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Question about 2nd order linear ODEs series solutions

by Tosh5457
Tags: linear, odes, order, series, solutions
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Tosh5457
#1
May12-12, 01:54 PM
P: 239
I got some questions about this topic...

[tex]y'' + p(z)y' + q(z)y=0[/tex]
where y (and its derivatives) is a function of z, z ∈ ℂ.

1) My books says this: In points where both p(z) and q(z) are analytic, y(z) is also analytic. But in points where p(z) or q(z) (or both) aren't analytic, y(z) may not be analytic.

Since y(z) is a differentiable complex function in an open set (it must be for the ODE to make sense, right?), it's holomorphic in that set, and so it's analytic. So I don't understand why y(z) may not be analytic when p or q aren't analytic. Why does y(z) being analytic or not even depends on the behavior of these functions?

2) This is a question about an example.
The problem is:
Find the series solutions, about z = 0, of
[tex]y''(z) + y(z) = 0[/tex]

y can be written as a power series (it's analytic), so:
[tex]y(z)=\sum_{n=0}^{\infty}a_{n}z^{n}[/tex]

Substituting the series in the EDO, we obtain the two-term recurrence relation:

[tex]a_{n+2}=-\frac{a_{n}}{(n+2)(n+1)}, n \geq 0[/tex]

So,

[tex]y(z)=1+z-\frac{z^{2}}{2!}+\frac{z^{3}}{3!}-...[/tex]

Which is equal to:

[tex]y(z) = cos(z) + sin(z)[/tex]

The solution to the EDO is:

[tex]y(z) = c_{1}cos(z) + c_{2}sin(z)[/tex]

Why isn't the solution this though?:
[tex]y(z) = cos(z) + sin(z)[/tex]

We stated that y(z) was a power series, and the power series turned out to be cos(z) + sin(z), so why do we take the linear combination of these 2 functions as the solution? Was the initial assumption that y(z) equals a power series wrong?

Thanks
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pasmith
#2
May12-12, 05:21 PM
HW Helper
Thanks
P: 945
To your first question: suppose p(z) = 1/z and q(z) = -1/z^2. These aren't analytic at z = 0 even though they are analytic everywhere else. Multiplying the ODE by z^2 then gives z^2y'' + zy' - y = 0, which has solution y(z) = Az + B/z for arbitrary constants A and B. We see that at z = 0, 1/z is not analytic. On the other hand, z is analytic at z = 0.

Points at which p(z) or q(z) have (essential) singularities are called singular points of the ODE, and typically at least one of the two linearly independent solutions will have a(n essential) singularity at such a point. Another example is Bessel's equation: p(z) = 1/z, q(z) = 1 - a^2/z^2 where a is a constant; the J solutions are bounded at z = 0, but the Y solutions have a singularity.

Quote Quote by Tosh5457 View Post
2) This is a question about an example.
The problem is:
Find the series solutions, about z = 0, of
[tex]y''(z) + y(z) = 0[/tex]
You already know (I hope) that the solution of that is ##y(0)\cos(z) + y'(0)\sin(z)##, so if you don't end up with that then you've made an error in your calculations.

Quote Quote by Tosh5457 View Post
y can be written as a power series (it's analytic), so:
[tex]y(z)=\sum_{n=0}^{\infty}a_{n}z^{n}[/tex]

Substituting the series in the EDO, we obtain the two-term recurrence relation:

[tex]a_{n+2}=-\frac{a_{n}}{(n+2)(n+1)}, n \geq 0[/tex]
So far so good.

Quote Quote by Tosh5457 View Post
So,

[tex]y(z)=1+z-\frac{z^{2}}{2!}+\frac{z^{3}}{3!}-...[/tex]
Here's your error: Why should a0 = a1 = 1? a0 and a1 are determined by the initial conditions, but you haven't been given any. So you leave a0 and a1 as arbitrary constants.

You're asked for series solutions, so now you need to solve this recurrence relation to find ##a_n## in terms of ##a_0## and ##a_1##.

You should satisfy yourself (ideally by a formal proof by induction) that if n = 2m (##m \geq 0##) then

[tex]a_{2m} = a_0 {{(-1)^m} \over (2m)!}[/tex]

and if n = 2m+1 (##m \geq 0##) then

[tex]a_{2m+1} = a_1 {{(-1)^m} \over (2m+1)!}[/tex]

so that

[tex]y(z) =
\sum_{n=0}^{\infty} a_n z^n =
a_0\sum_{m=0}^{\infty} {{(-1)^m z^{2m}} \over (2m)!}
+ a_1\sum_{m=0}^{\infty} {{(-1)^m z^{2m+1}} \over (2m+1)!}[/tex]

which gives us the series solutions you were asked to find. You should recognize this as the power series about ##z = 0## of ##y(z) = a_0 \cos(z) + a_1 \sin(z)##.
yus310
#3
May12-12, 09:52 PM
P: 81
Quote Quote by pasmith View Post
To your first question: suppose p(z) = 1/z and q(z) = -1/z^2. These aren't analytic at z = 0 even though they are analytic everywhere else. Multiplying the ODE by z^2 then gives z^2y'' + zy' - y = 0, which has solution y(z) = Az + B/z for arbitrary constants A and B. We see that at z = 0, 1/z is not analytic. On the other hand, z is analytic at z = 0.

