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Chapter 11: Problem 8

Deal with Hermite's equation: $$y^{\prime \prime}-2 x y^{\prime}+2 \alpha y=0, \quad-\infty

Short Answer

Expert verified
The two linearly independent series solutions of Hermite's equation $y'' - 2xy' + 2\alpha y = 0$ centered at $x = 0$ can be obtained by using the power series method. The solutions are given by: \(y_1(x) = a_0 \left(1 - \dfrac{\alpha x^2}{1!} + \dfrac{(\cdots) x^4}{2!} + \cdots\right)\) \(y_2(x) = a_1 \left(1 + \dfrac{2 x^3}{3!} + \cdots\right)\) Where $a_0$ and $a_1$ are arbitrary constants and the coefficients $a_n$ are obtained using the recurrence relation: \(a_{n+2} = \dfrac{2(n+1)a_{n+1} - 2\alpha a_n}{(n+2)(n+1)}\)

Step by step solution

01

Assume a power series solution

Assume a power series solution for \(y(x)\): \(y(x) = \sum_{n=0}^\infty a_nx^n\)
02

Compute the derivatives

Compute the first and second derivatives of \(y(x)\): \(y'(x) = \sum_{n=1}^\infty na_nx^{n-1}\) \(y''(x) = \sum_{n=2}^\infty n(n-1)a_nx^{n-2}\)
03

Plug the derivatives into the equation

Substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into Hermite's equation: \(\sum_{n=2}^\infty n(n-1)a_nx^{n-2} - 2x\sum_{n=1}^\infty na_nx^{n-1} + 2\alpha\sum_{n=0}^\infty a_nx^n = 0\)
04

Change the indices of the sums

Rewrite each sum with the same index and power to remove the sums and group terms according to the power of \(x\): \(\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - 2\sum_{n=0}^\infty (n+1)a_{n+1}x^n + 2\alpha\sum_{n=0}^\infty a_nx^n = 0\)
05

Equate the coefficients

Since the equation must hold for all \(x\), the coefficients of corresponding powers of \(x\) must be equal: \((n+2)(n+1)a_{n+2} - 2(n+1)a_{n+1} + 2\alpha a_n = 0\) Now we have a recurrence relation for the coefficients \(a_n\).
06

Simplify the recurrence relation

Solve for the coefficient \(a_{n+2}\) in terms of \(a_n\): \(a_{n+2} = \dfrac{2(n+1)a_{n+1} - 2\alpha a_n}{(n+2)(n+1)}\)
07

Compute the first few coefficients

Use the recurrence relation to compute the first few coefficients: \(a_0\), \(a_1\) are arbitrary constants. \(a_2 = \dfrac{-2\alpha a_0}{2}\) \(a_3 = \dfrac{2a_1}{3}\) \(a_4 = \dfrac{6a_2 - 8\alpha a_1}{12}\) \(\cdots\)
08

Determine the series solutions

Use the coefficients found above to determine two linearly independent series solutions for Hermite's equation: \(y_1(x) = a_0 \left(1 - \dfrac{\alpha x^2}{1!} + \dfrac{(\cdots) x^4}{2!} + \cdots\right)\) \(y_2(x) = a_1 \left(1 + \dfrac{2 x^3}{3!} + \cdots\right)\) Where \(y_1(x)\) and \(y_2(x)\) are two linearly independent series solutions to Hermite's equation centered at \(x = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series Solution
Understanding the power series solution is essential for tackling complex differential equations like Hermite's equation. A power series solution is an expression of a function as an infinite sum of terms, each of which is a multiple of a power of the variable. In this case, we express the solution for Hermite's equation as

define{y(x)}{power series} For Hermite's equation, the assumption is that the solution can be expressed as such a series centered at the point x = 0 (also called a Maclaurin series). This gives us a flexible, structured approach for finding a solution to the differential equation.
One key advantage of the power series solution method is that it allows us to address differential equations which may not have elementary function solutions. Once we have our series in the form of a power series, we can derive derivatives easily, which lends itself to simplifying the process of substituting into the original differential equation.
Recurrence Relation in Differential Equations
A recurrence relation in the context of differential equations is a relationship that defines each coefficient in a power series in terms of one or more previous coefficients. It鈥檚 a critical tool because it allows us to systematically determine each term of our series solution without having to solve the differential equation directly each time. In Hermite's equation, the recurrence relationship was established by matching coefficients from both sides of the equation after substituting the assumed power series and its derivatives. This relationship helps us determine the entire sequence of coefficients based on initial values, effectively building the solution piece by piece.

