Similarly, the terms corresponding to the $v_{i}$ degrees of freedom are calculated, and we obtain $$ \mathbf {S} _ {h} = \left[ \begin{array}{c c} \tilde {\mathbf {S}} _ {h} & \mathbf {0} \\ \mathbf {0} & \tilde {\mathbf {S}} _ {h} \end{array} \right]; \quad \tilde {\mathbf {S}} _ {h} = \frac {1}{6} \left[ \begin{array}{c c c c} 4 & - 1 & - 2 & - 1 \\ - 1 & 4 & - 1 & - 2 \\ - 2 & - 1 & 4 & - 1 \\ - 1 & - 2 & - 1 & 4 \end{array} \right] $$ Let us now consider the $u/p$ -c formulation. In this case the same expression as in (4.206) applies, but we need to use $\mathbf{G}_h = (\mathbf{K}_{pu})_h^T\mathbf{T}_h^{-1}(\mathbf{K}_{pu})_h$ , where $\mathbf{T}_h$ is the matrix of the $L^2$ -norm of $p_h$ (see Exercise 4.59); i.e., for any vector of pressure nodal values $\mathbf{P}_h$ , we have $\|p_h\| = \mathbf{P}_h^T\mathbf{T}_h\mathbf{P}_h$ . The form (4.206) of the inf-sup condition is effective because we can numerically evaluate the inf-sup value of the left-hand side and do so for a sequence of meshes. If the left-hand-side inf-sup value approaches (asymptotically) a value greater than zero (and there are no spurious pressure modes, further discussed below), the inf-sup condition is satisfied. In practice, only a sequence of about three meshes needs to be considered (see examples given below). The key is the evaluation of the inf-sup value of the expression in $(4.206)$ . We can show that this value is given by the square root of the smallest nonzero eigenvalue of the problem $$ \mathbf {G} _ {h} \boldsymbol {\phi} _ {h} = \lambda \mathbf {S} _ {h} \boldsymbol {\phi} _ {h} \tag {4.207} $$ Hence, if there are $(k - 1)$ zero eigenvalues (because $G_{h}$ is a positive semidefinite matrix) and we order the eigenvalues in ascending order, we find that the inf-sup value of the expression in (4.206) is $\sqrt{\lambda_{k}}$ . We prove this result in the following example. EXAMPLE 4.41: Consider the function $f(\mathbf{U}, \mathbf{V})$ defined as $$ f (\mathbf {U}, \mathbf {V}) = \frac {\mathbf {U} ^ {T} \mathbf {G} \mathbf {V}}{(\mathbf {U} ^ {T} \mathbf {G} \mathbf {U}) ^ {1 / 2} (\mathbf {V} ^ {T} \mathbf {S} \mathbf {V}) ^ {1 / 2}} \tag {a} $$ where $\mathbf{G}$ is an $n \times n$ symmetric positive semidefinite matrix, $\mathbf{S}$ is an $n \times n$ positive definite matrix, and $\mathbf{U}, \mathbf{V}$ are vectors of order $n$ . Show that $$ \inf _ {\mathbf {U}} \sup _ {\mathbf {V}} f (\mathbf {U}, \mathbf {V}) = \sqrt {\lambda_ {k}} \tag {b} $$ where $\lambda_{k}$ is the smallest nonzero eigenvalue of the problem $$ \mathbf {G} \boldsymbol {\phi} = \lambda \mathbf {S} \boldsymbol {\phi} \tag {c} $$ Let the eigenvalues of (c) be $$ \lambda_ {1} = \lambda_ {2} = \dots = \lambda_ {k - 1} = 0 < \lambda_ {k} \leq \lambda_ {k + 1} \dots \leq \lambda_ {n} $$ and the corresponding eigenvectors be $\phi_1, \phi_2, \ldots, \phi_n$ . To evaluate $f(\mathbf{U}, \mathbf{V})$ , we represent U and V as $$ \mathbf {U} = \sum_ {i = 1} ^ {n} \tilde {u} _ {i} \boldsymbol {\phi} _ {i}; \quad \mathbf {V} = \sum_ {i = 1} ^ {n} \tilde {v} _ {i} \boldsymbol {\phi} _ {i} $$ Therefore, for any U, $$ \sup _ {\mathbf {v}} f (\mathbf {U}, \mathbf {V}) = \sup _ {\tilde {v} _ {i}} \frac {\sum_ {i = 1} ^ {n} \lambda_ {i} \tilde {u} _ {i} \tilde {v} _ {i}}{\left(\sum_ {i = 1} ^ {n} \lambda_ {i} \tilde {u} _ {i} ^ {2}\right) ^ {1 / 2} \left(\sum_ {i = 1} ^ {n} \tilde {v} _ {i} ^ {2}\right) ^ {1 / 2}} $$ $$ = \frac {1}{\left(\sum_ {i = 1} ^ {n} \lambda_ {i} \tilde {u} _ {i} ^ {2}\right) ^ {1 / 2}} \sup _ {\tilde {v} _ {i}} \frac {\sum_ {i = 1} ^ {n} \lambda_ {i} \tilde {u} _ {i} \tilde {v} _ {i}}{\left(\sum_ {i = 1} ^ {n} \tilde {v} _ {i} ^ {2}\right) ^ {1 / 2}} \tag {d} $$ To evaluate the supremum value