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Chapter 5: Problem 21

The equation $$y^{\prime \prime}-2 x y^{\prime}+\lambda y=0, \quad-\infty

Short Answer

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Question: Determine the first four terms in each of two linearly independent solutions of the Hermite equation, using a power series method about \(x=0\). Obtain the polynomial solutions for given nonnegative even integer values of \(\lambda\), and find the Hermite polynomials \(H_0(x), \ldots, H_5(x)\). Answer: The first four terms of the two linearly independent solutions are: 1) \(y_1(x) = 1 - \frac{\lambda}{2}x^2 + \frac{(4-\lambda)(-\lambda)}{12}x^4 + \cdots\) 2) \(y_2(x) = x - \frac{\lambda}{6}x^3 + \cdots\) The polynomial solutions for given nonnegative even integer values of \(\lambda\) are: 1) for \(\lambda=0\): \(y(x) = 1\) 2) for \(\lambda=2\): \(y(x) = 1 - 2x^2\) 3) for \(\lambda=4\): \(y(x) = 1 - 4x^2 + 8x^4\) 4) for \(\lambda=6\): \(y(x) = x - 2x^3\) 5) for \(\lambda=8\): \(y(x) = x - 4x^3 + 16x^5\) 6) for \(\lambda=10\): \(y(x) = x - 6x^3 + 60x^5 - \frac{320}{3}x^7\) The Hermite Polynomials \(H_0(x), \ldots, H_5(x)\) are: 1) \(H_0(x) = 1\) 2) \(H_1(x) = 2x\) 3) \(H_2(x) = 4 - 8x^2\) 4) \(H_3(x) = 8x - 48x^3\) 5) \(H_4(x) = 16 - 192x^2 + 960x^4\) 6) \(H_5(x) = 32x - 640x^3 + 5120x^5\)

Step by step solution

01

(a) Finding Two Linearly Independent Solutions Using Power Series Method

To find the general solution of the given differential equation, we will use the Frobenius method (power series solution). Assume the solution in the form of a power series as follows: $$y(x)=\sum_{n=0}^{\infty} a_n x^n$$ Taking the first and second derivative, we have: $$y'(x)=\sum_{n=1}^{\infty} na_n x^{n-1}$$ $$y''(x)=\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}$$ Now, substitute these derivatives into the Hermite equation:
02

Substituting Power Series into Hermite Equation

$$\sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} - 2x\sum_{n=1}^{\infty}na_n x^{n-1}+\lambda\sum_{n=0}^{\infty}a_n x^n = 0$$ Now, we will perform term matching by first multiplying the x values, then shifting the indices to match the powers of x, and then equating the coefficients of the same powers of x:
03

Matching Powers of x and Equating Coefficients

Multiplying x and shifting indices: $$\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n-2\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n+\lambda\sum_{n=0}^{\infty}a_n x^n = 0$$ Equating coefficients of the same power of x: $$(n+2)(n+1)a_{n+2}-2(n+1)a_{n+1}+\lambda a_n=0$$ Solving this recursion relation, we can obtain the first four coefficients in a and b, and thus the first four terms of the \(y_1(x)\) and \(y_2(x)\):
04

Solving Recursion Relation and Obtaining Coefficients

Using the recursion relation, $$a_{n+2}=\frac{2(n+1)-\lambda}{(n+2)(n+1)}a_n$$ Selecting \(a_0=1, a_1=0\), we have: $$a_2=\frac{-\lambda}{2}a_0, a_3=0$$ $$a_4=\frac{(4-\lambda)(-\lambda)}{12}$$ Thus, \(y_1(x)=1-\frac{\lambda}{2}x^2+\frac{(4-\lambda)(-\lambda)}{12}x^4+\cdots\) Selecting \(a_0=0, a_1=1\), we have: $$b_1=1, b_2=0$$ $$b_3=\frac{-\lambda}{6}$$ $$b_4=0$$ Thus, \(y_2(x)=x-\frac{\lambda}{6}x^3+\cdots\) Therefore, we have found two linearly independent solutions: \(y_1(x)\) and \(y_2(x)\).
05

(b) Polynomial Solutions for Given Values of Lambda

For \(\lambda\) equal to nonnegative even integers, the power series terminates and becomes a polynomial solution. We can find the polynomial solutions for the given values of \(\lambda\): For \(\lambda=0\): Use \(y_1(x)\), then the polynomial solution is \(y(x)=1\). For \(\lambda=2\): Use \(y_1(x)\), then the polynomial solution is \(y(x)=1-2x^2\). For \(\lambda=4\): Use \(y_1(x)\), then the polynomial solution is \(y(x)=1-4x^2+8x^4\). For \(\lambda=6\): Use \(y_2(x)\), then the polynomial solution is \(y(x)=x-2x^3\). For \(\lambda=8\): Use \(y_2(x)\), then the polynomial solution is \(y(x)=x-4x^3+16x^5\) For \(\lambda=10\): Use \(y_2(x)\), then the polynomial solution is \(y(x)=x-6x^3+60x^5-\frac{320}{3}x^7\)
06

(c) Finding Hermite Polynomials \(H_0(x), \ldots, H_5(x)\)

Hermite polynomials are the polynomial solutions of the Hermite equation for \(\lambda=2n\) and have the coefficient of the highest power term equal to \(2^n\). We found polynomial solutions for various values of \(\lambda\). Now, we will scale them to obtain Hermite polynomials: For \(n=0\): \(H_0(x)=1\) For \(n=1\): \(H_1(x)=2x\) For \(n=2\): \(H_2(x)=4-8x^2\) For \(n=3\): \(H_3(x)=8x-48x^3\) For \(n=4\): \(H_4(x)=16-192x^2+960x^4\) For \(n=5\): \(H_5(x)=32x-640x^3+5120x^5\) Thus, we got the Hermite Polynomials \(H_0(x), \ldots, H_5(x)\).

