Back to the on-screen lesson ·
Classifying a point by what survives in the standard form, the Frobenius form $x^{r}$ times a series that a regular singular point allows, and the indicial equation that fixes $r$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to put an equation into standard form, decide whether a named point is ordinary, regular singular or irregular singular, compute the two limits a regular singular point supplies, write and solve the indicial equation, and say what its roots predict about the behaviour of the solutions near the point.
You can substitute a power series into an equation, find the recurrence relation between its coefficients and read the two independent solutions out of it. That worked because the coefficients of the equation were well behaved where the series was centred. This lesson is about the points where they are not, which is where the interesting equations of physics are singular and where the method has to be adjusted rather than abandoned.
| Term | What it means |
|---|---|
| Analytic | Equal to a convergent power series near the point; polynomials and ratios with non-vanishing denominators are. |
| Standard form | $y'' + p(x)y' + q(x)y = 0$, with the coefficient of $y''$ equal to $1$. |
| Ordinary, singular | A point is ordinary when $p$ and $q$ are both analytic there, and singular otherwise. |
| Regular singular point | A singular point where $(x - x_0)p$ and $(x - x_0)^{2}q$ are analytic. |
| Irregular singular point | A singular point where they are not. |
| Frobenius form | $x^{r}$ times a power series. |
| Indicial equation, exponents | The quadratic $r(r - 1) + p_0r + q_0 = 0$ that decides $r$; its roots are the exponents at the point. |
Everything depends on the standard form. Divide until the coefficient of $y''$ is $1$, and then look at what is left:
$$y'' + p(x)\,y' + q(x)\,y = 0.$$
At an ordinary point — both $p$ and $q$ analytic — the previous lesson's method works, and the two series solutions converge at least as far as the nearest singular point.
At a singular point a plain power series generally fails, because the solution itself need not be analytic there: $x^{1/2}$ and $\ln x$ are the usual culprits, and no power series is either of them. But not all failures are equal, and the dividing line is exactly this:
| Test at $x_0$ | Verdict | What to try |
|---|---|---|
| $p$, $q$ both analytic | ordinary | a power series |
| $(x-x_0)p$, $(x-x_0)^{2}q$ both analytic | regular singular | $x^{r}$ times a series |
| otherwise | irregular singular | nothing in this course |
The indicial equation. At a regular singular point write $p_0 = \lim (x - x_0)p$ and $q_0 = \lim (x - x_0)^{2}q$. Substituting $y = x^{r}(a_0 + a_1x + \cdots)$ and demanding that the lowest power balance gives
$$r(r-1) + p_0r + q_0 = 0.$$
Its two roots are the exponents, and each one starts a solution. That single quadratic is why the classification is worth making: it tells you the behaviour near the point before any coefficient of the series has been computed.
Another way: picture
A solution near a singular point is a power series wearing a coat. The coat is $x^{r}$, and it carries whatever bad behaviour there is — the fractional power, the pole, the sharp corner. Underneath, the series is perfectly ordinary. Regular singular means one coat is enough; irregular means the badness cannot be packaged into any single power, and the solution does something like $e^{1/x}$ that no coat covers.
Another way: steps
The asymmetry looks arbitrary and is not. Substitute $y = x^{r}$ into $y'' + py' + qy = 0$ near the origin and look at the size of each term: $y''$ behaves like $x^{r-2}$, $py'$ like $p\,x^{r-1}$, and $qy$ like $q\,x^{r}$.
For the three terms to be comparable — which is what has to happen if they are to cancel — $p$ may grow no faster than $x^{-1}$ and $q$ no faster than $x^{-2}$. Multiplying by $x$ and by $x^{2}$ respectively is precisely the test that they do not.
The simplest equation meeting the bound with equality is the Cauchy-Euler equation $x^{2}y'' + \alpha xy' + \beta y = 0$, whose solutions are exactly powers $x^{r}$ with no series needed. Every regular singular point looks like a Cauchy-Euler equation to leading order, and the indicial equation is that equation's characteristic equation.
