Boundary-value problems

Self-adjoint operators, Sturm-Liouville theory and eigenfunction expansions.

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Sturm-Liouville theory

Sturm-Liouville theory

Definition

Adjoint and self-adjoint differential operators

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The formal adjoint of a second-order differential operator, Lagrange's identity and Green's formula with the boundary term written out, the criterion for formal self-adjointness, and the integrating factor that puts any second-order equation into self-adjoint (Sturm-Liouville) form.

Prerequisites: Structure of linear equations and the Wronskian, Inner products and norms

Definition(Formal adjoint and conjunct)AdaptedAdapted from Stakgold & Holst (2011), §2.5, eqs. (2.5.8)–(2.5.9), p. 167, and §3.2, eqs. (3.2.17)–(3.2.20), p. 197; Hassani (2013), §14.5, eq. (14.21), p. 434

Work on a finite interval $[a,b]$ with the second-order operator

L = a_2(x)\,D^2 + a_1(x)\,D + a_0(x), \qquad D = \frac{\mathrm d}{\mathrm d x},

whose coefficients lie in $C^2[a,b]$ — the hypothesis under which Stakgold and Holst (Stakgold & Holst (2011), §2.5, p. 166) set up the identities below — and which we take to be real-valued. The formal adjoint of $L$ is the second-order operator

L^{*} = a_2\,D^2 + \bigl(2a_2' - a_1\bigr)D + \bigl(a_2'' - a_1' + a_0\bigr),

and the conjunct of two functions $u,v$ is the bilinear expression

J(u,v) = a_2\,\bigl(v u' - u v'\bigr) + \bigl(a_1 - a_2'\bigr)\,u v .

Hassani writes the same operator as $L^{\dagger}$ and the same expression as $p_2 v u' - (p_2 v)' u + p_1 u v$ , which expands to the display above. Note that $J$ involves only $u,v$ and their first derivatives: one order less than $L$ .

The word formal is deliberate. $L^{*}$ is produced by the coefficients of $L$ alone; no boundary conditions, no inner product and no domain have been chosen yet. Boundary conditions enter in the next definition, and the distinction is the subject of the remark below.

Definition(Lagrange's identity and Green's formula)AdaptedAdapted from Stakgold & Holst (2011), §2.5, eqs. (2.5.7) and (2.5.10)–(2.5.14), pp. 166–167; Hassani (2013), §14.5, Lagrange identities (14.22)–(14.23), pp. 434–435

For all $u,v \in C^2[a,b]$ ,

v\,Lu - u\,L^{*}v = \frac{\mathrm d}{\mathrm d x}\,J(u,v) ,

which is Lagrange's identity, and therefore, integrating from $a$ to $b$ ,

\int_a^b \bigl(v\,Lu - u\,L^{*}v\bigr)\,\mathrm d x \;=\; \Bigl[\,J(u,v)\,\Bigr]_a^b ,

which is Green's formula.

$L$ is called formally self-adjoint when $L^{*} = L$ . For a second-order operator this happens if and only if

a_2' = a_1 ,

in which case $L$ collapses to the divergence form $Lu = \bigl(a_2 u'\bigr)' + a_0 u$ and the conjunct loses its last term:

J(u,v) = a_2\,\bigl(v u' - u v'\bigr), \qquad \int_a^b \bigl(v\,Lu - u\,Lv\bigr)\,\mathrm d x = \Bigl[\,a_2\,\bigl(v u' - u v'\bigr)\,\Bigr]_a^b .

Idea of the proofOriginal

Both halves come from moving the derivatives off $u$ and onto $v$ , and we do it pointwise, before any integration. The product rule rewrites each term of $v\,Lu$ as the derivative of something plus a term carrying no derivative of $u$ : twice for $v\,a_2u''$ , once for $v\,a_1u'$ , not at all for $v\,a_0u$ . Collecting the two pieces, the coefficient that has accumulated in front of each derivative of $v$ is exactly the corresponding coefficient of $L^{*}$ , while what sits inside the total derivative is exactly $J(u,v)$ — and that identity of functions is Lagrange's. Green's formula is then one integration and the fundamental theorem of calculus, which is where the endpoints appear.

ProofAdaptedAdapted from Stakgold & Holst (2011), §2.5, the derivation of (2.5.7)–(2.5.11), pp. 166–167; Hassani (2013), §14.5, the computation leading to (14.22), p. 434

Move the derivatives off $u$ , one term at a time. Expand each term of $v\,Lu$ using the product rule, aiming to produce a derivative of something plus a term with no derivative on $u$ :
$v\,a_2 u'' = \bigl(a_2 v\,u' - (a_2 v)'\,u\bigr)' + (a_2 v)''\,u , \qquad v\,a_1 u' = \bigl(a_1 v\,u\bigr)' - (a_1 v)'\,u .$
For the first: differentiating the bracket gives $(a_2v)'u' + a_2 v u'' - (a_2v)''u - (a_2v)'u'$ , whose middle terms are what we want and whose outer terms cancel. The second is the ordinary product rule rearranged.
Assemble. Adding the two displays and $v\,a_0 u$ ,
$v\,Lu = \frac{\mathrm d}{\mathrm d x}\Bigl[\,a_2 v\,u' - (a_2 v)'\,u + a_1 v\,u \,\Bigr] \;+\; u\,\Bigl[\,(a_2 v)'' - (a_1 v)' + a_0 v\,\Bigr].$
Identify the two pieces. Expanding the second bracket with the product rule,
$(a_2 v)'' - (a_1 v)' + a_0 v = a_2 v'' + \bigl(2a_2' - a_1\bigr)v' + \bigl(a_2'' - a_1' + a_0\bigr) v = L^{*}v ,$
which is where the hypothesis $a_2 \in C^2$ , $a_1 \in C^1$ is used: $a_2''$ and $a_1'$ must exist. Expanding the first bracket the same way,
$a_2 v\,u' - (a_2 v)'\,u + a_1 v\,u = a_2\bigl(v u' - u v'\bigr) + \bigl(a_1 - a_2'\bigr)u v = J(u,v).$
Step 2 now reads $v\,Lu = \frac{\mathrm d}{\mathrm d x} J(u,v) + u\,L^{*}v$ , which is Lagrange's identity. Integrating over $[a,b]$ and using the fundamental theorem of calculus gives Green's formula. (Stakgold and Holst go the other way round: they integrate by parts to reach Green's formula first, then, since it holds for every upper limit, differentiate it with respect to that limit to obtain Lagrange's identity. The two routes give the same pair of statements.)
The criterion for formal self-adjointness. Two second-order operators agree as operators exactly when their three coefficients agree, because applying them to $1, x, x^2$ recovers the coefficients. Comparing $L^{*}$ with $L$ term by term, the $D^2$ coefficients already agree, the $D$ coefficients agree iff $2a_2' - a_1 = a_1$ , that is iff $a_2' = a_1$ , and this condition also forces the agreement of the last coefficients, since $a_2' = a_1$ gives $a_2'' - a_1' = 0$ . So $L^{*}=L$ iff $a_2'=a_1$ . In that case $a_2 u'' + a_1 u' = a_2 u'' + a_2' u' = (a_2 u')'$ , and $a_1 - a_2' = 0$ removes the last term of $J$ . $\square$

Definition(Symmetric boundary conditions and self-adjoint problems)AdaptedAdapted from Strauss (2008), §5.3, eqs. (4)–(5) and (9), pp. 119–121, and §11.4, the standing hypotheses on $p,q,m$, p. 314, eq. (8), p. 315, and the singular cases (i)–(iii), p. 316; Stakgold & Holst (2011), §3.2, pp. 197–198, including Example 4, p. 198; Riley, Hobson & Bence (2006), §17.2, eqs. (17.15)–(17.16), pp. 559–560; Hassani (2013), §19.2, the four conditions defining a singular S-L system, p. 572

A differential operator becomes an operator on a space of functions only once boundary conditions are imposed. Take a pair of homogeneous conditions on $[a,b]$ of the general form

\alpha_1 X(a) + \beta_1 X(b) + \gamma_1 X'(a) + \delta_1 X'(b) = 0, \qquad \alpha_2 X(a) + \beta_2 X(b) + \gamma_2 X'(a) + \delta_2 X'(b) = 0,

with eight real constants, and let $\mathcal M$ be the set of $C^2[a,b]$ functions satisfying both. Following Strauss, the pair is called symmetric (or hermitian) for the formally self-adjoint operator $Lu = (p u')' + a_0 u$ if

\Bigl[\,p\,\bigl(f' \overline{g} - f\, \overline{g}'\bigr)\,\Bigr]_{a}^{b} = 0 \qquad\text{for every pair } f,g \in \mathcal M .

Strauss states this with $p \equiv 1$ in §5.3 and with the factor $p$ for variable coefficients in §11.4; the bars are dropped for real-valued $f,g$ . Equivalently: the conjunct's boundary term vanishes for every admissible pair, so Green's formula reduces to $\int_a^b (\overline g\,Lf) = \int_a^b (f\,\overline{Lg})$ .

Stakgold and Holst organise the same fact differently. For an arbitrary $L$ and a given $\mathcal M$ , they call adjoint boundary conditions a pair of conditions whose solution set is

\mathcal M^{*} = \Bigl\{\, v \in C^2[a,b] \;:\; \bigl[J(u,v)\bigr]_a^b = 0 \ \text{for every } u \in \mathcal M \,\Bigr\},

and they call the problem $(L, B_1, B_2)$ self-adjoint when both $L^{*} = L$ and $\mathcal M^{*} = \mathcal M$ . The boundary functionals $B_1^*, B_2^*$ cutting out $\mathcal M^{*}$ are not unique — any independent linear combinations of them do as well — but $\mathcal M^{*}$ itself is determined by $\mathcal M$ .

Four standard cases make the boundary term vanish. With $Lu = (pu')' + a_0u$ and $J(u,v) = p(vu' - uv')$ :

Dirichlet, $u(a)=u(b)=0$ : each endpoint term has a factor $u$ or $v$ that vanishes.
Neumann, $u'(a)=u'(b)=0$ : each endpoint term has a factor $u'$ or $v'$ that vanishes.
Separated (unmixed) conditions $\alpha_1 u(a) + \beta_1 u'(a) = 0$ and $\alpha_2 u(b) + \beta_2 u'(b) = 0$ , with real $(\alpha_i,\beta_i) \neq (0,0)$ : see the proof below. Dirichlet and Neumann are the special cases $\beta_i = 0$ and $\alpha_i = 0$ ; Robin usually names the remaining case, with $\alpha_i$ and $\beta_i$ both nonzero, although Strauss attaches the word to the whole separated family (his Case 4, p. 119). Stakgold and Holst's Example 4 records that unmixed conditions with a formally self-adjoint $L$ give $\mathcal M^* = \mathcal M$ : "a typical self-adjoint BVP".
Periodic, $u(a) = u(b)$ and $u'(a) = u'(b)$ , provided $p(a) = p(b)$ : then the two endpoint contributions are equal and cancel.

A fifth case, important for the classical equations of physics, needs no boundary condition at all: if $p(a) = p(b) = 0$ then the boundary term vanishes for all $u,v$ that stay bounded with bounded derivatives, and $[a,b]$ is called the natural interval of the operator (Riley, Hobson & Bence (2006), §17.4.1, p. 565, and §17.4.2, p. 567).

This handbook calls a Sturm-Liouville problem singular when at least one of Hassani's four conditions holds (Hassani (2013), §19.2, p. 572): the interval stretches to infinity in one or both directions; $p$ or $w$ vanishes at one or both endpoints; $q$ is not continuous on $[a,b]$ ; or one of $p,q,w$ is singular at $a$ or $b$ . Strauss's list is the shorter one of three conditions — $p$ vanishes at an endpoint; one of $p$ , $q$ , $m$ becomes infinite there (his $m$ is our weight $w$ ); an endpoint is itself infinite (Strauss (2008), §11.4, p. 316). The two lists differ over a $q$ with a bounded jump: that makes the problem singular for us, while Strauss's list does not catch it — but neither does it leave the problem regular in his sense, because his standing hypotheses require $q$ continuous (Strauss (2008), §11.4, p. 314), so such an equation falls outside his setup altogether. A problem meeting none of the four, with the hypotheses listed in Sturm-Liouville eigenvalue problems, is regular.

Definition(Separated conditions are symmetric)AdaptedAdapted from Strauss (2008), §5.3, Case 4, p. 119, and Exercise 8, p. 123, and §11.4, Exercise 6, p. 317

Let $L u = (p u')' + a_0 u$ with $p, a_0$ real, and impose the separated conditions

\alpha_1 u(a) + \beta_1 u'(a) = 0, \qquad \alpha_2 u(b) + \beta_2 u'(b) = 0,

with $\alpha_i, \beta_i$ real and $(\alpha_1,\beta_1) \neq (0,0)$ , $(\alpha_2,\beta_2) \neq (0,0)$ . Then

\Bigl[\,p\,\bigl(v u' - u v'\bigr)\,\Bigr]_a^b = 0

for every pair $u,v$ of $C^1$ functions satisfying both conditions, and also when $v$ is replaced by $\overline v$ .

Idea of the proofOriginal

A separated condition at an endpoint is one linear equation on the two numbers $\bigl(u(a),u'(a)\bigr)$ , so it confines that pair to a line through the origin in the plane. Both $u$ and $v$ obey the same equation, so both pairs lie on that one line, hence are proportional and have vanishing determinant — and that determinant is precisely the factor $vu'-uv'$ in the endpoint term. The argument is run separately at $a$ and at $b$ , which is exactly what "separated" buys: a condition mixing the two endpoints would constrain the four numbers jointly and leave each determinant free.

ProofOriginal

One endpoint at a time. The bracket is a difference of a term at $b$ and a term at $a$ ; it suffices to show that $v u' - u v'$ vanishes at each endpoint separately.
At $x=a$ . The two row vectors $\bigl(u(a), u'(a)\bigr)$ and $\bigl(v(a), v'(a)\bigr)$ in $\mathbb R^2$ (or $\mathbb C^2$ ) are both annihilated by the linear functional $(y_0,y_1)\mapsto \alpha_1 y_0 + \beta_1 y_1$ . Because $(\alpha_1,\beta_1) \neq (0,0)$ , that functional is not identically zero, so its kernel is one-dimensional. Two vectors in a one-dimensional subspace are linearly dependent, so the determinant
$u(a)\,v'(a) - v(a)\,u'(a) = 0 ,$
which is $-\bigl(v u' - u v'\bigr)\big|_{x=a}$ . The same argument at $x=b$ uses $(\alpha_2,\beta_2) \neq (0,0)$ .
The conjugate. The constants $\alpha_i,\beta_i$ are real, so conjugating each boundary condition shows that $\overline v$ satisfies them too; step 2 applies verbatim to the pair $u,\overline v$ . $\square$

RemarkOriginal

Where the hypotheses are used.

$a_2 \in C^2$ , $a_1 \in C^1$ are needed for $L^{*}$ to be defined at all, since its coefficients involve $a_2''$ and $a_1'$ (step 3 of the first proof). We state the hypothesis as Stakgold and Holst do, $a_2,a_1,a_0 \in C^2$ , although only these two derivatives are differentiated.
Real coefficients. No conjugation appears anywhere in Lagrange's identity or Green's formula, so both hold verbatim for complex-valued coefficients. What needs reality is the second reading of $L^{*}$ , as a Hilbert-space adjoint. Put $\overline v$ in place of $v$ in Green's formula; with the inner product $(f,g)=\int_a^b f\,\overline g\,\mathrm d x$ it becomes
$(Lu,\,v) \;-\; \int_a^b u\,L^{*}\overline v\,\mathrm d x \;=\; \bigl[J(u,\overline v)\bigr]_a^b .$
Only when the coefficients of $L^{*}$ are real is $L^{*}\overline v = \overline{L^{*}v}$ , and only then does the middle term equal $(u, L^{*}v)$ , giving $(Lu,v) = (u,L^{*}v)$ once the boundary term is gone. Hassani takes exactly this step, remarking that all the operators and functions in play are real before concluding $L^{\dagger} = M$ .
$(\alpha_i,\beta_i)\neq(0,0)$ in the separated case is what makes the kernel of the boundary functional one-dimensional. Drop it — impose no condition at $a$ — and the boundary term survives: for $Lu = u''$ on $[0,1]$ with only $u(1)=v(1)=0$ imposed, $u = 1-x$ and $v = x^2-1$ give $v u' - u v'\big|_{0} = (-1)(-1) - (1)(0) = 1 \neq 0$ .
$p(a)=p(b)$ in the periodic case. Without it the two endpoint contributions need not cancel. Strauss gives a different non-example with $p\equiv 1$ : under the conditions $X(a)=X(b)$ , $X'(a)=2X'(b)$ the boundary term survives, with the value $f(b)g'(b) - f'(b)g(b)$ in the convention used here. (Strauss's own display is $f'(b)g(b)-f(b)g'(b)$ , the negative of ours, because his Green's second identity in §5.3 brackets $-f'g+fg'$ while his symmetry condition (5) and this unit bracket $f'g-fg'$ . The magnitude, and the conclusion that the term is generally nonzero, are the same either way.)

