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Use the results of Problem 21 to determine whether the point at infinity is an ordinary point, a regular singular point, or an irregular singular point of the given differential equation. \(y^{\prime \prime}-2 x y^{\prime}+\lambda y=0, \quad\) Hermite equation

Short Answer

Expert verified
Answer: The Point at infinity is a regular singular point for the Hermite equation.

Step by step solution

01

Substitute \(x=1/t\) in the Hermite equation

Start by substituting \(x=1/t\) into the given differential equation to analyze the behavior of the series solution at infinity. The Hermite equation will now be: \((1/t^2)y''-(2/t^2)y'+\lambda y=0\)
02

Transform the equation into a new equation with respect to \(t\)

We will rewrite the equation in terms of \(t\) by replacing \(y\) with \(y(t)\) and the derivatives accordingly: \((1/t^2)y''(t)-(2/t^2)y'(t)+\lambda y(t)=0\)
03

Multiply by \(t^2\) to simplify

Now, multiply both sides of the equation by \(t^2\) to get rid of the terms in the denominators: \(y''(t)-2ty'(t)+\lambda t^2 y(t)=0\)
04

Classify the point at infinity

Analyze the resulting equation as \(t \to 0\). Here, as \(t\) approaches zero, the terms involving \(t\) in the equation tend to zero: \(\lim_{t \to 0} y''(t)-2ty'(t)+\lambda t^2 y(t) =0\) Due to the coefficients approaching zero only in the linear term, based on the definitions of different types of points, we can conclude that the point at infinity is a regular singular point of the given differential equation.

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Key Concepts

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

Ordinary Point
In the study of differential equations, an ordinary point is where the equation behaves in a straightforward manner, meaning that solutions can be expressed as a power series. To better understand, when we refer to an ordinary point, it's where none of the coefficients of the equation become infinite. The coefficients in a differential equation tell us how the equation behaves or changes as the variable changes.

Consider a differential equation of the form: \[ P(x)y'' + Q(x)y' + R(x)y = 0 \] An ordinary point occurs when both \( P(x) \) and \( Q(x) \) are analytic at a particular point, meaning they can be represented as power series. More simply, the coefficients are nice and smooth and do not shoot up to infinity.

In such cases, because the coefficients behave well, you can reasonably expect the solutions to have a similar, well-behaved format. Often, this means you can solve the equation near this point using techniques like the method of Frobenius, which utilizes power series. If you can express the coefficients as neat series, things like complex numbers don't suddenly pop up out of nowhere to complicate conversions or solutions. This is very useful because it means you have strong chances of solving the equation using direct and simple methods.
  • Power series solutions are applicable here.
  • Coefficients do not have any singularities at the point.
Regular Singular Point
A regular singular point of a differential equation is a special kind of point where the equation does not behave quite as smoothly as at an ordinary point but is still manageable. Such points occur when the leading coefficient \( P(x) \) in the differential equation tends towards zero, but not in a severe way that disrupts the potential for solutions.

Let’s imagine a differential equation like: \[ P(x)y'' + Q(x)y' + R(x)y = 0 \] A regular singular point arises when the series representations of \( Q(x)/P(x) \) or \( R(x)/P(x) \) blow up or become infinite, but in a controlled manner, something typical of logarithmic singularities. Basically, this means that the irregularities in the coefficients are predictable and don’t wildly vary, so the solutions can still be tackled with some adjustments. It is important because knowing a point is a regular singular point allows us to predict and apply particular solution techniques, such as the Frobenius method.

In the exercise, the transformations simplify the Hermite equation such that the coefficients approach zero at the point at infinity in a manner typical of a regular singular point. This is a classic indicator that, although there are singularities, they can be treated systematically with known mathematical tools.
  • Leading coefficient becomes zero in a controlled manner.
  • Solution methods are adjusted but are still systematic.
Irregular Singular Point
An irregular singular point differs from the other point types mainly because of its unpredictability. At an irregular singular point, a differential equation's behavior is too wild or unrestrained to allow for straightforward power series expansions.

Returning to the general differential equation form:\[ P(x)y'' + Q(x)y' + R(x)y = 0 \] An irregular singular point occurs when expressions like \( Q(x)/P(x) \) or \( R(x)/P(x) \) become extremely unwieldy as you approach the point of interest. This means that attempting to solve the equation using simple power series representations is like trying to fit a square peg into a round hole because the series would diverge or become unpredictable.

Such points can feature in very complex equations where the coefficients shoot to infinity or oscillate without a clear pattern. Unlike regular singular points, these do not permit straightforward analytical methods to guarantee solutions without some major tuning or appropriations.
  • Values become wildly infinite, making series solutions impractical.
  • Requires more sophisticated techniques for a solution if possible.

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Most popular questions from this chapter

The Bessel equation of order zero is $$ x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0 $$ Show that \(x=0\) is a regular singular point; that the roots of the indicial equation are \(r_{1}=r_{2}=0 ;\) and that one solution for \(x>0\) is $$ J_{0}(x)=1+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n}}{2^{2 n}(n !)^{2}} $$ Show that the series converges for all \(x .\) The function \(J_{0}\) is known as the Bessel function of the first kind of order zero.

Show that the Bessel equation of order one-half, $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=0, \quad x>0 $$ can be reduced to the equation $$ v^{\prime \prime}+v=0 $$ by the change of dependent variable \(y=x^{-1 / 2} v(x)\). From this conclude that \(y_{1}(x)=\) \(x^{-1 / 2} \cos x\) and \(y_{2}(x)=x^{-1 / 2} \sin x\) are solutions of the Bessel equation of order one-half.

Use the results of Problem 21 to determine whether the point at infinity is an ordinary point, a regular singular point, or an irregular singular point of the given differential equation. \(y^{\prime \prime}+y=0\)

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}-x y^{\prime}+y=0\)

Show that the given differential equation has a regular singular point at \(x=0 .\) Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. Find the series solution \((x>0)\) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also. \(x y^{\prime \prime}+(1-x) y^{\prime}-y=0\)

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