in (d), let us define $\alpha_{i} = \lambda_{i}\tilde{u}_{i}$ ; then we note that $$ \sum_ {i = 1} ^ {n} \lambda_ {i} \tilde {u} _ {i} \tilde {v} _ {i} = \sum_ {i = 1} ^ {n} \alpha_ {i} \tilde {v} _ {i} \leq \sqrt {\sum_ {i = 1} ^ {n} \alpha_ {i} ^ {2} \sum_ {i = 1} ^ {n} \tilde {v} _ {i} ^ {2}} \tag {e} $$ (by the Schwarz inequality), and equality is reached when $\tilde{v}_i = \alpha_i$ . Substituting from (e) into (d) and using $\lambda_1 = \cdots = \lambda_{k-1} = 0$ , we thus obtain $$ \sup _ {\mathbf {V}} f (\mathbf {U}, \mathbf {V}) = \sqrt {\frac {\sum_ {i = 1} ^ {n} \lambda_ {i} ^ {2} \tilde {u} _ {i} ^ {2}}{\sum_ {i = 1} ^ {n} \lambda_ {i} \tilde {u} _ {i} ^ {2}}} = \sqrt {\frac {\sum_ {i = k} ^ {n} \lambda_ {i} ^ {2} \tilde {u} _ {i} ^ {2}}{\sum_ {i = k} ^ {n} \lambda_ {i} \tilde {u} _ {i} ^ {2}}} $$ If we now let $\sqrt{\lambda_i}\tilde{u}_i = \beta_i$ , we can write $$ \inf _ {\mathbf {U}} \sup _ {\mathbf {V}} f (\mathbf {U}, \mathbf {V}) = \inf _ {(\tilde {u} _ {i}) \tilde {r} = 1} \sqrt {\frac {\sum_ {i = k} ^ {n} \lambda_ {i} ^ {2} \tilde {u} _ {i} ^ {2}}{\sum_ {i = k} ^ {n} \lambda_ {i} \tilde {u} _ {i} ^ {2}}} = \inf _ {(\beta_ {i}) \tilde {r} = k} \sqrt {\frac {\sum_ {i = k} ^ {n} \lambda_ {i} \beta_ {i} ^ {2}}{\sum_ {i = k} ^ {n} \beta_ {i} ^ {2}}} \tag {f} $$ The last expression in (f) has the form of a Rayleigh quotient (see Section 2.6), and we know that the smallest value is $\sqrt{\lambda_k}$ , achieved for $\beta_k \neq 0$ and $\beta_i = 0$ , for $i \neq k$ , which gives the required result. In practice, to calculate the inf-sup value $\sqrt{\lambda_{k}}$ an eigenvalue solution routine should be used that can skip over all zero eigenvalues and then calculate $\lambda_{k}$ . A Sturm sequence test (see Section 11.4.3) will then also give the value of k, and then we can conclude directly whether the model contains spurious pressure modes. Namely, let $n_{p}$ be the number of pressure degrees of freedom and $n_{u}$ be the number of displacement degrees of freedom. Then the number of pressure modes, $k_{pm}$ , is $$ k _ {p m} = k - \left(n _ {u} - n _ {p} + 1\right) $$ If $k_{pm} > 0$ , the finite element discretization contains the constant pressure mode or spurious pressure modes [the inf-sup value in (4.193) is zero, although $\lambda_k$ (the first nonzero eigenvalue) may asymptotically approach a value greater than zero]. This formula follows because for there to be no pressure mode, the kernel $(\mathbf{K}_{up})_{h}$ must be zero [see (4.199)]. To demonstrate this inf-sup test, we show in Fig. 4.24 results obtained for the four-node and nine-node elements. We see that a sequence of three meshes used to calculate $\sqrt{\lambda_{k}}$ for each discretization was, in these cases, sufficient to identify whether the element locks. We note that, clearly, the four-node and the nine-node displacement-based elements do not satisfy the inf-sup condition and that the distortions of elements have a negligible effect on the results. In each of these tests $k_{pm}$ was zero, hence, as expected, the idealizations do not contain any pressure modes. Of course, a spurious pressure mode would be found for the 4/1 element if the boundary conditions of Example 4.38 were used. That is, in the general testing of elements for spurious modes the condition of zero displacements on the complete boundary should be considered [the smaller the dimension of $V_{h}$ , for a given $Q_{h}$ , the greater the possibility that (4.196) is satisfied]. The solutions in Fig. 4.24 are numerical results pertaining to only one problem and one mesh topology. However, if the inf-sup condition is not satisfied in these results, then we can conclude that it is not satisfied in general. 

