Dr Phill’s Science Made Simple

Mathematics

Legendre Equation

The Legendre equation is a second order differential equation that occurs in Physics. In particular, it is required to solve the Schrödinger equation for the Hydrogen atom. In the equation \(l\) is a non-negative integer.

\[(1 - x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + l(l + 1)y = 0\]

It can also be written negated.

\[(x^2 - 1)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} - l(l + 1)y = 0\]

We can look for a polynomial solution. As the equation equals zero, then the sum of the coefficients for each power of \(x\) must be zero. Start by assigning \(y\) to be a polynomial with \(n\) terms.

\[y = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4... a_nx^n\]

Differentiate Equation (3).

\[y' = a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 ... na_nx^{n-1}\]

Differentiate Equation (4).

\[y'' = 2a_2 + 6a_3x + 12a_4x^2 ... n(n - 1)a_nx^{n-2}\]

Collect the \(x^0\) terms form Equation (1).

\[2a_2 + l(l + 1)a_0 = 0 \quad\Rightarrow\quad a_2 = -\frac{l(l + 1)}{2}a_0\]

Collect the \(x^1\) terms from Equation (1).

\[6a_3 - 2a_1 + l(l + 1)a_1 = 0 \quad\Rightarrow\quad a_3 = -\frac{(l + 2)(l - 1)}{6}a_1\]

Collect the even \(x^n\) terms from Equation (1).

\[(n + 2)(n + 1)a_{n+2} - n(n - 1)a_n - 2na_n + l(l + 1)a_n = 0 \quad\Rightarrow\quad a_{n+2} = \frac{(l - n)(l + n + 1)}{(n + 2)(n + 1)}a_n\]

Now if \(n = l\) then \(a_{n+2} = 0\). Therefore every term with \(n\) plus an even number is also zero. There are clearly two possible solutions to the equation which are the even and odd polynomial terms. We can choose the solution based on whether \(l\) is even or odd. We can choose \(a_0 = 1\) and \(a_1 = 0\) if \(l\) is even or \(a_0 = 0\) and \(a_1 = 1\) if \(l\) is odd. The remaining \(a_i\) can be determined by iteration.

Leibnitz Equation (or Rule)

Gotfried Wilhelm Leibnitz independently discovered calculus. Isaac Newton independently discovered an alternative formulation. Leibnitz' notation is used today. He discovered a rule for differentiating the product of two functions any number of times.

If the two functions \(u(x)\) and \(v(x)\) are differentiable \(n\) times, then the nth derivative of their product can be defined.

\[\frac{d^n}{dx^n}(uv) = \sum_{k=0}^n {n\choose k} \frac{du^{n-k}}{dx^{n-k}}\frac{dv^k}{dx^k}\]

Where the binomial coefficient is:

\[{n\choose k} = \frac{n!}{k!(n - k)!}\]

Equation (9) can be proved by induction. First assume that the rule is correct and differentiate it.

\[\frac{d}{dx}\left(\frac{d^n}{dx^n}(uv)\right) = \frac{d^{n+1}}{dx^{n+1}}(uv) = \sum_{k=0}^n {n\choose k} \frac{d}{dx}\left(\frac{du^{n-k}}{dx^{n-k}}\frac{dv^k}{dx^k}\right) = \sum_{k=0}^n {n\choose k} \frac{du^{n-k+1}}{dx^{n-k+1}}\frac{dv^k}{dx^k} + \sum_{k=0}^n {n\choose k} \frac{du^{n-k}}{dx^{n-k}}\frac{dv^{k+1}}{dx^{k+1}}\]

In the second term we can replace \(k\) with \(k - 1\) and increment the limits with no effective change.

\[\frac{d^{n+1}}{dx^{n+1}}(uv) = \sum_{k=0}^n {n\choose k} \frac{du^{n-k+1}}{dx^{n-k+1}}\frac{dv^k}{dx^k} + \sum_{k=1}^{n+1} {n\choose k - 1} \frac{du^{n-k+1}}{dx^{n-k+1}}\frac{dv^{k}}{dx^{k}}\]

Split the first summation into the first term and the rest of the series. Also, split the second summation into the last term and the rest of the series.

