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430 lines
34 KiB
Markdown
<!-- source-page: 871 -->
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There are two zero diagonal elements in $\mathbf{M}$ ; hence, $\lambda_3 = \infty$ , $\lambda_4 = \infty$ . As corresponding eigenvectors we can use
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$$
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\boldsymbol {\Phi} _ {3} = \left[ \begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \end{array} \right] \quad \boldsymbol {\Phi} _ {4} = \left[ \begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \end{array} \right] \tag {a}
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$$
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Alternatively, any linear combination of $\Phi_3$ and $\Phi_4$ given in (a) would represent an eigenvector. We should note that $\Phi_i^T\mathbf{M}\Phi_i = 0$ for $i = 3,4$ , and therefore the magnitude of the elements in $\Phi_i$ cannot be fixed using the M-orthonormality condition.
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# 10.2.5 Transformation of the Generalized Eigenproblem
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# $K\phi = \lambda M\phi$ to a Standard Form
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The most common eigenproblems that are encountered in general scientific analysis are standard eigenproblems, and most other eigenproblems can be reduced to a standard form. For this reason, the solution of standard eigenproblems has attracted much attention in numerical analysis, and many solution algorithms are available. The purpose of this section is to show how the eigenproblem $K\phi = \lambda M\phi$ can be reduced to a standard form. The implications of the transformation are twofold. First, because the transformation is possible, use can be made of the various solution algorithms available for standard eigenproblems. We will see that the effectiveness of the eigensolution procedure employed depends to a large degree on the decision whether or not to carry out a transformation to a standard form. Second, if a generalized eigenproblem can be written in standard form, the properties of the eigenvalues, eigenvectors, and characteristic polynomials of the generalized eigenproblem can be deduced from the properties of the corresponding quantities of the standard eigenproblem. Realizing that the properties of standard eigenproblems are more easily assessed, it is to a large extent for the second reason that the transformation to a standard eigenproblem is important to be studied. Indeed, after presenting the transformation procedures, we will show how the properties of the eigenvectors (see Section 10.2.1) and the properties of the characteristic polynomials (see Section 10.2.2) of the problem $K\phi = \lambda M\phi$ are derived from the corresponding properties of the standard eigenproblem.
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In the following we assume that $\mathbf{M}$ is positive definite. This is the case when $\mathbf{M}$ is diagonal with $m_{ii} > 0$ , $i = 1, \ldots, n$ , or $\mathbf{M}$ is banded, as in a consistent mass analysis. If $\mathbf{M}$ is diagonal with some zero diagonal elements, we first need to perform static condensation on the massless degrees of freedom as described in Section 10.3.1. Assuming that $\mathbf{M}$ is positive definite, we can transform the generalized eigenproblem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ given in (10.4) by using a decomposition of $\mathbf{M}$ of the form
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$$
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\mathbf {M} = \mathbf {S} \mathbf {S} ^ {T} \tag {10.30}
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$$
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where S is any nonsingular matrix. Substituting for M into (10.4), we have
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$$
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\mathbf {K} \boldsymbol {\phi} = \lambda \mathbf {S} \mathbf {S} ^ {T} \boldsymbol {\phi} \tag {10.31}
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$$
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Premultiplying both sides of (10.31) by $\mathbf{S}^{-1}$ and defining a vector,
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$$
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\tilde {\boldsymbol {\phi}} = \mathbf {S} ^ {T} \boldsymbol {\phi} \tag {10.32}
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$$
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we obtain the standard eigenproblem,
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$$
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\tilde {\mathbf {K}} \tilde {\boldsymbol {\phi}} = \lambda \tilde {\boldsymbol {\phi}} \tag {10.33}
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$$
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where
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$$
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\tilde {\mathbf {K}} = \mathbf {S} ^ {- 1} \mathbf {K} \mathbf {S} ^ {- T} \tag {10.34}
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$$
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One of two decompositions of $\mathbf{M}$ is used in general: the Cholesky factorization or the spectral decomposition of $\mathbf{M}$ . The Cholesky factorization of $\mathbf{M}$ is obtained as described in Section 8.2.4 and yields $\mathbf{M} = \tilde{\mathbf{L}}_{\mathbf{M}}\tilde{\mathbf{L}}_{\mathbf{M}}^{T}$ . In (10.30) to (10.34) we therefore have
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$$
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\mathbf {S} = \tilde {\mathbf {L}} _ {\mathrm{M}} \tag {10.35}
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$$
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The spectral decomposition of M requires solution of the complete eigensystem of M. Denoting the matrix of orthonormal eigenvectors by R and the diagonal matrix of eigenvalues by $D^{2}$ , we have
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$$
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\mathbf {M} = \mathbf {R D} ^ {2} \mathbf {R} ^ {T} \tag {10.36}
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$$
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and we use in (10.30) to (10.34),
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$$
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\mathbf {S} = \mathbf {R D} \tag {10.37}
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$$
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It should be noted that when $\mathbf{M}$ is diagonal, the matrices $\mathbf{S}$ in (10.35) and (10.37) are the same, but when $\mathbf{M}$ is banded, they are different.
