Bulletin T.CXXIX de l’Académie serbe des sciences et des arts - 2004 Classe des sciences mathématiques et naturelles sciences mathématiques, N\underline{o} 29

LAPLACE TRANSFORM OF DISTRIBUTION-VALUED FUNCTIONS AND ITS APPLICATIONS

B. STANKOVIĆ

(Presented at the 2nd Meeting, held on March 26, 2004)

A b s t r a c t. For the distribution-valued functions the Laplace transform is defined and some properties of it are proved. In order to illustrate the elaborated results, the conditions are proved to have the unicity and existence of solutions to a mathematical model of the vibrating rod with boundary conditions.

AMS Mathematics Subject Classification (2000): 46F12, 35D05 Key Words: Distribution-valued function, Laplace transform.

1. Notation and notions

We repeat some definitions and facts, we need in our exposition but for special case.

Let Q be an open set belonging to ℝ^{n}. By D(Q) we denote the space {φ ∈ C^{∞}(Q); suppφ ⊂ K_{φ}}, K_{φ} is a compact set in Q which depends on φ. D′(Q) is the space of continuous linear functionals on D(Q) - the space of distributions. Every f ∈ L_{loc}(Q) defines a distribution, called regular distribution, denoted by [f],

⟨[f], φ⟩ = ∫_{Q} f(t)φ(t)dt, φ ∈ D(Q).

Important subspaces of D′(ℝ^{n}) we use, are: D′(\overline{ℝ}_{+}^{n}) = \{f ∈ D′(ℝ^{n}), suppf ⊂ \overline{ℝ}_{+}^{n}\}, the space of tempered distributions S′(ℝ^{n}) and S′(\overline{ℝ}_{+}^{n}) = \{f ∈ S′(ℝ^{n}), suppf ∈ \overline{ℝ}_{+}^{n}\}. For the space S′(ℝ^{n}) and its topology cf. [8].

Definition 1. The Laplace transform (in short LT) of f ∈ S′(\overline{ℝ}_{+}^{n}) is defined by

f̂(z) = L(f)(z) = ⟨f(t), η(t)e^{−zt}⟩, z ∈ ℝ_{+}^{n} + iℝ^{n},

where η ∈ C^{∞}(ℝ^{n}), |η^{(α)}(t)| ≤ C_{α}, α ∈ ℕ and η(t) = 1, t ∈ (−ε, ∞)^{n}; t ∈ ℝ^{n} \ (−∞, −2ε)^{n}, ε > 0; For the properties of so defined LT one can consult [8] and [9].

Let H^{α,β}(ℝ_{+}^{n}), α ≥ 0, β ≥ 0, denote the sets of holomorphic functions on ℝ_{+}^{n} + iℝ^{n} which satisfy the following growth condition:

|f(z)| ≤ M(1 + |z|^{2})^{α/2}(1 + |x|^{−β}), z = x + iy ∈ ℝ_{+}^{n} + iℝ^{n}.

For the applications of the LT the following is very important:

Proposition A. ([8] ,p. 191). The algebras S′(\overline{ℝ}_{+}^{n}) and H(ℝ_{+}^{n}) = ⋃_{α≥0} ⋃_{β≥0} H^{α,β}(ℝ_{+}^{n}) are isomorphic. This isomorphism is accomplished via the

LT.

We consider the space e^{ωt}S′(\overline{ℝ}_{+}^{n}) ⊂ D′(\overline{ℝ}_{+}^{n}) ⊂ D′(ℝ^{n}), ω > 0.

Definition 2. If f ∈ e^{ωt}S′(\overline{ℝ}_{+}^{n}), then the LT of it is

f̂(z) = L(f)(z) = ⟨f(t), η(t)e^{−(z−ω)t}⟩, z ∈ ω + ℝ_{+}^{n} + iℝ^{n}.

Let b > 0. We denote by A_{b} the space

A_{b} = \{f ∈ e^{ωt}D′(\overline{ℝ}_{+}^{n}), suppf ⊂ [b, ∞)^{n}\}.

Now we can define an equivalent relation in e^{ωt}S′(\overline{ℝ}_{+}^{n}) : f ∼ g ⟺ f − g ∈ A_{b}. Then the space B_{b} by definition is:

B_{b} = e^{ωt}S′(\overline{ℝ}_{+}^{n})/A_{b}.

Definition 3. The space D′_{ω}([0, b)^{n}) is by definition:

D′_{ω}([0, b)^{n}) = \{T ∈ D′([0, b)^{n}); ∃\overline{T} ∈ e^{ωt}S′(\overline{ℝ}_{+}^{n}), \overline{T} |_{(−ε,b)^{n}} = T \},

where \overline{T} |_{(−ε,b)^{n}} is the restriction of \overline{T} on (−ε, b)^{n}, ε > 0.

Proposition B. ([6], [7]) D′_{ω}([0, b)^{n}) is algebraically isomorphic to B_{b}.

2. Distribution-valued function and its Laplace transform

Our definition of the distribution-valued functions (in short d-v-f) (cf. [5]) is appropriate to the mathematical models in mechanics.

In this section Ω denotes an open set belonging to ℝ^{m}.

Definition 4. Let ω be a positive real number. The function [u(x, .)], x ∈ Ω, with values in e^{ωt}S′(\overline{ℝ}_{+}^{n}) ⊂ D′(\overline{ℝ}_{+}^{n}) :

Ω ∋ x → [u(x, .)] ∈ e^{ωt}S′(\overline{ℝ}_{+}^{n}), ωt = ωt_{1} + .... + ωt_{n}

will be refered to as distribution-valued function (d-v-f).

