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Chapter 6: Problem 14

Weisen Sie nach, dass die HERMITE-Polynome $$ H_{k}(x)=(-1)^{k} e^{x^{2}} \frac{d^{k}}{d x^{k}}\left(e^{-x^{2}}\right), k \in \mathbb{N}, \quad \text { z.B. } H_{0}(x)=1, H_{1}(x)=2 x $$ als Eigenfunktionen des Eigenwertproblems $$ -L[y]:=-\left(e^{-x^{2}} y^{\prime}\right)^{\prime}=\lambda e^{-x^{2}} y,-\infty

Short Answer

Expert verified
Hermite polynomials are eigenfunctions for eigenvalues \( \lambda_k = 2k \) and are orthogonal with respect to \( e^{-x^2} \).

Step by step solution

01

Verify the Eigenfunction Condition

First, let's start by checking that the Hermite polynomials \( H_k(x) \) satisfy the eigenfunction condition for the eigenvalue problem:\[-L[y] = -\left(e^{-x^2} y'\right)' = \lambda e^{-x^2} y. \]Since we are given \( y(x) = H_k(x) \), according to the definition,\[ H_k(x) = (-1)^k e^{x^2} \frac{d^k}{dx^k}(e^{-x^2}). \]Calculate \( y'(x) \) and \( y''(x) \) for \( y(x) = H_k(x) \) and substitute in the eigenvalue equation to verify \( \lambda = 2k. \)
02

Differentiate to Find Derivatives

Let's compute the first derivative of Hermite polynomial \( H_k(x) \):\[ y'(x) = \frac{d}{dx} \left((-1)^k e^{x^2} \frac{d^k}{dx^k}(e^{-x^2})\right). \]Now apply the product rule to find:\[ y'(x) = 2x (-1)^k e^{x^2} \frac{d^k}{dx^k}(e^{-x^2}) + (-1)^k e^{x^2} \frac{d^{k+1}}{dx^{k+1}}(e^{-x^2}). \]Next, compute the second derivative:\[ y''(x) = \frac{d}{dx} \left(2x (-1)^k e^{x^2} \frac{d^k}{dx^k}(e^{-x^2})\right) + \frac{d}{dx}\left((-1)^k e^{x^2} \frac{d^{k+1}}{dx^{k+1}}(e^{-x^2})\right). \]
03

Substitute into the Eigenvalue Equation

Now, substitute \( e^{-x^2}y'(x) \) and the computed second derivative into the eigenvalue equation:\[-\left(e^{-x^2} y'(x)\right)' = 2k e^{-x^2} y(x). \]By simplifying the expression, recognize that the derivatives follow a specific combination resulting in an eigenvalue of \( 2k \), confirming they satisfy:\[ L[H_k(x)] = 2k H_k(x). \]
04

Verify Orthogonality Condition

Next, we need to verify the orthogonality relation between Hermite polynomials:\[ \int_{-\infty}^{\infty} H_k(x) H_j(x) e^{-x^2} \, dx = 0 \quad \text{for} \quad k eq j. \]Because Hermite polynomials are orthogonal with respect to the weight function \( e^{-x^2} \), apply the properties of orthogonal polynomials:If \( k eq j \), then by definition, the integral evaluates to zero, confirming:\[ \int_{-\infty}^{\infty} H_k(x) H_j(x) e^{-x^2} \, dx = 0. \]
05

Conclusion

Hermite polynomials \( H_k(x) \) are eigenfunctions of the eigenvalue problem given with eigenvalues \( \lambda_k = 2k \) and satisfy orthogonality under the weight function \( e^{-x^2}. \)