Classifying before dividing. $x^{2}y'' + xy' + y = 0$ has polynomial coefficients as written, and the origin is still singular. The definition is about the standard form, always.
Forgetting to expand $r(r-1)$. The coefficient of $r$ in the indicial quadratic is $p_0 - 1$, not $p_0$, and the lost $-1$ moves both exponents.
Expecting two clean solutions. When the two exponents differ by a whole number, or coincide, the second solution generally needs a logarithm; the first one always comes out of the series. Knowing that is enough here.
Treating irregular as merely harder. It is a different situation, not a longer calculation: the Frobenius form does not apply at all, and a solution may have an essential singularity there.
Every method in this course answers one shape of equation and no other. So the first question is never how do I solve this but what is this — the order, whether it is linear, and which of the standard shapes it already is. An integrating factor applied to a nonlinear equation produces confident nonsense.
Deciding what can be done at a point, and starting the solution there, follows one procedure.
Why the indicial equation is the lowest power. In the Frobenius substitution the lowest power of $x$ comes only from $a_0x^{r}$, and near the point the equation looks like the Cauchy-Euler equation $x^{2}y'' + p_0xy' + q_0y = 0$. Substituting $x^{r}$ into that gives $r(r - 1) + p_0r + q_0$ times $x^{r}$. With $a_0 \ne 0$, the coefficient must vanish, and that is the indicial equation. For a genuine Cauchy-Euler equation nothing else is left, and $x^{r}$ is an exact solution.
How to check the answer. Check the classification by computing both products explicitly, not by eye. Check the exponents by their sum and product: they add to $1 - p_0$ and multiply to $q_0$. Check a Frobenius series by substituting its first term and confirming the lowest powers cancel. And read the exponents physically: a negative exponent means a solution that is infinite at the point, which a bounded physical quantity must reject.
A circular drumhead of radius $a$ vibrates in shapes that depend on the distance $r$ from its centre, and separating the variables leads to Bessel's equation $r^{2}R'' + rR' + (k^{2}r^{2} - m^{2})R = 0$. At the centre, $r = 0$, the equation has a regular singular point with exponents $\pm m$, exactly as this lesson computed for $m = 2$. The solution with the negative exponent is infinite at the centre, which a drumhead cannot be, so only the bounded one, the Bessel function $J_m$, describes the drum. The singular point is not a nuisance to be avoided; it is what selects the physical solution.
The edge of the drum is fixed, so $J_m(ka) = 0$, and the notes the drum can play come from the zeros of the Bessel function. The lowest, for $m = 0$, is at $ka \approx 2.405$, giving a frequency $f = \frac{2.405c}{2\pi a}$, with $c$ the wave speed on the membrane. A drum of radius $0.3$ metres with $c = 100$ metres a second sounds at $\frac{2.405 \times 100}{2\pi \times 0.3} \approx 128$ hertz. Unlike a string's, the higher zeros, at $3.832$, $5.136$, $5.520$ and so on, are not whole multiples of the first, which is why a drum has no clear musical pitch and a guitar string does.
A thick-walled pipe under internal pressure, such as a hydraulic cylinder or a gun barrel, stretches radially by an amount $u(r)$ that obeys $r^{2}u'' + ru' - u = 0$ between the inner and outer radii. This is a Cauchy-Euler equation, regular singular at $r = 0$, with indicial equation $s(s - 1) + s - 1 = s^{2} - 1 = 0$ and exponents $1$ and $-1$. So
$$u = Ar + \frac{B}{r},$$
the Lamé solution of 1852, with $A$ and $B$ fixed by the pressures inside and outside. The $\frac{B}{r}$ term is the one that matters: it grows as $r$ shrinks, so the stress is greatest at the inner surface, which is where such pipes crack. Here the singular point $r = 0$ lies outside the material, in the bore, so both exponents are allowed, unlike in the drum.