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Considering the effectiveness of the solution of the required eigenvalues and eigenvectors of (10.33), it is most important that $\tilde{K}$ has the same bandwidth as K when M is diagonal. However, when M is banded, $\tilde{K}$ in (10.33) is in general a full matrix, which makes the transformation ineffective in almost all large-order finite element analyses. This will become more apparent in Chapter 11 when various eigensystem solution algorithms are discussed.
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Comparing the Cholesky factorization and the spectral decomposition of M, it may be noted that the use of the Cholesky factors is in general computationally more efficient than the use of the spectral decomposition because fewer operations are involved in calculating $\tilde{L}_{M}$ than R and D. However, the spectral decomposition of M may yield a more accurate solution of $K\phi = \lambda M\phi$ . Assume that M is ill-conditioned with respect to inversion; then the transformation process to the standard eigenproblem is also ill-conditioned. In that case it is important to employ the more stable transformation procedure. Using the Cholesky factorization of M without pivoting, we find that $\tilde{L}_{M}^{-1}$ has large elements in many locations because of the coupling in M and $\tilde{L}_{M}^{-1}$ . Consequently, $\tilde{K}$ is calculated with little precision, and the lowest eigenvalues and corresponding eigenvectors are determined inaccurately.
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On the other hand, using the spectral decomposition of M, good accuracy may be obtained in the elements of R and $D^{2}$ , although some elements in $D^{2}$ are small in relation to the other elements. The ill-conditioning of M is now concentrated in only the small elements of $D^{2}$ , and considering $\tilde{K}$ , only those rows and columns that correspond to the small elements in D have large elements, and the eigenvalues of normal size are more likely to be preserved accurately.
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Consider the following examples of transforming the generalized eigenvalue problem $K\phi = \lambda M\phi$ to a standard form.
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EXAMPLE 10.8: Consider the problem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ , where
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$$
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\mathbf {K} = \left[ \begin{array}{r r r} 3 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{r r r} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{array} \right]
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$$
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Use the Cholesky factorization of M to calculate the matrix $\tilde{K}$ of a corresponding standard eigenproblem.
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We first calculate the $LDL^{T}$ decomposition of M,
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$$
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\mathbf {M} = \left[ \begin{array}{c c c} 1 & & \\ \frac {1}{2} & 1 & \\ 0 & \frac {2}{3} & 1 \end{array} \right] \left[ \begin{array}{c c c} 2 & & \\ & \frac {5}{2} & \\ & & \frac {8}{3} \end{array} \right] \left[ \begin{array}{c c c} 1 & \frac {1}{2} & 0 \\ & 1 & \frac {2}{3} \\ & & 1 \end{array} \right]
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$$
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Hence, the Cholesky factor of $\mathbf{M}$ is (see Section 8.2.4)
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$$
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\tilde {\mathbf {L}} _ {\mathbf {M}} = \left[ \begin{array}{c c c} \sqrt {2} & & \\ \frac {1}{\sqrt {2}} & \sqrt {\frac {5}{2}} & \\ 0 & \sqrt {\frac {2}{5}} & \sqrt {\frac {8}{5}} \end{array} \right]
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$$
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and $\tilde{\mathbf{L}}_{\mathbf{M}}^{-1} = \begin{bmatrix} \frac{1}{\sqrt{2}} & & \\ -\frac{1}{\sqrt{10}} & \sqrt{\frac{2}{5}} & \\ \frac{1}{\sqrt{40}} & -\frac{1}{\sqrt{10}} & \sqrt{\frac{5}{8}} \end{bmatrix}$
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The matrix of the standard eigenproblem $\tilde{\mathbf{K}} = \tilde{\mathbf{L}}_{\mathbf{M}}^{-1}\mathbf{K}\tilde{\mathbf{L}}_{\mathbf{M}}^{-T}$ is in this case,
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$$
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\tilde {\mathbf {K}} = \left[ \begin{array}{c c c} \frac {3}{2} & - \frac {\sqrt {5}}{2} & \frac {\sqrt {5}}{4} \\ - \frac {\sqrt {5}}{2} & \frac {3}{2} & - \frac {5}{4} \\ \frac {\sqrt {5}}{4} & - \frac {5}{4} & \frac {3}{2} \end{array} \right]
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$$
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EXAMPLE 10.9: Consider the generalized eigenproblem in Example 10.8. Use the spectral decomposition of $\mathbf{M}$ to calculate the matrix $\tilde{\mathbf{K}}$ of a corresponding standard eigenproblem.
