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  1. Show that a general second-order tensor $ a_{ij}$ can be decomposed into three tensors

    $\displaystyle a_{ij} = u_{ij} + v_{ij} + s_{ij},

    where $ u_{ij}$ is symmetric (i.e., $ u_{ji}=u_{ij}$ ) and traceless (i.e., $ u_{ii}=0$ ), $ v_{ij}$ is isotropic, and $ s_{ij}$ only has three independent components.

  2. Use tensor methods to establish the following vector identities:
    1. $ {\bf a}\cdot{\bf b}\times {\bf c}={\bf a}\times {\bf b}\cdot{\bf c}=-{\bf b}\cdot{\bf a}\times{\bf c}$ .
    2. $ ({\bf a}\times {\bf b})\times {\bf c} = ({\bf a}\cdot{\bf c})\,{\bf b} - ({\bf c}\cdot{\bf b})\,{\bf a}$ .
    3. $ ({\bf a}\times {\bf b})\cdot({\bf c}\times {\bf d})=
({\bf a}\cdot{\bf c})\,({\bf b}\cdot{\bf d}) - ({\bf a}\cdot{\bf d})\,({\bf b}\cdot{\bf c})$ .
    4. $ ({\bf a}\times {\bf b})\times ({\bf c}\times {\bf d})= ({\bf a}\times
{\bf b}\cdot{\bf d})\,{\bf c} - ({\bf a}\times {\bf b}\cdot{\bf c})\,{\bf d}$ .
    5. $ \nabla\cdot(\phi\,{\bf a}) = \phi\,\nabla\cdot {\bf a} + {\bf a}\cdot\nabla \phi$ .
    6. $ \nabla\times (\phi\,{\bf a})= \phi\,\nabla\times {\bf a} + \nabla\phi\times {\bf a}$ .
    7. $ \nabla\times({\bf a}\times {\bf b}) = ({\bf b}\cdot\nabla)\,{\bf a} - ({\bf a}...
...nabla)\,{\bf b} + (\nabla\cdot {\bf b})\,{\bf a}
-(\nabla\cdot{\bf a})\,{\bf b}$ .
    8. $ \nabla ({\bf a}\cdot{\bf a}) = 2\,{\bf a}\times (\nabla\times{\bf a}) + 2\,({\bf a}\cdot\nabla)\,{\bf a}$ .
    9. $ \nabla\times(\nabla\times {\bf a}) = \nabla(\nabla\cdot{\bf a}) - \nabla^{\,2} {\bf a}$ .
    Here, $ [({\bf b}\cdot{\nabla}){\bf a}]_i= b_j\,\partial a_i/\partial x_j$ , and $ (\nabla^{\,2} {\bf a})_i = \nabla^{\,2} a_i$ .

  3. A quadric surface has an equation of the form

    $\displaystyle a\,x_1^{\,2} + b\,x_2^{\,2}+c\,x_3^{\,2} + 2\,f\,x_1\,x_2+2\,g\,x_1\,x_3+2\,h\,x_2\,x_3=1.

    Show that the coefficients in the previous expression transform under rotation of the coordinate axes like the components of a symmetric second-order tensor. Hence, demonstrate that the equation for the surface can be written in the form

    $\displaystyle x_i\,T_{ij}\,x_j = 1,

    where the $ T_{ij}$ are the components of the aforementioned tensor.

  4. The determinant of a second-order tensor $ A_{ij}$ is defined

    $\displaystyle {\rm det}(A) = \epsilon_{ijk}\,A_{i1}\,A_{j2}\,A_{k3}.

    1. Show that

      $\displaystyle {\rm det}(A) = \epsilon_{ijk}\,A_{1i}\,A_{2j}\,A_{3k}

      is an alternative, and entirely equivalent, definition.
    2. Demonstrate that $ {\rm det}(A)$ is invariant under rotation of the coordinate axes.
    3. Suppose that $ C_{ij} = A_{ik}\,B_{kj}$ . Show that

      $\displaystyle {\rm det}(C) = {\rm det}(A)\,{\rm det}(B).

