First Connection Problem: From One Endpoint to Another

Suppose a differential equation has a solution that is regular at its left endpoint. Does that make it regular at its right endpoint? This first example gives a complete answer: the solution remains finite, but its derivative diverges. The coefficient of that singular behavior is a connection coefficient.

Before starting: you need second-order linear ODEs, differentiation of powers, and a two-by-two determinant. The required special-function notation is defined below. No monodromy, CFT, or gauge theory is needed.

First session: allow an estimated 60–90 minutes for the equation, matching calculation, endpoint conclusion, and Exercises 1–2. The numerical check, integral derivation, and branch exercise are optional extensions. This is a planning estimate, not a measured student completion time.

The equation and the question

On the real interval $0<z<1$, consider

$$12z(1-z)y''+(15-19z)y'-y=0.$$

This is the Gauss equation

$$z(1-z)y''+[c-(a+b+1)z]y'-aby=0, \qquad (a,b,c)=\left(\frac14,\frac13,\frac54\right).$$

Choose the solution analytic at zero with $y(0)=1$. Write it as $f_0(z)={}_2F_1(a,b;c;z)$. Here the notation means the convergent series

$${}_2F_1(a,b;c;z)=\sum_{n=0}^{\infty}d_nz^n, \qquad d_0=1,\quad d_{n+1}=\frac{(a+n)(b+n)}{(c+n)(n+1)}d_n.$$

Substitution into the ODE gives this recurrence. For our parameters, $f_0(z)=1+z/15+O(z^2)$. Its series converges for $|z|<1$, so it can be evaluated inside the interval. The endpoint $z=1$ sits on the boundary of that disc; termwise intuition there needs care.

Check your starting point: a different normalization $y(0)=2$ would double the entire solution. Every coefficient describing that solution must double too.

Choose a basis at the other endpoint

Let $s=1-z$ and $\delta=c-a-b=2/3$. Substituting $y\sim s^r$ near $s=0$ gives the indicial equation $r(r-\delta)=0$. There are two local behaviors: one analytic and one proportional to $s^{2/3}$. Unit-leading solutions are

$$\begin{aligned} f_1(z)&={}_2F_1\left(\frac14,\frac13;\frac13;1-z\right),\\ g_1(z)&=(1-z)^{2/3} {}_2F_1\left(1,\frac{11}{12};\frac53;1-z\right). \end{aligned}$$

Thus $f_1=1+O(s)$ and $g_1=s^{2/3}[1+O(s)]$. All powers are positive real on $0<z<1$. For complex continuation, use logarithms real on this interval and cuts on $(-\infty,0]$ and $[1,\infty)$.

Because $f_1,g_1$ are independent solutions of the same equation, there are constants $A,B$ such that

$$f_0=A f_1+B g_1.$$

This is the global question reduced to two numbers. The endpoint series fix the bases; the constants say how the solution selected at zero fits into the basis selected at one.

Match a value and a derivative

At any interior point $z_*$, solve

$$\begin{pmatrix}f_1&g_1\\f_1'&g_1'\end{pmatrix}_{z_*} \begin{pmatrix}A\\B\end{pmatrix} =\begin{pmatrix}f_0\\f_0'\end{pmatrix}_{z_*}.$$

Define the Wronskian by $\Wr[u,v]=uv'-u'v$. Cramer’s rule gives

$$A=\frac{\Wr[f_0,g_1]}{\Wr[f_1,g_1]}, \qquad B=\frac{\Wr[f_1,f_0]}{\Wr[f_1,g_1]}.$$

The signs depend on the order in each Wronskian. A useful independent check follows from Abel’s identity: if $y''+p y'+q y=0$, then $W'=-pW$. Here

$$\Wr[f_1,g_1] =-\frac23 z^{-5/4}(1-z)^{-1/3}.$$

The minus sign also follows directly from $g_1'\sim-\frac23(1-z)^{-1/3}$ near one. Although the Wronskians vary with $z$, their ratios are constant. That makes repeated matching a useful numerical test.

The analytic answer is

$$\begin{aligned} A&=\frac{\Gamma(5/4)\Gamma(2/3)}{\Gamma(1)\Gamma(11/12)} \simeq1.16278702942702179,\\ B&=\frac{\Gamma(5/4)\Gamma(-2/3)}{\Gamma(1/4)\Gamma(1/3)} =-\frac38. \end{aligned}$$

Here $\Gamma$ extends the factorial and satisfies $\Gamma(x+1)=x\Gamma(x)$. That recurrence alone simplifies $B$ to $-3/8$. The general connection formula is given in DLMF 15.10.21; an integral derivation for this particular case appears below.

Interpret the endpoint behavior

The connection relation gives

$$f_0(z)=A-\frac38(1-z)^{2/3}+O(1-z), \qquad f_0'(z)\sim\frac14(1-z)^{-1/3}.$$

So the analytic solution selected at zero has a finite limit at one, but its derivative diverges there. Boundedness is weaker than analyticity or differentiability at a singular endpoint.

If a boundary problem required the analytic branch at one, its condition would be $B=0$. Our fixed equation does not meet that condition. In an equation depending on a spectral parameter, zeros of the corresponding coefficient can select allowed parameter values. This is the basic idea behind many later spectral calculations.