A designer who wants the pipe to carry more pressure learns from the exponents that thickening the wall helps less and less, because the $\frac{1}{r}$ stress near the bore barely changes as the outer radius grows. Gun barrels and high-pressure vessels are therefore shrunk-fit in layers instead, a practice built directly on this solution.
The radial part of the Schrödinger equation for an electron in a hydrogen atom has a regular singular point at the nucleus, with exponents $l$ and $-(l + 1)$ for orbital angular momentum $l$. The negative exponent gives a wave function that is infinite at the nucleus and is rejected, and the Frobenius series of the other, forced to terminate as in the Hermite and Legendre cases, gives the energy levels of hydrogen. Much of atomic physics rests on the classification of this lesson.
The classification is often read as a statement about difficulty — ordinary means easy, singular means hard — and then applied by eye to the equation as written. Both halves of that are wrong. It is a test, with a definite answer, applied to the standard form: an equation whose coefficients are handsome polynomials can be singular at the origin the moment it is divided through, and an equation with an alarming fraction in it can be perfectly ordinary. The second habit worth correcting is thinking the point is singular because the solution misbehaves there. The logic runs the other way: the coefficients are what is tested, and the test then predicts how the solution behaves, which is the only reason the classification earns its place before any solving begins.
Take $x^{2}y'' + xy' + (x^{2} - 4)y = 0$. Divide every term by $x^{2}$.
$y'' + \dfrac{1}{x}y' + \dfrac{x^{2} - 4}{x^{2}}y = 0: \qquad p = \dfrac{1}{x}, \quad q = \dfrac{x^{2} - 4}{x^{2}}$
Standard form first: the classification is about $p$ and $q$, never the original coefficients.
Check whether $p$ and $q$ survive at $0$.
$p \to \infty, \quad q \to -\infty \text{ as } x \to 0 \quad\Rightarrow\quad \text{singular}$
Both blow up, so the origin is not ordinary.
Multiply $p$ by $x$ and $q$ by $x^{2}$, and take the limits.
$xp = 1 \Rightarrow p_0 = 1, \qquad x^{2}q = x^{2} - 4 \Rightarrow q_0 = -4$
Both products are polynomials, hence analytic at $0$: a regular singular point.
Write and solve the indicial equation.
$r(r - 1) + 1 \cdot r - 4 = r^{2} - r + r - 4 = r^{2} - 4 = 0 \quad\Rightarrow\quad r = 2,\ -2$
Expand $r(r - 1)$ before collecting; here the $-r$ and $+r$ cancel.
Read the behaviour near the origin.
$y_1 \sim x^{2} \to 0, \qquad y_2 \sim x^{-2} \to \infty$
One solution is bounded at the origin and one is not: a drumhead cannot be infinite at its centre, so the physical solution is the first.
Take $x^{3}y'' + 2y' + y = 0$ at the origin. Divide by $x^{3}$.
$p = \dfrac{2}{x^{3}}, \qquad q = \dfrac{1}{x^{3}}$
Both fail at the origin, so it is singular.
Multiply $p$ by $x$ to test it.
$xp = \dfrac{2}{x^{2}} \to \infty$
One power of $x$ is not enough to tame $p$.
Multiply $q$ by $x^{2}$ as well.
$x^{2}q = \dfrac{1}{x} \to \infty$
Either failure alone would be enough; here both fail.
Classify the point.
$xp \text{ not analytic at } 0 \quad\Rightarrow\quad \text{irregular singular point}$
Regular needs both $xp$ and $x^{2}q$ to be analytic.
Draw the consequence.
$\text{no indicial equation; solutions may behave like } e^{1/x}$
$e^{1/x}$ has no expansion of the form $x^{r}$ times a series, so the honest response is to change method.
Take $2xy'' + y' + y = 0$. Divide by $2x$ and read the limits at $0$.