\[\frac{d^{n+1}}{dx^{n+1}}(uv) = {n\choose 0}\frac{du^{n+1}}{dx^{n+1}}v + \sum_{k=1}^n {n\choose k} \frac{du^{n-k+1}}{dx^{n-k+1}}\frac{dv^k}{dx^k} + \sum_{k=1}^{n} {n\choose k - 1} \frac{du^{n-k+1}}{dx^{n-k+1}}\frac{dv^{k}}{dx^{k}} + {n\choose n}u\frac{dv^{n+1}}{dx^{n+1}}\]

If the bottom number of a binomial coefficient is zero then the value is always 1. Changing the top number makes no change.

\[{n\choose 0} = \frac{n!}{0!(n-0)!} = 1\quad{n\choose 0} = {n+1\choose 0} = 1\]

If the two number of a binomial coefficient are the same than the value is always 1. Changing the numbers makes no change.

\[{n\choose n} = \frac{n!}{n!(n-n)!} = 1\quad{n\choose n} = {n+1\choose n+1} = 1\]

The two summation terms have the same derivatives. Their binomial coefficients can be combined.

\[{n\choose k} + {n\choose k-1} = \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-k+1)!} = \frac{(n-k+1)n!}{k!(n-k+1)!} + \frac{kn!}{k!(n-k+1)!} = \frac{(n+1)!}{k!(n-k+1)!} = {n+1\choose k}\]

This gives the expected result.

\[\frac{d^{n+1}}{dx^{n+1}}(uv) = \sum_{k=0}^{n+1} {n+1\choose k} \frac{du^{n-k+1}}{dx^{n-k+1}}\frac{dv^k}{dx^k}\]

Rodrigues Formula

The Rodrigues formula is a solution to the Legendre equation.

\[P_n(x) = \frac{1}{2^nn!}\left(\frac{d}{dx}\right)^n(x^2 - 1)^n\]

It can be proved by defining \(v = x^2 - 1\) and differentiating.

\[\frac{dv}{dx} = 2xn(x^2 - 1)^{n-1}\]

Multiply by \(x^2 - 1\).

\[(x^2 - 1)\frac{dv}{dx} = 2xn(x^2 - 1)^n = 2nxv\]

We need to differentiate Equation (20) \(n+1\) times. This can be done using the Leibnitz rule. The left hand side only has three terms as \(x^2 - 1\) only has two non-zero derivatives. The summation is from \(n-1\) to \(n+1\). The right hand side only has two terms. The summation is from \(n\) to \(n+1\). Apply the Leibnitz rule.

\[{n+1\choose n-1}2\frac{d^{n}v}{dx^{n}} + {n+1\choose n}2x\frac{d^{n+1}v}{dx^{n+1}} + {n+1\choose n+1}(x^2 - 1)\frac{d^{n+2}v}{dx^{n+2}} = {n+1\choose n}2n\frac{d^{n}v}{dx^{n}} + {n+1\choose n+1}2nx\frac{d^{n+1}v}{dx^{n+1}}\]

Expand the binomial coefficients.

\[\frac{(n+1)!}{(n-1)!2!}2\frac{d^{n}v}{dx^{n}} + \frac{(n+1)!}{n!1!}2x\frac{d^{n+1}v}{dx^{n+1}} + (x^2 - 1)\frac{d^{n+2}v}{dx^{n+2}} = \frac{(n+1)!}{n!1!}2n\frac{d^{n}v}{dx^{n}} + 2nx\frac{d^{n+1}v}{dx^{n+1}}\]

Simplify the factorials.

\[n(n+1)\frac{d^{n}v}{dx^{n}} + 2x(n+1)\frac{d^{n+1}v}{dx^{n+1}} + (x^2 - 1)\frac{d^{n+2}v}{dx^{n+2}} = 2n(n+1)\frac{d^{n}v}{dx^{n}} + 2nx\frac{d^{n+1}v}{dx^{n+1}}\]

Set \(V = \frac{d^nv}{dx^n}\) and reorganise.

\[(x^2 - 1)\frac{d^2V}{dx^2} + 2x\frac{dV}{dx} - n(n+1)V = 0\]

This the the Legendre equation Equation (2). So \(V\) is a solution and \(P_n(x)\) from Equation (18) is also a solution.

\[P_n(x) = \frac{1}{2^nn!}\left(\frac{d}{dx}\right)^n(x^2 - 1)^n\]

Associated Legendre Equation

The associated Legendre equation has an additional term containing a second integer parameter \(m\).

