On Dissipative Solutions to a System Arising in Viscoelasticity

Abstract

We consider a model for an incompressible visoelastic fluid. It consists of the Navier–Stokes equations involving an elastic term in the stress tensor and a transport equation for the evolution of the deformation gradient. The novel feature of the paper is the introduction of the notion of a dissipative solution and its analysis. We show that dissipative solutions exist globally in time for arbitrary finite energy initial data and that a dissipative solution and a strong solution emanating from the same initial data coincide as long as the latter exists.

Appendix

The section is devoted to the construction of an orthonormal basis of ${\mathcal {H}}(\Omega )$ via the spectral decomposition of a certain symmetric and compact operator. Our intention is first to construct a certain operator on which we apply the following theorem summarizing results from [2, Section D.5.].

Theorem 5.1

Let H be an infinite dimensional Hilbert space with the scalar product $(\cdot ,\cdot )_H$, $K:H\rightarrow H$ be a linear compact operator, $\sigma (K)$ be the spectrum of K and $\sigma _p(K)\subset \sigma (K)$ be the discrete spectrum of K. Then

• 1. $0\in \sigma (K)$,

• 2. $\sigma (K){\setminus }\{0\}=\sigma _p(K){\setminus }\{0\}$,

• 3. Either $\sigma (K){\setminus }\{0\}$ is finite or $\sigma (K){\setminus }\{0\}$ is a sequence tending to 0.

• 4. If $\lambda \in \sigma (K){\setminus }\{0\}$ then the eigenspace $\ker (K-\lambda {\text {Id}})$ is finite dimensional.

• 5. If K is positive, i.e., $(Kv,v)_H\ge 0$ for all $v\in H$, then $\sigma (K)\subset [0,\infty )$.

• 6. If K is symmetric, i.e., $(Ku,v)_H=(u,Kv)_H$ for all $u,v\in H$, then $\sigma (K)\subset {\mathbf {R}}$. Additionaly, if H is separable then it possesses an orthonormal basis consisting of eigenvectors of K.

We start with the analysis of the following Neumann eigenvalue problem in $W^{1,2}(\Omega )^{d\times d}$: Find a function $\Phi$ (and an associated $\pi$) satisfying

$$\begin{aligned} -\Delta \Phi +(\nabla \pi )^\top +\Phi&=\lambda \Phi&\text { in }\Omega ,\\ \mathop {\mathrm {div}}\nolimits \Phi&=0&\text { in }\Omega ,\\ (\nabla \Phi -\pi \otimes {\text {I}})n&=0&\text { on }\partial \Omega \end{aligned}$$

understood in the sense: Find a $\Phi \in {\mathcal {W}}(\Omega )$ such that

$$\begin{aligned} (\nabla \Phi ,\nabla \Xi )+(\Phi ,\Xi )=\lambda (\Phi ,\Xi )\text { for all }\Xi \in {\mathcal {W}}(\Omega ). \end{aligned}$$

(5.1)

The results concerning the latter problem are summarized in the ensuing theorem.

Theorem 5.2

Let $\Omega \subset {\mathbf {R}}^d$ be a bounded Lipschitz domain. Then there is a sequence of eigenvalues $\{\lambda ^j\}_{j=1}^\infty$ such that

$$\begin{aligned} 1=\lambda ^1=\cdots =\lambda ^{d^2}\le \lambda ^{d^2+1}\le \cdots \le \lambda ^j\le \lambda ^{j+1}, \lambda ^j\rightarrow \infty \text { as }j\rightarrow \infty \end{aligned}$$

and a sequence of corresponding eigenfunctions $\{\Phi ^j\}_{j=1}^\infty$ satisfying (5.1). Moreover, $\{\Phi ^j\}_{j=1}^\infty$ forms an orthonormal basis of ${\mathcal {H}}(\Omega )$ as well as an orthogonal basis of ${\mathcal {W}}(\Omega )$.

Proof

We consider an operator $B:{\mathcal {W}}(\Omega )\rightarrow ({\mathcal {W}}(\Omega ))^*$ defined as

$$\begin{aligned} \left\langle B\Phi ,\Xi \right\rangle = (\nabla \Phi ,\nabla \Xi )+(\Phi ,\Xi ). \end{aligned}$$

Obviously, B is bounded and linear. As an immediate consequence of the Lax–Milgram theorem B is an isomorphism of ${\mathcal {W}}(\Omega )$ onto its dual. Let us denote by S the restriction of $B^{-1}$ on ${\mathcal {H}}(\Omega )$. As ${\mathcal {H}}(\Omega )\hookrightarrow ({\mathcal {W}}(\Omega ))^*$ and ${\mathcal {W}}(\Omega ){\mathop {\hookrightarrow }\limits ^{C}}{\mathcal {H}}(\Omega )$ we deduce that S is compact on ${\mathcal {H}}(\Omega )$. Next we get for arbitrary $\Phi ^1,\Phi ^2\in {\mathcal {H}}(\Omega )$

$$\begin{aligned} (S\Phi ^1,\Phi ^2)&=(S\Phi ^1,BS\Phi ^2)=(\nabla S\Phi ^1,\nabla S\Phi ^2)+(S\Phi ^1,S\Phi ^2)=(BS\Phi ^1,S\Phi ^2)\\&=(\Phi ^1,S\Phi ^2), \end{aligned}$$

i.e. S is symmetric. Since S is linear and bounded on ${\mathcal {H}}(\Omega )$, it is also self-adjoint. It is shown similarly that $(S\Phi ,\Phi )\ge 0$ for any $\Phi \in {\mathcal {H}}(\Omega )$. Therefore according to Theorem 5.1 there exists an orthonormal basis $\{\Phi ^j\}_{j=1}^\infty$ of the separable Hilbert space ${\mathcal {H}}(\Omega )$ consisting of eigenfunctions of the linear bounded compact symmetric and self-adjoint operator S with corresponding eigenvalues $\{\mu ^j\}_{j=1}^\infty \subset [0,\infty )$ such that $\mu ^j\rightarrow 0$ as $j\rightarrow \infty$. Obviously, as $S\Phi =0$ implies $\Phi =0$, 0 is not an eigenvalue of S. As an immediate consequence we observe that setting $\lambda ^j=\frac{1}{\mu ^j}$ for each $j\in {\mathbf {N}}$ we get that $\{\lambda ^j\}_{j=1}^\infty$ is a sequence of eigenvalues of B with corresponding eigenfunctions $\{\Phi ^j\}_{j=1}^\infty$. The fact that $\lambda ^1=\cdots =\lambda ^{d^2}=1$ follows from the observation that (5.1) is satisfied with $\lambda =1$ and $\Phi$ being equal to each basis element of ${\mathbf {R}}^{d\times d}$. Identity (5.1) with $\lambda =\lambda ^j$ and $\Phi =\Phi ^j$ and $\Xi =\Phi ^i$ also implies that $\{\Phi ^j\}_{j=1}^\infty$ is an orthogonal sequence in ${\mathcal {W}}(\Omega )$ since the left hand side of (5.1) is in fact the scalar product on ${\mathcal {W}}(\Omega )$. Moreover, it follows that $\overline{{\text {span}}\{\Phi ^j\}_{j=1}^\infty }$ has only the trivial orthogonal complement in ${\mathcal {W}}(\Omega )$. Hence $\{\Phi ^j\}_{j=1}^\infty$ is also an orthogonal basis in ${\mathcal {W}}(\Omega )$.

$\square$