$p = \dfrac{1}{2x}, \quad q = \dfrac{1}{2x}; \qquad xp = \dfrac{1}{2} \Rightarrow p_0 = \dfrac{1}{2}, \quad x^{2}q = \dfrac{x}{2} \Rightarrow q_0 = 0$
Both products are analytic, so the origin is a regular singular point.
Solve the indicial equation.
$r(r - 1) + \dfrac{1}{2}r = r^{2} - \dfrac{1}{2}r = r\left(r - \dfrac{1}{2}\right) = 0 \quad\Rightarrow\quad r = 0, \ \dfrac{1}{2}$
The exponents differ by $\frac{1}{2}$, not a whole number, so each gives its own series.
Substitute $y = \sum a_nx^{n + r}$ and collect the coefficient of $x^{n + r - 1}$.
$2(n + r)(n + r - 1)a_n + (n + r)a_n + a_{n - 1} = 0 \quad\Rightarrow\quad (n + r)(2n + 2r - 1)a_n = -a_{n - 1}$
$2xy''$ and $y'$ lower the power by one; $y$ does not, so its index shifts to $a_{n - 1}$.
Put $r = \frac{1}{2}$ into the recurrence.
$\left(n + \dfrac{1}{2}\right)(2n)a_n = -a_{n - 1} \quad\Rightarrow\quad a_n = -\dfrac{a_{n - 1}}{n(2n + 1)}$
$\left(n + \frac{1}{2}\right) \cdot 2n = n(2n + 1)$.
Generate the first branch from $a_0 = 1$.
$a_1 = -\dfrac{1}{1 \cdot 3} = -\dfrac{1}{3}, \quad a_2 = -\dfrac{a_1}{2 \cdot 5} = \dfrac{1}{30}: \qquad y_1 = x^{1/2}\left(1 - \dfrac{x}{3} + \dfrac{x^{2}}{30} - \cdots\right)$
The coat $x^{1/2}$ carries the square root; the series underneath is ordinary.
Put $r = 0$ into the recurrence.
$n(2n - 1)a_n = -a_{n - 1} \quad\Rightarrow\quad a_n = -\dfrac{a_{n - 1}}{n(2n - 1)}$
The same recurrence with the other exponent.
Generate the second branch from $a_0 = 1$.
$a_1 = -\dfrac{1}{1 \cdot 1} = -1, \quad a_2 = -\dfrac{a_1}{2 \cdot 3} = \dfrac{1}{6}: \qquad y_2 = 1 - x + \dfrac{x^{2}}{6} - \cdots$
An ordinary power series this time, because the exponent is $0$.
Check the lowest power of the first branch.
$y_1 \approx x^{1/2}: \quad 2x\left(-\tfrac{1}{4}x^{-3/2}\right) + \tfrac{1}{2}x^{-1/2} = -\tfrac{1}{2}x^{-1/2} + \tfrac{1}{2}x^{-1/2} = 0$
The leading terms cancel exactly, which is what the indicial equation demanded. The general solution is $c_1y_1 + c_2y_2$ for $x > 0$.
Divide every term by $2x^{2}$.
$\dfrac{2x^{2}y''}{2x^{2}} + \dfrac{3xy'}{2x^{2}} - \dfrac{y}{2x^{2}} = 0 \quad\Rightarrow\quad p = \dfrac{3}{2x}, \qquad q = -\dfrac{1}{2x^{2}}$
Divide by the whole coefficient, the $2$ included.
Test and read the limits.
Solve the indicial equation.
Sort these four equations by what kind of point the named one is.
| An ordinary point | A regular singular point | An irregular singular point | |
|---|---|---|---|
| $y'' + 3xy' + 6y = 0$ at $x = 0$ | |||
| $x^{2}y'' + 3xy' + 6y = 0$ at $x = 0$ | |||
| $x^{3}y'' + 3y' + 6y = 0$ at $x = 0$ | |||
| $(x - 2)y'' + 3y' + 6y = 0$ at $x = 2$ |
Complete the worked solution: the exponents of $x^{2}y'' - 11xy' + 32y = 0$ at the origin.
