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Orthogonality of Associated Legendre Polynomials 📂Functions

Orthogonality of Associated Legendre Polynomials

Theorem

The associated Legendre polynomials over the interval [−1,1][-1,1] for a fixed mm form an orthogonal set.

∫−11Plm(x)Pkm(x)dx=22l+1(l+m)!(l−m)!δlk \int_{-1}^{1} P_{l}^{m}(x)P_{k}^{m}(x)dx =\frac{ 2}{ 2l+1 }\frac{(l+m)!}{(l-m)!}\delta_{lk}

In the case of x=cos⁡θx=\cos \theta,

∫0πPlm(cos⁡θ)Pkm(cos⁡θ)sin⁡θdθ=22l+1(l+m)!(l−m)!δlk \int_{0}^{\pi} P_{l}^{m}(\cos \theta)P_ {k}^{m}(\cos\theta)\sin \theta d\theta =\frac{ 2}{ 2l+1 }\frac{(l+m)!}{(l-m)!}\delta_{lk}

Associated Legendre Polynomials Plm(x)=(1−x2)m212ll!dl+mdxl+m(x2−1)l P_{l}^{m}(x) = (1-x ^{2})^{\frac{m}{2}} \dfrac{1}{2^l l!} \dfrac{d^{l+m}}{dx^{l+m}}(x^2-1)^l

Proof

For convenience, let’s briefly denote it as Plm=Plm(x)P_{lm} = P_{l}^{m}(x). The associated Legendre differential equation is as follows.

ddx[(1−x2)Plm′]+[l(l+1)−m21−x2]Plm=0 \begin{equation} \frac{d}{dx} \left[ (1-x^{2})P_{lm}^{\prime} \right] +\left[ l(l+1)-\frac{m^{2}}{1-x^{2}} \right]P_{lm} = 0 \end{equation}

Case 1: l≠kl \ne k

The proof method is the same as showing the orthogonality of Legendre polynomials. Writing (1)(1) for ll and kk,

ddx[(1−x2)Plm′]+[l(l+1)−m21−x2]Plm=0ddx[(1−x2)Pkm′]+[k(k+1)−m21−x2]Pkm=0 \frac{d}{dx} \left[(1-x^{2}) P_{lm}^{\prime}\right] + \left[l(l+1) - \frac{m^{2}}{1-x^{2}}\right] P_{lm} = 0 \\ \frac{d}{dx} \left[(1-x^{2}) P_{km}^{\prime}\right] + \left[k(k+1) - \frac{m^{2}}{1-x^{2}}\right] P_{km} = 0

Multiply PkmP_{km} by the equation for ll, and PlmP_{lm} by the equation for kk, and subtract one from the other to get the following.

Pkmddx[(1−x2)Plm′]−Plmddx[(1−x2)Pkm′]+[l(l+1)−k(k+1)]PlmPkm=0 \begin{equation} P_{km} \frac{d}{dx} \left[ (1-x^{2})P_{lm}^{\prime} \right]-P_{lm}\frac{d}{dx} \left[ (1-x^{2})P_{km}^{\prime} \right]+\left[l(l+1)- k(k+1) \right]P_{lm}P_{km} = 0 \end{equation}

The first and second terms can be organized as follows.

 Pkmddx[(1−x2)Plm′]−Plmddx[(1−x2)Pkm′]=Pkm(1−x2)′Plm′+Pkm(1−x2)Plm′′−Plm(1−x2)′Pkm′−Plm(1−x2)Pkm′′=Pkm(1−x2)′Plm′+Pkm(1−x2)Plm′′−Plm(1−x2)′Pkm′−Plm(1−x2)Pkm′′+Pkm′(1−x2)Plm′−Pkm′(1−x2)Plm′=ddx[(1−x2)Plm′Pkm−(1−x2)PlmPkm′]=ddx[(1−x2)(Plm′Pkm−PlmPkm′)] \begin{align*} &\quad \ P_{km} \frac{d}{dx} \left[ (1-x^{2})P_{lm}^{\prime} \right]-P_{lm}\frac{d}{dx} \left[ (1-x^{2})P_{km}^{\prime} \right] \\ &= P_{km}(1-x^{2})^{\prime}P_{lm}^{\prime}+P_{km}(1-x^{2})P_{lm}^{\prime \prime}-P_{lm}(1-x^{2})^{\prime}P_{km}^{\prime}-P_{lm}(1-x^{2})P_{km}^{\prime \prime} \\ &= {\color{blue}P_{km}(1-x^{2})^{\prime}P_{lm}^{\prime}+P_{km}(1-x^{2})P_{lm}^{\prime \prime}}-{\color{orange}P_{lm}(1-x^{2})^{\prime}P_{km}^{\prime}-P_{lm}(1-x^{2})P_{km}^{\prime \prime} } \\ &\quad + {\color{blue}P_{km}^{\prime}(1-x^{2})P_{lm}^{\prime}}-{\color{orange}P_{km}^{\prime}(1-x^{2})P_{lm}^{\prime}} \\ &= \frac{d}{dx}\left[{\color{blue}(1-x^{2})P_{lm}^{\prime}P_{km}}-{\color{orange}(1-x^{2})P_{lm}P_{km}^{\prime} } \right] \\ &= \frac{d}{dx}\left[(1-x^{2})(P_{lm}^{\prime}P_{km}-P_{lm}P_{km}^{\prime}) \right] \end{align*}

Inserting this into (2)(2),

ddx[(1−x2)(Plm′Pkm−PlmPkm′)]+[l(l+1)−k(k+1)]PlmPkm=0 \frac{d}{dx}\left[(1-x^{2})(P_{lm}^{\prime}P_{km}-P_{lm}P_{km}^{\prime}) \right]+\left[l(l+1)- k(k+1) \right]P_{lm}P_{km}=0

Integrating both sides over the interval [−1,1][-1,1] yields the following.

