Quasistatic Limit of a Dynamic Viscoelastic Model with Memory

Abstract

We study the behaviour of the solutions to a dynamic evolution problem for a viscoelastic model with long memory, when the rate of change of the data tends to zero. We prove that a suitably rescaled version of the solutions converges to the solution of the corresponding stationary problem.

Appendix A.

Throughout this section we fix $a_0>0$, $b_0>0$, and $c_1\ge c_0>1$. For every a, b with

$$\begin{aligned} c_0a\le b\le c_1 a,\qquad b\ge b_0,\qquad a\ge a_0, \end{aligned}$$

(A.1)

we consider the polynomial $p(z):=\beta z^3+z^2+\beta b z+a$ depending on the complex variable z. The following result about the roots of this polynomial is used in the proof of Lemma 5.2 and Proposition 5.4.

Lemma A.1

There exists a positive constant $\alpha =\alpha (\beta ,a_0,b_0,c_0,c_1)$ such that, for every $a,b\in {\mathbb {R}}$ satisfying (A.1), the roots of the polynomial p have real parts in the interval $(-\frac{1}{\beta },-\alpha )$.

Proof

Let us set $z:=x+iy$ with $x,y\in {\mathbb {R}}$. Then $p(z)=0$ if and only if

$$\begin{aligned} {\left\{ \begin{array}{ll} \beta x^3+x^2+\beta b x-(3\beta x+1)y^2+a=0,\\ y(-\beta y^2+3\beta x^2+2x+\beta b)=0, \end{array}\right. } \end{aligned}$$

from which we derive

$$\begin{aligned}&{\left\{ \begin{array}{ll} q(x):=\beta x^3+x^2+\beta b x+a=0,\\ y=0, \end{array}\right. } \end{aligned}$$

(A.2)

$$\begin{aligned}&{\left\{ \begin{array}{ll} r(x):=8\beta x^3+8x^2+2\left( \frac{1}{\beta }+\beta b\right) x+b-a=0,\\ y^2=3x^2+\frac{2}{\beta }x+b. \end{array}\right. } \end{aligned}$$

(A.3)

By recalling $a>0$ and $b-a\ge (c_0-1)a>0$, for every $x\ge 0$ we have $q(x)>0$ and $r(x)>0$, and so the real part of the roots cannot be positive or zero. Moreover, since for every $x\le -\frac{1}{\beta }$ we have $\beta x^3+x^2\le 0$, we obtain

$$\begin{aligned}&q(x)\le -b+a\le (1-c_0)a<0\\&r(x)\le b-a-2\left( \tfrac{1}{\beta ^2}+b\right) =-b-a-\tfrac{2}{\beta ^2}<0, \end{aligned}$$

which imply that the real part of the roots does not belong to $(-\infty ,-\frac{1}{\beta }]$. Therefore, by calling $z_1,z_2,z_3\in {\mathbb {C}}$ the three roots of the polynomial p, we can say

$$\begin{aligned} \mathfrak {R}(z_i)\in (-\tfrac{1}{\beta },0)\quad \text {for }i=1,2,3. \end{aligned}$$

(A.4)

Case 1: there is only one real root. In this case by (A.3) there exists a unique $x_1\in (-\frac{1}{\beta },0)$ which satisfies $r(x_1)=0$ and $3x^2_1+\frac{2}{\beta }x_1+b>0$. Indeed by setting $y_1:=\sqrt{3x^2_1+\frac{2}{\beta }x_1+b}$ we obtain that $x_1+iy_1$ and $x_1-iy_1$ are two distinct non-real roots of p. Since

$$\begin{aligned} r(-\tfrac{1}{2\beta })&=-\tfrac{1}{\beta ^2}+\tfrac{2}{\beta ^2}-\tfrac{1}{\beta ^2}-b+b-a=-a<0,\\ r(-\tfrac{\beta (b-a)}{2\left( b\beta ^2+1 \right) })&=\tfrac{\beta ^2(b-a)^2((a+b)\beta ^2+2)}{(b\beta ^2+1)^3}>0, \end{aligned}$$

then $x_1\in (-\frac{1 }{2\beta },-\frac{\beta (b-a)}{2\left( b\beta ^2+1\right) })$. Moreover

$$\begin{aligned} q(-\tfrac{1}{\beta })&=-\tfrac{1}{\beta ^2}+\tfrac{1}{\beta ^2}-b+a=-b+a<0,\\ q(-\tfrac{a}{\beta b})&=-\tfrac{a^3}{b^3\beta ^2}+\tfrac{a^2}{b^2\beta ^2}-a+a=\tfrac{a^2(b-a)}{b^3\beta ^2}>0, \end{aligned}$$

