Differential equations – Series solutions of linear differential equations – The Bessel differential equation

Consider the Bessel differential equation \(x^2y''(x)+xy'(x)+(x^2-\nu^2)y(x)=0\) with \(\nu\geq0\). Note that \(x=0\) is the only singular point and that

\[\lim\limits_{x\to0}x\cdot\frac{x}{x^2}=1\quad\text{and}\quad\lim\limits_{x\to0}x^2\cdot\frac{x^2-\nu^2}{x^2}=-\nu^2.\]

So the indicial equation is \(r(r-1)+r-\nu^2=0\) or equivalently \(r^2-\nu^2=0\). So the indices are \(r=\nu\) and \(r=-\nu\). So for \(2\nu\notin\{0,1,2,\ldots\}\) there are two linearly independent solutions of the form \(y_1(x)=x^{\nu}\displaystyle\sum_{n=0}^{\infty}a_nx^n\) and \(y_2(x)=x^{-\nu}\displaystyle\sum_{n=0}^{\infty}b_nx^n\). For \(\nu=0\) there are two linearly independent solutions of the form \(y_1(x)=\displaystyle\sum_{n=0}^{\infty}a_nx^n\) and \(y_2(x)=y_1(x)\ln|x|+\displaystyle\sum_{n=0}^{\infty}b_nx^n\). For \(2\nu\in\{1,2,3,\ldots\}\) there are two linearly independent solutions of the form \(y_1(x)=x^{\nu}\displaystyle\sum_{n=0}^{\infty}a_nx^n\) and \(y_2(x)=Ay_1(x)\ln|x|+x^{-\nu}\displaystyle\sum_{n=0}^{\infty}b_nx^n\), where \(A\) might be equal to zero or not.

So in the case that \(\nu^2=\frac{1}{4}\) the indices are \(r_1=\frac{1}{2}\) and \(r_2=-\frac{1}{2}\). In that case for \(x > 0\) we have two linearly independent solutions

\[y_1(x)=x^{\frac{1}{2}}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n+1)!}=\frac{\sin(x)}{\sqrt{x}}\quad\text{and}\quad y_2(x)=x^{-\frac{1}{2}}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\frac{\cos(x)}{\sqrt{x}}.\]

This can easily be seen by substituting \(y(x)=x^{-\frac{1}{2}}u(x)\) into the differential equation with \(\nu^2=\frac{1}{4}\). Then we have:

\[y'(x)=x^{-\frac{1}{2}}u'(x)-\tfrac{1}{2}x^{-\frac{3}{2}}u(x)\quad\text{en}\quad y''(x)=x^{-\frac{1}{2}}u''(x)-x^{-\frac{3}{2}}u'(x)+\tfrac{3}{4}x^{-\frac{5}{2}}u(x).\]

Substitution then gives:

\[x^{\frac{3}{2}}u''(x)-x^{\frac{1}{2}}u'(x)+\tfrac{3}{4}x^{-\frac{1}{2}}u(x)+x^{\frac{1}{2}}u'(x)-\tfrac{1}{2}x^{-\frac{1}{2}}u(x) +x^{\frac{3}{2}}u(x)-\tfrac{1}{4}x^{-\frac{1}{2}}u(x)=0\quad\Longleftrightarrow\quad x^{\frac{3}{2}}\left(u''(x)+u(x)\right)=0.\]

This implies that \(u(x)=c_1\sin(x)+c_2\cos(x)\) and therefore: \(y(x)=x^{-\frac{1}{2}}u(x)=c_1\displaystyle\frac{\sin(x)}{\sqrt{x}}+c_2\frac{\cos(x)}{\sqrt{x}}\).

So for \(\nu\geq0\) there exists a solution of the form:

\[y(x)=x^{\nu}\sum_{n=0}^{\infty}c_nx^n=\sum_{n=0}^{\infty}c_nx^{n+\nu}\quad\Longrightarrow\quad y'(x)=\sum_{n=0}^{\infty}(n+\nu)c_nx^{n+\nu-1}\quad\Longrightarrow\quad y''(x)=\sum_{n=0}^{\infty}(n+\nu)(n+\nu-1)c_nx^{n+\nu-2}.\]

Substitution then gives:

\[x^2\sum_{n=0}^{\infty}(n+\nu)(n+\nu-1)c_nx^{n+\nu-2}+x\sum_{n=0}^{\infty}(n+\nu)c_nx^{n+\nu-1} +(x^2-\nu^2)\sum_{n=0}^{\infty}c_nx^{n+\nu}=0\]

or equivalently

\[\sum_{n=0}^{\infty}(n+\nu)(n+\nu-1)c_nx^{n+\nu}+\sum_{n=0}^{\infty}(n+\nu)c_nx^{n+\nu} +\sum_{n=0}^{\infty}c_nx^{n+\nu+2}-\nu^2\sum_{n=0}^{\infty}c_nx^{n+\nu}=0.\]

Hence:

\[\{\nu(\nu-1)+\nu-\nu^2\}c_0x^{\nu}+\{(\nu+1)\nu+\nu+1-\nu^2\}c_1x^{\nu+1} +\sum_{n=2}^{\infty}\left[\{(n+\nu)(n+\nu-1)+n+\nu-\nu^2\}c_n+c_{n-2}\right]x^{n+\nu}=0.\]

This implies that \((2\nu+1)c_1=0\) and \(\{(n+\nu)^2-\nu^2\}c_n+c_{n-2}=0\) for \(n=2,3,4,\ldots\). For \(\nu\geq0\) we then have

\[c_1=0\quad\text{and}\quad n(n+2\nu)c_n+c_{n-2}=0,\quad n=2,3,4,\ldots.\]

Hence:

\[c_{2k-1}=0\quad\text{en}\quad c_{2k}=-\frac{c_{2k-2}}{2k(2k+2\nu)}=\frac{(-1)^k}{2^{2k}k!\,(\nu+1)(\nu+2)\cdots(\nu+k)}c_0,\quad k=1,2,3,\ldots.\]

Hence: \(y(x)=x^{\nu}+x^{\nu}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^kx^{2k}}{2^{2k}k!\,(\nu+1)(\nu+2)\cdots(\nu+k)}\) is a solution.