Two conventions for the adjoint, and why they agree here. Stakgold and Holst, and Hassani, define $L^{*}$ by the bilinear pairing $\int v\,Lu$ , with no complex conjugate; Arfken and Riley-Hobson-Bence define it by the sesquilinear pairing $\int \overline{v}\,Lu$ (in their notation $\int v^{*}Lu$ ), which is the Hilbert-space adjoint of inner-product spaces. For real coefficients the two give the same operator, by the previous point, and this unit follows the bilinear route because the formulas are then free of bars. For complex coefficients they differ: the sesquilinear adjoint of $iD$ on $[a,b]$ is $iD$ , while its bilinear adjoint is $-iD$ .

Formal self-adjointness is not operator self-adjointness. Arfken states this plainly: because nothing is demanded of the boundary terms, the differential-equation notion is the weaker of the two, and whether $L$ is self-adjoint as a Hilbert-space operator turns on three things at once — the coefficients, the choice of scalar product, and the boundary conditions. Riley-Hobson-Bence encode the distinction in their vocabulary: $L$ is self-adjoint when $L^{\dagger}=L$ as a differential expression, and Hermitian over $[a,b]$ when in addition the boundary terms vanish. For unbounded operators even "Hermitian" is not the end of the story: Reed and Simon (Reed & Simon (1980), §VIII.2, pp. 255–256) call a densely defined $T$ symmetric if $T \subset T^{*}$ , that is $D(T)\subset D(T^{*})$ with $T\varphi = T^{*}\varphi$ there, and self-adjoint only if also $D(T)=D(T^{*})$ ; it is for self-adjoint operators alone that the spectral theorem holds and that $e^{-itT}$ is unitary. So "formally self-adjoint $+$ symmetric boundary conditions" delivers a symmetric operator, and the extra work of matching domains belongs to symmetric and self-adjoint operators and unbounded operators.

Sign conventions in the Sturm-Liouville form. This handbook writes the Sturm-Liouville equation as $-(pu')' + qu = \lambda w u$ , as Strauss and Hassani do. Riley-Hobson-Bence write $(py')' + qy + \lambda\rho y = 0$ , so their $q$ is the negative of ours, and they note that some authors flip the sign of the $\lambda\rho y$ term as well. Arfken writes $L = D(p_0 D) + p_2$ with $L\psi = \lambda\psi$ , so that his $p_2$ is $-q$ and his $\lambda$ is $-\lambda$ . Comparing formulas across books therefore requires checking two signs.

Method(Putting a second-order equation into self-adjoint form)AdaptedAdapted from Hassani (2013), §14.5, Thm 14.5.4, p. 435, and §19.2, Example 19.2.2(c), p. 571; Arfken, Weber & Harris (2013), §8.2, eqs. (8.14)–(8.19), pp. 385–386; Riley, Hobson & Bence (2006), §17.4.2, eqs. (17.38)–(17.40) and Table 17.1, pp. 566–567

Write $L = a_2 D^2 + a_1 D + a_0$ , as in the first definition, and start from the equation

L y + \lambda\,\rho(x)\,y \;=\; a_2(x)\,y'' + a_1(x)\,y' + a_0(x)\,y + \lambda\,\rho(x)\,y \;=\; 0

on an interval where $a_2$ does not vanish. Note the sign: the eigenvalue term is added to $Ly$ , as in the convention fixed in the remark above.

Test for self-adjointness. If $a_2' = a_1$ , the operator is already formally self-adjoint: take $h \equiv 1$ and skip to step 4.
Build the integrating factor.
$h(x) = \frac{1}{a_2(x)}\exp\!\int^{x} \frac{a_1(t)}{a_2(t)}\,\mathrm d t .$
Hassani derives this by demanding $h a_1 = (h a_2)'$ ; Riley-Hobson-Bence write the same factor as $F = \exp\int (a_1 - a_2')/a_2$ , which differs from $h$ by a constant and so serves equally well. Note that $h > 0$ wherever $a_2 > 0$ .
Multiply through. The equation becomes $(h a_2 y')' + h a_0 y + \lambda\, h\rho\, y = 0$ , because $h a_2 y'' + h a_1 y' = h a_2 y'' + (h a_2)' y' = (h a_2 y')'$ .
Read off the Sturm-Liouville data. In the form $-(p y')' + q y = \lambda w y$ used here,
$p = h\,a_2 , \qquad q = -\,h\,a_0 , \qquad w = h\,\rho .$
For an eigenvalue problem with no weight in the original equation — that is, $Ly + \lambda y = 0$ , or equivalently $Ly = -\lambda y$ — take $\rho \equiv 1$ , so that $w = h$ : the integrating factor is the weight, and the inner product consistent with hermiticity is $(f,g)_w = \int_a^b w f \overline g\,\mathrm d x$ (Arfken, Weber & Harris (2013), §8.2, eq. (8.19), p. 386). Arfken reaches the same weight from $L\psi = \lambda\psi$ ; his $\lambda$ is the negative of ours, by the remark above, but the weight is the same expression either way.
Fix the interval and the boundary conditions. Check that $p > 0$ and $w > 0$ on the interval, and choose boundary conditions from the symmetric list above. If $p$ vanishes at two points, take those two points as the endpoints of the interval — this is the natural interval — and boundedness of $y$ there replaces the boundary conditions.

Riley-Hobson-Bence's Table 17.1 is the result of running this recipe on the classical equations; they write the weight as $\rho$ , and their $q$ is the negative of ours. Legendre gives $p = 1-x^2$ , $w=1$ on $[-1,1]$ ; Chebyshev gives $p=(1-x^2)^{1/2}$ , $w=(1-x^2)^{-1/2}$ ; Laguerre gives $p = xe^{-x}$ , $w = e^{-x}$ on $[0,\infty)$ ; Hermite gives $p = w = e^{-x^2}$ on $\mathbb R$ ; Bessel gives $p = w = x$ with, in our sign convention, $q = \nu^2/x$ (their table prints $q = -\nu^2/x$ ; the value $\nu^2/x$ is Hassani's, (Hassani (2013), §19.2, Example 19.2.2(c), p. 571), whose convention is ours).

ExampleOriginal

An Euler operator. Take

L u = x^2 u'' + 3x\,u' + u \qquad\text{on } [1,2],

so $a_2 = x^2$ , $a_1 = 3x$ , $a_0 = 1$ .

Formal adjoint. $2a_2' - a_1 = 4x - 3x = x$ and $a_2'' - a_1' + a_0 = 2 - 3 + 1 = 0$ , so

L^{*}v = x^2 v'' + x\,v' = x\,(x v')' .

Since $a_2' = 2x \neq 3x = a_1$ , the operator is not formally self-adjoint, and indeed $L^{*}\neq L$ .

Conjunct. $a_1 - a_2' = 3x - 2x = x$ , so

J(u,v) = x^2\bigl(v u' - u v'\bigr) + x\,u v .

Checking Green's formula on a concrete pair. Take $u = v = x$ . Then $Lu = 0 + 3x + x = 4x$ and $L^{*}v = 0 + x = x$ , so

v\,Lu - u\,L^{*}v = 4x^2 - x^2 = 3x^2 ,

while $J(x,x) = x^2(x - x) + x\cdot x\cdot x = x^3$ , whose derivative is $3x^2$ : Lagrange's identity holds. Integrating,

\int_1^2 3x^2\,\mathrm d x = 7 = \bigl[x^3\bigr]_1^2 = 8-1 = 7 .

Self-adjoint form. The integrating factor is

h(x) = \frac{1}{x^2}\exp\!\int^x \frac{3}{t}\,\mathrm d t = \frac{x^3}{x^2} = x ,

and indeed

x\,Lu = x^3 u'' + 3x^2 u' + x u = \bigl(x^3 u'\bigr)' + x\,u .

(Check with $u=x$ : the left side is $4x^2$ , the right side is $(x^3)' + x^2 = 3x^2+x^2$ .)

As a Sturm-Liouville problem. The eigenvalue problem $Lu = -\lambda u$ on $[1,2]$ with $u(1)=u(2)=0$ becomes, after multiplication by $h = x$ ,

-\bigl(x^3 u'\bigr)' - x\,u = \lambda\,x\,u , \qquad\text{so}\qquad p = x^3,\quad q = -x,\quad w = x .

On $[1,2]$ we have $p \in C^1$ , $q, w$ continuous, $p \ge 1 > 0$ and $w \ge 1 > 0$ , and the boundary conditions are separated; this is a regular Sturm-Liouville problem, solved in the next unit. The weight $w = x$ is what makes the eigenfunctions orthogonal: without it they are not.

ApplicationOriginal

Why the classical polynomials are orthogonal, and with which weight. Legendre's, Chebyshev's, Hermite's and Laguerre's equations are not self-adjoint as written; the integrating factor of step 2 produces exactly the familiar weights $1$ , $(1-x^2)^{-1/2}$ , $e^{-x^2}$ , $e^{-x}$ on exactly the intervals where the boundary term dies. Arfken's worked case is Laguerre's: $h = x^{-1}e^{\ln x - x} = e^{-x}$ , and the factor $x$ in $p = xe^{-x}$ is what kills the boundary term at the origin. See orthogonal polynomials, Legendre polynomials and Bessel orthogonality.
Quantum mechanics. An observable must have real eigenvalues, so its operator must be hermitian, and for a differential operator that is two separate requirements: the right form of the coefficients, and boundary conditions that annihilate the conjunct. For $H = -\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm d x^2} + V(x)$ the first holds automatically ( $a_2$ constant, $a_1 = 0$ ), and the second is the real content of "the wavefunction must vanish at the walls" or "must be square-integrable on the line". Radial problems reach the same conclusion through the natural-interval mechanism: the measure $r^2\,\mathrm d r$ is the weight $w$ , and $p = r^2$ vanishing at $r=0$ removes the boundary term there.
Heat flow and diffusion in an inhomogeneous medium. Separating $c\rho\, u_t = (\kappa u_x)_x$ gives $-(\kappa X')' = \lambda\, c\rho\, X$ directly in self-adjoint form, with $p = \kappa$ the conductivity and $w = c\rho$ the heat capacity per unit length. The conjunct $\kappa(v X' - X v')$ is a difference of heat fluxes, and the symmetric boundary conditions are exactly the physically natural ones: fixed temperature (Dirichlet), insulated end (Neumann), Newton cooling $\kappa X' = -\sigma X$ (Robin), or a ring (periodic, with $\kappa$ matching at the join).
Green's functions. Stakgold and Holst's reason for introducing adjoint boundary conditions is that the Green's function of $(L,B_1,B_2)$ and that of the adjoint problem are related by $g^{*}(x,\xi) = g(\xi,x)$ ; when the problem is self-adjoint this says $g(x,\xi) = g(\xi,x)$ , the symmetry that underlies reciprocity theorems in acoustics and electrostatics. See Green's functions for boundary-value problems.

Theorem

Sturm-Liouville eigenvalue problems

Adapted

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For a regular Sturm-Liouville problem the eigenvalues are real and simple and form an increasing sequence tending to infinity, and eigenfunctions belonging to different eigenvalues are orthogonal with respect to the weight in the equation.

Prerequisites: Adjoint and self-adjoint differential operators, Structure of linear equations and the Wronskian, Inner products and norms

Definition(Regular Sturm-Liouville problem)AdaptedAdapted from Strauss (2008), §11.4, eq. (7) with the hypotheses on p. 314 and the symmetric condition (8), p. 315; Hassani (2013), §19.1, eq. (19.2) and the definition of a regular Sturm-Liouville system, pp. 566–567, §19.1, p. 569 (non-separated conditions may destroy regularity), and §19.2, the four conditions defining a singular system, p. 572; Riley, Hobson & Bence (2006), §17.1, eq. (17.6), p. 557

A regular Sturm-Liouville problem on a finite interval $[a,b]$ consists of the differential equation

-\bigl(p(x)\,u'\bigr)' + q(x)\,u \;=\; \lambda\,w(x)\,u , \qquad a < x < b ,

in which

$p \in C^1[a,b]$ and $q, w \in C[a,b]$ are real-valued,
$p(x) > 0$ and $w(x) > 0$ for every $x \in [a,b]$ ,

together with separated boundary conditions

\alpha_1\,u(a) + \beta_1\,u'(a) = 0 , \qquad \alpha_2\,u(b) + \beta_2\,u'(b) = 0 ,

with $\alpha_1,\beta_1,\alpha_2,\beta_2$ real and neither pair $(\alpha_1,\beta_1)$ nor $(\alpha_2,\beta_2)$ equal to $(0,0)$ ; Hassani imposes this last requirement by asking that the $2\times 2$ array of the four constants have no vanishing row. Strauss's hypotheses on $p$ , $q$ and $w$ (he writes $m$ for the weight) are the ones displayed above; Hassani sets up the theory for the Liouville normal form $u'' - q u = \lambda u$ and reduces the general case to it, as the note below records.

A number $\lambda \in \mathbb C$ is an eigenvalue if the problem has a solution $u \not\equiv 0$ in $C^2[a,b]$ , and such a $u$ is an eigenfunction belonging to $\lambda$ . The weighted inner product and norm are

(f,g)_w = \int_a^b w(x)\,f(x)\,\overline{g(x)}\,\mathrm d x , \qquad \lVert f\rVert_w = (f,f)_w^{1/2} ,

an inner product in the sense of inner products and norms precisely because $w$ is continuous and strictly positive: $\lVert f\rVert_w = 0$ then forces $f \equiv 0$ for continuous $f$ . Riley-Hobson-Bence require of their weight that it be real and non-negative on $[a,b]$ ; the strict positivity assumed here is Strauss's hypothesis.

By the self-adjoint form of a second-order operator, every second-order linear equation with a non-vanishing leading coefficient can be brought to this shape by an integrating factor, so the theorem below applies far beyond equations that arrive already in divergence form.

Following Hassani (Hassani (2013), §19.2, p. 572), the problem is called singular when at least one of four conditions on the equation holds: the interval stretches to infinity in one or both directions; $p$ or $w$ vanishes at one or both endpoints; $q$ is not continuous on $[a,b]$ ; or one of $p,q,w$ is singular at $a$ or $b$ . Nothing about the boundary conditions appears in that list, and nothing about them appears in Strauss's shorter list either (Strauss (2008), §11.4, p. 316); of non-separated conditions Hassani says only that they may destroy the regularity of the system (Hassani (2013), §19.1, p. 569). So a problem with smooth positive $p$ and $w$ but periodic boundary conditions is not regular while being singular in neither author's sense — and it is exactly such a problem that breaks part 3 below, as the remark records. For any problem that is not regular, the conclusions of parts 3 and 4 may fail.

Theorem(Sturm-Liouville theorem, regular case)AdaptedAdapted from Strauss (2008), §5.3, Thms 1–2, pp. 120–121, together with §11.4, Thm 1, p. 314; Riley, Hobson & Bence (2006), §17.3.1–17.3.2 and §17.3.4, pp. 561–564; Hassani (2013), §19.1, the argument at (19.3), p. 567, and Thm 19.1.9, pp. 568–569; Arfken, Weber & Harris (2013), §8.2, eq. (8.20), p. 386

For a regular Sturm-Liouville problem:

The eigenvalues are real, and every eigenfunction is a constant multiple of a real-valued eigenfunction.
Orthogonality with weight. If $\lambda \neq \mu$ are eigenvalues with eigenfunctions $u$ and $v$ , then $(u,v)_w = 0$ . Each eigenfunction may therefore be scaled so that the eigenfunctions form an orthonormal family in $(\cdot,\cdot)_w$ .
The eigenvalues are simple. The solutions of the problem for a fixed eigenvalue $\lambda$ form a one-dimensional space: any two eigenfunctions belonging to the same eigenvalue are proportional.
The spectrum. There are countably many eigenvalues and no more; they are bounded below, and arranged in increasing order they form a strictly increasing sequence $\lambda_1 < \lambda_2 < \lambda_3 < \cdots \longrightarrow +\infty .$ Part 4 is stated here without proof; see the note below.

Parts 1-3 use only Green's formula and the separated boundary conditions. Part 4 is where the compactness of the inverse operator enters.

