A SOLUTION OF THE CAUCHY PROBLEM IN THE CLASS OF ABSOLUTELY CONTINUOUS DISTRIBUTION–VALUED

FUNCTIONS

by Margareta Wiciak

Abstract. The aim of the paper is to give an exact formula for the solution of an evolution problem with matrix coefficients, and initial condition and external forces being tempered distributions.

1. Introduction. LetS be the Schwartz space,

S:={ϕ∈ C^{∞}(R^{n},C) : P ·D^{α}ϕis bounded ∀α∈N^{n} ∀P ∈ P(R^{n})},
where P(R^{n}) denotes the set of all polynomialsR^{n}→R. S is a Fr´echet space
with the topology induced by the family of seminorms

q_{m}(ϕ) := sup

x∈R^{n}

sup

|α|≤m

(1 +|x|^{2})^{m}|D^{α}ϕ(x)|, m= 0,1,2, . . .

Write snS for the cone of all continuous seminorms onS. LetB be a complex
Banach space and D(R^{n}) stand for the space of all complex test functions on
R^{n}. A distribution T :D(R^{n})→B (T ∈ D^{0}(R^{n};B)) is tempered when it has
the unique continuous extension T :S →B (T ∈ D^{0}_{temp}).

Let J be an interval inR. We will consider the Cauchy problem (1)

( _{d}

dtu(t) = P

α

Aα(t)◦D^{α}u(t) +f(t) for a.e. t∈J
u(t0) = u0

with given t0 ∈ J, initial condition u0 ∈ D^{0}_{temp} and external forces f : J 7→

D_{temp}^{0} . In order to prove the main theorem (Th. 17) on the existence and
uniqueness of (1) we use the method used by K. Holly in the case of scalar
coefficients Aα:J 7→C, [4]. This method is based on the integration of func-
tions whose values are tempered distributions. In the second part of Section
2 we give a brief exposition of Holly’s theory ([3]) of absolutely continuous

distribution-valued functions and their integrals (Def. 7 – Th. 16). In the first part of Section 2 we will be concerned with the notion of the product of a distribution and a vector-valued function. Having this product and a product integral in a Fr´echet space (see [6]), we obtain the exact formula (11) for the solution of problem (1), being an extension of a similar formula in the scalar case.

In the case of time-independed coefficientsA_{α}, we rewrite the main theorem
as Theorem 22, replacing abstract assumption (9) with easy to check spectral
condition (25).

In Section 4, some applications are indicated.

2. Preliminaries. We begin by recalling the notion of the product of a
distribution and a scalar-valued function. Let Ω ∈ topR^{n}, T ∈ D^{0}(Ω;B),
ψ∈ C^{∞}(Ω). Then

ψT : D(Ω)3ϕ7−→^{def} T(ϕψ)∈B.

Now we extend the product to the case of a vector-valued function. Assume
that B is a complex vector space. Let End B denote the space of all linear
endomorphisms on B and{b_{1}, . . . , b_{k}}be the basis of End B.

Definition 1. LetT ∈ D^{0}(Ω;B),η ∈ C^{∞}(Ω, End B). Then
η·T :=

k

X

j=1

(b^{?}_{j} ◦η) (bj◦T).

The above definition does not depend on the choice of a basis of End B.

Moreover,η·T ∈ D^{0}(Ω;B)) and for u∈L^{1}_{loc}(Ω, B) there isη·[u] = [ηu], where
[u] denotes the regular distribution

[u](ϕ) :=

Z

Ω

u(x)ϕ(x)dx forϕ∈ D(Ω).

The product has properties analogous to those in the scalar case. We have pointed out two of them which are strickly connected to a vector-valued function.

Remark 2. Let T ∈ D^{0}(Ω;B), η ∈ C^{∞}(Ω, End B) and ψ ∈ C^{∞}(Ω,K).

Then

η·T =

k

P

j=1

bj ◦((b^{?}_{j} ◦η)T),
ψ(ηT) = (ψη)T =η(ψT).

Proof. We prove the first equality. The proof of the second is analogous.

Let ϕ∈ D(Ω).

(ηT)(ϕ) =

k

X

j=1

(b^{?}_{j} ◦η)(bj ◦T)

(ϕ) =

k

X

j=1

(bj◦T)((b^{?}_{j} ◦η)ϕ) =

=

k

X

j=1

bj(((b^{?}_{j} ◦η)T)(ϕ)) =

k

X

j=1

(bj◦(b^{?}_{j} ◦η)T)(ϕ).

As in the scalar case the following theorem holds.

Theorem 3. Let T ∈ D_{temp}^{0} , η∈ C^{∞}(R^{n}, End B) be polynomially bounded
together with all of its derivatives. Then ηT ∈ D_{temp}^{0} and

ηT =

k

X

j=1

(b^{?}_{j} ◦η) (bj◦T).

Proof. Since bj ◦T :S →B is linear continuous and b^{?}_{j} ◦η :R^{n} → Kis
polynomially bounded together with all of its derivatives for anyj∈ {1, . . . , k},

k

P

j=1

(b^{?}_{j} ◦η) (b_{j}◦T) :S →B is also linear continuous. Clearly,

k

X

j=1

(b^{?}_{j}◦η) (b_{j} ◦T)_{|D(R}^{n}_{)}=ηT.

Let Hol:=Hol(C^{n}, End B) stand for the space of all End B-valued holo-
morphic functions with the topology of local uniform convergence. Setting for
any h, f ∈ Hol: h·f : Ω 3 z 7−→^{def} h(z)◦f(z) ∈ End B, Hol is a Fr´echet
algebra (see [6]).

Definition 4. Letη∈ Hol,T ∈ D^{0}_{temp}. Then
η·T :=η|R^{n}·T.

Let K be a compact subset ofR^{n}. FixN ∈Nand consider the space
(2) X :={T ∈ D_{temp}^{0} : suppT ⊂K, T isq_{N}-continuous }.

Then X is a Banach space with the norm|T|_{X} :=|T|_{q}_{N}.

Theorem 5.

(i) For any h∈ Hol, the mapping
(3) h˜:X 3T 7−→^{def} X

β∈N^{n}

1

β!D^{β}h(0)◦x^{β}T ∈X
is correctly defined, linear and continuous.