hence there exists $x_0\in (-\frac{1}{\beta },-\frac{a}{\beta b})$ such that $q(x_0)=0$. As a consequence of this, $(x_0,0)$ satisfies the system in (A.2), which implies that $x_0$ is the real root of p, hence we have $\mathfrak {R}(z_i)\in (-\frac{1}{\beta },\max \{-\frac{a}{\beta b},-\frac{\beta (b-a)}{2\left( b \beta ^2+1\right) }\})$. Thanks to (A.1) we can say $-\tfrac{a}{\beta b}\le -\tfrac{1}{c_1\beta }$ and $-\tfrac{\beta (b-a)}{2\left( b\beta ^2+1\right) }\le \tfrac{\beta (1-c_0)a}{2(c_1a\beta ^2+1)}\le \tfrac{\beta (1-c_0)a_0}{2\left( c_1a_0\beta ^2+1\right) }$, where in the last inequality we use the decreasing property of the function $a\mapsto \tfrac{\beta (1-c_0)a}{2(c_1a\beta ^2+1)}$. This implies

$$\begin{aligned} \mathfrak {R}(z_i)\in (-\tfrac{1}{\beta },\max \{-\tfrac{1}{c_1\beta },\tfrac{\beta (1-c_0)a_0}{2\left( c_1a_0\beta ^2+1\right) }\})\quad \text {for }i=1,2,3. \end{aligned}$$

(A.5)

Case 2: there are only real roots. In this case we have $b\le \frac{1}{3\beta ^2}$, otherwise $q'(x)>0$ for every $x\in {\mathbb {R}}$, which forces p to have also non-real roots. Thanks to (A.1) we have also $a<b\le \frac{1}{3\beta ^2}$. By setting ${\tilde{b}}_0:=1-\sqrt{1-3b_0\beta ^2}$, we can write

$$\begin{aligned} -{\tilde{b}}_0a_0\beta \ge -{\tilde{b}}_0a\beta \ge -(1-\sqrt{1-3b\beta ^2})a\beta>\tfrac{-1+\sqrt{1-3 b\beta ^2}}{3\beta }>-\tfrac{1}{\beta }, \end{aligned}$$

which implies

$$\begin{aligned} q'(x)>0\quad \text { for every } x\in [-{\tilde{b}}_0a_0\beta ,+\infty ). \end{aligned}$$

(A.6)

Since

$$\begin{aligned} q(-{\tilde{b}}_0a_0\beta )&\ge \beta ^2{\tilde{b}}_0^2a_0^2(1-\beta ^2{\tilde{b}}_0a_0)+a_0(1-\beta ^2{\tilde{b}}_0b)\\&>a_0(1+\beta ^2{\tilde{b}}_0^2 a_0)(1-\beta ^2{\tilde{b}}_0b)>0, \end{aligned}$$

thanks to (A.1), (A.4), and (A.6) we get

$$\begin{aligned} \mathfrak {R}(z_i)\in (-\tfrac{1}{\beta },-{\tilde{b}}_0a_0\beta ),\quad \text {for }i=1,2,3. \end{aligned}$$

(A.7)

By combining (A.5) and (A.7), we obtain the conclusion with

$$\begin{aligned} \alpha :=\min \{{\tilde{b}}_0a_0\beta ,\frac{1}{c_1\beta },\frac{\beta (c_0-1)a_0}{2\left( c_1a_0\beta ^2+1\right) }\}. \end{aligned}$$

$\square$

The following easy estimate is used in the proof of Lemma 5.2.

Lemma A.2

For every $z,w\in {\mathbb {C}}$ with $\mathfrak {R}(z)>0$ and $\mathfrak {R}(w)<0$ the following inequality holds:

$$\begin{aligned} |(z-w)(z-{{\bar{w}}})|\ge |\mathfrak {R}(w)||\mathfrak {I}(w)|. \end{aligned}$$

Proof

Without loss of generality we can suppose $\mathfrak {I}(w)>0$, otherwise we exchange the role of w with ${{\bar{w}}}$. If $\mathfrak {I}(z)>0$, then

$$\begin{aligned}&|z-w|\ge |\mathfrak {R}(z-w)|=|\mathfrak {R}(z)+\mathfrak {R}(-w)|=\mathfrak {R}(z)+\mathfrak {R}(-w)\ge |\mathfrak {R}(w)|, \\&|z-{{\bar{w}}}|\ge |\mathfrak {I}(z-{{\bar{w}}})|=|\mathfrak {I}(z)+\mathfrak {I}(w)|=\mathfrak {I}(z)+\mathfrak {I}(w)\ge |\mathfrak {I}(w)|, \end{aligned}$$

which give the conclusion in this case. If $\mathfrak {I}(z)<0$, then

$$\begin{aligned}&|z-w|\ge |\mathfrak {I}(z-w)|=|-\mathfrak {I}(-z)-\mathfrak {I}(w)|=\mathfrak {I}(-z)+\mathfrak {I}(w)\ge |\mathfrak {I}(w)|, \\&|z-{{\bar{w}}}|\ge |\mathfrak {R}(z-{{\bar{w}}})|=|\mathfrak {R}(z)+\mathfrak {R}(-w)|=\mathfrak {R}(z)+\mathfrak {R}(-w)\ge |\mathfrak {R}(w)|, \end{aligned}$$

which conclude the proof. $\square$