Idea of the proofOriginal

Everything in parts 1 and 2 comes from one identity. Green's formula for the formally self-adjoint operator $Lu = -(pu')' + qu$ says that $\int_a^b(\overline v\,Lu - u\,L\overline v)$ equals a boundary term built from $p$ , $u$ , $v$ and their first derivatives; separated boundary conditions annihilate that term. Feeding in $Lu = \lambda w u$ and $L v = \mu w v$ turns the left-hand side into $(\lambda - \overline\mu)(u,v)_w$ . Taking $v=u$ gives $(\lambda-\overline\lambda)\lVert u\rVert_w^2 = 0$ , hence $\lambda$ real; taking $\lambda \neq \mu$ gives $(u,v)_w = 0$ . It is the same one-line argument that makes the eigenvalues of a Hermitian matrix real and its eigenvectors orthogonal; the boundary conditions are what supply the "Hermitian" part.

Simplicity is a different mechanism, and it is the one place where separated conditions matter rather than merely symmetric ones. Two eigenfunctions for the same $\lambda$ both satisfy one and the same linear condition on the pair $(u(a),u'(a))$ , which pins that pair down to a line in the plane; so their Wronskian vanishes at $a$ , and a Wronskian of two solutions of the same second-order equation that vanishes once vanishes identically.

ProofAdaptedAdapted from Strauss (2008), §5.3, proofs of Thms 1 and 2, pp. 120–121; Riley, Hobson & Bence (2006), §17.3.1, eqs. (17.18)–(17.23), p. 561, and §17.3.4, p. 563; Hassani (2013), §19.1, eq. (19.3), p. 567; Arfken, Weber & Harris (2013), §8.2, eqs. (8.18)–(8.20), p. 386

Write $Lu = -(pu')' + qu$ , so that the equation is $Lu = \lambda w u$ .

The master identity. For any $f,g \in C^2[a,b]$ , the product rule gives
$g\,Lf - f\,Lg = -g\,(pf')' + f\,(pg')' = \frac{\mathrm d}{\mathrm d x}\Bigl[\,p\,\bigl(f\,g' - g\,f'\bigr)\Bigr] ,$
because differentiating the bracket produces $f'(pg') + f(pg')' - g'(pf') - g(pf')'$ , whose first and third terms cancel. Integrating,
$\int_a^b \bigl(g\,Lf - f\,Lg\bigr)\,\mathrm d x = \Bigl[\,p\,\bigl(f g' - g f'\bigr)\,\Bigr]_a^b .$
This is Green's formula of the previous unit. The sign is worth pinning down: our $L = -(pu')' + qu$ is that unit's divergence-form operator $(a_2u')' + a_0u$ with $a_2 = -p$ and $a_0 = q$ , so its conjunct is $J(f,g) = a_2\,(gf' - fg') = p\,(fg' - gf')$ , which is exactly the bracket displayed above. Here $q$ has dropped out because it enters both terms symmetrically.
The boundary term dies. Let $u,v$ be eigenfunctions, with eigenvalues $\lambda,\mu$ . Because the coefficients $\alpha_i,\beta_i$ are real, $\overline v$ satisfies the same two boundary conditions as $v$ . Apply step 1 with $f = u$ , $g = \overline v$ . At $x=a$ the vectors $\bigl(u(a),u'(a)\bigr)$ and $\bigl(\overline v(a),\overline v{}'(a)\bigr)$ are both annihilated by the nonzero functional $(y_0,y_1)\mapsto \alpha_1 y_0 + \beta_1 y_1$ , whose kernel is a line; two vectors on a line have vanishing determinant, so $u\overline v{}' - \overline v u' = 0$ at $a$ . The same holds at $b$ , using $(\alpha_2,\beta_2)\neq(0,0)$ . Hence
$\int_a^b \bigl(\overline v\,Lu - u\,L\overline v\bigr)\,\mathrm d x = 0 .$
Insert the eigenvalue equations. Since $p,q,w$ are real, conjugating $Lv = \mu w v$ gives $L\overline v = \overline\mu\, w\,\overline v$ . So the integrand in step 2 is $\lambda\,w\,u\overline v - \overline\mu\,w\,u\overline v$ , and
$\bigl(\lambda - \overline{\mu}\bigr)\,(u,v)_w = 0 .$
This is the master identity. (Arfken's eq. (8.20) is the same statement with the boundary term still displayed on the right.)
Part 1, reality. Put $v = u$ , $\mu = \lambda$ : then $(\lambda - \overline\lambda)\lVert u\rVert_w^2 = 0$ . Here the hypothesis $w > 0$ is used: $u$ is continuous and not identically zero, so $\lVert u\rVert_w^2 = \int_a^b w|u|^2 > 0$ . Therefore $\lambda = \overline\lambda$ .
Part 1, real eigenfunctions. Let $u = y + iz$ with $y,z$ real. Since $\lambda$ , $p$ , $q$ and $w$ are real, taking real and imaginary parts of $Lu = \lambda w u$ gives $Ly = \lambda w y$ and $Lz = \lambda w z$ ; and since the eight boundary constants are real, $y$ and $z$ satisfy the boundary conditions too. At least one of $y,z$ is not identically zero, so $\lambda$ has a real eigenfunction. By part 3, proved below, every eigenfunction for $\lambda$ is then a multiple of that one.
Part 2, orthogonality. With $\lambda \neq \mu$ , both real by part 1, step 3 reads $(\lambda-\mu)(u,v)_w = 0$ with $\lambda - \mu \neq 0$ , so $(u,v)_w = 0$ . Normalising is possible because $\lVert u\rVert_w > 0$ , again by $w>0$ ; replacing $u$ by $u/\lVert u\rVert_w$ makes the family orthonormal.
Part 3, simplicity. Let $u_1,u_2$ be eigenfunctions for the same $\lambda$ . Both satisfy $\alpha_1 y(a) + \beta_1 y'(a) = 0$ , so as in step 2 their Wronskian
$W(x) = u_1(x)\,u_2'(x) - u_2(x)\,u_1'(x)$
vanishes at $x=a$ . Both solve the same second-order equation, which in the form $u'' = -\frac{p'}{p}u' + \frac{q-\lambda w}{p}u$ has continuous coefficients on $[a,b]$ — this is where $p > 0$ and $p \in C^1$ are needed. By Abel's identity for the Wronskian, $W$ is either identically zero or nowhere zero; since $W(a)=0$ it is identically zero, so $u_1$ and $u_2$ are linearly dependent. (Directly: $(pW)' = u_1(pu_2')' - u_2(pu_1')' = u_1(q-\lambda w)u_2 - u_2(q-\lambda w)u_1 = 0$ , so $pW$ is constant, and it is $0$ at $a$ ; as $p>0$ , $W\equiv 0$ .) Hence the eigenspace is one-dimensional. $\square$

Note(Part 4 is not proved here)AdaptedAdapted from Hassani (2013), §19.1, Thm 19.1.9, pp. 568–569, with the construction of the compact resolvent on p. 567, and §19.2, eq. (19.7), p. 569; Strauss (2008), §5.4, Thm 1, p. 125, and §11.4, p. 316; Stakgold & Holst (2011), §6.3, Thms 6.3.1–6.3.4 and Thm 6.3.6, pp. 371–375

That there are infinitely many eigenvalues and that they tend to $+\infty$ cannot be reached by the elementary identity above; every proof goes through an auxiliary compactness argument, and this handbook quotes the result rather than proving it.

Hassani's route: shift $\lambda$ so that $0$ is not an eigenvalue, invert the operator by an explicit Green's kernel built from two solutions, observe that the resulting integral operator is Hilbert-Schmidt and hence compact, and read the spectrum of the differential operator off the spectrum of its compact inverse. His Theorem 19.1.9 states, for a regular Sturm-Liouville system, that there are countably many eigenvalues, that they can be arranged in an increasing sequence having infinity as its limit, that the eigenvectors are nondegenerate and form a complete orthogonal set, and that the eigenfunction $u_n$ attached to the $n$ th eigenvalue vanishes at exactly $n$ points of the interval on which the system is posed. He states it for the Liouville normal form $u'' - qu = \lambda u$ with separated conditions, and then shows in §19.2 that the Liouville substitution

u(x) = v(t)\,\bigl[p(x)w(x)\bigr]^{-1/4}, \qquad t = \int_a^x \sqrt{\tfrac{w(s)}{p(s)}}\,\mathrm d s ,

carries the general equation $-(pu')'+qu = \lambda w u$ into that form, so that the theorem still holds.

Stakgold and Holst run the same strategy in the language of integral equations: the Green's function of a self-adjoint boundary-value problem is a real symmetric Hilbert-Schmidt kernel, its compact self-adjoint integral operator has eigenvalues of finite multiplicity tending to $0$ , and their Theorem 6.3.6 concludes that the eigenfunctions of the boundary-value problem form an orthonormal basis. Strauss records both conclusions for Sturm-Liouville problems — unbounded growth of $\lambda_n$ , and a complete eigenfunction family — pointing the reader to Coddington-Levinson and Courant-Hilbert for an ODE-style proof, and proves the model case $-X''=\lambda X$ by a minimisation argument in Chapter 11.

The relevant ingredients are developed in compact operators, the spectral theorem for compact self-adjoint operators and Hilbert-Schmidt kernels; the zero count is the subject of Sturm comparison and oscillation theorems. Completeness itself is taken up in eigenfunction expansions.

RemarkOriginal

Where the hypotheses are used.

$w > 0$ is used twice in step 4 and step 6, to know that $\lVert u\rVert_w \neq 0$ . If $w$ were allowed to change sign, $\int w|u|^2$ could vanish for a nonzero $u$ and the argument for reality would collapse; such indefinite problems genuinely can have complex eigenvalues.
$p > 0$ and $p \in C^1$ make the equation a genuine second-order equation with continuous coefficients on all of $[a,b]$ , which is what the Wronskian argument of step 7 needs. Where $p$ vanishes the problem is singular and simplicity can fail.
Separated boundary conditions are needed only for part 3. Parts 1 and 2 hold for any symmetric boundary conditions, periodic ones included. Simplicity genuinely fails for periodic conditions: Hassani's Example 19.2.2(b) takes $u'' + \omega^2 u = 0$ on $[-T,T]$ with $u(-T)=u(T)$ , $u'(-T)=u'(T)$ and finds the eigenvalue $(n\pi/T)^2$ carried by the two independent eigenfunctions $\sin(n\pi t/T)$ and $\cos(n\pi t/T)$ — so, by Theorem 19.1.9, that system is not regular. When an eigenvalue is degenerate, orthogonality within its eigenspace is not automatic but can be arranged by Gram-Schmidt, as Riley-Hobson-Bence and Arfken both note.
Real coefficients are used in step 2 (so that $\overline v$ obeys the boundary conditions), in step 3 (so that $L\overline v = \overline\mu w \overline v$ ) and in step 5. With complex $q$ the operator is not symmetric and nothing survives.

The sign of the eigenvalues. Nothing above says the eigenvalues are positive, and in general they are not: when $w \equiv 1$ , adding a constant $c$ to $q$ leaves the eigenfunctions alone and shifts every $\lambda_n$ by $c$ , so any finite number of them can be pushed below zero. Strauss gives a sufficient condition in the model case $-X''=\lambda X$ with symmetric boundary conditions: if

\Bigl[\,f(x)\,f'(x)\,\Bigr]_a^b \le 0

for every real $f$ satisfying the boundary conditions, there is no negative eigenvalue. This holds for Dirichlet, Neumann and periodic conditions, and for Robin conditions when the two Robin constants have the right sign; it fails for others, and the second example below exhibits a Robin problem with one negative eigenvalue. What is always true in the regular case is that the eigenvalues are bounded below, since they increase to $+\infty$ with only finitely many below any level.

Two conventions. This unit writes the equation as $-(pu')' + qu = \lambda w u$ , following Strauss and Hassani. Riley-Hobson-Bence write $(py')' + qy + \lambda\rho y = 0$ , so their $q$ is the negative of ours and their operator is $Ly = -(py')'-qy$ ; their $\rho$ is our $w$ . They record that some authors also flip the sign of the $\lambda\rho y$ term. Arfken writes $L = D(p_0D) + p_2$ with $L\psi = \lambda\psi$ and a weight $w$ in the scalar product, so his $\lambda$ is the negative of ours. The mathematics is the same; only the labels move.

ExampleOriginal

A weighted problem with closed-form eigenvalues. Take the Euler equation of the previous unit as an eigenvalue problem on $[1,2]$ :

x^2 u'' + 3x\,u' + u = -\lambda\,u , \qquad u(1) = u(2) = 0 .

Putting it in Sturm-Liouville form. The integrating factor is $h = x$ , and multiplying through gives

-\bigl(x^3 u'\bigr)' - x\,u = \lambda\,x\,u , \qquad p = x^3, \quad q = -x, \quad w = x .

On $[1,2]$ : $p\in C^1$ with $p \ge 1 > 0$ , $q$ and $w$ continuous with $w \ge 1 > 0$ , and the conditions $u(1)=u(2)=0$ are separated with $(\alpha_1,\beta_1)=(\alpha_2,\beta_2)=(1,0)$ . The problem is regular, so parts 1-4 apply.

Solving it. The equation $x^2u'' + 3xu' + (1+\lambda)u = 0$ is of Euler-Cauchy type; substituting $u = x^{m}$ gives $m(m-1) + 3m + 1 + \lambda = (m+1)^2 + \lambda = 0$ , so $m = -1 \pm i\sqrt{\lambda}$ for $\lambda > 0$ and

u(x) = \frac{1}{x}\Bigl[A\cos\bigl(\sqrt\lambda\,\ln x\bigr) + B\sin\bigl(\sqrt\lambda\,\ln x\bigr)\Bigr].

The condition $u(1)=0$ forces $A = 0$ . Then $u(2)=0$ needs $\sin(\sqrt\lambda\,\ln 2) = 0$ , that is $\sqrt\lambda\,\ln 2 = n\pi$ . Checking separately that $\lambda = 0$ gives $u = (A + B\ln x)/x$ , killed by the two conditions, and that $\lambda < 0$ gives real exponents $m = -1 \pm \sqrt{-\lambda}$ and again only $u \equiv 0$ , the full list is

\lambda_n = \Bigl(\frac{n\pi}{\ln 2}\Bigr)^{2} , \qquad \varphi_n(x) = \frac{1}{x}\,\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr) , \qquad n = 1,2,3,\dots

Numerically $\lambda_1 = 20.5423$ , $\lambda_2 = 82.1692$ , $\lambda_3 = 184.881$ , and $\lambda_n = 20.5423\,n^2 \to \infty$ : real, simple, increasing without bound, as parts 1, 3 and 4 promise.

Orthogonality, with the weight and without it. Substituting $s = \ln x/\ln 2$ , so that $\mathrm d x / x = \ln 2\,\mathrm d s$ and $s$ runs over $[0,1]$ ,

(\varphi_n,\varphi_m)_w = \int_1^2 x\cdot\frac{1}{x^2}\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr)\sin\!\Bigl(\frac{m\pi\ln x}{\ln 2}\Bigr)\mathrm d x = \ln 2\int_0^1 \sin(n\pi s)\sin(m\pi s)\,\mathrm d s = \frac{\ln 2}{2}\,\delta_{nm} ,

so $\lVert\varphi_n\rVert_w^2 = \tfrac12\ln 2 = 0.346574$ for every $n$ , and $\sqrt{2/\ln 2}\,\varphi_n$ is an orthonormal family. The weight is doing real work: without it,

\int_1^2 \varphi_1(x)\,\varphi_2(x)\,\mathrm d x = 0.0307804 \neq 0 .

Simplicity, concretely. For $\lambda = \lambda_n$ the solution space of the ODE is spanned by $x^{-1}\cos(\sqrt{\lambda_n}\ln x)$ and $\varphi_n$ , a two-dimensional space; the single condition $u(1)=0$ cuts it down to the one-dimensional span of $\varphi_n$ , and the condition at $x=2$ is then satisfied automatically. This is part 3 in action: one separated condition already removes one dimension.

Zeros. $\varphi_n$ vanishes at $x = 2^{k/n}$ for $k = 0,1,\dots,n$ , hence exactly $n-1$ times inside $(1,2)$ and twice on the boundary. The zero count quoted in the note — " $u_n$ vanishes at exactly $n$ points" — therefore matches only after the index shift $n \mapsto n-1$ : our $\varphi_n$ , with $n-1$ interior zeros, is Hassani's $u_{n-1}$ , so his numbering of the eigenvalues would have to begin at $n=0$ . His own Example 19.2.2(a) numbers the Dirichlet eigenfunctions $\sin(n\pi t/T)$ from $n=1$ , as we do. We report the theorem as he prints it.

