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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 righthand inequality of (4.6.3). The triangular inequality gives
h3 < h1 + h2. Thus, at least one edge, say, h2 > h3 =2. This yields the lefthand
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 lefthand 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. ThreeDimensional 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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 Spring '14
 JosephE.Flaherty
 The Land

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