Solution to Legendre's Differential Equation

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Theorem

The solution of Legendre's differential equation:

$\paren {1 - x^2} \dfrac {\d^2 y} {\d x^2} - 2 x \dfrac {\d y} {\d x} + p \paren {p + 1} y = 0$

can be obtained by Power Series Solution Method.

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Proof

Let:

$\ds y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$

such that:

$a_0 \ne 0$

Differentating with respect to $x$:

$\ds \dot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1}$

$\ds \ddot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$

Substituting in the original equation:

\(\ds \paren {1 - x^2} \ddot y - 2 x \dot y + p \paren {p + 1} y\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {1 - x^2} \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - x^2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}\)

\(\)

\(\ds \)

\(\ds {} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}\)

\(=\)

\(\ds 0\)

The summations are dependent upon $n$ and not $x$.

Therefore it is a valid operation to multiply the $x$'s into the summations, thus:

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}\)

\(\)

\(\ds \)

\(\ds {} - \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n} - 2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}\)

\(\)

\(\ds \)

\(\ds {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n - 1} - 2 \paren {k - n} }\)

\(=\)

\(\ds 0\)

Increasing the summation variable $n$ to $n + 2$:

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}\)

\(\)

\(\ds \)

\(\ds {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }\)

\(=\)

\(\ds 0\)

Taking the first 2 terms of the second summation out:

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}\)

\(\)

\(\ds \)

\(\ds {} + a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }\)

\(\)

\(\ds \)

\(\ds {} + \sum_{n \mathop = 2}^\infty a_{n - 2} x^{k - n} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {\paren {p + 1} - k \paren {k - 1} }\)

\(\)

\(\ds \)

\(\ds {} + \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }\)

\(=\)

\(\ds 0\)

Equating each term to $0$:

\(\text {(1)}: \quad\)

\(\ds a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} }\)

\(=\)

\(\ds 0\)

\(\text {(2)}: \quad\)

\(\ds a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }\)

\(=\)

\(\ds 0\)

\(\text {(3)}: \quad\)

\(\ds \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }\)

\(=\)

\(\ds 0\)

Take equation $(1)$:

$a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } = 0$

It is assumed that $a_0 \ne 0$ and $x^k$ can never be zero for any value of $k$.

Thus:

\(\ds p \paren {p + 1} - k \paren {k + 1}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds p^2 - k^2 + p - k\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {p - k} \paren {p + k} + \paren {p - k}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {p - k} \paren {p + k + 1}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds p - k\)

\(=\)

\(\ds 0\)

\(\, \ds \lor \, \)

\(\ds p + k + 1\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds k\)

\(=\)

\(\ds p\)

\(\, \ds \lor \, \)

\(\ds k\)

\(=\)

\(\ds p - 1\)

Take equation $(2)$:

$a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} } = 0$

As before, it is assumed that $x^{k - 1}$ can never be zero for any value of $k$.

Thus:

\(\ds a_1 \paren {p \paren {p + 1} - k \paren {k - 1} }\)

\(=\)

\(\ds 0\)

\(\ds a_1 k\)

\(=\)

\(\ds 0\)

substituting the value of $k$ from $(1)$

\(\ds a_1\)

\(=\)

\(\ds 0\)

as $2 k \ne 0$

Take equation $(3)$:

$\ds \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } } = 0$

As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$.

Thus:

\(\ds 0\)

\(=\)

\(\ds a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }\)

\(\ds \leadsto \ \ \)

\(\ds a_n\)

\(=\)

\(\ds -\frac {\paren {k - n + 2} \paren {k - n + 1} } {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } a_{n - 2}\)

Since Legendre's differential equation is a second order ODE, it has two independent solutions.

Solution 1 (for $k = p$)

\(\ds y\)

\(=\)

\(\ds x^p \sum_{n \mathop = 0}^\infty a_n x^{-n}\)

\(\ds \)

\(=\)

\(\ds x^k \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\)

\(\ds a_n\)

\(=\)

\(\ds -\frac {\paren {p - n + 2} \paren {p - n + 1} } {p \paren {p + 1} - \paren {p - n} \paren {p - n + 1} } a_{n - 2}\)

from equation $(4)$

\(\ds \)

\(=\)

\(\ds -\frac {\paren {p - n + 2} \paren {p - n + 1} } {n \paren {2 p - n + 1} } a_{n - 2}\)

Thus:

\(\ds a_2\)

\(=\)

\(\ds -\frac {p \paren {p - 1} } {2 \paren {2 p - 1} } a_0\)

for $n = 2$

\(\ds a_4\)

\(=\)

