Frobenius Method: Finding Smaller Root with Larger

[asdf1]

for the forbenius method,
if the roots to the indical equation differ by an integer,
why do you always have to take the larger root to find the smaller root?

 

[HallsofIvy]

?? I'm sorry, this makes no sense to me at all. You find the roots of the indicial equation by solving it! It is not necessary to "take the larger root to find the smaller"!

I expect you mean that you find a solutions by "plugging in" the larger root in order to find the coefficients of the power series expansion and the second, independent solution, will involve that first solution as well as the second, smaller root.

Specifically, if y₁(x) is the series solution to the equation using r₁, the larger root, then another, independent, solution is of the form
$$ay_1(x)ln x+ x^{r_2}\left[1+ \Sigma_{n-1}^\infty c_n x^n\right]$$
That is from Elementary Differential Equations and Boundary Value Equations by Boyce and DiPrima. Unfortunately they say "For this case the derivation is considerably more complicated and will not be given here"! They do note that a derivation is given in An Introduction to Ordinary Differential Equations by E.A. Coddington.

 

[asdf1]

?? I expect you mean that you find a solutions by "plugging in" the larger root in order to find the coefficients of the power series expansion and the second, independent solution, will involve that first solution as well as the second, smaller root.
Specifically, if y₁(x) is the series solution to the equation using r₁, the larger root, then another, independent, solution is of the form
$$ay_1(x)ln x+ x^{r_2}\left[1+ \Sigma_{n-1}^\infty c_n x^n\right]$$

yes, that's what i mean... but why does that happen?

 

[srknkcyrk]

How it can be decided if the natural logarithm function will be used for the second solution(for the case 3 in which the roots of indicial eqn. differ by an integer)?

Here is a quotation from Erwin Kreyszig, Advanced Eng. Mathematics 9th ed.,
"Indicial equation r(r - 1) + r - 1 = 0. Hence r1 = 1, r2 = -1. These roots differ
by an integer; this is Case 3. It turns out that no logarithm will appear."

 

[HallsofIvy]

Well, that says "this is Case 3". What are "Case 1", "Case 2", and "Case 3"?

The simplest way to get the idea, though not the proof, is to look at the Euler type equation, which is, in a sense, the "critical case" here.

An Euler type equation, also called "equipotential" has xⁿ as coefficient of the n^th derivative. It can be shown that the substitution t= ln(x) converts an Euler type equation, in x, to an equation with constant coefficients, in t. The two equations have the same "characteristic equation". Of course, if an equation with constant coefficients has a double root, say r as double root, then two independent solutions to the differential equation are e^rt and te^rt. Replacing t with ln(x) gives e^rln(x)= x^r and ln(x)x^r as independent solutions to the original Euler type equation.

 

[chwie]

i don't remember the proof, but in practice when you work out the problems with the small root normally the ak in recursion relation diverge. for example the bessell solution is like

ak+2=- ak 1/((k+2)(2n+k+1) in the case that the root are n and -n both integer and n is a natural number.

if we use -n, then for the case a2n-3=a2n-1 1/((2n-3)(-2n +(2n-1)+1)) then a2n-3 is not define.

the proof of why for the larger root is always possible, should be a special case of the fuch' s theorem.