Integral equations

Fredholm and Volterra equations and their solution.

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Linear integral equations

Linear integral equations

Definition

Classification of integral equations

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The vocabulary of linear integral equations — first and second kind, Fredholm and Volterra, the kernel and the classes of kernel that make the integral operator bounded or compact — together with the recipe that turns a second-order linear ODE with its data into an equivalent integral equation, Volterra for initial conditions and Fredholm for boundary conditions.

Prerequisites: Inner products and norms, Bounded linear operators, Compact operators

Definition(Linear integral equations: the four kinds)AdaptedAdapted from Riley, Hobson & Bence (2006), §23.2, eq. (23.4), pp. 804–805; Hassani (2013), §18.1, pp. 543–544; Arfken, Weber & Harris (2013), §21.1, eqs. (21.1)–(21.4), pp. 1047–1048

An integral equation is an equation for an unknown function in which that function occurs under an integral sign. Riley, Hobson and Bence take as the general linear integral equation

g(x)\,y(x) \;=\; f(x) \;+\; \lambda\int_{a}^{b} K(x,z)\,y(z)\,\mathrm{d}z ,

in which $y$ is the unknown and the functions $g$ , $f$ and $K$ , the limits $a$ and $b$ , and the scalar $\lambda$ are given. The function $K$ of two variables is the kernel of the equation, and $\lambda$ is a parameter whose role is explained below. Both limits may depend on $x$ .

Four names classify the cases. The kind records where the unknown appears:

$g\equiv 0$ : the unknown occurs only inside the integral, and the equation is of the first kind.
$g\equiv 1$ : the unknown occurs both inside and outside, and the equation is of the second kind.

The other name records the limits:

both limits are constants: a Fredholm equation.
the upper limit is the variable $x$ : a Volterra equation.

Spelling out the four combinations with $g\equiv0$ or $g\equiv1$ , and absorbing $\lambda$ into $K$ for the equations of the first kind, where it may be divided out:

\int_{a}^{x} K(x,t)\,u(t)\,\mathrm{d}t = v(x) \qquad\text{Volterra, first kind,}

\int_{a}^{b} K(x,t)\,u(t)\,\mathrm{d}t = v(x) \qquad\text{Fredholm, first kind,}

u(x) = v(x) + \lambda\int_{a}^{x} K(x,t)\,u(t)\,\mathrm{d}t \qquad\text{Volterra, second kind,}

u(x) = v(x) + \lambda\int_{a}^{b} K(x,t)\,u(t)\,\mathrm{d}t \qquad\text{Fredholm, second kind.}

Independently of all this, the equation is homogeneous if $v\equiv 0$ and inhomogeneous otherwise. It is singular if either limit is infinite, or if $K$ becomes infinite somewhere in the range of integration.

Definition(The integral operator and its kernel classes)AdaptedAdapted from Stakgold & Holst (2011), §6.1, eqs. (6.1.16)–(6.1.17), p. 355, the three kernel definitions, pp. 356–357, and eqs. (6.1.23), (6.1.25)–(6.1.27), pp. 358–359; Riley, Hobson & Bence (2006), §23.3, eq. (23.5) and the formula for K†, p. 805

Let $-\infty\le a<b\le\infty$ . A kernel $K$ on the square $[a,b]\times[a,b]$ defines the integral operator $\mathcal{K}$ by

(\mathcal{K}u)(x) \;=\; \int_{a}^{b} K(x,t)\,u(t)\,\mathrm{d}t , \qquad x\in(a,b) .

(The open interval, because $a$ and $b$ are allowed to be infinite and are then not points at which $\mathcal{K}u$ is evaluated; for a finite interval the endpoints may be included or not, since a set of measure zero does not change an element of $L^{2}$ .) It is linear in $u$ , so every second-kind Fredholm equation reads $u = v+\lambda\mathcal{K}u$ , that is $(1-\lambda\mathcal{K})u = v$ , and every first-kind one reads $\mathcal{K}u = v$ . Throughout we use the inner product

(u,w) \;=\; \int_{a}^{b} u(x)\,\overline{w(x)}\,\mathrm{d}x

on $L^{2}(a,b)$ , linear in its first slot, and $\|u\|^{2}=(u,u)$ . Stakgold and Holst name the following classes of kernel; $\mathcal{K}$ is called separable, Hilbert–Schmidt or Holmgren when its kernel is.