The space of d-v-f defined on Ω we denote by (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω}. A d-v-f [u(x, .)] = [u_{0}(x, .)]e^{ωt}, where [u_{0}(x, .)] ∈ S′(\overline{ℝ}_{+}^{n}) for every x ∈ Ω, hence [u(x, .)]e^{−ωt} = [u_{0}(x, .)] ∈ S′(\overline{ℝ}_{+}^{n}).

It is easily seen that [u(x, .)] ∈ (e^{ω_{1}t}S′(\overline{ℝ}_{+}^{n}))^{Ω} for every ω_{1} ≥ ω.

If [u_{1}(x, .)] and [u_{2}(x, .)] belong to (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω}, then [u_{1}(x, .)] = [u_{2}(x, .)] if and only if ⟨[u_{1}(x, t) − [u_{2}(x, t)], φ(t)⟩ = 0 for every φ ∈ D(ℝ^{n}), x ∈ Ω.

Definition 5. A d-v-f (1) is called continuous if for every φ ∈ S(ℝ^{n}) the function x → ⟨[u(x, t)]e^{−ωt}, φ(t)⟩ is continuous.

Let x_{0} ∈ ∂Ω (∂Ω is the boundary of Ω). By definition:

lim_{x→x_{0},x∈Ω}[u(x, .)] = [u_{0}(t)]e^{ωt} ∈ e^{ωt}S′(\overline{ℝ}_{+}^{n})

if [u(x, .)]e^{−ωt} converges to [u_{0}(t)] in S′(ℝ^{n}).

Remarks: a) The definition of the continuity and the limit do not depend on ω_{1} ≥ ω.

b) The defined continuity of a d-v-f is equivalent to the statement that the mapping x → [u(x, .)] of Ω into e^{ωt}S′(ℝ^{n}) is continuous (S′(ℝ^{n}) provided with the usual topology).

Definition 6. A distribution u ∈ D′(Ω × ℝ^{n}) is called the distribution induced by the d-v-f (1) if

⟨u, ψ⟩ = ∫_{Ω} ⟨[u(x, .)], ψ(x, .)⟩dx, ψ ∈ D(Ω × ℝ^{n}),

provided the integral exists for every ψ ∈ D(Ω × ℝ^{n}).

The next Proposition gives conditions that a distribution, induced by a d-v-f, has been induced by a unique d-v-f.

Proposition 1. Let u ∈ D′(Ω × ℝ_{+}^{n}) ⊂ D′(Ω × ℝ^{n}) be the distribution induced by a continuous d-v-f [u(x, .)], x ∈ Ω. Then [u(x, .)], x ∈ Ω, is uniquely determined by u in the class of continuous d-v-fs: Ω → e^{ωt}S′(\overline{ℝ}_{+}^{n}).

P r o o f. Assume that there are two d-v-fs [u_{1}(x, .)] and [u_{2}(x, .)] which induce the distribution u. Put [w(x, .)] = [u_{1}(x, .)] − [u_{2}(x, .)] and let

ψ(x, t) ∈ D(Ω × ℝ^{n}), ψ(x, t) = α(x)β(t), α ∈ D(Ω) and β ∈ D(ℝ^{n}).

By Definition 6 we have for every α and β :

∫_{Ω} ⟨[w(x, .)], β(t)⟩α(x)dx = 0.

It follows that

⟨[w(x, .)]e^{−ωt}, e^{ωt}β(t)⟩ = 0 for a.a. x ∈ Ω.  (1)

Since D(ℝ^{n}) is dense in S(ℝ^{n}), (1) is also true for every φ ∈ S(ℝ^{n}), i.e.,

⟨[w(x, .)]e^{−ωt}, φ(t)⟩ = 0 for a.a. x ∈ Ω.

From the Definition on continuity it follows that the distribution [w(x, .)] = 0 for x ∈ Ω. Consequently, [u_{1}(x, .)] = [u_{2}(x, .)], x ∈ Ω.

Definition 7. Let η = (η_{1}, ..., η_{m}), η_{s} = 0 for s ≠ j and η_{j} = 1. Let ε_{0} > 0 be such that for x_{0} ∈ Ω, x_{0} + ηε ∈ Ω, |ε| < ε_{0}. If for [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω} the limit

lim_{ε→0} \frac{1}{ε}([u(x_{0} + εη, .)] − [u(x_{0}, .)])e^{−ωt} = [v(x_{0}, .)]

exists in S′(ℝ^{n}) for every x_{0} ∈ Ω, we put by definition

\frac{∂}{∂x_{j}} [u(x, .)] = [v(x, .)]e^{ωt} ∈ (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω}.

If the function Ω ∋ x → ⟨[u(x, t)]e^{−ωt}, φ(t)⟩ is of class C^{k}(Ω) for every φ ∈ S(ℝ^{n}), then the d-v-f [u(x, .)] is by definition of class C^{k}(Ω).

Remark. Let U(x, t) denote U(x, t) = ⟨[u(x, t)]e^{−ωt}, φ(t)⟩. Then

\frac{∂}{∂x}[u(x, .)] exists if and only if there exists v(x, t) ∈ S′(\overline{ℝ}_{+}^{n}), x ∈ Ω, such that

\frac{∂}{∂x}U(x, t) = \frac{∂}{∂x}⟨[u(x, t)]e^{−ωt}, φ(t)⟩ = ⟨v(x, t), φ(t)⟩

for every φ ∈ S(ℝ^{n}) and x ∈ Ω.

The next two propositions refer to the connection between some operations on a distribution induced by a d-v-f [u(x, .)] and this d-v-f [u(x, .)].