ExampleOriginal

A Robin problem with a negative eigenvalue. Take

-u'' = \lambda u \quad\text{on } [0,1], \qquad u(0) = 0, \qquad u'(1) = 2\,u(1) ,

so $p = w = 1$ , $q = 0$ , and the conditions are separated with $(\alpha_1,\beta_1) = (1,0)$ and $(\alpha_2,\beta_2) = (-2,1)$ . The problem is regular.

Positive eigenvalues. For $\lambda = k^2 > 0$ the condition at $0$ gives $u = \sin kx$ , and the condition at $1$ gives $k\cos k = 2\sin k$ , that is $\tan k = k/2$ . The roots are

k_1 = 4.27478,\quad k_2 = 7.59655,\quad k_3 = 10.81267,\quad k_4 = 13.99522,\dots

with $\lambda = k^2$ : $18.2738$ , $57.7075$ , $116.914$ , $195.866$ . (For large $n$ , $k_n \approx (n+\tfrac12)\pi - 2/k_n$ ; the fourth root is $4.4548\pi$ .)

A negative eigenvalue. For $\lambda = -\mu^2 < 0$ the condition at $0$ gives $u = \sinh\mu x$ , and the condition at $1$ gives $\mu\cosh\mu = 2\sinh\mu$ , that is $\mu\coth\mu = 2$ . Since $\mu\coth\mu$ increases from $1$ at $\mu = 0^{+}$ to $\infty$ , there is exactly one root,

\mu_0 = 1.915008 , \qquad \lambda_0 = -\mu_0^2 = -3.667256 ,

with eigenfunction $\varphi_0(x) = \sinh(1.915008\,x)$ . And $\lambda = 0$ is not an eigenvalue: $u = cx$ then needs $c = 2c$ , so $c=0$ .

Why Strauss's criterion does not forbid it. For a real $f$ with $f(0)=0$ and $f'(1)=2f(1)$ ,

\Bigl[f f'\Bigr]_0^1 = f(1)f'(1) = 2\,f(1)^2 \;\ge\; 0 ,

which is the wrong sign for the hypothesis of Strauss's Theorem 3; the negative eigenvalue is what the criterion was guarding against. Flipping the condition to $u'(1) = -2u(1)$ gives $[ff']_0^1 = -2f(1)^2 \le 0$ , so Theorem 3 applies and there is no negative eigenvalue — that, and no more, is what it concludes. To get strict positivity one rules out $\lambda = 0$ separately: it would need $u = cx$ with $c = -2c$ , hence $c = 0$ . With both facts in hand, every eigenvalue of the flipped problem is positive.

Orthogonality across the sign change. With $w \equiv 1$ the inner product is the plain integral, and

(\varphi_0,\varphi_1)_w = \int_0^1 \sinh(\mu_0 x)\,\sin(k_1 x)\,\mathrm d x = 1.2\times10^{-44}

when $\mu_0$ and $k_1$ are taken to $40$ digits: zero to the working precision. The residual is a measure of the root-finding, not of the mathematics: with the six-digit values $\mu_0 = 1.915008$ and $k_1 = 4.27478$ printed above, the same integral evaluates to $1.48\times10^{-6}$ instead. Almost all of that comes from the rounding of $k_1$ , which is low by $2.3\times10^{-6}$ ; the integral's sensitivity to $k_1$ near the root is about $0.65$ , and $0.65\times 2.3\times10^{-6} = 1.5\times10^{-6}$ . Orthogonality holds exactly at the exact roots, and the truncated roots are not eigenvalues. Meanwhile $\lVert\varphi_0\rVert_w^2 = 2.50531$ and $\lVert\varphi_1\rVert_w^2 = 0.455104$ , both comfortably of order one. Orthogonality does not care that one eigenvalue is negative and the other positive; it only cares that they differ.

ApplicationOriginal

Normal modes. Separating a linear wave equation $w(x)\,u_{tt} = (p(x)u_x)_x - q(x)u$ on $[a,b]$ gives exactly the regular Sturm-Liouville problem above, with $\lambda = \omega^2$ . Part 1 says the squared frequencies are real — the system oscillates rather than growing or decaying; part 2 says the modes do not exchange energy, so the total energy is the sum of the modal energies; part 3 says that for a one-dimensional system with ends held in these ways there are no degenerate modes, which is why a string has no accidental mode doubling while a drumhead does. See normal modes and the vibrating string.
Quantum mechanics on an interval. For $-\psi'' + V(x)\psi = E\psi$ on $[a,b]$ with any separated boundary condition — an infinite well, a finite box with a Robin matching condition at the wall — part 1 is the statement that the measured energies are real, part 2 that stationary states of different energy are orthogonal, and part 3 that in one dimension bound states are non-degenerate, the theorem behind the statement that a one-dimensional bound-state spectrum is a simple ladder. Negative eigenvalues are the bound states; the Robin example above is a one-state model of exactly that situation.
Heat conduction and relaxation times. For $w u_t = (pu_x)_x - qu$ , separating variables gives $u = \sum c_n e^{-\lambda_n t}\varphi_n(x)$ . Part 4 says the decay rates $\lambda_n$ run off to infinity, so all but the first few modes die almost at once and the long-time behaviour is governed by $\lambda_1$ alone: it is the slowest decay rate, and its reciprocal $1/\lambda_1$ is the relaxation time of a rod, a fin, or a diffusing tracer. Bounds for it without solving the equation come from the variational characterisation and the Rayleigh-Ritz method.
Where the classical special functions come from. Legendre's, Bessel's, Hermite's and Laguerre's equations are Sturm-Liouville equations whose problems are singular, so the conclusions above are not automatic; the orthogonality relations nevertheless survive whenever the boundary term vanishes, which is the content of Hassani's Box 19.2.3 and the reason each family comes with its own weight and its own interval. See general orthogonal polynomials.

Theorem

Eigenfunction expansions

Adapted

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The eigenfunctions of a regular Sturm-Liouville problem form a complete orthogonal family, so every square-integrable function has a mean-square convergent expansion in them with coefficients given by weighted inner products, and the expansion turns a forced boundary-value problem into a term-by-term division by $\lambda_n-\sigma$ .

Prerequisites: Sturm-Liouville eigenvalue problems, Inner products and norms, L^p spaces

Definition(Expansion coefficients)AdaptedAdapted from Strauss (2008), §5.4, eq. (4) and the three definitions of convergence (5)–(7), pp. 125–126, and the definition of the $L^2$ norm (12), p. 131; Riley, Hobson & Bence (2006), §17.1, eqs. (17.6), (17.9) and (17.11), pp. 557–558

Keep the data of a regular Sturm-Liouville problem on $[a,b]$ from the previous unit: the operator $Lu = -(pu')' + qu$ , the weight $w > 0$ , the separated boundary conditions, the weighted inner product

(f,g)_w = \int_a^b w\,f\,\overline g\,\mathrm d x , \qquad \lVert f\rVert_w = (f,f)_w^{1/2} ,

and the eigenvalues $\lambda_1 < \lambda_2 < \cdots$ with real eigenfunctions $\varphi_1,\varphi_2,\dots$ . By part 2 of the Sturm-Liouville theorem these may be, and from now on are, normalised:

(\varphi_m,\varphi_n)_w = \delta_{mn} .

Write $L^2_w(a,b)$ for the set of measurable $f$ on $(a,b)$ with $\lVert f\rVert_w < \infty$ , that is $\int_a^b w\,|f|^2\,\mathrm d x < \infty$ — a space of the kind studied in $L^p$ spaces. For $f \in L^2_w(a,b)$ the expansion coefficients of $f$ and its partial sums are

c_n = (f,\varphi_n)_w = \int_a^b w\,f\,\varphi_n\,\mathrm d x , \qquad S_N = \sum_{n=1}^{N} c_n\,\varphi_n ,

and $\sum_n c_n \varphi_n$ is the eigenfunction expansion, or generalised Fourier series, of $f$ . Before the eigenfunctions are normalised the coefficient is $c_n = (f,\varphi_n)_w / (\varphi_n,\varphi_n)_w$ , which is the form in which Strauss and Riley-Hobson-Bence write it.

A series $\sum_n f_n$ converges in mean square (or in $L^2_w$ ) to $f$ on $(a,b)$ if $\lVert f - \sum_{n\le N} f_n\rVert_w \to 0$ as $N \to\infty$ ; this is weaker than uniform convergence and neither implies nor is implied by pointwise convergence. Strauss sets out all three notions with examples separating them.

Theorem(Least squares and Bessel's inequality)AdaptedAdapted from Strauss (2008), §5.4, Thm 5 and eqs. (15)–(18), pp. 131–133

Let $\{\varphi_n\}$ be any orthonormal family in $L^2_w(a,b)$ — completeness is not assumed — and let $f \in L^2_w(a,b)$ . Fix $N$ .

Least-square approximation. Among all choices of constants $d_1,\dots,d_N$ , the quantity $\bigl\lVert f - \sum_{n\le N} d_n\varphi_n\bigr\rVert_w$ is smallest exactly when $d_n = c_n$ for every $n \le N$ . The best approximation to $f$ from $\operatorname{span}\{\varphi_1,\dots,\varphi_N\}$ is the partial sum of its own expansion.
The error. $\lVert f - S_N\rVert_w^2 = \lVert f\rVert_w^2 - \sum_{n\le N}|c_n|^2 .$
Bessel's inequality. $\sum_{n=1}^{\infty}|c_n|^2 \le \lVert f\rVert_w^2$ ; in particular the series on the left converges.

Idea of the proofAdaptedAdapted from Strauss (2008), §5.4, proof of Thm 5, pp. 132–133

Expand the squared error. Orthonormality kills every cross term between different $\varphi$ 's, so what is left is a quadratic polynomial in the coefficients $d_n$ with no products $d_nd_m$ , $n\neq m$ . Completing the square in each $d_n$ separately isolates the coefficients inside a sum of squares plus a remainder that does not involve them at all; the minimum is therefore at the point where every square vanishes, and the remainder is the minimal error. Since that remainder is a norm it is non-negative, which is Bessel's inequality.

ProofAdaptedAdapted from Strauss (2008), §5.4, proof of Thm 5 and the derivation of (16)–(18), pp. 132–133

Expand. For any constants $d_n$ , writing $E_N$ for the squared error and using sesquilinearity,
$E_N = \Bigl\lVert f - \sum_{n\le N} d_n\varphi_n\Bigr\rVert_w^2 = \lVert f\rVert_w^2 - \sum_{n\le N}\bigl[\,\overline{d_n}\,(f,\varphi_n)_w + d_n\,\overline{(f,\varphi_n)_w}\,\bigr] + \sum_{m,n\le N} d_m\overline{d_n}\,(\varphi_m,\varphi_n)_w .$
Use orthonormality. By $(\varphi_m,\varphi_n)_w = \delta_{mn}$ the double sum collapses to $\sum_{n\le N}|d_n|^2$ . With $c_n = (f,\varphi_n)_w$ the middle bracket is $\overline{d_n}c_n + d_n\overline{c_n}$ , so
$E_N = \lVert f\rVert_w^2 + \sum_{n\le N}\bigl(|d_n|^2 - \overline{d_n}c_n - d_n\overline{c_n}\bigr) .$
Complete the square, one coefficient at a time. For each $n$ , $|d_n|^2 - \overline{d_n}c_n - d_n\overline{c_n} = |d_n - c_n|^2 - |c_n|^2$ . Hence
$E_N = \lVert f\rVert_w^2 - \sum_{n\le N}|c_n|^2 + \sum_{n\le N}\bigl|d_n - c_n\bigr|^2 .$
The coefficients $d_n$ now appear only in the last sum, which is a sum of non-negative terms.
Parts 1 and 2. The last sum is zero exactly when $d_n = c_n$ for all $n \le N$ , and positive otherwise; so that choice, and only that choice, minimises $E_N$ , and the minimum value is $\lVert f\rVert_w^2 - \sum_{n\le N}|c_n|^2$ . Putting $d_n = c_n$ gives part 2.
Part 3. Since $E_N$ is a squared norm, $\lVert f\rVert_w^2 - \sum_{n\le N}|c_n|^2 \ge 0$ , so the partial sums of $\sum|c_n|^2$ are bounded above by $\lVert f\rVert_w^2$ . A series of non-negative terms with bounded partial sums converges, and its sum obeys the same bound. $\square$

Theorem(Completeness of the eigenfunctions)AdaptedAdapted from Strauss (2008), §5.4, Thm 3, Thm 6 and the definition of completeness (19), pp. 128 and 133, with §11.3, Thm 2, p. 311, and §11.4, Thm 1, p. 314; Hassani (2013), §19.1, Thm 19.1.9, pp. 568–569; Stakgold & Holst (2011), §6.3, Thm 6.3.6, p. 375

Let $\{\varphi_n\}$ be the normalised eigenfunctions of a regular Sturm-Liouville problem on $[a,b]$ .

Mean-square convergence. For every $f \in L^2_w(a,b)$ , $\lVert f - S_N\rVert_w \longrightarrow 0 \qquad (N\to\infty) .$ No continuity or differentiability of $f$ is required, and no boundary condition.
Parseval's equality. Equivalently, $\sum_{n=1}^{\infty} \bigl|c_n\bigr|^2 \;=\; \lVert f\rVert_w^2 , \qquad\text{and more generally}\qquad \sum_{n=1}^{\infty} c_n\,\overline{d_n} = (f,g)_w$ for $f,g \in L^2_w(a,b)$ with coefficients $c_n$ and $d_n$ . An orthonormal family for which the first identity holds for every $f\in L^2_w(a,b)$ is called complete; so the statement is that the eigenfunctions of a regular Sturm-Liouville problem are complete.
Uniqueness of the expansion. If $f\in L^2_w(a,b)$ has $(f,\varphi_n)_w = 0$ for every $n$ , then $\lVert f\rVert_w = 0$ . Two functions of $L^2_w(a,b)$ with the same coefficients therefore agree except on a set of measure zero, and if both are continuous they agree everywhere.

Part 1 is quoted, not proved here; see the note. Parts 2 and 3 are deduced from it below.

Idea of the proofOriginal

Part 2 is a restatement of part 1 and nothing more: by part 2 of the least-squares statement, the squared error $\lVert f - S_N\rVert_w^2$ is the gap $\lVert f\rVert_w^2 - \sum_{n\le N}|c_n|^2$ in Bessel's inequality, so the error tends to zero precisely when the gap closes. The polarised form then follows because an inner product is determined by the norms of $f+g$ and $f-g$ . Part 3 is Parseval read backwards: if all the coefficients vanish, Parseval leaves $\lVert f\rVert_w^2 = 0$ .

ProofAdaptedAdapted from Strauss (2008), §5.4, proof of Thm 6 and eq. (16), p. 133

Parts 1 and 2 are the same statement. Part 2 of the least-squares statement, which needed only orthonormality, gives the exact identity
$\lVert f - S_N\rVert_w^2 = \lVert f\rVert_w^2 - \sum_{n\le N}|c_n|^2$
for every $N$ . The left side tends to $0$ if and only if the subtracted sum tends to $\lVert f\rVert_w^2$ , that is, if and only if Parseval's equality holds. So part 1 and the first identity of part 2 are equivalent, and part 1 gives both.
The polarised form. Fix $f,g \in L^2_w(a,b)$ with coefficients $c_n,d_n$ . The map $f \mapsto (c_n)$ is linear, so $f \pm g$ has coefficients $c_n \pm d_n$ . Applying Parseval to $f+g$ and to $f-g$ and subtracting,
$\sum_n \bigl(|c_n+d_n|^2 - |c_n-d_n|^2\bigr) = \lVert f+g\rVert_w^2 - \lVert f-g\rVert_w^2 .$
Both sides expand by sesquilinearity: the left is $2\sum_n(c_n\overline{d_n} + \overline{c_n}d_n)$ and the right is $2\bigl((f,g)_w + \overline{(f,g)_w}\bigr)$ , so the real parts of $\sum_n c_n\overline{d_n}$ and $(f,g)_w$ agree. Repeating with $ig$ in place of $g$ — whose coefficients are $i d_n$ — matches the imaginary parts as well. (For real-valued $f,g$ the first computation alone suffices.) The series converges absolutely, by Cauchy-Schwarz in $\ell^2$ together with Bessel's inequality for $f$ and for $g$ .
Part 3. If every $c_n = 0$ , Parseval gives $\lVert f\rVert_w^2 = 0$ , hence $\int_a^b w|f|^2 = 0$ ; since $w > 0$ , $|f|^2 = 0$ almost everywhere. Applied to a difference $f - g$ , this gives the uniqueness claim, and for continuous $f-g$ a continuous non-negative function with zero integral against the positive weight $w$ vanishes identically. $\square$

Note(Part 1 is quoted, not proved here)AdaptedAdapted from Strauss (2008), §11.3, Thm 2 with the proof for trial functions, pp. 311–313, and §11.4, Thm 1, p. 314, with the remark on p. 316; Hassani (2013), §19.1, Thm 19.1.9 and the compact-resolvent construction, pp. 567–569; Stakgold & Holst (2011), §6.3, Thm 6.3.4 and Thms 6.3.5–6.3.6, pp. 373–375

Completeness is the one deep statement in Sturm-Liouville theory, and like part 4 of the Sturm-Liouville theorem it is quoted here with pointers rather than proved. Three routes appear in the sources.