$K$ is separable (or degenerate) if $K(x,t) = \sum_{i=1}^{n} p_i(x)\,\overline{q_i(t)}$ with $n$ finite and every $p_i,q_i$ in $L^{2}(a,b)$ .
$K$ is a Hilbert–Schmidt kernel if it is square-integrable on the square, $\int_{a}^{b}\!\!\int_{a}^{b} \lvert K(x,t)\rvert^{2}\,\mathrm{d}x\,\mathrm{d}t < \infty .$
$K$ is a Holmgren kernel if $\sup_{a\le \eta\le b}\ \int_{a}^{b}\!\!\int_{a}^{b} \lvert K(x,t)\rvert\,\lvert K(x,\eta)\rvert\,\mathrm{d}x\,\mathrm{d}t < \infty .$

The adjoint kernel is $K^{*}(x,t) = \overline{K(t,x)}$ ; it generates the Hilbert-space adjoint $\mathcal{K}^{*}$ , so that $(\mathcal{K}u,w) = (u,\mathcal{K}^{*}w)$ . Riley, Hobson and Bence write $\mathsf{K}^{\dagger}$ for the same operator. The kernel is symmetric if $K(x,t)=K(t,x)$ and Hermitian if $K(x,t)=\overline{K(t,x)}$ ; for a Hermitian kernel $K^{*}=K$ , so $\mathcal{K}^{*}=\mathcal{K}$ and the operator is self-adjoint. In this handbook symmetric always has the kernel-level meaning just given and never the operator-theoretic one: functional analysis often uses "symmetric operator" as a synonym for self-adjoint in the bounded case, but a complex symmetric kernel gives a self-adjoint operator only when $K$ is real, so the two words would conflict here. Finally, the iterated kernels are

K_{1}=K, \qquad K_{m}(x,t) \;=\; \int_{a}^{b} K(x,s)\,K_{m-1}(s,t)\,\mathrm{d}s \quad (m\ge2),

and $K_{m}$ is the kernel of $\mathcal{K}^{m}$ ; equivalently $K_m$ is the $(m-1)$ -fold integral of a product of $m$ copies of $K$ along a chain of intermediate variables. Both symmetries pass to the iterates: if $K$ is symmetric, so is every $K_{m}$ , and if $K$ is Hermitian, so is every $K_{m}$ . For the Hermitian case and $m=2$ , conjugating and relabelling gives

\overline{K_{2}(t,x)} = \int_{a}^{b}\overline{K(t,s)}\,\overline{K(s,x)}\,\mathrm{d}s = \int_{a}^{b} K(x,s)\,K(s,t)\,\mathrm{d}s = K_{2}(x,t) ,

and for general $m$ the same computation works on $K_{m}(x,t)=\int_{a}^{b}K_{m-1}(x,s)K(s,t)\,\mathrm{d}s$ (the kernel of $\mathcal{K}^{m-1}\mathcal{K}$ , which is the same operator as $\mathcal{K}\mathcal{K}^{m-1}$ ), with $K_{m-1}$ Hermitian by induction. Replacing every conjugate bar by nothing gives the symmetric case.

One notational point, fixed here once: $u$ and $v$ name the unknown and the given function in the abstract statements, and $y$ and $f$ name them wherever the equation comes from a differential equation — so the second-kind Fredholm equation is written $u=v+\lambda\mathcal{K}u$ in the definitions and statements above and below, and $y=f+\lambda\mathcal{K}y$ in the method, the example and the applications, where $y$ is the solution of an ODE and $f$ carries its data. The two are the same equation under $u\leftrightarrow y$ , $v\leftrightarrow f$ .