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The eigenvalues and corresponding eigenvectors of the problem $\mathbf{M}\phi = \lambda \phi$ can be calculated as shown in Example 10.4. We obtain $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 4$ , and
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$$
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\boldsymbol {\phi} _ {1} = \left[ \begin{array}{c} \frac {1}{\sqrt {3}} \\ - \frac {1}{\sqrt {3}} \\ \frac {1}{\sqrt {3}} \end{array} \right]; \quad \boldsymbol {\phi} _ {2} = \left[ \begin{array}{c} \frac {1}{\sqrt {2}} \\ 0 \\ - \frac {1}{\sqrt {2}} \end{array} \right]; \quad \boldsymbol {\phi} _ {3} = \left[ \begin{array}{c} \frac {1}{\sqrt {6}} \\ \frac {2}{\sqrt {6}} \\ \frac {1}{\sqrt {6}} \end{array} \right]
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$$
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<!-- source-page: 874 -->
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Hence, the decomposition $\mathbf{M} = \mathbf{R}\mathbf{D}^2\mathbf{R}^T$ is
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$$
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\mathbf {M} = \left[ \begin{array}{c c c} \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {2}} & \frac {1}{\sqrt {6}} \\ - \frac {1}{\sqrt {3}} & 0 & \frac {2}{\sqrt {6}} \\ \frac {1}{\sqrt {3}} & - \frac {1}{\sqrt {2}} & \frac {1}{\sqrt {6}} \end{array} \right] \left[ \begin{array}{c c c} 1 & & \\ & 2 & \\ & & 4 \end{array} \right] \left[ \begin{array}{c c c} \frac {1}{\sqrt {3}} & - \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {3}} \\ \frac {1}{\sqrt {2}} & 0 & - \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {6}} & \frac {2}{\sqrt {6}} & \frac {1}{\sqrt {6}} \end{array} \right]
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$$
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Noting that $\mathbf{S} = \mathbf{RD}$ and $\mathbf{S}^{-1} = \mathbf{D}^{-1}\mathbf{R}^T$ because $\mathbf{RR}^T = \mathbf{I}$ , we obtain
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$$
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\mathbf {S} ^ {- 1} = \left[ \begin{array}{c c c} \frac {1}{\sqrt {3}} & - \frac {1}{\sqrt {3}} & \frac {1}{\sqrt {3}} \\ \frac {1}{2} & 0 & - \frac {1}{2} \\ \frac {1}{2 \sqrt {6}} & \frac {1}{\sqrt {6}} & \frac {1}{2 \sqrt {6}} \end{array} \right]
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$$
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The matrix of the standard eigenproblem is $\tilde{\mathbf{K}} = \mathbf{S}^{-1}\mathbf{K}\mathbf{S}^{-T}$ ; i.e.,
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$$
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\tilde {\mathbf {K}} = \left[ \begin{array}{c c c} \frac {1 0}{3} & \frac {1}{\sqrt {3}} & - \frac {1}{3 \sqrt {2}} \\ \frac {1}{\sqrt {3}} & 1 & \frac {1}{2 \sqrt {6}} \\ - \frac {1}{3 \sqrt {2}} & \frac {1}{2 \sqrt {6}} & \frac {1}{6} \end{array} \right]
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$$
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We should note that the matrix $\tilde{K}$ obtained here is different from the matrix $\tilde{K}$ derived in Example 10.8.
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In the above discussion we considered only the factorization of M into $M = SS^{T}$ and then the transformation of $K\phi = \lambda M\phi$ into the form given in (10.33). We pointed out that this transformation can yield inaccurate results if M is ill-conditioned. In such a case it seems natural to avoid the decomposition of M and instead use a factorization of K. Rewriting $K\phi = \lambda M\phi$ in the form $M\phi = (1/\lambda)K\phi$ , we can use an analogous procedure to obtain the eigenproblem
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$$
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\tilde {\mathbf {M}} \tilde {\boldsymbol {\phi}} = \frac {1}{\lambda} \tilde {\boldsymbol {\phi}} \tag {10.38}
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$$
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where $\tilde{\mathbf{M}} = \mathbf{S}^{-1}\mathbf{MS}^{-T}$ (10.39)
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$$
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\mathbf {K} = \mathbf {S} \mathbf {S} ^ {T} \tag {10.40}
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$$
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$$
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\tilde {\boldsymbol {\phi}} = \mathbf {S} ^ {T} \boldsymbol {\phi} \tag {10.41}
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$$
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and S is obtained using the Cholesky factor or spectral decomposition of K. If K is well-conditioned, the transformation is also well-conditioned. However, since K is always banded, $\tilde{M}$ is always a full matrix, and the transformation is usually inefficient for the solution of $K\phi = \lambda M\phi$ .
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<!-- source-page: 875 -->
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As we pointed out earlier, the possibility of actually solving a generalized eigenproblem by first transforming it into a standard form is only one reason why we considered the above transformations. The second reason is that the properties of the eigensolution of the generalized eigenproblem can be deduced from the properties of the solution of the corresponding standard eigenproblem. Specifically, we can derive the orthogonality properties of the eigenvectors as given in (10.10) and (10.11), and the Sturm sequence property of the characteristic polynomials of the eigenproblem $K\phi = \lambda M\phi$ and of its associated constraint problems as given in (10.22). In both cases no fundamentally new concepts need be proved; instead, the corresponding properties of the standard eigenproblem, which is obtained from the generalized eigenproblem, are used. We give the proofs in the following examples as an application of the transformation of a generalized eigenproblem to a standard form.
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EXAMPLE 10.10: Show that the eigenvectors of the problem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ are $\mathbf{M}$ - and $\mathbf{K}$ -orthogonal and discuss the orthogonality of the eigenvectors of the problems $^t\mathbf{K}\phi = \lambda^{t - \Delta t}\mathbf{K}\phi$ given in (10.6) and $\mathbf{K}\phi = \lambda \mathbf{C}\phi$ given in (10.7).
