Problem 9 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

For the given Hermite's equation with \(\alpha = N\) (positive integer), we have the polynomial solutions as follows: 1. For \(\alpha = 0\): \(y(x) = a_0 + x^2\) 2. For \(\alpha = 1\): \(y(x) = a_1x + \frac{1}{3}x^3\) 3. For \(\alpha = 2\): \(y(x) = a_0 + x^2 + \frac{1}{6}x^4\) 4. For \(\alpha = 3\): \(y(x) = a_1x + \frac{1}{3}x^3 + \frac{1}{20}x^5\) Where \(a_0\) and \(a_1\) are constants.

Step 1: Assume a Polynomial Solution

Let's assume that there exists a polynomial solution for Hermite's equation in the form: \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) where \(a_n\) are the coefficients of the polynomial.

Step 2: Find \(y'\) and \(y''\)

We need to compute the first and second derivatives of the assumed solution to be able to plug it into the Hermite's equation and study the coefficients. For the first derivative we have: \(y'(x) = \sum_{n=1}^{\infty} na_n x^{n-1}\) And for the second derivative: \(y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}\)

Step 3: Replace y, \(y'\) and \(y''\) in Hermite's Equation

Now, let's replace our polynomial and its derivatives in Hermite's Equation: \(\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} - 2x\sum_{n=1}^{\infty} na_n x^{n-1} + 2\alpha\sum_{n=0}^{\infty}a_n x^n = 0\)

Step 4: Shift the Indices

We can now change the indices to make the powers equal so that we can simplify the equation further. For \(y''\), we can substitute \(n-2 = m\), which gives: \(\sum_{m=0}^{\infty} (m+2)(m+1)a_{m+2} x^m\) For \(y'\), we can substitute \(n-1 = m\), which gives: \(\sum_{m=0}^{\infty} 2(m+1)a_{m+1} x^{m}\) Rewriting the Hermite equation, we have: \(\sum_{m=0}^{\infty}[(m+2)(m+1)a_{m+2} - 2(m+1)a_{m+1}x + 2\alpha a_m]x^m = 0\)

Step 5: Find the Recursion Relation

As the sum must be zero for all m, we have: \((m+2)(m+1)a_{m+2} - 2(m+1)a_{m+1}x + 2\alpha a_m = 0\) Now, we find the recursion relation: \(a_{m+2} = \frac{2x(m+1)a_{m+1} - 2\alpha a_m}{(m+2)(m+1)}\) If \(\alpha = N\) (a positive integer), then the recursion relation can terminate, yielding a polynomial solution.

Step 6: Determine the Polynomial Solutions

We will now find the polynomial solutions for \(\alpha = 0, 1, 2, 3\). #### For \(\alpha = 0\): We have the recursion relation as follows: \(a_{m+2} = \frac{2x(m+1)a_{m+1}}{(m+2)(m+1)}\) For \(m = 0\), we have \(a_2 = 1\), and for \(m > 0\), we have \(a_{m+2} = 0\). Thus, the polynomial solution becomes: \(y(x) = a_0 + a_2 x^2 = a_0 + x^2\) #### For \(\alpha = 1\): With the same recursion relation, we can determine that for \(m = 1\), we have \(a_3 = \frac{1}{3}\) and for \(m > 1\), \(a_{m+2} = 0\). The polynomial solution for this case is: \(y(x) = a_1x + a_3 x^3 = a_1x + \frac{1}{3}x^3\) #### For \(\alpha = 2\): Again, using the recursion relation, we can determine that for \(m = 2\), we have \(a_4 = \frac{2}{12}\) and for \(m > 2\), \(a_{m+2} = 0\). The polynomial solution for this case is: \(y(x) = a_0 + a_2 x^2 + a_4 x^4 = a_0 + x^2 + \frac{1}{6}x^4\) #### For \(\alpha = 3\): Finally, using the recursion relation, we can determine that for \(m = 3\), we have \(a_5 = \frac{3}{60}\) and for \(m > 3\), \(a_{m+2} = 0\). The polynomial solution for this case is: \(y(x) = a_1x + a_3 x^3 + a_5x^5 = a_1x + \frac{1}{3}x^3 + \frac{1}{20}x^5\) Thus, we have found the polynomial solutions for \(\alpha = 0, 1, 2, 3\).

Key Concepts

Polynomial Solutions

Polynomial solutions are a type of solution for differential equations where the function is expressed as a polynomial. In the context of Hermite's equation, we hope to find solutions that are polynomials. This means that the function \( y(x) \) is expressed as a finite sum:

• \( y(x) = a_0 + a_1 x + a_2 x^2 + \, \dots\,, a_n x^n \)

When \( \alpha = N \), a positive integer, Hermite’s equation can be solved such that \( y(x) \) results in a polynomial. By substituting these solutions back into the original equation, it simplifies to an identity, confirming that our assumptions were correct.
Polynomial solutions are important because they represent cases where the behavior of the function is simple and well-understood, as polynomials are among the most fundamental expressions in mathematics.

Recursion Relation

The recursion relation is a critical step in solving certain types of equations, as it helps to determine the coefficients of the polynomial solution iteratively. In this exercise, after substituting the assumed polynomial and its derivatives into Hermite's equation, we arrived at such a relation. The recursion relation used here is:

• \( a_{m+2} = \frac{2x(m+1)a_{m+1} - 2\alpha a_m}{(m+2)(m+1)} \)

This formula allows us to compute each coefficient \( a_{m+2} \) based on the previously found coefficients \( a_{m+1} \) and \( a_m \).
Using the recursion relation, we can find each coefficient step by step, beginning from a known starting point, typically \( a_0 \) or \( a_1 \), which is often determined by boundary or initial conditions. Moreover, when \( \alpha \) is a positive integer, the recursion terminates, signifying that only a finite number of terms remain, confirming the polynomial nature of the solution.

Differential Equations

Differential equations involve functions and their derivatives, describing how a particular function changes in relation to one or more variables. Hermite's equation is a second-order linear differential equation. It takes the form:

• \( y'' - 2xy' + 2\alpha y = 0 \)

Such equations are used extensively in various fields like physics, engineering, and mathematics to model complex systems and processes. Solving a differential equation like Hermite's gives insights into the system it's modeling.
In Hermite's equation, if \( \alpha \) is a positive integer, solving the equation yields solutions that are simple polynomials. This is particularly advantageous because it leads to easily understandable solutions that are practical in mathematical modeling.

Second Derivative

The second derivative, symbolized as \( y'' \), is a major component of Hermite's equation. It represents the rate of change of the rate of change of a function or simply the curvature of the function \( y(x) \).
In solving Hermite's equation, the second derivative is calculated from the assumed polynomial function \( y(x) \). We found:

• \( y''(x) = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} \)

Computing the second derivative is essential because it allows us to substitute back into the differential equation and proceed with finding the polynomial solutions.
In terms of applications, understanding the second derivative helps in analyzing the behavior of the function, especially in physical systems where it might represent acceleration or curvature. In the context of Hermite's equation, it plays a pivotal role in obtaining polynomial solutions.