) = A ^m (v(t)) +

m−1 X

j=1

A ^j (B ^(m−j−1) (t)) + B ^(m−1) (t) for all t ∈ R and m ∈ N with m ≥ 2. Now, fix any r ≥ 0. We get kv ^(m) (t)k _E

≤ kA ^m k _L(E) kv(t)k _E +

m−1 X

j=1

kA ^j k _L(E) kB ^(m−j−1) (t)k _E + kB ^(m−1) (t)k _E

≤ sup

n∈N

kA ⁿ k L(E) sup

t∈[−r,r]

kv(t)k E +  X ^∞

n=1

kA ⁿ k L(E) + 1 

kBk [−r,r],E

for each t ∈ [−r, r] and m ≥ 2. It is easy to see that the last inequality also holds for t ∈ [−r, r] and m = 1. Hence

kv ^(m) (t)k _E

≤ max{1, sup

n∈N

kA ⁿ k L(E) } sup

t∈[−r,r]

kv(t)k E +  X ^∞

n=0

kA ⁿ k L(E)



kBk [−r,r],E

for each t ∈ [−r, r] and m ∈ N 0 . Consequently, v ∈ V (R, E) and (4) kvk [−r,r],E

≤ sup

n∈N ₀

kA ⁿ k _L(E) sup

t∈[−r,r]

kv(t)k _E +  X ^∞

n=0

kA ⁿ k _L(E) 

kBk _[−r,r],E for each r ≥ 0. In particular, by (4) we get

kvk 0,E ≤ ( sup

n∈N ₀

kA ⁿ k L(E) )kv(0)k _E +  X ^∞

n=0

kA ⁿ k L(E)



kBk 0,E .

By Proposition 2 of [4] we have kvk [−r,r],E ≤ e ^r kvk 0,E , hence

(5) kvk _[−r,r],E

≤ e ^r  ( sup

n∈N 0

kA ⁿ k _L(E) )kv(0)k E +  X ^∞

n=0

kA ⁿ k _L(E)  kBk 0,E

 . On the other hand, since v(t) = v(0) +

T

t

0 A(v(τ )) dτ +

T

t

0 B(τ ) dτ we get kv(t)k _E ≤ kv(0)k _E + rkBk _[−r,r],E + kAk _L(E) 

t

\

0

kv(τ )k _E dτ   

for every t ∈ [−r, r]. By Gronwall’s lemma, we get

(6) kv(t)k E ≤ (kv(0)k E + rkBk [−r,r],E ) e ^{rkAk L(E)} for all t ∈ [−r, r]. By (4) and (6) we get

(7) kvk _[−r,r],E

≤ ( sup

n∈N 0

kA ⁿ k L(E)

kv(0)k E e ^{rkAk L(E)}

+  ( sup

n∈N ₀

kA ⁿ k _L(E) )re ^{rkAk L (E)} + X ∞ n=0

kA ⁿ k _L(E) 

kBk _[−r,r],E . Our claim follows easily from (5) and (7).

We point out that the operator A satisfies P ∞

n=1 kA ⁿ k L(E) < ∞ if, for instance, kAk _L(E) < 1 or if there is some m ∈ N such that A ^m = 0 for all m ≥ m. When the former situation occurs, Proposition 2 reduces to Proposition 4 of [4], while in the latter case from Proposition 2 we get (8) kvk _[−r,r],E

≤ min n

( max

0≤j≤ ¯ m−1 kA ^j k _L(E) )kv(0)k _E e ^{rkAk L (E)} + 

( max

0≤j≤ ¯ m−1 kA ^j k _L(E) )re ^{rkAk L (E)} +

¯ m−1 X

j=0

kA ^j k _L(E) 

kBk _[−r,r],E ,

e ^r 

( max

0≤j≤ ¯ m−1 kA ^j k _L(E) )kv(0)k _E +  ^m−1 X ^¯

j=0

kA ^j k _L(E)  kBk 0,E

o . Proposition 3. Let k ∈ N with k ≥ 2, and let A ₀ , A ₁ , . . . , A _k−1 ∈ L(E) be pairwise commuting operators. Assume that there exists m ^∗ ∈ N such that A ^m _j ^∗ = 0 for each j = 0, 1, . . . , k − 1. Let λ > 0, and consider the operator A : E ^k → E ^k defined by

A(y) = 

λy ₁ , λy ₂ , . . . , λy _k−1 ,

k−1 X

j=0

λ ^j−k+1 A _j (y _j ) 

for each y = (y 0 , y 1 , . . . , y k−1 ) ∈ E ^k . Then A ^{m ¯} = 0 for m = k ² m ^∗ .

