MHB Problem of the Week #96 - January 27th, 2014

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: For $n\geq 0$, let $P_n(x)$ denote the Legendre polynomial. Show that
\[\int_{-1}^1 P_m(x)P_n(x)\,dx = \begin{cases}0 & m\neq n\\ \dfrac{2}{2n+1} & m=n\end{cases}\]

-----

Hints for the $m\neq n$ case: [sp]Start with Legendre's differential equation \[(1-x^2)y^{\prime\prime} -2xy^{\prime}+\lambda(\lambda+1)y = 0\]
and rewrite it in the form
\[[(1-x^2)y^{\prime}]^{\prime} = -\lambda(\lambda+1)y.\tag{1}\] Then use the fact that $P_m(x)$ and $P_n(x)$ are solutions to this equation with $\lambda=m$ and $\lambda=n$ respectively, and substitute them into $(1)$. From there, find a way to combine these two equations in order to make $P_m(x)P_n(x)$ appear and then integrate to get the result.[/sp]Hints for the $m=n$ case: [sp]First find a way to combine the two recurrence relations (which follow from Bonnet's recurrence relation)
\[(n+1)P_{n+1}(x)+nP_{n-1}(x) = (2n+1)xP_n(x)\]
and
\[nP_n(x) + (n-1)P_{n-2}(x) = (2n-1)xP_{n-1}(x)\]
and then use the $m\neq n$ integral result to come up with a recurrence relation involving $\displaystyle\int_{-1}^1 P_n^2(x)\,dx$ and $\displaystyle\int_{-1}^1 P_{n-1}^2(x)\,dx$. Finally, use the fact that $P_0(x)=1$ and induction to show that $\displaystyle\int_{-1}^1 P_n^2(x)\,dx = \frac{2}{2n+1}$[/sp]

 

[Chris L T521]

This week's problem was correctly answered by MarkFL and chisigma (I have no idea why I didn't see that the first time). You can find Mark's solution below.

[sp]The first case - $m\ne n$:

Let's begin with Legendre's differential equation for $P_m(x)$ in the form:

$$\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]+m(m+1)P_m(x)=0$$

Multiplying through by $$P_n(x)$$ and integrating with respect to $x$ using the given limits $-1$ and $1$, we obtain:

(1) $$\int_{-1}^{1} P_n(x)\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx+m(m+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0$$

The first integral may be integrated using integration by parts, where:

$$u=P_n(x)\,\therefore\,du=\frac{d}{dx}P_n(x)\,dx$$

$$dv=\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx\,\therefore\,v=\left(1-x^2 \right)\frac{d}{dx}P_m(x)$$

And so we may state:

$$\int_{-1}^{1} P_n(x)\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx=\left[P_n(x)\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]_{-1}^{1}-\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx$$

We see that the first term on the right evaluates to zero, hence we are left with:

$$\int_{-1}^{1} P_n(x)\frac{d}{dx}\left[\left(1-x^2 \right)\frac{d}{dx}P_m(x) \right]\,dx=-\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx$$

Thus (1) becomes:

(2) $$-\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx+m(m+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0$$

If we had begun with the Legendre differential equation for $$P_n(x)$$ instead, and multiplied through by $P_m(x)$, and followed the sames steps as above, we would of course obtain:

(3) $$-\int_{-1}^{1} \left(1-x^2 \right)\frac{d}{dx}P_m(x)\frac{d}{dx}P_n(x)\,dx+n(n+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0$$

Subtracting (3) from (2), there results:

$$\left(m(m+1)-n(n+1) \right)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0$$

Now, if we observe that:

$$m(m+1)-n(n+1)=m^2+m-n^2-n=m^2+mn+m-mn-n^2-n=(m-n)(m+n+1)$$

we may now write

$$(m-n)(m+n+1)\int_{-1}^{1} P_m(x)P_n(x)\,dx=0$$

Since $$m\ne n$$ and we must have $$m+n+1>0$$ we may divide through by $$(m-n)(m+n+1)$$ to obtain the desired result:

(4) $$\int_{-1}^{1} P_m(x)P_n(x)\,dx=0$$

The second case - $m=n$:

Let's begin with the forms of Bonnet's recurrence relation:

$$nP_n(x)+(n-1)P_{n-2}(x)=(2n-1)xP_{n-1}(x)$$

$$(n+1)P_{n+1}(x)+nP_{n-1}(x)=(2n+1)xP_n(x)$$

Solving both for $x$, and equating the results, we obtain:

$$\frac{nP_n(x)+(n-1)P_{n-2}(x)}{(2n-1)P_{n-1}(x)}=\frac{(n+1)P_{n+1}(x)+nP_{n-1}(x)}{(2n+1)P_n(x)}$$

Cross-multiplication yields:

$$n(2n+1)P_n^2(x)+(n-1)(2n+1)P_{n-2}(x)P_n(x)=(n+1)(2n-1)P_{n-1}P_{n+1}(x)+n(2n-1)P_{n-1}^2(x)$$

Integrating with respect to $x$, and using the result in (4), we obtain after dividing through by $$1\le n$$:

(5) $$(2n+1)\int_{-1}^{1}P_n^2(x)\,dx=(2n-1)\int_{-1}^{1} P_{n-1}^2(x)\,dx$$

Now, let's define:

$$A_{n}\equiv \int_{-1}^{1}P_n^2(x)\,dx$$

Thus, with this definition, (5) may be expressed as the recursion:

(6) $$A_{n}=\frac{2n-1}{2n+1}A_{n-1}$$

Using the fact that $$P_0(x)=1$$, we obtain:

$$A_{0}= \int_{-1}^{1} \,dx=2$$

Hence, we find the following:

$$A_{1}=\frac{1}{3}\cdot2=\frac{2}{3}=\frac{2}{2(1)+1}$$

$$A_{2}=\frac{3}{5}\cdot\frac{2}{3}=\frac{2}{5}= \frac{2}{2(2)+1}$$

Thus, based on the developing pattern, we may state the following induction hypothesis $P_{n}$ (having already shown the base case $P_1$ is true):

$$A_{n}=\frac{2}{2n+1}$$

Now, the recursion we found in (6) may be written with $n+1$ instead of $n$ as:

(7) $$A_{n+1}=\frac{2n+1}{2n+3}A_{n}$$

Thus, as our induction step, we may multiply through the induction hypothesis $P_{n}$ by $$\frac{2n+1}{2n+3}$$ to obtain:

$$\frac{2n+1}{2n+3}A_{n}=\frac{2}{2n+1}\cdot\frac{2n+1}{2n+3}$$

Using (7) on the left side, and simplifying the right side, there results:

$$A_{n+1}=\frac{2}{2(n+1)+1}$$

We have derived $P_{n+1}$ from $P_{n}$, thereby completing the proof by induction. Thus, we may state:

(8) $$A_n=\int_{-1}^{1}P_n^2(x)\,dx=\frac{2}{2n+1}$$

Combining the two cases:

Combining our two cases, $m\ne n$ with (4) and $m=n$ with (8), we may now state:

(9) $$\int_{-1}^{1} P_m(x)P_n(x)\,dx=\begin{cases}0 & m\neq n\\ \dfrac{2}{2n+1} & m=n\end{cases}$$ [/sp]