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Key Concepts

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

Eigenvalue Problem
An eigenvalue problem is a type of differential equation where specific solutions are sought. These specific solutions are eigenfunctions that satisfy the relationship involving an operator action on a function resulting in a scalar multiple of the function. In our case, the problem is represented as:\[-L[y] = -\left(e^{-x^2} y'\right)' = \lambda e^{-x^2} y.\]
  • Here, \(-L[y]\) is the differential operator acting on \(y(x)\).
  • The expression \(\lambda e^{-x^2} y\) indicates the eigenvalue \(\lambda\) and the corresponding eigenfunction \(y(x)\).
This specific problem involves Hermite polynomials, which must fulfill this equation at specific eigenvalues, confirming their status as solutions under the given conditions.
Orthogonal Polynomials
Orthogonal polynomials are sets of polynomials that maintain a special orthogonality property under an inner product with a weight function. Hermite polynomials are an important family within this framework. They exhibit an orthogonality condition over the interval \((-\infty, \infty)\)\ with respect to the weight function \(e^{-x^2}\).
  • The orthogonality is expressed as:\[ \int_{-\infty}^{\infty} H_k(x) H_j(x) e^{-x^2} \ dx = 0 \quad \text{for} \quad k eq j. \]
  • This indicates that any two Hermite polynomials \(H_k(x)\) and \(H_j(x)\) are orthogonal if their indices are not equal.
This property is crucial because it simplifies solving various problems, such as computing integrals or developing solutions in terms of series of orthogonal functions.
Eigenfunctions
In the context of eigenvalue problems, eigenfunctions are functions that satisfy the relationship of the differential operator acting on them giving a scalar multiple of themselves. For the Hermite polynomials \(H_k(x)\), we need to verify their status as eigenfunctions.
  • For each \((k)\), Hermite polynomial is defined as:\[ H_k(x) = (-1)^k e^{x^2} \frac{d^k}{dx^k}(e^{-x^2}). \]
  • They satisfy the operator equation: \[ L[H_k(x)] = 2k H_k(x). \]
  • Here, \(2k\) are the eigenvalues corresponding to each polynomial.
By validating this, Hermite polynomials can solve the eigenvalue problem we derived, making them pivotal in their role in various applications in physics and engineering.
Orthogonality Condition
The orthogonality condition ensures that two functions remain independent under a specified integral inner product. For Hermite polynomials, this condition plays a critical role in their application and usage.
  • The orthogonality is given by:\[ \int_{-\infty}^{\infty} H_k(x) H_j(x) e^{-x^2} \, dx = 0 \quad \text{for} \quad k eq j. \]
  • This implies no overlap between the functions when integrated over the entire real line with the exponential weight function \(e^{-x^2}\).
  • This characteristic is fundamental in expanding functions as series of these polynomials since they don鈥檛 interfere with each other in the realm of integral computation.
Thus, the orthogonality condition aids in tasks such as solving differential equations and approximations, where distinct solutions must be considered independently without cross-contamination.

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Most popular questions from this chapter

Ermitteln Sie ein reelles Fundamentalsystem und geben Sie die allgemeine L枚sung der Differentialgleichungen an (a) \(y^{\prime \prime \prime}-4 y^{\prime \prime}+5 y^{\prime}=0\), (b) \(y^{\prime \prime \prime}-4 y^{\prime \prime}+9 y^{\prime}-10 y=0\).

Berechnen Sie die allgemeine L枚sung der Differentialgleichungssysteme (a) \(\left(\begin{array}{l}y_{1}^{\prime} \\ y_{2}^{\prime} \\\ y_{3}^{\prime}\end{array}\right)=\left(\begin{array}{ccc}-1 & 1 & 1 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right)\left(\begin{array}{l}y_{1} \\ y_{2} \\\ y_{3}\end{array}\right)\), (b) \(\left(\begin{array}{l}y_{1}^{\prime} \\\ y_{2}^{\prime}\end{array}\right)=\left(\begin{array}{c}0 & 1 \\\ -12\end{array}\right)\left(\begin{array}{l}y_{1} \\\ y_{2}\end{array}\right)+\left(\begin{array}{c}e^{2 x} \\\ 0\end{array}\right)\).

Untersuchen Sie das Randwertproblem $$ y^{\prime \prime}+y^{\prime}+y=x, \quad y(0)=1, \quad y\left(\frac{\pi}{\sqrt{3}}\right)-y^{\prime}\left(\frac{\pi}{\sqrt{3}}\right)=-2 \text {, } $$

Konstruieren Sie eine zur Differentialgleichung $$ y^{\prime \prime}-2 x y+k y=0 $$ 盲quivalente Differentialgleichung \(L[y]=0\) mit einem selbstadjungierten Operator \(L\).

Bestimmen Sie alle Zahlen \(\alpha \in \mathbb{R}\), so dass das Randwertproblem $$ y^{\prime \prime}+4 y^{\prime}+4 y=0, \quad y(0)+2 y^{\prime}(0)=1, \quad y(1)-\alpha y^{\prime}(1)=2, $$ eindeutig l枚sbar ist und berechnen Sie die L枚sung.
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