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The eigenvector orthogonality is proved by transforming the generalized eigenproblem to a standard form and using the fact that the eigenvectors of a standard eigenproblem with a symmetric matrix are orthogonal. Consider first the problem $K\phi = \lambda M\phi$ and assume that M is positive definite. Then we can use the transformation in (10.30) to (10.34) to obtain, as an equivalent eigenproblem,
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$$
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\tilde {\mathbf {K}} \tilde {\boldsymbol {\phi}} = \lambda \tilde {\boldsymbol {\phi}}
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$$
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where $\mathbf{M} = \mathbf{SS}^T;$ $\tilde{\mathbf{K}} = \mathbf{S}^{-1}\mathbf{KS}^{-T};$ $\tilde{\phi} = \mathbf{S}^T\phi$
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But since the eigenvectors $\tilde{\Phi}_i$ of the problem $\tilde{\mathbf{K}}\tilde{\Phi} = \lambda \tilde{\Phi}$ have the properties (see Section 2.7)
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$$
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\tilde {\boldsymbol {\Phi}} _ {i} ^ {T} \tilde {\boldsymbol {\Phi}} _ {j} = \delta_ {i j}; \quad \tilde {\boldsymbol {\Phi}} _ {i} ^ {T} \tilde {\mathbf {K}} \tilde {\boldsymbol {\Phi}} _ {j} = \lambda_ {i} \delta_ {i j}
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$$
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we have, substituting $\tilde{\Phi}_i = \mathbf{S}^T\phi_i$ , $\tilde{\Phi}_j = \mathbf{S}^T\phi_j$ ,
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$$
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\boldsymbol {\phi} _ {i} ^ {T} \mathbf {M} \boldsymbol {\phi} _ {j} = \delta_ {i j}; \quad \boldsymbol {\phi} _ {i} ^ {T} \mathbf {K} \boldsymbol {\phi} _ {j} = \lambda_ {i} \delta_ {i j} \tag {a}
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$$
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If $\mathbf{M}$ is not positive definite, we consider the eigenproblem $\mathbf{M}\phi = (1 / \lambda)\mathbf{K}\phi$ (with $\mathbf{K}$ positive definite or a shift must be imposed; see Section 10.2.3). We now use the transformation
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$$
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\tilde {\mathbf {M}} \tilde {\boldsymbol {\phi}} = \left(\frac {1}{\lambda}\right) \tilde {\boldsymbol {\phi}}
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$$
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where $\mathbf{K} = \mathbf{SS}^T;$ $\tilde{\mathbf{M}} = \mathbf{S}^{-1}\mathbf{MS}^{-T};$ $\tilde{\phi} = \mathbf{S}^T\phi$
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and the properties $\tilde{\Phi}_i^T\tilde{\Phi}_j = \delta_{ij};\qquad \tilde{\Phi}_i^T\tilde{\mathbf{M}}\tilde{\Phi}_j = \left(\frac{1}{\lambda_i}\right)\delta_{ij}$
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Substituting for $\tilde{\Phi}_i$ and $\tilde{\Phi}_j$ , we obtain
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$$
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\boldsymbol {\phi} _ {i} ^ {T} \mathbf {K} \boldsymbol {\phi} _ {j} = \delta_ {i j}; \quad \boldsymbol {\phi} _ {i} ^ {T} \mathbf {M} \boldsymbol {\phi} _ {j} = \left(\frac {1}{\lambda_ {i}}\right) \delta_ {i j} \tag {b}
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$$
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with the eigenvectors now being K-orthonormalized, because the problem $\mathbf{M}\phi = (1/\lambda)\mathbf{K}\phi$ was considered. To obtain the same vectors as in the problem $K\phi = \lambda M\phi$ , we need to multiply the eigenvectors $\phi_i$ of the problem $\mathbf{M}\phi = (1/\lambda)\mathbf{K}\phi$ by the factors $\sqrt{\lambda_i}, i = 1, \ldots, n$ .
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Considering these proofs, we note that we arranged the eigenproblems in such a way that the matrix associated with the eigenvalue, i.e., the matrix on the right-hand side of the eigenvalue
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<!-- source-page: 876 -->
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problem, is positive definite. This is necessary to be able to carry out the transformation of the generalized eigenproblem to a standard form and thus derive the eigenvector orthogonality properties in (a) and (b). However, considering the problems $K\phi = \lambda^{t-\Delta t}K\phi$ and $K\phi = \lambda C\phi$ given in (10.6) and (10.7), we can proceed in a similar manner. The results would be the eigenvector orthogonality properties given in (10.14) and (10.15).
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EXAMPLE 10.11: Prove the Sturm sequence property of the characteristic polynomials of the problem $K\phi = \lambda M\phi$ and the associated constraint problems. Demonstrate the proof for the following matrices:
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$$
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\mathbf {K} = \left[ \begin{array}{r r r} 3 & - 1 & \\ - 1 & 2 & - 1 \\ & - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{r r r} 4 & 4 & \\ 4 & 8 & 4 \\ & 4 & 8 \end{array} \right] \tag {a}
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$$
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The proof that we consider here is based on the transformation of the eigenproblems $K\phi = \lambda M\phi$ and $\mathbf{K}^{(r)}\phi^{(r)} = \lambda^{(r)}\mathbf{M}^{(r)}\phi^{(r)}$ to standard eigenproblems, for which the characteristic polynomials are known to form a Sturm sequence (see Sections 2.6 and 8.2.5).