P r o o f. We divide the proof into several steps.

F i r s t s t e p. Let y = (y 0 , y 1 , . . . , y k−1 ) ∈ E ^k and s ∈ {1, . . . , k} be fixed. Let us show that if one puts A ^s (y) = (x 0 , x 1 , . . . , x k−1 ), then the vector x j can be represented in the following way:

(9) x j =

m j

X

m=1

µ j,m (λ)A ⁿ ₀ ^j,m,0 A ⁿ ₁ ^j,m,1 . . . A ⁿ _k−1 ^j,m,k−1 (y r(j,m) ) if j = k − s, . . . , k − 1 (with m _j ≥ 1, and P k−1

l=0 n _j,m,l ≥ 1, r(j, m) ∈ {0, 1, . . . , k − 1} for each m = 1, . . . , m _j ), and

x j = λ ^s y j+s

if j = 0, . . . , k − s − 1 (if s < k). To prove our claim, we argue by induction on s. Of course, our claim is true for s = 1. Now, assume that it is true for s = i (with i < k) and let us show that it remains true for s = i + 1. By assumption, if we put A ⁱ (y) = (e x 0 , e x 1 , . . . , e x k−1 ), then we have

x e _j =

˜ m j

X

m=1

µ e _j,m (λ)A ⁿ ₀ ^{˜ j,m,0} A ⁿ ₁ ^{˜ j,m,1} . . . A ⁿ _k−1 ^{˜ j,m,k−1} (y _{˜ r(j,m)} ) if j = k − i, . . . , k − 1 (with e m j ≥ 1, and P k−1

l=0 n e j,m,l ≥ 1, e r(j, m) ∈ {0, 1, . . . , k − 1} for each m = 1, . . . , e m _j ), and

e

x _j = λ ⁱ y j+i

if j = 0, . . . , k − i − 1. Put A ⁱ⁺¹ (y) = (w ₀ , w ₁ , . . . , w _k−1 ). We get w k−1 =

k−1 X

d=0

λ ^d−k+1 A _d (e x _d )

=

k−i−1 X

d=0

λ ^d−k+1 A _d (λ ⁱ y d+i )

+

k−1 X

d=k−i

λ ^d−k+1 A _d  X ^{m ˜ d}

m=1

e µ _d,m (λ)A ⁿ ₀ ^{˜ d,m,0} A ⁿ ₁ ^{˜ d,m,1} . . .A ^{˜ n} _k−1 ^d,m,k−1 (y _{r(d,m) ˜} ) 

=

k−i−1 X

d=0

λ ^d−k+i+1 A _d (y _d+i )

+

k−1 X

d=k−i m ˜ d

X

m=1

λ ^d−k+1 µ e _d,m (λ)A _d (A ^{˜ n} ₀ ^d,m,0 A ^{˜ n} ₁ ^d,m,1 . . .A ⁿ _k−1 ^{˜ d,m,k−1} (y _{˜ r(d,m)} )),

hence it is easily seen that w k−1 is of the form (9). Now, let j ∈ {k − i − 1,

. . . , k − 2}. We get w _j = λe x _j+1 = λ

˜ m X j+1

m=1

µ e _j+1,m (λ)A ⁿ ₀ ^{˜ j+1,m,0} A ^{˜ n} ₁ ^j+1,m,1 . . . A ⁿ _k−1 ^{˜ j+1,m,k−1} (y _{r(j+1,m) ˜} ), hence w _j is of the form (9) even for j = k −i−1, . . . , k −2. Thus, if i = k −1, our claim follows. If i < k − 1, for each j = 0, . . . , k − i − 2 we have

w j = λe x j+1 = λ ⁱ⁺¹ y j+i+1 , as desired.