(ii) The map

˜:Hol3h7−→^{def} ˜h∈End X
is a continuous homomorphism of algebras.

Proof. Leth∈ Hol. Fixλ∈ D(R^{n}) such thatλ= 1 in the neighbourhood
of K. There is r > 0 such that suppλ ⊂ {|x| < r}. Fix r < R < ∞. Set
a_{β} := 1

β!D^{β}h(0)∈End B. Let T ∈X andϕ∈ S.

(4) |a_{β} ◦x^{β}T(ϕ)|=|(a_{β} ◦x^{β}T)(ϕ)| ≤ |a_{β}| · |T|_{q}_{N} ·qN(λx^{β}ϕ).

By the Cauchy inequalities,

(5) |a_{β}| ≤ p_{R}(h)

R^{|β|} ,
where p_{R}(h) := sup

|z_{i}|<R,∀i=1,...,n

kh(z)k_{End B}<∞. And by Leibniz’s formula,
(6) q_{N}(λx^{β}ϕ)≤ max

|α|≤N

X

γ≤α

α γ

kD^{γ}(λx^{β})k_{L}∞(suppλ)·q_{N}(ϕ).

Combining (4)–(6) we conclude that X

β∈N^{n}

1 β!

D^{β}h(0)◦x^{β}T

(ϕ) is convergent for all ϕ∈ S and

X

β∈N^{n}

1 β!

D^{β}h(0)◦x^{β}T

(ϕ)

≤M ·pR(h)· |T|_{q}_{N} ·qN(ϕ),

where the constant M depends on N, R and r. This gives ˜h(T) ∈ X and

|˜h(T)|_{X} ≤M·p_{R}(h)· |T|_{X}. In cosequence the map
X3T 7→˜h(T)∈X

is continuous and |˜h|_{End X} ≤ M ·pR(h), which proves that also h 7→ ˜h is
continuous.

An easy computation shows that ifh_{1},h_{2} ∈ Holthen (h_{1}·h_{2})˜= ˜h_{1}·˜h_{2}.

Theorem 6. If T ∈ D^{0}(R^{n};B), suppT is compact and if η∈ Hol, η_{|}_{R}^{n} is
polynomially bounded together with all of its derivatives, then

η·T = ˜η(T)

Proof. Let (bj)j=1,...,k be a basis ofEnd B. Thenb^{?}_{j}◦η ∈ Hol(C^{n},C) and
(b^{?}_{j}◦η)(x) = X

β∈N^{n}

1

β!(b^{?}_{j} ◦D^{β}η(0))x^{β}

for any x∈R^{n}. Let ϕ∈ S. Since η|R^{n} is polynomially bounded with all of its
derivatives and suppT is compact, it follows thatηT is tempered and

(7) ηT(ϕ) =

k

X

j=1

X

β∈N^{n}

1

β!(b^{?}_{j}◦D^{β}η(0))·bj(T(x^{β}ϕ)).

On the other hand, in the proof of Theorem 5, we have proved that X

β∈N^{n}

1

β!|(D^{β}η(0)◦x^{β}T)(ϕ)|is convergent, and, in consequence,

(8) X

β∈N^{n}

1

β!(D^{β}η(0)◦x^{β}T)(ϕ) =

k

X

j=1

X

β∈N^{n}

1

β!(b^{?}_{j} ◦D^{β}η(0))·bj(T(x^{β}ϕ)).

Combining (7) and (8), by the arbitrariness of ϕ∈ S, we finally obtain η·T = ˜η(T).

Now turn to distribution-valued functions. Let B be a complex Banach space. Let L(S, B) denote the space of all C-linear continuous operators of S intoB. Fixing q ∈snS, we consider the space L((S, q), B) of all C-linear, q-continuous S →B mappings. Note that it is a Banach space with the norm

|T|_{q} := sup_{q(ϕ)≤1}|T(ϕ)|. Clearly,
L(S, B) = [

q∈snS

L((S, q), B) =

∞

[

m=0

L((S, q_{m}), B).

We equipped the space L(S, B) with an inductive topology, i.e. the strongest locally convex topology such that for all q ∈snS, canonical inclusions

L((S, q), B),→ L(S, B) are continuous.

Let I be a closed interval inR.

Definition 7. A function χ : I → L(S, B) is summable iff there is q ∈ snS such that χ(I) ⊂ L((S, q), B) and the function χ : I → L((S, q), B) is summable in the Bochner sense.

Definition 8. Let χ : I → L(S, B) be summable and q ∈ snS be the seminorm from Definition 7. If

Z

I

χ(t)dt

q

denotes the Bochner integral of χ:I → L((S, q), B), then

L(S, B)3 Z

I

χ(t)dt:=

Z

I

χ(t)dt

q

.

Obviously, the definition does not depend on the choice ofq.

Definition 9. A function χ:I → D_{temp}^{0} is summable iff
χ:I 3t7→χ(t)∈ L(S, B)

is summable.

Clearly, Z

I

χ(t)dt∈ D_{temp}^{0} and
Z

I

χ(t)dt⊂ Z

I

χ(t)dt.

Let c(·) =F :D_{temp}^{0} → D_{temp}^{0} denote the Fourier transform.

Theorem 10. Let χ : I → D^{0}_{temp} be summable. Consider a multi-index
α ∈ N^{n} and a function η ∈ C^{∞}(R^{n},C) which is polynomially bounded with
all of its derivatives. Then the functions D^{α}χ : J 3 t 7→ D^{α}χ(t) ∈ D^{0}_{temp},
ηχ:J 3t7→ηχ(t)∈ D_{temp}^{0} , χˆ:J 3t7→χ(t)d ∈ D_{temp}^{0} are summable and

a):

Z

I

D^{α}χ(t)dt=D^{α}
Z

I

χ(t)dt⊂ Z

I

D^{α}χ(t)dt,
b):

Z

I

(ηχ)(t)dt=η Z

I

χ(t)dt⊂ Z

I

ηχ(t)dt, c):

Z

I

ˆ

χ(t)dt=F Z

I

χ(t)dt

⊂ Z

I

ˆ χ(t)dt.

Definition 11. A function u : I → L(S, B) is absolutely continuous iff there is a locally summamble χ:I → L(S, B) such that

∀t_{1}, t2 ∈I u(t2)−u(t1) =
Z _{t}_{2}

t1

χ(τ)dτ.