Compact resolvent (Hassani). Shift $\lambda$ so that $0$ is not an eigenvalue and invert the operator by an explicit kernel built from two solutions; the inverse is a Hilbert-Schmidt, hence compact, operator whose null space is trivial, so by the spectral theory of compact self-adjoint operators its eigenfunctions span the whole space. Hassani's Theorem 19.1.9 lists completeness among the conclusions for a regular Sturm-Liouville system, stated for the normal form $u''-qu = \lambda u$ and transported to the general equation by the Liouville substitution of §19.2.
Green's function and integral equations (Stakgold and Holst). Their Theorem 6.3.4, the expansion theorem, says that every function in the range of a compact self-adjoint integral operator $K$ expands in the eigenfunctions belonging to its nonzero eigenvalues; Theorem 6.3.5 adds that all the eigenfunctions together, including those of the zero eigenvalue, form a basis. Applied to the Green's function of a self-adjoint boundary-value problem — a real symmetric kernel whose integral operator has trivial null space — this yields their Theorem 6.3.6: for a self-adjoint boundary-value problem, its own eigenfunctions are an orthonormal basis. They state it for a formally self-adjoint $L$ of even order with self-adjoint boundary functionals, on the unweighted $L^2(a,b)$ .
Minimisation (Strauss). His Theorem 11.3.2 derives completeness from two facts: the eigenvalues are successive minima of the Rayleigh quotient $\lVert\nabla w\rVert^2/\lVert w\rVert^2$ , and $\lambda_N\to\infty$ . The remainder $r_N = f - S_N$ is orthogonal to $\varphi_1,\dots,\varphi_N$ , so it is admissible in the minimum principle, which therefore gives $\lambda_{N+1} \le \lVert\nabla r_N\rVert^2/\lVert r_N\rVert^2$ ; and an energy identity, using the eigenvalue equation and the boundary condition, bounds the numerator by $\lVert\nabla f\rVert^2$ . His conclusion (12) is $\lVert r_N\rVert^2 \;\le\; \frac{\lVert\nabla r_N\rVert^2}{\lambda_N} \;\le\; \frac{\lVert\nabla f\rVert^2}{\lambda_N} ,$ whose right-hand side has a fixed numerator and so tends to $0$ . (His minimum principle (6) supplies the divisor $\lambda_{N+1}$ while (12) is printed with $\lambda_N$ ; as $\lambda_N \le \lambda_{N+1}$ the printed form is the weaker of the two, and either serves once $\lambda_N\to\infty$ .) Strauss carries this out for $f$ a trial function and refers to advanced texts for a general $f\in L^2$ ; his Theorem 11.4.1 extends the conclusion to the variable-coefficient problem with the weighted inner product, and he records on p. 316 that the same conclusions hold for Sturm-Liouville problems, citing Coddington-Levinson and Courant-Hilbert.

The machinery lives in compact operators, compact self-adjoint operators, orthonormal bases and Hilbert-Schmidt kernels; the variational route is the variational characterisation of eigenvalues. Arfken's summary puts the same conclusion in operator language: the eigenfunctions of a Hermitian operator on a Hilbert space form a complete family, spanning the space and usable as an expansion basis.

Theorem(Forced problems and the Fredholm alternative)AdaptedAdapted from Strauss (2008), §11.5, Thm 1 with eqs. (1), (5) and (7), pp. 318–319, specialised to one dimension by §11.4, eq. (7), p. 315

Consider the forced problem on $[a,b]$ ,

-\bigl(p\,u'\bigr)' + q\,u \;=\; \sigma\,w\,u + F ,

with the same separated boundary conditions, where $\sigma$ is a given constant and $F$ is a given real function. Strauss states the result for the elliptic equation $-\nabla\cdot(p\nabla u) + qu = \sigma m u + f$ with homogeneous Dirichlet, Neumann or Robin conditions; in one dimension it reads:

If $\sigma$ is not an eigenvalue of the homogeneous problem, there is exactly one solution — existence is quoted from Strauss, whose Theorem 1(a) asserts it; the proof below establishes only uniqueness, the coefficients and $L^2_w$ convergence, as the remark sets out — for every $F$ with $\int_a^b \frac{F(x)^2}{w(x)}\,\mathrm d x < \infty ,$ and it is $u(x) = \sum_{n=1}^{\infty} \frac{1}{\lambda_n - \sigma}\Bigl(\int_a^b F\,\varphi_n\,\mathrm d x\Bigr)\varphi_n(x) ,$ the series converging in $L^2_w(a,b)$ .
If $\sigma = \lambda_N$ is an eigenvalue, then depending on $F$ the problem has no solution whatever, or else infinitely many. Precisely, a solution exists if and only if $\int_a^b F\,\varphi_N\,\mathrm d x = 0 ,$ and in that case the solutions are $u(x) = \sum_{n \neq N} \frac{1}{\lambda_n - \lambda_N}\Bigl(\int_a^b F\,\varphi_n\,\mathrm d x\Bigr)\varphi_n(x) \;+\; C\,\varphi_N(x) , \qquad C \in \mathbb R .$

Note that $\int_a^b F\varphi_n\,\mathrm d x = (F/w,\,\varphi_n)_w$ : the raw forcing $F$ is paired with $\varphi_n$ without the weight, and the hypothesis $\int F^2/w < \infty$ says exactly that $F/w \in L^2_w(a,b)$ . Writing $F = wG$ turns the two into $(G,\varphi_n)_w$ and $G \in L^2_w(a,b)$ .

Idea of the proofAdaptedAdapted from Strauss (2008), §11.5, proof of Thm 1, pp. 318–319

Pair the equation with one eigenfunction at a time. Green's formula moves the operator from $u$ onto $\varphi_n$ with no boundary term left over, because $u$ and $\varphi_n$ both satisfy the boundary conditions; the eigenvalue equation then replaces $L\varphi_n$ by $\lambda_n w\varphi_n$ , and what survives is a single scalar equation $(\lambda_n - \sigma)(u,\varphi_n)_w = \int F\varphi_n$ . So the differential equation becomes, mode by mode, a division by $\lambda_n - \sigma$ . Completeness is what licenses reassembling the quotients into $u$ ; and when some $\lambda_N$ equals $\sigma$ the division is by zero, which is where the dichotomy comes from.

ProofAdaptedAdapted from Strauss (2008), §11.5, proof of Thm 1, eqs. (3)–(7), pp. 318–319

One scalar equation per mode. Let $u \in C^2[a,b]$ solve the problem. Multiply the equation by $\varphi_n$ and integrate:
$\int_a^b \varphi_n\,Lu\,\mathrm d x = \sigma\int_a^b w\,u\,\varphi_n\,\mathrm d x + \int_a^b F\,\varphi_n\,\mathrm d x .$
By Green's formula for $L$ (step 1 of the proof in the previous unit), $\int\varphi_n Lu - \int u L\varphi_n$ is a boundary term, and it vanishes because $u$ and $\varphi_n$ satisfy the same separated boundary conditions. Replacing $L\varphi_n = \lambda_n w\varphi_n$ gives $\int \varphi_n Lu = \lambda_n (u,\varphi_n)_w$ , so
$\bigl(\lambda_n - \sigma\bigr)\,(u,\varphi_n)_w = \int_a^b F\,\varphi_n\,\mathrm d x \qquad\text{for every } n. \tag{$\ast$}$
Part 1, uniqueness and the formula. If no $\lambda_n$ equals $\sigma$ , $(\ast)$ determines every coefficient: $(u,\varphi_n)_w = (\lambda_n-\sigma)^{-1}\int F\varphi_n$ . By part 3 of the completeness statement a function of $L^2_w$ is determined by its coefficients, so there is at most one solution, and by part 1 of that statement it is the sum of its own expansion — which is the displayed series.
Part 1, convergence. Put $\delta = \inf_n|\lambda_n - \sigma|$ , which is positive: the $\lambda_n$ increase to $+\infty$ , so only finitely many lie within distance $1$ of $\sigma$ , and none equals it. Let $b_n = \int_a^b F\varphi_n\,\mathrm d x = (F/w,\varphi_n)_w$ . Parseval applied to $F/w$ , which lies in $L^2_w$ by hypothesis, gives
$\sum_n b_n^2 = \lVert F/w\rVert_w^2 = \int_a^b \frac{F^2}{w}\,\mathrm d x < \infty ,$
and therefore, by orthonormality,
$\Bigl\lVert \sum_{n\le N}\frac{b_n}{\lambda_n - \sigma}\varphi_n\Bigr\rVert_w^2 = \sum_{n\le N}\frac{b_n^2}{(\lambda_n-\sigma)^2} \le \frac{1}{\delta^2}\int_a^b\frac{F^2}{w}\,\mathrm d x .$
The partial sums form a Cauchy sequence in $L^2_w$ by the same estimate applied to the tail, so the series converges in $L^2_w$ . This is where the hypothesis $\int F^2/w < \infty$ is used, and where $\lambda_n \to\infty$ is used.
Part 2, the obstruction. If $\sigma = \lambda_N$ , then $(\ast)$ with $n=N$ has a vanishing left-hand side, so $\int F\varphi_N\,\mathrm d x = 0$ is forced: without it there is no solution. The condition is the orthogonality of the forcing to the resonant mode.
Part 2, the solutions. Suppose $\int F\varphi_N = 0$ . The eigenvalue is simple, so $\varphi_N$ spans the solutions of the homogeneous problem with $\sigma = \lambda_N$ , and $L(C\varphi_N) - \lambda_N w (C\varphi_N) = 0$ : adding $C\varphi_N$ to a solution gives a solution. Conversely $(\ast)$ fixes $(u,\varphi_n)_w$ for every $n \neq N$ and says nothing about $(u,\varphi_N)_w$ , which is therefore the free constant $C$ . The displayed family is thus exactly the set of solutions, and it is infinite. $\square$

RemarkOriginal

Where the hypotheses are used.

Orthonormality alone carries the least-squares statement and Bessel's inequality; those hold for any orthonormal family in any inner-product space, eigenfunctions or not.
Completeness is what upgrades Bessel's inequality to Parseval's equality and so converts the least-squares property into actual convergence. Drop it and the conclusions fail outright: $\{\varphi_2,\varphi_3,\dots\}$ is still orthonormal, but $f = \varphi_1$ has all coefficients zero against it, so the partial sums converge to $0 \neq f$ .
$w>0$ is used in part 3 of the completeness statement, to turn $\int w|f|^2 = 0$ into $f = 0$ .
$\int F^2/w < \infty$ and $\lambda_n\to\infty$ are both used in step 3 of the last proof, the first to make $\sum b_n^2$ finite and the second to make $\delta > 0$ . Without the growth of the eigenvalues, $\sigma$ could be a limit of eigenvalues and the series could diverge.
Separated boundary conditions on $u$ . Step 1 needs the boundary term of Green's formula to vanish for the pair $u,\varphi_n$ , so $u$ must satisfy the same homogeneous conditions. Inhomogeneous boundary data must first be removed by subtracting any function that carries it.

What the proof does and does not deliver. The computation above establishes uniqueness, produces the coefficients and shows that the series converges in $L^2_w$ ; it does not verify by itself that the sum is twice differentiable and satisfies the equation pointwise. Strauss states existence as part of his Theorem 1 and argues as above; smoothness of the sum requires extra decay of $b_n$ , which extra smoothness of $F$ supplies. The same caution applies to term-by-term differentiation of the series in the separation-of-variables argument, which Strauss carries out under explicit differentiability assumptions and defers to the weak-derivative framework of his Chapter 12.

Three modes of convergence. Mean-square convergence holds for every $f \in L^2_w$ and says nothing at a single point: Strauss notes that there exist integrable functions whose Fourier series diverge at every point, and continuous functions whose series diverge at many points. For better convergence one pays with hypotheses, and Strauss's two further theorems are stated for the model problem $X'' + \lambda X = 0$ with symmetric boundary conditions:

Uniform convergence on $[a,b]$ if $f$ , $f'$ and $f''$ exist and are continuous on $[a,b]$ and $f$ satisfies the boundary conditions. The second half is not decoration: the sine series of $f\equiv 1$ on $(0,\pi)$ cannot converge uniformly, since every partial sum vanishes at the endpoints while $f$ does not.
Pointwise convergence on $(a,b)$ , for the classical full, sine and cosine series, if $f$ is continuous on $[a,b]$ and $f'$ piecewise continuous on $[a,b]$ . If $f$ itself is only piecewise continuous, with piecewise continuous derivative, the series still converges at every point of the whole line; its sum is $\tfrac12[f(x+)+f(x-)]$ for $a<x<b$ , and $\tfrac12[f_{\mathrm{ext}}(x+)+f_{\mathrm{ext}}(x-)]$ everywhere, $f_{\mathrm{ext}}$ being the periodic, odd periodic or even periodic extension that goes with the series. So at a jump inside the interval the series picks the midpoint of the two one-sided limits, and outside the interval it describes the extension, not $f$ .

The closure relation. Substituting the coefficient formula into the expansion and exchanging sum and integral gives, formally,

w(z)\sum_n \varphi_n(x)\,\overline{\varphi_n(z)} = \delta(x-z) ,

which Riley-Hobson-Bence call the closure or completeness relation, noting that $w(z)$ may be replaced by $w(x)$ since the right side is supported at $x=z$ . It is an identity between distributions, not between functions, and it is the same object as the bilinear expansion of a Green's function; see eigenfunction expansions of Green's functions.

The Fredholm alternative. Part 2 of the last statement is the differential-equation form of a pattern that recurs throughout linear analysis: for $(L-\sigma)u = F$ either $L-\sigma$ is invertible and the solution is unique, or $\sigma$ is an eigenvalue and solvability requires $F$ to be orthogonal to the corresponding null space, in which case the solution is unique only up to that null space. See the Fredholm alternative.

Method(Solving a forced Sturm-Liouville problem by eigenfunction expansion)AdaptedAdapted from Riley, Hobson & Bence (2006), §17.5, pp. 569–570; Strauss (2008), §11.5, eqs. (2)–(5), pp. 318–319

To solve $-(pu')' + qu - \sigma\,w\,u = F$ on $[a,b]$ with homogeneous separated boundary conditions:

Normalise the problem. Put the equation into the form $-(pu')' + qu = \lambda w u$ for the homogeneous problem, by the integrating-factor recipe of the self-adjoint form if necessary. Read off $p$ , $q$ , $w$ .
Solve the eigenvalue problem. Find $\lambda_n$ and $\varphi_n$ , and normalise so that $(\varphi_m,\varphi_n)_w = \delta_{mn}$ .
Check the forcing. Confirm $\int_a^b F^2/w < \infty$ . If the boundary data are inhomogeneous, subtract any convenient function satisfying them and move the result into $F$ .
Project. Compute $b_n = \int_a^b F\,\varphi_n\,\mathrm d x$ . (If the forcing is naturally written as $F = wG$ , this is $(G,\varphi_n)_w$ .)
Divide mode by mode. If $\sigma$ is not an eigenvalue, the solution is $u = \sum_n b_n\varphi_n/(\lambda_n - \sigma)$ . If $\sigma = \lambda_N$ , first test $b_N$ : if $b_N \neq 0$ there is no solution; if $b_N = 0$ , drop the $n=N$ term and add an arbitrary multiple of $\varphi_N$ .
Judge the truncation. The coefficients decay like $b_n/\lambda_n$ , so the series converges faster than the expansion of $F$ itself; the smoother $F$ is, the fewer terms are needed. For a definite answer at a point, compute partial sums until successive values agree to the accuracy wanted.