S e c o n d s t e p. We prove that for each fixed y = (y 0 , y 1 , . . . , y k−1 ) ∈ E ^k and s ∈ N, if we put A ^sk (y) = (z ₀ , z ₁ , . . . , z _k−1 ), then for each j ∈ {0, 1, . . . , k − 1} the vector z _j can be represented in the following way:

z _j =

b j

X

m=1

σ _j,m (λ)A ^p ₀ ^j,m,0 A ^p ₁ ^j,m,1 . . . A ^p _k−1 ^j,m,k−1 (y _v(j,m) ) with b j ≥ 1 and P k−1

l=0 p j,m,l ≥ s, v(j, m) ∈ {0, 1, . . . , k − 1} for each m = 1, . . . , b _j . Again, we argue by induction. First, we observe that by the first part of the proof, if we put A ^k (y) = (u 0 , u 1 , . . . , u k−1 ), then for each j = 0, 1, . . . , k − 1 the vector u j can be represented in the form

u j =

¯ m j

X

m=1

µ _j,m (λ)A ⁿ ₀ ^{¯ j,m,0} A ⁿ ₁ ^{¯ j,m,1} . . . A ⁿ _k−1 ^{¯ j,m,k−1} (y ¯ r(j,m) ) with m _j ≥ 1 and P k−1

l=0 n _j,m,l ≥ 1, r(j, m) ∈ {0, 1, . . . , k − 1} for each m = 1, . . . , m _j , hence our claim is true for s = 1. Assume that it is true for s = i, and let us show that it is true for s = i + 1. Thus, if we put A ^ik (y) = (b z 0 , b z 1 , . . . , b z k−1 ), then for each j = 0, 1, . . . , k − 1 the vector b z _j can be represented in the following way:

(10) b z j =

ˆ b j

X

m=1

b

σ j,m (λ)A ^p ₀ ^{ˆ j,m,0} A ^p ₁ ^{ˆ j,m,1} . . . A ^p _k−1 ^{ˆ j,m,k−1} (y v(j,m) ˆ ) with bb _j ≥ 1 and P k−1

l=0 p b _j,m,l ≥ i, bv(j, m) ∈ {0, 1, . . . , k − 1} for each m = 1, . . . ,bb _j . From the first part of the proof, if we put A ^(i+1)k (y) = A ^k (A ^ik (y)) = ( b w 0 , b w 1 , . . . , b w k−1 ), then for each j = 0, 1, . . . , k − 1 we have

w b _j =

ˆ m j

X

m=1

µ b _j,m (λ) A ^{ˆ n} ₀ ^j,m,0 A ^{ˆ n} ₁ ^j,m,1 . . . A ⁿ _k−1 ^{ˆ j,m,k−1} (b z _{ˆ r(j,m)} ) with b m j ≥ 1 and P k−1

l=0 b n j,m,l ≥ 1, b r(j, m) ∈ {0, 1, . . . , k − 1} for each

m = 1, . . . , b m _j . By (10), for each j = 0, 1, . . . , k − 1 we get

b w j =

m ˆ j

X

m=1 ˆ b ˆ r(j,m)

X

d=1

b

µ j,m (λ) b σ r(j,m),d ˆ (λ)

·A ^{ˆ n} ₀ ^{j,m,0 +ˆ p r(j,m),d,0 ˆ} A ^{ˆ n} ₁ ^{j,m,1 +ˆ p r(j,m),d,1 ˆ} . . . A ^{n ˆ j,m,k−1 +ˆ p ˆ} r(j,m),d,k−1

k−1 (y v(ˆ ˆ r(j,m),d) ).

Since for each m ∈ {1, . . . , b m _j } and d ∈ {1, . . . , b _{r(j,m) ˆ} } we have P k−1 l=0 b n _j,m,l + b p ˆ r(j,m),d,l ≥ i + 1, our claim follows.

T h i r d s t e p. We claim that A ^{k 2 m ∗} = 0. To see this, choose any y = (y 0 , y 1 , . . . , y k−1 ) ∈ E ^k . If we put A ^{k 2 m ∗} (y) = ( e w 0 , e w 1 , . . . , e w k−1 ), from the second part of the proof we see that for each j = 0, 1, . . . , k − 1 the vector

e

w j can be represented in the following way:

w e _j =

˜ b j

X

m=1

σ e _j,m (λ)A ^p ₀ ^{˜ j,m,0} A ^p ₁ ^{˜ j,m,1} . . . A ^p _k−1 ^{˜ j,m,k−1} (y _{v(j,m) ˜} )

with eb _j ≥ 1 and P k−1

l=0 p e _j,m,l ≥ km ^∗ , ev(j, m) ∈ {0, 1, . . . , k − 1} for each m = 1, . . . ,eb _j . Now, it is easy to see that for each fixed j ∈ {0, 1, . . . , k − 1}

and m ∈ {1, . . . ,eb j } there exists e l ∈ {0, 1, . . . , k − 1} such that e p _{j,m,˜ l} ≥ m ^∗ . Hence, we conclude that e w _j = 0 for all j = 0, 1, . . . , k − 1. This completes the proof.

