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then follows from the theorem below.
8.2. PARITY 161

Table 8.1: Symmetry classes for trigonometric basis functions.
“Even” parity with respect to x = π/2 means that the functions are symmetric with respect
to that point, that is,
←’ Even parity about x =
f (’x + π) = f (x)
or equivalently, f (π/2 ’ y) = f (π/2 + y) for the shifted variable y = x ’ π/2.
Similarly, antisymmetry with respect to π/2 implies that
f (’x + π) = ’f (x) ←’ Odd parity about x =

Parity with Parity with
respect to respect to
Trig. Functions x=0 x=
Even Even
Even Odd
cos([2n + 1]x)
Odd Odd
Odd Even
sin([2n + 1]x)

These four symmetry classes are illustrated in Fig. 8.1. The dotted lines on each graph
denote the symmetry planes at the origin and at x = π/2.


(i) All EVEN powers of x are SYMMETRIC about the origin:

{1, x2 , x4 , x6 , . . . } are of EVEN PARITY

(ii) All ODD powers of x are ANTISYMMETRIC:

{x, x3 , x5 , x7 , . . . } are of ODD PARITY

PROOF: Replace xn by (’x)n and see what happens.
Although trivial to prove, this theorem justi¬es both Theorem 22 and the following.

A function of EVEN parity, i. e. f (x) = f (’x) for all x, has a power series expansion contain-
ing only EVEN powers of x. A function of ODD parity, that is, one such that f (x) = ’f (’x),
can be expanded in a power series that contains only ODD powers of x.

PROOF: Odd powers of x change sign under the replacement x ’ (’x), so it is impos-
sible for a function to be symmetric about the origin unless its power series contains only
even powers of x. (Note that the powers of x are all linearly independent; there is no way

Figure 8.1: Schematic of the four symmetry classes of the terms of a general Fourier series
along with the simplest member of each class. All Fourier functions can be classi¬ed ac-
cording to their symmetry with respect to (i) x=0 and (ii) x = π/2. These symmetry points
are marked by the dashed vertical lines.

that possible cancellations among various odd powers can keep them from violating the
symmetry condition for all x.). Similarly, even powers of x would wreck the condition of
antisymmetry except perhaps at a few individual points.

Differentiating a function f (x) which is of de¬nite parity REVERSES the parity if the differ-
entiation is performed an ODD number of times and leaves the parity UNCHANGED if f (x) is
differentiated an EVEN number of times.

PROOF: A function of even parity has a power series that contains only even powers of
x by Theorem 24. Differentiating x2n gives (2n) x2n’1 which is an odd power of x for any
n. Similarly, differentiating x raised to an odd number gives x raised to an even power.
Differentiating twice, however, merely restores the original parity.
These four theorems are elementary, but useful. One way of determining the parity
of a function with respect to x = 0 is to calculate its power series expansion about that
point. One can show rigorously that a function f (x) must retain its parity, if any, outside
the radius of convergence of the power series.
8.2. PARITY 163

Figure 8.2: Schematic illustrating the decomposition of an arbitrary function into its sym-
metric [S(x)] and antisymmetric [A(x)] parts.

It is usually possible to determine if the solution to a differential equation has parity
without ¬rst solving it by inspecting the coef¬cients of the equation and the boundary con-
ditions. We need a few preliminary results ¬rst.

An arbitrary function f (x) can be decomposed into the sum of two functions of de¬nite parity,

f (x) = S(x) + A(x)

where S(x) is symmetric about the origin [S(x) = S(’x) for all x] and A(x) is antisymmetric
[A(x) = ’A(’x)]:

f (x) + f (’x)
S(x) ≡ [Symmetric] (8.4)
f (x) ’ f (’x)
A(x) ≡ [Antisymmetric] (8.5)
PROOF: It is trivial to verify that the sum of S(x) and A(x) correctly adds up to f (x)
since the terms in f (’x) cancel. It is just as simple to verify that S(x) is symmetric since
the subsitution x ’ (’x) gives us back the same function except that f (x) and f (’x) have
swapped places with respect to the plus sign.
Fig. 8.2 gives √ graphical proof of the theorem for the unsymmetric function f (x) =
(cos(x) + sin(x))/ 2.

Consider an ordinary differential equation in the form
dn u
an (x) n = f (x)

Then u(x) is a symmetric function if and only if the boundary conditions are compatible with sym-
metry and if also either

(i) every even coef¬cient in (8.6) is even and the coef¬cient of every odd derivative has
odd parity and also f (x) is symmetric or
(ii) every even coef¬cient in (8.6) is odd and the coef¬cient of every odd derivative has
even parity and also f (x) is antisymmetric.