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

We have an ansatz solution.

\[P^m_l(x) = (-1)^m(1 - x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}P_l(x)\]

To prove this we note that for \(m=0\) this is the regular Legendre equation with solutions \(P_l(x)\).

\[(1 - x^2)\frac{d^2P_l}{dx^2} - 2x\frac{dP_l}{dx} + l(l + 1)P_l = 0\]

Now, differentiate Equation (28) \(m\) times using the Leibnitz rule.

Start with the second order term.

\[\frac{d^m}{dx^m}\left[(1 - x^2)\frac{d^2P_l}{dx^2}\right] = (1-x^2)\frac{d^{m+2}P_l}{dx^{m+2}} - 2mx\frac{d^{m+1}P_l}{dx^{m+1}} - m(m-1)\frac{d^mP_l}{dm^m}\]

Now, the first order term.

\[\frac{d^m}{dx^m}\left[-2x\frac{dP_l}{dx}\right] = -2x\frac{d^{m+1}P_l}{dx^{m+1}}- 2m\frac{d^mP_l}{dm^m}\]

The final term.

\[\frac{d^m}{dx^m}\left[l(l+1)P_l\right] = l(l+1)\frac{d^mP_l}{dm^m}\]

Add the three terms to complete the equation.

\[(1-x^2)\frac{d^{m+2}P_l}{dx^{m+2}} - 2mx\frac{d^{m+1}P_l}{dx^{m+1}} - m(m-1)\frac{d^mP_l}{dm^m} -2x\frac{d^{m+1}P_l}{dx^{m+1}}- 2m\frac{d^mP_l}{dm^m} +l(l+1)\frac{d^mP_l}{dm^m}=0\]

Combine terms.

\[(1-x^2)\frac{d^{m+2}P_l}{dx^{m+2}} - 2(m+1)x\frac{d^{m+1}P_l}{dx^{m+1}} +[l(l+1) - m(m+1)]\frac{d^mP_l}{dm^m}=0\]

Define.

\[u = \frac{d^mP_l}{dm^m}\]

\[(1 - x^2)\frac{d^2u}{dx^2} - 2x(m+1)\frac{du}{dx} + [l(l+1) - m(m+1)]u = 0\]

Now another ansatz! Define a new function.

\[u = v(1 - x^2)^{-\frac{m}{2}}\]

Differentiate.

\[\frac{du}{dx} = \frac{dv}{dx}(1 - x^2)^{-\frac{m}{2}} + mxv(1 - x^2)^{-\frac{m+2}{2}} = \left(\frac{dv}{dx} + \frac{mvx}{1 - x^2}\right)(1 - x^2)^{-\frac{m}{2}}\]

Differentiate again.

\[\frac{d^2u}{dx^2} = \left(\frac{d^2v}{dx^2} + \frac{dv}{dx}\frac{mx}{1 - x^2} + \frac{mv}{1 - x^2} + \frac{2mvx^2}{(1 - x^2)^2}\right)(1 - x^2)^{-\frac{m}{2}} + \left(\frac{dv}{dx} + \frac{mvx}{1 - x^2}\right)\frac{mx}{1 - x^2}(1 - x^2)^{-\frac{m}{2}}\]

Combine terms.

\[\frac{d^2u}{dx^2} = \left(\frac{d^2v}{dx^2} + \frac{dv}{dx}\frac{2mx}{1 - x^2} + \frac{mv}{1 - x^2} + \frac{m(m+2)vx^2}{(1 - x^2)^2}\right)(1 - x^2)^{-\frac{m}{2}}\]

Substitute Equation (36), Equation (37), and Equation (39) into Equation (35) and divide by the common term.

\[(1 - x^2)\frac{d^2v}{dx^2} + 2mx\frac{dv}{dx} + mv + \frac{m(m+2)vx^2}{1 - x^2} - 2x(m+1)\frac{dv}{dx} - \frac{2x^2vm(m+1)}{1 - x^2} + [l(l+1) - m(m+1)]v = 0\]

Combine terms.

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

This is the associated Legendre equation! So \(v\) is a solution. So, from Equation (36).

\[v = u(1 - x^2)^{\frac{m}{2}}\]

The solution is as expected.

\[P^m_l(x) = (-1)^m(1 - x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}P_l(x)\]

The \((-1)^m\) term separates the even terms that have polynomial solutions from the odd term that do not.

That was quite a journey!