\(\ds -\frac {\paren {p - 2} \paren {p - 3} } {4 \paren {2 p - 3} } a_2\)

for $n = 4$

\(\ds \)

\(=\)

\(\ds \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } a_0\)

\(\ds a_6\)

\(=\)

\(\ds -\frac {\paren {p - 4} \paren {p - 5} } {6 \paren {2 p - 5} } a_4\)

for $n = 6$

\(\ds \)

\(=\)

\(\ds -\frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \paren {p - 4} \paren {p - 5} } {2 \cdot 4 \cdot 6 \paren {2 p - 1} \paren {2 p - 3} \paren {2 p - 5} } a_0\)

Similarly:

\(\ds a_{2 n}\)

\(=\)

\(\ds \paren {-1}^n \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \dotsm \paren {p - 2 n - 1} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p - 1} \paren {2 p - 3} \paren {2 p - 5} \dotsm \paren {2 p - 2 n + 1} } a_0\)

Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$

Substituting for $a_n$:

\(\ds y\)

\(=\)

\(\ds a_0 x^k \leftparen {1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } x^{-4} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \dotsb\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \rightparen {\paren {-1}^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \paren {p - 2 n + 1} } {\paren {2 \cdot 4 \cdot 6 \dots 2 n} \paren {2 p - 1} \paren {2 p - 3} \dotsm \paren {p - 2 n + 1} } x^{-2 n} + \dotsb}\)

$\Box$

Solution 2 (for $k = p - 1$)

\(\ds y\)

\(=\)

\(\ds x^{-p - 1} \sum_{n \mathop = 0}^\infty a_n x^{-n}\)

\(\ds \)

\(=\)

\(\ds x^{-k - 1} \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\)

The validity of the material on this page is questionable. In particular: Should it be $x^{-k - 1}$ in the above, as $k = p - 1$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

\(\ds a_n\)

\(=\)

\(\ds -\frac {\paren {-p - 1 - n + 2} \paren {-p - 1 - n + 1} } {p \paren {p + 1} - \paren {- p - 1 - n} \paren {-p - n} } a_{n - 2}\)

from equation $(4)$

\(\ds \)

\(=\)

\(\ds \frac {\paren {p + n - 1} \paren {p + n} } {n \paren {2 p + n + 1} } a_{n - 2}\)

Thus:

\(\ds a_2\)

\(=\)

\(\ds \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } a_0\)

for $n = 2$

\(\ds a_4\)

\(=\)

\(\ds \frac {\paren {p + 3} \paren {p + 4} } {4 \paren {2 p + 5} } a_2\)

for $n = 4$

\(\ds \)

\(=\)

\(\ds \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } a_0\)

\(\ds a_6\)

\(=\)

\(\ds \frac {\paren {p + 5} \paren {p + 6} } {6 \paren {2 p + 7} } a_4\)

for $n = 6$

\(\ds \)

\(=\)

\(\ds \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} \paren {p + 5} \paren {p + 6} } {2 \cdot 4 \cdot 6 \paren {2 p + 3} \paren {2 p + 5} \paren {2 p + 7} } a_0\)

Similarly:

\(\ds a_{2 n}\)

\(=\)

\(\ds \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \paren {2 p + 7} \dotsm \paren {2 p + 2 n + 1} } a_0\)

Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$

Substituting for $a_n$:

\(\ds y\)

\(=\)

\(\ds a_0 x^{-k - 1} \leftparen {1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } x^{-4} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \dotsb\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \rightparen {\paren {-1}^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \dotsm \paren {p - 2 n} } x^{-2 n} + \dotsb}\)

$\Box$

In summary, the two particular solutions are:

\(\ds y\)

\(=\)

\(\ds a_0 x^k \leftparen {1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } x^{-4} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \dotsb\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \rightparen {\paren {-1}^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \paren {p - 2 n + 1} } {\paren {2 \cdot 4 \cdot 6 \dots 2 n} \paren {2 p - 1} \paren {2 p - 3} \dotsm \paren {p - 2 n + 1} } x^{-2 n} + \dotsb}\)

and:

\(\ds y\)

\(=\)

\(\ds a_0 x^{-k - 1} \leftparen {1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } x^{-4} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \dotsb\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \rightparen {\paren {-1}^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \dotsm \paren {p - 2 n} } x^{-2 n} + \dotsb}\)

Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: Express in summation form You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$\blacksquare$

Also see

The solutions of Legendre's differential equation are known as Legendre polynomials, which are functions of the parameter $p$.

Source of Name

This entry was named for Adrien-Marie Legendre.

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 18$: Basic Differential Equations and Solutions: $18.12$: Legendre's Equation