What is the other solution for the Bessel differential equation of order 0?

[fluidistic]

Homework Statement

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.

Homework 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}.

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.

 

[jackmell]

Homework Statement

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.

Homework 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}.

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.

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.

 

[fluidistic]

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.

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.

 

[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.