Question

The Laguerre \({ }^{10}\) differential equation \(t y^{\prime \prime}+(1-t) y^{\prime}+\alpha y=0\) has a regular singular point at \(t=0\). (a) Determine the indicial equation and show that the roots are \(\lambda_{1}=\lambda_{2}=0\). (b) Find the recurrence relation. Show that if \(\alpha=N\), where \(N\) is a nonnegative integer, then the series solution reduces to a polynomial. Obtain the polynomial solution when \(N=5\). The polynomial solutions of this differential equation, when properly normalized, are called Laguerre polynomials. (c) Is the polynomial obtained in part (b) for \(\alpha=N=5\) an even function, an odd function, or neither? Would you expect even and odd solutions of the differential

Short answer

The indicial equation for the given Laguerre differential equation is: $$\lambda^2-\lambda+\alpha = 0$$ b) What is the polynomial solution for the provided differential equation when α = 5? The polynomial solution when α = 5 is: $$y(t) = a_0 \left(1 - \frac{1}{6}t^2 + \frac{1}{180}t^4 \right)$$ c) Is the polynomial solution found in part (b) even, odd, or neither? The polynomial solution found in part (b) is an even function.

Step-by-step solution

Rewrite the differential equation

First, we can rewrite the given Laguerre differential equation in a more compact form, and also write it in terms of the Frobenius method: $$t y'' + (1-t)y' + \alpha y = 0$$ $$\Rightarrow t\frac{\mathrm{d}^2y}{\mathrm{d}t^2}+(1-t)\frac{\mathrm{d}y}{\mathrm{d}t}+\alpha y=0$$ We know that the Frobenius method uses a series solution of the form: $$y(t) = \sum_{n=0}^{\infty} a_n t^{n+\lambda}$$

Calculate derivatives and plug into the differential equation

To apply the Frobenius method, we differentiate y(t) with respect to t: $$y'(t) = \sum_{n=0}^{\infty} (n+\lambda) a_n t^{n+\lambda-1}$$ $$y''(t) = \sum_{n=0}^{\infty} (n+\lambda)(n+\lambda-1) a_n t^{n+\lambda-2}$$ Now plug these expressions into the differential equation: $$t\sum_{n=0}^{\infty} (n+\lambda)(n+\lambda-1) a_n t^{n+\lambda-2} +(1-t)\sum_{n=0}^{\infty} (n+\lambda) a_n t^{n+\lambda-1}+\alpha\sum_{n=0}^{\infty} a_n t^{n+\lambda}=0$$

Determine the indicial equation and roots

In order to determine the indicial equation, we need to examine the coefficient of the lowest power of t in the equation. We start by simplifying the equation and then analyzing the terms: $$\sum_{n=0}^{\infty} (n+\lambda)(n+\lambda-1) a_n t^{n+\lambda-1} +(1-t)\sum_{n=0}^{\infty} (n+\lambda) a_n t^{n+\lambda-1}+\alpha\sum_{n=0}^{\infty} a_n t^{n+\lambda}=0$$ For the lowest power of t, consider n=0: $$\lambda(\lambda-1)a_0t^\lambda + \lambda a_0t^\lambda - \lambda a_0t^{\lambda+1}+\alpha a_0t^\lambda = 0$$ Since a_0 is not zero, we can divide the equation by a_0t^\lambda: $$\lambda(\lambda-1)+\lambda-\lambda t+\alpha = 0$$ This is our indicial equation, setting t = 0: $$\lambda(\lambda-1)+\lambda+\alpha=0 \Rightarrow \lambda^2-\lambda+\alpha = 0$$ To show that the roots are both zero, we can use the quadratic formula, and see that this is a perfect square: $$\lambda_1=\lambda_2=0$$

Find the recurrence relation

Now that we have the indicial equation and the roots, we can find the recurrence relation: Comparing the coefficients of the powers of t: $$((n+\lambda)(n+\lambda-1)-\lambda) a_n + (n+\lambda+1-\alpha) a_{n+1} = 0$$ Solving for a_(n+1): $$a_{n+1} = -\frac{(n+\lambda)(n+\lambda-1)-\lambda}{n+\lambda+1-\alpha} a_n$$

Find the polynomial solution for α=5

When α is a nonnegative integer, the series solution reduces to a polynomial. For α=5=N, the recurrence relation is: $$a_{n+1} = -\frac{n(n-1)}{n+6} a_n$$ We can now generate the polynomial solution: $$y(t) = a_0 + a_1t + a_2t^2 + a_3t^3 + a_4t^4 + a_5t^5$$ Using the recurrence relation, we can find a_1, a_2, ..., a_5: $$a_1 = 0$$ $$a_2 = -\frac{1}{6}a_0$$ $$a_3 = 0$$ $$a_4 = \frac{1}{180}a_0$$ $$a_5 = 0$$ So the polynomial solution is: $$y(t) = a_0 \left(1 - \frac{1}{6}t^2 + \frac{1}{180}t^4 \right)$$ These are called Laguerre polynomials.

Determine if the polynomial solution is even, odd or neither

To check if the polynomial solution is even, odd or neither, we test the function for evenness and oddness: $$y(-t)=a_0 \left(1 - \frac{1}{6}(-t)^2 + \frac{1}{180}(-t)^4 \right) = y(t)$$ Since y(-t) = y(t), the polynomial solution is even. As the polynomial solution with α=N=5 is even, we expect solutions to oscillate between even and odd as α increases. However, we cannot conclude that all solutions will be even or odd, as this depends on the specific conditions of the problem.

Key concepts

Differential Equations

Understanding differential equations is crucial when studying mathematical models that describe real-world phenomena. These equations involve an unknown function and its derivatives, and they can be used to determine the behavior of various systems, from physics to finance. The equation presented in the exercise, the Laguerre differential equation, is a second-order linear differential equation, where the solution describes an important class of orthogonal polynomials in mathematics. The equation contains both the function itself, represented by y, and its derivatives y' and y''. Solving such an equation typically requires finding a function y(t) that satisfies the given relationship between the function and its derivatives for all values within a certain domain.

Frobenius Method

The Frobenius method offers a way to tackle differential equations with irregular singular points by using a series solution approach. It is particularly useful when the solution around the singular point can be expressed as a power series with a variable coefficient. The essence of the method is expanding the solution in terms of an infinite sum of powers, as shown in the step by step solution. The key aspect to notice is the introduction of a shift by λ in the power series, which is determined later by solving the indicial equation. The method provides a systematic procedure to obtain series solutions, making it a powerful tool in the analysis of differential equations where traditional methods might fail.

Series Solution

A series solution to a differential equation is an expression of the solution as an infinite sum of terms that are functions of the independent variable. In this case, the series solution for the Laguerre differential equation takes the form of a power series with coefficients a_n. The series solution is particularly helpful because it allows us to approximate the solution to any desired degree of accuracy by calculating enough terms of the series. When coefficients of higher powers vanish, as shown in the solution when α=N, we obtain a finite polynomial, which offers a precise and simplified representation of the solution.

Recurrence Relation

In mathematics, a recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms. In the context of series solutions, they help us express coefficients (like a_n) in terms of previous coefficients in the series. The recurrence relation derived from the Laguerre differential equation allows the calculation of all the coefficients in the series based on initial values. As mentioned in the exercise, for certain values of α, especially nonnegative integers, the recurrence relation can simplify significantly, leading to a termination of the series and the emergence of a polynomial solution, known as a Laguerre polynomial. This recurrence relation is central to generating the terms of the polynomial and understanding the nature of the solution.