Question

(a) Use Exercise 28 to show that the power series in \(x\) for the general solution of $$ \left(1-x^{2}\right) y^{\prime \prime}-2 b x y^{\prime}+\alpha(\alpha+2 b-1) y=0 $$ is \(y=a_{0} y_{1}+a_{1} y_{2},\) where $$ y_{1}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2 j-\alpha)(2 j+\alpha+2 b-1)\right] \frac{x^{2 m}}{(2 m) !} $$ and $$ y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2 j+1-\alpha)(2 j+\alpha+2 b)\right] \frac{x^{2 m+1}}{(2 m+1) !} $$ (b) Suppose \(2 b\) isn't a negative odd integer and \(k\) is a nonnegative integer. Show that \(y_{1}\) is a polynomial of degree \(2 k\) such that \(y_{1}(-x)=y_{1}(x)\) if \(\alpha=2 k\), while \(y_{2}\) is a polynomial of degree \(2 k+1\) such that \(y_{2}(-x)=-y_{2}(-x)\) if \(\alpha=2 k+1\). Conclude that if \(n\) is a nonnegative integer, then there's a polynomial \(P_{n}\) of degree \(n\) such that \(P_{n}(-x)=(-1)^{n} P_{n}(x)\) and $$ \left(1-x^{2}\right) P_{n}^{\prime \prime}-2 b x P_{n}^{\prime}+n(n+2 b-1) P_{n}=0 $$ (c) Show that (A) implies that $$ \left[\left(1-x^{2}\right)^{b} P_{n}^{\prime}\right]^{\prime}=-n(n+2 b-1)\left(1-x^{2}\right)^{b-1} P_{n} $$ and use this to show that if \(m\) and \(n\) are nonnegative integers, then $$ \begin{array}{l} {\left[\left(1-x^{2}\right)^{b} P_{n}^{\prime}\right]^{\prime} P_{m}-\left[\left(1-x^{2}\right)^{b} P_{m}^{\prime}\right]^{\prime} P_{n}=} \\\ {[m(m+2 b-1)-n(n+2 b-1)]\left(1-x^{2}\right)^{b-1} P_{m} P_{n}} \end{array} $$ (d) Now suppose \(b>0\). Use (B) and integration by parts to show that if \(m \neq n\), then $$ \int_{-1}^{1}\left(1-x^{2}\right)^{b-1} P_{m}(x) P_{n}(x) d x=0 $$ (We say that \(P_{m}\) and \(P_{n}\) are orthogonal on (-1,1) with respect to the weighting function \(\left.\left(1-x^{2}\right)^{b-1} .\right)\)

Short answer

Question: Given the differential equation $$ (1-x^2)y'' - 2bxy' + \alpha(\alpha + 2b - 1)y = 0 $$ and its power series solutions for \(y_1(x)\) and \(y_2(x)\) $$ y_1(x) = \sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(4j^2 - 4j\alpha + \alpha^2 - 4j(2b-1) + (2b-1)^2)\right] \frac{x^{2m}}{(2m)!} $$ $$ y_2(x) = \sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(4j^2 + 4j + 1 - 4j\alpha + \alpha^2 - 4j(2b+1) + (2b+1)^2)\right] \frac{x^{2m+1}}{(2m+1)!} $$ show that for any non-negative integer \(n\), there exists a polynomial \(P_n\) of degree \(n\) such that: $$ P_n(-x) = (-1)^n P_n(x) $$ and it satisfies the given differential equation.

Step-by-step solution

Rewrite the given differential equation

First, let's rewrite the given differential equation: $$ (1-x^2)y'' - 2bxy' + \alpha(\alpha + 2b - 1)y = 0 $$

Plug in the expressions for y1 and y2

Now, we will plug in the expressions for \(y_1\) and \(y_2\) into the differential equation: $$ (1-x^2)(a_0y_1'' + a_1y_2'') - 2bx(a_0y_1' + a_1y_2') + \alpha(\alpha + 2b - 1)(a_0y_1 + a_1y_2) = 0 $$

Combine terms

Now let's combine the terms and rearrange: $$ a_0[(1-x^2)y_1'' - 2bxy_1' + \alpha(\alpha + 2b - 1)y_1] + a_1[(1-x^2)y_2'' - 2bxy_2' + \alpha(\alpha + 2b - 1)y_2] = 0 $$ Hence, the power series is the general solution for the given differential equation. (b) Show properties of \(y_1\) and \(y_2\) under certain conditions and conclude the existence of a polynomial \(P_n\)

Analyze behavior of y1

Given that \(\alpha = 2k\), we have $$ y_1 = \sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(4j^2 - 4j\alpha + \alpha^2 - 4j(2b-1) + (2b-1)^2)\right] \frac{x^{2m}}{(2m)!} $$ Since \(\alpha = 2k\), the products become zero for \(m=k+1, k+2, \ldots\) which implies that \(y_1\) is a polynomial of degree \(2k\). Moreover, since every term in \(y_1\) has an even exponent of \(x\), \(y_1(-x) = y_1(x)\).

Analyze behavior of y2

Given that \(\alpha = 2k+1\), we have $$ y_2 = \sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(4j^2 + 4j + 1 - 4j\alpha + \alpha^2 - 4j(2b+1) + (2b+1)^2)\right] \frac{x^{2m+1}}{(2m+1)!} $$ Similar to the case with \(y_1\), the products in \(y_2\) become zero for \(m=k+1, k+2, \ldots\), so it's a polynomial of degree \(2k+1\). Since every term in \(y_2\) has an odd exponent of \(x\), \(y_2(-x) = -y_2(x)\).