Theorem 12. Let u : I → L(S, B). The following conditions are equi- valent

(i) u is absolutely continuous;

(ii) there isq ∈snS such that u(I)⊂ L((S, q), B) and u:I → L((S, q), B) is absolutely continuous and a.e. differentiable;

(iii) there isn∈Nsuch that u(I)⊂ L((S, q_{N}), B)andu:I → L((S, qN), B)
is absolutely continuous and a.e. differentiable.

Definition 13. A function u:I → D_{temp}^{0} is absolutely continuous iff
u:I 3t7→u(t)∈ L(S, B)

is absolutely continuous.

Let J be an arbitrary interval in R.

Definition 14. A function u:J → D_{temp}^{0} is absolutely continuous iff the
restriction u_{I} is absolutely continuous for every closed subintervalI ⊂J.

Remark 15. A function u :J → D^{0}_{temp} is absolutely continuous iff there
are t_{0}∈J and locally summableχ:J → D^{0}_{temp} such that

u(t) =u(t0) + Z t

t0

χ(τ)dτ ∀t∈J.

Moreover, the set {t ∈ domu^{0} : u^{0}(t) 6= χ(t)} is Borelian and of measure
zero.

Theorem 16. Let u : J → D^{0}_{temp} be absolutely continuous. Consider
a multi-index α ∈ N^{n} and a function η ∈ C^{∞}(R^{n},C) polynomially bounded
with all its derivatives. Then the functions D^{α}u :J 3 t 7→ D^{α}u(t) ∈ D^{0}_{temp},
ηu : J 3 t 7→ ηu(t) ∈ D^{0}_{temp}, uˆ : J 3 t 7→ u(t)d ∈ D_{temp}^{0} are absolutely
continuous and for a.e. t∈J:

a) (D^{α}u)^{0}(t) =D^{α}u^{0}(t),
b) (ηu)^{0}(t) =ηu^{0}(t),
c) ˆu^{0}(t) =ud^{0}(t).

3. Results. Let B be a complex vector space and J be an interval inR.
Theorem 17. Given are t_{0} ∈ J, u_{0} ∈ D_{temp}^{0} and a locally summable
f :J → D_{temp}^{0} . Consider a family of locally summable coefficients Aα : J →
End B, (α∈N^{n}), such that #{α:A_{α}6= 0}<∞ and

(9)

∀t_{1} ∈J R^{n}3ξ 7→

t

Q

s

e

P

α

(iξ)^{α}Aα(τ)dτ

∈End B

is polynomially bounded together with all of its
derivatives, uniformly in (s, t)∈[t_{0}, t]×[t_{0}, t_{1}].

Then there is the unique absolutely continuous function u : J → D_{temp}^{0} such
that

(10)

( _{d}

dtu(t) = P

α

A_{α}(t)◦D^{α}u(t) +f(t) for a.e. t∈J
u(t0) = u0.

Moreover, for each t∈J (11)

u(t) =F^{−1}

t

Y

t0

e

P

α

(iξ)^{α}Aα(τ)dτ

·cu_{0}

! +

Z t t0

F^{−1}

t

Y

s

e

P

α

(iξ)^{α}Aα(τ)dτ

·fb(s)

! ds.

t

Y

s

e

P

α

(iξ)^{α}Aα(τ)dτ

denotes the product integral. Let us denote
(12) J^{2}(t0) :={(s, t)∈J×J : s∈[t0, t]}.

By the Leibniz formula, the following lemmas hold:

Lemma 18. Let η ∈ C^{∞}(R^{n}, End B) be polynomially bounded with all its
derivatives. For any N ∈Nthere is N such that for all j= 1, . . . , k the map

(S, q_{N})3ϕ7−→^{def} (b^{?}_{j} ◦η)ϕ∈(S, q_{N})
is linear and continuous.

Lemma 19. Fix (s, t) ∈J^{2}(t0). Let us consider η_{s}^{t} ∈ C^{∞}(R^{n}, End B) such
that

(13)

∀t_{1}∈J R^{n}3ξ 7→η^{t}_{s}(ξ)∈End B

is polynomially bounded together with all of its derivatives, uniformly in(s, t)∈[t0, t]×[t0, t1].

Additionally assume that

(14) if (s_{ν}, t_{ν})^{ν→∞}−→ (s, t) in J^{2}(t_{0}), then for every γ ∈N^{n}
D^{γ}η_{s}^{t}^{ν}_{ν} ^{ν→∞}−→ D^{γ}η_{s}^{t} almost uniformly in R^{n}.

Then for any N ∈Nthere is N ∈Nsuch that for all j = 1, . . . , k the map
lj :J^{2}(t0)3(s, t)7−→^{def} l^{s,t}_{j} ∈ L (S, q_{N}),(S, q_{N})

,
l^{s,t}_{j} (ϕ) := (b^{?}_{j} ◦η^{t}_{s})ϕ for ϕ∈ S
is continuous.

By the continuous dependence on the limits of integration for a product integral (see [6]) we obtain

Remark 20. LetA_{α} :J −→ End B (α ∈N^{n}) be a finite family of locally
summable functions and

η_{s}^{t}:C3z7−→^{def}

t

Y

s

e

P

α

(iz)^{α}Aα(τ)dτ

∈ Hol ∀(s, t)∈J^{2}(t_{0}).

Then η_{s}^{t}|_{R}n satisfies assumption (14).

Corollary 21. If Aα ∈End B (α∈N^{n}) then
η_{s}^{t}(ξ) :=e^{(t−s)}

P

α

(iξ)^{α}Aα

∀ξ∈R^{n} ∀(s, t)∈J^{2}(t_{0})
satisfies (14).

Proof of Theorem 17.

Step1^{◦} We begin by proving that the right-hand side of (11) makes sense.

First, note that for any t∈J,s∈[t_{0}, t] the mapping
(15) η_{s}^{t}: R^{n}3ξ 7−→^{def}

t

Y

s

e

P

α

(iξ)^{α}Aα(τ)dτ

∈End B

is polynomially bounded with all its derivatives, and so η_{t}^{t}_{0} ·uˆ_{0}, η_{s}^{t}·fˆ(s) are
tempered distributions.

