Exercise 5

Velocity decomposition

Recall the decomposition of a velocity field \(\mathbf{v} \left( \mathbf{x} + \mathbf{r} \right)\) into translation, rotation and internal deformation:

\[ \mathbf{v} \left( \mathbf{x} + \mathbf{r} \right) = \mathbf{v} \left( \mathbf{x}\right) + \mathbf{w} \times \mathbf{r} + \mathbf{D} \cdot \mathbf{r} \]

Consider the following flow fields that describe plane motions (2-D):

A) \(\mathbf{v} \left( \mathbf{x} \right) = \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix}\), B) \(\mathbf{v} \left( \mathbf{x} \right) = \frac{1}{x^2+y^2} \begin{pmatrix}x \\ y \\ 0 \end{pmatrix}\), C) \(\mathbf{v} \left( \mathbf{x} \right) = \begin{pmatrix}-y \\ x \\ 0 \end{pmatrix}\) and

D) \(\mathbf{v} \left( \mathbf{x} \right) = \frac{1}{x^2+y^2} \begin{pmatrix}-y \\ x \\ 0 \end{pmatrix}\).

Tasks

Task 1

Decompose the given velocity fields and calculate for each field: \(\mathbf{w}\) and \(\mathbf{D}\)

Tip

  • Deformation tensor:
    \[ \mathbf{D} = \frac{1}{2}(\nabla \mathbf{v} + \nabla \mathbf{v}^T) \]

  • Angular velocity vector: \[ \mathbf{w} = \frac{1}{2} \nabla \times \mathbf{v} \]

For 2‑D planar motion in \((x,y)\) with zero z-component,
\[ \mathbf{w} = (0,0,\omega_z) \] with \[ \omega_z = \frac{1}{2}\left(\frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y}\right). \]

Field \(\mathbf{w}\) \(\mathbf{D}\)
A \(\mathbf{0}\) \(\mathbf{0}\)
B \(\mathbf{0}\) \(D_{xx}=\tfrac{y^2-x^2}{r^4},\; D_{yy}=\tfrac{x^2-y^2}{r^4},\; D_{xy}=\tfrac{-2xy}{r^4}\)
C \((0,0,1)\) \(\mathbf{0}\)
D \(\mathbf{0}\) \(D_{xx}=\tfrac{2xy}{r^4},\; D_{yy}=\tfrac{-2xy}{r^4},\; D_{xy}=\tfrac{y^2-x^2}{r^4}\)

where \(r^2 = x^2 + y^2\).

Task 2

Describe the motion produced by each velocity field.

  • A: pure uniform translation (no rotation, no deformation)
  • B: radial expansion — a source flow; fluid stretches outward with no rotation
  • C: rigid body rotation — every point rotates about the origin with angular velocity 1; no deformation
  • D: irrotational vortex — fluid circulates around the origin but \(\mathbf{w}=0\); pure straining

Task 3

Sketch each flow field within a domain given by \(x,y \in [-4, 4]\times [-4, 4]\)

The hint shows field B. Apply the same quiver-plot approach for A (uniform arrows), C (circular arrows), and D (circular arrows decaying with radius).

Solution for B)

Task 4

Within the flow field, sketch how a volume element having vertices \((1, 1), (1, 2), (2, 2), (2,1)\) deforms.

See flow field sketches from task 3.

Solution

See Solution to Task 3

Wave propagation - plane waves

The momentum balance for an isotropic, linear elastic body with \(\rho \in \mathbb{R}\) constant is given by

\[ \rho \frac{\partial^2}{\partial t^2}\mathbf{w} = \mu \Delta \mathbf{w} + (\mu + \lambda) \nabla ( \nabla \cdot \mathbf{w}) \tag{1}\]

where \(\mathbf{w}(\boldsymbol{\zeta}, t)\) is the displacement field and \(\lambda, \mu \in \mathbb{R}\) are the Lamé constants.

We consider plane wave solutions of the form

\[ \mathbf{w}(\boldsymbol{\zeta}, t) = A \mathbf{n}_d \psi (k \mathbf{n}_p \cdot \boldsymbol{\zeta} - \omega t) \tag{2}\]

where \(\psi\) is an arbitrary scalar function describing the displacement profile, \(\mathbf{n}_d\) and \(\mathbf{n}_p\) are unit vectors corresponding to the direction of displacement and progapation. \(A, k\) and \(\omega\) are constants for the amplitude, wave number and angular frequency.

Note

The acceleration of the displacement is given in Lagrangian coordinates. However, for the small displacement limit, this is equivalent to the Eulerian description.

  1. Show that Equation 1 satisfies Equation 2 if and only if the condition

\[ A \, k^2 \left( \mu \mathbf{n}_d + (\mu + \lambda) (\mathbf{n}_d \cdot \mathbf{n}_p)\mathbf{n}_p \right) = A\rho\omega^2 \mathbf{n}_d \]

is met.

