Variational formulation for fractional hyperbolic problems in the theory of viscoelasticity

Abstract

In this article, a theoretical framework for problems involving fractional equations of hyperbolic type arising in the theory of viscoelasticity is presented. Based on the Galerkin method, a variational problem of the fractionary viscoelasticity is studied. An appropriate functional setting is introduced in order to establish the existence, uniqueness and a priori estimates for weak solutions. This framework is developed in close concordance with important physical quantities of the theory of viscoelasticity.

Appendix A. Appendix

In this appendix, we collect several useful properties of fractional operators used throughout the article; we provide some of their proofs for completeness. For more details, see [10]. In what follows $\alpha \in (0,1)$ and (a, b) is a bounded interval.

Let us start by recalling the Fourier transform and its inverse:

$$\begin{aligned} \mathcal {F}(u)(\omega ) = \hat{u} (\omega ) {:=} \int \limits _{-\infty }^\infty u(t) e^{-i \omega t} \, dt, \\ \mathcal {F}^{-1}(u)(t) = \check{u} (t) {:=} \int \limits _{-\infty }^\infty u(\omega ) e^{-i \omega t} \, d\omega , \end{aligned}$$

where $i = \sqrt{-1}$, and their following well known properties:

$$\begin{aligned} \int \limits _{-\infty }^\infty \hat{u} v = \int \limits _{-\infty }^\infty u \hat{v}, \quad \hat{u}(\omega ) = \check{u} (-\omega ). \end{aligned}$$

(A.1)

The proof of the following important theorem can be found in Section 7.1 of [10].

Property A.1

Given a function $\varphi \in C_c^\infty (\mathbb {R})$, the Fourier transform of its Riemann–Liouville fractional derivatives and integrals satisfy:

• i if $\alpha \ge 0$,

  $$\begin{aligned} \mathcal {F}({}^{}{-\infty }{D}_t^\alpha \varphi )(\omega ) = (i \omega )^\alpha \hat{\varphi }(\omega ), \quad \mathcal {F}({}^{}{t}{D}_\infty ^\alpha \varphi )(\omega ) = (-i \omega )^\alpha \hat{\varphi }(\omega ); \end{aligned}$$

• ii if $\alpha \in [0,1)$,

  $$\begin{aligned} \mathcal {F}({}^{}{-\infty }{I}_t^{\alpha /2} \varphi )(\omega ) = \frac{\hat{\varphi }(\omega )}{(i \omega )^{\alpha /2}}, \quad \mathcal {F}({}^{}{t}{D}_\infty ^{\alpha /2} \varphi )(\omega ) = \frac{\hat{\varphi }(\omega )}{(-i \omega )^{\alpha /2}}. \end{aligned}$$

It is important to mention that restricting $\alpha$ to [0, 1) in (ii) guarantees that all the expressions remain in $L^2(\mathbb {R})$. We recall that

$$\begin{aligned} (i \omega )^\alpha = |\omega |^\alpha e^{\frac{\alpha \pi i}{2} {{\,\mathrm{sgn}\,}}\omega }, \end{aligned}$$

(A.2)

where ${{\,\mathrm{sgn}\,}}$ is the sign function.

The proof of the following result can be found in [10], Theorem 2.6.

Property A.2

For all $\beta \ge 0$, ${}^{}{a}{I}_t^\beta$ and ${}^{}{t}{I}_b^\beta$ are bounded linear operators in $L^2(a,b)$.

Property A.3

(Inverse) For all $u \in H_0^{\alpha /2}(a,b)$ we have

$$\begin{aligned}&{}^{}{a}{I}_t^{\alpha /2} {}^{}{a}{D}_t^{\alpha /2} u = {}^{}{a}{D}_t^{\alpha /2} {}^{}{a}{I}_t^{\alpha /2} u = u, \end{aligned}$$

(A.3)

$$\begin{aligned}&{}^{}{t}{I}_b^{\alpha /2} {}^{}{t}{D}_b^{\alpha /2} u = {}^{}{t}{D}_b^{\alpha /2} {}^{}{t}{I}_b^{\alpha /2} u = u. \end{aligned}$$

(A.4)

Proof

Observe that it suffices to prove the result for $u \in C_c^\infty (a,b)$, as by definition all the other elements in $H_0^{\alpha /2}(a,b)$ are limits of sequences of these elements (recall also Lemma 3.4 and Property A.2). So let $u \in C_c^\infty (a,b)$. To prove (A.3), we just have to take Fourier transform in both sides of the equations and use Property A.1. $\square$