text_image

Uniform mesh


text_image

Distorted mesh

(a) Problem considered in inf-sup test. $N =$ number of elements along each side; we show $N = 4$ , plane strain case 

text_image

4-node element


text_image

9-node element

(b) Elements used Figure 4.24 The inf-sup test applied in a simple problem 

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| N | log(1/N) | log(inf-sup value) | |-----|----------|---------------------| | 8 | -0.8 | -0.6 | | 4 | -0.6 | -0.4 | | 2 | -0.4 | -0.2 | | 8 | -0.2 | -0.0 | | 4 | -0.0 | 0.0 | | 2 | 0.0 | 0.0 | | 8 | 0.2 | 0.0 | | 4 | 0.4 | 0.0 | | 2 | 0.6 | 0.0 | | 8 | 0.8 | 0.0 | | 4 | 1.0 | 0.0 | | 2 | 1.2 | 0.0 | | 8 | 1.4 | 0.0 |

(c) 4-node elements 

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| N | log(1/N) | log(inf-sup value) | |-----|----------|---------------------| | 8 | -0.8 | -1.6 | | 4 | -0.6 | -1.4 | | 2 | -0.4 | -1.2 | | 2 | -0.2 | -1.0 | | 2 | 0.0 | -0.8 |

(d) 9-node elements Figure 4.24 (continued) Figure 4.25 shows results pertaining to the three-node triangular constant pressure element, formulated as a u/p element (see Exercise 4.50). The results show that the inf-sup condition is not satisfied by this element. Further, it is interesting to note that the meshes with pattern B do not contain spurious pressure modes, whereas the other meshes in general do contain spurious pressure modes. 

