# time and space dependent velocities: the case of suddenly applied wall stress

Posted in math on Tuesday, June 17 2014

Previous examples in fluid mechanics assumed steady state. This time lets try something else and imagine a simple non-steady state scenario. Imagine a semi-infinite fluid bounded by a wall at y=0, what happens if the shear stress at the wall undergoes a step-change, at time t=0, to some constant value τ0? (Transport Phenomena 4B.1)

Suppose the flow is laminar, Newtonian, and incompressible, and that we can neglect gravity or any pressure gradients. If the shear stress is such that the flow moves in the x direction only, it is safe to say: $$v_{x}=v(y,t), v_y = v_z = 0$$

The equations of motion for a Newtonian fluid are: $$\rho \frac{D}{Dt} \mathbf{v} = -\nabla p + \mu \nabla^2 \mathbf{v} + \rho \mathbf{g}$$

In terms of x: $$\rho \frac{\partial}{\partial t} v = \mu \frac{\partial^2}{\partial y^2} v$$

Differentiating both sides with respect to y: $$\rho \frac{\partial}{\partial t} \frac{\partial}{\partial y} v = \mu \frac{\partial^2}{\partial y^2} \frac{\partial}{\partial y} v$$

Substituting: $$\tau_{yx} = \tau = -\mu \frac{\partial}{\partial y} v$$

$$\frac{\partial}{\partial t} \tau = \nu \frac{\partial^2}{\partial y^2} \tau$$

The initial condition is that at t=0, τ=0 and the two boundary conditions are that at y=0, τ=τ0 and as y→∞, τ→0

Taking the Laplace transform of both sides, and re-arranging: $$\nu T^{\prime \prime} - s T = 0$$ which is a straight forward ODE with solution: $$T (y,s) = A(s) \exp \left( \sqrt{\frac{s}{v}} y \right) + B(s) \exp \left( - \sqrt{\frac{s}{v}} y \right)$$

Using the second boundary condition, that T → 0 as y→∞, clearly A(s) must be zero.

Using the first boundary condition, at y=0, τ=τ0, or in Laplace space T=τ0/s $$T (0,s) = B(s) \exp \left( - \sqrt{\frac{s}{v}} 0 \right) = B(s) = \frac{\tau_0}{s}$$

So.. $$T (y,s) = \frac{\tau_0}{s} \exp \left( - \sqrt{\frac{s}{v}} y \right)$$

Taking a look through tables of Laplace transforms, you find the following identity: $$\mathcal{L} \left\{ \mbox{erfc} \left( \frac{k}{2 \sqrt{2}} \right) \right\} = \frac{1}{s} \exp \left( -k \sqrt{s} \right) , k > 0$$

Allowing for the easy inverse Laplace of T(y,s) back to τ(y,t) $$\tau (y,t) = \tau_0 \cdot \mbox{erfc} \left( \frac{y}{2 \sqrt{\nu t}} \right)$$

Recall Newton's law of viscosity: $$\tau_{yx} = -\mu \frac{\partial}{\partial y} v_x$$

Or $$\tau_0 \cdot \mbox{erfc} \left( \frac{y}{2 \sqrt{\nu t}} \right) = -\mu \frac{\partial}{\partial y} v$$

Integrating both sides from y to ∞ $$\int \limits_{y}^{\infty} \tau_0 \mbox{erfc} \left( \frac{y}{2 \sqrt{\nu t}} \right) dy = \int \limits_{y}^{\infty} \left( -\mu \frac{\partial}{\partial y} v \right) dy$$

Looking at the right hand side: $$\int \limits_{y}^{\infty} \left( -\mu \frac{\partial}{\partial y} v \right) dy = -\mu \left( v(\infty,t) - v(y,t) \right) = \mu v(y,t)$$ since we assume everything decays out in the limits, the shear stress drops to zero and the velocity of the fluid remains zero.

Looking at the left hand side, we first need the identity: $$\int \limits_{x}^{\infty} \mbox{erfc} (u) du = \frac{1}{\sqrt{\pi}} \exp \left( -x^2 \right) - x \cdot \mbox{erfc}(x)$$

Which leads to: $$v(y,t) = \tau_0 \sqrt{\frac{4 t}{\mu \rho}} \left[ \frac{1}{\sqrt{\pi}} \exp \left( \frac{-y^2}{4 \nu t} \right) - \frac{y}{\sqrt{4 \nu t}} \mbox{erfc}\left( \frac{y}{\sqrt{4 \nu t}} \right) \right]$$