Proposition 2. Let u ∈ D′(Ω × ℝ^{n}) be the distribution induced by a d-v-f [u(x, .)] ∈ (e^{ωt}S(\overline{ℝ}_{+}^{n}))^{Ω} of class C^{1}. Then the distribution derivative D_{x_{j}}u, j ∈ (1, ..., m), is indiced by the d-v-f:

Ω ∋ x → \frac{∂}{∂x_{j}} [u(x, .)] = [q_{j}(x, .)]e^{ωt} ∈ e^{ωt}S′(\overline{ℝ}_{+}^{n}),

i.e., for every ψ ∈ D(Ω × R^{n}) we have

⟨D_{x_{j}}u, ψ⟩ = ∫_{Ω} ⟨\frac{∂}{∂x_{j}} u(x, .), ψ(x, .)⟩dx.

P r o o f. By Definition 6 for every ψ ∈ D(Ω × R^{n}) :

\begin{aligned}⟨D_{x_{j}}u, ψ⟩ & = −⟨u, \frac{∂}{∂x_{j}} ψ⟩ \\ & = − ∫_{Ω} ⟨u(x, .)], \frac{∂}{∂x_{j}} ψ(x, .)⟩dx. \end{aligned}

It is well-knowm (cf. [4, Chapter IV, Theorem III]) that the subspace of functions of the form ∑_{i} v_{i}(x)w_{i}(t) is dense in D(Ω × ℝ^{n}), where v_{i} ∈ D(Ω)

and w_{i} ∈ D(ℝ^{n}). For every ψ ∈ D(Ω × ℝ^{n}) there exists a sequence ψ_{ν}(t) =

∑_{i}^{ν} v_{i}(x)w_{i}(t) which converges in D(Ω × ℝ^{n}) to ψ(x, t).

Hence, there exist compact sets K_{v} ⊂ Ω and K_{w} ⊂ ℝ^{n} such that suppψ ⊂ K_{v} ×K_{w}, v_{i} ∈ C_{0}^{∞}(K_{v}) and w_{i} ∈ C_{0}^{∞}(K_{w}) for i ∈ ℕ; {ψ_{ν}(x, t)}_{ν∈ℕ} converges uniformly on K_{v} × K_{w}. Thus

⟨D_{x_{j}}u, ψ⟩ = − lim_{ν→∞} ∑_{i}^{ν} ∫_{Ω}⟨[u(x, .)], w_{i}(.)⟩\frac{∂}{∂x_{j}} v_{i}(x)dx

= lim_{ν→∞} ∑_{i}^{ν} ∫_{Ω} \frac{∂}{∂x_{j}} ⟨[u_{0}(x, .)], e^{ωt}w_{i}(.)⟩v_{i}(x)dx

= lim_{ν→∞} ∑_{i}^{ν} ∫_{Ω}⟨[q(x, .)]e^{ωt}, w_{i}(.)⟩v_{i}(x)dx

= ∫_{Ω}⟨[q(x, .)]e^{ωt}, ψ(x, .)⟩dx

= ∫_{Ω} ⟨\frac{∂}{∂x_{j}} [u(x, .)], ψ(x, .)⟩dx.

Proposition 3. Let Ω be an open interval (a, b) ⊂ ℝ. Suppose that u ∈ e^{ωt}S′((a, b) × ℝ^{n}) satisfies the equation D_{x}u = 0. The distribution u is induced by the constant d-v-f defined on (a, b). This is the unique continuous d-v-f on Ω which induces u.

P r o o f. If D_{x}u = 0, then u does not depend on x ∈ (a, b). Let ψ be any function belonging to D((a, b) × ℝ^{n}). As in the previous proof we use

the sequence ψ_{ν}(x, t) = ∑_{i}^{ν} v_{i}(x)w_{i}(t) and we have

⟨u, ψ⟩ = lim_{ν→∞} ∑_{i}^{ν}⟨⟨u(.), w_{i}(.)⟩, v_{i}(x)⟩

= lim_{ν→∞} ∑_{i}^{ν} ∫_{Ω}⟨u(.), w_{i}(.)⟩v_{i}(x)dx

= ∫_{Ω}⟨u(.), ψ(x, .)⟩dx.

By Definition 6 u is defined by the d-v-f: (a, b) ∋ x → [u(.)]. By Proposition 1 this is the unique continuous d-v-f which induces u.

We are going now to prove some properties of the regular d-v-f.

Let C_{Ω,\overline{ℝ}_{+}^{n}} denote the class of functions f(x, t), x ∈ Ω, t ∈ ℝ^{n} such that:

1. suppf(x, t) ⊂ \overline{ℝ}_{+}^{n}, x ∈ Ω;

2. |f(x, t)e^{−ωt}/(1 + |t|^{2})^{m/2}| ≤ g(t) ∈ L^{1}(\overline{ℝ}_{+}^{n}) for an m ∈ ℕ_{0} and for x ∈ Ω.

Proposition 4. Let Ω be an open set in ℝ^{m}.

a) If f ∈ C_{Ω,\overline{ℝ}_{+}^{n}}, then f defines a d-v-f [f(x, .)] which is called regular d-v-f.

b) If f ∈ C_{Ω,\overline{ℝ}_{+}^{n}} and f is continuous in x ∈ Ω for almost all t ∈ \overline{ℝ}_{+}^{n}, then [f(x, .)] is a continuous d-v-f on Ω. It defines a regular distribution on Ω × ℝ^{n}, as well.

c) If f is a function defined on Ω × ℝ^{n} such that \frac{∂^{i}}{∂x_{j}} f(x, t), i = 0, ..., k,

are continuous in x ∈ Ω for a.a. t ∈ \overline{ℝ}_{+}^{n}. and belong to C_{Ω,\overline{ℝ}_{+}^{n}}, then

\frac{∂^{i}}{∂x_{j}^{i}} f(x, t) defines a regular continuous d-v-f on Ω :

[\frac{∂^{i}}{∂x_{j}^{i}} f(x, t)], i = 0, ..., k, j ∈ (1, ..., m) and

\frac{∂^{i}}{∂x_{j}^{i}} [f(x, .)] = [\frac{∂^{i}}{∂x_{j}^{i}} f(x, .)], x ∈ Ω.