Points at which p(z) or q(z) have (essential) singularities are called singular points of the ODE, and typically at least one of the two linearly independent solutions will have a(n essential) singularity at such a point. Another example is Bessel's equation: p(z) = 1/z, q(z) = 1 - a^2/z^2 where a is a constant; the J solutions are bounded at z = 0, but the Y solutions have a singularity.



You already know (I hope) that the solution of that is ##y(0)\cos(z) + y'(0)\sin(z)##, so if you don't end up with that then you've made an error in your calculations.



So far so good.



Here's your error: Why should a0 = a1 = 1? a0 and a1 are determined by the initial conditions, but you haven't been given any. So you leave a0 and a1 as arbitrary constants.

You're asked for series solutions, so now you need to solve this recurrence relation to find ##a_n## in terms of ##a_0## and ##a_1##.

You should satisfy yourself (ideally by a formal proof by induction) that if n = 2m (##m \geq 0##) then

[tex]a_{2m} = a_0 {{(-1)^m} \over (2m)!}[/tex]

and if n = 2m+1 (##m \geq 0##) then

[tex]a_{2m+1} = a_1 {{(-1)^m} \over (2m+1)!}[/tex]

so that

[tex]y(z) =
\sum_{n=0}^{\infty} a_n z^n =
a_0\sum_{m=0}^{\infty} {{(-1)^m z^{2m}} \over (2m)!}
+ a_1\sum_{m=0}^{\infty} {{(-1)^m z^{2m+1}} \over (2m+1)!}[/tex]

which gives us the series solutions you were asked to find. You should recognize this as the power series about ##z = 0## of ##y(z) = a_0 \cos(z) + a_1 \sin(z)##.
"excellent job and your analysis point on, pasmith. You are correct on all notes,

a_0 and a_1 are arbitary, unless given specifications. "

yus310

Tosh5457
#4
May13-12, 12:40 PM
P: 239
Question about 2nd order linear ODEs series solutions

Quote Quote by pasmith View Post
To your first question: suppose p(z) = 1/z and q(z) = -1/z^2. These aren't analytic at z = 0 even though they are analytic everywhere else. Multiplying the ODE by z^2 then gives z^2y'' + zy' - y = 0, which has solution y(z) = Az + B/z for arbitrary constants A and B. We see that at z = 0, 1/z is not analytic. On the other hand, z is analytic at z = 0.

Points at which p(z) or q(z) have (essential) singularities are called singular points of the ODE, and typically at least one of the two linearly independent solutions will have a(n essential) singularity at such a point. Another example is Bessel's equation: p(z) = 1/z, q(z) = 1 - a^2/z^2 where a is a constant; the J solutions are bounded at z = 0, but the Y solutions have a singularity.



You already know (I hope) that the solution of that is ##y(0)\cos(z) + y'(0)\sin(z)##, so if you don't end up with that then you've made an error in your calculations.



So far so good.



Here's your error: Why should a0 = a1 = 1? a0 and a1 are determined by the initial conditions, but you haven't been given any. So you leave a0 and a1 as arbitrary constants.

You're asked for series solutions, so now you need to solve this recurrence relation to find ##a_n## in terms of ##a_0## and ##a_1##.

You should satisfy yourself (ideally by a formal proof by induction) that if n = 2m (##m \geq 0##) then

[tex]a_{2m} = a_0 {{(-1)^m} \over (2m)!}[/tex]

and if n = 2m+1 (##m \geq 0##) then

[tex]a_{2m+1} = a_1 {{(-1)^m} \over (2m+1)!}[/tex]

so that

[tex]y(z) =
\sum_{n=0}^{\infty} a_n z^n =
a_0\sum_{m=0}^{\infty} {{(-1)^m z^{2m}} \over (2m)!}
+ a_1\sum_{m=0}^{\infty} {{(-1)^m z^{2m+1}} \over (2m+1)!}[/tex]

which gives us the series solutions you were asked to find. You should recognize this as the power series about ##z = 0## of ##y(z) = a_0 \cos(z) + a_1 \sin(z)##.
On the 2nd part, I didn't do the exercise, that was the resolution from the book. Thanks for the reply, I understand now


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