Unformatted text preview: j kW k0kV k0)
to obtain kE ( t)k kW ( t)k0k ( t)k0 + 2
0 Zt
0 kW k0 k k0d : (9.4.15b) N
Introduce a basis on S0 and write W in the standard form W (x y ) = N
X
j =0 cj ( ) j (x y): (9.4.16) Substituting (9.4.16) into (9.4.13a) and following the steps introduced in Section 9.2, we
are led to
_
;Mc + Kc = 0 (9.4.17a) where Mij = (
Kij = A( i j i j ) (9.4.17b) i j = 1 2 : : : N: ) (9.4.17c) Assuming that the sti ness matrix K is independent of , (9.4.17a) may be solved exactly
to show that (cf. Lemmas 9.4.1 and 9.4.2 which follow) kW ( Zt
0 )k0 kE ( kW k0 d t)k0 0< t (9.4.18a) C (1 + j log ht2 j)kE ( t)k0: (9.4.18b) Equation (9.4.18a) is used in conjunction with (9.4.15b) to obtain kE ( t)k 2
0 (kE ( t)k0 + Zt
0 kW k0d ) maxt] k (
2(0 )k0: Now, using (9.4.18b) kE ( t)k0 C (1 + j log ht2 j) maxt] k (
2(0
Writing
and taking an L2 norm ~~
u;U =u;U +U ;U = ;E ku ; U k0 k k0 + kE k0: )k0: (9.4.19) 26 Parabolic Problems Using (9.4.19) ku ; U k0 C (1 + j log ht2 j) maxt] k (
2(0 )k0: (9.4.20a) Finally, since satis es the elliptic problem (9.4.13c), we can use Theorem 7.2.4 to
write k( )k0 C h2ku( )k2: (9.4.20b) Combining (9.4.20a) and (9.4.20b) yields the desired result (9.4.12).
The two results that were used without proof within Theorem 9.4.2 are stated as
Lemmas.
Lemma 9.4.1. Under the conditions of Theorem 9.4.2, there exists a constant C > 0
such that C
A(V V ) h2 kV k2
0 8V 2 S0N : (9.4.21) Proof. The result can be inferred from Example 9.2.1 however, a more formal proof is
given by Johnson 9], Chapter 7. Instead of establishing (9.4.18b), we'll examine a slightly more general situation. Let
c be the solution of
_
Mc + Kc = 0
t>0
c(0) = c0:
(9.4.22)
The mass and sti ness matrices M and K are positive de nite, so we can diagonalize
(9.4.22). In particular, let be a diagonal matrix containing the eigenvalues of M;1K
and R be a matrix whose columns are the eigenvectors of the same matrix, i.e.,
M;1KR = R :
(9.4.23a)
Further let d(t) = R;1c(t):
Then (9.4.22) can be written in the diagonal form
_
d+ d=0 (9.4.23b)
(9.4.24a) by multiplying it by (MR);1 and using (9.4.23a,b). The initial conditions generally
remain coupled through (9.4.23a,b), i.e., d(0) = d0 = R;1c0 :
With these preliminaries, we state the desired result. (9.4.24b) 9.5. ConvectionDi usion Systems 27 Lemma 9.4.2. If d(t) is the solution of (9.4.24) then
0
_
jdj + j dj C jtd j
t>0
p
where jdj = dT d. If, in addition,
max j j C2
h
6=0 j j
then ZT
0 _
(jdj + j dj)dt C (1 + j log T2 j)jd0j: h Proof. cf. Problem 1. (9.4.25a)
(9.4.25b)
(9.4.25c) Problems
1. Prove Lemma 9.4.2. 9.5 ConvectionDi usion Systems
Problems involving convection and di usion arise in uid ow and heat transfer. Let us
consider the model problem ut + ! ru = r (pru) (9.5.1a) where ! = !1 !2]T is a velocity vector. Written is scalar form, (9.5.1a) is ut + !1 ux + !2uy = (pux)x + (puy )y : (9.5.1b) The vorticity transport equation of uid mechanics has the form of (9.5.1). In this case,
u would represent the vorticity of a twodimensional ow.
If the magnitude of ! is small relative to the magnitude o...
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 Spring '14
 JosephE.Flaherty
 The Land, Tn, Boundary value problem, Numerical ordinary differential equations, nite element, parabolic problems

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