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

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

Stumbling Robot

Prove that the Bessel functions are solutions of the Bessel equation

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