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| Pattern | log(1/N) | log(inf-sup value) | | --------- | -------- | ------------------ | | Pattern A | -1.0 | -0.6 | | Pattern A | -0.8 | -0.4 | | Pattern A | -0.6 | -0.2 | | Pattern A | -0.4 | 0.0 | | Pattern B | -1.0 | -0.8 | | Pattern B | -0.8 | -0.6 | | Pattern B | -0.6 | -0.4 | | Pattern B | -0.4 | -0.2 | | Pattern C | -1.0 | -1.0 | | Pattern C | -0.8 | -0.8 | | Pattern C | -0.6 | -0.6 | | Pattern C | -0.4 | -0.4 | | Pattern C | -0.2 | -0.2 |

Figure 4.25 Inf-sup test of triangular elements, using problem of Fig.4.24(a). The patterns A and C result in spurious modes. Additional results are given in Table 4.8 (see D. Chapelle and K. J. Bathe [A]). This table gives a summary of the results of the numerical evaluations of the inf-sup condition and analytical results, given, for example, by F. Brezzi and M. Fortin [A]. The numerical evaluation is useful because the same procedure applies to all u/p and u/p-c elements, in uniform or distorted meshes, and, in general, elements and formulations can be evaluated for which no analytical results are (yet) available, see K. J. Bathe [E] and, for example, D. Pantuso and K. J. Bathe [A] (but also C. Lovadina [A]), K. J. Bathe, C. Nitikitpaiboon, and X. Wang [A] and X. Wang and K. J. Bathe [A], A. Iosilevich, K. J. Bathe, and F. Brezzi [A] and S. De and K. J. Bathe [C]. However, we considered linear analysis, and while an element may satisfy the ellipticity and inf-sup conditions in linear analysis, the element may not be stable in large deformation formulations, see for example P. Wriggers and S. Reese [A] and D. Pantuso and K. J. Bathe [B], and even when then only small displacements and strains are measured, see T. Sussman and K. J. Bathe [E]. TABLE 4.8 Inf-sup numerical predictions

	$Element^†$	Inf-sup condition		Remarks
		Analytical proof	Numerical prediction
	$3/1^†$	Fail	Fail	See Fig. 4.25
	$4/1^†$	Fail	Fail	See Fig. 4.24
	8/3	Fail	Fail
	8/1	Pass	Pass
	9/4	Fail	Fail
	9/3	Pass	Pass	See Example 4.36, Fig. 4.24
	4/3-c	Pass	Pass
	9/9-c	Fail	Fail
	9/8-c	Fail	Fail
	9/5-c	?	Fail
	9/4-c	Pass	Pass
	9/(4-c + 1)	?	Pass	For the element see P. M. Gresho, R. L. Lee, S. T. Chan, and J. M. Leone, Jr. [A]

