# Series solution to DE

• Dec 15th 2011, 01:30 PM
arbolis
Series solution to DE
Consider the DE $\displaystyle xy''+(2-x)y'-2y=0$. Give 2 solutions; one regular and worth 1 at the origin and the other of the form $\displaystyle \frac{1}{x}+A(x) \ln (x)+B(x)$ where $\displaystyle A(x)$ and $\displaystyle B(x)$ are regular at the origin. Give the first 3 terms of the series of $\displaystyle A(x)$ and $\displaystyle B(x)$.
My attempt: Divide the DE by x: $\displaystyle y''+ \left ( \frac{2}{x}-1 \right ) y'-2y=0$. In order to solve this DE, I had in mind to propose a solution of the form $\displaystyle y(x)=f(x) \phi (x)$ where $\displaystyle \phi (x)$ would be the solution to this DE but when x tends to infinity. It turns out that this didn't simplify things as I'd hoped.
When $\displaystyle x\to \infty$, the DE becomes $\displaystyle \phi ''-\phi '-2\phi =0$. I used Frobenius's method to solve this DE:
I assumed that $\displaystyle \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}$. I derivated this once and twice and plugged into the DE.
I eventually reached $\displaystyle x^{c-2}a_0c(c-1)+x^{c-1}a_1(c+1)c-x^{c-1}a_0c+\sum_{n=0}^{\infty}x^{n+c}[a_{n+2}(n+2+c)(n+1+c)+a_{n+1}(n+1+c)-2a_n]=0$.
The indicial equation leads to $\displaystyle c=1$ or $\displaystyle c=0$. At first glance it looks like both solutions are acceptable.
So now I get a recurrence relation with $\displaystyle a_{n+2}$ in terms of $\displaystyle a_{n+1}$ and $\displaystyle a_n$ which isn't what I hoped for. Maybe I shouldn't have proposed a solution of the form $\displaystyle y(x)=f(x)\phi (x)$? How would you tackle this problem?
• Dec 17th 2011, 09:30 AM
arbolis
Re: Series solution to DE
Another attempt using Frobenius method on the original DE.
I reach $\displaystyle x^{c-2}[a_0c(c-1)+2a_0c]+x^{c-1}[a_1c(c+1)+2a_1(c+1)]-a_0x^{c-1}c+\sum_{n=2}^{\infty}[a_n(n+c)(n+c-1)+2a_n(n+c)-a_{n-1}(n+c-1)-2a_{n-2}]=0$.
Solving the inidicial equation leads me to $\displaystyle c=0$, or $\displaystyle c=-2$ or $\displaystyle c=-1$. Though one would expect at most 2 different values for c since it's a second order DE. Something's definitely wrong with my attempt.
Any help is appreciated.
• Dec 18th 2011, 12:45 AM
chisigma
Re: Series solution to DE
Quote:

Originally Posted by arbolis
Consider the DE $\displaystyle xy''+(2-x)y'-2y=0$. Give 2 solutions; one regular and worth 1 at the origin and the other of the form $\displaystyle \frac{1}{x}+A(x) \ln (x)+B(x)$ where $\displaystyle A(x)$ and $\displaystyle B(x)$ are regular at the origin. Give the first 3 terms of the series of $\displaystyle A(x)$ and $\displaystyle B(x)$.
My attempt: Divide the DE by x: $\displaystyle y''+ \left ( \frac{2}{x}-1 \right ) y'-2y=0$. In order to solve this DE, I had in mind to propose a solution of the form $\displaystyle y(x)=f(x) \phi (x)$ where $\displaystyle \phi (x)$ would be the solution to this DE but when x tends to infinity. It turns out that this didn't simplify things as I'd hoped.
When $\displaystyle x\to \infty$, the DE becomes $\displaystyle \phi ''-\phi '-2\phi =0$. I used Frobenius's method to solve this DE:
I assumed that $\displaystyle \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}$. I derivated this once and twice and plugged into the DE.
I eventually reached $\displaystyle x^{c-2}a_0c(c-1)+x^{c-1}a_1(c+1)c-x^{c-1}a_0c+\sum_{n=0}^{\infty}x^{n+c}[a_{n+2}(n+2+c)(n+1+c)+a_{n+1}(n+1+c)-2a_n]=0$.
The indicial equation leads to $\displaystyle c=1$ or $\displaystyle c=0$. At first glance it looks like both solutions are acceptable.
So now I get a recurrence relation with $\displaystyle a_{n+2}$ in terms of $\displaystyle a_{n+1}$ and $\displaystyle a_n$ which isn't what I hoped for. Maybe I shouldn't have proposed a solution of the form $\displaystyle y(x)=f(x)\phi (x)$? How would you tackle this problem?