Property A.4

(Fractional integration by parts) For every $\varphi , \psi \in C_c^\infty (\mathbb {R})$, we have

$$\begin{aligned}&\int \limits _{-\infty }^\infty {}^{}{-\infty }{D}_t^{\alpha /2} \varphi \psi = \int \limits _{-\infty }^\infty \varphi {}^{}{t}{D}_\infty ^{\alpha /2} \psi , \end{aligned}$$

(A.5)

$$\begin{aligned}&\int \limits _{-\infty }^\infty {}^{}{-\infty }{I}_t^{\alpha /2} \varphi \psi = \int \limits _{-\infty }^\infty \varphi {}^{}{t}{I}_\infty ^{\alpha /2} \psi . \end{aligned}$$

(A.6)

Proof

Let us prove (A.5), the proof of (A.6) is similar. Indeed, by Property A.1 and (A.1) we have

$$\begin{aligned} \int \limits _{-\infty }^\infty {}^{}{-\infty }{D}_t^{\alpha /2} \varphi \psi&= \int \limits _{-\infty }^\infty \varphi \left( (it)^{\alpha /2} \check{\psi }\right) ^\wedge \\&= \int \limits _{-\infty }^\infty \varphi \left( (-it)^{\alpha /2} \check{\psi }(-t) \right) ^\vee = \int \limits _{-\infty }^\infty \varphi {}^{}{t}{D}_\infty ^{\alpha /2} \psi . \end{aligned}$$

$\square$

We observe that (A.5) is still true for every $\alpha > 0$.

Property A.5

(Semigroup property)

• i Given $\beta , \gamma \ge 0$, if $u \in L^1(a,b)$ then

  $$\begin{aligned} {}^{}{a}{I}_t^\beta {}^{}{a}{I}_t^\gamma u = {}^{}{a}{I}_t^{\beta + \gamma } u, \quad {}^{}{t}{I}_b^\beta {}^{}{t}{I}_b^\gamma u = {}^{}{t}{I}_b^{\beta +\gamma } u. \end{aligned}$$

  (A.7)

• ii If $u \in C^1[a,b]$ and $u(a)=0$, then

  $$\begin{aligned} {}^{}{a}{D}_t^{\alpha /2} {}^{}{a}{D}_t^{\alpha /2} u = {}^{}{a}{D}_t^\alpha u; \end{aligned}$$

  (A.8)

Proof

The proof of (i) can be found in Section 2.3 of [10]. To prove (ii) we use the fact that $u(a)=0$ together with (2.5) and (i) to find that

$$\begin{aligned} {}^{}{a}{D}_t^{\alpha /2} {}^{}{a}{D}_t^{\alpha /2} u&= \frac{d}{dt} {}^{}{a}{I}_t^{1- \alpha /2} {}^{}{a}{I}_t^{1- \alpha /2} u' \\&= \frac{d}{dt} {}^{}{a}{I}_t^{2- \alpha } u' \\&= {}^{}{a}{I}_t^{1- \alpha } u' = {}^{}{a}{D}_t^\alpha u. \end{aligned}$$

$\square$

Property A.6

Given $T>0$, vectors $a,b,c,d \in \mathbb {R}^m$ and a vector field $f \in L^1(0,T)^m$, there exists a unique vector field $u \in C^2[0,T]^m$, solution of the fractional Cauchy problem

$$\begin{aligned} \left\{ \begin{matrix} u'' + b {}^{C}{0}{D}_t^\alpha u + au = f(t), \quad t \in (0,T), \\ u(0) = c, \quad u'(0)= d. \end{matrix} \right. \end{aligned}$$

(A.9)

Proof

To solve (A.9), we use a classical approach, to transform the equation in a Volterra integral equation of the second kind: indeed, after integrating (A.9) once and using (2.6) we obtain

$$\begin{aligned} u' = d + {}^{}{0}{I}_t f - b {}^{}{0}{I}_t^{1-\alpha }(u-c) - a {}^{}{0}{I}_t u. \end{aligned}$$

(A.10)

After integrating once more and using (A.7), we deduce that

$$\begin{aligned} u(t) = c + dt + \frac{bc}{\Gamma (3-\alpha )} t^{2-\alpha } + {}^{}{0}{I}_t^2 f(t) - b {}^{}{0}{I}_t^{2-\alpha } u(t) - a {}^{}{0}{I}_t^2 u(t), \end{aligned}$$

(A.11)

where we have used the identity (see Section 2.5 of [10])

$$\begin{aligned} {}^{}{0}{I}_t^\beta 1 = \frac{t^\beta }{\Gamma (\beta +1)}, \quad \beta \ge 0. \end{aligned}$$

The Volterra integral equation (A.11) can be solved by a classical fixed point argument, we omit the details. To conclude, observe that this solution belongs to the space $C^2[0,T]^m$, by (A.10) and (A.11).

$\square$