What is the Proof for the Bessel Differential Equation Solution?

[DivGradCurl]

I'm supposed to show that

$$J_0 (x) = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 }$$

satisfies the differential equation

$$x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = 0$$


Here's what I've got:

$$x^2 J_0 ^{\prime \prime} (x) = x^2 \sum _{n=2} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) \left( 2n-1 \right) x^{2n-2}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+2} \left( 2n+4 \right) \left( 2n+3 \right) x^{2n+4}}{2^{2n+4}\left[ \left( n+2 \right) ! \right] ^2 } = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x)$$

$$x J_0 ^{\prime} (x) = x \sum _{n=1} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) x^{2n-1}}{2^{2n}\left( n! \right) ^2} = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+1} \left( 2n+2 \right) x^{2n+2}}{2^{2n+4}\left[ \left( n+1 \right) ! \right] ^2 } = - x J_1 (x)$$

$$x^2 J_0 (x) = x^2 \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n+2}}{2^{2n} \left( n! \right) ^2 } = x^2 J_0 (x)$$

Then, I get

$$x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x) - x J_1 (x) + x^2 J_0 (x) = \frac{x^2}{2}$$

which is not correct, except for when $$x=0$$. Can anyone help me find where I made a mistake?

Thank you very much.

 

[arildno]

This is most easily done the other way around (it's easier to find the correct coefficients in a series solution):
Consider the initial value problem:
$$x^{2}J_{0}''(x)+xJ_{0}'(x)+x^{2}J_{0}(x)=0,J_{0}(0)=1,J_{0}'(0)=0$$
Assume a series solution:
$$J_{0}(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$
Then,
$$x^{2}J_{0}''(x)=\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n}$$
$$xJ_{0}'(x)=\sum_{n=1}^{\infty}na_{n}x^{n}$$
$$x^{2}J_{0}(x)=\sum_{n=2}^{\infty}a_{n-2}x^{n}$$
Or:
$$a_{1}x+\sum_{n=2}^{\infty}(n^{2}a_{n}+a_{n-2})x^{n}=0$$
for all x.
Hence, all odd "a"'s must be zero, while we must have for the even powers:
$$a_{2m}=-\frac{1}{(2m)^{2}}a_{2(m-1)}$$
This can be rewritten in the form you seek..

 

[DivGradCurl]

If

$$J_0 (x) = \sum _{n=0} ^{\infty} a_n x^n$$

then

$$x^2 J_0 (x) = \sum _{n=2} ^{\infty} a_{n-2} x^n$$

$$x J_0 ^{\prime} (x) = \sum _{n=2} ^{\infty} \left( n-1 \right) a_{n-1} x^{n-1} = \sum _{n=2} ^{\infty} \left( n a_{n-1} x^{n-1} \right) - \sum _{n=2} ^{\infty} \left( a_{n-1} x^{n-1} \right)$$

$$x^2 J_0 ^{\prime \prime} (x) = \sum _{n=2} ^{\infty} n \left( n-1 \right) a_{n-1} x^{n} = \sum _{n=2} ^{\infty} \left( n^2 a_n x^n \right) - \sum _{n=2} ^{\infty} \left( n a_n x^n \right)$$

and so

$$\sum _{n=2} ^{\infty} \left[ \left( n^2 a_n - n a_n + a_{n-2} \right) x^{n} + \left( n a_n - a_{n-1} \right) x^{n-1} \right] = 0$$

I now see that the $$a_n$$'s should all be zero, although I cannot obtain

$$a_{1}x+\sum_{n=2}^{\infty}(n^{2}a_{n}+a_{n-2})x^{n}=0$$

from what I've written based on your input. Furthermore, I don't see how you get

$$a_{2m}=-\frac{1}{(2m)^{2}}a_{2(m-1)}$$

Also, it's not clear how I can work my way back from there.

To be honest, I really need to present a solution in the original form, because the directions from the book are specific about using the bessel function of order zero as the basis to obtain the proof. Thanks, anyway.

 

[arildno]

Do NOT change the index on your first derivative series!

Secondly, J0 is DEFINED to be the solution of the initial value problem.

 

[DivGradCurl]

I'm sorry. I see my remark is not clear enough. Well, my book says "show that J_0 (the Bessel function of order 0 given in Example 4) satisfies the differential equation"

$$x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = 0$$

So, the kind of answer expected is to follow from the definition provided by example 4, namely

$$J_0 (x) = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 }$$

which, of course, could alternatively be written as

$$J_{0}(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$

However, we do run into a problem, since it does not provide the solution the way requested... do you know what I mean?