Let t ∈ J. The function f_{[t}_{0}_{,t]} : [t_{0}, t] → D^{0}_{temp} is summable, thus by
Theorem 10, so is ˆf_{[t}_{0}_{,t]} : [t0, t] → D_{temp}^{0} . According to Definition 9 and
Definition 7 there is N ∈ N such that ˆf([t_{0}, t]) ⊂ L((S, q_{N}), B) and ˆf_{[t}_{0}_{,t]} :
[t_{0}, t]→ L((S, q_{N}), B) is summable.

Fixj∈ {1, . . . , k}. Due to Remark 20 and Lemma 19, there isN such that
l_{j}(J^{2}(t_{0}))⊂ L((S, q_{N}),(S, q_{N})) and the map l_{j} :J^{2}(t_{0})→ L((S, q_{N}),(S, q_{N}))
is continuous.

Consider the continuous bilinear mapping
(16) G_{N ,N} :

( L((S, q_{N}),(S, q_{N}))× L((S, q_{N}), B) −→ L((S, q_{N}), B)
(E, T) 7−→^{def} T ◦E.

Then the function

[t0, t]3s7→ G_{N ,N}(l^{s,t}_{j} ,fˆ(s))∈ L((S, q_{N}), B)
is summable, and for s∈[t_{0}, t], ϕ∈ S there is

G_{N ,N}

l_{j}^{s,t},f(s)ˆ

(ϕ) =

fˆ(s)◦l^{s,t}_{j}

(ϕ) = ˆf(s)((b^{?}_{j}◦η^{t}_{s})ϕ) = (b^{?}_{j} ◦η^{t}_{s}) ˆf(s)(ϕ).

Consequently,

[t_{0}, t]3s7→

k

X

j=1

b_{j}◦(b^{?}_{j}◦η_{s}^{t}) ˆf(s)∈ L((S, q_{N}), B)
is summable, and by Definitions 7, 9, so is

[t0, t]3s7→η^{t}_{s}·fˆ(s)∈ D^{0}_{temp}(R^{n};B).

Finally according to the Schwartz theorem and Theorem 10, the function
[t0, t]3s7→ F^{−1}(η^{t}_{s}·fˆ(s))∈ D_{temp}^{0}

is well defined and summable.

Step2^{◦} Uniqueness.

Let u :J → D_{temp}^{0} be an absolutely continuous solution of problem (10). Fix
ψ∈ D(R^{n}). Then on account of Theorem 16 the functionv:J 3t7−→^{def} ψˆu(t)∈
D_{temp}^{0} is also absolutely continuous and for a.e. t∈J

(17) v^{0}(t) =X

α

A_{α}(t)◦(iξ)^{α}v(t) +ψfˆ(t).

Fixt_{?} ∈J and denoteI := [t_{0}, t_{?}]. Similarly as in Step 1^{o}, there is N ∈Nsuch
that

v(I)⊂ L((S, q_{N}), B) and v_{I}:I → L((S, q_{N}), B) is absolutely
continuous, a.e. differentiable,
ψfˆ(I)⊂ L((S, q_{N}), B) and ψfˆ:I → L((S, q_{N}), B) is summable,
ψˆu0 ∈ L((S, q_{N}), B).

Consider the Banach space

X :={T ∈ D_{temp}^{0} : suppT ⊂suppψ, T is q_{N} −continuous}

with the norm |T|_{X} := |T|_{q}_{N} (see (2)). Let ι : X 3 T 7→ T ∈ L((S, qN), B)
be a canonical injection. ι(X) is a closed subspace ofL((S, q_{N}), B). v(I)⊂X
hence v(I) ⊂ι(X). Similarlyψfˆ(I)⊂ι(X). Thus vI :I → ι(X) is absolutely
continuous, a.e. differentiable and ψfˆ_{I} :I →ι(X) is summable.

Setting x := vI, we derive that the function x : I → X is absolutely
continuous, a.e. differentiable. Similarly ψfˆ_{I} : I → X is summable and
ψuˆ0 ∈X. Clearly,

(18) x(t) =˙ v^{0}(t)

a.e. in I, where the derivative on the left-hand side is a derivative in topX
while the one on the right-hand side is a derivative in D_{temp}^{0} . Therefore by
(17), for a.e. t∈I, there is

˙

x(t) =X

α

A_{α}(t)◦(iξ)^{α}x(t) +ψfˆ(t) = ˜A_{t}x(t) +ψfˆ(t),
where A_{t}(z) :=X

α

(iz)^{α}A_{α}(t) for z∈C^{n}. Obviously,A_{t}∈ Hol. The mapping

˜:Hol→End Xis defined as in Theorem 5 (ii), in particular ˜A_{t}T =X

α

A_{α}(t)◦

(iξ)^{α}T forT ∈X.

Thus we have obtained the Cauchy problem in the Banach spaceX, treated as a module over the Banach algebra End X:

(19)

x(t)˙ = A˜_{t}x(t) +ψfˆ(t)
x(t0) = ψuˆ0.

On account of Theorem 41, [6]

x(t) =

t

Y

t0

e^{A}^{˜}^{τ}^{dτ}(ψˆu_{0}) +
Z t

t0

t

Y

s

e^{A}^{˜}^{τ}^{dτ}(ψfˆ(s))ds

is the unique absolutely continuous solution of problem (19). By Lemma 44, [6] (comp. Th. 5 (ii)) and using notation (15)

x(t) =

t

Y

t0

e^{A}^{τ}^{dτ}

!e

(ψˆu_{0}) +
Z t

t0

t

Y

s

e^{A}^{τ}^{dτ}

!e

(ψfˆ(s))ds=

=ηf_{t}^{t}_{0}(ψˆu0) +
Z t

t0

ηe^{t}_{s}(ψfˆ(s))ds.

Functions η^{t}_{t}_{0}, η^{t}_{s} are polynomially bounded together with all its derivatives,
distributions ψˆu0,ψfˆ(s) have compact supports, so by Theorem 6

x(t) =η^{t}_{t}_{0} ·(ψuˆ0) +
Z t

t0

η_{s}^{t}·(ψfˆ(s))ds,
with

Z _{t}

t0

η_{s}^{t}·(ψfˆ(s))ds being an integral inX. The function

[t0, t]3s7→ι(η^{t}_{s}·(ψf(s))) =ˆ η_{s}^{t}·(ψfˆ(s))∈ L((S, q_{N}), B)
is summable in the sense of Bochner and

ι Z t

t0

η^{t}_{s}·(ψfˆ(s))ds

= Z t

t0

η_{s}^{t}·(ψfˆ(s))ds.