P r o o f. a) By Proposition 3, p. 158 in [5], [f(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω}.

b) By definition of a regular d-v-f defined by f(x, t) depending on the parameter x, we have for x ∈ Ω and φ ∈ S(ℝ^{n}) :

⟨[f(x, .)]e^{−ωt}, φ(·)⟩ = ∫_{\overline{ℝ}_{+}^{n}} f(x, t)e^{−ωt}φ(t)dt.

The proof of b) follows now by the properties of the integral (cf. [4], Proposition 45).

c) We prove only if i = 1. For i > 1 we have only to repeat the procedure. By a) f defines a regular d-v-f

[f(x, .)] = e^{ωt}[e^{−ωt}f(x, t)].

Following Definition 7 of the derivative in x_{j} of [f(x, .)] and the Remark after it we analyse the limit:

lim_{ε→0} \frac{1}{ε}⟨[e^{−ωt}f(x + εη, t)] − [e^{−ωt}f(x, t)], φ(t)⟩

= lim_{ε→0} \frac{1}{ε} ∫_{\overline{ℝ}_{+}^{n}} e^{−ωt}(f(x + εη, t) − f(x, t))φ(t)dt

= ∫_{\overline{ℝ}_{+}^{n}} e^{−ωt} \frac{∂}{∂x_{j}} f(x, t)φ(t)dt

= ⟨e^{−ωt} \frac{∂}{∂x_{j}} f(x, t), φ(t)⟩, φ ∈ S(ℝ^{n})

(cf. Proposition 46, p. 62 in [4]).

We apply a) once more to \frac{∂}{∂x_{j}} f(x, t). Then this partial derivative defines

a d-v-f, as well, and

e^{−ωt}[\frac{∂}{∂x_{j}} f(x, .)] ∈ S′(\overline{ℝ}_{+}^{n}).

Hence

\frac{∂}{∂x_{j}} [f(x, .)] = [\frac{∂}{∂x_{j}} f(x, .)].

Proposition 5. Let Ω ⊂ ℝ^{m} be an open set and Ω_{1} any open subset of Ω, \overline{Ω}_{1} ⊂ Ω.. Let f ∈ C_{Ω,\overline{ℝ}_{+}^{n}}. If for any x_{0} ∈ ∂Ω_{1} we have:

a) There exist

lim_{x→x_{0},x∈Ω_{1}} f(x, t) = v_{1}(t) and lim_{x→x_{0},x∈Ω\\overline{Ω}_{1}} f(x, t) = v_{2}(t)

for a.a. t ∈ \overline{ℝ}_{+}^{n}; b) there exist p_{i} ∈ ℕ_{0}, such that e^{−ωt}v_{i}(t)/(|t|^{p_{i}} + 1) ∈ L^{1}(\overline{ℝ}_{+}^{n}), i = 1, 2;

c) [f(x, .)] is a continuous d-v-f on Ω.

Then f(x, t) is continuous on Ω for a.a. t ∈ \overline{ℝ}_{+}^{n}.

P r o o f. By Definition 5

lim_{x→x_{0},x∈Ω_{1}}[f(x, .)] = lim_{x→x_{0},x∈Ω_{1}} ∫_{\overline{ℝ}_{+}^{n}} f(x, t)e^{−ωt}φ(t)dt

= ∫_{\overline{ℝ}_{+}^{n}} v_{1}(t)e^{−ωt}φ(t)dt

and

lim_{x→x_{0},x∈Ω\\overline{Ω}_{1}}[f(x, .)] = lim_{x→x_{0},x∈Ω\\overline{Ω}_{1}} ∫_{\overline{ℝ}_{+}^{n}} f(x, t)e^{−ωt}φ(t)dt

= ∫_{\overline{ℝ}_{+}^{n}} v_{2}(t)e^{−ωt}φ(t)dt

for every φ ∈ S(ℝ^{n}). Since [f(x, .)] is continuous, we have:

∫_{\overline{ℝ}_{+}^{n}} e^{−ωt}(v_{1}(t) − v_{2}(t))φ(t)dt

for every φ ∈ S(ℝ^{n}). Consequently, v_{1}(t) = v_{2}(t) for a.a. t ∈ \overline{ℝ}_{+}^{n}.

Proposition 6. Let ∂Ω denote the boundary of Ω. If f ∈ C_{Ω,\overline{ℝ}_{+}^{n}} and for an x_{0} ∈ ∂Ω there exists

lim_{x→x_{0},x∈Ω} f(x, t) = f_{0}(t) for a.a. t ∈ \overline{ℝ}_{+}^{n},

then we have

lim_{x→x_{0},x∈Ω}[f(x, .)] = [f_{0}(t)e^{−ωt}]e^{ωt} = [f(·)].

P r o o f. By Definition 5 we have first to find

lim_{x∈Ω^{x→x_{0}}}⟨[f(x, t)]e^{−ωt}, φ(t)⟩ = lim_{x∈Ω^{x→x_{0}}} ∫_{\overline{ℝ}_{+}^{n}} f(x, t)e^{−ωt}φ(t)dt

= ∫_{\overline{ℝ}_{+}^{n}} f_{0}(t)e^{−ωt}φ(t)dt

because f ∈ C_{Ω,\overline{ℝ}_{+}^{n}}. This proves the assertion of Proposition 6.

