Unformatted text preview: kin Problems 3 As usual, we apply the divergence theorem to the secondderivative terms in L to reduce
the continuity requirements on u. When L has the form of (9.2.1b), the Galerkin problem
1
consists of determining u 2 HE (t > 0) such that
(v ut) + A(v u) = (v f )+ < v ; u > 8v 2 H01 t > 0: (9.2.2a) The L2 inner product, strain energy, and boundary inner product are, as with elliptic
problems,
(v f ) = A(v u) = ZZ ZZ vfdxdy (9.2.2b) p(vxux + vy uy ) + vqu]dxdy (9.2.2c) and < v pun >= Z
@N vpunds: (9.2.2d) The natural boundary condition (9.2.1e) has been used to replace pun in the boundary
inner product. Except for the presence of the (v ut) term, the formulation appears to
the same as for an elliptic problem.
Initial conditions for (9.2.2a) are usually determined by projection of the initial data
(9.2.1c) either in L2
(v u) = (v u0) 8v 2 H01 t=0 (9.2.3a) or in strain energy A(v u) = A(v u0) 8v 2 H01 t = 0: (9.2.3b) Example 9.2.1. We analyze the onedimensional heat conduction problem ut = (pux)x + f (x t) 0<x<1 t>0 u(x 0) = u0(x)
0x1
u(0 t) = u(1 t) = 0
t>0
thoroughly in the spirit that we did in Chapter 1 for a twopoint boundary value problem.
A Galerkin form of this heatconduction problem consists of determining u 2 H01
satisfying
t>0
(v ut) + A(v u) = (v f )
8v 2 H01 4 Parabolic Problems
U(x,t) cj c1 cN1 x
0 = x0 x1 xj xN1 xN = 1 Figure 9.2.1: Mesh for the nite element solution of Example 9.2.1.
(v u) = (v u0)
where A(v u) = Z 8v 2 H01
1
0 t=0 vxpuxdx: Boundary terms of the form (9.2.2d) disappear because v = 0 at x = 0 1 with Dirichlet
data.
We introduce a mesh on 0 x 1 as shown in Figure 9.2.1 and choose an approxiN
mation U of u in a nitedimensional subspace S0 of H01 having the form U (x t) = X N ;1
j =1 cj (t) j (x): Unlike steady problems, the coe cients cj , j = 1 2 : : : N ;1, depend on t. The Galerkin
N
nite element problem is to determine U 2 S0 such that
( j Ut ) + A( j U ) = ( j f ) t>0 ( j U ) = ( j u0 )
t=0
j = 1 2 : : : N ; 1:
Let us chose a piecewiselinear polynomial basis 8 x;x ;
> x ;x ;
<;
(x) = > xx ;xx
k
:0 k1
k
k1
k+1
k+1
k if xk;1 < x xk
if xk < x xk+1 :
otherwise This problem is very similar to the onedimensional elliptic problem considered in Section
1.3, so we'll skip several steps and also construct the discrete equations by vertices rather
than by elements. 9.2. SemiDiscrete Galerkin Problems
Since j 5 has support on the two elements containing node j we have A( j U ) = Zx j 0 pU
j xj ;1 x dx + where ( )0 = d( )=dx. Substituting for j and Ux
Z xj 1 cj ; cj;1
Z
A( U ) =
p(x)(
)dx +
j xj ;1 hj hj where Zx j +1 0 pU
j xj xj +1 x dx ; h 1 p(x)( cj+1 ; cj )dx
h
j +1 xj j +1 hj = xj ; xj;1: Using the midpoint rule to evaluate the integrals, we have j
;
A( j U ) pjh1=2 (cj ; cj;1) ; ph+1=2 (cj+1 ; cj )
j
j +1 where pj;1=2 = p(xj;1=2 ).
Similarly,
(
or ( j Ut ) = Zx j xj ;1 j Ut ) = Zx j xj ;1 U dx + jt c
_
j (_j ;1 j ;1 + cj j )dx + Zx j +1 U dx jt xj Zx j +1 j xj (_j j + cj+1
c
_ j +1 )dx where (_) = d( )=dt. Since the integrands are quadratic functions of x they may be
integrated exactly using...
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Full Document
 Spring '14
 JosephE.Flaherty
 The Land, Tn, Boundary value problem, Numerical ordinary differential equations, nite element, parabolic problems

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