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As in Example 10.10, we assume first that $\mathbf{M}$ is positive definite. In this case we can transform the problem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ into the form
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$$
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\tilde {\mathbf {K}} \tilde {\boldsymbol {\phi}} = \lambda \tilde {\boldsymbol {\phi}}
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$$
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where $\tilde{\mathbf{K}} = \tilde{\mathbf{L}}_{\mathbf{M}}^{-1}\mathbf{K}\tilde{\mathbf{L}}_{\mathbf{M}}^{-T};\qquad \mathbf{M} = \tilde{\mathbf{L}}_{\mathbf{M}}\tilde{\mathbf{L}}_{\mathbf{M}}^{\mathrm{T}};\qquad \tilde{\boldsymbol{\phi}} = \tilde{\mathbf{L}}_{\mathbf{M}}^{\mathrm{T}}\boldsymbol{\phi}$
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and $\tilde{\mathbf{L}}_{\mathbf{M}}$ is the Cholesky factor of $\mathbf{M}$ .
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Considering the eigenproblems $\tilde{\mathbf{K}}\tilde{\boldsymbol{\Phi}} = \lambda \tilde{\boldsymbol{\Phi}}$ and $\tilde{\mathbf{K}}^{(r)}\tilde{\boldsymbol{\Phi}}^{(r)} = \lambda^{(r)}\tilde{\boldsymbol{\Phi}}^{(r)}, r = 1, \ldots, n - 1$ [see (8.37)], we know that the characteristic polynomials form a Sturm sequence. On the other hand, if we consider the eigenproblem $\mathbf{K}\boldsymbol{\Phi} = \lambda \mathbf{M}\boldsymbol{\Phi}$ and the eigenproblems of its associated constraint problems, i.e., $\mathbf{K}^{(r)}\boldsymbol{\Phi}^{(r)} = \lambda^{(r)}\mathbf{M}^{(r)}\boldsymbol{\Phi}^{(r)}$ [see (10.20)], we note that the problems $\tilde{\mathbf{K}}^{(r)}\tilde{\boldsymbol{\Phi}}^{(r)} = \lambda^{(r)}\tilde{\boldsymbol{\Phi}}^{(r)}$ and $\mathbf{K}^{(r)}\boldsymbol{\Phi}^{(r)} = \lambda^{(r)}\mathbf{M}^{(r)}\boldsymbol{\Phi}^{(r)}$ have the same eigenvalues. Namely, $\tilde{\mathbf{K}}^{(r)}\boldsymbol{\Phi}^{(r)} = \lambda^{(r)}\tilde{\boldsymbol{\Phi}}^{(r)}$ is a standard eigenproblem corresponding to $\mathbf{K}^{(r)}\boldsymbol{\Phi}^{(r)} = \lambda^{(r)}\mathbf{M}^{(r)}\boldsymbol{\Phi}^{(r)}$ ; i.e., instead of eliminating the $r$ rows and columns from $\tilde{\mathbf{K}}$ (to obtain $\tilde{\mathbf{K}}^{(r)}$ ), we can also calculate $\tilde{\mathbf{K}}^{(r)}$ as follows:
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$$
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\tilde {\mathbf {K}} ^ {(r)} = \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(r) - 1} \mathbf {K} ^ {(r)} \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(r) - T}; \quad \mathbf {M} ^ {(r)} = \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(r)} \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(r) T}; \quad \tilde {\boldsymbol {\phi}} ^ {(r)} = \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(r) T} \boldsymbol {\phi} ^ {(r)} \tag {b}
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$$
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Note that $\tilde{\mathbf{L}}_{\mathbf{M}}^{(r)}$ and $\tilde{\mathbf{L}}_{\mathbf{M}}^{(r)-1}$ can be obtained simply by deleting the last $r$ rows and columns of $\tilde{\mathbf{L}}_{\mathbf{M}}$ and $\tilde{\mathbf{L}}_{\mathbf{M}}^{-1}$ , respectively.
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Hence, the Sturm sequence property also holds for the characteristic polynomials of $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ and the associated constraint problems.