The product of a numerical function c(x), x ∈ Ω, and a d-v-f [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω}, c(x)[u(x, .)], we define in every point x_{0} ∈ Ω :

x_{0} → c(x_{0})[u(x_{0}, .)].  (2)

It is easily seen that if c(x) is continuous on Ω, then c(x) [u(x, .)] is a continuous d-v-f on Ω.

We cite the next Proposition which gives the relation between the distributional and classical derivatives of a function.

Proposition C. (cf. [3]) Let f ∈ C^{(p)}((−∞, b)), p ∈ ℕ_{0} ≡ ℕ ∪ {0}, and H_{a} be a function such that H_{a}(x) = 0, −∞ < x < a < b ≤ ∞; H_{a}(x) = 1, 0 ≤ a ≤ x < b. Denote by [ℍ_{a}f] the regular distribution defined by H_{a}f. Hence, [H_{a}f] ∈ D′((−∞, b)), supp[H_{a}f] ⊂ [a, b) or [H_{a}f] ∈ D′([a, b)), as well. By [f_{a}^{(p)}], p ∈ ℕ, we denote the distribution defined by the function f_{a}^{(p)} equals f^{(p)}(x), x ∈ (a, b) and equals zero for x ∈ (−∞, a) and is not defined for x = a.

Since the function (H_{a}f)^{(k)} has in general a discontinuity of the first kind in x = a, k = 0, 1, ..., p, by the well known formula (cf. [51])

\begin{aligned}D^{p}[H_{a}f] & = [f_{a}^{(p)}] + f^{(p−1)}(a)δ(x − a) + ... + f(a)δ^{(p−1)}(x − a) \\ & = [f_{a}^{(p)}] + R_{p,a}(f) = [H_{a}f^{(p)}] + R_{p,a}(f), \end{aligned}

where D^{p}[H_{a}f] is the derivative of order p in the sense of distributions, and

R_{a,p}(f) = f^{(p−1)}(a)δ(x − a) + ... + f(a)δ^{(p−1)}(x − a).

We define now the Laplace transform (in short LT) of a d-v-f.

Definition 8. Let [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{Ω}. The LT of [u(x, .)] is defined by

L[u(x, .)](z) ≡ û(x, z) =
= ⟨[u(x, t)]e^{−ωt}, η(t)e^{−(z−ω)t}⟩, z ∈ ω + ℝ_{+}^{n} + iℝ^{n}.

For the function η(t) cf. Definition 1. In this way the LT of a d-v-f has been deduced to the LT of tempered distributions.

Definition 9. The LT of elements belonging to D′_{ω} is defined by

L(D′_{ω}([0, b)^{n}) = L(e^{ωt}S(\overline{ℝ}_{+}^{n}))/L(A_{b}).

Proposition 7. If f(x, t) belongs to C_{Ω,\overline{ℝ}_{+}^{n}} and for every x ∈ Ω has the classical LT in t denoted by F (x, z), Re z > ω, then the regular d-v-f [f(x, .)] has the LT f̂(x, z), Re z > ω, and f̂(x, z) = F (x, z), Re z > ω.

P r o o f. By Proposition 1. a) f(x, t) defines the regular d-v-f [f(x, .)]. Then

\begin{aligned}f̂(x, z) & = ⟨[f(x, t)]e^{−ωt}, η(t)e^{−(z−ω)t}⟩ \\ & = ∫_{\overline{ℝ}_{+}^{n}} f(x, t)e^{−zt}dt = F (x, z), x ∈ Ω, Re z > ω. \end{aligned}

Proposition 8. Let x_{0} ⊂ ∂Ω. If f(x, t) belongs to C_{Ω,\overline{ℝ}_{+}^{n}} and converges for almost all t ∈ \overline{ℝ}_{+}^{n} to f_{0}(t) when x → x_{0}, x ∈ Ω, then

lim_{x→x_{0},x∈Ω} L[f(x, t)](z) = L([f_{0}])(z), z ∈ ω + ℝ_{+}^{n} + iℝ^{n}.

Since η(t)e^{−(z−ω)t} ∈ S(ℝ^{n}), Re z > ω, the proof of Proposition 8 follows from Proposition 6 and Proposition 7.

3. Application to mathematical models appeared in mechanics

In this Section 3 our aim is only to illustrate the method elaborated in Section 2 analysing the generalized solutions to a partial differential equation.

The mathematical model of the vibrating rod is

\frac{∂^{4}}{∂x^{4}} u(x, t) + \frac{∂^{2}}{∂t^{2}} u(x, t) = 0, 0 < x < 1, t ≥ 0.  (3)

We shall examine first the uniqueness of generalized solutions to (3) with boundary conditions (4) of clamped ends:

lim_{x→0^{+}} u(x, t) = lim_{x→1^{−}} u(x, t) = lim_{x→0^{+}} \frac{∂}{∂x}u(x, t)

= lim_{x→1^{−}} \frac{∂}{∂x}u(x, t) = 0, t ∈ \overline{ℝ}_{+}.

(4)

First we have to find the equation in (e^{−ωt}S′(\overline{ℝ}_{+})^{(0,1)} which corresponds to equation (3) in such a way that if there exists a solution [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}))^{(0,1)} of this equation and is given by a function u(x, t) which

has the continuous classical partial derivatives \frac{∂^{4}}{∂x^{4}} u(x, t) and \frac{∂^{2}}{∂t^{2}} u(x, t), then u(x, t) has to be a solution to (3). Let us do it.

The function u(x, t), a solution to (3), has its support in (0, 1) × [0, ∞). To use the Laplace transform we have to extend it on (0, 1) × (−∞, ∞) in such a way that u(x, t) = 0 on (0, 1) × (−∞, 0). Then by Proposition C

[\frac{∂^{2}}{∂t^{2}} u(x, t)] = D_{t}^{2}[u(x, t)] − u(x, 0)δ^{(1)}(t) − \frac{∂}{∂t}u(x, t)|_{t=0}δ(t).