Definition(Boundedness and compactness)AdaptedAdapted from Stakgold & Holst (2011), §6.1, eq. (6.1.20), p. 357, and the theorem that an H.-S. operator is compact, with its proof, pp. 357–358; Reed & Simon (1980), Thm VI.23, p. 210; Hassani (2013), §18.1, p. 544

Let $K$ be a Hilbert–Schmidt kernel on $[a,b]\times[a,b]$ . Then $\mathcal{K}$ maps $L^{2}(a,b)$ into itself, is bounded with

\|\mathcal{K}\| \;\le\; \Bigl(\int_{a}^{b}\!\!\int_{a}^{b}\lvert K(x,t)\rvert^{2}\,\mathrm{d}x\,\mathrm{d}t\Bigr)^{1/2},

and is compact. A separable kernel is Hilbert–Schmidt, and a Holmgren kernel also generates a bounded operator.

Reed and Simon state the converse as well: an operator on $L^{2}(M,\mathrm{d}\mu)$ is Hilbert–Schmidt — that is, has $\operatorname{tr}A^{*}A<\infty$ — if and only if it is the integral operator of a kernel $K\in L^{2}(M\times M,\mathrm{d}\mu\otimes\mathrm{d}\mu)$ , and then $\|A\|_{2}^{2}=\iint\lvert K\rvert^{2}$ . So "Hilbert–Schmidt kernel" and "Hilbert–Schmidt operator" name the same objects from the two sides.

Compactness is what makes the second-kind Fredholm equation behave like a finite linear system. Hassani records the consequence for the parameter $\lambda$ : a value of $\lambda$ for which the homogeneous equation $u=\lambda\mathcal{K}u$ has a nonzero solution is called a characteristic value of the equation, and $\lambda$ is a characteristic value exactly when $\lambda^{-1}$ is an eigenvalue of $\mathcal{K}$ . His Theorem 18.1.1 then says how they are distributed in the second-kind Fredholm case: there are either finitely many of them, or infinitely many, and in the second case they can be listed as a sequence $\lambda_1,\lambda_2,\dots$ of complex numbers whose moduli grow without bound. Equivalently — and this is how Hassani derives it — the eigenvalues $\lambda_n^{-1}$ of the compact operator $\mathcal{K}$ either are finite in number or accumulate only at $0$ (Compact operators).

Remark(Two conventions for the parameter, and what to watch)AdaptedAdapted from Hassani (2013), §18.1, eqs. (18.1)–(18.3), pp. 543–544; Stakgold & Holst (2011), §6.2, eqs. (6.2.1)–(6.2.2), p. 360; Kreyszig (1978), §5.4, eq. (1) and footnote 2, p. 319, and p. 323; Riley, Hobson & Bence (2006), §23.2, p. 805