In consequence, [t0, t]3s7→η_{s}^{t}·(ψf(s))ˆ ∈ D^{0}_{temp}is summable and the integral
Z t

t0

η_{s}^{t}·(ψfˆ(s))ds in the space of tempered distributionsD^{0}_{temp} is equal to the
integral in X.

In Step 1^{◦} we have proved that the function
[t_{0}, t]3s7→η^{t}_{s}·fˆ(s)∈ D^{0}_{temp}

is summable. By the arbitrariness of ψ∈ D(R^{n}) and Remark 2, there is
ˆ

u(t) =η^{t}_{t}_{0} ·uˆ0+
Z t

t0

η_{s}^{t}·f(s)dsˆ

for all t∈I (in particular fort?). Finally, by the Schwartz theorem,
u(t?) =F^{−1} η_{t}^{t}^{?}_{0} ·uˆ0

+F^{−1}
Z t?

t0

η^{t}_{s}^{?}·f(s)dsˆ

.
As t_{?}∈J is arbitrary and on account of Theorem 10

u(t) =F^{−1} η_{t}^{t}_{0}·uˆ_{0}
+

Z t t0

F^{−1}

η^{t}_{s}·fˆ(s)ds

∀t∈J.

Since every solution of (10) is of the form (11), we have proved the uniqueness.

Step 3^{◦} Existence.

We prove that the function

(20) u:J 3t7−→ F^{def} ^{−1} η^{t}_{t}_{0} ·uˆ_{0}
+

Z t t0

F^{−1}

η_{s}^{t}·fˆ(s)

ds∈ D_{temp}^{0}
is an absolutely continuous solution of (10).

Let I ⊂J be a closed interval such that t_{0}∈I. Clearly, u(t_{0}) =u_{0}.
In fact, we shall prove that for allt∈I u(t) = ˆˆ u0+

Z t t0

χ(τ)dτ, where
(21) χ:I 3t 7−→^{def} A_{t}·u(t) + ˆˆ f(t)∈ D_{temp}^{0} .

Applying the argument fromStep 1^{◦}, one can deduce thatχis summable.

Let t ∈ I and ψ ∈ D(R^{n}). There is N ∈N such that ˆu_{0} ∈ L((S, q_{N}), B)
and the function I 3t7→fˆ(t)∈ L((S, q_{N}), B) is summable. Having N and ψ
we consider the spaceX, as in Step 1^{◦}. Thenψˆu_{0}∈XandI 3t7→ψfˆ(t)∈X
is summable.

Similarly as in Step 2^{◦}, the problem
(22)

x(t)˙ = A˜_{t}x(t) +ψfˆ(t)
x(t0) = ψuˆ0

has the unique solution given for allt∈I by the formula x(t) =

t

Y

t0

e^{A}^{τ}^{dτ}

!

·(ψˆu_{0}) +
Z t

t0

t

Y

s

e^{A}^{τ}^{dτ}

!

·(ψfˆ(s))ds.

Thus remembering Remark 2 and definition (20) we obtain

(23) x(t) =ψu(t).ˆ

On the other hand, combining (22), (23) and (21) we obtain
(24) x(t) =˙ A_{t}·x(t) +ψfˆ(t) =ψχ(t).

The function x : I → X is absolutely continuous, a.e. differentiable, so by (23), (24), for allt∈I, there holds

ψˆu(t) =x(t) =x(t_{0}) +
Z t

t0

˙

x(τ)dτ =ψˆu_{0}+
Z t

t0

ψχ(τ)dτ.

Consequently,

∀t∈I : u(t) = ˆˆ u_{0}+
Z t

t0

χ(τ)dτ,

since ψ∈ D(R^{n}) is arbitrary. Therefore, by Remark 15, the function ˆuI :I →
D_{temp}^{0} is absolutely continuous and for a.e. t∈I

(ˆu_{I})^{0}(t) =χ(t).

In consequence, so is ˆu:J → D_{temp}^{0} , and for a.e. t∈J,
ˆ

u^{0}(t) =At·u(t) + ˆˆ f(t) =X

α

Aα(t)◦(iξ)^{α}u(t) + ˆˆ f(t),

since I is arbitrary. On account of Theorem 16, we finally have that u :J →
D_{temp}^{0} is absolutely continuous and for a.e. t∈J

u^{0}(t) =F^{−1}(ˆu^{0}(t)) =X

α

A_{α}(t)◦D^{α}u(t) +f(t).

Thus u defined by (20) is a solution of (10).

In the case of time-independent coefficients A_{α} ∈End B, let us define
Λ_{1}:= sup

(

Reλ: λ∈σ X

α

(iξ)^{α}A_{α}

!

, ξ∈R^{n}
)

,

Λ−1 := inf (

Reλ: λ∈σ X

α

(iξ)^{α}Aα

!

, ξ∈R^{n}
)

, with σ(A) denoting the spectrum of A∈End B.

Theorem 22. Given are t_{0} ∈ J, u_{0} ∈ D^{0}_{temp}, locally summable f : J →
D_{temp}^{0} and a finite family of coefficientsAα∈End B (α∈N^{n}). Suppose that
(25) for all t∈J\ {t_{0}} Λ_{sgn(t−t}_{0}_{)}∈R.

Then there is the unique absolutely continuous function u : J → D_{temp}^{0} such
that

(26)

( _{d}

dtu(t) = P

α

A_{α}◦D^{α}u(t) +f(t) for a.e. t∈J
u(t0) = u0.

Moreover for each t∈J (27)

u(t) =F^{−1}

e^{(t−t}^{0}^{)}

P

α

(iξ)^{α}Aα

·cu0

+

Z t t0

F^{−1}

e^{(t−s)}

P

α

(iξ)^{α}Aα

·f(s)b

ds.