Now to equation (3) it corresponds in the space (e^{ωt}S′(\overline{ℝ}_{+}^{n}))^{(0,1)} the equation (cf. Proposition 4 c) and Proposition C):

\frac{∂^{4}}{∂x^{4}} [u(x, .)] + D_{t}^{2}[u(x, .)] = B_{1}(x)δ(t) + B_{0}(x)δ^{(1)}(t),  (5)

where

B_{0}(x) = u(x, 0) and B_{1}(x) = \frac{∂}{∂t}u(x, t)|_{t=0}.  (6)

The functions B_{0} and B_{1} are continuous in x ∈ (0, 1) which can be continuously extended to [0, 1]. D_{t}^{2} is the second partial derivative in the sense of distributions.

Because of the equality in the space (e^{ωt}S′(\overline{ℝ}_{+}))^{(0,1)} equation (5) can be written in the form:

⟨\frac{∂^{4}}{∂x^{4}} [u(x, t)] + D_{t}^{2}[u(x, t)] − B_{1}(x)δ(t) − B_{0}(x)δ^{(1)}(t), φ(t)⟩ = 0

for every φ ∈ S(ℝ). Since η(t)e^{−zt} ∈ S(ℝ), z ∈ ω + ℝ_{+} + iℝ, by Definition 8 and the Remark after Definition 7 the last equation gives

\frac{∂^{4}}{∂x^{4}} ⟨u(x, t)e^{−ωt}, e^{−(z−ω)t}⟩ + ⟨D_{t}^{2}u(x, t), e^{−zt}⟩ = B_{0}(x)z + B_{1}(x),

z ∈ ω + ℝ_{+} + iℝ, and finally

\frac{∂^{4}}{∂x^{4}} û(x, z) + z^{2}û(x, z) = B_{1}(x) + B_{0}(x)z.  (7)

Let us remark that (7) is a classical differential equation in x with a parameter z. Hence we can apply the well known methods to solve it.

3.1 Uniqueness of a solution to (3)

Suppose that we have two solutions u_{1}(x, t) and u_{2}(x, t) to (3), (4) with the same initial condition (6). Then W (x, t) = u_{1}(x, t) − u_{2}(x, t) satisfies the initial condition with B_{0}(x) = 0 and B_{1}(x) = 0, the boundary condition (4) and the homogenuous part of equation (3). Then the corresponding equation (5) is also homogeneous and the LT of it is

\frac{∂^{4}}{∂x^{4}} Ŵ(x, z) + z^{2}Ŵ(x, z) = 0, 0 < x < 1, Re z > ω.  (8)

The general solution to (8) is

Ŵ(x, z) = ∑_{j=1}^{4} C_{j}(z)e^{r_{j}x}, 0 < x < 1, Re z > ω,  (9)

where C_{j}(z), j = 1, ..., 4, are functions in z, Re z > ω and r_{j}, j = 1, ..., 4 are solutions to equation r^{4} + z^{2} = 0 :

r_{1} = \sqrt{\frac{z}{2}}(1+i), r_{2} = \sqrt{\frac{z}{2}}(−1+i), r_{3} = \sqrt{\frac{z}{2}}(−1−i), r_{4} = \sqrt{\frac{z}{2}}(1−i). (10)

In order to apply Proposition 8 we give an other form to Ŵ(x, z) given in (9):

\begin{aligned}Ŵ(x, z) & = D_{1}(z)e^{−r_{1}(1+a−x)} + D_{2}(z)e^{r_{2}(x+a)} \\ & +D_{3}(z)e^{r_{3}(x+a)} + D_{4}(z)e^{−r_{4}(1+a−x)} , \end{aligned}

where a > 0 and D_{i}(z), i = 1, ..., 4 are holomorphic functions for Re z > ω > 0. Now we can apply Proposition 8 when we use boundary conditions (4).

In this way the functions D_{j}, j = 1, ..., 4 have to satisfy the homogeneous system:

\begin{aligned}D_{1}(z)e^{−r_{1}(1+a)} + D_{2}(z)e^{r_{2}a} + D_{3}(z)e^{r_{3}a} + D_{4}(z)e^{−r_{4}(1+a)} & = 0 \\ D_{1}(z)e^{−r_{1}a} + D_{2}(z)e^{r_{2}(1+a)} + D_{3}(z)e^{r_{3}(1+a)} + D_{4}(z)e^{−r_{4}a} & = 0 \\ D_{1}(z)r_{1}e^{−r_{1}(1+a)} + D_{2}(z)r_{2}e^{r_{2}a} + D_{3}(z)r_{3}e^{r_{3}a} + D_{4}(z)r_{4}e^{−r_{4}(1+a)} & = 0 \\ D_{1}(z)r_{1}e^{−r_{1}a} + D_{2}(z)r_{2}e^{r_{2}(1+a)} + D_{3}(z)r_{3}e^{r_{3}(1+a)} + D_{4}(z)r_{4}e^{−r_{4}a} & = 0. \end{aligned}

(11)

We will prove that there is no number ω > 0 such that D_{i}(z), i = 1, ..., 4, are solutions to the system (11) and are not identically zero for Re z > ω.

The determinant of the system (11) has the form

Δ(z) = | \begin{matrix}1 & 1 & 1 & 1 \\ e^{r_{1}} & e^{r_{2}} & e^{−r_{1}} & e^{−r_{2}} \\ r_{1} & r_{2} & −r_{1} & −r_{2} \\ r_{1}e^{r_{1}} & r_{2}e^{r_{2}} & −r_{1}e^{−r_{1}} & −r_{2}e^{−r_{2}} \end{matrix} | e^{−r_{1}(1+2a)+r_{2}(1+2a)}  (12)

= Δ′(z)e^{−r_{1}(1+2a)+r_{2}(1+2a)}.

