# Bessel differential equation

1. Dec 13, 2011

### fluidistic

1. The problem statement, all variables and given/known data
The Bessel DE of order 0 is $x^2y''+xy'+x^2y=0$. A solution is $J_0(x)-\left ( \frac{x}{2} \right ) ^2+\frac{1}{4}\left ( \frac{x}{2} \right ) ^4+...$
Show that there's another solution for $x\neq 0$ that has the form $J_0(x)\ln (|x|)+Ax^2+Bx^4+Cx^6+...$ and find the coefficients A, B and C.

2. Relevant equations
I checked out in Boas's book check out what was worth $J_0(x)$ and to my surprise $J_0(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{\Gamma (n+1) \Gamma (n+1)}\left( \frac{x}{2} \right ) ^{2n}$. Now I'm looking at the problem statement and I'm not really sure how the solution looks. It doesn't look beautiful at all to me.

I've been digging into my professor's notes (unfortunately in Spanish) and there's a theorem that states that if the inditial equation leads to 2 roots such that $\mu _1-\mu _2 \in \mathbb{N}$ then there's another linearly independent solution and he gives the form.
He takes as example precisely the Bessel DE with $p=0$, hence our case.
He says that the theorem leads to a solution of the form (knowing that $J_0(x)$ is a solution): $K_0(x)=J_0(x)\ln (|x|)+\sum _{j=1}^{\infty} \frac{(-1)^{j+1}}{(j!)^2} H_j \left ( \frac{x}{2} \right ) ^{2j}$ where $H_k=\sum _{j=1}^{k} j^{-1}$.

3. The attempt at a solution
If I take for granted his results, I get that another solution is $K_0(x)=J_0(x)\ln (|x|)+\frac{x^2}{4}+\frac{3x^4}{128}+\frac{11}{13824}x^6-...$ where it's very easy to see all the coefficient they ask for.
I wonder if it's right and I'm also not understanding well the point of the problem statement when they give $J_0(x)-\left ( \frac{x}{2} \right ) ^2+\frac{1}{4}\left ( \frac{x}{2} \right ) ^4+...$.
I'd appreciate any comment.

2. Dec 13, 2011

### jackmell

I think you have that not quite stated correctly. A solution of the equation is Bessel's function given by:

$$J_0(x)=1-\frac{x^2}{4}+\frac{x^4}{64}+\cdots$$

Ok, that's one. The other solution contains a logarithm term. I think the question is asking you to go through that derivation for the second solution. That is, how is the second solution determined? Gotta' take partials with respect to c right? Or with respect to whatever you're calling the roots of the indical equation.

3. Dec 14, 2011

### fluidistic

Ah ok thank you. Well that's surprising, I've checked out and this exercise is assigned to students at least since 1995 in my university; I'll ask a friend if he did this exercise.
Ok so for now I assume they meant "Assume that a solution is $J_0(x)$".
I checked out my professor's class notes and it confuses me a lot.
Basically he states a theorem and then apply it (without all the details, which confuses me even more) to the Bessel DE of order 0.
He gives the form of both solutions when the indicial equation leads to 2 different roots that differ by an integer. If the real part of $c_1$ is greater than the real part of $c_2$, then $y_1(x)=|x-x_0|^{c_1} \sum _{n \in \mathbb{N}} a_n (x-x_0)^n$ and the other solution (the one I'm looking for in the exercise) is of the form $y_2 (x)=|x-x_0|^{c_2} \sum _{n \in \mathbb{N}} b_n (x-x_0)^n +\gamma y_1(x) \ln |x|$. He then gives an enormous formula to calculate the $b_n$ and $\gamma$ coefficients.
In my exercise the inidicial equation leads to both roots are worth 0. So a solution would be of the form $y_1(x)=\sum _{n\in \mathbb{N}}a_n x^n$. My professor says this lead to $a_n=\frac{-a_{n-2}}{n(2c+n)}$ for $n\geq 2$. $a_{2n+1}=0$ and $a_{2n}=-\frac{a_{2(n-1)}}{4n(c+n)}$ for $n \geq 1$. However when I try to derive this, I do not reach any of this.
So $y_1(x)=\sum _{n=0}^{\infty }a_n x^n$, $y_1'(x)=\sum _{n=0}^{\infty }na_n x^{n-1}$, $y_1''(x)=\sum _{n=0}^{\infty }n(n-1)a_n x^{n-2}$.
When I plug this into the original DE (divided by $x^2$), I get $\sum _{n=0}^{\infty }n(n-1)a_n x^{n-2}+\sum _{n=0}^{\infty }a_n x^{n-2}+\sum _{n=0}^{\infty }a_n x^{n}=0$. Which imply $\sum _{n=0}^{\infty }a_n x^{n-2}[n(n+1)+1]+\sum _{n=2}^{\infty }a_{n-2} x^{n-2}=0$. Thus $\sum _{n=0}^{\infty }a_n x^{n-2} (n^2+n+1)+\sum _{n=2}^{\infty }a_{n-2} x^{n-2}=0$. So $a_0x^{-2}+3a_1x^{-1}+\sum _{n=2}^{\infty }x^{n-2}[a_n(n^2+n+1)+a_{n-2}]=0 \Rightarrow a_n=-\frac{a_{n-2}}{n^2+n+1}$ which differ from the professor result.