 

[arildno]

Well, if you find that the solution of the initial value problem coincides with the expression for J0, then you've given a fully valid proof.

Secondly, I'll continue:
If
$$a_{2m}=-\frac{a_{2m-1}}{(2m)^{2}}=\frac{a_{2(m-1)}}{2^{2}m^{2}}$$
Then, by induction, you may prove that:
$$a_{2m}=\frac{(-1)^{m}}{2^{2m}(m!)^{2}}=-\frac{1}{2^{2}m^{2}}\frac{(-1)^{m-1}}{2^{2(m-1)}((m-1)!)^{2}}=-\frac{a_{2m-1}}{(2m)^{2}}$$

 

[DivGradCurl]

What really puzzles me from my work is that the part I did myself was checked n times... and later I get to results such as

$$x^2 J_0 ^{\prime \prime} (x) = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x)$$

and

$$x J_0 ^{\prime} (x) = - x J_1 (x)$$

with the aid of Mathematica.

So, chances are... there is some mistake I made, and that ultimately gave Mathematica the wrong series. So, my question is: where is the mistake up there?

 

[DivGradCurl]

What I'm trying to say is that although math gives the freedom to find the same solution through geometry and algebra, let's say, there is also the need to be consistent with the conditions you are given by the problem, as I explained you earlier. Please, don't get me wrong by saying this. I really need your help, but we do need to speak the same language.

 

[DivGradCurl]

What I can say is that there is something wrong with

$$x^2 J_0 ^{\prime \prime} (x) = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x)$$

the $$\frac{x^2}{2}$$ should not be there.

So, something must be wrong with

$$x^2 J_0 ^{\prime \prime} (x) = x^2 \sum _{n=2} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) \left( 2n-1 \right) x^{2n-2}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+2} \left( 2n+4 \right) \left( 2n+3 \right) x^{2n+4}}{2^{2n+4}\left[ \left( n+2 \right) ! \right] ^2 }$$

which I cannot find myself.

 

[arildno]

Your basic problem is that you don't use the same coefficients for x.
One place you use (2n+4) as your power, other places (2n+2).
This will most often result in miscalculations.
You need to bring the powers in x to the same power, otherwise you'll get a mess out of it.

 

[arildno]

Let:
$$J_{0}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{2^{2n}(n!)^{2}}$$
Then:
$$x^{2}J_{0}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2(n+1)}}{2^{2n}(n!)^{2}}=-\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2^{2(n-1)}((n-1)!)^{2}}$$
$$xJ_{0}'(x)=\sum_{n=1}^{\infty}\frac{2n(-1)^{n}x^{2n}}{2^{2n}(n!)^{2}}$$
$$x^{2}J_{0}''(x)=\sum_{n=1}^{\infty}\frac{2n(2n-1)(-1)^{n}x^{2n}}{2^{2n}(n!)^{2}}$$
Hence:
$$xJ_{0}'(x)+x^{2}J_{0}''(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}x^{2n}}{2^{2(n-1)}((n-1)!)^{2}}$$
That is:
$$xJ_{0}'(x)+x^{2}J_{0}''(x)+x^{2}J_{0}(x)=0$$

 

[DivGradCurl]

There is just one thing I cannot understand. When you differentiate the second time, doesn't the index change from $$n=1$$ to $$n=2$$, making it slightly different:

$$x^2 J^{\prime \prime} _0 (x) = \sum _{n=2} ^{\infty} \frac{\left( 2n \right) \left( 2n -1 \right) \left( -1 \right) ^n x^{2n}}{2^{2n} \left( n ! \right) ^2}$$

If so, wouldn't you need to change its index back to $$n=1$$? I hope my guess is wrong, because I can see the cancellations you get otherwise.

Thanks.

 

[arildno]

Let's take it once more:
$$J_{0}'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}2nx^{2n-1}}{2^{2n}(n!)^{2}}$$
Your lowest power in "x" is therefor $$x^{1}$$
Hence, when you differentiate that one you're left with a non-zero constant:
$$J_{0}''(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}2n(2n-1)x^{2n-2}}{2^{2n}(n!)^{2}}$$
This is a perfectly acceptable representation of the series.

Note that your own flawed series representation of $$x^{2}J_{0}''(x)$$ starts at $$x^{4}$$
and not at $$x^{2}$$ as it should.
This is probably the reason why you got a wrong result earlier.

 

[DivGradCurl]

I get your point. Now it's perfectly clear to me.

Thank you very much.