Proposition 4. Let k ∈ N and let A 0 , A 1 , . . . , A k−1 ∈ L(E) be pairwise commuting operators. Assume that there exists m ^∗ ∈ N such that A ^m _j ^∗ = 0 for each j = 0, 1, . . . , k − 1. Then for each B ∈ V (R, E) and for each w 0 , w 1 , . . . , w k−1 ∈ E, there exists a unique v ∈ V (R, E) such that

(11)

(

v ^(k) (t) = P k−1

j =0 A j (v ^(j) (t)) + B(t) for all t ∈ R,

v ^(j) (0) = w _j for j = 0, 1, . . . , k − 1.

Moreover , if m := k ² m ^∗ , then for each fixed r ≥ 0 and λ > 0, if one puts c _A = max n

λ,

k−1 X

j=0

λ ^j−k+1 kA _j k _L(E) o , one has

(12) max

0≤i≤k−1 λ ⁻ⁱ kv ⁽ⁱ⁾ k _[−r,r],E

≤ min n

( max

0≤j≤ ¯ m−1 c ^j _A ) max

0≤i≤k−1 λ ⁻ⁱ kw _i k _E e ^{rc A}

+ 

( max

0≤j≤ ¯ m−1 c ^j _A )re ^{rc A} +

¯ m−1 X

j=0

c ^j _A 

λ ^1−k kBk _[−r,r],E ,

e ^r 

( max

0≤j≤ ¯ m−1 c ^j _A ) max

0≤i≤k−1 λ ⁻ⁱ kw _i k _E +  ^{m−1 ¯} X

j=0

c ^j _A 

λ ^1−k kBk _0,E o . P r o o f. If k = 1, our claim follows by Picard–Lindel¨of’s theorem and Proposition 2. Now, let k ≥ 2, and consider the space E ^k endowed with the norm

kyk _E ^k = max

0≤j≤k−1 ky _i k _E ,

where y = (y ₀ , y ₁ , . . . , y _k−1 ). Fix λ > 0, and let A : E ^k → E ^k be defined by setting

A(y ₀ , y ₁ , . . . , y _k−1 ) = 

λy ₁ , λy ₂ , . . . , λy _k−1 ,

k−1 X

j=0

λ ^j−k+1 A _j (y _j ) 

for each y ∈ E ^k . Now, observe that for each y ∈ E ^k one has kA(y)k _E k ≤ max n

λ,

k−1 X

j=0

λ ^{j −k+1} kA _j k _L(E) o

kyk _E k = c _A kyk _E k . Thus, A ∈ L(E ^k ) and kAk _L(E ^k ₎ ≤ c A . By Proposition 3 one has A ^{m ¯} = 0, where m = k ² m ^∗ . By Picard–Lindel¨of’s theorem, there exists a unique v ∈ C ^k (R, E) such that

(

v ^(k) (t) = P k−1

j =0 A _j (v ^(j) (t)) + B(t) for t ∈ R,

v ^(j) (0) = w _j for all j = 0, 1, . . . , k − 1.

Let Γ : R → E ^k and ω : R → E ^k be defined by setting for each t ∈ R, Γ (t) = (0, 0, . . . , λ ^1−k B(t)),

ω(t) = (v(t), λ ⁻¹ v ^′ (t), λ ⁻² v ^′′ (t), . . . , λ ^1−k v ^(k−1) (t)).

It is easy to see that Γ ∈ V (R, E ^k ), ω ∈ C ¹ (R, E ^k ) and ω ^′ (t) = A(ω(t)) + Γ (t) for all t ∈ R.

By Proposition 2 we get ω ∈ V (R, E ^k ), hence v ∈ V (R, E). Moreover, (8) gives

0≤i≤k−1 max λ ⁻ⁱ kv ⁽ⁱ⁾ k _[−r,r],E = kωk _[−r,r],E k

≤ min n

( max

0≤j≤ ¯ m−1 kA ^j k _L(E ^k ₎ )kω(0)k _E ^k e ^{rkAk L(Ek )}

+ 

( max

0≤j≤ ¯ m−1 kA ^j k _L(E k ) )re ^{rkAk L (Ek )} +

¯ m−1 X

j =0

kA ^j k _L(E k )

 kΓ k _[−r,r],E k ,

e ^r 

( max

0≤j≤ ¯ m−1 kA ^j k _L(E k ) )kω(0)k _E k +  ^m−1 X ^¯

j=0

kA ^j k _L(E k )

 kΓ k _0,E k

o

for every r ≥ 0. Since kA ^j k _L(E ^k ₎ ≤ kAk ^j _L(E k ) ≤ c ^j _A for each j = 1, . . . , m−1, our claim follows at once.