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For the example to be considered, we have
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$$
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\tilde {\mathbf {L}} _ {\mathrm{M}} = \left[ \begin{array}{l l l} 2 & 0 & 0 \\ 2 & 2 & 0 \\ 0 & 2 & 2 \end{array} \right] \tag {c}
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$$
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Hence,
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$$
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\tilde {\mathbf {K}} = \left[ \begin{array}{c c c} \frac {1}{2} & 0 & 0 \\ - \frac {1}{2} & \frac {1}{2} & 0 \\ \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} \end{array} \right] \left[ \begin{array}{c c c} 3 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \left[ \begin{array}{c c c} \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} \\ 0 & \frac {1}{2} & - \frac {1}{2} \\ 0 & 0 & \frac {1}{2} \end{array} \right] = \left[ \begin{array}{c c c} \frac {3}{4} & - 1 & 1 \\ - 1 & \frac {7}{4} & - 2 \\ 1 & - 2 & \frac {5}{2} \end{array} \right] \tag {d}
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$$
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Using $\tilde{\mathbf{K}}$ in (d) to obtain $\tilde{\mathbf{K}}^{(1)}$ and $\tilde{\mathbf{K}}^{(2)}$ , we have
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$$
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\tilde {\mathbf {K}} ^ {(1)} = \left[ \begin{array}{c c} \frac {3}{4} & - 1 \\ - 1 & \frac {7}{4} \end{array} \right]; \quad \tilde {\mathbf {K}} ^ {(2)} = \left[ \frac {3}{4} \right]
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$$
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On the other hand, we can obtain the same matrices $\tilde{\mathbf{K}}^{(1)}$ and $\tilde{\mathbf{K}}^{(2)}$ using the relations in (b),
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$$
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\tilde {\mathbf {K}} ^ {(1)} = \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(1) - 1} \mathbf {K} ^ {(1)} \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(1) - T}; \quad \tilde {\mathbf {K}} ^ {(2)} = \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(2) - 1} \mathbf {K} ^ {(2)} \tilde {\mathbf {L}} _ {\mathbf {M}} ^ {(2) - T}
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$$
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where $\mathbf{K}^{(r)}$ and $\mathbf{M}^{(r)}$ (to calculate $\tilde{\mathbf{L}}_{\mathbf{M}}^{(r)}$ ) are obtained from $\mathbf{K}$ and $\mathbf{M}$ in (a).
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In the preceding discussion we assumed that $\mathbf{M}$ is positive definite. If $\mathbf{M}$ is positive semidefinite, we can consider the problem $\mathbf{M}\phi = (1 / \lambda)\mathbf{K}\phi$ instead, in which $\mathbf{K}$ is positive definite (this may mean that a shift has to be imposed; see Section 10.2.3) and thus show that the Sturm sequence property still holds.
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It may be noted that it follows from this discussion that the characteristic polynomials of the eigenproblems $K\phi = \lambda'^{-\Delta'}K\phi$ and $K\phi = \lambda C\phi$ given in (10.6) and (10.7) and of their associated constraint problems also form a Sturm sequence.
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# 10.2.6 Exercises
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10.1. Consider the generalized eigenproblem
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$$
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\left[ \begin{array}{r r r} 6 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right] \boldsymbol {\Phi} = \lambda \left[ \begin{array}{r r r} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{array} \right] \boldsymbol {\Phi}
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$$
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(a) Calculate the eigenvalues and eigenvectors and show explicitly that the eigenvectors are M-orthogonal.
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(b) Find two vectors that are M-orthogonal but are not eigenvectors.
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10.2. Calculate the eigenvalues of the eigenproblem in Exercise 10.1 and of its associated constraint problems. Show that the eigenvalues satisfy the separation property (10.22).
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10.3. Consider the eigenproblem
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$$
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\left[ \begin{array}{c c c} 2 & - 1 & 0 \\ - 1 & 2 & 0 \\ 0 & 0 & 3 \end{array} \right] \boldsymbol {\phi} = \lambda \left[ \begin{array}{c c c} 1 & & \\ & 2 & \\ & & \frac {3}{2} \end{array} \right] \boldsymbol {\phi}
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$$
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(a) Calculate the eigenvalues and eigenvectors of the problem. Also, calculate the eigenvalues of the associated constraint problems [see (10.20)].
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(b) Establish two vectors that are M-orthogonal but are not eigenvectors.
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10.4. Calculate the eigenvalues and eigenvectors of the problem
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$$
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\left[ \begin{array}{c c} 6 & - 1 \\ - 1 & 4 \end{array} \right] \boldsymbol {\phi} = \lambda \left[ \begin{array}{c c} 2 & 0 \\ 0 & 0 \end{array} \right] \boldsymbol {\phi}
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$$
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Then apply a shift $\rho = 3$ on $\mathbf{K}$ and calculate the eigenvalues and eigenvectors of the new problem [see (10.25)].
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10.5. Transform the generalized eigenproblem in Exercise 10.1 into a standard form.
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10.6. (a) The eigenvalues and eigenvectors of the problem
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$$
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\mathbf {K} \phi = \lambda \phi
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$$
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are $\lambda_{1} = 1;\qquad \phi_{1} = \frac{1}{\sqrt{2}}\left[ \begin{array}{l}1\\ 1 \end{array} \right]$
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$$
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\lambda_ {2} = 4; \quad \phi_ {2} = \frac {1}{\sqrt {2}} \left[ \begin{array}{c} 1 \\ - 1 \end{array} \right]
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$$
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Calculate K.
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(b) The eigenvalues and eigenvectors of the problem
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$$
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\mathbf {K} \boldsymbol {\phi} = \lambda \mathbf {M} \boldsymbol {\phi}
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$$
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are $\lambda_{1} = 1;\qquad \phi_{1} = \frac{1}{\sqrt{3}}\left[ \begin{array}{l}1\\ 1 \end{array} \right]$
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$$
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\lambda_ {2} = 4; \quad \phi_ {2} = \sqrt {\frac {2}{3}} \left[ \begin{array}{c} - \frac {1}{2} \\ 1 \end{array} \right]
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$$
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Calculate K and M. Are the K and M matrices in (a) and (b) unique?
