Problem #1

UCI. MAE200B, winter 2006. by nasser Abbasi.

Show from Rodrigues formula that the legendre polynomials MATH are orthogonal with unit weight function over $[-1,1]$ and MATH

Solution

Rodrigues formula is MATH

We need to show (for the first part of the problem) that MATH

We start the proof by assuming that $n<m$ (if $n>m$ then we simply switch the indices, and if $n=m$ then this will be shown in part 2 of the problem).

We know MATH is a polynomial in $x$. This can be seen from Rodrigues formula. This poynomial starts with a term which is either a constant, or a constant times $x$ depending if $n$ is odd or even as can be seen from these few examples


ivssa500.png

Hence, let us first assume that $n$ is odd. Then we can write MATH and if $n$ was even, then we write MATH

Now instead of showing the solution for $n$ is odd and $n$ is even, we show the solution for $n$ is even only since the solution in this case will encompase the case for $n$ is odd as will be explained more below.

Write LHS of equation (2) as

MATH

The strong form of the proof requires then that we show that each term (integral) above is zero. i.e. we need to show thatMATH

where in the above $n$ can be zero (this covers the case for $n$ is odd)

Now since $a_{n}\neq0\,$, then the above means we need to show that

MATH

equation (3) above means that MATH

(the above relation will be used in the proof for part 2).

Now to proof equation (3), use integration by parts.




Differntiate the $x^{n}$ term and integrate the $P_{m}$ terms. Let $u=x^{n}$, MATH. Hence $du=nx^{n-1}$ and MATH

But MATH

However, now we see that MATH since the expression (after repeated differentiation) will come out to be a polynomial in MATH, and when we substitute for $x=1$ or $x=-1$ we will get zero as the value of each term, hence MATH

Hence the result of integration by parts is as follows

MATH

Now we repeate the process. We perform integration by part on the integral in the RHS above. This will result in MATH

And again, we perform integration by parts. We see that each time the exponent of $x$ becomes one less than before, and the order of differentiation becomes one less than before. So it is a race between $n$ and $m$ to reach the value zero. However we started by assuming that $n<m$, hence by repeated integration by parts we will eventually reach the following form

MATH

where $r=m-n$

But we have shown above in equation 4 that MATH hence this means that MATH

which is what we wanted to proof. This implies that MATH QED.

Now we start part 2, which is to show that when $n=m$ then MATH

Start by using the recurive formula for Legendre polynomials. For clarity I will write $P_{n}$ instead of MATH.

The recurive formula to use is MATH

Multiply eq (6) by $P_{n}$ we obtain

MATH

Integrate both sides from $-1,1$

MATH

But $P_{n-1}^{\prime}$ is a polynomial of degree $<\,n$ , as can be seen from expansion of Rodrigues formula MATH , hence the second integral shown above on the LHS side is zero, since it is as if we have MATH which we showed in part 1 that this is zero. Hence eq (7) simplifies to

MATH

Now use integration by parts on the integral on the LHS.

MATH

But since MATH hence MATH and MATH, hence the above becomes MATH

Now substitute (9) into LHS of (8)

MATH

Which is what we are asked to show.