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The elastic wave equation or momentum equation is given by:

$$\rho \frac {\partial^2 u_i} {\partial t^2} = \ \frac {\partial \sigma_{ij}}{\partial x_j}$$

This equation has remarkable similarities with the 1D and 2D wave equations which are given by

$$\frac {\partial^2 u} {\partial x^2} = \frac 1 {c^2} \frac {\partial^2 u} {\partial t^2}$$

and

$$\frac {\partial^2 u} {\partial x^2}+\frac {\partial^2 u} {\partial y^2} = \frac 1 {c^2} \frac {\partial^2 u} {\partial t^2}$$

For instance, expanding one term of the momentum equation we obtain the following:

$$\rho \frac {\partial ^2 u_x} {\partial t^2} = \frac {\partial \sigma_{xx}}{\partial x}+\frac {\partial \sigma_{xy}}{\partial y}+\frac {\partial \sigma_{xz}}{\partial z}$$

Substituting in the values of $\sigma$ based on the constitutive elastic matrix and the definitions for strain we obtain:

$$\rho \frac {\partial ^2 u_x} {\partial t^2} = \frac{E}{(1+\nu)(1-2 \nu)} \left( \frac \partial {\partial x}\left[ (1-\nu)\frac {\partial u_x}{\partial x} + \nu \frac {\partial u_y} {\partial y} + \nu \frac {\partial u_z}{\partial z}\right] + \frac {\partial} {\partial y} \left[\frac 1 2 (1-2 \nu) (\frac {\partial u_x}{\partial y} + \frac {\partial u_y} {\partial x})\right]+\frac {\partial} {\partial z} \left[\frac 1 2 (1-2 \nu) (\frac {\partial u_x}{\partial z} + \frac {\partial u_z} {\partial x})\right] \right)$$

Which bears the following similarities with the 1D and 2D wave equations:

  • The left hand side is the second partial derivative with respect to time.
  • The right hand side contains several spatial derivatives taken twice.
  • dividing both sides by $\rho$ we see emerge a coefficient not unlike the $1/c^2$ term in the 1D and 2D wave equations.

QUESTIONS

I have 2 questions:

  • Does there exist a relationship between the 1D/2D wave equation and the elastic wave equation? Can this relationship be demonstrated?
  • What analytical methods can handle a PDE with 4 independent variables and 3 dependent solutions (such as the momentum equation)?
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  • $\begingroup$ looks like im going to have to figure this out on my own... $\endgroup$ – user32882 Aug 20 '17 at 13:17
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Let us use the index notation (Einstein convention). In linear elasticity, the momentum equation $\rho \ddot{u}_i = \sigma_{ij,j}$ is combined with the constitutive law $\sigma_{ij} = C_{ijk\ell}u_{k,\ell}$, where the dot denotes $\partial_t$ and $C_{ijk\ell} = \lambda\delta_{ij}\delta_{k\ell}+\mu(\delta_{ik}\delta_{j\ell} + \delta_{i\ell}\delta_{jk})$ is the stiffness tensor of Hooke's law. Thus, the constitutive law reads $\sigma_{ij} = \lambda u_{k,k}\delta_{ij} + \mu (u_{i,j}+u_{j,i})$, where $\lambda$, $\mu$ are the Lamé parameters and $\delta_{ij}$ is the Kronecker delta. The momentum equation rewrites as $$ \rho \ddot{u}_i = C_{ijk\ell}u_{k,\ell j} = (\lambda+\mu) u_{j,ji} + \mu u_{i,jj} \, , $$ or in intrinsic notation, \begin{aligned} \rho \ddot{\bf u} &= (\lambda+\mu) \nabla\,(\nabla\cdot {\bf u}) + \mu \nabla^2{\bf u} \, , \\ &= (\lambda+2\mu) \nabla\,(\nabla\cdot {\bf u}) - \mu\nabla\times\nabla\times {\bf u} \, . \end{aligned} In linear elasticity, compressional waves (P-waves) with speed $\sqrt{(\lambda+2\mu)/\rho}$ and shear waves (S-waves) with speed $\sqrt{\mu/\rho}$ are decoupled. Their corresponding wave equations are obtained by applying $\nabla\cdot$ and $\nabla\times$ to the previous equations, respectively. Setting $\psi = \nabla\cdot{\bf u}$ and ${\bf a} = \nabla\times {\bf u}$, one has \begin{aligned} \rho \ddot\psi - (\lambda+2\mu) \nabla^2\psi &= 0 \, ,\\ \rho\ddot{\bf a} - \mu\nabla^2 {\bf a} &= 0 \, . \end{aligned} Dedicated analytical methods are the classical methods for wave equations (Fourier transform, normal modes decomposition, etc.), based on a decomposition in scalar and vector potentials ${\bf u} = \nabla \varphi + \nabla\times {\bf b}$. The relationship with Young's modulus $E$ and Poisson's ratio $\nu$ in OP is $$ E = \frac{\mu (3\lambda+2\mu)}{\lambda+\mu}\, , \nu = \frac{\lambda}{2(\lambda+\mu)} \quad\text{or}\quad \lambda = \frac{E\nu}{(1+\nu)(1-2\nu)}\, , \mu = \frac{E}{2(1+\nu)}\, . $$

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