$^{\dagger}$ O, Continuous pressure degree of freedom; X, discontinuous pressure degree of freedom. $^{\dagger}$ 3/1 and 4/1 element discretizations can contain spurious pressure modes. EXAMPLE 4.42: Assume that the inf-sup condition (4.175) holds. Prove that (4.166) follows. Let the eigenvectors and corresponding eigenvalues of (4.207) with $G_{h}$ corresponding to $D_{h}$ [and not $P_{h}(D_{h})$ because in (4.175) we consider $D_{h}$ ] be $\phi_{i}$ and $\lambda_{i}, i = 1, \ldots, n$ . The vectors $\phi_{i}$ form an orthonormal basis of $V_{h}$ . Then we can write any vector $w_{h}$ in $V_{h}$ as $$ \mathbf {w} _ {h} = \sum_ {i = 1} ^ {n} w _ {h} ^ {i} \boldsymbol {\phi} _ {i} \tag {a} $$ and we have by use of the eigenvalue and vector properties (see Section 2.5) $$ \| \operatorname{div} \mathbf {w} _ {h} \| ^ {2} = \sum_ {i = 1} ^ {n} \lambda_ {i} (w _ {h} ^ {i}) ^ {2} $$ Let us now pick any $q_h$ and any $\tilde{\mathbf{w}}_h$ satisfying $\operatorname{div} \tilde{\mathbf{w}}_h = q_h$ . We can decompose $\tilde{\mathbf{w}}_h$ in the form of (a), $$ \tilde {\mathbf {w}} _ {h} = \sum_ {i = 1} ^ {k - 1} \tilde {w} _ {h} ^ {i} \boldsymbol {\phi} _ {i} + \sum_ {i = k} ^ {n} \tilde {w} _ {h} ^ {i} \boldsymbol {\phi} _ {i} \tag {b} $$ The first summation sign in (b) defines a vector that belongs to $K_{h}(0)$ and may be a large component. However, we are concerned only with the component that is not an element of $K_{h}(0)$ , which we call $w_{h}$ , $$ \mathbf {w} _ {h} = \sum_ {i = k} ^ {n} \tilde {w} _ {h} ^ {i} \phi_ {i} $$ $$ \begin{array}{l} \geq \lambda_ {k} \\ = \beta_ {h} ^ {2} \\ \geq \beta^ {2} \\ \end{array} $$ With this $\mathbf{w}_h$ , we have $\frac{\|q_h\|^2}{\|\mathbf{w}_h\|^2} = \frac{\sum_{i=k}^n \lambda_i(\tilde{w}_h^i)^2}{\sum_{i=k}^n (\tilde{w}_h^i)^2}$ and (4.166) follows with $c' = 1 / \beta$ . # 4.5.7 An Application to Structural Elements: The Isoparametric Beam Elements In the above discussion we considered the general elasticity problem (4.151) and the corresponding variational discrete problem (4.154) subject to the constraint of (near or total) incompressibility. However, the ellipticity and inf-sup conditions are also the basic conditions to be considered in the development of beam, plate, and shell elements that are subject to shear and membrane strain constraints (see Section 5.4). We briefly introduced a mixed two-node beam element in Example 4.30 and we consider this and higher-order elements of the same kind in Section 5.4.1. Let us briefly discuss the ellipticity and inf-sup conditions for mixed interpolated and pure displacement-based beam elements. # General Considerations The variational discrete problem of the displacement-based formulation is $$ \min _ {v _ {h} \in V _ {h}} \left\{\frac {E I}{2} \int_ {0} ^ {L} \left(\beta_ {h} ^ {\prime}\right) ^ {2} d x + \frac {G A k}{2} \int_ {0} ^ {L} \left(\gamma_ {h}\right) ^ {2} d x - \int_ {0} ^ {L} p w _ {h} d x \right\} \tag {4.208} $$ where EI and GAk are the flexural and shear rigidities of the beam (see Section 5.4.1), L is the length of the beam, p is the transverse load per unit length, $\beta_{h}$ is the section rotation, $\gamma_{h}$ is the transverse shear strain, $$ \gamma_ {h} = \frac {\partial w _ {h}}{\partial x} - \beta_ {h} \tag {4.209} $$ $w_{h}$ is the transverse displacement, and an element of $V_{h}$ is $$ \mathbf {v} _ {h} = \left[ \begin{array}{l} w _ {h} \\ \beta_ {h} \end{array} \right] \tag {4.210} $$ The constraint to be dealt with is now the shear constraint, $$ \gamma_ {h} = \frac {\partial w _ {h}}{\partial x} - \beta_ {h} \rightarrow 0 \tag {4.211} $$ In practice, $\gamma_{h}$ is usually very small and can of course also be zero. Hence we have, using our earlier notation, the spaces $$ K _ {h} (q _ {h}) = \{\mathbf {v} _ {h} \mid \mathbf {v} _ {h} \in V _ {h}, \gamma_ {h} (\mathbf {v} _ {h}) = q _ {h} \} \tag {4.212} $$ $$ D _ {h} = \left\{q _ {h} \mid q _ {h} = \gamma_ {h} \left(\mathbf {v} _ {h}\right) \text { for some } \mathbf {v} _ {h} \in V _ {h} \right\} \tag {4.213} $$ and the norms $$ \left\| \mathbf {v} _ {h} \right\| ^ {2} = \int_ {\mathrm{Vol}} \left[ \left(\frac {\partial w _ {h}}{\partial x}\right) ^ {2} + L ^ {2} \left(\frac {\partial \beta_ {h}}{\partial x}\right) ^ {2} \right] d \mathrm{Vol}; \quad \left\| \gamma_ {h} \right\| ^ {2} = \int_ {\mathrm{Vol}} \left(\gamma_ {h}\right) ^ {2} d \mathrm{Vol} \tag {4.214} $$ The ellipticity condition is satisfied in this problem formulation because $$ E I \int_ {0} ^ {L} \left(\beta_ {h} ^ {\prime}\right) ^ {2} d x \geq \alpha \| \mathbf {v} _ {h} \| ^ {2} \quad \forall \mathbf {v} _ {h} \in K _ {h} (0) \tag {4.215} $$ with $\alpha > 0$ and independent of $h$ . To prove this relation we need only to note that $$ \int_ {0} ^ {L} \left(\frac {\partial w _ {h}}{\partial x}\right) ^ {2} d x = \int_ {0} ^ {L} \left(\beta_ {h}\right) ^ {2} d x \leq \int_ {0} ^ {L} L ^ {2} \left(\frac {\partial \beta_ {h}}{\partial x}\right) ^ {2} d x \tag {4.216} $$ and therefore, $\| \mathbf{v}_h\|^2 \leq 2L^2 \int_{\mathrm{Vol}} \left(\frac{\partial \beta_h}{\partial x}\right)^2 d\mathrm{Vol}$ (4.217) giving $\alpha = EI / 2L^2$ The inf-sup condition for this formulation is $$ \inf _ {\gamma_ {h} \in D _ {h}} \sup _ {\mathbf {v} _ {h} \in V _ {h}} \frac {\int_ {\mathrm{Vol}} \gamma_ {h} \left[ \left(\partial w _ {h} / \partial x\right) - \beta_ {h} \right] d \mathrm{Vol}}{\left\| \gamma_ {h} \right\| \left\| \mathbf {v} _ {h} \right\|} \geq c > 0 \tag {4.218} $$ in which the constant $c$ is independent of $h$ .  Figure 4.26 Analysis of cantilever beam using two-node beam elements. Four equal length elements are used. (Shear correction factor k of (5.57) is taken equal to 1.0.) The two-node element. Let us first consider the two-node displacement-based element for which $w_{h}$ and $\beta_{h}$ are assumed linear over each element [see Fig. 4.26(a) for an example solution]. A comparison of the computed results with the Bernoulli beam theory solution given in Fig. 4.26 shows that the element performs quite badly. In this case $K_{h}(0) = \{0\}$ , and so the inf-sup condition in (4.218) is not satisfied. Referring to (4.164), we can also see that a good convergence behavior is not possible; namely, $d(\mathbf{u}, V_{h}) \to 0$ as we increase the space $V_{h}$ , whereas $d[\mathbf{u}, K_{h}(0)] = \|\mathbf{u}\|$ (a constant value). Next, consider the two-node mixed interpolated element for which $w_{h}$ and $\beta_{h}$ are linear and $\gamma_{h}$ is constant over each element. Figure 4.26(b) shows the results obtained in the cantilever analysis and indicates the good predictive capability of this element. The ellipticity condition is again satisfied (see Exercise 4.61), and in addition we now need to investi- 