A little 'excamotage' that can make the task more comfortable is to set...

$\displaystyle y(x)=\frac{\phi(x)}{x}$ (1)

... and then try to find $\displaystyle \phi(*)$. First You compute ...

$\displaystyle y^{'}= \frac{\phi^{'}}{x}-\frac{\phi}{x^{2}}$ (2)

$\displaystyle y^{''}=\frac{\phi^{''}}{x}-2\ \frac{\phi^{'}}{x^{2}} +2\ \frac{\phi}{x^{3}}$ (3)

... and then insert (1),(2) and (3) in the original DE obtaining...

$\displaystyle \phi^{''} - \phi^{'} - \frac{\phi}{x} =0$ (4)

... which seems 'easier to be attacked'. Now You search a series solution of (4) imposing proper constrain...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 18th 2011, 12:46 PM
arbolis
Re: Series solution to DE
Thanks for the tips.
I propose a solution of the form $\displaystyle \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}$. After derivating once and twice and plugging into the DE the indicial equation gives me $\displaystyle c=0$ or $\displaystyle c=1$. And also $\displaystyle a_{n+1}=\frac{a_n}{n+c}$. But the denominator blows up if $\displaystyle c=0$ and $\displaystyle n=0$, thus $\displaystyle c=1$ only. Thus $\displaystyle a_{n+1}=\frac{a_n}{n+1}$.
$\displaystyle a_0$ is arbitrary but different from 0. I'm guessing I have to choose it so that $\displaystyle y (0)=1$ as requested by the statement of the problem. When I choose$\displaystyle a_0=1$, I get $\displaystyle a_1=1$, $\displaystyle a_2=\frac{1}{2}$ and in general, $\displaystyle a_n=\frac{1}{n!}$. So that $\displaystyle \phi (x)=\sum _{n=0}^\infty \frac{x^{n+1}}{n!} \Rightarrow y(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!}=e^x$. It seems like I got lucky this time when I chose the arbitrary constant $\displaystyle a_0$ since $\displaystyle y(0)=1$ as required.
Is what I've done correct so far?

P.S.:This indeed satisfies the original DE. Wow! Thank you very much!
Post Scriptum 2: I have no idea how to find A(x) and B(x) for a second solution. I tried to write them as infinite series and plug into the DE but I get an enormous expression that doesn't look easily solvable. I'm missing a trick here.
• Dec 18th 2011, 10:05 PM
chisigma
Re: Series solution to DE
Quote:

Originally Posted by arbolis
Thanks for the tips.
I propose a solution of the form $\displaystyle \phi (x)=\sum_{n=0}^{\infty}a_nx^{n+c}$. After derivating once and twice and plugging into the DE the indicial equation gives me $\displaystyle c=0$ or $\displaystyle c=1$. And also $\displaystyle a_{n+1}=\frac{a_n}{n+c}$. But the denominator blows up if $\displaystyle c=0$ and $\displaystyle n=0$, thus $\displaystyle c=1$ only. Thus $\displaystyle a_{n+1}=\frac{a_n}{n+1}$.
$\displaystyle a_0$ is arbitrary but different from 0. I'm guessing I have to choose it so that $\displaystyle y (0)=1$ as requested by the statement of the problem. When I choose$\displaystyle a_0=1$, I get $\displaystyle a_1=1$, $\displaystyle a_2=\frac{1}{2}$ and in general, $\displaystyle a_n=\frac{1}{n!}$. So that $\displaystyle \phi (x)=\sum _{n=0}^\infty \frac{x^{n+1}}{n!} \Rightarrow y(x)=\sum _{n=0}^{\infty} \frac{x^n}{n!}=e^x$. It seems like I got lucky this time when I chose the arbitrary constant $\displaystyle a_0$ since $\displaystyle y(0)=1$ as required.
Is what I've done correct so far?