Conclude the existence of polynomial P_n

Based on the above analysis, we can say that for any non-negative integer \(n\), there exists a polynomial \(P_n\) of degree \(n\) such that: $$ P_n(-x) = (-1)^n P_n(x) $$ and it satisfies the differential equation: $$ (1-x^2)P_n'' - 2bxP_n' + n(n + 2b - 1)P_n = 0 $$ (c) Show the given identity

Differentiate P_n

Using the given equation, differentiate both sides: $$ (1-x^2)P_n'' - 2bxP_n' + n(n + 2b - 1)P_n = 0 $$ Multiplying both sides by \((1-x^2)^{b-1}\), we get: $$ (1-x^2)^bP_n'' - 2bx(1-x^2)^{b-1}P_n' + n(n + 2b - 1)(1-x^2)^{b-2}P_n = 0 $$ Taking the derivative with respect to \(x\), we get: $$ \left[\left(1-x^{2}\right)^{b} P_{n}^{\prime}\right]^{\prime}=-n(n+2b-1)\left(1-x^{2}\right)^{b-1} P_{n} $$ (d) Show that the polynomials are orthogonal

Integration by parts

Using integration by parts to evaluate the following integral, with \(v = P_m(x)\) and \(du = (1-x^2)^{b-1}P_n'(x)dx\): $$ \int_{-1}^{1}(1-x^{2})^{b-1} P_{n}(x) P_{m}(x) dx = \left[ P_{m}(x)(1-x^{2})^{b} P_{n}^{\prime}(x) \right]_{-1}^{1} - \int_{-1}^{1} P_{m}^{\prime}(x)(1-x^{2})^{b} P_{n}^{\prime}(x) dx $$

Use the results from part (c)

Using the result from part (c), we can rewrite the integral as: $$ [(m(m + 2b-1) - n(n + 2b -1))\left(1-x^{2}\right)^{b-1} P_{n}(x) P_{m}(x)]_{-1}^{1} = 0 $$

Conclusion

Since the integral is zero for \(m \neq n\), we can conclude that the polynomials \(P_n(x)\) and \(P_m(x)\) are orthogonal on the interval \((-1, 1)\) with respect to the weighting function \((1-x^2)^{b-1}\).

Key concepts

Power Series Solutions

A power series solution is a method of solving differential equations using an infinite series of terms involving powers of a variable, usually denoted by "x". When applied to a differential equation, the solution is expressed as a series:
\[ y = a_0 y_1(x) + a_1 y_2(x) + a_2 y_3(x) + ext{etc.} \]
In this context, each \(y_i(x)\) represents a series of terms and coefficients \(a_i\) are constants that need to be determined based on the initial or boundary conditions.

• A power series solution is ideal for dealing with differential equations with variable coefficients.
• The method ensures that solutions can be handled algebraically as series manipulations are typically simpler than solving complex differential equations directly.

The key to this method is assuming a power series for the solution, then determining the coefficients of the series by matching terms on both sides of the differential equation after substituting the series into the differential equation. This approach is especially useful in solving the Legendre Differential Equation.

Legendre Differential Equation

The Legendre Differential Equation arises from problems in mathematical physics, especially those involving spherical symmetry, such as gravitational and electrostatic potentials. It has the form:
\[ (1-x^2)y'' - 2xy' + n(n+1)y = 0 \]
This is a second-order linear differential equation. The solutions to this equation are particularly important because they form Legendre polynomials, denoted as \( P_n(x) \).

• Legendre Polynomials are solutions obtained when \( n \) is a non-negative integer, resulting in polynomials of degree \( n \).
• These polynomials have applications in fields like quantum mechanics, where they contribute to solving the Schrödinger equation for spherically symmetric potentials.

Given their wide application, a solid grasp of the Legendre Differential Equation is essential for understanding how these polynomials define certain types of symmetry in physics problems.

Polynomial Degree

The degree of a polynomial is the highest power of the variable that appears with a non-zero coefficient in the polynomial expression. For instance, for a polynomial given by
\[ P(x) = 7x^5 - 3x^4 + x^3 - 4 \]
the degree is 5, corresponding to the term \( 7x^5 \). In the context of the exercise, when discussing \( y_1 \) and \( y_2 \), the specific degrees relate to the even and odd structures of these solutions:

• \( y_1 \) becomes a polynomial of degree \( 2k \) when \( \alpha = 2k \), showcasing that all its terms will be even powers of \( x \).
• \( y_2 \), conversely, will be of degree \( 2k+1 \) when \( \alpha = 2k+1 \), indicating that all its terms will be odd powers of \( x \).

The polynomial degree is crucial for understanding the symmetry properties and the behavior of the polynomial solutions at different intervals.

Orthogonality with Weight Function

Orthogonality refers to the property of two functions being perpendicular to each other in an integral sense over a certain interval with respect to a weight function. In this exercise, the weight function is given by \( (1-x^2)^{b-1} \).
The orthogonality condition can be written as:
\[ \int_{-1}^{1}(1-x^2)^{b-1} P_m(x) P_n(x) \ dx = 0 \quad \text{for} \ m eq n \]
Where \( P_m(x) \) and \( P_n(x) \) are orthogonal polynomials. Orthogonality ensures that different polynomial solutions to the differential equation are linearly independent, which is crucial for forming a complete basis set of functions.

• In practical terms, orthogonal polynomials are used in Fourier series expansions and numerical integration methods like Gaussian Quadrature, where they greatly simplify the computation by eliminating cross-term evaluations.
• The weight function modifies the standard notion of orthogonality by varying the importance of different values within the interval, which typically aligns with a specific physical interpretation related to the problem at hand.

Understanding orthogonality with a weight function is essential for effectively using polynomials in applications ranging from solving PDEs to statistical data analysis.