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Page 1: 8.6 Method of Frobenius

8.6 Method of Frobenius

Regular and Irregular Singular Points A singular point π‘₯0 of

𝑦" + 𝑝(π‘₯)𝑦′ + π‘ž(π‘₯)𝑦 = 0

Is said to be a regular singular point if the functions

𝑃(π‘₯) = (π‘₯ βˆ’ π‘₯0)𝑝(π‘₯)

𝑄(π‘₯) = (π‘₯ βˆ’ π‘₯0)

2π‘ž(π‘₯)

Are both analytic at π‘₯0. Otherwise, π‘₯0 is called an irregular singular point.

Example

1. Determine the singular points of the differential equation. Classify them as regular or irregular.

π‘₯(π‘₯ + 3)3𝑦′′ βˆ’ 𝑦 = 0

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Frobenius’ Theorem

If π‘₯0 is a regular singular point of the differential equation

π‘Ž2(π‘₯)𝑦" + π‘Ž1(π‘₯)𝑦′ + π‘Ž0(π‘₯)𝑦 = 0

Then there exists at least one series solution of the form

𝑦 = βˆ‘π‘π‘›(π‘₯ βˆ’ π‘₯0)𝑛+π‘Ÿ

∞

𝑛=0

Where the number r is a constant referred to as the indicial root or exponent..

The series converges at least on some interval 0 < π‘₯ βˆ’ π‘₯0 < 𝑅, where R is the distance from π‘₯0 to the nearest

other singular point of the differential equation.

The Method of Frobenius

This method is similar to the method of 8.3, only instead of using a power series for y we will use

𝑦 = βˆ‘π‘π‘›(π‘₯ βˆ’ π‘₯0)𝑛+π‘Ÿ

∞

𝑛=0

Now, in addition to finding π‘Žπ‘› we will also have to find r.

Note: As in previous sections we will look at examples with the regular singular point π‘₯0 = 0.

Indicial Equation If π‘₯0 is a regular singular point of

𝑦" + 𝑝(π‘₯)𝑦′ + π‘ž(π‘₯)𝑦 = 0

Then the indicial equation for this point is

π‘Ÿ(π‘Ÿ βˆ’ 1) + 𝑝0π‘Ÿ + π‘ž0 = 0

where

𝑝0 = limπ‘₯β†’π‘₯0

(π‘₯ βˆ’ π‘₯0)𝑝(π‘₯)

π‘ž0 = lim

π‘₯β†’π‘₯0(π‘₯ βˆ’ π‘₯0)

2π‘ž(π‘₯)

The roots of the indicial equation are called the exponents (indices) of the singularity π‘₯0.

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Examples

2. Find the indicial equation and the exponents at the singularity π‘₯ = 0.

π‘₯2𝑦′′ + 4π‘₯𝑦′ + 2𝑦 = 0

3. Find a series solution about the regular singular point π‘₯ = 0 of

2π‘₯2𝑦′′ βˆ’ π‘₯𝑦′ + (1 + π‘₯)𝑦 = 0

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4. Find a series solution about the regular singular point π‘₯ = 0 of

2π‘₯2𝑦′′ βˆ’ π‘₯𝑦′ + (π‘₯2 + 1)𝑦 = 0

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