CHAPTER 4
Series Solutions, Frobenius, Bessel and Legendre
When an ODE has variable coefficients like \(xy''+y'+xy=0\), no elementary formula solves it. So we guess a power series, plug it in, and let the equation tell us every coefficient. This topic appears in every past paper, worth 25 marks.
4.1 Ordinary, regular singular and irregular singular points
Definition 4.1 for \(P_0(x)y''+P_1(x)y'+P_2(x)y=0\). Asked in 2021 Q2(a) and 2022 Q2(a).
Ordinary point\(P_0(a)\ne0\). Plain power series \(\sum a_n(x-a)^n\) works.
Regular singular\(P_0(a)=0\) but both
\[p=(x-a)\frac{P_1}{P_0},\quad q=(x-a)^2\frac{P_2}{P_0}\]
stay finite (analytic) at \(a\). Use Frobenius.
Irregular singular\(P_0(a)=0\) and \(p\) or \(q\) blows up at \(a\). Beyond the course.
"Analytic" in plain wordsIt can be written as a convergent power series \(\sum c_n(x-a)^n\) near \(a\). For a ratio of polynomials, that just means the factor \((x-a)\) cancels completely and nothing is left in the denominator.
Singular point detective: does (x − a)P₁/P₀ or (x − a)²P₂/P₀ explode?
p(x) = (x−a)P₁/P₀q(x) = (x−a)²P₂/P₀
4.2 Power series about an ordinary point
The 4-step recipe, with the step everyone gets wrong (aligning powers) made visual
① Assume \(y=\sum a_nx^n\)
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② Differentiate and substitute
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③ Collect each power \(x^k\) and set it to 0
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④ Recurrence gives \(a_2,a_3,\dots\) in terms of \(a_0,a_1\)
The index-shift trick (the red circles in your notes)
\(y'=\sum_{n\ge1}na_nx^{n-1}=\sum_{n\ge0}(n+1)a_{n+1}x^n\) and \(y''=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n\).
More generally, a term \(x^p\,y^{(d)}\) puts \(a_{k+d-p}\) into the column \(x^k\). The table below does this for you.
More generally, a term \(x^p\,y^{(d)}\) puts \(a_{k+d-p}\) into the column \(x^k\). The table below does this for you.
Power-series engine: pick an equation, click a column
y₁ series (a₀ = 1, a₁ = 0)y₂ series (a₀ = 0, a₁ = 1)exact numerical solution (RK4)
Exam pattern"Solve about x = a" with \(a\ne0\): substitute \(t=x-a\) first. Then it is the same method in \(t\). The engine has the 2021, 2022 and 2025 versions, already shifted for you.
Theorem 4.5About an ordinary point \(x=a\), solutions are \(y=\sum a_n(x-a)^n\), valid in an interval around \(a\). Watch the graph: for \((1-x^2)y''\dots\) the series stops matching past \(|x|=1\), where \(P_0=0\).
4.3 Frobenius method about a regular singular point
Theorem 4.6: \(y=x^m\sum a_nx^n\) with \(a_0\ne0\). The new unknown \(m\) comes from the indicial equation.
Assume \(y=\sum a_nx^{n+m}\), so \(y'=\sum(n+m)a_nx^{n+m-1}\) and \(y''=\sum(n+m)(n+m-1)a_nx^{n+m-2}\).
Substitute and look at the lowest power of x. Its coefficient is \(a_0\cdot I(m)\). Since \(a_0\ne0\), \(I(m)=0\). This quadratic is the indicial equation.
The rest of the powers give the recurrence, so \(a_1,a_2,\dots\) follow in terms of \(a_0\).
The two roots \(m_1\ge m_2\) decide the form of the second solution (cards below).
