Bessel differential equation and it's series solution

Bessel differential equation:

We encounter Bessel differential equation while solving Helmoltz equation in cylindrical polar co-ordinate system. The form of Bessel differential equation is

$$x^2\frac{d^2y}{dx^2}+ x\frac{dy}{dx}+(x^2-n^2)y=0\dotsm(1)$$

Note:

\(y''+p(x)y'+Q(x)y=0\)

\(\frac{d^2y}{dx^2}+\frac1x \frac{dy}{dx}+(1-\frac{n^2}{x^2})y=0\)

Dividing both side by \(x^2\)

\(p(x)=\frac1x\) , \(Q(x)=1-\frac{n^2}{x^2}\)

\(\lim\limits_{x\to0} p(x)= \infty\)

\(\lim\limits_{x\to0} Q(x)=\infty\)

Singular point (x\(\to\) 0)

\(\lim\limits_{x\to0} x^2 Q(x)=1\)

\(\lim\limits_{x\to0} x^2\times\biggl(\frac{x^2-n^2}{x^2}\biggr)\)

$$=-n^2\;\;(finite)$$

Regular single point.

Since x\(\to\)0 is the regular singular point of equation (1) the solution of equation (1) can be written in power series solution. The solution of equation (1) will be in the form.

$$y=\sum_{r=0}^\infty a_r x^{k+r}, a_\circ=0\dotsm(2)$$

Since equation (2) is solution of equation (1) so, it should satisfy equation (1) , then

$$y'= \sum_{r=0}^\infty a_r(k+r)x^{(k+r-1)}$$

$$y''=\sum_{r=0}^\infty a_r(k+r)(k+r-1) x^{k+r-2}$$

Substituting y,y' and y''' in equation (1) , we get

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

$$or,\;\; \sum_{r=0}^\infty a_r(k+r)(k+r-1)x^{k+r}+\sum_{r=0}^\infty a_r(k+r)x^{k+r}+\sum_{r=0}^\infty a_r x^{k+r+2}-n^2\sum_{r=0}^\infty a_r x^{k+r}=0$$

$$or,\;\; \sum_{r=0}^\infty a_r[ (k+r)(k+r-1)+(k+r)-n^2] x^{k+r}+\sum_{r=0}^\infty a_r x^{k+r+2}=0$$

$$or,\;\; \sum_{r=0}^\infty a_r[(k+r)^2-n^2] x^{k+r}+\sum_{r=0}^\infty a_r x^{k+r+2}=0$$

Now, equating the coeffiencient of \(x^k\) [ i.e. set r=0] to zero,

We get,

$$ a_0[ k^2-n^2]=0$$

Since, \( a_0\ne0\)

$$\therefore k^2-n^2=0\to k^2=n^2\to k=\pm n$$

Now equating the coefficient of \(x^{k+4}\)(i.e r=1 ) to zero we get,

$$ a_1 [ (k+1)^2 – n^2]=0$$

The bracketed term is never zero for k=+n,-n. So, \(a_1\) must be equal to zero

$$ a_1= a_3= a_5= a_7=\dotsm=0$$

Now equating the coefficient of \(x^{k+r}\) to zero, we get,

$$ a_r[(k+r)^2-n^2]+ a_{r-2}=0$$

$$or\;\; a_r=\frac{-1}{[(k+r)^2-n^2]}a_{r-2}\dotsm(3)$$

This is the recurrence relation.

Case I:

\(k=+n, a_0=0, a_1=a_3=a_5=\dotsm=0\)

$$a_r=\frac{-1}{[(x+r)^2-n^2]} a_{r-2}$$

$$=\frac{-1}{[2nr+r^2]}a_{r-2}$$

$$=\frac{-1}{r(2n+r)}a_{r-2}$$

Put r=2,

$$ a_2=\frac{-1}{2(2n+2)}a_0 $$

$$=\frac{-1}{2\cdot 2 (n+1)}a_{r-2}$$

Put r=4,

$$ a_4=\frac{-1}{4(2n+4)} a_2$$

$$=\frac{-1}{4\cdot 2 (n+2)}\cdot \frac{-1}{2\cdot 2 (n+1)}a_0$$

$$=(-1)^2\cdot \frac{1}{2^2\cdot 2\cdot 4(n+1) (n+2)}$$

Similarly,

$$ a_{2J}=\frac{(-1)^J}{2^J(2.4…2J) (n+1)(n+2)…(n+J)} a_0$$

$$=\frac{(-1)^J a_0}{2^{2J}J! (n+1) (n+2)… (n+j)}\times \frac{n(n-1)(n-2)(n-3)…3.2.1}{n(n-1)(n-2)(n-3)…3.2.1}$$

$$=\frac{(-1)^J a_0 n!}{2^{2J} J! (n+J)!}$$

The solution of Bessel from equation (2) is

$$y=\sum_{r=0}^\infty a_r x^{k+r}= \sum_{r=0}^\infty a_r x^{n+4}= a_0 x^n + a_1 x^{n+1}+ a_2 x^{n+2}+ a_3 x^{n+3}+…$$

$$or,\;\; y= a_0 x^n+ a_2 x^{n+2}+ a_4 x^{n+4}+…$$

$$=\sum_{j=0}^\infty a_{2J} x^{n+2J}$$

$$=\sum_{J=0}^\infty \frac{(-1)^J n! a_0 x^{n+2J}}{J! (n+J)! 2^{2J}}$$

Put, \(a_0=\frac{1}{2^n n!}\) [ For polynomial ]

$$y= \sum_{J=0}^\infty \frac{(-1)^j x^{n+2j}}{2^{n+2J}J! (n+J)!}$$

$$=\sum_{J=0}^\infty \frac{(-1)^j (x/2)^{x+2J}}{J! (n+J)!}$$

$$\therefore\;\;\;\; J_n(x)= y= \sum_{J=0}^\infty \frac{(-1)^J (x/2)^{n+2J}}{J! (n+J)!}$$

Case II:

For k=-n, \(a_0\ne0, a_1= a_3= a_5= a_7=…=0\)

$$\therefore\;\; J_{-n}(x)=\sum_{J=0}^\infty \frac{(-1)^J (x/2)^{-n+2J}}{J! (-n+J)!}$$

Hence the complete solution is

$$y= AJ_n(x) + BJ_{-n}(x)$$

Prove:

\(J_{-n}(x)= (-1)^n J_n(x)\)

L.H.S

\( J_{-n}(x)=\sum_{j=0}^\infty\frac{(-1)^J(x/2)^{-n+2j}}{j! (-n+j)!}\)

As, n>J,

\((-n+j)!\to\infty\), then the solution vanishes, we have to correct limit.

\(J_{-n}(x)=\sum_{j=n}^\infty \frac{(-1)^j\cdot (x/2)^{-n+2j}}{j!(-n-j)!}\)

Put j= m + n

Whe J=n,m=0 and when j=\(\infty , m= \infty\)

\(=\sum_{m=0}^\infty \frac{(-1)^{m+n} (x/2)^{-n+2m+2n}}{(m+n)! m!}\)

\(=(-1)^n\sum_{m=0}^\infty \frac{(-1)^m (x/2)^{n+2m}}{m! (m+n)!}\)

\(=(-1)^nJ_n(x)\)

\(\therefore \;\;\; J_{-n}(x)= (-1)^n J_n(x)\)

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.