4. Dec 14, 2011

### jackmell

Ok, that's really confussing what you writing up there. I tell you what, I'll give you my version and if you can follow it, then I want one of those silver stars they give out in here for really brilliant exposition:

Do not assign a value to c until the very end. So if we go through the process we obtain
$$\sum_{n=0}^{\infty}a_n(n+c)^2 x^{n+c}+\sum_{n=2}^{\infty}a_{n-2}x^{n+c}=0$$
Now, by the usual process, we can choose the $a's$ so that all but the first two coefficients vanish and so obtain

$$a_n(c)=-\frac{a_{n-2}}{(n+c)^2},\quad n\geq 2$$
To arrive at a particular solution, we chose $a_0=1$ and $a_1=0$ since those are arbitrary and therefore we can write as one solution:

$$y_1(x,c)=x^c+\sum_{n=2}^{\infty}a_n(c)x^{n+c}$$

and that means if we substitute $y_1(x,c)$ into the differential equation, all but the first term, the $a_0$ term, will remain since we choose the remaining coefficients such that when substituted into the DE, the left side of the DE would be zero. So writing the left side of the DE as an operator $L$, and substituting $y_1(x,c)$, we can write

$$L(y)=x^2 y''+xy'+x^2y$$
and

$$L(y_1(x,c))=c^2 x^c$$

That's because we've not yet choosen the value of $c$. One solution however can be obtained by letting c=0 in that last expression. That is

$$L(y_1(x,0))=0$$

or the function $y_1(x,0)$ satisfies the differential equation $L(y)=0$ right?

To obtain the second, we differentiate with respect to c:

$$\frac{\partial}{\partial c} L(y_1(x,c))=\frac{\partial}{\partial c} c^2 x^c$$
or switching the order of differentiation:

$$L\left(\frac{\partial}{\partial c} y_1(x,c)\right)=c^2 \ln(x) x^c+2c x^c$$

and now if we set c=0 in this expression we obtain:

$$L\left(\frac{\partial}{\partial c} y_1(x,c)\biggr|_{c=0}\right)=0$$

and therefore a second solution must be that expression in parenthesis, that is:

$$\frac{\partial}{\partial c} y_1(x,c)=x^c \ln(x)+\sum_{n=2}^{\infty} a_n(c) x^{n+c} \ln(x)+\sum_{n=2}^{\infty} \frac{d a_n(c)}{dc} x^{n+c}$$

or:

$$y_2(x,c)=\left(y_1(x,c)\ln(x)+\sum_{n=2}^{\infty} \frac{da_n(c)}{dc} x^{n+c}\right)_{c=0}$$

Ok then, bingo-bango. Just need to differentiate $a_n(c)$, get that expression, and we obtain the second solution.

Last edited: Dec 14, 2011