Proof. First note that the theorem is a corollary of Theorem 17. Namely,
denoting as in the proof of Theorem 17, A_{t} ≡ A ∈ Hol for t ∈ J, with
A(z) := X

α

(iz)^{α}A_{α} for z ∈C^{n}, it suffices to note that A_{t}◦A_{s} =A_{s}◦A_{t} for
all s, t∈J and in consequence

t

Y

s

e^{A}^{τ}^{dτ} =e

Rt

sAτdτ =e^{(t−s)A} ∀s, t∈J.

The task is now to show that the family Aα satisfies condition (9). The proof will be devided into 3 steps.

Step1^{◦} First we prove that, for anyt1 ∈J, the function
(28) R^{n}3ξ 7→e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α} ∈End B
is polynomially bounded, uniformly in (s, t)∈[t_{0}, t]×[t_{0}, t_{1}].

Fix t1∈J. Let (s, t)∈[t0, t]×[t0, t1], ξ∈R^{n}. According to the lemma in
[2], p. 78, there is

(29)

e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

≤e^{Λ}^{(t−s),ξ}·

dimB−1

X

j=0

2|t−s|

X

α

(iξ)^{α}Aα

!j

,

with Λ_{(t−s),ξ} := sup
(

Reλ: λ∈σ (t−s)X

α

(iξ)^{α}A_{α}

!) . Since

Λ(t−s),ξ ≤ |t_{1}−t0|max{|Λ_{1}|,|Λ_{−1}|}<∞,

there is a polynomialQ:R^{n}→]0,∞[ such that∀(s, t)∈[t_{0}, t]×[t_{0}, t_{1}] ∀ξ ∈R^{n}

e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

≤Q(ξ).

Step 2^{◦} We will prove polynomial boundedness of derivatives of function
(28) by induction. Let j = 1, . . . , n. We will estimate

∂

∂ξj

e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}
.
Let ξ∈R^{n}. There is R >0 such that|ξ|< R. The mapping

A: [s, t]×G3(τ, ξ)7−→^{def} X

α

(iξ)^{α}A_{α}∈End B

is τ-summable for any ξ ∈G:={ξ ∈R^{n} : |ξ|< R}. Its derivative _{∂ξ}^{∂}

jA(τ, ξ) is dominated by summable function ϕ, namely

ϕ: [s, t]3τ 7→ sup

|ξ|≤R

∂

∂ξ_{j}
X

α

(iξ)^{α}Aα

. Therefore, according to Theorem 45, [6]

(30)

∂

∂ξj

e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

= Z t

s

e^{(t−r)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}◦ ∂

∂ξ_{j}
X

α

(iξ)^{α}A_{α}◦e^{(r−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}dr.

Since both (r, t) and (s, r) are elements of [t0, t]×[t0, t1], it follows from Step
1^{◦} that both ξ 7→ e^{(t−r)}^{P}^{α}^{(iξ)}^{α}^{A}^{α} and ξ 7→ e^{(r−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α} are polynomially
bounded. Finally, there is a polynomial Qej :R^{n} →]0,∞[ such that ∀(s, t) ∈
[t0, t]×[t0, t1] ∀ξ ∈R^{n}

∂

∂ξ_{j}e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

≤Qej(ξ).

Step3^{◦}Assume that the functionR^{n}3ξ7→D^{γ}

e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

∈End B
is polynomially bounded, uniformly in (s, t)∈ [t_{0}, t]×[t_{0}, t_{1}] for γ ∈ N^{n} such
that |γ| ≤k. Let β ∈N^{n} be a multi-index such that |β|=k+ 1. On account
of Theorem 49, [6]

(31)

D^{β}e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α} =
X

β6=γ≤β

β γ

Z t s

e^{(t−r)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}◦D^{β−γ}X

α

(iξ)^{α}A_{α}◦D^{γ}e^{(r−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}dr.

For every γ ≤β and γ 6=β, there is a polynomial Qγ such that

D^{γ}e^{(r−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}dr

≤Qγ(ξ)

for any (s, r)∈[t0, t]×[t0, t1] andξ ∈R^{n}. Moreover, as in Step 1^{o}, there isQ0

such that

e^{(t−r)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

≤Q_{0}(ξ) for all (r, t)∈[t_{0}, t]×[t_{0}, t_{1}] andξ∈R^{n}.
Thus

D^{β}e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

≤ X

β6=γ≤β

β γ

Q_{0}(ξ)· |D^{β−γ}X

α

(iξ)^{α}A_{α}| ·Q_{γ}(ξ)· |t_{1}−t_{0}|.

By induction we finally conclude that ∀t_{1} ∈J ∀β ∈N^{n} ∃Q_{β} ∈ P(R^{n}), Q_{β} >0

∀(s, t)∈[t_{0}, t]×[t_{0}, t_{1}] ∀ξ ∈R^{n}

D^{β}e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}

≤Q_{β}(ξ).

4. Examples.

4.1. Parabolic systems in the sense of Pietrowski.

Let B be a complex vector space and J be an interval in R. Let t_{0} ∈ J
and b∈N. Let us recall that the system

(32) d

dtu(t) = X

|α|≤2b

A_{α}◦D^{α}u(t) +f(t)

is parabolic in the sense of Pietrowski iff ∃δ >0 ∀ζ ∈R^{n},|ζ|= 1:

Reλ(ζ)≤ −δ, whereλ(ζ)∈σ

X

|α|=2b

(iζ)^{α}Aα

, see, e.g., [1].

We will prove that every system which is parabolic in the sense of Pietrowski satisfies condition (25).

X

|α|≤2b

(iξ)^{α}Aα=|ξ|^{2b}

X

|α|<2b

(iξ)^{α}

|ξ|^{2b}Aα+ X

|α|=2b

(iξ)^{α}

|ξ|^{2b} Aα

.

There is δ >0 such that Reλ < −δ for anyλ∈σ

X

|α|=2b

(iξ)^{α}

|ξ|^{2b} A_{α}

. On the

other hand, there is R > 0 such that σ

X

|α|≤2b

(iξ)^{α}

|ξ|^{2b} Aα

⊂ {Reλ < −δ} ⊂
{Reλ ≤ 0} for |ξ| ≥ R (see, e.g., [5], Th. 10.20). Therefore Λ_{1} < ∞, and
according to Theorem 22, there is the unique solution of system (32).

