# Thread: question about neighboorhood at the origin for a solution to a ODE

1. ## question about neighboorhood at the origin for a solution to a ODE

Find the solution to $\displaystyle x^2y'' + xy' + y = 0$ in terms of power series in the neighbourhood of the origin.

The only real solution that I find is $\displaystyle y = 0$. Do you know if a solution of the form $\displaystyle y = c_1cos(ln x) + c_2sin(ln x) , x > 0$ is considered to be in the neighbourdhood of 0 ?

2. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by Fractalus
Find the solution to $\displaystyle x^2y'' + xy' + y = 0$ in terms of power series in the neighbourhood of the origin.

The only real solution that I find is $\displaystyle y = 0$. Do you know if a solution of the form $\displaystyle y = c_1cos(ln x) + c_2sin(ln x) , x > 0$ is considered to be in the neighbourdhood of 0 ?
The ODE is 'Euler's type' and its solution is...

$\displaystyle y(x)= c_{1}\ e^{i\ \ln x} + c_{2}\ e^{-i\ \ln x}$ (1)

Neither $\displaystyle e^{i\ \ln x}$ nor $\displaystyle e^{-i\ \ln x}$ is analytic in $\displaystyle x=0$ so that the only 'analytic solution' is for $\displaystyle c_{1}=c_{2}=0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by chisigma
The ODE is 'Euler's type' and its solution is...

$\displaystyle y(x)= c_{1}\ e^{i\ x} + c_{2}\ e^{-i\ x}$ (1)

Neither $\displaystyle e^{i\ x}$ nor $\displaystyle e^{-i\ x}$ is analytic in $\displaystyle x=0$ so that the only 'analytic solution' is for $\displaystyle c_{1}=c_{2}=0$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Thank you for your answer chisigma. I knew this was Euler type but I have to find the solution in terms of $\displaystyle \underline{power series}$ in the neighbourhood of 0.

The solution I gave in my last post is correct for $\displaystyle ]0,\infty[$. But I don't know if it is considered to be in the neighbourhoud of 0.

Do you understand how I got my solution? You have to use the Euler formula.

Is it helping you?

Regards,
Fractalus

4. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by Fractalus
Thank you for your answer chisigma. I knew this was Euler type but I have to find the solution in terms of $\displaystyle \underline{power series}$ in the neighbourhood of 0.

The solution I gave in my last post is correct for $\displaystyle ]0,\infty[$. But I don't know if it is considered to be in the neighbourhoud of 0.

Do you understand how I got my solution? You have to use the Euler formula.

Is it helping you?

Regards,
Fractalus
First I must apologize because the errors contained in first stesure of post, errors caused by the 'hurry' for the presence of 'competitors' ...

The 'general solution' of ODE is...

$\displaystyle y(x)= c_{1}\ (\cos \ln x + i\ \sin \ln x) + c_{2}\ (\cos \ln x - i\ \sin \ln x)$ (1)

Now function like $\displaystyle \cos x$ or $\displaystyle \sin x$ are analytic in $\displaystyle x=0$ so that their Taylor series expansion 'somewhere around' $\displaystyle x=0$ exists. But function like $\displaystyle \cos \ln x$ or $\displaystyle \sin \ln x$ aren't analytic around $\displaystyle x=0$ and that is why $\displaystyle \ln x$ doesn't have neither Taylor nor even Laurent expansion around $\displaystyle x=0$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by Fractalus
The solution I gave in my last post is correct for $\displaystyle ]0,\infty[$. But I don't know if it is considered to be in the neighbourhoud of 0.
As $\displaystyle x=0$ is a regular singular point of $\displaystyle y''+y'/x+y/x^2=0$ then, by a well known theorem there exists a solution valid in $\displaystyle 0<x<R\leq +\infty$. The name for $\displaystyle (0,R)$ it is irrelevant. For example "right punctured neighborhood" of $\displaystyle 0$ ?

6. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by FernandoRevilla
As $\displaystyle x=0$ is a regular singular point of $\displaystyle y''+y'/x+y/x^2=0$ then, by a well known theorem there exists a solution valid in $\displaystyle 0<x<R\leq +\infty$. The name for $\displaystyle (0,R)$ it is irrelevant. For example "right punctured neighborhood" of $\displaystyle 0$ ?
I definetely agree with you. That's what I thought and I wondered if we had to find a neighborhood where 0 is included. So thank you, now I know that my answer is correct.

7. ## Re: question about neighboorhood at the origin for a solution to a ODE

An 'old wolf' may lose his teeth and also his memory so that I remember only now a problem I analysed two years ago. The 'general solution' of the ODE...

$\displaystyle x^{2}\ y^{''} + x\ y^{'} + y = 0$ (1)

... is...

$\displaystyle y(x)= c_{1}\ \text{icos} (x) + c_{2}\ \text{isin} (x)$ (2)

... where icos(*) and isin(*) are what I called 'I-functions' and their properties I briefly described in ...

http://digilander.libero.it/luposaba...-functions.pdf

In particular the Laurent expansion's coefficients of the I-functions around $\displaystyle s=0$ are computed...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by chisigma
An 'old wolf' may lose his teeth and also his memory so that I remember only now a problem I analysed two years ago. The 'general solution' of the ODE...

$\displaystyle x^{2}\ y^{''} + x\ y^{'} + y = 0$ (1)

... is...

$\displaystyle y(x)= c_{1}\ \text{icos} (x) + c_{2}\ \text{isin} (x)$ (2)

... where icos(*) and isin(*) are what I called 'I-functions' and theis properties I briefly described in ...

http://digilander.libero.it/luposaba...-functions.pdf

In particular the Laurent expansion's coefficients of the I'functions around $\displaystyle s=0$ are computed...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Thanks anyway chisigma, but I think other answer I got is enought. What you gave me is overkill because I'm not doing a complex functions course.

9. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by chisigma
An 'old wolf' may lose his teeth and also his memory so that I remember only now a problem I analysed two years ago. The 'general solution' of the ODE...

$\displaystyle x^{2}\ y^{''} + x\ y^{'} + y = 0$ (1)

... is...

$\displaystyle y(x)= c_{1}\ \text{icos} (x) + c_{2}\ \text{isin} (x)$ (2)

... where icos(*) and isin(*) are what I called 'I-functions' and their properties I briefly described in ...

http://digilander.libero.it/luposaba...-functions.pdf

In particular the Laurent expansion's coefficients of the I-functions around $\displaystyle s=0$ are computed...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
But your article is still very interesting!

10. ## Re: question about neighboorhood at the origin for a solution to a ODE

Originally Posted by chisigma
An 'old wolf' may lose his teeth and also his memory so that I remember only now a problem I analysed two years ago. The 'general solution' of the ODE...

$\displaystyle x^{2}\ y^{''} + x\ y^{'} + y = 0$ (1)

... is...

$\displaystyle y(x)= c_{1}\ \text{icos} (x) + c_{2}\ \text{isin} (x)$ (2)

... where icos(*) and isin(*) are what I called 'I-functions' and their properties I briefly described in ...

http://digilander.libero.it/luposaba...-functions.pdf

In particular the Laurent expansion's coefficients of the I-functions around $\displaystyle s=0$ are computed...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
The paper about the 'I-functions' written two years ago had never been controlled... today I discovered some errors [I apologize for that ...] and they have been corrected...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$