Reproduce the calculation

Download gauss-starter-check.py. The supported baseline is Python 3.10 or later with mpmath==1.3.0. From the directory containing the downloaded file, run

python3 -m pip install mpmath==1.3.0
python3 gauss-starter-check.py

The program forms local series and their derivatives at $z=1/3,1/2,2/3$, then extracts $A,B$ through the Wronskians. It does not use the gamma answer to construct those series. It checks the gamma coefficients, Abel’s identity, the differential equation, and the compensating basis phase from Exercise 3.

With 240 terms and 60 working digits, the largest checked error was about $8.5\times10^{-41}$, below the declared $10^{-30}$ tolerance. The expected summary is PASS, with $A\simeq1.16278702942702179$ and $B=-0.375$. This is numerical evidence on the stated inputs, not a rigorous enclosure.

Try an intentionally insufficient truncation:

python3 gauss-starter-check.py --terms 24

It must report FAIL and exit with nonzero status. Then compare 80, 160, and 240 terms. With fixed arithmetic precision, the errors should decrease until rounding limits further improvement. Matching at a point outside a series’ convergence disc would be a different error; more terms would not repair it. Use --json to save the complete numerical record.

Optional derivation from one integral

For our positive parameters, Euler’s integral is

$$f_0(z)=K\int_0^1 u^{b-1}(1-u)^{c-b-1}(1-zu)^{-a}\,\dd u, \qquad K=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}.$$

The integral converges for $0<z<1$. Since $\delta>0$, setting $z=1$ also gives a convergent beta integral. Thus

$$A=f_0(1)=K\,\mathrm B(b,\delta) =\frac{\Gamma(c)\Gamma(\delta)}{\Gamma(c-a)\Gamma(c-b)}.$$

To recover $B$, differentiate the integral. Its singular part comes from $u$ near one. Put $1-u=s v$, where $s=1-z$. The exact change of variables gives

$$f_0'(z)=aK s^{\delta-1} \int_0^{1/s} \frac{(1-sv)^b v^{c-b-1}}{[1+(1-s)v]^{a+1}}\,\dd v.$$

For $0<s\leq1/2$, extend the integrand by zero past $1/s$. It is bounded by $v^{c-b-1}(1+v/2)^{-a-1}$, which is integrable at zero because $c-b>0$ and at infinity because $\delta<1$. Dominated convergence therefore justifies the limiting integral:

$$\begin{aligned} f_0'(z) &\sim aK s^{\delta-1} \int_0^\infty v^{c-b-1}(1+v)^{-a-1}\,\dd v\\ &=\frac{\Gamma(c)\Gamma(1-\delta)}{\Gamma(a)\Gamma(b)} s^{\delta-1}. \end{aligned}$$

Comparing with $B g_1'\sim-\delta B s^{\delta-1}$ yields the stated coefficient because $\Gamma(1-\delta)=-\delta\Gamma(-\delta)$. The inequalities $b>0$, $c-b>0$, and $0<\delta<1$ justify this particular integral argument. More general parameters require analytic continuation and attention to resonant limits.

Exercises with progressive help

1. Change the normalization

Replace $g_1$ by $\widetilde g_1=2g_1$, keeping $f_0$ and $f_1$ fixed. Find the new coefficient and decide whether the endpoint derivative changes.

Hint

Write the same function in both bases; the product of a basis vector and its coefficient must agree.

Solution

$\widetilde B=B/2=-3/16$. The physical function and its derivative are unchanged because $\widetilde B\widetilde g_1=B g_1$. A connection coefficient needs its basis normalization to have meaning.

2. Diagnose an endpoint mistake

A calculation finds that both $f_1$ and $g_1$ are bounded at one and concludes that they are both analytic there. Identify the mistaken step and formulate the condition selecting the analytic branch.

Hint

Differentiate the leading noninteger power. An analytic function has an ordinary Taylor expansion in integer nonnegative powers.

Solution

$g_1\sim(1-z)^{2/3}$ is bounded but has a divergent derivative. It has a branch point rather than a Taylor expansion at one. For $y=A f_1+B g_1$, analyticity requires $B=0$ in this nonresonant basis.

3. Transfer: change the branch convention

At the same matching point, define a new basis by replacing $\Log(1-z)$ with $\Log(1-z)+2\pi\ii$ in $g_1$ while keeping the same continued solution $f_0$. Find the compensating coefficient. Explain why this is different from continuing $f_0$ around the branch point.

Hint 1

Only the noninteger-power factor changes in the new local basis.

Hint 2

Use $\widetilde g_1=\ee^{2\pi\ii\delta}g_1$ in the unchanged connection relation.

Solution

$\widetilde B=\ee^{-4\pi\ii/3}B$ and $\widetilde A=A$. This is a change of representation of a fixed solution. If instead the solution itself is continued once counterclockwise around one, its nonanalytic part acquires the phase $\ee^{4\pi\ii/3}$ relative to the original local branch. That is monodromy. A basis change and a change of continuation path answer different questions.

Continue when this calculation is clear

You are ready to continue if you can extract $B$ with the correct sign, explain why it changes under a basis rescaling, and distinguish boundedness from the endpoint condition you actually want.

Read the full hypergeometric benchmark for the entire matrix, resonant limits, and a spectral application. Then use scalar equations and systems to understand the matrix structure, or move to the Heun laboratories to see why an extra singularity makes the same connection problem harder.

The source identities used here are the hypergeometric local bases and Wronskians, Euler integral, and Gauss value at one.