ExampleOriginal

An expansion and a forced problem for the Euler operator. Continue with the regular problem of the previous unit on $[1,2]$ :

-\bigl(x^3u'\bigr)' - x\,u = \lambda\,x\,u , \quad u(1)=u(2)=0 , \qquad \lambda_n = \Bigl(\frac{n\pi}{\ln 2}\Bigr)^2 , \quad \varphi_n(x) = \sqrt{\frac{2}{\ln 2}}\;\frac{1}{x}\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr) ,

now normalised: $\lVert x^{-1}\sin(n\pi\ln x/\ln 2)\rVert_w^2 = \tfrac12\ln 2$ , so the factor $\sqrt{2/\ln 2} = 1.69864$ makes $(\varphi_m,\varphi_n)_w = \delta_{mn}$ . Here $w = x$ , and the substitution $s = \ln x/\ln 2$ , $\mathrm d x/x = \ln 2\,\mathrm d s$ , turns every integral into one over $(0,1)$ .

Part A: expanding $f(x) = 1/x$ . It is in $L^2_w(1,2)$ , since $\lVert f\rVert_w^2 = \int_1^2 x\cdot x^{-2}\,\mathrm d x = \ln 2 = 0.693147$ . Its coefficients are

c_n = (f,\varphi_n)_w = \sqrt{\tfrac{2}{\ln 2}}\int_1^2 \frac{1}{x}\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr)\mathrm d x = \sqrt{\tfrac{2}{\ln 2}}\,\ln 2\int_0^1\!\sin(n\pi s)\,\mathrm d s = \sqrt{2\ln 2}\;\frac{1-\cos n\pi}{n\pi} ,

so $c_n = 0$ for even $n$ and $c_n = 2\sqrt{2\ln 2}/(n\pi)$ for odd $n$ ; numerically $c_1 = 0.749563$ , $c_3 = 0.249854$ , $c_5 = 0.149913$ . Writing the expansion without the normalisation constants,

\frac{1}{x} \;=\; \sum_{n\ \mathrm{odd}} \frac{4}{n\pi}\,\frac{1}{x}\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr) \qquad\text{in } L^2_w(1,2) ,

which under $s = \ln x/\ln 2$ is the classical sine series of the constant $1$ on $(0,1)$ . That is no accident: the Liouville substitution of Hassani's §19.2 is here $u = v(t)/x$ with $t = \int_1^x\sqrt{w/p}\,\mathrm d s = \int_1^x \mathrm d s/s = \ln x$ and $[pw]^{-1/4} = (x^4)^{-1/4} = 1/x$ , and it turns the problem into $-v'' = \lambda v$ on $(0,\ln 2)$ with $v(0)=v(\ln 2)=0$ . Our $s$ is that $t$ rescaled to $(0,1)$ . Checking Parseval:

\sum_{n\ \mathrm{odd}} |c_n|^2 = \frac{8\ln 2}{\pi^2}\sum_{n\ \mathrm{odd}}\frac{1}{n^2} = \frac{8\ln 2}{\pi^2}\cdot\frac{\pi^2}{8} = \ln 2 = \lVert f\rVert_w^2 . \checkmark

Read the other way, Parseval gives $\sum_{n\ \mathrm{odd}} n^{-2} = \pi^2/8$ .

Part B: a forced problem with $\sigma=0$ . Solve

-\bigl(x^3 u'\bigr)' - x\,u = 1 , \qquad u(1) = u(2) = 0 .

Here $F \equiv 1$ and $\sigma = 0$ , which is not an eigenvalue since $\lambda_1 = 20.5423 > 0$ . The hypothesis holds: $\int_1^2 F^2/w\,\mathrm d x = \int_1^2 \mathrm d x/x = \ln 2 < \infty$ . The projections are the same integrals as in Part A, because $F/w = 1/x = f$ :

b_n = \int_1^2 1\cdot\varphi_n(x)\,\mathrm d x = (f,\varphi_n)_w = c_n .

So the solution is

u = \sum_{n}\frac{c_n}{\lambda_n}\varphi_n = \sum_{n\ \mathrm{odd}} \frac{4}{n\pi}\Bigl(\frac{\ln 2}{n\pi}\Bigr)^{2}\frac{1}{x}\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr) = \sum_{n\ \mathrm{odd}} \frac{4(\ln 2)^2}{n^3\pi^3}\,\frac{1}{x}\sin\!\Bigl(\frac{n\pi\ln x}{\ln 2}\Bigr) .

Checking against a closed form. Set $t = \ln x$ and $u = V(t)/x$ . Since $x^2u''+3xu'+u = (x\,\mathrm d/\mathrm d x+1)^2u$ and $(x\,\mathrm d/\mathrm d x+1)\bigl(e^{-t}V\bigr) = e^{-t}V'$ , applying the operator twice gives $x^2u''+3xu'+u = V''(t)/x$ . The equation $-(x^3u')' - xu = 1$ is $-x\,(x^2u''+3xu'+u) = 1$ , that is $V'' = -1$ , with $V(0)=V(\ln 2)=0$ . Hence

V(t) = \tfrac12\,t\,(\ln 2 - t) , \qquad u(x) = \frac{\ln x\,\bigl(\ln 2 - \ln x\bigr)}{2x} .

Its sine series in $s = t/\ln 2$ is $V = \tfrac12(\ln 2)^2 s(1-s)$ with $s(1-s) = \sum_{n\ \mathrm{odd}} 8\,\sin(n\pi s)/(n^3\pi^3)$ , reproducing the coefficients above exactly.

A number. At $x = \sqrt 2$ the closed form gives $u = (\ln 2)^2/(8\sqrt 2) = 0.0424664$ . The partial sums of the series are $0.0438275$ (one term), $0.0422042$ , $0.0425548$ , $0.0424271$ , $0.0424872$ , $0.0424543$ (six terms, $n$ up to $11$ ), bracketing the exact value as the alternating $1/n^3$ terms shrink.

Part C: resonance. Now take $\sigma = \lambda_1$ , so the problem is

-\bigl(x^3u'\bigr)' - x\,u - \lambda_1\,x\,u = F , \qquad u(1)=u(2)=0 .

With $F \equiv 1$ the test integral is $b_1 = c_1 = 0.749563 \neq 0$ , so there is no solution: the forcing has a component along the resonant mode, and nothing can absorb it. With $F = x\,\varphi_2(x)$ instead, $b_1 = (\varphi_2,\varphi_1)_w = 0$ and $b_n = \delta_{2n}$ , so solutions exist and are

u = \frac{\varphi_2}{\lambda_2 - \lambda_1} + C\varphi_1 = \frac{\varphi_2}{61.6269} + C\varphi_1 , \qquad C\in\mathbb R ,

a one-parameter family. This is directly verifiable: applying $-(x^3\cdot)' - x - \lambda_1 x$ to $\varphi_2/(\lambda_2-\lambda_1)$ gives $(\lambda_2-\lambda_1)x\varphi_2/(\lambda_2-\lambda_1) = x\varphi_2 = F$ , and it annihilates $C\varphi_1$ .

Application(Separation of variables, justified)AdaptedAdapted from Strauss (2008), §11.5, the separation-of-variables argument, eqs. (8)–(11), pp. 319–320

For $w\,u_t = (p u_x)_x - q u$ on $[a,b]$ with homogeneous separated boundary conditions, expand $u(\cdot,t)$ in the eigenfunctions at each fixed $t$ : completeness gives $u = \sum_n a_n(t)\varphi_n(x)$ for some coefficients, as yet unknown functions of $t$ . Substituting and matching coefficients turns the partial differential equation into the decoupled scalar equations $a_n' = -\lambda_n a_n$ , whence $a_n(t) = a_n(0)e^{-\lambda_n t}$ with $a_n(0) = (u(\cdot,0),\varphi_n)_w$ read off from the initial data. This is Strauss's point, and it is a point about logic: the usual expansion is forced rather than guessed, since the expansion exists because the eigenfunctions are complete and not because the solution was assumed to be a sum of products. The step from the series to the term-by-term differentiation needs its own hypotheses, which Strauss states (square-integrability of $u$ and $\nabla u$ , plus differentiability enough to differentiate the series termwise). See separation of variables and the heat equation on a bounded domain.

ApplicationOriginal

Steady forcing and resonance. Part C above is the mathematics of resonance. A driven string, a driven circuit, a cavity pumped at a mode frequency: if the drive has any overlap with the resonant eigenfunction, the steady problem has no bounded steady solution and the response grows in time; if the drive is orthogonal to it, a steady solution exists but is determined only up to a free oscillation in that mode. The denominators $\lambda_n - \sigma$ are the amplification factors, largest for the modes nearest resonance. See the damped and driven oscillator.
Quantum mechanics. Completeness of the eigenfunctions of a Hermitian Hamiltonian is the superposition principle: any state expands in energy eigenstates, $|c_n|^2$ is the probability of measuring $E_n$ , and Parseval's equality is the statement that these probabilities sum to one. The forced problem with $\sigma$ not an eigenvalue is the resolvent $(H-\sigma)^{-1}$ , and the expansion $\sum_n \varphi_n\varphi_n^{*}/(\lambda_n - \sigma)$ is its spectral representation — the object perturbation theory expands. See the quantum-mechanical framework.
Green's functions. Setting $F = \delta(x-\xi)$ in the solution formula gives $g(x,\xi) = \sum_n \varphi_n(x)\varphi_n(\xi)/(\lambda_n - \sigma)$ , the bilinear or eigenfunction expansion of the Green's function; its symmetry in $x$ and $\xi$ is manifest, and its poles in $\sigma$ sit exactly at the eigenvalues. See Green's functions for boundary-value problems and their eigenfunction expansions.
Least squares and truncation in practice. Part 1 of the least-squares statement is why modal truncation is the standard reduced-order model in structural dynamics and in climate and ocean models: once a set of modes is chosen, the expansion coefficients are the best possible coefficients in the energy-weighted norm, and no fitting is needed to find them. Which modes to keep is a separate question, and the answer is not automatically "the first $N$ ": for a set $S$ of indices the squared error is $\lVert f\rVert_w^2 - \sum_{n\in S}|c_n|^2$ , so the best $N$ -mode subspace is the one spanned by the $N$ eigenfunctions with the largest $|c_n|$ . The first $N$ are optimal exactly when $|c_1| \ge |c_2| \ge \cdots$ , which is typical but not guaranteed — in Part A above $c_2 = 0$ while $c_3 = 0.249854$ , so $\operatorname{span}\{\varphi_1,\varphi_3\}$ gives squared error $0.068876$ against $0.131303$ for $\operatorname{span}\{\varphi_1,\varphi_2\}$ . What is always true, and what makes truncation cheap to assess, is that the error is exactly the discarded sum $\sum_{n\notin S}|c_n|^2$ .

Theorem · advanced

Sturm comparison and oscillation theorems

Adapted

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The zeros of two independent solutions of a second-order linear equation interlace, a solution of the equation with the larger coefficient must vanish between consecutive zeros of a solution of the equation with the smaller one, and these two facts bound the spacing of zeros, decide whether solutions oscillate, and count the zeros of Sturm-Liouville eigenfunctions.

Prerequisites: Sturm-Liouville eigenvalue problems, Adjoint and self-adjoint differential operators, Structure of linear equations and the Wronskian

Definition(Zeros, the normal form, and oscillation)AdaptedAdapted from Hassani (2013), §14.4, Def. 14.4.1, p. 425, and Prop. 14.4.11, p. 431; Coddington (1989), Ch. 3, §4, Exercise 8, p. 118

Throughout, $I$ is an interval of the real line and all functions on it are real-valued. The two shapes of second-order equation used here are

y'' + P(x)\,y' + Q(x)\,y = 0 \qquad\text{and}\qquad \bigl(p(x)\,u'\bigr)' + G(x)\,u = 0 ,

with $P,Q,G$ continuous on $I$ , $p \in C^1(I)$ and $p > 0$ . The second is the divergence shape of the previous chapter's self-adjoint form; dividing it by $p$ turns it into the first with $P = p'/p$ and $Q = G/p$ , so both are genuine second-order equations with continuous coefficients, and the structure theory applies to each.

A zero of a solution is a point of $I$ at which it vanishes. A solution is nontrivial if it is not identically zero. A zero $x_0$ of a solution $y$ is simple when $y'(x_0) \neq 0$ ; since $y(x_0) = y'(x_0) = 0$ would force $y \equiv 0$ by the uniqueness of the solution of an initial-value problem for a linear equation with continuous coefficients, every zero of a nontrivial solution is simple. A simple zero is isolated, because $y(x) = y'(x_0)(x-x_0) + o(x - x_0)$ is nonzero for $x$ near $x_0$ and $x \neq x_0$ ; so a nontrivial solution has only finitely many zeros in any compact subinterval of $I$ , and the zeros in such a subinterval can be listed in increasing order. Two zeros $x_1 < x_2$ are consecutive when the solution does not vanish anywhere in $(x_1,x_2)$ .

The normal form (or Liouville form) of $y'' + Py' + Qy = 0$ is obtained by the substitution

u(x) = y(x)\,\exp\!\Bigl(\tfrac12\!\int_\alpha^x P(t)\,\mathrm d t\Bigr) , \qquad\text{which satisfies}\qquad u'' + S\,u = 0 , \quad S = Q - \tfrac14 P^2 - \tfrac12 P' ,

for any base point $\alpha \in I$ (Hassani (2013), Prop. 14.4.11, p. 431); the formula requires $P \in C^1(I)$ . The exponential factor never vanishes, so $y$ and $u$ have exactly the same zeros, and every question about zeros may be moved to the normal form. The same remark applies to the divergence shape as soon as $p \in C^2(I)$ , which is what makes $P = p'/p$ continuously differentiable: the normalising factor is then $\exp\bigl(\frac12\int_\alpha^x p'/p\bigr) = \sqrt{p(x)/p(\alpha)} > 0$ , so $(pu')' + Gu = 0$ and its normal form share their zeros too. Under the standing assumption $p \in C^1$ alone the normal form need not exist, and the comparison theorem below is proved directly in the divergence shape for exactly that reason.

Finally, a nontrivial solution on $[a,\infty)$ is called oscillatory if it has infinitely many zeros there and nonoscillatory otherwise. By the separation theorem below, applied on each compact subinterval of $[a,\infty)$ , this is a property of the equation and not of the particular solution: if one nontrivial solution has infinitely many zeros then so has every other, since an independent partner must vanish between consecutive ones.

The Wronskian of two solutions is $W[f,g] = f g' - g f'$ , as in the structure theory; for the divergence shape the useful quantity is the weighted Wronskian $p\,W[f,g]$ , which is the conjunct of the self-adjoint unit.

Theorem(Sturm separation theorem)AdaptedAdapted from Hassani (2013), §14.4.3, Thm 14.4.8, p. 430; Coddington (1989), Ch. 3, §4, Exercise 5, p. 117

Let $f$ and $g$ be real-valued linearly independent solutions of $y'' + Py' + Qy = 0$ on $[a,b]$ , with $P$ and $Q$ real and continuous there. Then

$f$ and $g$ have no common zero in $[a,b]$ ;
between two consecutive zeros of $g$ in $[a,b]$ there lies exactly one zero of $f$ , and between two consecutive zeros of $f$ there lies exactly one zero of $g$ .

The zeros of the two solutions therefore interlace. Hassani records the conclusion as the statement that the zeros of two independent solutions alternate; Coddington puts it as the existence of a zero of one between successive zeros of the other, and states it for real-valued solutions.

Theorem(Sturm comparison theorem)AdaptedAdapted from Hassani (2013), §14.4.3, Thm 14.4.10, p. 431; Coddington (1989), Ch. 3, §4, Exercise 4(a), p. 117; do Carmo (2016), §5-5, Exercise 6, p. 376

Divergence shape. Let $p \in C^1[a,b]$ with $p > 0$ , and let $G$ and $H$ be continuous and real on $[a,b]$ with

H(x) \;\ge\; G(x) \qquad (a \le x \le b).

Let $u$ be a nontrivial real solution of $(pu')' + Gu = 0$ and $v$ a nontrivial real solution of $(pv')' + Hv = 0$ , and let $x_1 < x_2$ be two zeros of $u$ . Then $v$ vanishes somewhere in the open interval $(x_1,x_2)$ , unless

H \equiv G \text{ on } [x_1,x_2] \quad\text{and}\quad v \text{ is a constant multiple of } u \text{ there},

in which case $v$ vanishes at $x_1$ and at $x_2$ instead. In particular $v$ always has a zero in $(x_1, x_2]$ .

