Lecture Note
An Overview of the Frobenius Method
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University:
Massachusetts Institute of Technology -
Course:
18.075 | Advanced Calculus for Engineers Academic year:
2024
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Views:
34
Pages:
9
Author:
Carmen Vargas
An Overview of the Frobenius method
Subheading:
A. Classification of singular points of linear, 2nd-order ODE's
Content:
Consider the linear, 2nd-order homogeneous ordinary differential equation (ODE)
[y''(x) + a_1(x)y'(x) + a_2(x)y(x) = 0]
where (y'(x) = \frac{dy}{dx})
(1)
(i) A point x0 is called ordinary if both a1(x) and a2(x) are analytic at x0.
(ii) A point x0 is called a singular point of the ODE if a1(x) or a2(x) is NOT analytic at x0. In this case, x0 is either
(a) a regular singular point if:
(x - x0)a1(x): analytic at x0
and (x - x0)2a2(x): analytic at x0
(b) an irregular singular point otherwise.
WARNING: To check the analyticity of a function at a point (x = x0), say of the functions (x - x0)a1(x) and (x - x0)2a2(x), it is NOT sufficient to show that the limits (lim_{x \rightarrow x_{0}}[(x - x_{0})a_{1}(x)]) and (lim_{x \rightarrow x_{0}}[(x - x_{0})^{2}a_{2}(x)]) exist and are finite. One HAS to show that (x - x0)a1(x) and (x - x0)2a2(x) have a TAYLOR SERIES EXPANSION around x0.
Examples:
(x^2y'' + xy' + (x^2 - p^2)y = 0)
Divide both sides by x²:
y'' + (1/x)y' + (1 - p²/x²)y = 0; a1(x) = 1/x, a2(x) = 1 - p²/x²
Hence, a1(x) and a2(x) are analytic everywhere except at x = 0.
Hence, x0 = 0 is a singular point of this ode (the sole one).
(x - x0)a1(x) = x(1/x) = 1; (x - x0)²a2(x) = x²(1 - p²/x²) = x² - p²; both analytic at x0 = 0.
Thus, x0 = 0 is a regular singular point of this ode.
Example 2:
[x(x-1)y'' + (x² - 1)y' + (x + x²)y = 0]
Clearly, a1(x) = (x² - 1)/x(x - 1) = (x + 1)/x; a2(x) = (x + x²)/x(x - 1) = (1 + x)/(x - 1)
a1(x) has a (simple) pole at x0 = 0 while a2(x) has a simple pole at x0 = 1.
Hence, the points x0 = 0,1 are singular points of this ode.
(x - x0)a1(x) = (x - 0)(x + 1)/x = x + 1; analytic at x0 = 0
(x - x0)²a2(x) = (x - 0)²(1 + x)/(x - 1) = x²(1 + x)/(x - 1); analytic at x0 = 0
(x - x1)a1(x) = (x - 1)(x + 1)/x = x - 1; analytic at x0 = 1
(x - x1)²a2(x) = (x - 1)²(1 + x)/(x - 1) = (x - 1)(1 + x) = x² - 1; analytic at x0 = 1
Hence, the points x0 = 0,1 are regular singular points of this ode.
Example 3:
[(1 - cos x)y'' + (sin²x)y' + xy = 0]
a1(x) = sin²x/(1 - cos x)
a2(x) = x/(1 - cos x)
Possible singular points: cos x = 1 ⇒ x = 2nπ, n = 0, ±1, ±2, ...
Let w = x - 2nπ; a1(x) = sin²(w + 2nπ)/(1 - cos(w + 2nπ)) = sin²(w)/(1 - cos w) = w²/(1 - (1 - w²/2! + ...)) = w/(1 - w/2! + ...) as w → 0
a2(x) = (w + 2nπ)/(1 - cos(w + 2nπ)) = (w + 2nπ)/(1 - (1 - w²/2! + ...)) = w + 2nπ / (w²/2! - ...) = (w + 2nπ)/w² * (2!/w - 1 + ...) as w → 0
Hence, a1(x) has a Taylor expansion at w = x - 2nπ, so a1(x) is analytic at x = 2nπ.
Notice that a2(x) has a double pole at w = 0 if n ≠ 0, and a simple pole if n = 0 (x0 = 0)
Hence, xn = 2nπ are singular points of this ode.
It is easy to check that (x - xn)a1(x) and (x - xn)2a2(x) are both analytic at xn.
Thus, xn = 2nπ are regular singular points of this ode.
Example 4:
[\sqrt{x}y'' + (ln x)\sqrt{x}y' + (sin \sqrt{x})y = 0.]
a1(x) = ln x has a branch point at x0 = 0 because ln x is a multiple-valued function.
a2(x) = sin√x = (√x) - (√x)³/3! + ... = (x - x²/2! + ...) : analytic at x = 0.
Thus, x0 = 0 is a singular point of the ODE (because of a1(x)).
Notice that (x - x0)a1(x) = x ln x cannot be analytic at x0 = 0.
Hence, x0 = 0 is an irregular singular point of this ode.
B. General remarks on the ode y'' + a1(x)y' + a2(x)y = 0
The following results (theorems) are known about the 2 independent solutions of this ode:
(i) If x0 is an ordinary point, then both independent solutions are analytic at x0 and can be expanded in a Taylor series around x = x0:
[y(x) = \sum_{n=0}^{\infty} A_n (x - x_0)^n]
[Later, we choose x0 = 0 without loss of generality.]
Detail on convergence: The radius of convergence of this series is at least as large as the distance to the nearest singularity of 1 a1(x), a2(x) in the complex z plane (x + iy: complex).
1.
(ii) If x0 is a regular singular point, there is at least one solution of
An Overview of the Frobenius Method
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