Nonconforming and Other Methods - Finite Elements: Theory, Fast Solvers, and Applications in Elasticity Theory - page 140

Proof. We consider three cases.
Case 1. Let u X + λ M ≤ δ − 1 t | λ | c , where δ> 0 will be chosen later. Then

A(u, λ ; u, − λ) = a(u, u) + t 2 c(λ, λ)

1 1

2 t 2

2
c

2 t 2

2
c

≥ | λ | + | λ |

1 1

2 δ 2

{ ( u X + λ M ) 2 + t 2

2
c

4 δ 2

2 .

≥ | λ | }≥ ||| (u, λ) |||

Dividing through by ||| (u, λ) ||| ,wehave

1 A(u, λ ; u, − λ)

||| (u, λ) |||

A(u, λ ; v, µ)
||| (v, µ) |||

4 δ 2

||| (u, λ) ||| ≤ ≤ sup

(v,µ)

.

β

u X + λ M >δ − 1 t | λ | c and

Case 2. Let u X ≤

a λ M . By the inf-sup

2

condition (4.8),

b(v, λ)
v X =

A(u, λ ; v, 0 ) − a(u, v)
v X

β λ M ≤ sup

v

sup
v

A(u, λ ; v, µ)
||| (v, µ) |||

≤ sup

(v,µ)

+ a u X

A(u, λ ; v, µ)
||| (v, µ) |||

1
2 β λ M .

≤ sup

(v,µ)

+

Now we can estimate λ M , and in view of the case distinction u X and t | λ | c as

well, by the first term on the right-hand side.
Case 3.

β

>δ − 1 t | λ | c

a λ M . Then
δ ||| (u, λ) |||≤ u X , where δ depends only on α, β and δ . By hypothesis (4.25),

Let u X + λ M and u X ≥

2

αδ ||| (u, λ) ||| ≤ α u X

0 ,µ) + t 2 c(λ, µ)
µ M + t | µ | c

a(u, u)
u X +

A(u, λ ;

≤ su µ

A(u, λ ; u, − λ)
u X

A(u, λ ; 0 ,µ)

||| ( 0 ,µ) |||

≤ + su µ + t | λ | c

1
δ +

1 sup
(v,µ)

A(u, λ v, µ)

||| (v, µ) |||

;

≤ + t | λ | c .

αβ 1

With δ ≤ we have t | λ | c ≤

2 α u X , and the second term in the sum can 4 a + 2 β

be absorbed by a factor of 2.
This establishes the assertion in all cases.