DISCUSSION
5. Further studies
that are not pictured in Figure A-2) are not neighbors of
p
0 and hence those coecients must vanish.To compute
c
1, note that (1;0) =A
[G
. Over the regionA
, the gradients arer
1 = 1h
(1;
1);
r0 = 1h
(0;
;1);
(A.32) and over the regionG
, they arer
1 = 1h
(0;
1);
r0 = 1h
(;1;
;1):
(A.33) The area of each element is 12h
2, so the coecientc
1 is simply -2. Similarly,c
3=c
5=c
7=;2 and
c
2 =c
6 = 0. Finally,c
0 = 1+1+1+1+2+2 = 8, so thep
0 equation of the systemA x
=b
is 8c
0;2c
1;2c
3;2c
5;2c
7 = 0, which upon rearrangement yieldsc
0=c
1+c
3+c
5+c
74
:
(A.34)Equation (A.34) is simply the standard nite dierence approximation for Laplace's equa-tion, and similar computations yield the same equations for the case when
p
0 is on the boundary@
.over a complete basis, the problem reduces to the determination of the unknown coecients.
Using completeness, this is equivalent to
Z
(
D
12u
+D
22u
)i = 0;i
= 1;
2;
3;::::
(A.38) Expandingu
in its innite series, the above equation becomes1
X
i=1
a
iZ
(
D
21i+D
22i)j = 0;j
= 1;
2;
3;::::
(A.39) Galerkin's method generalizes this procedure to the case when the basis is nite, and there-fore not complete.More specically, let f
i;i
= 1;
2;:::;M
g and fi;i
= 1;
2;:::;N
g now denote the nite element basis functions considered in xA.2.3, where, as before,M
nodes lie in the interior of andN
nodes lie on the boundary. Since the nite element basis is nite, it cannot be a complete basis for the solution space (which is generally innite-dimensional). However, by analogy with Equation (A.39), one can still require that the residual be orthogonal to the basis functions, producingM
X
i=1
a
iZ
(
D
12i+D
22i)j =;XNi=1
b
iZ
(
D
21i+D
22i)j;
(A.40) withj
= 1;
2;:::;M
. Integrating by parts and noting that each basis functions vanishes outside a bounded region, the equations become;
M
X
i=1
a
iZ
(
D
1iD
1j+D
2iD
2j) =XNi=1
b
iZ
(
D
1iD
1j+D
2iD
2j);
(A.41) again forj
= 1;
2;:::;M
. These equations are identical, up to a sign, to (A.28).Let
u
denote the approximate solution given by Galerkin's method. Galerkin's method only requires that the residualD
12u
+D
22u
lies in the orthogonal complement of the span of the basis. Thus, without some other criterion to justify the equations, Galerkin's method does not actually guarantee that the approximate solution satises the dierential equation in any sense. Notice the resemblance between the orthogonality condition and least-squares approximations: Recall that ifis a function to be approximated, andu
is linear combina-tion of basis funccombina-tionsi, then the orthogonality conditionh;u;
ii= 0 indeed produces the least-squares approximation. But in this case, the exact solutionis not available to us.Thus, Galerkin's method does not actually produce the least squares approximation. Or-thogonalizing the residual does not minimize it. Indeed, since the basis is not complete, the error residual can be arbitrarily large while still being orthogonal to all the basis functions.
Integration of Dierential Forms on Manifolds
This appendix briey sketches the construction of dierential forms, which are mathematical objects that can be integrated on oriented manifolds. While they are of less importance in the theory of dierential equations on manifolds, they are very much essential in the study of dierential topology. However, those applications would take us too far aeld and will not be discussed here. For more information, please see either Guillemin and Pollack [14]
or Warner [28].
Recall the change of variables theorem (3.19):
Z
y2V2
f
(y
)dy
=Zx2V1
f
((x
))jdetD
(x
)jdx
(B.1) wheref
is a function onB
and :A
!B
is a smooth bijection. In x3.4.1, this the-orem is used to dene integrals of scalar-valued functions on compact Riemannian man-ifolds. Another possible approach, which we will briey sketch here, involves assigning\determinant-like" functions to each tangent space of the manifold. Such an assignment is called a \dierential form."
Let
V
be a nite-dimensional vector space. A scalar-valued functionT
onV
:::
V
is multilinear if it is linear in each of its components, and is alternating if exchanging any two arguments changesT
to ;T
. The degree ofT
is the number of argumentsT
has. It is a theorem of linear algebra that such functions, called alternating tensors, are always proportional to the determinant function onV
with respect to some basis.Now, let
!
be a function that assigns to each pointp
2M
an alternating tensor!
p onT
pM
. One can show (although it is not done here) that the usual notions of smoothness also apply to these alternating tensor elds. A dierential form on a manifoldM
is then a smooth alternating tensor eld onM
.Because alternating tensors are proportional to the determinant function, it is not very dicult to show that one could obtain a consistent denition of integration for dierential forms. To do this, rst choose a partition of unity so that the problem is reduced to a local one. Then, note that tensors transform naturally in the following way:
T
0(v
1;:::;v
k) =T
(Lv
1;:::;Lv
k);
(B.2) whereT
0 is a tensor on some vector spaceW
,T
is a tensor onV
,L
:V
!W
is a linear transformation, andk
is the degree ofT
0. Generalizing this to dierential forms onmanifolds, we can simply replace
L
by the dierentiald
of the transition map. A little bit of linear algebra shows that this almost gives us the change of variables theorem:Z
y2V2
f
(y
)dy
=Zx2V1
f
((x
))detD
(x
)dx;
(B.3) wheref
det is a (local) dierential form onV
2 and (f
)det is a dierential form onV
1. (Note that both their degrees have to agree with the dimension of the space,n
, because of the dimensions ofD
as a matrix.) As one can see, this is the change of variables theorem except for the absolute value. Thus, if one could choose charts so that all the transition maps have positive determinants:det
D
(x
)>
0;
(B.4)then we can dene integrals consistently. Manifolds for which such atlases exist are called orientable manifolds, and we can thus dene integration of dierential forms of degree
n
on compact orientablen
-manifolds.B.1 Stokes's theorem
One of the most important things one can do with dierential forms is to generalize Stokes's theorem to compact orientable manifolds. This is done through a map called the exterior derivative, which takes a dierential form
!
of degreek
to another dierential formd!
of degreek
+1. Deningd
takes a little bit of work and will not be done here. But to show how much Stokes's theorem is simplied, here is the statement of the theorem using dierential forms:Z
M
d!
=Z@M!:
(B.5)This is actually so abstract that it does not say much, unless one has studied dierential forms in some depth. However, note that the boundary operator on manifolds satises:
@
(M
N
) = (@M
N
)[(M
@N
);
(B.6) just like the product rule. As there is a corresponding product rule for exterior derivatives, this shows that there is a rather deep duality between geometric objects on the one hand and algebraic structures (such as dierential forms) on the other.Bibliography
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