In this calculations we used the relations between r_{j} : r_{3} = −r_{1} and r_{4} = −r_{2}.

Let k_{0} ∈ ℕ be such that 8k_{0}^{2}π^{2} > ω for a given ω > 0. Then if z = 8k^{2}π^{2}, then \sqrt{\frac{z}{2}} = 2kπ. It is easily seen that Δ′(8k^{2}π^{2}) ≠ 0, k > k_{0}, k ∈ ℕ. Since Δ′(ξ) is continuous function for ξ > 0, then Δ′(ξ) ≠ 0 in a sequence of open neighbourhoods Ω_{k} of the points ξ = 8k^{2}π^{2}. Consequently, every function D_{i}(z), i = 1, ..., 4, equals zero in each open set in which it is holomorphic, if this open set contains an Ω_{k}. Hence, there is no ω > 0 such that Δ(z) ≠ 0 for Re z > ω. This prove that the system (11) has the unique solution D_{j}(z) ≡ 0, j = 1, ..., 4, Re z > ω.

Theorem 1. A generalized solution [u(x, .)] to (3), (4), (6), such that

the d-v-f \frac{∂^{i}}{∂x^{i}} [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}))^{(0,1)}, i = 0, 1, ..., 4, and u_{x}^{(i)}(x, t) ∈

C_{(0,1)×\overline{ℝ}_{+}}, i = 0, 1, if it exists, it is unique.

Remark. It is well known that (3), (4) has a family of solutions of the form u_{k}(x, t) = V_{k}(x)T_{k}(t), k ∈ ℕ, (cf. [1]). Then B_{0}(x) = V (x)T (0) and B_{1}(x) = V (x)T^{(1)}(0) for a fixed k ∈ ℕ. With such initial condition the corresponding solution is unique. Hence the boundary conditions of the form (4) is not sufficient for unicity.

3.2. Existence of the solution to (3), (13), (6)

We have seen that the existence of a solution to (3) depends not only on the boundary conditions but it also depends on the initial condition. We are interested in finding the form of the initial condition (6) such that equation (3) has a solution in case of supported ends of the vibrating rod. Then the boundary condition is

lim_{x→0^{+}} u(x, t) = lim_{x→1^{−}} u(x, t) = lim_{x→0^{+}} \frac{∂^{2}}{∂x^{2}} u(x, t) =

= lim_{x→1^{−}} \frac{∂^{2}}{∂x^{2}} u(x, t) = 0, t ∈ ℝ_{+}.

(13)

For this purpose we have first to find a particular solution û_{p}(x, z) to differential equation (7) which satisfies condition (13) (cf. Proposition 8). This is very easy if B_{0}(x) and B_{1}(x) are of some special form. Then by Proposition A we have to show that û_{p}(x, z) is the LT of a solution to (3).

To reduce the routine work, let us suppose that B_{0} = 0 in (6). The solution of the homogeneous part of equation (7) is

û_{0}(x, z) = ∑_{j=1}^{4} C_{j}(z)e^{r_{j}x}, 0 < x <, Re z > ω,  (14)

where C_{j}(z), j = 1, ..., 4, have the same properties as in (9) and r_{j}, j = 1, ..., 4 have been given in (10).

In order to find a particular solution û_{1}(x, z) to (7) with B_{0} = 0, we use the method named ”Variation of Constants”.

Suppose that C_{j}(x, z), j = 1, ..., 4 are functions with the continuous first partial derivative in x for every z, Re z > ω. Then we have to solve the system

\begin{aligned}∑_{j=1}^{4} \frac{∂}{∂x}C_{j}(x, z)e^{r_{j}x} & & & = 0 \\ ∑_{j=1}^{4} \frac{∂}{∂x}C_{j}(x, z)r_{j}e^{r_{j}x} & & & = 0 \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ ∑_{j=1}^{4} \frac{∂}{∂x}C_{j}(x, z)r_{j}^{3}e^{r_{j}x} & & & = B_{1}(x). \end{aligned}  (15)

The determinant of the system (15) is

Δ(z) = | \begin{matrix}e^{r_{1}x} & & ... & & e^{r_{4}x} \\ r_{1}e^{r_{1}x} & & ... & & r_{4}e^{r_{4}x} \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ r_{1}^{3}e^{r_{1}x} & & ... & & r_{4}^{3}e^{r_{4}x} \end{matrix} | \begin{aligned}= & −2(\sqrt{2z})^{6} ≠ 0, \\ & Re z > ω > 0. \end{aligned}  (16)

Hence we can find

\frac{∂}{∂x}C_{j}(x, z) = \frac{1}{2(\sqrt{2z})^{3}} A_{j}B_{1}(x)e^{−r_{j}x}, j = 1, ..., 4,

where A_{1} = (−1)(1 + i), A_{2} = −1 + i, A_{3} = 1 + i and A_{4} = 1 − i. The looked - for functions C_{j}(x, z), j = 1, ..., 4 are

C_{j}(x, z) = \frac{1}{2(\sqrt{2z})^{3}} A_{j}( ∫_{0}^{x} B_{1}(ξ)e^{−r_{j}ξ}dξ + E_{j}(z)),  (17)

where E_{j}(z) are undefined functions.