Normal form. This is the case $p \equiv 1$ , and it needs no new letters: if $H \ge G$ on $[a,b]$ , $u$ is a nontrivial real solution of $u'' + Gu = 0$ and $v$ a nontrivial real solution of $v'' + Hv = 0$ , then $v$ has a zero between any two zeros of $u$ , except in the stated degenerate case. The solution of the equation with the larger coefficient is the one forced to oscillate faster. In the notation $u'' + S_u u = 0$ , $v'' + S_v v = 0$ of the normal form, the hypothesis is $S_v \ge S_u$ . (Hassani writes the two zeroth-order coefficients as $p$ and $q$ ; here $P$ and $Q$ are already committed, by the definition block and the separation theorem above, to the coefficients of $y'' + Py' + Qy = 0$ , so the two coefficients being compared are called $G$ and $H$ throughout this unit.)

Hassani states the normal-form version and refers to Birkhoff and Rota for a proof; his exceptional clause reads as a condition on the whole of $[a,b]$ , and the remark below explains why it has to be read on $[x_1,x_2]$ . Coddington's version assumes the strict inequality $H > G$ and then has no exceptional case. Do Carmo's "Sturm oscillation theorem" is the conclusion in the closed-interval form $(x_1,x_2]$ given above.

Idea of the proofOriginal

One identity does all the work. For two solutions of two equations in divergence shape with the same $p$ , the weighted Wronskian $W = p\,(uv' - vu')$ has derivative

W' = u\,(pv')' - v\,(pu')' = (G - H)\,u\,v ,

because the cross terms $p u'v'$ cancel. So the sign of $G - H$ controls whether $W$ rises or falls. Suppose $u$ vanishes at $x_1$ and $x_2$ and $v$ did not vanish between: normalise both to be positive inside. At a zero of $u$ the weighted Wronskian collapses to $-p\,v\,u'$ , which is $\le 0$ at $x_1$ (where $u$ is increasing) and $\ge 0$ at $x_2$ (where $u$ is decreasing). So $W$ has increased across $[x_1,x_2]$ . But $H \ge G$ makes $W' \le 0$ there, so $W$ cannot have increased. The only escape is that every inequality is an equality, which pins down the degenerate case.

The separation theorem is the same computation with $H = G$ , where the identity says that $W$ is constant; Abel's formula then says that the Wronskian of two independent solutions never vanishes, and a nonvanishing Wronskian at a zero of $g$ forces $f$ to be nonzero and $g'$ to change sign between consecutive zeros, so that $f$ changes sign too.

Proof(Separation)AdaptedAdapted from Hassani (2013), §14.4.3, proof of Thm 14.4.8, pp. 430–431, using Prop. 14.4.2, p. 425

Write $W(x) = f(x)g'(x) - g(x)f'(x)$ .

The Wronskian never vanishes. By Abel's formula, $W(x) = W(c)\exp\bigl(-\int_c^x P\bigr)$ for any $c \in [a,b]$ . Linear independence of $f$ and $g$ makes $W(c) \neq 0$ , and an exponential is never zero, so $W$ is nowhere zero on $[a,b]$ and keeps one sign there. This is the step that uses the continuity of $P$ , which is what makes Abel's formula available.
No common zeros. If $f(x_0) = g(x_0) = 0$ then $W(x_0) = 0$ , contradicting step 1. This proves part 1.
At a zero of $g$ . Let $x_0$ be a zero of $g$ . Then $W(x_0) = f(x_0)\,g'(x_0)$ , and since $W(x_0) \neq 0$ both factors are nonzero: $f(x_0) \neq 0$ , and $g$ has a simple zero at $x_0$ .
Sign change of $g'$ between consecutive zeros. Let $x_1 < x_2$ be consecutive zeros of $g$ . On $(x_1,x_2)$ the continuous function $g$ has no zero, so it keeps one sign; say $g > 0$ there, the other case being the same with $g$ replaced by $-g$ . Then $g'(x_1) \ge 0$ and $g'(x_2) \le 0$ as one-sided limits of difference quotients, and by step 3 neither is zero, so $g'(x_1) > 0 > g'(x_2)$ .
Hence a zero of $f$ in between. By step 3, $f(x_1) = W(x_1)/g'(x_1)$ and $f(x_2) = W(x_2)/g'(x_2)$ . The numerators have the same sign by step 1 and the denominators have opposite signs by step 4, so $f(x_1)$ and $f(x_2)$ have opposite signs. As $f$ is continuous, the intermediate value theorem gives a point of $(x_1,x_2)$ at which $f$ vanishes. Here the hypothesis that $f$ and $g$ are real-valued is used twice: a complex-valued function has no sign and obeys no intermediate value theorem.
Exactly one, and the symmetric statement. Suppose $f$ had two zeros $y_1 < y_2$ in $(x_1,x_2)$ ; shrinking, we may take them consecutive zeros of $f$ . Steps 3–5 apply verbatim with the roles of $f$ and $g$ exchanged (the Wronskian of the exchanged pair is $-W$ , which also never vanishes), so $g$ would vanish somewhere in $(y_1,y_2) \subset (x_1,x_2)$ — impossible, since $x_1$ and $x_2$ are consecutive zeros of $g$ . So there is exactly one. Exchanging the roles once more gives the statement with $f$ and $g$ interchanged. $\square$

Proof(Comparison)AdaptedAdapted from Coddington (1989), Ch. 3, §4, Exercise 4(a) with its hint, p. 117; Hassani (2013), §14.4.3, Thm 14.4.10, p. 431

It is enough to treat $x_1 < x_2$ consecutive zeros of $u$ , and for a non-consecutive pair the exceptional clause is vacuous. Indeed, suppose $u$ has a zero $z \in (x_1,x_2)$ and apply the consecutive case to a pair $x_1' < x_2'$ of consecutive zeros of $u$ inside $[x_1,z]$ : either $v$ vanishes in $(x_1',x_2') \subset (x_1,x_2)$ , or the degenerate case puts a zero of $v$ at $x_2' \le z < x_2$ , which again lies in the open interval $(x_1,x_2)$ . So between non-consecutive zeros of $u$ the solution $v$ vanishes in the open interval with no exception, and the degenerate clause of the statement has content only for a consecutive pair, which is the case treated from here on. Suppose, for contradiction, that $v$ has no zero in $(x_1,x_2)$ .

Normalise the signs. On $(x_1,x_2)$ neither $u$ nor $v$ vanishes, so each keeps one sign; replacing $u$ by $-u$ and $v$ by $-v$ if necessary — which changes neither equation nor the set of zeros — we may take $u > 0$ and $v > 0$ on $(x_1,x_2)$ . By continuity $v(x_1) \ge 0$ and $v(x_2) \ge 0$ .
The endpoint derivatives of $u$ . The zeros of the nontrivial solution $u$ are simple, so $u'(x_1) \neq 0 \neq u'(x_2)$ . Since $u(x_1) = 0$ and $u > 0$ immediately to the right, $u'(x_1) > 0$ ; since $u(x_2) = 0$ and $u > 0$ immediately to the left, $u'(x_2) < 0$ . This step uses $p > 0$ , through the fact that $(pu')' + Gu = 0$ is a second-order equation with continuous coefficients, so that its initial-value problems have unique solutions.
The identity. Put $W = p\,(uv' - vu')$ . Expanding and cancelling $p\,u'v'$ ,
$W' = p'(uv' - vu') + p(uv'' - vu'') = u\,(pv')' - v\,(pu')' = u\,(-Hv) - v\,(-Gu) = (G-H)\,u\,v .$
Integrate. Because $H \ge G$ on $[x_1,x_2]$ and $uv \ge 0$ there by step 1, the integrand is $\le 0$ , so
$W(x_2) - W(x_1) = \int_{x_1}^{x_2} (G-H)\,u\,v\,\mathrm d x \;\le\; 0 .$
Evaluate the endpoints. At a zero of $u$ the first term of $W$ drops out: $W(x_i) = -\,p(x_i)\,v(x_i)\,u'(x_i)$ . With $p > 0$ , $v(x_i) \ge 0$ and the signs of step 2,
$W(x_1) = -\,p(x_1)\,v(x_1)\,u'(x_1) \;\le\; 0 , \qquad W(x_2) = -\,p(x_2)\,v(x_2)\,u'(x_2) \;\ge\; 0 ,$
so $W(x_2) - W(x_1) \ge 0$ .
Both inequalities are equalities. Steps 4 and 5 force $W(x_1) = W(x_2) = 0$ and $\int_{x_1}^{x_2}(H-G)uv = 0$ . From $W(x_1) = 0$ with $p(x_1) > 0$ and $u'(x_1) \neq 0$ we get $v(x_1) = 0$ , and likewise $v(x_2) = 0$ : so $v$ does vanish at both endpoints, which is the first half of the degenerate case. From the vanishing integral, with $(H-G)uv$ continuous, nonnegative and with $uv > 0$ on $(x_1,x_2)$ , we get $H = G$ throughout $[x_1,x_2]$ — if $H(x_\ast) > G(x_\ast)$ at some interior $x_\ast$ the integrand would be strictly positive on a neighbourhood and the integral could not vanish, and the endpoints are then covered by continuity.
Identify the degenerate case. On $[x_1,x_2]$ the functions $u$ and $v$ now solve one and the same equation $(py')' + Gy = 0$ , and both vanish at $x_1$ . Their weighted Wronskian is constant there by step 3, and it is $0$ at $x_1$ ; since $p > 0$ , $uv' - vu' \equiv 0$ , so $u$ and $v$ are linearly dependent on $[x_1,x_2]$ and $v = cu$ with $c \neq 0$ .

So either $v$ vanishes in $(x_1,x_2)$ , or the degenerate case holds and $v(x_2) = 0$ ; in both cases $v$ has a zero in $(x_1,x_2]$ . Setting $p \equiv 1$ gives the normal-form statement verbatim. $\square$

RemarkOriginal

Where the hypotheses are used.

Real-valuedness is used in step 5 of the separation proof and in step 1 of the comparison proof, both of which argue from signs. It cannot be dropped. Hassani's Theorem 14.4.8 does not carry the word, but $y'' + y = 0$ has the independent solutions $f = \sin x$ and $g = \mathrm e^{\mathrm i x}$ , and $g$ has no zero at all, so nothing interlaces with the zeros of $f$ at $0$ and $\pi$ . Coddington's Exercise 5 states the separation property for real-valued solutions, and that is the form given above.
Linear independence is what makes the Wronskian nonzero in step 1 of the separation proof. For $f = g$ it fails completely: a solution does not vanish between its own consecutive zeros, by definition.
$p > 0$ is used twice in the comparison proof: in step 2, to know that a zero of a nontrivial solution is simple, and in steps 5–7, where $W(x_i) = 0$ is converted into $v(x_i) = 0$ by dividing by $p(x_i)$ . At a point where $p$ vanishes the equation degenerates and zeros can pile up; this is one of the ways a Sturm-Liouville problem becomes singular, as the previous unit records.
The same $p$ in both equations is essential to the identity of step 3. Two equations in divergence shape with different leading coefficients are compared by the Sturm-Picone theorem, which is not treated here; in practice one first reduces both to normal form, where the leading coefficient is $1$ .
$H \ge G$ is used twice. In step 4 it makes the integrand $(G-H)uv$ nonpositive, which is what gives $W(x_2) - W(x_1) \le 0$ ; and in step 6 it is used again, because a vanishing integral forces $H = G$ only for an integrand of one sign. Reversing the hypothesis reverses step 4 alone: the two inequalities displayed in step 5 contain neither $G$ nor $H$ — they come from the signs of $u'$ and $v$ at the endpoints — and so continue to give $W(x_2) - W(x_1) \ge 0$ . With both conclusions now pointing the same way there is no contradiction left, and the argument proves nothing. This is why the theorem is one-sided: only the solution of the equation with the larger coefficient is forced to vanish.

The exceptional case is local, not global. Hassani's Theorem 14.4.10 states the exception as the equality of the two coefficients together with one solution being a constant multiple of the other, without restricting the equality of the coefficients to the interval between the two zeros. Read on the whole of $[a,b]$ that clause is too narrow. Take $[a,b] = [0,2\pi]$ , $p \equiv 1$ , $G \equiv 1$ and

H(x) = \begin{cases} 1, & 0 \le x \le \pi, \\ 1 + (x-\pi)^2, & \pi \le x \le 2\pi, \end{cases}

so that $H$ is continuous, $H \ge G$ , and $H \neq G$ on $[a,b]$ . Then $u = \sin x$ solves $u'' + Gu = 0$ with consecutive zeros $0$ and $\pi$ , and the solution $v$ of $v'' + Hv = 0$ with $v(0) = 0$ , $v'(0) = 1$ equals $\sin x$ on $[0,\pi]$ , so it has no zero in $(0,\pi)$ — while the exception as quoted does not apply, because $H$ and $G$ differ somewhere in $[a,b]$ . With the clause read on $[x_1,x_2]$ , as in the statement above, the theorem is correct and is what the proof gives. Coddington sidesteps the question by assuming $H > G$ strictly, which removes the exception altogether.

Why the conclusion cannot be improved to an open interval. Keeping $H \ge G$ rather than $H > G$ really does cost the open interval: for $H = G$ and $v = u$ the only zeros of $v$ in $[x_1,x_2]$ are the endpoints. The closed-interval form $(x_1,x_2]$ of do Carmo's Exercise 6 is the sharp statement.

Corollary(Oscillation, zero spacing, and the zeros of eigenfunctions)AdaptedAdapted from Coddington (1989), Ch. 3, §4, Exercises 4(b) and 7, pp. 117–118; Hassani (2013), §19.1, Thm 19.1.9, pp. 568–569; Strauss (2008), §10.4, p. 278 and Thm 1, p. 281

Let $u$ be a nontrivial real solution of $u'' + S u = 0$ with $S$ continuous and real.

Nonoscillation. If $S \le 0$ on an interval $J$ , then $u$ has at most one zero in $J$ .
Zeros are not too close. If $S \le M^2$ on $[\alpha,\beta]$ with $M > 0$ , then any two zeros of $u$ in $[\alpha,\beta]$ differ by at least $\pi/M$ .
Zeros are not too far apart. If $S \ge m^2 > 0$ on $[\alpha,\alpha+\pi/m]$ , then $u$ has a zero in $(\alpha,\alpha+\pi/m]$ , whatever $u$ is.
Oscillation. If $S \ge m^2 > 0$ on $[a,\infty)$ then every nontrivial solution is oscillatory, with consecutive zeros at most $\pi/m$ apart. If $S \le 0$ on $[a,\infty)$ every nontrivial solution is nonoscillatory.
Counting. Let $u(\alpha) = 0$ and let $N$ be the number of zeros of $u$ in $(\alpha,\beta]$ . Then $S \le M^2 \text{ on } [\alpha,\beta] \ (M>0) \;\Longrightarrow\; N \le \Bigl\lfloor \frac{(\beta-\alpha)M}{\pi} \Bigr\rfloor , \qquad S \ge m^2 \text{ on } [\alpha,\beta] \ (m>0) \;\Longrightarrow\; N \ge \Bigl\lfloor \frac{(\beta-\alpha)m}{\pi} \Bigr\rfloor .$ The two halves are independent: neither needs the bound in the other direction.

Now let $-(p\varphi')' + q\varphi = \lambda w \varphi$ on $[a,b]$ be a regular Sturm-Liouville problem in the sense of the previous unit, with Dirichlet conditions $\varphi(a) = \varphi(b) = 0$ and eigenvalues $\lambda_1 < \lambda_2 < \cdots$ .

More eigenvalue, more zeros. If $\lambda < \mu$ are eigenvalues with eigenfunctions $\varphi$ and $\psi$ , then $\psi$ vanishes somewhere strictly between any two consecutive zeros of $\varphi$ in $[a,b]$ . Consequently, if $\varphi$ has $k$ zeros in the open interval $(a,b)$ , then $\psi$ has at least $k+1$ : the number of interior zeros increases strictly with the eigenvalue.
The exact count is quoted, not proved here: Hassani's Theorem 19.1.9 states that the eigenfunction $u_n$ belonging to $\lambda_n$ has exactly $n$ zeros in the interval on which the system is posed, and Strauss records the same count in the form that the $n$ th Dirichlet eigenfunction on an interval has $n-1$ nodes, dividing the interval into $n$ pieces. The two agree after the shift of index discussed in the previous unit; the numbering used below is Strauss's, so that $\varphi_n$ has $n-1$ interior zeros.

Proof(The corollary)Original

Parts 1–5 are the comparison theorem with a constant-coefficient equation in the other slot, and part 6 is the comparison theorem with $p$ the same and the coefficient moved by $\lambda$ .