P r o o f o f T h e o r e m 1. First, we denote by F 1 , . . . , F _p the elements of the set {A j,α : A j,α 6= 0, j = 0, . . . , k − 1, α ∈ N ⁿ ₀ , |α| ≤ m}. Fix f ∈ V (R ⁿ⁺¹ , E) and ϕ 0 , ϕ 1 , . . . , ϕ k−1 ∈ V (R ⁿ , E), where the space V (R ⁿ , E) will be considered with any norm k · k _Ω,E . For each j = 0, 1, . . . , k − 1, v ∈ V (R ⁿ , E) and x ∈ R ⁿ , let

T j (v)(x) = − X

|α|≤m

A j,α (D ^α v(x)).

By Proposition 6 of [4] we have T j ∈ L(V (R ⁿ , E)) and kT j k _{L(V (R} ⁿ _,E)) ≤ P

|α|≤m kA _j,α k _L(E) . Consider the problem (13)

( ω ^(k) (t) = P k−1

j=0 T _j (ω ^(j) (t)) + Ψ ⁻¹ (f )(t) in R,

ω ^(j) (0) = ϕ j for j = 0, 1, . . . , k − 1, where Ψ : V (R, V (R ⁿ , E)) → V (R ⁿ⁺¹ , E) is the function defined as in Proposition 3 of [4]. Namely, Ψ(g)(t, x) = g(t)(x) for g ∈ V (R, V (R ⁿ , E)), t ∈ R, and x ∈ R ⁿ . Now, it is easily seen that the operators T j are pairwise commuting. We claim that

kT _j ^{p¯ q} k _{L(V (R} ⁿ _,E)) = 0 for each j = 0, 1, . . . , k − 1.

To see this, let j ∈ {0, 1, . . . , k − 1}, v ∈ V (R ⁿ , E) and x ∈ R ⁿ be fixed, and let {A _j,α(j,i) } ^s _i=1 ^j be the elements of the family {A _j,α } _α∈N ⁿ

0 ,|α|≤m that are different from the origin of L(E). Thus, we have

T j (v)(x) = −

s j

X

i=1

A j,α(j,i) (D ^α(j,i) v(x)).

Now we show that for each h ∈ N the vector T _j ^h (v)(x) can be represented as follows:

(14) T _j ^h (v)(x) = (−1) ^h

b(h) X

l=1

A ^r(h,l,1) _j,α(j,1) A ^r(h,l,2) _j,α(j,2) . . . . . . A ^r(h,l,s _j,α(j,s ^{j )}

j ) (D r(h,l,1)α(j,1)+r(h,l,2)α(j,2)+...+r(h,l,s j )α(j,s j ) v(x))

for suitable b(h) ∈ N and r(h, l, 1), r(h, l, 2), . . . , r(h, l, s _j ) ∈ N with

s j

X

i=1

r(h, l, i) = h for each l ∈ {1, . . . , b(h)}.

We argue by induction. Of course, our claim is true for h = 1, with r(1, l, d) = 1 if d = l, while r(1, l, d) = 0 if d 6= l. Now, assume that our claim is true for some h ∈ N. We have

T _j ^h+1 (v)(x)

= T j (T _j ^h (v))(x) = −

s j

X

i=1

A j,α(j,i) (D ^α(j,i) T _j ^h (v)(x))

=

s j

X

i=1

(−1) ^h+1

b(h) X

l=1

A ^r(h,l,1) _j,α(j,1) A ^r(h,l,2) _j,α(j,2) . . .

. . . A ^r(h,l,s _j,α(j,s ^{j )}

j ) A j,i (D ^α(j,i) D r(h,l,1)α(j,1)+r(h,l,2)α(j,2)+...+r(h,l,s j )α(j,s j ) v(x)).