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# 10.3 APPROXIMATE SOLUTION TECHNIQUES
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It is apparent from the nature of a dynamic problem that a dynamic response calculation must be substantially more costly than a static analysis. Whereas in a static analysis the solution is obtained in one step, in dynamics the solution is required at a number of discrete time points over the time interval considered. Indeed, we found that in a direct step-by-step integration solution, an equation of statics, which includes the effects of inertia and damping forces, is considered at the end of each discrete time step (see Section 9.2). Considering a mode superposition analysis, the main computational effort is spent in the calculation of the required frequencies and mode shapes, which also requires considerably more effort than a static analysis. It is therefore natural that much attention has been directed toward effective algorithms for the calculation of the required eigensystem in the problem $K\phi = \lambda M\phi$ . In fact, because the “exact” solution of the required eigenvalues and corresponding eigenvectors can be prohibitively expensive when the order of the system is large and a “conventional” technique is used, approximate techniques of solution have been developed. The purpose of this section is to present the major approximate methods that have been designed and are currently still in use.
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The approximate solution techniques have primarily been developed to calculate the lowest eigenvalues and corresponding eigenvectors in the problem $K\phi = \lambda M\phi$ when the order of the system is large. Most programs use exact solution techniques in the analysis of small-order systems. However, the problem of calculating the few lowest eigenpairs of relatively large-order systems is very important and is encountered in all branches of structural engineering and in particular in earthquake response analysis. In the following sections we present three major techniques. The aim in the presentation is not to advocate the implementation of any one of these methods but rather to describe their practical use, their limitations, and the assumptions employed. Moreover, the relationships between the approximate techniques are described, and in Section 11.6 we will, in fact, find that the approximate techniques considered here can be understood to be a first iteration (and may be used as such) in the subspace iteration algorithm.
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# 10.3.1 Static Condensation
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We have already encountered the procedure of static condensation in the solution of static equilibrium equations, where we showed that static condensation is, in fact, an application of Gauss elimination (see Section 8.2.4). In static condensation we eliminated those degrees
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of freedom that are not required to appear in the global finite element assemblage. For example, the displacement degrees of freedom at the internal nodes of a finite element can be condensed out because they do not take part in imposing interelement continuity. We mentioned in Section 8.2.4 that the term “static condensation” was actually coined in dynamic analysis.
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The basic assumption of static condensation in the calculation of frequencies and mode shapes is that the mass of the structure can be lumped at only some specific degrees of freedom without much effect on the accuracy of the frequencies and mode shapes of interest. In the case of a lumped mass matrix with some zero diagonal elements, some of the mass lumping has already been carried out. However, additional mass lumping is in general required. Typically, the ratio of mass degrees of freedom to the total number of degrees of freedom may be somewhere between $\frac{1}{2}$ and $\frac{1}{10}$ . The more mass lumping is performed, the less computer effort is required in the solution; however, the more probable it is also that the required frequencies and mode shapes are not predicted accurately. We shall have more to say about this later.
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Assume that the mass lumping has been carried out. By partitioning the matrices, we can then write the eigenproblem in the form
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$$
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\left[ \begin{array}{l l} \mathbf {K} _ {a a} & \mathbf {K} _ {a c} \\ \mathbf {K} _ {c a} & \mathbf {K} _ {c c} \end{array} \right] \left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right] = \lambda \left[ \begin{array}{l l} \mathbf {M} _ {a} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} \end{array} \right] \left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right] \tag {10.42}
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$$
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where $\phi_{a}$ and $\phi_{c}$ are the displacements at the mass and the massless degrees of freedom, respectively, and $\mathbf{M}_{a}$ is a diagonal mass matrix. The relation in (10.42) gives the condition
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$$
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\mathbf {K} _ {c a} \boldsymbol {\phi} _ {a} + \mathbf {K} _ {c c} \boldsymbol {\phi} _ {c} = \mathbf {0} \tag {10.43}
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$$
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which can be used to eliminate $\phi_{c}$ . From (10.43) we obtain
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$$
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\phi_ {c} = - \mathbf {K} _ {c c} ^ {- 1} \mathbf {K} _ {c a} \phi_ {a} \tag {10.44}
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$$
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and substituting into (10.42), we obtain the reduced eigenproblem
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$$
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\mathbf {K} _ {a} \boldsymbol {\phi} _ {a} = \lambda \mathbf {M} _ {a} \boldsymbol {\phi} _ {a} \tag {10.45}
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$$
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where $\mathbf{K}_a = \mathbf{K}_{aa} - \mathbf{K}_{ac}\mathbf{K}_{cc}^{-1}\mathbf{K}_{ca}$ (10.46)
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The solution of the generalized eigenproblem in (10.45) is in most cases obtained by transforming the problem first into a standard form as described in Section 10.2.5. Since $M_{a}$ is a diagonal mass matrix with all its diagonal elements positive and probably not small, the transformation is in general well-conditioned.
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The analogy to the use of static condensation in static analysis should be noted. Realizing that the right-hand side of (10.42) may be understood to be a load vector R, where
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$$
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\mathbf {R} = \left[ \begin{array}{c} \lambda \mathbf {M} _ {a} \boldsymbol {\phi} _ {a} \\ \mathbf {0} \end{array} \right] \tag {10.47}
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$$
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we can use Gauss elimination on the massless degrees of freedom in the same way as we do on the degrees of freedom associated with the interior nodes of an element or a substructure (see Section 8.2.4).
