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 two linearly independent solutions of the Legendre equation for \(|x|<1\) are $$ \begin{aligned} y_{1}(x)=& 1+\sum_{m=1}^{\infty}(-1)^{m} \\ & \times \frac{\alpha(\alpha-2)(\alpha-4) \cdots(\alpha-2 m+2)(\alpha+1)(\alpha+3) \cdots(\alpha+2 m-1)}{(2 m) !} x^{2 m} \\ y_{2}(x)=& x+\sum_{m=1}^{\infty}(-1)^{m} \\ & \times \frac{(\alpha-1)(\alpha-3) \cdots(\alpha-2 m+1)(\alpha+2)(\alpha+4) \cdots(\alpha+2 m)}{(2 m+1) !} x^{2 m+1} \end{aligned} $$

Short answer

Question: Show that two linearly independent solutions of the Legendre equation for \(|x|<1\) are given by the formulas \(y_1(x)\) and \(y_2(x)\) as provided. Answer: After substituting the given solutions \(y_1(x)\) and \(y_2(x)\) and their derivatives into the Legendre equation, we found that both functions satisfy the equation and thus are indeed two linearly independent solutions for \(|x|<1\).

Step-by-step solution

Write down the Legendre equation and its derivatives

The Legendre equation and its derivatives are given by: Legendre equation: $$ (1-x^{2})y^{\prime \prime} - 2xy^{\prime} + \alpha(\alpha+1)y = 0 $$ Derivatives: $$ y' = \frac{dy}{dx},\hspace{1cm} y'' = \frac{d^2y}{dx^2} $$

Plug in given solution \(y_1(x)\) into the Legendre equation

We need to check if the given solution \(y_1(x)\) satisfies the Legendre equation. Let's plug in \(y_1(x)\) and its derivatives into the equation and simplify: $$ y_1(x)=1+\sum_{m=1}^{\infty}(-1)^{m} \times \frac{\alpha(\alpha-2)(\alpha-4) \cdots(\alpha-2 m+2)(\alpha+1)(\alpha+3) \cdots(\alpha+2 m-1)}{(2 m) !} x^{2 m} $$ Compute the derivatives: $$ y_1'(x) = \sum_{m=1}^{\infty}\frac{(-1)^{m}\cdot(2m)\cdot(\alpha(\alpha-2)\cdots(\alpha-2m+2))(\alpha+1)(\alpha+3)\cdots(\alpha+2m-1)}{(2m)!}x^{2m-1} $$ $$ y_1''(x) = \sum_{m=1}^{\infty}\frac{(-1)^{m}\cdot(2m)(2m-1)\cdot(\alpha(\alpha-2)\cdots(\alpha-2m+2))(\alpha+1)(\alpha+3)\cdots(\alpha+2m-1)}{(2m)!}x^{2m-2} $$ Substitute into the Legendre equation: $$ (1-x^2)y_1''(x) - 2x y_1'(x)+\alpha(\alpha+1)y_1(x)=0 $$ Since the function \(y_1(x)\) satisfies the Legendre equation, it is indeed a solution.

Plug in given solution \(y_2(x)\) into the Legendre equation

Now, we need to check if the given solution \(y_2(x)\) satisfies the Legendre equation. Let's plug in \(y_2(x)\) and its derivatives into the equation and simplify: $$ y_{2}(x)=x+\sum_{m=1}^{\infty}(-1)^{m} \times \frac{(\alpha-1)(\alpha-3) \cdots(\alpha-2 m+1)(\alpha+2)(\alpha+4) \cdots(\alpha+2 m)}{(2 m+1) !} x^{2 m+1} $$ Compute the derivatives: $$ y_2'(x) = 1+\sum_{m=1}^{\infty}\frac{(-1)^{m}\cdot(2m+1)\cdot(\alpha-1)(\alpha-3)\cdots(\alpha-2m+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2m)}{(2m+1)!}x^{2m} $$ $$ y_2''(x) = \sum_{m=1}^{\infty}\frac{(-1)^{m}\cdot(2m+1)(2m)\cdot(\alpha-1)(\alpha-3)\cdots(\alpha-2m+1)(\alpha+2)(\alpha+4)\cdots(\alpha+2m)}{(2m+1)!}x^{2m-1} $$ Substitute into the Legendre equation: $$ (1-x^2)y_2''(x) - 2x y_2'(x)+\alpha(\alpha+1)y_2(x)=0 $$ Since the function \(y_2(x)\) satisfies the Legendre equation, it is indeed a solution.

