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Prove that the Bessel functions are solutions of the Bessel equation

In the previous exercise (Section 11.16, Exercise #10) we defined the Bessel functions of the first kind of orders zero and one by,

    \[ J_0 (x) = \sum_{n=0}^{\infty} (-1)^n  \frac{x^{2n}}{(n!)^2 2^{2n}}, \qquad J_1(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}}. \]

Prove that these Bessel functions are solutions of the differential equation

    \[ x^2 y'' + xy' + (x^2 - n^2) y = 0 \]

when n = 0 and n = 1, respectively.


Proof. In the previous exercise (linked above) we proved the following

    \[ J'_0(x) = -J_1(x), \qquad \text{and} \qquad xJ_0(x) = \left( x J_1(x) \right)'. \]

For the case n = 0 we have the differential equation

    \[ x^2 y'' + xy' + x^2 y = 0. \]

Plugging in y = J_0(x) we then have

    \begin{align*}  x^2 y'' + xy' + x^2y &= x^2 J_0''(x) + xJ_0'(x) + x^2 J_0(x) \\[9pt]  &= x^2 \left( -J_1'(x) \right) + x \left(-J_1(x)\right) + x^2 J_0(x) \\[9pt]  &= -x \left( xJ_1'(x) \right)  - x J_1(x) + x^2 J_0(x) \\[9pt]  &= -x \left( \left(xJ_1(x) \right)' - J_1(x) \right) - x J_1(x) + x^2 J_0(x) \\[9pt]  &= -x \left( xJ_0(x) - J_1(x)  \right) - xJ_1(x) + x^2 J_0(x) \\[9pt]  &= -x^2 J_0(x) + xJ_1(x) - xJ_1(x) + x^2 J_0(x) \\[9pt]  &= 0. \end{align*}

So, J_0(x) is indeed a solution in the case n =0.

Now, from the previous exercise we have the relations

    \[ J_0'(x) = -J_1(x), \qquad xJ_0(x) = xJ_1'(x) + J_1(x). \]

Starting with the n = 0 case we differentiate,

    \begin{align*}   && x^2 J_0''(x) + xJ_0'(x) + x^2 J_0(x) &= 0 \\[9pt]  \implies && x^2 J_0'''(x) + 2x J_0''(x) + xJ_0''(x) + J_0'(x) + x^2 J_0'(x) + 2x J_0(x) &= 0 \\[9pt]  \implies && x^2 J_0'''(x) + 3x J_0''(x) + (1+x^2)J_0'(x) + 2x J_0(x) &= 0. \end{align*}

Using the relations above from the previous problem, we substitute

    \begin{align*}  && x^2 J_0'''(x) + 3x J_0''(x) + (1+x^2)J_0'(x) + 2x J_0(x) &= 0 \\[9pt]  \implies && -x^2 J_1''(x) -3x J_1'(x) - (1+x^2) J_1(x) + 2 \left( xJ_0(x) \right) &= 0 \\[9pt]  \implies && -x^2 J_1''(x) - 3x J_1'(x) - (1+x^2) J_1(x) + 2 \left( xJ_1'(x) + J_1(x) \right) &= 0 \\[9pt]  \implies && -x^2 J_1''(x) - 3x J_1'(x) - J_1 (x) - x^2 J_1(x) + 2x J_1'(x) + 2 J_1(x) &= 0 \\[9pt]  \implies && -x^2 J_1''(x) - x J_1'(x) - x^2 J_1(x) + J_1(x) &= 0 \\[9pt]  \implies && x^2 J_1''(x) + x J_1'(x) + x^2 J_1(x) - J_1(x) &= 0 \\[9pt]  \implies && x^2 J_1''(x) + x J_1'(x) + (x^2 - 1) J_1(x) &= 0. \end{align*}

Hence, J_1(x) is indeed a solution of the differentiation equation

    \[ x^2 y'' + xy' + (x^2 - 1)y = 0. \qquad \blacksquare \]

Prove some properties of the Bessel functions of the first kind of orders zero and one

We define the Bessel functions of the first kind of orders zero and one by

    \[ J_0 (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}}, \qquad J_1 (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}}. \]

  1. Prove that both J_0(x) and J_1(x) converge for all x \in \mathbb{R}.
  2. Prove that J'_0(x) = -J_1(x).
  3. If we define two new functions

        \[ j_0(x) = xJ_0(x), \qquad j_1 (x) = x J_1(x) \]

    prove that j_0(x) = j_1'(x).


  1. Proof. For the order zero Bessel function of the first kind J_0 we have a_n = \frac{x^{2n}}{(n!)^2 2^{2n}} and so using the ratio test we have

        \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+2}}{(n+1)!^2 2^{2n+2}} \right) \left( \frac{(n!)^2 2^{2n}}{x^{2n}} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{x^2}{4 (n+1)^2} \\[9pt]  &= 0. \end{align*}

    Hence, J_0 (x) converges for all x \in \mathbb{R}.

    For J_1(x) we have a_n = \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}} and so

        \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+3}}{(n+1)! (n+2)! 2^{2n+3}} \right) \left( \frac{n!(n+1)! 2^{2n+1}}{x^{2n+1}} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{x^2}{4(n+1)(n+2)} \\[9pt]  &= 0. \end{align*}

    Hence, J_1(x) converges for all x \in \mathbb{R}. \qquad \blacksquare

  2. Proof. We compute the derivative of J_0(x) directly,

        \begin{align*}  && J_0(x) &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}} \\[9pt]  \implies && J_0'(x) &= \sum_{n=1}^{\infty} (-1)^n \frac{2nx^{2n-1}}{(n!)^2 2^{2n}} \\[9pt]  &&&= \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(n-1)!n! 2^{2n-1}} \\[9pt]  &&&= \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2(n+1)-1}}{(n+1-1)!(n+1)! 2^{2(n+1)-1}} \\[9pt]  &&&= - \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt]  &&&= - J_1(x). \qquad \blacksquare \end{align*}

  3. Proof. First, we have

        \[ j_0 (x) = x J_0(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(n!)^2 2^{2n}}. \]

    On the other hand we have,

        \[ j_1 (x) = xJ_1(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+2}}{n!(n+1)! 2^{2n+1}}. \]

    Therefore,

        \begin{align*}   j_1'(x) &= \sum_{n=0}^{\infty} (-1)^n \frac{(2n+2)x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt]  &= \sum_{n=0}^{\infty} (-1)^n \frac{2(n+1)x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt]  &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(n!)^2 2^{2n}}. \end{align*}

    Hence, we indeed have j_0(x) = j_1'(x). \qquad \blacksquare