Amath-Math 586/Atm S 581 Spring 2005
Final Assignment Solution
1) Consider a FEM using an arbitrary monotone increasing set of nodes xj, j = 0,…, N, based on
piecewise linear chapeau functions ()
j
x
ϕ
, each of which is 1 at xj and 0 at all other nodes:
.
0
(,) () ()
N
jj
j
qxt q t x
ϕ
=
=∑
The node x0 = 0 is chosen to be the left boundary, so its expansion coefficient is
determined by the left BC to be = sin(50t). The node x
0()qt
0()qt N is chosen to be the right
boundary; its expansion coefficient is unknown. Let ∆xj+1/2 = xj+1 - xj. As in class, the FEM
equations are found by zeroing the projection of the residual onto each basis function that has
an unknown expansion coefficient:
00
0(),(,) ,1,,
NN
n
jjnjnn
nn
da
x
Rxt I J a j N
dt
ϕ
==
==+=
∑∑ ….
The required inner products are easily computed for the interior nodes:
1/2
1/2 1/ 2
1/2
1
1
(), () 2( ) , 1, , 1
61
j
jn j n j j
j
xnj
Ixx xxnjjN
xnj
ϕϕ
−
−+
+
∆=−
==∆+∆==
∆=+
…−
,
1/ 2 1
(), 0 , 1, , 1
1/2 1
n
jn j
nj
d
Jx njjN
dx nj
ϕ
ϕ
−=−
====
=+
…−
N
.
For the right boundary node, only the projections with the node to its left and the self-
projection between 1N
x
xx
−<< contribute, altering the inner products to:
1/2
1/2
,1
1
(), () 2,
6
N
Nn N n
N
xnN
Ixx
x
nN
ϕϕ
−
−
∆=
==
∆=
−
,
1/2, 1
(), 1/ 2,
n
Nn N
nN
d
Jx nN
dx
ϕ
ϕ
−=
==
=
−
.
Defining the solution vector q(t) = {qj(t), j = 1,…,N}, the tridiagonal inner product matrices
I, J = {Ijn, Jjn, j, n = 1,…, N), and the vectors i0, j0 = {I0n, J0n, j, n = 1,…, N), we can write the
FEM in matrix form as
10 0 10 0
/1
01
I
dq dt J q j
d+= j
dt
−−
>
q
IJq =
.
Using trapezoidal time differencing, we obtain the desired time-discretized FEM:
