Problem #1
UCI. MAE200B, winter 2006. by nasser Abbasi.
Show from Rodrigues formula that the legendre polynomials
are orthogonal with unit weight function over
and
Solution
Rodrigues formula is
We need to show (for the first part of the problem) that
We start the proof by assuming that
(if
then we simply switch the indices, and if
then this will be shown in part 2 of the problem).
We know
is a polynomial in
.
This can be seen from Rodrigues formula. This poynomial starts with a term
which is either a constant, or a constant times
depending if
is odd or even as can be seen from these few examples
Hence, let us first assume that
is odd. Then we can write
and if
was even, then we write
Now instead of showing the solution for
is odd and
is even, we show the solution for
is even only since the solution in this case will encompase the case for
is odd as will be explained more below.
Write LHS of equation (2) as
The strong form of the proof requires then that we show that each term
(integral) above is zero. i.e. we need to show
that
where in the above
can be zero (this covers the case for
is odd)
Now since
,
then the above means we need to show that
equation (3) above means that
(the above relation will be used in the proof for part 2).
Now to proof equation (3), use integration by parts.
Differntiate the
term and integrate the
terms. Let
,
.
Hence
and
But
However, now we see that
since the expression (after repeated differentiation) will come out to be a
polynomial in
,
and when we substitute for
or
we will get zero as the value of each term, hence
Hence the result of integration by parts is as follows
Now we repeate the process. We perform integration by part on the integral in
the RHS above. This will result in
And again, we perform integration by parts. We see that each time the exponent
of
becomes one less than before, and the order of differentiation becomes one
less than before. So it is a race between
and
to reach the value zero. However we started by assuming that
,
hence by repeated integration by parts we will eventually reach the following
form
where
But we have shown above in equation 4 that
hence this means that
which is what we wanted to proof. This implies that
QED.
Now we start part 2, which is to show that when
then
Start by using the recurive formula for Legendre polynomials. For clarity I
will write
instead of
.
The recurive formula to use is
Multiply eq (6) by
we obtain
Integrate both sides from
But
is a polynomial of degree
, as can be seen from expansion of Rodrigues formula
, hence the second integral shown above on the LHS side is zero, since it is
as if we have
which we showed in part 1 that this is zero. Hence eq (7) simplifies to
Now use integration by parts on the integral on the LHS.
But since
hence
and
,
hence the above becomes
Now substitute (9) into LHS of (8)
Which is what we are asked to show.