Orthogonality relation-Associated Legendre Equation

Orthogonality of Associated Legendre Polynomials \(P_\ell^{\,m}(x)\)

Teaching-style derivation (Step-by-step substitutions + exam-ready flow)  |  Fixed \(m\), integrate over \([-1,1]\)

ODE in self-adjoint form Define \(y_\ell, y_n\) Cross-multiply & subtract Total derivative Integrate & boundary kills Normalization (where factorials appear)

Goal / Final Result

\[ \boxed{ \int_{-1}^{1} P_\ell^{\,m}(x)\,P_n^{\,m}(x)\,dx = \frac{2}{2\ell+1}\,\frac{(\ell+m)!}{(\ell-m)!}\,\delta_{\ell n} } \]

• Orthogonality: integral is 0 when \(\ell\neq n\).
• Normalization: when \(\ell=n\), the integral equals the constant shown.

Step 1 — Start from the Associated Legendre differential equation (self-adjoint form)

Given ODE

\[ \frac{d}{dx}\left[(1-x^2)\frac{dy}{dx}\right] +\left[\ell(\ell+1)-\frac{m^2}{1-x^2}\right]y=0. \]

This is already in Sturm–Liouville (self-adjoint) form. Orthogonality comes from “cross-multiply + subtract + integrate”.

Teaching cue: Whenever you see \(\dfrac{d}{dx}[p(x)y'] + \lambda w(x)y=0\), think orthogonality. Here, \(p(x)=1-x^2\).

Step 2 — Make the substitution clearly: \(y_\ell=P_\ell^{\,m}(x)\), \(y_n=P_n^{\,m}(x)\)

Define the two solutions

\[ y_\ell(x)=P_\ell^{\,m}(x), \qquad y_n(x)=P_n^{\,m}(x). \]

Both have the same order \(m\), but different degrees \(\ell\) and \(n\).

Equation for \(y_\ell\) (Eℓ)

\[ \frac{d}{dx}\Big[(1-x^2)y_\ell'\Big] +\left[\ell(\ell+1)-\frac{m^2}{1-x^2}\right]y_\ell=0. \]

Equation for \(y_n\) (En)

\[ \frac{d}{dx}\Big[(1-x^2)y_n'\Big] +\left[n(n+1)-\frac{m^2}{1-x^2}\right]y_n=0. \]

Step 3 — Cross-multiply (show every term) and subtract

Multiply (Eℓ) by \(y_n\)

\[ y_n\frac{d}{dx}\Big[(1-x^2)y_\ell'\Big] +\left[\ell(\ell+1)-\frac{m^2}{1-x^2}\right]y_\ell y_n=0. \]

\[ y_n\frac{d}{dx}\Big[(1-x^2)y_\ell'\Big] +\ell(\ell+1)\,y_\ell y_n -\frac{m^2}{1-x^2}y_\ell y_n=0. \]

Call this (A)

Multiply (En) by \(y_\ell\)

\[ y_\ell\frac{d}{dx}\Big[(1-x^2)y_n'\Big] +\left[n(n+1)-\frac{m^2}{1-x^2}\right]y_\ell y_n=0. \]

\[ y_\ell\frac{d}{dx}\Big[(1-x^2)y_n'\Big] +n(n+1)\,y_\ell y_n -\frac{m^2}{1-x^2}y_\ell y_n=0. \]

Call this (B)

Now subtract: (A) − (B)

\[ \Bigg[ y_n\frac{d}{dx}\Big((1-x^2)y_\ell'\Big) - y_\ell\frac{d}{dx}\Big((1-x^2)y_n'\Big) \Bigg] + \Big[\ell(\ell+1)-n(n+1)\Big]y_\ell y_n =0. \]

✅ The two terms containing \(\displaystyle -\frac{m^2}{1-x^2}y_\ell y_n\) cancel perfectly.

Step 4 — Convert the first bracket into a total derivative

Define a Wronskian-like quantity

\[ W(x)=y_n y_\ell' - y_\ell y_n'. \]

This is chosen because it matches the bracket from the subtraction step.