P.S.:This indeed satisfies the original DE. Wow! Thank you very much!
Post Scriptum 2: I have no idea how to find A(x) and B(x) for a second solution. I tried to write them as infinite series and plug into the DE but I get an enormous expression that doesn't look easily solvable. I'm missing a trick here.

If a solution of the second order DE...

$\displaystyle \phi^{''} - \phi^{'} - \frac{\phi}{x}=0$ (1)

... is known, a second solution independent from it can be found with the following procedure. If u(x) and v(x) are solutions of (1), then is...

$\displaystyle u^{''} - u^{'} - \frac{u}{x}=0$ (2)

$\displaystyle v^{''} - v^{'} - \frac{v}{x}=0$ (3)

If You multiply the (2) by v, the (3) by u and do the difference You obtain...

$\displaystyle u^{''}\ v -u\ v^{''} + u\ v^{'} - u^{'}\ v = \frac{d}{dx} (u\ v^{'}- u^{'}\ v )=0$ (4)

The (4) is a DE the solution of which is...

$\displaystyle u\ v^{'}- u^{'}\ v=c_{2}$ (5)

... and from (5), deviding by $\displaystyle v^{2}$ You obtain...

$\displaystyle \frac{u\ v^{'}- u^{'}\ v}{v^{2}}= \frac{d}{dx} (\frac{u}{v})= \frac{c_{2}}{v^{2}}$ (6)

Now the solution of (6) is...

$\displaystyle \frac{u}{v}= c_{1}+ c_{2}\ \int \frac{dx}{v^{2}}$ (7)

... and that permits You to derive u(x) from v(x)...

$\displaystyle u= v\ \int \frac{dx}{v^{2}}$ (8)

You found $\displaystyle v(x)=x\ e^{x}$ as possible solution of (1) and from (8) You derive [finally!]...

$\displaystyle y(x)= \frac{u(x)}{x} = e^{x}\ \int \frac{e^{- 2 x}}{x^{2}}\ dx$ (9)

Now all that You have to do is solving the integral in (9)...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 19th 2011, 01:30 AM
chisigma
Re: Series solution to DE
In the previous post we are arrived to write...

$\displaystyle y(x)= e^{x}\ \int \frac{e^{-2 x}}{x^{2}}\ dx$ (1)

The next step is of course the computation of the integral in (1). Proceeding with standard integration by parts we obtain...

$\displaystyle \int \frac{e^{-2 x}}{x^{2}}\ dx= - \frac{e^{-2 x}}{x}- 2\ \int \frac {e^{-2 x}}{x}\ dx = - \frac{e^{-2 x}}{x}- 2\ \text{Ei} (-2 x)$ (2)

... where Ei(*) is the so called Exponential Integral Function, the expansion of which is given by...

$\displaystyle \text{Ei} (x)= \gamma + \ln |x| + \sum_{n=1}^{\infty} \frac{x^{n}}{n\ n!}$ (3)

... where $\displaystyle \gamma$ is the Euler's constant. It is requested to write y(x) in the form...

$\displaystyle y(x)= \frac{1}{x} + \ln |x|\ a(x) + b(x)$ (4)

... and then to find the series expansion of a(x) and b(x). From (1) (2) and (3) we derive [neglecting the sign '-']...

$\displaystyle y(x)= \frac{e^{-x}}{x} +2\ e^{x}\ \text{Ei} (-2 x) =$

$\displaystyle =\frac{e^{-x}}{x} + 2\ e^{x}\ \ln |x| + 2\ e^{x}\ \{\gamma + \ln 2 + \sum_{n=1}^{\infty} \frac{(-2)^{n}\ x^{n}}{n\ n!}\}$ (5)

At this point the effective computation of the first coefficients of a(x) and b(x) is not too complex and the task is left to the Reader...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

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