A general solution to (7) with B_{0}(x) = 0 is:

û(x, z) = \frac{1}{2(\sqrt{2z})^{3}} ∑_{j=1}^{4} A_{j}( ∫_{0}^{x} B_{1}(ξ)e^{−r_{j}ξ}dξ + E_{j}(z))e^{r_{j}x}.  (18)

We have two limitations on û(x, z). One of them comes from the boundary condition (13) and the other requires that there exists [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}))^{(0,1)} such that L([u(x, .)]) = û(x, z). This limitations give the conditions for B_{1}(x) to have a solution to (5), (13). For practical use we shall prove

Theorem 2. A sufficient condition that the differential problem (3),

(13) has a solution [u(x, .)] such that the d-v-f

\frac{∂^{i}}{∂x^{i}} [u(x, .)] ∈ (e^{ωt}S′(\overline{ℝ}_{+}))^{(0,1)}, i = 0, 1, ..., 4,

and u^{(i)}(x, t) ∈ C_{(0,1)×\overline{ℝ}_{+}}, i = 1, 2, is that we can find functions E_{i}(z) such that:

1. ∫_{0}^{x} B_{1}(ξ)e^{−r_{i}ξ}dξ + E_{i}(z) = F̂_{i}(x, z)e^{−r_{i}x}, i = 1, ..., 4, for 0 ≤ x ≤ 1,

where

E_{1}(z) = E_{3}(z), E_{2}(z) = E_{4}(z), Re z > ω.

2. F̂_{1}(x, z) − F̂_{3}(x, z) = F̂_{2}(x, z) − F̂_{4}(x, z) = 0 for x = 0, and x = 1.

3. F̂_{i}(x, z)/(\sqrt{z})^{3} is holomorphic for Re z > ω, for an ω > 0 and tends to zero when z tends to infinity belonging to the half plane {Re z ≥ ω+δ, δ > 0} uniformly in x ∈ [0, 1]. Also

∫_{−∞}^{∞} |F̂(x, z)/(\sqrt{z})^{3}|d(Imz) < ∞, 0 < x < 1, Re z > ω.

P r o o f. Let us suppose that conditions 1., 2. and 3. are satisfied. Then û(x, z), given by (8) can have the form

û(x, z) = \frac{1}{2(\sqrt{2z})^{3}} ∑_{j=1}^{4} A_{j}F̂(x, z).  (19)

By Theorem 3 in [2, I, p. 263] there exist functions F_{i}(x, t) such that L^{−1}(F̂_{i}(x, z) /(2\sqrt{2z})^{2})(t) = F_{i}(x, t), i = 1, ..., 4, and the function u(x, t) = L^{−1}û(x, z)(t),

u(x, t) = ∑_{j=1}^{4} A_{j}F_{j}(x, t), 0 ≤ x ≤ 1, t > 0.  (20)

Now we can use Proposition 8 to satisfy boundary condition (13). We start with (19) and

\frac{∂^{2}}{∂x^{2}} û(x, z) = \frac{1}{2(\sqrt{2z})^{3}} ∑_{j=1}^{4} A_{j}(r_{j}^{−2}F̂_{j}(x, z) + r_{1}^{−1}B_{1}(x) + B_{1}^{(1)}(x)).

In order to satisfy boundary condition (13) we obtain the following system:

\begin{aligned}∑_{j=1}^{4} A_{j}F̂_{j}(0, z) & = 0 \\ ∑_{j=1}^{4} A_{j}F̂_{j}(1, z) & = 0 \\ ∑_{j=1}^{4} A_{j}(r_{j}^{−2}F̂_{j}(0, z) + B_{1}^{(1)}(0)) & = 0 \\ ∑_{j=1}^{4} A_{j}(r_{j}^{−2}F̂_{j}(1, z) + B_{1}^{(1)}(1)) & = 0. \end{aligned}  (21)

Since

∑_{j=1}^{4} A_{j} = 0 and ∑_{j=1}^{4} A_{j}r_{j}^{−2} = (−1)^{j+1}A_{j},

system (21) becomes:

\begin{aligned}∑_{j=1}^{4} A_{j}F̂_{j}(0, z) & = 0 \\ ∑_{j=1}^{4} A_{j}F̂_{j}(1, z) & = 0 \\ ∑_{j=1}^{4}(−1)^{j+1}A_{j}F̂_{j}(0, z) & = 0 \\ ∑_{j=1}^{4}(−1)^{j+1}A_{j}F̂_{j}(1, z) & = 0. \end{aligned}  (22)

From (22) it follows that E_{1}(z) = E_{3}(z), E_{2}(z) = E_{4}(z) and

F̂_{1}(x, z) − F̂_{3}(x, z) = F̂_{2}(x, z) − F̂_{4}(x, z) = 0

for x = 0 and x = 1 which is our supposition 2. This completes the proof of Theorem 2.

As an example we consider the case B_{1}(x) = sin kπx, for a k ∈ ℕ. Then equation (7) becomes

\frac{∂^{4}}{∂x^{4}} û(x, z) + z^{2}û(x, z) = sin kπx, 0 < x < 1, Re z > ω  (23)

In order to find a solution to (23), we suppose that û(x, z) = A(z) sin kπx. In that case (23) gives

(kπ)^{4}A(z) sin kπx + z^{2}A(z) sin kπx = sin kπx.

Hence,

A(z) = 1/((kπ)^{4} + z^{2}), Re z > 0, and û(x, z) = sin kπx/((kπ)^{4} + z^{2}).

It is easily seen that û(x, z) satisfies the boundary condition (13). A solution to (3), (13) with initial condition (6), expressed by B_{0}(x) = 0, B_{1}(x) = sin kπx is u(x, t) = (kπ)^{−2} sin kπx sin(kπ)^{2}t. That is a well known result. It is also easy to verify that B_{1}(x) = sin kπx satisfies the cited conditions in Theorem 2.

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Department of Mathematics University of Novi Sad Trg Dositeja Obradovića 4 21000 Novi Sad Serbia and Montenegro