[(1−x2)(Plm′Pkm−PlmPkm′)]−11+[l(l+1)−k(k+1)]∫−11PlmPkmdx=0 \left[(1-x^{2})(P_{lm}^{\prime}P_{km}-P_{lm}P_{km}^{\prime}) \right]_{-1}^{1}+\left[l(l+1)- k(k+1) \right]\int_{-1}^{1}P_{lm}P_{km}dx=0

The first term is 00, so it goes as follows.

[l(l+1)−k(k+1)]∫−11PlmPkmdx=0 \left[l(l+1)- k(k+1) \right]\int_{-1}^{1}P_{lm}P_{km}dx=0

Since l≠kl \ne k, it results in l(l+1)−k(k+1)≠0l(l+1)-k(k+1)\ne 0. Thus, it follows that.

∫−11Plm(x)Pkm(x)dx=0 \int_{-1}^{1}P_{l}^{m}(x)P_{k}^{m}(x)dx=0

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Case 2: l=kl=k

Auxiliary Theorem

The following holds.

dl+mdxl+m(x2−1)l=(l+m)!(l−m)!(x2−1)−mdl−mdxl−m(x2−1)l(3) \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}=\frac{(l+m)!}{(l-m)!}(x^{2}-1)^{-m}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \tag {3}

By applying the above formula to the associated Legendre polynomials,

Plm(x)=(1−x2)m212ll!dl+mdxl+m(x2−1)l=(1−x2)m212ll!(l+m)!(l−m)!(x2−1)−mdl−mdxl−m(x2−1)l=(−1)m2ll!(l+m)!(l−m)!(1−x2)−m2dl−mdxl−m(x2−1)l \begin{align*} P_{l}^{m}(x) &= (1-x ^{2})^{\frac{m}{2}} \dfrac{1}{2^l l!} \dfrac{d^{l+m}}{dx^{l+m}}(x^2-1)^l \\ &= (1-x ^{2})^{\frac{m}{2}}\dfrac{1}{2^l l!} \frac{(l+m)!}{(l-m)!}(x^{2}-1)^{-m}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \\ &= \dfrac{(-1)^{m}}{2^l l!} \frac{(l+m)!}{(l-m)!}(1-x^{2})^{-\frac{m}{2}}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \end{align*}

Squaring both sides of the above equation yields the following.

[Plm(x)]2=122l(l!)2[(l+m)!(l−m)!]2(1−x2)−mdl−mdxl−m(x2−1)ldl−mdxl−m(x2−1)l [P_{l}^{m}(x)]^{2} =\dfrac{1}{2^{2l}(l!)^{2}} \left[ \frac{(l+m)!}{(l-m)!} \right]^{2}(1-x^{2})^{-m}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l}

By substituting (3)(3) into the above equation,

 [Plm(x)]2=122l(l!)2[(l+m)!(l−m)!]2(1−x2)−m[(l−m)!(l+m)!(x2−1)mdl+mdxl+m(x2−1)l]dl−mdxl−m(x2−1)l=(−1)m22l(l!)2(l+m)!(l−m)!dl+mdxl+m(x2−1)ldl−mdxl−m(x2−1)l \begin{align*} &\quad \ [P_{l}^{m}(x)]^{2} \\ &= \dfrac{1}{2^{2l}(l!)^{2}} \left[ \frac{(l+m)!}{(l-m)!} \right]^{2}(1-x^{2})^{-m}\left[ \frac{(l-m)!}{(l+m)!}(x^{2}-1)^{m}\frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l} \right]\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \\ &= \dfrac{(-1)^{m}}{2^{2l}(l!)^{2}} \frac{(l+m)!}{(l-m)!} \frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \end{align*}

Now, integrating both sides over the interval [−1,1][-1,1] yields the following.