For \(\nu=0\) this ist \(y(x)=\displaystyle\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{2^{2k}(k!)^2}=J_0(x)\), the Bessel function of the first kind of order \(0\). Then a second solution has the form

\begin{align*} &y(x)=J_0(x)\ln|x|+\sum_{n=0}^{\infty}d_nx^n\quad\Longrightarrow\quad y'(x)=J_0'(x)\ln|x|+\frac{J_0(x)}{x}+\sum_{n=1}^{\infty}nd_nx^{n-1}\\[2.5mm] &{}\hspace{25mm}\Longrightarrow\quad y''(x)=J_0''(x)\ln|x|+\frac{2J_0'(x)}{x}-\frac{J_0(x)}{x^2}+\sum_{n=2}^{\infty}n(n-1)d_nx^{n-2}. \end{align*}

Substitution into the differential equation \(x^2y''(x)+xy'(x)+x^2y(x)=0\) then gives:

\begin{align*} &x^2J_0(x)\ln|x|+2xJ_0'(x)-J_0(x)+\sum_{n=2}^{\infty}n(n-1)d_nx^n+xJ_0'(x)\ln|x|+J_0(x)+\sum_{n=1}^{\infty}nd_nx^n +x^2J_0(x)\ln|x|+\sum_{n=0}^{\infty}d_nx^{n+2}=0 \end{align*}

or equivalently

\[\left(x^2J_0''(x)+xJ_0'(x)+x^2J_0(x)\right)\ln|x|+2xJ_0'(x)+d_1x+\sum_{n=2}^{\infty}\left[\{n(n-1)+n\}d_n+d_{n-2}\right]x^n=0.\]

Now we have \(x^2J_0''(x)+xJ_0'(x)+x^2J_0(x)=0\) and

\[J_0(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{2^{2k}(k!)^2}\quad\Longrightarrow\quad J_0'(x)=\sum_{k=1}^{\infty}\frac{(-1)^kx^{2k-1}}{2^{2k-1}k!(k-1)!} \quad\Longrightarrow\quad 2xJ_0'(x)=\sum_{k=1}^{\infty}\frac{(-1)^kx^{2k}}{2^{2k-2}k!(k-1)!}.\]

Hence:

\[\sum_{k=1}^{\infty}\frac{(-1)^kx^{2k}}{2^{2k-2}k!(k-1)!}+d_1x+\sum_{n=2}^{\infty}\left(n^2d_n+d_{n-2}\right)x^n=0\]

or equivalently

\[\sum_{k=1}^{\infty}\frac{(-1)^kx^{2k}}{2^{2k-2}k!(k-1)!}+d_1x +\sum_{k=1}^{\infty}\left(4k^2d_{2k}+d_{2k-2}\right)x^{2k}+\sum_{k=1}^{\infty}\left((2k+1)^2d_{2k+1}+d_{2k-1}\right)x^{2k+1}.\]

This implies that

\[d_1=0\quad\text{and}\quad(2k+1)^2d_{2k+1}+d_{2k-1},\quad k=1,2,3\ldots\quad\Longrightarrow\quad d_{2k+1}=0,\quad k=0,1,2,\ldots\]

and

\[\sum_{k=1}^{\infty}\frac{(-1)^k}{2^{2k-2}k!(k-1)!}+\sum_{k=1}^{\infty}\left(4k^2d_{2k}+d_{2k-2}\right)=0 \quad\Longrightarrow\quad\frac{(-1)^k}{2^{2k-2}k!(k-1)!}+4k^2d_{2k}+d_{2k-2}=0,\quad k=1,2,3,\ldots.\]

Using the harmonic numbers \(H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}\) for \(n=1,2,3,\ldots\) the solution can be written as \(d_{2k}=\displaystyle\frac{(-1)^{k+1}H_k}{2^{2k}(k!)^2}\). Indeed:

\begin{align*} 4k^2d_{2k}+d_{2k-2}&=\frac{(-1)^{k+1}H_k}{2^{2k-2}((k-1)!)^2}+\frac{(-1)^kH_{k-1}}{2^{2k-2}((k-1)!)^2} =\frac{(-1)^{k+1}}{2^{2k-2}((k-1)!)^2}\left(H_k-H_{k-1}\right)\\[2.5mm] &=\frac{(-1)^{k+1}}{2^{2k-2}((k-1)!)^2}\cdot\frac{1}{k}=\frac{(-1)^{k+1}}{2^{2k-2}k!(k-1)!},\quad k=1,2,3,\ldots \quad\text{with}\quad d_0=0. \end{align*}

Hence: \(y(x)=J_0(x)\ln|x|+\displaystyle\sum_{k=1}^{\infty}d_{2k}x^{2k}=J_0(x)\ln|x|-\sum_{k=1}^{\infty}\frac{(-1)^kH_k}{2^{2k}(k!)^2}x^{2k}\) is also a solution.

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Last modified on May 13, 2021