Now, it is easy to see that T _j ^h+1 (v)(x) is also of the form (14). Hence, our claim is true for h + 1, hence it is true for all h ∈ N. In particular, if h = pq, the representation (14) holds, with P s j

i=1 r(h, l, i) = pq for each fixed l ∈ {1, . . . , b(pq)}. Observe that, for each fixed l ∈ {1, . . . , b(pq)}, we have

A ^r(p¯ _j,α(j,1) ^q,l,1) A ^r(p¯ _j,α(j,2) ^q,l,2) . . . A ^r(p¯ _j,α(j,s ^{q,l,s j )}

j ) = F ₁ ^{q 1} . . . F _p ^{q p} with

X p i=1

q _i =

s j

X

d=1

r(pq, l, d) = pq.

Of course, this implies that there exists i ∈ {1, . . . , p} such that q i ≥ q, hence F _i ^{q i} = 0. Therefore, for each fixed l ∈ {1, . . . , b(qh)}, the operator

A ^r(p¯ _j,α(j,1) ^q,l,1) A ^r(p¯ _j,α(j,2) ^q,l,2) . . . A ^r(p¯ _j,α(j,s ^{q,l,s j )}

j )

identically vanishes, hence T _j ^h (v)(x) = 0 E . The arbitrariness of v ∈ V (R ⁿ , E) and x ∈ R ⁿ gives

kT _j ^{p¯ q} k L(V (R ⁿ ,E)) = 0 for each j = 0, 1, . . . , k − 1.

By Proposition 4 there exists a unique ω ∈ V (R, V (R ⁿ , E)) satisfying (13).

Now, if t ∈ R and x ∈ R ⁿ , we have

ω ^(k) (t)(x) = Ψ (ω ^(k) )(t, x) =

k−1 X

j=0

Ψ (T j ◦ ω ^(j) )(t, x) + f (t, x)

= −

k−1 X

j=0

X

|α|≤m

A j,α (D ^α ω ^(j) (t)(x)) + f (t, x), ω ^(j) (0)(x) = ϕ _j (x) for all j = 0, 1, . . . , k − 1.

If we put u = Ψ (ω), observing that by Proposition 3 of [4] we have D _t ^j Ψ (ω) = Ψ (D ^j ω), for each t ∈ R and x ∈ R ⁿ we get

( D _t ^k u(t, x) + P k−1 j=0

P

|α|≤m A _j,α (D _t ^j D ^α _x u(t, x)) = f (t, x), D _t ^j u(0, x) = ϕ _j (x) for all j = 0, 1, . . . , k − 1.

Hence, u is a solution of problem (2). Conversely, reasoning as in [4] one can show that if e u solves (2), then e u = u. By Proposition 4, if s := k ² pq, r ≥ 0 and λ > 0, we get

0≤i≤k−1 max λ ⁻ⁱ kω ⁽ⁱ⁾ k [−r,r],V (R ⁿ ,E)

≤ min n ( max

0≤j≤¯ s−1 σ ^j ) max

0≤i≤k−1 (λ ⁻ⁱ kϕ _i k _{V (R} ⁿ _,E) )e ^rσ + 

( max

0≤j≤¯ s−1 σ ^j )re ^rσ +

¯

X s−1 j=0

σ ^j 

λ ^1−k kΨ ⁻¹ (f )k [−r,r],V (R ⁿ ,E) , e ^r 

( max

0≤j≤¯ s−1 σ ^j ) max

0≤i≤k−1 (λ ⁻ⁱ kϕ _i k _{V (R} ⁿ _,E) ) +  ^s−1 X ^¯

j =0

σ ^j 

λ ^1−k kΨ ⁻¹ (f )k _{0,V (R} n ,E)

o

≤ min n ( max

0≤j≤¯ s−1 σ ^j ) max

0≤i≤k−1 (λ ⁻ⁱ kϕ _i k _Ω,E )e ^rσ + 

( max

0≤j≤¯ s−1 σ ^j )re ^rσ +

¯

X s−1 j=0

σ ^j 

λ ^1−k kf k _{[−r,r]×Ω,E} , e ^r 

( max

0≤j≤¯ s−1 σ ^j ) max

0≤i≤k−1 (λ ⁻ⁱ kϕ _i k _Ω,E ) +  ^s−1 X ^¯

j =0

σ ^j 

λ ^1−k kf k _{0}×Ω,E o ,

where Ω is any non-empty bounded subset of R ⁿ . Since

kD _t ⁱ uk _{[−r,r]×Ω,E} = kω ⁽ⁱ⁾ k [−r,r],V (R ⁿ ,E) ,

our claim follows.