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One important aspect should be observed when comparing the static condensation procedure on the massless degrees of freedom in (10.42) to (10.46) on the one side, with Gauss elimination or static condensation in static analysis on the other side. Considering (10.47), we find that the loads at the $\phi_{a}$ degrees of freedom depend on the eigenvalue
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(free-vibration frequency squared) and eigenvector (mode shape displacements). This means that in (10.45) a further reduction of the number of degrees of freedom to be considered is not possible. This is a basic difference to static condensation as applied in static analysis, where the loads are given explicitly and their effect can be carried over to the remaining degrees of freedom.
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EXAMPLE 10.12: Use static condensation to calculate the eigenvalues and eigenvectors of the problem $K\phi = \lambda M\phi$ , where
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$$
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\mathbf {K} = \left[ \begin{array}{r r r r} 2 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{c c c c} 0 & & & \\ & 2 & & \\ & & 0 & \\ & & & 1 \end{array} \right]
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$$
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First we rearrange columns and rows to obtain the form given in (10.42), which is
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$$
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\left[ \begin{array}{c c c c} 2 & 0 & - 1 & - 1 \\ 0 & 1 & 0 & - 1 \\ - 1 & 0 & 2 & 0 \\ - 1 & - 1 & 0 & 2 \end{array} \right] \left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right] = \lambda \left[ \begin{array}{c c c c} 2 & & & \\ & 1 & & \\ & & 0 & \\ & & & 0 \end{array} \right] \left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right]
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$$
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Hence, $\mathbf{K}_{\alpha}$ given in (10.46) is in this case,
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$$
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\mathbf {K} _ {a} = \left[ \begin{array}{c c} 2 & 0 \\ 0 & 1 \end{array} \right] - \left[ \begin{array}{c c} - 1 & - 1 \\ 0 & - 1 \end{array} \right] \left[ \begin{array}{c c} \frac {1}{2} & 0 \\ 0 & \frac {1}{2} \end{array} \right] \left[ \begin{array}{c c} - 1 & 0 \\ - 1 & - 1 \end{array} \right] = \left[ \begin{array}{c c} 1 & - \frac {1}{2} \\ - \frac {1}{2} & \frac {1}{2} \end{array} \right]
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$$
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The eigenproblem $\mathbf{K}_a\phi_a = \lambda \mathbf{M}_a\phi_a$ is, therefore,
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$$
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\left[ \begin{array}{c c} 1 & - \frac {1}{2} \\ - \frac {1}{2} & \frac {1}{2} \end{array} \right] \boldsymbol {\phi} _ {a} = \lambda \left[ \begin{array}{c c} 2 & \\ & 1 \end{array} \right] \boldsymbol {\phi} _ {a}
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$$
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and we have $\operatorname{det}(\mathbf{K}_a - \lambda \mathbf{M}_a) = 2\lambda^2 - 2\lambda + \frac{1}{4}$
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Hence, $\lambda_{1} = \frac{1}{2} -\frac{\sqrt{2}}{4};\qquad \lambda_{2} = \frac{1}{2} +\frac{\sqrt{2}}{4}$
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The corresponding eigenvectors are calculated using
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$$
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(\mathbf {K} _ {a} - \lambda_ {i} \mathbf {M} _ {a}) \boldsymbol {\phi} _ {a _ {i}} = \mathbf {0}; \quad \boldsymbol {\phi} _ {a _ {i}} ^ {T} \mathbf {M} _ {a} \boldsymbol {\phi} _ {a _ {i}} = 1
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$$
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Hence $\pmb{\Phi}_{a_1} = \left[ \begin{array}{c}\frac{1}{2}\\ \frac{\sqrt{2}}{2} \end{array} \right];\qquad \Phi_{a_2} = \left[ \begin{array}{c}\frac{-1}{2}\\ \frac{\sqrt{2}}{2} \end{array} \right]$
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Using (10.44), we obtain
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$$
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\boldsymbol {\Phi} _ {c _ {1}} = - \left[ \begin{array}{l l} \frac {1}{2} & 0 \\ 0 & \frac {1}{2} \end{array} \right] \left[ \begin{array}{c c} - 1 & 0 \\ - 1 & - 1 \end{array} \right] \left[ \begin{array}{l} \frac {1}{2} \\ \frac {\sqrt {2}}{2} \end{array} \right] = \left[ \begin{array}{l} \frac {1}{4} \\ \frac {1 + \sqrt {2}}{4} \end{array} \right]
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$$
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$$
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\boldsymbol {\Phi} _ {c _ {2}} = - \left[ \begin{array}{l l l} \frac {1}{2} & & 0 \\ 0 & & \frac {1}{2} \end{array} \right] \left[ \begin{array}{l l} - 1 & 0 \\ - 1 & - 1 \end{array} \right] \left[ \begin{array}{l} - \frac {1}{2} \\ \frac {\sqrt {2}}{2} \end{array} \right] = \left[ \begin{array}{l} - \frac {1}{4} \\ \frac {- 1 + \sqrt {2}}{4} \end{array} \right]
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$$
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