 ∫−11[Plm(x)]2dx=(−1)m22l(l!)2(l+m)!(l−m)!∫−11[dl+mdxl+m(x2−1)ldl−mdxl−m(x2−1)l]dx \begin{align*} &\quad \ \int_{-1}^{1}[P_{l}^{m}(x)]^{2}dx \\ &= \dfrac{(-1)^{m}}{2^{2l}(l!)^{2}} \frac{(l+m)!}{(l-m)!}\int_{-1}^{1} \left[ \frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \right]dx \tag{4} \end{align*}

Let’s focus only on the integral part of the right-hand side. By resolving it with integration by parts, it becomes,

 ∫−11[dl+mdxl+m(x2−1)ldl−mdxl−m(x2−1)l]dx=∫−11[dl+m−1dxl+m−1(x2−1)l]′dl−mdxl−m(x2−1)ldx=[dl+m−1dxl+m−1(x2−1)ldl−mdxl−m(x2−1)l]−11−∫−11dl+m−1dxl+m−1(x2−1)ldl−m+1dxl−m+1(x2−1)ldx \begin{align*} &\quad \ \int_{-1}^{1} \left[ \frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} \right]dx \\ &= \int_{-1}^{1} \left[ \frac{ d ^{l+m-1}}{ dx^{l+m-1} }(x^{2}-1)^{l}\right]^{\prime}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} dx \\ &= \left[ \frac{ d ^{l+m-1}}{ dx^{l+m-1} }(x^{2}-1)^{l}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l}\right]_{-1}^{1}-\int_{-1}^{1}\frac{ d ^{l+m-1}}{ dx^{l+m-1} }(x^{2}-1)^{l}\frac{ d ^{l-m+1}}{ dx^{l-m+1} }(x^{2}-1)^{l}dx \end{align*}

Here, the first term is 00. (x2−1)l(x^{2}-1)^{l} is a polynomial of the degree 2l2land since ∣m∣<l|m| \lt l, both l+m−1l+m-1 and l−ml-m are smaller than ll, leaving at least (x2−1)(x^{2}-1) not differentiated remaining. Inserting ±1\pm 1 results in 00. Continuing to solve the remaining terms by integration by parts,

 −∫−11dl+m−1dxl+m−1(x2−1)ldl−m+1dxl−m+1(x2−1)ldx=[−dl+m−2dxl+m−2(x2−1)ldl−m+1dxl−m+1(x2−1)l]+∫−11dl+m−2dxl+m−2(x2−1)ldl−m+2dxl−m+2(x2−1)ldx \begin{align*} &\quad \ -\int_{-1}^{1}\frac{ d ^{l+m-1}}{ dx^{l+m-1} }(x^{2}-1)^{l}\frac{ d ^{l-m+1}}{ dx^{l-m+1} }(x^{2}-1)^{l}dx \\ &= \left[- \frac{ d ^{l+m-2}}{ dx^{l+m-2} }(x^{2}-1)^{l}\frac{ d ^{l-m+1}}{ dx^{l-m+1} }(x^{2}-1)^{l} \right]+\int_{-1}^{1} \frac{ d ^{l+m-2}}{ dx^{l+m-2} }(x^{2}-1)^{l}\frac{ d ^{l-m+2}}{ dx^{l-m+2} }(x^{2}-1)^{l}dx \end{align*}

Here, the first term is 00 for the reason mentioned above. By repeating this process of integration by parts mm times, the following equation is obtained.

∫−11dl+mdxl+m(x2−1)ldl−mdxl−m(x2−1)ldx=(−1)m∫−11dldxl(x2−1)ldldxl(x2−1)ldx \int_{-1}^{1} \frac{ d ^{l+m}}{ dx^{l+m} }(x^{2}-1)^{l}\frac{ d ^{l-m}}{ dx^{l-m} }(x^{2}-1)^{l} dx=(-1)^{m}\int_{-1}^{1}\frac{ d ^{l}}{ dx^{l} }(x^{2}-1)^{l}\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}dx

Therefore, (4)(4) is as follows.

∫−11[Plm(x)]2dx=122l(l!)2(l+m)!(l−m)!∫−11[dldxl(x2−1)l]2dx \int_{-1}^{1}[P_{l}^{m}(x)]^{2}dx= \dfrac{1}{2^{2l}(l!)^{2}} \frac{(l+m)!}{(l-m)!}\int_{-1}^{1}\left[ \frac{ d ^{l}}{ dx^{l} }(x^{2}-1)^{l}\right]^{2}dx

Rodrigues’ Formula

Pl(x)=12ll!dldxl(x2−1)l P_{l}(x)=\dfrac{1}{2^l l!} \dfrac{d^l}{dx^l}(x^2-1)^l

Then, by Rodrigues’ formula, it follows that.

∫−11[Plm(x)]2dx=122l(l!)2(l+m)!(l−m)!22l(l!)2∫−11[Pl(x)]2dx=(l+m)!(l−m)!∫−11[Pl(x)]2dx \begin{align*} \int_{-1}^{1}[P_{l}^{m}(x)]^{2}dx &= \dfrac{1}{2^{2l}(l!)^{2}} \frac{(l+m)!}{(l-m)!}2^{2l}(l!)^{2}\int_{-1}^{1}\left[ P_{l}(x)\right]^{2}dx \\ &= \frac{(l+m)!}{(l-m)!}\int_{-1}^{1}\left[ P_{l}(x)\right]^{2}dx \end{align*}

Finally, by the orthogonality of the Legendre polynomials, the following is concluded.

∫−11[Plm(x)]2dx=22l+1(l+m)!(l−m)! \int_{-1}^{1}[P_{l}^{m}(x)]^{2}dx = \frac{2}{2l+1}\frac{(l+m)!}{(l-m)!}

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