  5. If

    $\displaystyle A_{ij}\,x_j = \lambda\,x_i

    then $ \lambda$ and $ x_j$ are said to be eigenvalues and eigenvectors of the second-order tensor $ A_{ij}$ , respectively. The eigenvalues of $ A_{ij}$ are calculated by solving the related homogeneous matrix equation

    $\displaystyle (A_{ij}-\lambda\,\delta_{ij})\,x_j = 0.

    Now, it is a standard result in linear algebra that an equation of the previous form only has a non-trivial solution when (Riley 1974)

    $\displaystyle {\rm det}(A_{ij}-\lambda\,\delta_{ij}) = 0.

    Demonstrate that the eigenvalues of $ A_{ij}$ satisfy the cubic polynomial

    $\displaystyle \lambda^{\,3} - {\rm tr}(A) \,\lambda^{\,2} + {\mit\Pi}(A)\,\lambda - {\rm det}(A) = 0,

    where $ {\rm tr}(A) = A_{ii}$ and $ {\mit\Pi}(A)= (A_{ii}\,A_{jj}-A_{ij}\,A_{ji})/2$ . Hence, deduce that $ A_{ij}$ possesses three eigenvalues--$ \lambda_1$ , $ \lambda_2$ , and $ \lambda_3$ (say). Moreover, show that

    $\displaystyle {\rm tr}(A)$ $\displaystyle =\lambda_1+\lambda_2+\lambda_3,$    
    $\displaystyle {\rm det}(A)$ $\displaystyle =\lambda_1\,\lambda_2\,\lambda_3.$    

  6. Suppose that $ A_{ij}$ is a (real) symmetric second-order tensor: that is, $ A_{ji}=A_{ij}$ .
    1. Demonstrate that the eigenvalues of $ A_{ij}$ are all real, and that the eigenvectors can be chosen to be real.
    2. Show that eigenvectors of $ A_{ij}$ corresponding to different eigenvalues are orthogonal to one another. Hence, deduce that the three eigenvectors of $ A_{ij}$ are, or can be chosen to be, mutually orthogonal.
    3. Demonstrate that $ A_{ij}$ takes the diagonal form $ A_{ij}=\lambda_i\,\delta_{ij}$ (no sum) in a Cartesian coordinate system in which the coordinate axes are each parallel to one of the eigenvectors.

  7. In an isotropic elastic medium under stress, the displacement $ u_i$ satisfies

    $\displaystyle \frac{\partial\sigma_{ij}}{\partial x_j}$ $\displaystyle =\rho\,\frac{\partial^{\,2} u_i}{\partial t^{\,2}},$    
    $\displaystyle \sigma_{ij}$ $\displaystyle =c_{ijkl}\,\frac{1}{2}\!\left(\frac{\partial u_k}{\partial x_l}+ \frac{\partial u_l}{\partial x_k}\right),$    

    where $ \sigma_{ij}$ is the stress tensor (note that $ \sigma_{ji}=\sigma_{ij}$ ), $ \rho$ the mass density (which is a uniform constant), and

    $\displaystyle c_{ijkl} = K\,\delta_{ij}\,\delta_{kl} + \mu\,[\delta_{ik}\,\delta_{jl}+\delta_{il}\,\delta_{jk} - (2/3)\,\delta_{ij}\,\delta_{kl}].

    the isotropic stiffness tensor. Here, $ K$ and $ \mu$ are the bulk modulus and shear modulus of the medium, respectively. Show that the divergence and the curl of $ {\bf u}$ both satisfy wave equations. Furthermore, demonstrate that the characteristic wave velocities of the divergence and curl waves are $ [(K+4\mu/3)/\rho]^{1/2}$ and $ (\mu/\rho)^{1/2}$ , respectively.

next up previous
Next: Non-Cartesian Coordinates Up: Cartesian Tensors Previous: Isotropic Tensors
Richard Fitzpatrick 2016-03-31