Question about 2nd order linear ODEs series solutions

[Tosh5457]

I got some questions about this topic...

$$y'' + p(z)y' + q(z)y=0$$
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
$$y''(z) + y(z) = 0$$

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

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

$$a_{n+2}=-\frac{a_{n}}{(n+2)(n+1)}, n \geq 0$$

So,

$$y(z)=1+z-\frac{z^{2}}{2!}+\frac{z^{3}}{3!}-...$$

Which is equal to:

$$y(z) = cos(z) + sin(z)$$

The solution to the EDO is:

$$y(z) = c_{1}cos(z) + c_{2}sin(z)$$

Why isn't the solution this though?:
$$y(z) = cos(z) + sin(z)$$

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

 

[pasmith]

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.

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

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.

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

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

$$a_{n+2}=-\frac{a_{n}}{(n+2)(n+1)}, n \geq 0$$

So far so good.

So,

$$y(z)=1+z-\frac{z^{2}}{2!}+\frac{z^{3}}{3!}-...$$

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

$$a_{2m} = a_0 {{(-1)^m} \over (2m)!}$$

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

$$a_{2m+1} = a_1 {{(-1)^m} \over (2m+1)!}$$

so that

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

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]

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

$$a_{2m} = a_0 {{(-1)^m} \over (2m)!}$$

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

$$a_{2m+1} = a_1 {{(-1)^m} \over (2m+1)!}$$

so that

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

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]

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

$$a_{2m} = a_0 {{(-1)^m} \over (2m)!}$$

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

$$a_{2m+1} = a_1 {{(-1)^m} \over (2m+1)!}$$

so that

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

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

 

[blue_raver22]

I am unable to provide specific answers to your questions as they pertain to mathematics and not my area of expertise. I suggest consulting with a mathematician or referring to a textbook for further clarification on these topics. However, I can offer some general advice on approaching mathematical problems and understanding solutions.

1) It is important to note that a function being analytic means it can be expressed as a power series. This does not necessarily mean it is holomorphic or differentiable in all points. In the context of the given ODE, the behavior of p(z) and q(z) can affect the analyticity of y(z). This is because the coefficients of the power series may not be well-defined or may not converge if p(z) or q(z) are not analytic. Therefore, the behavior of these functions does play a role in the analyticity of y(z).

2) In the given example, the solution is a linear combination of cos(z) and sin(z) because these functions are linearly independent solutions of the given ODE. This means that any linear combination of these two functions will also be a solution. Therefore, the assumption that y(z) is a power series was not wrong, but it is important to note that the solution is not just a single power series, but a combination of two power series.

I hope this helps clarify some of your questions. It is always important to carefully consider the assumptions and behavior of functions when solving mathematical problems.