61 we nd the determinant of this jacobian as detje x2

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Unformatted text preview: ngles as 1 2 3 and the lengths of the edges opposite these angles as h1, h2, and h3 (Figure 4.6.1). With 1 = e being the smallest angle of Element e, write the determinant of the Jacobian as det(Je) = h2h3 sin e: Using the law of sines we have h1 h2 h3 = he. Replacing h2 by h3 in the above expression yields the right-hand inequality of (4.6.3). The triangular inequality gives h3 < h1 + h2. Thus, at least one edge, say, h2 > h3 =2. This yields the left-hand inequality of (4.6.3). Theorem 4.6.1. Let (x y) 2 H s( e) and ~( ) 2 H s( 0 ) be such that (x y) = ~( ) where e is the domain of element e and 0 is the domain of the canonical element. Under the linear transformation (4.6.1), there exist constants cs and Cs, independent of , ~, he , and e such that cs sins;1=2 ehs;1j js e j ~js 0 Cs sin;1=2 ehs;1j js e (4.6.4a) e e where the Sobolev seminorm is j j2 s e = X ZZ j j=s (D )2 dxdy e with D u being a partial derivative of order j j = s (cf. Section 3.2). Proof. Let us begin with s = 0, where ZZ 2 dxdy = det(J or e e) ZZ ~2 d d 0 j j2 e = det(Je )j ~j2 0 : 0 0 Dividing by det(Je) and using (4.6.3) j j2 e 2j j2 e 0 0: j ~j2 0 0 sin eh2 sin eh2 e e (4.6.4b) 32 Finite Element Approximation p Taking a square root, we see that (4.6.4a) is satis ed with c0 = 1 and C0 = 2. With s = 1, we use the chain rule to get x Then, j j2 1 e = ZZ ( = ~ x+~ x y 2 + 2 )dxdy = det(J x e) y ZZ = ~ y + ~ y: (g1 e ~2 + 2g2 e ~ ~ + g3 e ~2 )d d 0 e where 2 2 2 2 g1 e = x + y g2 e = x x + y y g3 e = x + y : Applying the inequality ab (a2 + b2 )=2 to the center term on the right yields j j2 e 1 det(Je) Letting ZZ g1 e ~2 + g2 e( ~2 + ~2 ) + g3 e ~2 ]d d : 0 = max(jg1 e + g2 ej jg3 e + g2 ej) and using (4.6.4b), we have j j2 1 e det(Je) j ~j2 0 : 1 (4.6.5a) Either by using the chain rule above with = x and y or by inverting the mapping (4.6.1), we may show that y x y x: y=; x=; y=; x= det(Je) det(Je) det(Je) det(Je) From (4.6.2), jx j, jx j, jy j, jy j he thus, using (4.6.3), we have j xj, j y j, j xj, j y j 2=(he sin e). Hence, 16 : (h sin )2 e e Using this result and (4.6.3) with (4.6.5a), we nd 16 j ~j2 : j j2 e (4.6.5b) 1 sin e 1 0 Hence, the left-hand inequality of (4.6.4a) is established with c1 = 1=4. To establish the right inequality, we invert the transformation and proceed from 0 to e to obtain ~2 ~j2 0 j j1 e j1 (4.6.6a) det(Je) 4.4. Three-Dimensional Shape Functions with 33 ~ = max(jg1 e + g2 ej jg3 e + g2 ej) ~ ~~ ~ g1 e = x2 + x2 ~ g2 e = x y + x y ~ g3 e = y2 + y2: ~ We've indicated that jx j, jx j, jy j, jy j he. Thus, ~ 4h2 and, using (4.6.3), we nd e 8 j j2 : j ~j2 0 (4.6.6b) 1 sin e 1 e p Thus, the right inequality of (4.6.4b) is established with C1 = 2 2. The remainder of the proof follows the same lines and is described in Axelsson and Barker 2]. With Theorem 4.6.1 established, we can concentrate on estimating interpolation errors on the canonical triangle. For simplicity, we'll use the Lagrange interpolating polynomial n X ~ U ( ) = u( j j )Nj ( ) ~ (4.6.7) j =1 with n...
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