# Strain/displacement field dynamics

I am trying to model a viscoelastic fluid and I am having a difficulty understanding the equations.

Basically what I want to model is the dynamics of the system, which is confined between two infinite parallel plates (the problem is consequently 1-dimensional). Initially the system is at rest. At $t=0$ I displace the upper plate by a finite (small) distance.

What I want to find is the solution for the displacement field $u_i(z)$. The problem that I encounter is that the equations seem to contradict each other and I do not know how to keep the boundary condition of fixed displacement at the upper plate (and fixed zero displacement at the bottom plate).

The equations governing a viscoelastic fluid are the following:

1.) Navier-Stokes equation: $\rho \dot{v}_i + \partial_j \sigma_{ij}=0$, with $\sigma_{ij}$ the total stress tensor, which is a function of the strain rate tensor $A_{ij}=\frac{1}{2}(\partial_i v_j + \partial_j v_i)$, and the elastic strain $\varepsilon_{ij}=\frac{1}{2}(\partial_i u_j + \partial_j u_i)$.

2.) Strain diffusion and relaxation: $\dot{\varepsilon}_{ij}+f_{ij}=0$, with $f_{ij}$ being a function of $\varepsilon_{ij}$ (strain relaxation), $\partial_{i}\partial_{j} \varepsilon_{ij}$, and other combinations of the derivatives (strain diffusion)

The way I tried to solve this was to put $\dot{u}_{i}=v_i$ and plug this in to the Navier-Stokes equation. This equation then becomes a wave equation for the displacement field $u_i(z)$. And this can be in principle solved with the two (fixed displacement) boundary conditions. The problem with this is that the second equation seems redundant, and this should not be since I want to describe viscoelasticity.

Then I thought I'd start with the second equation for the strain field and then solve the Navier-Stokes in parallel. But this time I wouldn't put $\dot{u}_i=v_i$ and keep $v_i$ as independent from $u_i$. The problem I encounter here is how to keep the fixed displacement boundary condition, since the second equation is only written for the strain field, not the displacement field.

• This seems like a bit of a tricky problem. What exactly is the rheological constitutive equation that you are using? You seem to be employing some sort of a creep model, but the exact nature isn't clear. Can you elaborate? The only component of the stress tensor that really matters is the shear component, so the rheological equation you can use is scalar. – Chet Miller Feb 10 '18 at 13:25
• The total stress tensor $\sigma_{ij}$ is composed on one hand of the elastic stress tensor $\psi_{ij}=\frac{\delta f}{\delta \varepsilon_{ij}}=C_{ijkl}\varepsilon_{kl}$, with $f$ the free energy density and $C_{ijkl}$ the elastic tensor. The second component is the viscous tensor, $\sigma_{ij}=-\psi_{ij}+\psi_{ik}\varepsilon_{kj} + \nu_{ijkl}A_{kl}$. For the viscous tensor you can use the isotropic form. For the $f_{ij}$ in the second equation of my question you can just take strain relaxation, i.e. $f_{ij}=\alpha \varepsilon_{ij}$, where $\alpha$ is a positive constant. – EntropicFluid Feb 10 '18 at 13:57
• For the stress tensor there is also ofcourse the contribution of the pressure: $\sigma_{ij}=-p\delta_{ij}$ and for the strain the equation should read $\dot{\varepsilon}_{ij}-A_{ij}+f_{ij}=0$ to eliminate the effects of translation. – EntropicFluid Feb 10 '18 at 14:11
• Have you solved this problem for a Newtonian fluid and for a purely elastic solid yet? Aside from this, the pressure (and other normal stresses) do not figure in the solution to this shear problem. – Chet Miller Feb 10 '18 at 15:31
• Consider 1D motion of a particle. Newton says $m\ddot{x}=f$. Here, $f=-x-\dot{x}$ would give damped oscillation, but the analog of viscoelastic relaxation is $f=-\dot{f}-\dot{x}$. Restoring forces abate. In your problem, stress should be $\sigma (\varepsilon ,{{\nabla }^{2}}\sigma )$. – Bert Barrois Feb 11 '18 at 14:48