The recurrence relation also provides insights into the nature of the solution and the behavior of the coefficients as they progress. Understanding this relationship is crucial, as specific properties or patterns within the recurrence can greatly simplify the work necessary to find the complete solution, especially for large values of n.
Series Solutions to Differential Equations
Series solutions are potent as they can provide us with a way to express solutions to differential equations that cannot be solved by standard methods. The approach involves finding the coefficients of a power series that satisfy the equation. We use the power series and its derivatives, substituting them into the given differential equation. The result is then organized by powers of x, leading us to establish a recurrence relation for the coefficients.

In the case of Hermite's equation, through careful manipulation, we were able to find a recurrence relation that allowed us to compute coefficients for the entire power series systematically. The fact that there are two arbitrary constants (a_0 and a_1) indicates that we would get a family of solutions, representing the general solution to the differential equation. By choosing different values for a_0 and a_1, we can generate two linearly independent solutions, which is crucial for covering the spectrum of potential solutions to the original differential equation.

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Most popular questions from this chapter

(a) Determine all values of \(x\) at which the function \(f(x)=\frac{1}{x^{2}-1}\) is analytic. (b) Determine the radius of convergence of a power series representation of the function \((11.1 .6)\) centered at \(x=x_{0} .\) (You will need to consider the cases \(-11\) separately.

Consider the differential equation $$x^{2} y^{\prime \prime}+x(1+b x) y^{\prime}+\left[b(1-N) x-N^{2}\right] y=0, x>0, \quad \quad (11.7.13)$$ where \(N\) is a positive integer and \(b\) is a constant. (a) Show that the roots of the indicial equation are \(r=\pm N.\) (b) Show that the Frobenius series solution corresponding to \(r=N\) is $$y_{1}(x)=a_{0} x^{N} \sum_{n=0}^{\infty} \frac{(2 N) !(-b)^{n}}{(2 N+n) !} x^{n}$$ and that by an appropriate choice of \(a_{0},\) one solution to ( 11.7 .13 ) is $$y_{1}(x)=x^{-N}\left[e^{-b x}-\sum_{n=0}^{2 N-1} \frac{(-b x)^{n}}{n !}\right].$$ (c) Show that Equation (11.7.13) has a second linearly independent Frobenius series solution that can be taken as $$y_{2}(x)=x^{-N} \sum_{n=0}^{2 N-1} \frac{(-b x)^{n}}{n !}.$$ Hence, conclude that Equation (11.7.13) has linearly independent solutions $$y_{1}(x)=x^{-N} e^{-b x}, y_{2}(x)=x^{-N} \sum_{n=0}^{2 N-1} \frac{(-b x)^{n}}{n !}.$$

By integrating the recurrence relation for derivatives of the Bessel functions of the first kind, show that (a) \(\int x^{p} J_{p-1}(x) d x=x^{p} J_{p}(x)+C\) (b) \(\int x^{-p} J_{p+1}(x) d x=-x^{-p} J_{p}(x)+C\)

(a) Determine a series solution to the initial-value problem $$4 y^{\prime \prime}+x y^{\prime}+4 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0$$ (b) Find a polynomial that approximates the solution to Equation (11.2.19) with an error less than \(10^{-5}\) on the interval [-1,1] [Hint: The series obtained is a convergent alternating series.

Determine two linearly independent power series solutions to the given differential equation centered at \(x=0 .\) Give a lower bound on the radius of convergence of the series solutions obtained. $$\left(1-4 x^{2}\right) y^{\prime \prime}-20 x y^{\prime}-16 y=0.$$
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