Substituting the plane wave ansatz into Equation 1 yields the condition: \[ Ak^2\!\left(\mu\mathbf{n}_d + (\mu+\lambda)(\mathbf{n}_d\cdot\mathbf{n}_p)\mathbf{n}_p\right) = A\rho\omega^2\mathbf{n}_d \]

Let \(\theta=k\mathbf{n}_p\cdot\boldsymbol{\zeta}-\omega t\). Then \[ \mathbf{w}=A\mathbf{n}_d\psi(\theta), \quad \frac{\partial\theta}{\partial t}=-\omega, \quad \frac{\partial\theta}{\partial{\zeta}_j}=k(\mathbf{n}_p)_j. \] So \[ \frac{\partial\mathbf{w}}{\partial t} =A\mathbf{n}_d\psi'(\theta)\frac{\partial\theta}{\partial t} =-A\omega\mathbf{n}_d\psi' \] and \[ \frac{\partial^2\mathbf{w}}{\partial t^2} =A\omega^2\mathbf{n}_d\psi''. \]

  • Gradient of \(\mathbf{w}\)

The components are \[ w_i =A(\mathbf{n}_d)_i\psi(kn_p^j\zeta_j-\omega t), \] so \[ (\nabla\mathbf{w})_{ij} =\frac{\partial w_i}{\partial\zeta_j} =A(\mathbf{n}_d)_i\psi'\frac{\partial\theta}{\partial\zeta_j} =Ak(\mathbf{n}_d)_i(\mathbf{n}_p)_j\psi'. \]

  • Divergence \(\nabla\cdot\mathbf{w}\)

\[ \nabla\cdot\mathbf{w} =\frac{\partial w_i}{\partial\zeta_i} =Ak(\mathbf{n}_d\cdot\mathbf{n}_p)\psi'. \]

  • Gradient of the divergence

\[ \nabla(\nabla\cdot\mathbf{w}) =Ak(\mathbf{n}_d\cdot\mathbf{n}_p)\nabla\psi' =Ak^2(\mathbf{n}_d\cdot\mathbf{n}_p)\mathbf{n}_p\psi''. \]

  • Laplacian of \(\mathbf{w}\)

The Laplacian acts componentwise: \[ \Delta w_i =\frac{\partial}{\partial\zeta_j}\frac{\partial w_i}{\partial\zeta_j} =Ak^2(\mathbf{n}_d)_i\psi''. \] So \[ \Delta\mathbf{w}=Ak^2\mathbf{n}_d\psi''. \]

  1. Using the formula from the first question. In the following, we want to show that for a linear elastic, isotropic solid with \(\lambda + \mu \neq 0\), plane waves only admit two types of solutions. The first wave type are longitudinal waves in direction \(\mathbf{n}_p\) where \(\mathbf{n}_p \, \parallel \, \mathbf{n}_d\). The second wave type are transverse waves where \(\mathbf{n}_p \, \bot \, \mathbf{n}_d\).

Assume \(\lambda + \mu \neq 0\). For any given \(\mathbf{n}_p \neq \mathbf{0}\) and \(A, k \neq 0\) show that the only independent solutions \((\mathbf{n}_d, \omega)\) are:

\[ \begin{aligned} & \mathbf{n}_d = \pm \mathbf{n}_p, \quad \omega^2 = k^2 (\lambda + 2 \mu) / \rho, \\ & \mathbf{n}_d \cdot \mathbf{n}_p = 0, \quad \omega^2 = k^2 \mu / \rho. \end{aligned} \]

The only two independent solutions are: \[ \begin{aligned} &\text{Longitudinal:} \quad \mathbf{n}_d = \pm\mathbf{n}_p, \quad \omega^2 = \frac{k^2(\lambda+2\mu)}{\rho} \\ &\text{Transverse:} \quad \mathbf{n}_d\cdot\mathbf{n}_p = 0, \quad \omega^2 = \frac{k^2\mu}{\rho} \end{aligned} \]

Start from the condition we proved in part 1:

\[ A k^2 \Big(\mu\mathbf{n}_d +(\mu+\lambda)(\mathbf{n}_d\cdot\mathbf{n}_p)\mathbf{n}_p \Big) =A\rho\omega^2\mathbf{n}_d. \]

Since \(A\neq 0\), divide by \(A k^2\):

\[ \mu\mathbf{n}_d +(\mu+\lambda)(\mathbf{n}_d\cdot\mathbf{n}_p)\mathbf{n}_p =\frac{\rho\omega^2}{k^2}\mathbf{n}_d. \]

Two cases satisfy the above equation:

  • Case 1: longitudinal wave \(\mathbf{n}_d \parallel \mathbf{n}_p\), \(\mathbf{n}_d = \pm \mathbf{n}_p\)

  • Case 2: transverse wave \(\mathbf{n}_d\perp\mathbf{n}_p\), \(\mathbf{n}_d \cdot \mathbf{n}_p = 0\)

What happens in the case \(\lambda + \mu = 0\).