Suppose $u$ had two zeros in $J$ ; shrinking, take them consecutive, $x_1 < x_2$ . Compare $u$ (coefficient $S$ ) with the solution $v \equiv 1$ of $v'' + 0\cdot v = 0$ (coefficient $0 \ge S$ ). The comparison theorem puts a zero of $v$ in $(x_1,x_2]$ — but $v$ is never zero. The degenerate case is no escape either: it would make $v$ a nonzero multiple of $u$ , and $u(x_1) = 0$ while $v(x_1) = 1$ .
Let $x_1 < x_2$ be zeros of $u$ in $[\alpha,\beta]$ ; shrinking as before, take them consecutive, and suppose $x_2 - x_1 < \pi/M$ . Compare $u$ (coefficient $S \le M^2$ ) with $v(x) = \sin\bigl(M(x-x_1)\bigr)$ , a nontrivial solution of $v'' + M^2 v = 0$ . Since $M^2 \ge S$ , the theorem gives a zero of $v$ in $(x_1,x_2]$ . But the zeros of $v$ are $x_1 + j\pi/M$ , and the first one after $x_1$ lies beyond $x_2$ . Contradiction.
The function $\sin\bigl(m(x-\alpha)\bigr)$ is a nontrivial solution of $y'' + m^2y = 0$ with $\alpha$ and $\alpha+\pi/m$ as consecutive zeros. Apply the comparison theorem on $[\alpha,\alpha+\pi/m]$ with this function in the slot of the smaller coefficient $m^2$ and $u$ in the slot of the larger coefficient $S \ge m^2$ : the conclusion is a zero of $u$ in $(\alpha, \alpha+\pi/m]$ . No hypothesis on $u$ is needed.
Apply part 3 repeatedly from $\alpha = a$ , then from each zero produced, to get zeros $z_1 < z_2 < \cdots$ with $z_{j+1} \le z_j + \pi/m$ ; they do not accumulate, since zeros are isolated, so there are infinitely many. The second half is part 1.
Let $z_0 = \alpha < z_1 < z_2 < \cdots$ be the zeros of $u$ in $[\alpha,\beta]$ in increasing order, and put $J = \lfloor(\beta-\alpha)m/\pi\rfloor$ . Claim: $z_j$ exists and satisfies $z_j \le \alpha + j\pi/m$ for every $j \le J$ . Induct: the case $j=0$ is the hypothesis $u(\alpha)=0$ , and if it holds for some $j < J$ then $z_j + \pi/m \le \alpha + (j+1)\pi/m \le \beta$ , so the window $[z_j, z_j+\pi/m]$ lies in $[\alpha,\beta]$ and part 3 applies there, producing a further zero at distance at most $\pi/m$ . Each of $z_1, \dots, z_J$ lies in $(\alpha,\beta]$ , so $N \ge J$ , the lower bound. For the upper bound, part 2 applied to the consecutive pairs $z_0,z_1,\dots$ gives $z_j - z_{j-1} \ge \pi/M$ and hence $z_j \ge \alpha + j\pi/M$ ; a zero of index $j$ can satisfy $z_j \le \beta$ only if $j \le (\beta-\alpha)M/\pi$ , so $N \le \lfloor(\beta-\alpha)M/\pi\rfloor$ .
Write the eigenvalue equation in divergence shape: $(p\varphi')' + (\lambda w - q)\varphi = 0$ and $(p\psi')' + (\mu w - q)\psi = 0$ , with the same $p$ . Their coefficients differ by
$(\mu w - q) - (\lambda w - q) = (\mu - \lambda)\,w > 0 ,$
strictly, because $\mu > \lambda$ and $w > 0$ is part of regularity. The comparison theorem applies with $u = \varphi$ , $v = \psi$ , and the strict inequality rules out the degenerate case, so $\psi$ vanishes strictly between any two consecutive zeros of $\varphi$ . For the count: with $\varphi(a) = \varphi(b) = 0$ and $k$ zeros inside, the zeros of $\varphi$ in $[a,b]$ are $a = z_0 < z_1 < \cdots < z_k < z_{k+1} = b$ , which is $k+1$ consecutive pairs; $\psi$ vanishes inside each of the $k+1$ open intervals $(z_j, z_{j+1})$ , and these are disjoint, so $\psi$ has at least $k+1$ zeros in $(a,b)$ . $\square$

MethodOriginal

To locate, bound or count the zeros of solutions of a second-order linear equation:

Normalise. If the equation is $y'' + Py' + Qy = 0$ , pass to the normal form $u'' + Su = 0$ with $S = Q - \frac14P^2 - \frac12P'$ ; the zeros are unchanged. If two different equations are to be compared, put both in normal form, or bring both to divergence shape with the same $p$ — the comparison theorem needs a common leading coefficient.
Bracket the coefficient. On the window $[\alpha,\beta]$ of interest find constants with $m^2 \le S \le M^2$ . Take $m^2$ to be the infimum and $M^2$ the supremum of $S$ there; a tighter window gives a tighter bracket.
Read off the spacing. Consecutive zeros are at least $\pi/M$ and at most $\pi/m$ apart. If $S \le 0$ anywhere on the window, there is at most one zero in that part of it.
Count. For a solution vanishing at $\alpha$ , part 5 of the corollary brackets the number of zeros in $(\alpha,\beta]$ between $\lfloor(\beta-\alpha)m/\pi\rfloor$ and $\lfloor(\beta-\alpha)M/\pi\rfloor$ .
For an eigenvalue problem, the coefficient is $\lambda w - q$ , which increases with $\lambda$ at every point. So running step 4 backwards turns a known zero count into a bracket for the eigenvalue: if the $n$ th Dirichlet eigenfunction is known to have $n-1$ interior zeros, then the eigenvalue must be large enough to fit $n$ half-oscillations into the interval and small enough not to fit $n+1$ .

ExampleOriginal

The spacing of the zeros of $J_0$ . Bessel's equation of order zero, $x^2y'' + xy' + x^2y = 0$ , is $y'' + x^{-1}y' + y = 0$ on $x > 0$ . Here $P = 1/x$ and $Q = 1$ , so the normalising factor is $\exp\bigl(\frac12\int \mathrm dx/x\bigr) = \sqrt x$ and

S = Q - \tfrac14 P^2 - \tfrac12 P' = 1 - \frac{1}{4x^2} + \frac{1}{2x^2} = 1 + \frac{1}{4x^2} ,

so $u(x) = \sqrt x\,J_0(x)$ satisfies $u'' + \bigl(1 + \tfrac1{4x^2}\bigr)u = 0$ (this is the substitution Coddington uses in Chapter 4, Exercise 3, p. 171). Because $\sqrt x > 0$ for $x > 0$ , the positive zeros of $J_0$ are exactly the zeros of $u$ . See Bessel functions of the first kind.

Upper bound on the gaps. On $(0,\infty)$ we have $S > 1$ , so part 3 of the corollary with $m = 1$ gives a zero of $u$ in every interval $(\alpha, \alpha+\pi]$ ; and since the inequality $S > m^2$ is strict, the degenerate case cannot occur and the zero lies in the open interval. Hence consecutive positive zeros of $J_0$ are less than $\pi = 3.141593$ apart, and part 4 says there are infinitely many of them.

Lower bound on the gaps. Write $j_{0,1} = 2.404826$ for the first positive zero. On $[\,j_{0,1},\infty)$ ,

S(x) \;\le\; 1 + \frac{1}{4\,j_{0,1}^2} \;=\; 1.043229 \;=\; M^2 , \qquad M = 1.021386 ,

so part 2 says consecutive zeros beyond $j_{0,1}$ are at least $\pi/M = 3.075814$ apart.

Check. The first six positive zeros of $J_0$ are

2.404826,\quad 5.520078,\quad 8.653728,\quad 11.791534,\quad 14.930918,\quad 18.071064 ,

with gaps

3.115253,\quad 3.133650,\quad 3.137807,\quad 3.139383,\quad 3.140146 .

Every gap lies in $[3.075814,\ 3.141593]$ , as predicted, and the smallest of them, $3.115253$ , clears the lower bound by $0.039$ . The gaps increase towards $\pi$ because $S(x) \to 1$ ; this is the same statement as the asymptotic form $J_0(x) \sim \sqrt{2/\pi x}\,\cos(x - \pi/4)$ of the Bessel asymptotics, obtained here without any asymptotic analysis.

Separation, concretely. $J_0$ and $Y_0$ are independent solutions of the same equation, so their zeros must interlace. The first few are

\underbrace{0.893577}_{Y_0} < \underbrace{2.404826}_{J_0} < \underbrace{3.957678}_{Y_0} < \underbrace{5.520078}_{J_0} < \underbrace{7.086051}_{Y_0} < \underbrace{8.653728}_{J_0} ,

alternating exactly as the separation theorem requires.

ExampleOriginal

Bracketing the eigenvalues of a tilted well. Take the regular Sturm-Liouville problem

-u'' + 10\,x\,u = \lambda\,u \quad \text{on } [0,1], \qquad u(0) = u(1) = 0 ,

so $p = w = 1$ , $q(x) = 10x$ and the normal-form coefficient is $S(x) = \lambda - 10x$ , with

\lambda - 10 \;\le\; S(x) \;\le\; \lambda \qquad (0 \le x \le 1).

Let $\varphi_n$ be the $n$ th eigenfunction, with $k = n-1$ interior zeros by part 7 of the corollary, so that $\varphi_n$ has $N = k+1 = n$ zeros in $(0,1]$ counting the one at $x=1$ .

Every eigenvalue is positive. If $\lambda \le 0$ then $S \le 0$ on $[0,1]$ , and part 1 allows at most one zero; but $\varphi_n$ vanishes at both endpoints. So $\lambda_n > 0$ and we may take $M = \sqrt{\lambda_n}$ .

Lower bound. By the counting bound of part 5 in the direction that needs only $S \le M^2$ , taken with $\alpha=0$ , $\beta=1$ and $M = \sqrt{\lambda_n}$ ,

n \;=\; N \;\le\; \Bigl\lfloor \frac{\sqrt{\lambda_n}}{\pi} \Bigr\rfloor \;\le\; \frac{\sqrt{\lambda_n}}{\pi} , \qquad\text{so}\qquad \lambda_n \;\ge\; n^2\pi^2 .

Upper bound. Suppose $\lambda_n > n^2\pi^2 + 10$ and set $m = \sqrt{\lambda_n - 10} > n\pi$ . Then $S(x) = \lambda_n - 10x > m^2$ for $x < 1$ , so each application of part 3 gives a zero strictly inside the window, and the zeros $0 = z_0 < z_1 < \cdots$ of $\varphi_n$ satisfy $z_j < j\pi/m < j/n$ . In particular $z_n < 1$ , so $z_1, \dots, z_n$ are $n$ zeros in the open interval $(0,1)$ — one more than the $n-1$ that $\varphi_n$ has. Hence

n^2\pi^2 \;\le\; \lambda_n \;\le\; n^2\pi^2 + 10 .

The width of the bracket is the oscillation of $q$ across the interval, which is the right answer: adding a constant to $q$ shifts every eigenvalue by that constant.

Check against the exact values. Substituting $s = 10^{1/3}(x - \lambda/10)$ turns the equation into Airy's equation (see Airy functions), so the eigenvalues are the roots of

\operatorname{Ai}(s_0)\operatorname{Bi}(s_1) - \operatorname{Ai}(s_1)\operatorname{Bi}(s_0) = 0 , \qquad s_0 = -\lambda\,10^{-2/3}, \quad s_1 = 10^{1/3} - \lambda\,10^{-2/3} .

Solving numerically:

$n$	$n^2\pi^2$	$\lambda_n$	$n^2\pi^2 + 10$	interior zeros of $\varphi_n$
1	9.86960	14.76010	19.86960	none
2	39.47842	44.51092	49.47842	$0.484206$
3	88.82644	93.84593	98.82644	$0.327158,\ 0.660335$
4	157.91367	162.92561	167.91367	$0.247063,\ 0.496043,\ 0.747000$

Every eigenvalue sits inside its bracket, and $\varphi_n$ has exactly $n-1$ interior zeros.

The spacing bound, checked on $\varphi_4$ . Here $m = \sqrt{162.925610-10} = 12.366309$ and $M = \sqrt{162.925610} = 12.764232$ . Part 2 asks only for $S \le M^2$ on $[0,1]$ , so every gap is at least $\pi/M = 0.246125$ . Part 3 is the weaker tool: it needs $S \ge m^2$ on a whole window $[z, z+\pi/m]$ , so it bounds a gap above by $\pi/m = 0.254044$ only for a zero $z$ with $z + \pi/m \le 1$ . That covers the gaps starting at $0$ , $0.247063$ and $0.496043$ , but not the last one, whose window $[0.747000,\ 1.001044]$ leaves the interval; there the upper bound is unavailable, and the gap is instead closed by the Dirichlet zero at $x=1$ . The four gaps between the zeros $0,\ 0.247063,\ 0.496043,\ 0.747000,\ 1$ are

0.247063,\quad 0.248980,\quad 0.250958,\quad 0.253000 ,

all at least $0.246125$ , the first three below the predicted $0.254044$ and the fourth below it as well although nothing above forces that; they increase from left to right because $S(x) = \lambda_4 - 10x$ decreases as $x$ grows: the eigenfunction oscillates faster where the potential is lower.

Interlacing. Part 6 predicts a zero of $\varphi_4$ strictly between consecutive zeros of $\varphi_3$ . The zeros of $\varphi_3$ in $[0,1]$ are $0,\ 0.327158,\ 0.660335,\ 1$ , and each of the three gaps does contain one of $0.247063$ , $0.496043$ , $0.747000$ .

ApplicationOriginal

Nodes of a vibrating string. For $w u_{tt} = (pu_x)_x - qu$ the modes are the Sturm-Liouville eigenfunctions, and their interior zeros are the points of the string that never move. Part 6 is the statement that each higher harmonic has at least one more node than the one below it, and part 7 fixes the count at $n-1$ for the $n$ th mode — the fact a player uses when touching the string at a node to sound a harmonic (Strauss (2008), §10.4, pp. 278–279). Part 5 says where the nodes are: closer together where $p/w$ is small, which is why the nodes of a string of varying density crowd into the heavy sections. See the vibrating string and normal modes.
Counting bound states. For $-\psi'' + V(x)\psi = E\psi$ the normal-form coefficient is $S = E - V$ , positive exactly in the classically allowed region. Part 3 says the wavefunction completes a half-oscillation in every stretch of length $\pi/\sqrt{E - V_{\max}}$ there, and part 1 says it cannot oscillate at all where $E < V$ ; counting the half-oscillations that fit is the node-counting argument behind the statement that the $n$ th bound state has $n-1$ nodes, and, read backwards as in the second example, it brackets the energies. The refinement in which $\sqrt{E-V(x)}$ is integrated instead of bounded is the WKB approximation.
Conjugate points and the stability of extremals. Along a geodesic on a surface, a Jacobi field normal to the geodesic obeys $v'' + K(s)v = 0$ with $K$ the Gaussian curvature, and the conjugate points are the zeros of $v$ (do Carmo (2016), §5-5, Exercise 3, p. 375). The comparison theorem then reads: the more positively curved the surface, the sooner geodesics refocus. Do Carmo draws the two standard conclusions — if $K \le K_0$ with $K_0 > 0$ there is no conjugate point before $\pi/\sqrt{K_0}$ , and if $K \ge K_1 > 0$ there is one by $\pi/\sqrt{K_1}$ — which are parts 2 and 3 of the corollary in geometric dress. The same computation in the calculus of variations is Jacobi's condition for a weak minimum: see conjugate points and geodesics.
Zeros of the special functions. Every classical orthogonal family satisfies a second-order equation, so its zeros are governed by the corollary rather than by its series. The first example gives the spacing of the zeros of $J_0$ from the equation alone; the same argument applied to $x^{1/2}J_\nu(x)$ , whose normal-form coefficient is $S_\nu(x) = 1 + (1-4\nu^2)/(4x^2)$ , shows that every $J_\nu$ with $\nu \ge 0$ has infinitely many positive zeros (Coddington (1989), Ch. 4, §8, Exercise 6, p. 179, with the case of order zero in Ch. 4, §7, Exercise 3, p. 171). For $\nu > 1/2$ one gets more: there $S_\nu < 1$ , so part 2 with $M = 1$ separates consecutive zeros by at least $\pi$ , and a gap of exactly $\pi$ is impossible — comparing with $\sin(x - x_1)$ , whose only zero in $(x_1, x_1+\pi]$ is the right endpoint, would put us in the degenerate case and force $S_\nu \equiv 1$ on the gap. So for $\nu > 1/2$ the zeros are strictly more than $\pi$ apart. Those zeros are what the Dirichlet condition on a circular drum selects, and they index the Fourier-Bessel series of Bessel orthogonality.