Similarly, u(x) is an antisymmetric function if and only if the boundary conditions change sign
under the replacement of x by (’x) and also if either

(iii) every coef¬cient of an even derivative in (8.6) is odd and the coef¬cient of every odd
derivative has even parity and also f (x) is symmetric or if
(iv) every even coef¬cient in (8.6) is even and the coef¬cient of every odd derivative has
odd parity and also f (x) is antisymmetric.

PROOF: We will illustrate the argument with a second order equation, but the theorem
is true for general N . The ODE is

a2 (x) uxx + a1 (x) ux + a0 (x) u = f (x)

Split all the functions into their symmetric and antisymmetric parts as

u(x) = S(x) + A(x); f (x) = s(x) + a(x)

a2 (x) = S2 + A2 ; a1 (x) = S1 + A1 ; a0 (x) = S0 + A0

De¬ne u(x) ≡ u(’x), which need not have any de¬nite parity; this solves

a2 (’x) uxx ’ a1 (’x) ux + a0 (’x) u = f (’x) (8.10)
˜ ˜ ˜

If we add (8.10) to (8.7) and divide by 2, similar to the way the symmetric function S(x) is
created from an arbitrary f (x) in Theorem 26, then we obtain an equation in which all the
terms are symmetric. Similarly, subtracting (8.10) from (8.7) gives an equation in which all
terms are antisymmetric. These two coupled equations are

S2 Sxx + A2 Axx + S1 Ax + A1 Sx + S0 S + A0 A = s(x) [Symm. Eqn.]
S2 Axx + A2 Sxx + S1 Sx + A1 Ax + S0 A + A0 S = a(x) [Antisymm.]

We can solve this coupled pair of equations in no more operations than for the original
single differential equation (8.6) ” but no fewer unless the coef¬cients satisfy the symme-
try conditions of the theorem. If we assume that u(x) is purely symmetric so that A(x) ≡ 0,

then (8.11) implies that S(x) must simultaneously satisfy two equations. This is an impos-
sible task for a single function unless one of the pair of equations degenerates into 0 = 0.
With A(x) ≡ 0, (8.11) is

S2 (x) Sxx + A1 (x) Sx + S0 (x) S = s(x)
A2 (x) Sxx + S1 (x) Sx + A0 (x) S = a(x)

There are only two possibilities. One is that {S2 , A1 , S0 } are non-zero while {A2 , S1 , A0 ,
a(x)} are all 0. In words, this means that the coef¬cients of the second derivative and the
undifferentiated term are symmetric about x = 0 while the coef¬cient of the ¬rst derivative
is antisymmetric; the forcing function f (x) must also be symmetric or 0. This is (i) of the
theorem. The alternative is that all the coef¬cients and the forcing function in (8.12) are
zero, and this is (ii) of the theorem. The proof for antisymmetric u(x) is similar. Q. E. D.
Thus, we can predict in advance whether or not a linear differential equation with a
particular set of boundary or initial conditions will have solutions with de¬nite parity. The
same line of reasoning can be extended to nonlinear ordinary differential equations and to
partial differential equations, too.

8.3 Modifying the Grid to Exploit Parity
When the basis set is halved because the solution has de¬nite parity, the grid must be
modi¬ed, too, as illustrated in Fig. 8.3
For example, if we use a “half-basis” of cosines only, applying collocation conditions
on x ∈ [’π, π] is disastrous. The reason is that because all the included basis functions
have de¬nite parity and so, by assumption, does the residual, the collocation conditions at
x = ’jh where h is the grid spacing are identical to the collocation conditions at x = jh
(except for a sign change if the residual is antisymmetric). It follows that the pseudospectral
matrix will be singular because it has only N/2 linearly independent rows; each row has
the same elements (or the negative of the elements) of one of the other rows in the square
This disaster can be avoided by restricting the collocation points to half of the original
interval as shown by middle panel in Fig. 8.3. Similarly, if u(x) has double parity, one must
restrict the collocation points to x ∈ [0, π/2], one quarter of the spatial period, in order to
avoid redundant collocation conditions.

8.4 Other Discrete Symmetries
Parity is by far the most useful of symmetries because it is the simplest. More exotic exam-
ples are possible, however.
In the language of group theory, symmetry with respect to the origin means that f (x)
is “invariant under the actions of the group C2 in the complex plane”. This is a highbrow
way of saying that we can rotate a symmetric function f (x) through 180 degrees in the
complex x-plane, which is equivalent to replacing x by ’x, without changing anything.
Functions may also be invariant under Cn , the group of rotations through any multiple
of 360 degrees/n. For example, if a function is invariant under C4 , then its power series
expansion is of the form

a4n x4n [invariant under C4 ] (8.14)
f (x) =

No symmetry
Full Basis: All sines & cosines

-π 0

Half Basis: Parity
cosines ONLY
OR sines ONLY

Quarter Basis:
ODD cosines ONLY
OR EVEN cosines ONLY
OR EVEN sines ONLY 0


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