Hey guys! (I am not sure if I should post this thread in Physics or Mathematics)
I have had some issues with developing expressions for the polarizations (material displacement) of waves propagating in anisotropic media. To bring you guys up to speed I have to start a few steps before the problem appears. We start with Christoffel's equation (http://en.wikipedia.org/wiki/Linear_...offel_equation):
\begin{equation}
(C_{ijkl}n_jn_l - \rho c^2\delta_{ik})u_k = 0,
\end{equation}
where Cijkl is the stiffness tensor, ni is the normal to the plane wave, ρ is the density, c is the phase velocity, δik is the Kroenecker delta and uk is the polarization of the wave. This is obviously and eigenvalue problem where ρc2 is the eigenvalue. Introducing reduced notation: 11, 22, 33, 23, 13 and 12 is replaced by 1, 2, 3, 4, 5 and 6, respectively. Now, implementing a transverse isotropic material (http://en.wikipedia.org/wiki/Linear_...ogeneous_media) and looking in a direction in the x1-x3 plane (ni = [sin(θ), 0 , cos(θ)]T), the eigenvalue problem becomes
\begin{equation}
\begin{vmatrix}
C_{11}n_1^2+C_{44}n_3^2-\rho c^2 & 0 & (C_{13}+C_{44})n_1n_3 \\
0 & \frac{1}{2}(C_{11}-C_{12})n_1^2+C_{44}n_3^2 -\rho c^2 & 0 \\
(C_{13}+C_{44})n_1n_3 & 0 & C_{44}n_1^2+C_{33}n_3^2 -\rho c^2
\end{vmatrix} = 0.
\end{equation}
To avoid working with many symbols, we can insert values for the stiffness coefficients: C11 = 5, C33 = 6, C44 = 2, C12 = 4 and C13 = 3. The eigenvalues then become:
\begin{eqnarray}
\rho c^2_1 &=& cos(\theta)^2+1 \\
\rho c^2_2 &=& 0.5cos(\theta)^2+3.5+0.5\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9} \\
\rho c^2_3 &=& 0.5cos(\theta)^2+3.5-0.5\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}
\end{eqnarray}
and the corresponding eigenvectors are:
\begin{eqnarray}
\mathbf{u}_1 &=& \begin{Bmatrix}
0\\
1 \\
0
\end{Bmatrix}, \\
\mathbf{u}_2 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3+\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9})\\
0 \\
1
\end{Bmatrix},\\
\mathbf{u}_3 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3-\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}) \\
0 \\
1
\end{Bmatrix}.
\end{eqnarray}
For those interested: the eigensolutions tell us that we can have three different types of plane waves propagating in direction n, each with its characteristic wave speed. One is termed quasi-longitudinal wave and the two other are called quasi-transverse waves. The eigenvectors tells us what the polarization (material displacement direction) of the wave is.
Here is the issue: u2 is not defined for θ = ∏/2 and u3 is not defined for θ=0 (zero divided by zero). If I insert these values for θ before I solve the eigenvalue problem, I obtain the correct eigenvectors. Why does this happen? If I use Maple to take the limit of θ goes to ∏/2 on u2, I get the answer 'undefined' (I guess this means 'infinity'). Is it possible to avoid this division by zero for certain θs through some mathematical trick? I have tried scaling up the eigenvectors etc. Also, can I say that the vector ['infinity',0,1] is equivalent with [1, 0, 0]?
I have had some issues with developing expressions for the polarizations (material displacement) of waves propagating in anisotropic media. To bring you guys up to speed I have to start a few steps before the problem appears. We start with Christoffel's equation (http://en.wikipedia.org/wiki/Linear_...offel_equation):
\begin{equation}
(C_{ijkl}n_jn_l - \rho c^2\delta_{ik})u_k = 0,
\end{equation}
where Cijkl is the stiffness tensor, ni is the normal to the plane wave, ρ is the density, c is the phase velocity, δik is the Kroenecker delta and uk is the polarization of the wave. This is obviously and eigenvalue problem where ρc2 is the eigenvalue. Introducing reduced notation: 11, 22, 33, 23, 13 and 12 is replaced by 1, 2, 3, 4, 5 and 6, respectively. Now, implementing a transverse isotropic material (http://en.wikipedia.org/wiki/Linear_...ogeneous_media) and looking in a direction in the x1-x3 plane (ni = [sin(θ), 0 , cos(θ)]T), the eigenvalue problem becomes
\begin{equation}
\begin{vmatrix}
C_{11}n_1^2+C_{44}n_3^2-\rho c^2 & 0 & (C_{13}+C_{44})n_1n_3 \\
0 & \frac{1}{2}(C_{11}-C_{12})n_1^2+C_{44}n_3^2 -\rho c^2 & 0 \\
(C_{13}+C_{44})n_1n_3 & 0 & C_{44}n_1^2+C_{33}n_3^2 -\rho c^2
\end{vmatrix} = 0.
\end{equation}
To avoid working with many symbols, we can insert values for the stiffness coefficients: C11 = 5, C33 = 6, C44 = 2, C12 = 4 and C13 = 3. The eigenvalues then become:
\begin{eqnarray}
\rho c^2_1 &=& cos(\theta)^2+1 \\
\rho c^2_2 &=& 0.5cos(\theta)^2+3.5+0.5\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9} \\
\rho c^2_3 &=& 0.5cos(\theta)^2+3.5-0.5\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}
\end{eqnarray}
and the corresponding eigenvectors are:
\begin{eqnarray}
\mathbf{u}_1 &=& \begin{Bmatrix}
0\\
1 \\
0
\end{Bmatrix}, \\
\mathbf{u}_2 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3+\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9})\\
0 \\
1
\end{Bmatrix},\\
\mathbf{u}_3 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3-\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}) \\
0 \\
1
\end{Bmatrix}.
\end{eqnarray}
For those interested: the eigensolutions tell us that we can have three different types of plane waves propagating in direction n, each with its characteristic wave speed. One is termed quasi-longitudinal wave and the two other are called quasi-transverse waves. The eigenvectors tells us what the polarization (material displacement direction) of the wave is.
Here is the issue: u2 is not defined for θ = ∏/2 and u3 is not defined for θ=0 (zero divided by zero). If I insert these values for θ before I solve the eigenvalue problem, I obtain the correct eigenvectors. Why does this happen? If I use Maple to take the limit of θ goes to ∏/2 on u2, I get the answer 'undefined' (I guess this means 'infinity'). Is it possible to avoid this division by zero for certain θs through some mathematical trick? I have tried scaling up the eigenvectors etc. Also, can I say that the vector ['infinity',0,1] is equivalent with [1, 0, 0]?