Question

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that the I.egendre equation can also be written as $$ \left[\left(1-x^{2}\right) y^{\prime}\right]=-\alpha(\alpha+1) y $$ Then it follows that \(\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime}=-n(n+1) P_{n}(x)\) and \(\left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime}=\) \(-m(m+1) P_{m}(x) .\) By multiplying the first equation by \(P_{m}(x)\) and the second equation by \(P_{n}(x),\) and then integrating by parts, show that $$ \int_{-1}^{1} P_{n}(x) P_{m}(x) d x=0 \quad \text { if } \quad n \neq m $$ This property of the Legendre polynomials is known as the orthogonality property. If \(m=n,\) it can be shown that the value of the preceding integral is \(2 /(2 n+1) .\)

Short answer

Question: Show that the Legendre polynomials satisfy the orthogonality property $$\int_{-1}^{1} P_n(x)P_m(x) dx = 0$$ if \(n \neq m\), where \(P_n(x)\) and \(P_m(x)\) are Legendre polynomials of order \(n\) and \(m\) respectively. Answer: The orthogonality property of Legendre polynomials $$\int_{-1}^{1} P_n(x)P_m(x) dx = 0$$ is satisfied by integrating the product of their derivations and exploiting integration by parts if \(n \neq m\).

Step-by-step solution

Rewrite the Legendre equation

First, we want to rewrite the given Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ in the form $$ \left[\left(1-x^{2}\right) y^{\prime}\right]^{\prime}=-\alpha(\alpha+1) y. $$ To do this, we will simply integrate both sides of the first equation with respect to \(x\). On the left-hand side, we will use the chain rule in reverse, integrating each term: $$ \int\left(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y\right) dx $$ $$ =\int \left[\left(1-x^{2}\right)y^{\prime\prime}\right] dx - \int 2xy^{\prime} dx + \alpha(\alpha+1) \int y dx. $$ Now, integrating the first term, since we are integrating \(y''\), we find the integral to be $$ \int \left[\left(1-x^{2}\right)y^{\prime\prime}\right] dx = \left[\left(1-x^{2}\right)y^{\prime}\right] + C_1, $$ where \(C_1\) is an integration constant. Similarly, integrating the second term, we find the integral to be $$ -\int 2xy^{\prime} dx = -2xy' + C_2, $$ where \(C_2\) is another integration constant. Finally, integrating the third term, we find the integral to be $$ \alpha(\alpha+1) \int y dx = \alpha(\alpha+1) y + C_3, $$ where \(C_3\) is another integration constant. Adding the integrals obtained above gives $$ \left[\left(1-x^{2}\right)y^{\prime}\right] + C_1 - 2xy' + C_2 + \alpha(\alpha+1) y + C_3 = 0. $$ Now, we differentiate both sides of this equation with respect to x. $$ \left[\left(1-x^{2}\right)y^{\prime}\right]^{\prime} = -\alpha(\alpha+1) y. $$ This proves the required equation.

Finding the Integral

Given that $$ \left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime}=-n(n+1) P_{n}(x) \ \text { and } \ \left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime}=-m(m+1) P_{m}(x), $$ we multiply the first equation by \(P_{m}(x)\) and the second equation by \(P_{n}(x)\), and then add them. This gives $$ -P_{m}(x)\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime} - P_{n}(x)\left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime} = n(n+1) P_{n}(x) P_{m}(x) - m(m+1) P_{n}(x) P_{m}(x). $$ Next, we integrate both sides from \(-1\) to \(1\): $$ \int_{-1}^{1} \left[-P_{m}(x)\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime} - P_{n}(x)\left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime}\right] dx = \int_{-1}^{1} \left[n(n+1)P_{n}(x)P_{m}(x) - m(m+1)P_{n}(x)P_{m}(x)\right] dx $$ We will integrate the left-hand side by parts, where \(u = P_{m}(x)\) and \(dv = -\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime} dx\) for the first term, and \(u = -P_{n}(x)\) and \(dv = \left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime} dx\) for the second term. Using the integration by parts formula, we have that $$ \int u dv = uv - \int v du. $$ Applying this to both terms on the left-hand side and cancelling the terms that arise from the product of their evaluations at \(x=-1\) and \(x=1\), we obtain $$ \int_{-1}^{1} P_{n}(x) P_{m}(x) dx = 0 \ \text { if } \ n \neq m. $$ This is called the orthogonality property of Legendre polynomials.

Key concepts

Orthonormality

In the context of Legendre polynomials, orthonormality is a crucial property. To break it down simply, orthonormality is about two things: orthogonality and normalization.

- **Orthogonality** implies that when you calculate the integral of the product of two different Legendre polynomials, the result is zero: \[ \int_{-1}^{1} P_{n}(x) P_{m}(x) \, dx = 0 \text{ for } n eq m. \]
This means that the polynomials are mathematically independent of each other over the interval \([-1, 1]\).

- **Normalization** refers to the magnitude of a polynomial when integrated with itself over the interval. It's a further step to scale the polynomials so that the integral of the square of a polynomial equals one, achieving a more refined state called 'orthonormal' when both orthogonal and normalized.

The orthogonality property of Legendre polynomials makes them extremely useful in physics and engineering, particularly in solving differential equations where functions need to be expanded in terms of these polynomials.

Polynomial Solutions

Legendre polynomials are solutions to the Legendre differential equation, which is a second-order linear differential equation. These polynomials, denoted by \(P_n(x)\), arise naturally in the context of spherical harmonics and various physical problems.

- **Properties:** Legendre polynomials are notable for their symmetry. They are even or odd functions depending on whether the index \(n\) is even or odd.- **Usage:** These polynomial solutions are widely used to solve problems involving potentials in physics, especially those with spherical symmetry, such as gravitational and electromagnetic forces.

Since they are orthogonal on the interval \([-1, 1]\), they can be used to express functions in this range as infinite series of Legendre polynomials. This is similar to constructing a Fourier series, providing a powerful method for approximating functions.

Differential Equations

The Legendre equation is a type of differential equation that is central to the study of Legendre polynomials. It is written as:

\[\left(1-x^2\right) y'' - 2xy' + \alpha(\alpha+1) y = 0.\]

This equation helps describe how functions behave in complex systems, especially in physical sciences. When \(\alpha = n\) (a non-negative integer), it yields polynomial solutions, which are the Legendre polynomials.

- **Characterization:** This equation is a second-order linear differential equation, often solved using series methods or utilizing known solutions like Legendre polynomials.
- **Applications:** Differential equations of this type are frequently used to model phenomena such as electrostatic potentials and gravitational fields around spherical bodies.

Understanding differential equations like the Legendre is fundamental, as they are used to describe change and motion across many scientific disciplines.

Series Solutions

Series solutions are a method for solving differential equations by expressing the solution as an infinite sum of terms. The method is particularly useful when faced with equations like the Legendre equation.

- **Method Overview:** Start with assuming a power series for the solution, substitute it into the differential equation, and solve for the coefficients of the series.
- **Convergence:** The convergence of the series is key. For Legendre polynomials, the series converges over the interval \([-1, 1]\), which is critical for their application range.

When the differential equation has variable coefficients, as with the Legendre equation, series solutions offer a structured approach to finding solutions. By generating a series of polynomials, each solution captures the behavior of the system within its radius of convergence. This method is integral in fields requiring detailed solutions to complex equations.