4.2. Dirac equation.

Consider the Dirac equation in Weyl’s representation

(33) 1

i

∂ψ

∂t =

3

X

j=1

γ_{j}
1

i

∂

∂x_{j} −A_{j}(t)

ψ+m_{0}βψ+V(t)ψ,
where the matrices

γj =

σ_{j} 0
0 −σ_{j}

, β =

0 I_{2}
I_{2} 0

,

satisfy the relations

γjγk+γkγj = 2δjkI4 j, k= 1,2,3,4, γ4 =β,
I_{N} is the N×N identity matrix,

σ_{1} =

0 1 1 0

σ_{2} =

0 −i i 0

, σ_{3} =

1 0 0 −1

are Pauli matrices, m0 is the rest mass, A1, A2, A3, V are components of an external electromagnetic potential. Units were chosen so that c=h= 1.

Consider equation (33) together with the initial condition

(34) ψ(t0) =ψ0

where ψ_{0} ∈ D_{temp}^{0} (R^{3};C^{4}). Observe that

A(ξ) =

iξ_{3} iξ_{1}+ξ_{2} im_{0} 0
iξ_{1}−ξ_{2} −iξ_{3} 0 im_{0}

im_{0} 0 −iξ_{3} −iξ_{1}−ξ_{2}
0 im0 −iξ_{1}+ξ2 iξ3

and compute

det(A(ξ)−λI) = |ξ|^{2}+m^{2}_{0}+λ^{2}2

. Thus eigenvalues of A(ξ) are

λ=±i q

m^{2}_{0}+|ξ|^{2}
and ∀ξ ∈R^{n} Reλ= 0. Therefore, Λ_{1},Λ−1 ∈R.

Suppose first that Aj ≡0≡V forj = 1,2,3. Then, according to Theorem 22, the only solution of problem {(33), (34)} is

ψ(t) =F^{−1}(e^{(t−t}^{0}^{)A(ξ)}·ψˆ0).

In the case of non-zero potentials, supposing that Aj, V :J → C are locally summable (j= 1,2,3), Theorem 22 gives the following integral formula for ψ:

ψ(t) =F^{−1}(e^{(t−t}^{0}^{)A(ξ)}·ψˆ0)
+

Z t t0

F^{−1}

e^{(t−s)A(ξ)}·i(V(s)ψ(s)b −

3

X

j=1

γ_{j}A_{j}(s)ψ(s))b

ds.

4.3. Second-order time equations.

Let B, B_{1}, B_{2} be complex Banach spaces. J is an interval in R. For
T1 ∈ D^{0}(R^{n};B1), T2 ∈ D^{0}(R^{n};B2), let us consider the distribution T1^{4}T2 ∈
D^{0}(R^{n};B_{1}×B_{2})

T1^{4}T2 : D(R^{n})3ϕ7−→^{def} (T1(ϕ), T2(ϕ))∈B1×B2.

Clearly, if T_{1} ∈ D^{0}_{temp}(R^{n};B_{1}), T_{2} ∈ D^{0}_{temp}(R^{n};B_{2}) then T_{1}4T_{2} is tempered
and

T1^{4}T2 =T1^{4}T2.

Let functions x : J → D_{temp}^{0} (R^{n};B1), y : J → D^{0}_{temp}(R^{n};B2) be absolutely
continuous. Then

x^{4}y : J 3t7−→^{def} x(t)^{4}y(t)∈ D_{temp}^{0} (R^{n};B1×B2)

is absolutely continuous and for anyt∈dom ˙x∩dom ˙y(the set of full measure in J) there is

(x4y)^{0}(t) = ˙x(t)4y(t).˙

Definition 23. Given a mapF :R× D_{temp}^{0} (R^{n};B×B)−→ D◦ ^{0}_{temp}(R^{n};B).

A function x:J → D^{0}_{temp}(R^{n};B) is a solution of the equation

(35) x¨=F(t, x4x)˙

iff

1^{◦} x : J → D^{0}_{temp}(R^{n};B) is absolutely continuous, differentiable for all
t∈J and ˙x:J → D_{temp}^{0} (R^{n};B) is absolutely continuous,

2^{◦} (t, x(t)^{4}x(t))˙ ∈domF for any t∈J,
3^{◦} x(t) =¨ F(t, x(t)4x(t))˙ for a.e. t∈J.

Theorem 24. Let x : J → D^{0}_{temp}(R^{n};B) satisfy condition 1^{◦}. Denote
z:=x4x.˙ x is a solution of (35) iffz is a solution of the equation z˙=v(t, z),
where v: domF→ D^{0}_{temp}(R^{n}; B×B) and v(t, y^{4}y) := ˙˙ y^{4}F(t, y^{4}y).˙

Proof. If x : J → D_{temp}^{0} (R^{n};B) is a solution of (35) then z : J →
D_{temp}^{0} (R^{n};B×B) is absolutely continuous, and for a.e. t∈J, ˙z(t) = ˙x(t)^{4}x(t).¨
Moreover,

v(t, x(t)^{4}x(t)) = ˙˙ x(t)^{4}F(t, x(t)^{4}x(t)) = ˙˙ x(t)^{4}x(t) = ˙¨ z(t)
for a.e. t∈J.

Conversely, ifzis a solution of the equation ˙z=v(t, z) then (t, z(t))∈domv
for any t ∈ J. Thus (t, x(t)4x(t))˙ ∈ domF for any t ∈ J. Let us denote
Q(T) :=T2 for T =T1^{4}T2 ∈ D^{0}_{temp}(R^{n};B×B). Then

¨

x(t) =Q( ˙z(t)) =Q( ˙x(t)^{4}F(t, x(t)^{4}x(t))) =˙ F(t, x(t)^{4}x(t))˙
for a.e. t∈J.

Let us denote P(T) :=T1,Q(T) :=T2 forT =T1^{4}T2 ∈ D^{0}_{temp}(R^{n};B×B).

Corollary 25. Let v(t, x4y) := y4F(t, x4y). Suppose that z : J →
D_{temp}^{0} (R^{n};B×B)is a solution of the equation z˙=v(t, z) andx=P(z). Then
z=x^{4}x˙ and x is a solution of (35).

Proof. Since z:J → D^{0}_{temp}(R^{n};B×B) is absolutely continuous, so are
x=P◦z,y:=Q◦z:J → D_{temp}^{0} (R^{n};B). Let t∈dom ˙z.

