Method of Applied Math

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Lecture 1 Sujin Khomrutai – 1 / 10

Method of Applied Math

Lecture 3: Frobenius Method

Sujin Khomrutai, Ph.D.

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Series Solutions: Regular Singular Points

Series Sol: Sing Pts Motivation

Def: Reg Sing Point EX 1

EX 2

Meth: Frobenius EX 3

EX 4

Now we solve

y^′′ + p(x)y^′ + q(x)y = 0

when a is a singular point of the ODE.

We will develop the Frobenius method in the case that a is not too singular. See below.

One of the most important applications of the Frobenius method is to solve the Bessel equation in the next section.

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Motivation

EX 2

EX 4

EX. Consider the equation x²y^′′ + 3xy^′ + 3

4y = 0.

Let us choose a = 0. Observe that 0 is a singular point of the above equation.

Applying the power series method: y = P^∞

n=0 b_nxⁿ, then we get b_n = 0 for all n. So we obtain y = 0 as the solution by this

method.

However, as one can check directly, x⁻^1/2 and x⁻^3/2 are two non-zero solutions! The Frobenius method will help us finding these two solutions.

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Regular Singular Point

EX 2

EX 4

Assume a is a singular point of the ODE y^′′ + p(x)y^′ + q(x)y = 0.

a is called a regular singular point for this ODE if

xlim→a(x − a)p(x) and lim

x→a(x − a)²q(x) exist and are finite.

Otherwise, a is an irregular singular point for this ODE.

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Example 1

EX 2

EX 4

EX. Consider the ODE

x²(x − 1)y^′′ + y^′ − y = 0.

Show that 0,1 are singular points for this equation. Show that 1 is regular whereas 0 is irregular.

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Example 2

EX 2

EX 4

EX. (Bessel’s equation)

x²y^′′ + xy^′ + (x² − ν²)y = 0.

Show that 0 is a regular singular point of the ODE and find p₀, q₀.

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Frobenius Method

EX 2

EX 4

Theorem (Frobenius). Let a be a regular singular point for y^′′ + p(x)y^′ + q(x)y = 0.

Then there is a constant r such that the ODE has a solution of the form

y = (x − a)^r

X∞ n=0

b_n(x − a)ⁿ

= b₀(x − a)^r + b₁(x − a)^r⁺¹ + b₂(x − a)^r⁺² + · · · .

The process of obtaining a solution of the above form is called the Frobenius method.

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Frobenius Method

EX 2

EX 4

Theorem. The constant r in the Frobenius method is the larger root of the quadratic equation

r(r − 1) + p₀r + q₀ = 0 (indicial equation) where

p₀ = lim

x→a(x − a)p(x), q₀ = lim

x→a(x − a)²q(x).

Remark. The smaller root of the indicial equation may lead to the zero solution in some cases! Having the Frobenius solution, one can get the other linearly independent solution by the

Reduction of Order.

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Example 3

EX 2

EX 4

EX. Find a nonzero solution of the ODE x²y^′′ − 3xy^′ + 3y = 0,

using Frobenius method about a = 0. Use reduction of order to find the other linearly independent solution.

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Example 4

EX 2

EX 4

EX. Use the Frobenius method about a = 0 to solve the ODE x(x − 1)y^′′ + (3x − 1)y^′ + y = 0.

Sol. r₁ = r₂ = 0,

y₁ =

X∞ n=0

xⁿ = 1 1 − x. Reduction of Order:

y₂ = y₁

Z e⁻^R ^p(x)dx

y₁² dx = ln x 1 − x.