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| x | w_h/W | |---|---| | 16.4 | 16.4 | | 60.9 | 60.9 | | 124 | 124 | | 197 | 197 | Bernoulli beam theory solution = 200


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| x | βₕ/W | | ---- | ----- | | 13.1 | 13.1 | | 22.5 | 22.5 | | 28.1 | 28.1 | | 30.0 | 30.0 |


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| x | Gγh/W | | --- | --- | | 0 | 100 | | 1 | 100 | | 2 | 100 | | 3 | 100 | | 4 | 100 | | 5 | 100 | | 6 | 100 | | 7 | 100 | | 8 | 100 | | 9 | 100 | | 10 | 100 | | 11 | 100 | | 12 | 100 | | 13 | 100 | | 14 | 100 | | 15 | 100 | | 16 | 100 | | 17 | 100 | | 18 | 100 | | 19 | 100 | | 20 | 100 | | 21 | 100 | | 22 | 100 | | 23 | 100 | | 24 | 100 | | 25 | 100 | | 26 | 100 | | 27 | 100 | | 28 | 100 | | 29 | 100 | | 30 | 100 | | 31 | 100 | | 32 | 100 | | 33 | 100 | | 34 | 100 | | 35 | 100 | | 36 | 100 | | 37 | 100 | | 38 | 100 | | 39 | 100 | | 40 | 100 | | 41 | 100 | | 42 | 100 | | 43 | 100 | | 44 | 100 | | 45 | 100 | | 46 | 100 | | 47 | 100 | | 48 | 100 | | 49 | 100 | | 50 | 100 | | 51 | 100 | | 52 | 100 | | 53 | 100 | | 54 | 100 | | 55 | 100 | | 56 | 100 | | 57 | 100 | | 58 | 100 | | 59 | 100 | | 60 | 100 | | 61 | 100 | | 62 | 100 | | 63 | 100 | | 64 | 100 | | 65 | 100 | | 66 | 100 | | 67 | 100 | | 68 | 100 | | 69 | 100 | | 70 | 100 | | 71 | 100 | | 72 | 100 | | 73 | 100 | | 74 | 100 | | 75 | 100 | | 76 | 100 | | 77 | 100 | | 78 | 100 | | 79 | 100 | | 80 | 100 | | 81 | 100 | | 82 | 100 | | 83 | 100 | | 84 | 100 | | 85 | 100 | | 86 | 100 | | 87 | 100 | | 88 | 100 | | 89 | 100 | | 90 | 100 | | 91 | 100 | | 92 | 100 | | 93 | 100 | | 94 | 100 | | 95 | 100 | | 96 | 100 | | 97 | 100 | | 98 | 100 | | 99 | 100 | | 100 | 100 |

(b) Analysis with mixed interpolated element; $w_{h}$ and $\beta_{h}$ vary linearly over each element, and $\gamma_{h}$ is constant in each element Figure 4.26 (continued) gate whether the following inf-sup condition is satisfied: $$ \inf _ {\gamma_ {h} \in P _ {h} (D _ {h})} \sup _ {\mathbf {v} _ {h} \in V _ {h}} \frac {\int_ {\mathrm{Vol}} \gamma_ {h} \left[ \left(\partial w _ {h} / \partial x\right) - \beta_ {h} \right] d \mathrm{Vol}}{\left\| \gamma_ {h} \right\| \left\| \mathbf {v} _ {h} \right\|} \geq c > 0 \tag {4.219} $$ Now $K_{h}(0) \neq \{0\}$ , and we test for the inf-sup condition by considering a typical $\gamma_{h}$ (where $\gamma_{h}$ is thought of as a variable). Then with a typical $\gamma_{h}$ given, we choose $$ \hat {\mathbf {v}} _ {h} = \left[ \begin{array}{c} \hat {w} _ {h} \\ \hat {\beta} _ {h} \end{array} \right] \tag {4.220} $$ with $\hat{\beta}_h = 0$ and $\partial \hat{w}_h / \partial x = \gamma_h$ . Now consider $$ \frac {\int_ {\mathrm{Vol}} \gamma_ {h} \left[ \left(\partial \hat {w} _ {h} / \partial x\right) - \hat {\beta} _ {h} \right] d \mathrm{Vol}}{\| \hat {\mathbf {v}} _ {h} \|} = \sqrt {\int_ {\mathrm{Vol}} \left(\gamma_ {h}\right) ^ {2} d \mathrm{Vol}} \tag {4.221} $$