Reverse product rule identity

\[ \frac{d}{dx}\Big[(1-x^2)W(x)\Big] = y_n\frac{d}{dx}\Big[(1-x^2)y_\ell'\Big] - y_\ell\frac{d}{dx}\Big[(1-x^2)y_n'\Big]. \]

Therefore the subtracted equation becomes

\[ \frac{d}{dx}\Big[(1-x^2)(y_n y_\ell' - y_\ell y_n')\Big] + \big[\ell(\ell+1)-n(n+1)\big]y_\ell y_n =0. \]

Step 5 — Integrate from \(-1\) to \(1\) and kill the boundary term

Integrate

\[ \int_{-1}^{1} \frac{d}{dx}\Big[(1-x^2)(y_n y_\ell' - y_\ell y_n')\Big]dx + \big[\ell(\ell+1)-n(n+1)\big] \int_{-1}^{1} y_\ell y_n\,dx =0. \]

The first integral is a boundary term:

\[ \Big[(1-x^2)(y_n y_\ell' - y_\ell y_n')\Big]_{-1}^{1}. \]

Boundary term vanishes (“boundary killer”)

\[ (1-x^2)=0 \quad \text{at } x=\pm 1 \;\Rightarrow\; \Big[(1-x^2)(\cdots)\Big]_{-1}^{1}=0. \]

Hence

\[ \big[\ell(\ell+1)-n(n+1)\big]\int_{-1}^{1} y_\ell y_n\,dx=0. \]

Substitute back \(y_\ell=P_\ell^{\,m}\), \(y_n=P_n^{\,m}\)

\[ \big[\ell(\ell+1)-n(n+1)\big] \int_{-1}^{1} P_\ell^{\,m}(x)\,P_n^{\,m}(x)\,dx =0. \]

\[ \ell(\ell+1)-n(n+1)=(\ell-n)(\ell+n+1). \]

Orthogonality for \(\ell\neq n\)

\[ (\ell-n)(\ell+n+1)\neq 0 \;(\ell\neq n) \Rightarrow \boxed{ \int_{-1}^{1} P_\ell^{\,m}(x)\,P_n^{\,m}(x)\,dx=0 }. \]

6-mark tip: For the orthogonality part, never expand Rodrigues. Use only the differential equation method (fast + clean).

Step 6 — Normalization for \(\ell=n\): where \((\ell+m)!\) and \((\ell-m)!\) come from

Start from the definition (do not skip this)

\[ P_\ell^{\,m}(x)=(-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}P_\ell(x), \qquad P_\ell(x)=\frac{1}{2^\ell \ell!}\frac{d^\ell}{dx^\ell}(x^2-1)^\ell. \]

So the derivative inside \(P_\ell^{\,m}\) becomes:

\[ \frac{d^m}{dx^m}P_\ell(x) = \frac{1}{2^\ell \ell!}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^\ell. \]

Define the normalization integral

\[ I_{\ell m}=\int_{-1}^{1}\left[P_\ell^{\,m}(x)\right]^2dx = \int_{-1}^{1}(1-x^2)^m\left(\frac{d^m}{dx^m}P_\ell(x)\right)^2dx. \]

(The factor \((-1)^m\) disappears because we square.)

Why boundary terms vanish (this justifies integration by parts)

• \((1-x^2)^m = 0\) at \(x=\pm 1\).
• \((x^2-1)^\ell\) has zeros of order \(\ell\) at \(x=\pm 1\).
• So, during repeated integration by parts, all boundary terms contain a factor that goes to zero at \(\pm 1\).

This is the reason we are allowed to “move derivatives” between factors safely.

Where the factorial ratio comes from (counting derivatives)

After performing integration by parts repeatedly (exact steps are in standard textbooks), the integral reduces to:

\[ \boxed{ \int_{-1}^{1}\left[P_\ell^{\,m}(x)\right]^2dx = \frac{(\ell+m)!}{(\ell-m)!} \int_{-1}^{1}\left[P_\ell(x)\right]^2dx }. \]

Key explanation (teach this clearly):
The ratio appears because repeated differentiation produces a product of consecutive integers:

\[ (\ell+m)(\ell+m-1)\cdots(\ell-m+1) = \frac{(\ell+m)!}{(\ell-m)!}. \]

• Numerator \((\ell+m)!\) comes from “more derivatives → more factors”.
• Denominator \((\ell-m)!\) is what remains after cancellations.

Memory cue: “More derivatives → bigger factorial on top.”  and  “Minus stays in the bottom: \((\ell-m)!\)”

Finish using the known normalization of \(P_\ell(x)\)

\[ \int_{-1}^{1}[P_\ell(x)]^2dx=\frac{2}{2\ell+1}. \]

Therefore,

\[ \boxed{ \int_{-1}^{1}\left[P_\ell^{\,m}(x)\right]^2dx = \frac{2}{2\ell+1}\frac{(\ell+m)!}{(\ell-m)!} }. \]

Final Orthogonality Relation (Boxed)

\[ \boxed{ \int_{-1}^{1} P_\ell^{\,m}(x)\,P_n^{\,m}(x)\,dx = \frac{2}{2\ell+1}\frac{(\ell+m)!}{(\ell-m)!}\,\delta_{\ell n} } \]

One-line teaching summary: Orthogonality from DE (cross-subtract-integrate), normalization from derivative-counting (factorials).