Example 10.1. Let \(\omega >0\). We use the power series method to obtain two linearly independent solutions of

\[ u''(x)+\omega ^2 u(x)=0. \]

We assume that \(u(x)=\sum _{k=0}^\infty a_kx^k\), so that

\[ u'(x)=\sum _{k=0}^\infty a_kkx^{k-1},\qquad u''(x)=\sum _{k=0}^\infty a_kk(k-1)x^{k-2}. \]

Substituting this in the differential equation gives

\[ \sum _{k=0}^\infty a_kk(k-1)x^{k-2}+\sum _{k=0}^\infty a_k\omega ^2x^k=0. \]

To make the exponents line up, we “shift the index” in the first sum. Explicitly, we define \(n=k-2\) (so that \(k=n+2\)) and we obtain (noting that the terms \(k=0\) and \(k=1\) (i.e. \(n=-2\) and \(n=-1\)) are zero, so we can omit them)

\[ \sum _{k=0}^\infty a_kk(k-1)x^{k-2} = \sum _{n=0}^\infty a_{n+2}(n+2)(n+1)x^n. \]

We then change the name for the index back to \(k\) and obtain for the differential equation

\[ \sum _{k=0}^\infty a_{k+2}(k+2)(k+1)x^k+\sum _{k=0}^\infty a_k\omega ^2x^k=0, \]

which we can rewrite as

\[ \sum _{k=0}^\infty \left (a_{k+2}(k+2)(k+1)+a_k\omega ^2\right )x^k=0. \]

This equality of functions means that the coefficients must all equal zero:

\[ a_{k+2}(k+2)(k+1)+a_k\omega ^2=0. \]

This we can re-write as

\[ a_{k+2}=-\omega ^2\,\frac {a_k}{(k+2)(k+1)}. \]

From this we see that \(a_0\) and \(a_1\) uniquely determine the whole sequence \((a_k)\). If we choose \(a_0=1\) and \(a_1=0\), then \(a_k=0\) for all \(k\) odd and therefore we obtain an even function. If we choose \(a_0=0\) and \(a_1=1\), then \(a_k=0\) for all \(k\) even and therefore we obtain an odd function. We see that in the first (even) case we have

\[ a_{2k}=\frac {(-1)^k\omega ^{2k}}{(2k)!},\qquad a_{2k+1}=0, \]

so that the power series is

\[ \sum _{k=0}^\infty \frac {(-1)^k}{(2k)!}(\omega x)^{2k}, \]

which is the power series of \(\cos (\omega x)\). In the second (odd) case we have

\[ a_{2k}=0,\qquad a_{2k+1}=\frac {(-1)^k\omega ^{2k+1}}{(2k+1)!}, \]

so that the power series is

\[ \sum _{k=0}^\infty \frac {(-1)^k}{(2k+1)!}(\omega x)^{2k+1}, \]

which is the power series of \(\sin (\omega x)\). Therefore the general solution is (as we of course already knew)

\[ C_1\cos (\omega x)+C_2\sin (\omega x). \]

Example 10.2. We consider a special case of Bessel’s equation (see Chapter 11 for the more general case; in the notation used there, the current case has \(\sigma =0\)). This equation is

\[ xu''+u'+xu=0. \]

We again make the Ansatz

\[ u(x)=\sum _{k=0}^\infty a_kx^k,\qquad u'(x)=\sum _{k=0}^\infty a_kkx^{k-1},\qquad u''(x)=\sum _{k=0}^\infty a_kk(k-1)x^{k-2}. \]

Substituting this in the differential equation gives

\[ x\sum _{k=0}^\infty a_kk(k-1)x^{k-2} +\sum _{k=0}^\infty a_kkx^{k-1} +x\sum _{k=0}^\infty a_kx^k=0, \]

which simplifies to

\[ \sum _{k=0}^\infty a_kk(k-1)x^{k-1} +\sum _{k=0}^\infty a_kkx^{k-1} +\sum _{k=0}^\infty a_kx^{k+1}=0. \]

To make the exponents line up, we shift the index in the third term: we define \(n:=k+2\) so that \(k=n-2\) and we obtain

\[ \sum _{k=0}^\infty a_kx^{k+1}=\sum _{n=2}^\infty a_{n-2}x^{n-1}. \]

Changing the name for the index back to \(k\), we obtain for the differential equation

\[ \sum _{k=0}^\infty a_kk(k-1)x^{k-1} +\sum _{k=0}^\infty a_kkx^{k-1} +\sum _{k=2}^\infty a_{k-2}x^{k-1}=0. \]

We note that the \(k\in \{0,1\}\) terms in the first sum are zero and that the \(k=0\) term in the second sum equals zero; therefore we obtain

\[ \sum _{k=2}^\infty a_kk(k-1)x^{k-1} +\sum _{k=1}^\infty a_kkx^{k-1} +\sum _{k=2}^\infty a_{k-2}x^{k-1}=0. \]

We can write this as

\[ \sum _{k=1}^\infty b_kx^{k-1}=0,\qquad b_k=\begin {cases} a_kk&k=1\\ a_kk^2+a_{k-2}&k>1. \end {cases} \]

Because a power series equals zero if and only if all its coefficients are zero, we obtain \(b_k=0\). From \(k=1\) we therefore obtain \(a_1=0\) and for \(k\geq 2\) we obtain

\[ a_kk^2+a_{k-2}=0, \]

which we can alternatively write as

\[ a_{k}=-\frac {a_{k-2}}{k^2},\qquad k\geq 2. \]

Since \(a_1=0\), this recurrence shows that \(a_k=0\) for all odd \(k\). Therefore \(u\) has a power series with only even terms (and is therefore an even function). We can explicitly solve the recurrence to obtain

\[ a_{2k}=\frac {(-1)^k}{2^{2k}(k!)^2}a_0, \]

so that we obtain (with the convenient choice \(a_0=1\))

\[ u(x)=\sum _{k=0}^\infty \frac {(-1)^k}{2^{2k}(k!)^2}x^{2k}. \]

This function is traditionally denoted by \(J_0\) and is called the Bessel function of order zero (of the first kind).

Remark. Note that in Example 10.1 we obtain two linearly independent solutions whereas in Example 10.2 we obtain only one linearly independent solution. This is because the second linearly independent solution (which does exist) is not given by a power series.