˙

x(t) =P( ˙z(t)) =P(v(t, z(t))) =P(y(t)^{4}F(t, z(t))) =y(t).

Thus xis differentiable for anyt∈J, ˙x is absolutely continuous andz=x4x.˙ On account of Theorem 24, x is a solution of (35).

4.4. Wave equation.

Let c >0,t0 = 0∈J. Given u0, u1 ∈ D^{0}_{temp}(R^{n};C) and locally summable
f : J → D_{temp}^{0} (R^{n};C). Let us consider the Cauchy problem for the wave
equation

(36)

d^{2}

dt^{2}u(t) = c^{2}∆u(t) +f(t) for a.e. t∈J
u(0) = u_{0}

u^{0}(0) = u1.
Setting

w (=w1^{4}w2) : J 3t7−→^{def} u(t)^{4}u^{0}(t)∈ D^{0}_{temp}(R^{n};C^{2})
we can rewrite (36) as the following first-order Cauchy problem:

(37)

w^{0}(t) = X

|α|≤2

A_{α}◦D^{α}w(t) + ˜f(t) for a.e. t∈J
w(0) = w0,

where ˜f(t) := 04f(t), w_{0}:=u_{0}4u_{1} ∈ D_{temp}^{0} (R^{n};C^{2}),
A_{0}:=

0 1 0 0

, A_{2e}_{j} :=c^{2}

0 0 1 0

,
(ej)^{n}_{j=1} is the canonical basis in R^{n},

Aα:= 0 for all other multi-indices α. The eigenvalues of the matrix X

|α|≤2

(iξ)^{α}A_{α}=

0 1 0 0

−c^{2}|ξ|^{2}

0 0 1 0

are: −ic|ξ|,ic|ξ|, thus Λ_{1}, Λ−1= 0∈Rand according to Theorem 22, there is
the unique solution of problem (37) given by the formula

w(t) =F^{−1}

e^{t}^{P}^{α}^{(iξ)}^{α}^{A}^{α} ·wˆ_{0}
+

Z t 0

F^{−1}

e^{(t−s)}^{P}^{α}^{(iξ)}^{α}^{A}^{α}·fb˜(s)
ds.

Note that

X

α

(iξ)^{α}A_{α}

!2ν

= (−c^{2}|ξ|^{2})^{ν}·I,

X

α

(iξ)^{α}Aα

!2ν+1

= (−c^{2}|ξ|^{2})^{ν} · X

α

(iξ)^{α}Aα

!

for ν= 0,1,2, . . ., hence
e^{τ}^{P}^{α}^{(iξ)}^{α}^{A}^{α} =

cos(τ c|ξ|) sin(τ c|ξ|) c|ξ|

−c|ξ|sin(τ c|ξ|) cos(τ c|ξ|)

. On account of Cor. 25, we obtain the formula for the solution of (36):

(38)

u(t) = F^{−1}(cos(tc|ξ|)·uˆ0) +F^{−1}

sin(tc|ξ|) c|ξ| ·uˆ1

+ +

Z t 0

F^{−1}

sin((t−s)c|ξ|) c|ξ| ·fˆ(s)

ds.

Fixingf ≡0,u_{0} = 0,u_{1} =δin (38), we obtain the formula for the fundamental
solution of the d’Alembert operator:

u(t) =F^{−1}

sin(tc|ξ|) c|ξ| ·δˆ

= 1

2π
^{n}_{2}

F^{−1}

sin(tc|ξ|) c|ξ|

,

where [ψ] denotes the regular distribution generated byψ∈ L^{1}_{loc}(R^{n}), i.e.,
[ψ](ϕ) :=

Z

R^{n}

ψ(x)ϕ(x)dx ∀ϕ∈ D(R^{n}).

4.5. Navier–Lam´e equation.

Let t0 ∈ J; positive numbers λ, µ are the Lam´e constants. Given u^{0},
u^{1} ∈ D^{0}_{temp}(R^{3};C^{3}) and locally summable f : J → D^{0}_{temp}(R^{3};C^{3}). We are
looking for an absolutely continuous u:J → D^{0}_{temp}(R^{3};C^{3}) such that

(39)

d^{2}

dt^{2}u(t) = µ∆u(t) + (λ+µ)∇divu(t) +f(t) for a.e. t∈J
u(t_{0}) = u^{0}

u^{0}(t0) = u^{1}.

For t∈J, denote

u(t) =u1(t)^{4}u2(t)^{4}u3(t), f(t) =f1(t)^{4}f2(t)^{4}f3(t).

The Navier–Lam´e equation can now be written in the form
(40) d^{2}

dt^{2}uj(t) =µ∆uj(t) + (λ+µ) ∂

∂xj 3

X

k=1

∂

∂x_{k}uk(t)

!

+fj(t) for j= 1,2,3 and a.e. t∈J. Settings

w^{1}_{j}(t) :=uj(t) w_{j}^{2}(t) :=u^{0}_{j}(t), w(t) := (w_{1}^{1}^{4}w^{1}_{2}^{4}w^{1}_{3}^{4}w^{2}_{1}^{4}w^{2}_{2}^{4}w^{2}_{3})(t)
for t∈J transform (40) into the following first-order system:

(41)

d

dtw^{1}_{j}(t) = w_{j}^{2}
d

dtw^{2}_{j}(t) = µ∆w^{1}_{j}(t) + (λ+µ) ∂

∂x_{j}

3

X

k=1

∂

∂x_{k}w^{1}_{k}(t)

!

+fj(t)
for j = 1,2,3 and a.e. t ∈ J. Denoting ˜f(t) := 0^{4}f(t) ∈ D^{0}_{temp}(R^{3};C^{6})
(0∈ D^{0}_{temp}(R^{n};C^{3})),

A_{0} :=

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

Ajk :=

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

. . .

. . .(3 +j)-th row .

. ...

k-th column

A_{e}_{j}_{+e}_{k} :=A_{jk} forj6=kandA_{2e}_{j} :=µA^{T}_{0} + (λ+µ)A_{jj}, whereA^{T}_{0} is transposed
to A0, (ej)^{3}_{j=1} is the canonical basis in R^{3},Aα = 0 for other multi-indices α,