With \(\lambda = -\mu\): \(\lambda + 2\mu = \mu\), so \(\omega^2 = k^2\mu/\rho\) for both modes — longitudinal and transverse waves travel at the same speed.

Since \[ \lambda+2 \mu=-\mu+2 \mu=\mu. \]

Therefore, \(\omega^2=\frac{k^2 \mu}{\rho}\)

Asymptotics of a visoelatic material

A very simple linear viscoelastic material model yields the following equation:

\[ \rho \frac{\partial^2}{\partial t^2} u = \alpha \frac{\partial^2}{\partial x^2} u + \beta \frac{\partial}{\partial t} \frac{\partial^2}{\partial x^2} u \quad, \tag{3}\]

with constants \(\rho, \alpha, \beta \in \mathbb{R}\) and unknown displacement field \(u = u(t, x) \in \mathbb{R}\). We want to study solution of this equation that obey the form

\[ u(t, x) = A \, exp(i (x-c \, t)) \tag{4}\]

with \(i^2 = -1\), \(A \in \mathbb{R}\), and \(A, c \neq 0\).

Tip

Recall Euler’s formula \[ exp(i \, \theta) = cos(\theta) + i \, sin(\theta) \]

Tip

A wave solution is damped if \(u \rightarrow 0\) for \(t \rightarrow \infty\).

Tasks

Task 1

The elastic limit is given for \(\alpha > 0\) and \(\beta = 0\). Show that Equation 3 admits solutions of type Equation 4 which are not dampened in time.

\[ c = \pm\sqrt{\frac{\alpha}{\rho}} \] Since \(c\) is real the solution is not damped: \(|u| = |A|\) remains constant.

\[ \rho c^2 = \alpha \implies c = \pm \sqrt{\frac{\alpha}{\rho}} \]

Since \(c\) is real, the magnitude \(|u| = |A|\) remains constant and does not decay to zero.

Task 2

The viscous limit is given for \(\alpha = 0\) and \(\beta > 0\). Show that Equation 3 admits stationary solutions of type Equation 4 which are dampened exponentially in time.

\[ c = -\frac{i\beta}{\rho}, \qquad u(t,x) = A\,e^{ix}\,e^{-(\beta/\rho)\,t} \] The solution decays exponentially in time.

\[ \rho(-c^2)u=\beta(ic)u. \]

Then,

\[ c=-\,\frac{i\beta}{\rho}. \]

so the solution is

\[ u(t,x)=A e^{ix}e^{-(\beta/\rho)t}. \]

Task 3

How does the solution look like for the case \(\alpha > 0\) and \(\beta > 0\)? Differentiate the cases

  1. \(4 \, \rho \alpha > \beta^2\)
  2. \(4 \, \rho \alpha < \beta^2\)

Both cases are damped.

  1. \(4\rho\alpha > \beta^2\): damped oscillation, \[ u(t,x) = A\,e^{-\frac{\beta}{2\rho}t}\,e^{i\!\left(x \mp \frac{\sqrt{4\rho\alpha-\beta^2}}{2\rho}t\right)} \]

  2. \(4\rho\alpha < \beta^2\): pure exponential decay, \[ u(t,x) = A\,e^{ix}\,e^{-\frac{\beta\mp\sqrt{\beta^2-4\rho\alpha}}{2\rho}\,t} \]

  1. if \(4 \rho \alpha>\beta^2\) \[ \begin{aligned} \rightarrow c & = \pm \frac{\sqrt{4 \rho \alpha-\beta^2}}{2 \rho}-i \frac{\beta}{2 \rho} \\ \rightarrow u(t, x) & =A e^{i x} e^{-i c t}=A e^{-\frac{\beta}{2 \rho} t} e^{i\left(x \mp \frac{\sqrt{4 \rho \alpha-\beta^2}}{2 \rho} t\right)} \end{aligned} \] Therefore it is damping.
  2. if \(4 \rho \alpha<\beta^2\) \[ c=\frac{i \beta \pm i \sqrt{\beta^2-4 \rho \alpha}}{2 \rho}=-i \frac{\beta \mp \sqrt{\beta^2-4 \rho \alpha}}{2 \rho} \] assume: \[ \lambda=\frac{\beta \mp \sqrt{\beta^2-4 \rho \alpha}}{2 \rho} \] \[ \begin{aligned} & \lambda_1=\frac{\beta+\sqrt{\beta^2-4 \rho \alpha}}{2 \rho}>0 \\ & \lambda_2=\frac{\beta-\sqrt{\beta^2-4 \rho 2}}{2 \rho}>0 \end{aligned} \] Then \[ \begin{aligned} \mu(t, x)= & A e^{i x} e^{-\frac{\beta \mp \sqrt{\beta^2-4 \rho \alpha}}{2 \rho} t} \\ = & A e^{\frac{-\beta \mp \sqrt{\beta^2-4 \rho \alpha}}{2 \rho} t} e^{i x}\end{aligned}\]

Therefore it is damping.