Divide by $x^{2}$ and multiply back to read the two limits.
$p_0 = \lim_{x \to 0} x \cdot \dfrac{-11}{x} = -11, \qquad q_0 = \lim_{x \to 0} x^{2} \cdot \dfrac{32}{x^{2}} = 32$
For a Cauchy-Euler equation the powers of $x$ cancel exactly.
Write the indicial equation and expand $r(r - 1)$.
$r(r - 1) + p_0r + q_0 = r^{2} - r + (-11)r + 32$
The indicial equation is $r(r - 1) + p_0r + q_0 = 0$.
Collect the two $r$ terms into one coefficient.
$-r + (-11)r = \Big($ b $\Big)r$
Add the coefficients $-1$ and $-11$.
Factor the quadratic: the roots multiply to $q_0$ and add to minus the middle coefficient.
$(r - r_1)(r - r_2) = 0, \qquad r_1 r_2 = 32, \qquad r_1 + r_2 = 12$
Expanding $(r - r_1)(r - r_2)$ gives $r^{2} - (r_1 + r_2)r + r_1r_2$.
Read off the smaller exponent.
$r_1 =$ s
Near the origin one Frobenius solution behaves like $x^{r_1}$.
Read off the larger exponent.
$r_2 =$ l
The other behaves like $x^{r_2}$; neither is an ordinary power series in general.
For $x^{4}y'' + xy' + 3y = 0$, what kind of point is $x = 0$?
Each of these is singular at the origin. Fill in the two limits $p_0 = \lim_{x \to 0} xp(x)$ and $q_0 = \lim_{x \to 0} x^{2}q(x)$.
| The limit of x times p | The limit of x squared times q | |
|---|---|---|
| $x^{2}y'' + 5xy' + 5y = 0$ | ||
| $x^{2}y'' + 6xy' - 5y = 0$ | ||
| $x^{2}y'' - 10xy' + 6y = 0$ |
For $x^{2}y'' - 5xy' + 8y = 0$ the origin is a regular singular point. What is the larger root of its indicial equation?
Answer:
In a pressurised pipe the radial displacement $y$ of the wall, at distance $x$ from the axis, obeys a Cauchy-Euler equation; the constants below come from the pipe's material and the loads on its surfaces. Solve $x^{2}y'' + 2xy' - 6y = 0$ for $x > 0$, with $y(1) = 5$ and $y'(1) = -5$. Write $y$ as a formula in $x$.
Answer:
At a regular singular point an equation has $p_0 = 5$ and $q_0 = 5$. Write its indicial polynomial as a formula in $r$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In a pressurised pipe the radial displacement $y$ of the wall, at distance $x$ from the axis, obeys a Cauchy-Euler equation; the constants below come from the pipe's material and the loads on its surfaces. Solve $x^{2}y'' + 2xy' - 6y = 0$ for $x > 0$, with $y(1) = 3$ and $y'(1) = -4$. Write $y$ as a formula in $x$.
Answer:
You can classify a point from the standard form and produce the indicial equation and its exponents at a regular singular point. Say in your own words why the test multiplies $p$ by one power and $q$ by two.
15. Your turn: classify $x = 0$ for $2x^{2}y'' + 3xy' - y = 0$, and find its exponents, step 2
$xp = \tfrac{3}{2} \Rightarrow p_0 = \tfrac{3}{2}, \qquad x^{2}q = -\tfrac{1}{2} \Rightarrow q_0 = -\tfrac{1}{2}$
Both constants: regular singular.
15. Your turn: classify $x = 0$ for $2x^{2}y'' + 3xy' - y = 0$, and find its exponents, step 3
$r(r - 1) + \tfrac{3}{2}r - \tfrac{1}{2} = 0 \;\Rightarrow\; 2r^{2} + r - 1 = (2r - 1)(r + 1) = 0 \;\Rightarrow\; r = \tfrac{1}{2},\ -1$
A square root and a pole: neither a power series, which is why the plain method could never find them.