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2. Two-Point Boundary Value Problems
58
which we recognize as the finite difference approximation of the second
derivative. Notice that this definition is valid both for discrete and contin-
uous functions.
Now we can formulate the discrete problem (2.13) as follows: Find a
discrete function v ∈ D h,0 such that
(L h v)(x j )= f(x j )
(2.27)
for all j =1,... ,n.
In this formulation, we take care of the boundary conditions in the re-
quirement that v ∈ D h,0 . This is exactly how we did it in the continuous
case.
Some of the properties of the two operators L and L h that we shall
derive are connected to the inner product of functions. These inner products
are defined by integration for continuous functions and by summation for
discrete functions. For two continuous functions u and v, we define the
inner product of the functions by

1

u, v = u(x)v(x) dx. (2.28)
0
Similarly, for two discrete functions, i.e. for u and v in D h , we define the
inner product to be
n

u, v h = h( u 0 v 0 + u n+1 v n+1 + u j v j ), (2.29)
2
j=1
where we have used the shorthand notation v j for v(x j ). Clearly, (2.29) is
an approximation of (2.28). In the language of numerical integration, this is
referred to as the trapezoidal rule; you will find more about this in Exercise
2.20.
Having established a suitable notation for the continuous and the discrete
problem, we are in position to start deriving some properties.
2.3.2 Symmetry
The first property we will show is that both the operators L and L h are
symmetric. For matrices we are used to saying that a matrix A ∈ R n,n is
symmetric if the transpose of the matrix equals the matrix itself, i.e. if
T
A = A.
It turns out that this is equivalent to the requirement that
7
(Ax, y)=(x, Ay)

7 Note that (·, ·) denotes the usual Euclidean inner product of vectors in R ; see
n
Exercise 2.21 on page 79 or Project 1.2 on page 31.