Case 1 · \(m_1-m_2\notin\mathbb Z\)Two plain Frobenius series:
\(y_1=x^{m_1}\sum a_nx^n\), \(y_2=x^{m_2}\sum b_nx^n\)
\(y_1=x^{m_1}\sum a_nx^n\), \(y_2=x^{m_2}\sum b_nx^n\)
Case 2 · \(m_1=m_2\)A log is guaranteed:
\(y_2=y_1\ln|x|+x^{m_1}\sum b_nx^n\)
\(y_2=y_1\ln|x|+x^{m_1}\sum b_nx^n\)
Case 3 · \(m_1-m_2=N\in\mathbb Z^+\)\(y_2=C\,y_1\ln|x|+x^{m_2}\sum b_nx^n\).
Sometimes \(C=0\), giving no log.
Sometimes \(C=0\), giving no log.
Frobenius engine
y₁ with N termsy₂ with N terms (when it is a plain series)40-term reference
Remark 4.9If \(m\) is not an integer, \(x^m\) is only real for \(x>0\). Work on \(x>0\), and treat \(x<0\) separately if needed. The same applies to \(x>a\) for a point \(x=a\).
4.4 Bessel's equation and Bessel functions
\(x^2y''+xy'+(x^2-\nu^2)y=0\): drums, circular pipes, anything round
Bessel function of the first kind
\[J_n(x)=\sum_{r=0}^\infty\frac{(-1)^r}{r!\,(n+r)!}\left(\frac x2\right)^{2r+n}\]
\(x=0\) is regular singular: \(p=1\) and \(q=x^2-\nu^2\) are both polynomials.
Indicial: \(m(m-1)+m-\nu^2=0\Rightarrow m=\pm\nu\).
Recurrence: \(a_k\big((m+k)^2-\nu^2\big)=-a_{k-2}\). With \(m=\nu\): \(a_k=-\dfrac{a_{k-2}}{k(k+2\nu)}\), and the odd coefficients are 0.
So \(a_{2r}=\dfrac{(-1)^ra_0}{2^{2r}r!(\nu+1)(\nu+2)\cdots(\nu+r)}\). Choosing \(a_0=\frac1{2^\nu\nu!}\) gives exactly \(J_\nu\).
\(\nu=0\) gives \(xy''+y'+xy=0\) (divide by \(x\)). Equal roots, so Case 2:
\[y_1=J_0=1-\frac{x^2}{4}+\frac{x^4}{64}-\cdots,\quad y_2=J_0\ln x+\frac{x^2}{4}-\frac{3x^4}{128}+\frac{11x^6}{13824}-\cdots\]
Bessel functions Jₙ(x)
Jₙ(x)partial sum, R termszeros
Exercise 4.17(a) \(x^2y''+xy'+(x^2-1)y=0\) is Bessel with \(\nu=1\): roots \(m=\pm1\) differ by 2, which gives Case 3 with a log, and \(y_1=J_1\). (b) About \(x=1\), the point is ordinary. Put \(t=x-1\) and use the power-series engine above (preset "Ex 4.17(b)").
4.5 Legendre polynomials
\((1-x^2)y''-2xy'+n(n+1)y=0\): one of the two series stops and becomes a polynomial
Rodrigues' formula\[P_n(x)=\frac1{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n\]normalised so that \(P_n(1)=1\).
Orthogonality on [−1, 1]\[\int_{-1}^1P_mP_n\,dx=0\ (m\ne n),\qquad\int_{-1}^1P_n^2\,dx=\frac{2}{2n+1}\]
Legendre polynomials and the Rodrigues machine
Pₙ(x²−1)ⁿ and its derivatives
\((x^2-1)^5=x^{10}-5x^8+10x^6-10x^4+5x^2-1\)
Differentiate 5 times: \(\frac{10!}{5!}x^5-5\frac{8!}{3!}x^3+10\frac{6!}{1!}x=30240x^5-33600x^3+7200x\)
Divide by \(2^5\cdot5!=3840\): \(\boxed{P_5(x)=\tfrac18\left(63x^5-70x^3+15x\right)}\)
The first few\(P_0=1,\ P_1=x,\ P_2=\frac12(3x^2-1),\ P_3=\frac12(5x^3-3x),\ P_4=\frac18(35x^4-30x^2+3)\)