Past exam of the advanced fluid dynamics course of the University of Oxford / 2016 / 2 / b

When deformation is stopped, the rate-of-strain tensor and velocity gradient vanish, leaving $\mathsf{T}+\tau {\partial }_t\mathsf{T}=0$. HenceThe positive constant $\tau$ is the viscoelastic relaxation time. For slow weak deformation, all the transport and quadratic terms are smaller than $\mathsf{T}$, so $\mathsf{T}\simeq 2\mu \mathsf{E}$. Thus $\mu >0$ is the zero-rate dynamic viscosity, measured in stress times time. The corresponding short-time shear modulus of the Maxwell fluid response is $G_{\mathrm{el}}=\mu /\tau$. The source's $\alpha$ has units of time, since $\alpha \mathsf{TE}$ must have units of stress.

Linearize about a relaxed state $\mathsf{T}=0$ with constant mass density ${\rho }_0$. In the unforced momentum conservation equation ${\rho }_0{D}_t\mathbf{u}=-\nabla p+\nabla \cdot \mathsf{T}$, the $x$ component of the pressure gradient is zero for a freely evolving unidirectional flow independent of $x$. A prescribed external pressure gradient would add a forcing term.

For $\mathbf{u}=u(y,t)\hat{\mathbf{x}}$, incompressibility is automatic and the only nonzero entries of the rate-of-strain tensor are $E_{xy}=E_{yx}=\frac{1}{2}{\partial }_{y}u$. Products of stress and velocity gradient are quadratic in the small amplitude. The constitutive equation consequently becomes $\mathsf{T}+\tau {\partial }_t\mathsf{T}=2\mu \mathsf{E}$ to first order. With $T=T_{xy}=T_{yx}$, its shear component and the $x$ component of momentum conservation areThe diagonal stress components have only homogeneous relaxation at this order and vanish if initially relaxed. In particular, $\alpha$ does not enter this linear response.

Eliminating the shear stress from the two linear equations gives the Maxwell shear-wave dispersion equation in physical space,where $\nu$ is the kinematic viscosity. Substitution of a Fourier mode $e^{iky+\sigma t}$ givesThese are temporal exponents: the oscillation angular frequency is the imaginary part of $\sigma$, with the sign determined by the selected time convention. For $4\tau \nu k^2<1$ both exponents are real and nonpositive. At equality the repeated exponent is $-1/(2\tau )$; generic solutions include $t{e}^{-t/(2\tau )}$. Above the threshold the modes are damped propagating shear disturbances:At large $|k|$ their phase velocity tends to $\sqrt{\nu /\tau }=\sqrt{G_{\mathrm{el}}/{\rho }_0}$.

For the limit $\tau \to 0$ at fixed $\mu ,{\rho }_0$ and $k$, expansion of the square root givesAfter a short initial relaxation, the surviving branch obeys the diffusion equation $u_t=\nu u_{yy}$ of a Newtonian fluid. The fast branch disappears on the vanishing viscoelastic relaxation time; at $k=0$ the two exponents are exactly $0$ and $-1/\tau$. This limit is not uniform in $k$: modes with $k$ of order $(\tau \nu {)}^{-1/2}$ still sample the oscillatory transition. If instead $G_{\mathrm{el}}=\mu /\tau$ is held fixed as $\tau \to 0$, then $\mu \to 0$ and the fixed-$k$ slow exponent tends to zero. The ordinary Newtonian fluid limit therefore specifies which material parameter is held fixed.