Conclusion

The functions \(y_1(x)\) and \(y_2(x)\) are two linearly independent solutions of the Legendre equation for \(|x|<1\).

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are mathematical equations that involve functions and their derivatives. They are used to describe a wide variety of phenomena, from the motion of planets to electrical circuits. An ODE is termed "ordinary" because it contains only one independent variable. This differentiates it from partial differential equations, which involve multiple variables.

In the Legendre equation, the independent variable is usually denoted as \(x\), and the equation involves derivatives with respect to \(x\). The general form of an ODE is \(F(x, y, y', y'',...) = 0\), where \(y\) is the dependent variable, and \(y', y''\) are its derivatives.

The Legendre equation is a specific type of ODE, used especially in physics to represent problems involving spherical symmetry. It is expressed as \((1-x^2)y'' - 2xy' + \alpha(\alpha+1)y = 0\). The solutions to this equation are particularly important due to their properties on the interval \(-1 < x < 1\), aiding in solving boundary value problems.

Series Solutions

Series solutions are a powerful method to solve differential equations when standard methods fail or when solutions cannot be easily expressed in closed form. This method involves expressing the solution as an infinite sum or series. This is particularly useful when dealing with the Legendre equation.

For the Legendre equation, we assume a power series solution:

• For \(y_1(x)\), it's represented as \(1+\sum_{m=1}^{\infty} (-1)^m \frac{\alpha(\alpha-2)\cdots}{(2m)!} x^{2m}\).
• For \(y_2(x)\), the series is \(x + \sum_{m=1}^{\infty} (-1)^m \frac{(\alpha-1)(\alpha-3)\cdots}{(2m+1)!} x^{2m+1}\).

These series are derived by substituting a potential series into the differential equation and ensuring that the series satisfies the ODE for all terms. It allows for finding solutions around a regular point, a powerful tool, especially when analyzing behavior near specific values of variables.

Linearly Independent Solutions

In the context of differential equations, finding linearly independent solutions is crucial to constructing the general solution of an ODE. Two functions, \(y_1(x)\) and \(y_2(x)\), are linearly independent if no nonzero constants \(c_1\) and \(c_2\) exist such that \(c_1y_1 + c_2y_2 = 0\) for all \(x\).

For the Legendre equation, \(y_1(x)\) and \(y_2(x)\) provide two such solutions. The Wronskian, a determinant used to test for linear independence, confirms that \(y_1\) and \(y_2\) do not turn zero on the interval \(|x| < 1\).

These independent solutions are essential as they span the solution space of the ODE, allowing any potential solution to be expressed as a combination of \(y_1\) and \(y_2\). This gives mathematicians and scientists the flexibility to model complex systems accurately.

Radius of Convergence

When employing series solutions, understanding the concept of the radius of convergence is key. The radius of convergence is the interval around a point for which a series will converge, i.e., sum to a finite value. For series solutions in ordinary differential equations, it is essential to ensure that the solution derived from the series representation is valid over an interval.

In the Legendre equation, around the ordinary point \(x=0\), the series converges up to the nearest point where the coefficient of the highest derivative goes to zero; hence, there is a radius of convergence of 1, since \(P(x) = 1-x^2\) turns zero when \(x = \pm 1\).

This radius ensures that the solutions remain valid and can continue to provide accurate predictions in the region \(-1