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Key Concepts

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

Power Series Method
The power series method is a powerful technique for finding solutions to differential equations. This method assumes that the solution can be represented as an infinite sum involving powers of a variable, in this case, the variable is \(x\). In general, we express the solution as:
  • \(y(x) = \sum_{n=0}^{\infty} a_n x^n\)
This method involves calculating the derivatives of the power series and substituting them back into the differential equation.
From there, you align powers of \(x\) and equate coefficients for each power to solve for unknown coefficients \(a_n\).
This method is used to tackle the hermite differential equation, allowing us to generate two series solutions which are often infinite in nature.
Linearly Independent Solutions
In the context of differential equations, linearly independent solutions are two functions whose linear combination cannot be reduced to a single function. This is a critical concept because it ensures that you have a complete set of solutions to express any general solution of the equation.
For the Hermite differential equation, through the power series method, we find two such solutions, \(y_1(x)\) and \(y_2(x)\).
  • \(y_1(x) = 1 - \frac{\lambda}{2}x^2 + \cdots\)
  • \(y_2(x) = x - \frac{\lambda}{6}x^3 + \cdots\)
These two solutions are linearly independent because no constant \(c\) exists where \(y_1(x) = c \cdot y_2(x)\). This ensures that every possible solution to the Hermite differential equation can be represented as \(C_1y_1(x) + C_2y_2(x)\), where \(C_1\) and \(C_2\) are constants.
Hermite Polynomials
Hermite Polynomials, denoted as \(H_n(x)\), are a specific class of orthogonal polynomials that arise in many physics and engineering applications, notably in quantum mechanics. These polynomials are solutions to the Hermite differential equation when the parameter \(\lambda = 2n\), where \(n\) is a non-negative integer.
  • For example, the polynomial solution for \(\lambda=2\) becomes \(H_1(x)=2x\)
  • The main characteristic is that the coefficient of the highest power of \(x\) in \(H_n(x)\) is \(2^n\).
Such properties make these polynomials useful in computations for problems that involve normal distributions and other symmetric functions over a defined range. Hermite polynomials also play an important role in numerical analysis and approximation theory due to their orthogonality properties.
Polynomial Solutions
Polynomial solutions for differential equations are those that stop at a finite degree, unlike infinite series solutions. For the Hermite equation, if \(\lambda\) is a non-negative even integer, one of the series solutions becomes a polynomial.
This polynomial termination occurs because, beyond a certain degree, the coefficients \(a_n\) become zero, halting further terms.
  • For \(\lambda=0\), the solution simply becomes \(y(x) = 1\).
  • For \(\lambda=4\), the polynomial is \(y(x) = 1 - 4x^2 + 8x^4\).
These polynomial solutions are significant in physics and engineering problems where solutions need to be expressed in closed form, particularly in dealing with quantum mechanics, vibrations, and other wave phenomena where Hermite polynomials arise naturally.

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Most popular questions from this chapter

Find all singular points of the given equation and determine whether each one is regular or irregular. \((x+3) y^{\prime \prime}-2 x y^{\prime}+\left(1-x^{2}\right) y=0\)

Consider the differential equation $$ x^{3} y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0 $$ where \(\alpha\) and \(\beta\) are real constants and \(\alpha \neq 0\). (a) Show that \(x=0\) is an irregular singular point. (b) By attempting to determine a solution of the form \(\sum_{n=0}^{\infty} a_{n} x^{r+n},\) show that the indicial equation for \(r\) is linear, and consequently there is only one formal solution of the assumed form. (c) Show that if \(\beta / \alpha=-1,0,1,2, \ldots,\) then the formal series solution terminates and therefore is an actual solution. For other values of \(\beta / \alpha\) show that the formal series solution has a zero radius of convergence, and so does not represent an actual solution in any interval.

Show that the given differential equation has a regular singular point at \(x=0,\) and determine two linearly independent solutions for \(x>0 .\) $$ x^{2} y^{\prime \prime}+x y^{\prime}+2 x y=0 $$

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 \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) It can be shown that the general formula for \(P_{n}(x)\) is $$ P_{n}(x)=\frac{1}{2^{n}} \sum_{k=0}^{\ln / 2-} \frac{(-1)^{k}(2 n-2 k) !}{k !(n-k) !(n-2 k) !} x^{n-2 k} $$ where \([n / 2]\) denotes the greatest integer less than or equal to \(n / 2 .\) By observing the form of \(P_{n}(x)\) for \(n\) even and \(n\) odd, show that \(P_{n}(-1)=(-1)^{n} .\)

Use the method of Problem 23 to solve the given equation for \(x>0 .\) \(x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=3 x^{2}+2 \ln x\)
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