To conclude, we now present a simple example of application of The- orem 1 to integro-differential equations. Let Y ⊆ R ^m (m ∈ N) be a non- empty compact set. Following [4], denote by V ₀ (R ⁿ × Y ) the space of all functions u : R ⁿ × Y → R such that u( · , y) ∈ C ^∞ (R ⁿ ) for each y ∈ Y , D ^α _x u ∈ C ⁰ (R ⁿ × Y ) for each α ∈ N ⁿ ₀ and

sup

α∈N ⁿ ₀

sup

(x,y)∈Ω×Y

|D _x ^α u(x, y)| < ∞

for each bounded set Ω ⊆ R ⁿ . Also recall ([4], Proposition 8) that if, for u ∈ V (R ⁿ , C ⁰ (Y )) (C ⁰ (Y ) is endowed with the usual sup-norm), Ψ n (u) denotes the function, from R ⁿ × Y into R, defined by

Ψ n (u)(x, y) = u(x)(y) (x ∈ R ⁿ , y ∈ Y ),

then Ψ _n (u) ∈ V 0 (R ⁿ × Y ), the mapping u → Ψ _n (u) is surjective, and, for each α ∈ N ⁿ ₀ , one has D ^α _x Ψ n (u) = Ψ n (D ^α u).

Theorem 2. Let k, n, m ∈ N and let {g j,α } j=0,...,k−1, α∈N ⁿ ₀ ,|α|≤m be a family of continuous real functions on Y each of which satisfies

T

Y g j,α (y) dy

= 0. Then, for each f ∈ V 0 (R ⁿ⁺¹ × Y ) and ϕ 0 , ϕ 1 , . . . , ϕ k−1 ∈ V 0 (R ⁿ × Y ), there exists a unique function u ∈ V 0 (R ⁿ⁺¹ × Y ) such that, for each t ∈ R, x ∈ R ⁿ and y ∈ Y , one has

( D _t ^k u(t, x, y) + P k−1 j=0

P

|α|≤m



Y D ^j _t D ^α _x u(t, x, ξ) dξ 

g _j,α (y) = f (t, x, y), D _t ^j u(0, x, y) = ϕ j (x, y), j = 0, . . . , k − 1.

P r o o f. For each j = 0, . . . , k − 1, α ∈ N ⁿ ₀ , let A j,α be the element of L(C ⁰ (Y )) defined by putting

A _j,α (v)(y) = 

Y

v(ξ) dξ  g _j,α (y)

for all v ∈ C ⁰ (Y ) and y ∈ Y . Clearly, A _j,α ◦A _j ^′ _,α ^′ = A _j ^′ _,α ^′ ◦A _j,α = 0. Conse- quently, by Theorem 1, there exists a unique function w ∈ V (R ⁿ⁺¹ , C ⁰ (Y )) such that, for each t ∈ R and x ∈ R ⁿ , one has

( D _t ^k w(t, x) + P k−1 j =0

P

|α|≤m A _j,α (D _t ^j D _x ^α w(t, x)) = Ψ _n+1 ⁻¹ (f )(t, x), D _t ^j w(0, x) = Ψ _n ⁻¹ (ϕ _j )(x), j = 0, . . . , k − 1.

Then the function u = Ψ n+1 (w) satisfies the conclusion.

References

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Rend. Sem. Mat. Univ. Politec. Torino (1983), special issue on “Linear partial and

pseudodifferential operators”, 81–89.

[2] L. C a t t a b r i g a and E. D e G i o r g i, Una dimostrazione diretta dell’esistenza di soluzioni analitiche nel piano reale di equazioni a derivate parziali a coefficienti costanti , Boll. Un. Mat. Ital. (4) 4 (1971), 1015–1027.

[3] B. M a l g r a n g e, Existence et approximation des solutions des ´equations aux d´eriv´ees partielles et des ´ equations de convolution, Ann. Inst. Fourier (Grenoble) 6 (1955–56), 271–355.

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IA Math. 38 (1991), 623–640.

[5] F. T r`ev e s, Linear Partial Differential Equations with Constant Coefficients, Gordon and Breach, 1966.

Department of Mathematics University of Messina

98166 Sant’Agata-Messina, Italy E-mail: cubiotti@mat520.unime.it

Re¸ cu par la R´ edaction le 30.9.1995