Where $\lambda$ sits. This handbook follows Hassani, Arfken–Weber–Harris and Riley–Hobson–Bence in writing the second-kind equation as $u=v+\lambda\mathcal{K}u$ . Then $\lambda$ multiplies the integral, the homogeneous equation is $u=\lambda\mathcal{K}u$ , and the special values of $\lambda$ are reciprocals of the eigenvalues of $\mathcal{K}$ ; Hassani calls them characteristic values, Riley, Hobson and Bence call them simply the eigenvalues $\lambda_i$ of the integral equation. Stakgold and Holst instead write the equation as $\mathcal{K}u-\lambda u = f$ , so that their $\lambda$ is an eigenvalue of $\mathcal{K}$ and their first kind is the case $\lambda=0$ . The two conventions are related by $\lambda_{\text{here}} = 1/\lambda_{\text{S\&H}}$ , and formulas transcribed between them without that substitution are wrong: for instance "the series converges for $\lvert\lambda\rvert$ small" in the first convention is "the series converges for $\lvert\lambda\rvert$ large" in the second. Where a result below is taken from Stakgold and Holst, it is restated in the convention of this handbook and the translation is flagged.
Kreyszig's sign. Kreyszig writes the second-kind Fredholm equation as $x(t)-\mu\int_a^b k(t,\tau)x(\tau)\,\mathrm{d}\tau = v(t)$ , which is ours with $\lambda=\mu$ and the integral moved to the left. He notes in a footnote that it is the presence of the term $x(t)$ outside the integral that makes iteration possible, and that is exactly why the second kind is the tractable one.
Volterra is a special Fredholm. A Volterra equation is the Fredholm equation with the same constant limits whose kernel has been extended by $K(x,t)=0$ for $t>x$ , as both Kreyszig (p. 323) and Riley, Hobson and Bence (p. 805) point out. The extended kernel need not be continuous across the diagonal $t=x$ . Everything proved for Fredholm equations therefore applies to Volterra equations, but not conversely: the triangular support gives the Volterra case much stronger conclusions (Volterra equations).
The first kind is the hard one. For the second kind, $1-\lambda\mathcal{K}$ is a small perturbation of the identity when $\lambda$ is small, and it is invertible. For the first kind one must invert $\mathcal{K}$ itself, and a compact operator on an infinite-dimensional space has no bounded inverse: its range is not all of $L^2$ and, as Stakgold and Holst remark, the range need not even be closed. So a first-kind equation may have no solution for a perfectly reasonable $v$ , and the solution, where it exists, need not depend continuously on $v$ . This is the mathematical content of the statement that deconvolution and inverse problems are ill-posed.

Method(Turning a second-order linear ODE into an integral equation)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.1, eqs. (21.7)–(21.11), pp. 1049–1050, Example 21.1.2 and the summary, pp. 1050–1052; Riley, Hobson & Bence (2006), §23.1, eqs. (23.1)–(23.3), pp. 803–804

Given

y'' + A(x)\,y' + B(x)\,y = g(x)

on an interval starting at $a$ , with $A$ continuously differentiable and $B$ , $g$ continuous, proceed as follows. (Arfken, Weber and Harris carry out the computation without stating any hypotheses on $A$ , $B$ and $g$ ; the ones named here are ours, chosen so that every step below is justified, and the remark says where each is used. Continuity of $A'$ , not merely its existence, is what the integration by parts, the exchange of the order of integration and the continuity of the resulting kernel all need.)

Initial data. If the data are $y(a)=y_{0}$ and $y'(a)=y_{0}'$ , integrate the equation once from $a$ to $x$ , integrate the $A y'$ term by parts to remove the derivative of the unknown, and integrate a second time, converting the resulting double integral $\int_a^x\mathrm{d}u\int_a^u \mathrm{d}t$ into $\int_a^x (x-t)\,\mathrm{d}t$ by exchanging the order of integration. The result is the Volterra equation of the second kind $y(x) = f(x) + \int_{a}^{x} K(x,t)\,y(t)\,\mathrm{d}t$ with $K(x,t) = (t-x)\bigl[B(t)-A'(t)\bigr] - A(t),$ $f(x) = \int_{a}^{x}(x-t)\,g(t)\,\mathrm{d}t \;+\; \bigl[A(a)\,y_{0}+y_{0}'\bigr](x-a) \;+\; y_{0}.$ The initial conditions have been absorbed into $f$ ; no side conditions remain to be imposed.
Two-point boundary data. If instead the data are $y(a)=y(b)=0$ , the quantity $y'(a)$ is not given, so carry it as an unknown constant through the two integrations and then fix it by imposing the condition at $x=b$ . Substituting the resulting expression back and splitting the range at $x$ produces a Fredholm equation of the second kind on $[a,b]$ . Arfken, Weber and Harris summarise the outcome thus: boundary conditions at both ends lead to a Fredholm-type equation whose kernel is the Green's function belonging to the differential operator with the boundary conditions imposed (Green's function for a boundary-value problem). They carry the computation out for the oscillator $y''+\omega^{2}y=0$ on $[0,b]$ with $y(0)=y(b)=0$ — the case $A\equiv0$ , $B\equiv\omega^{2}$ of the equation above, with the constant $\omega^{2}$ taken outside the integral as the parameter — and obtain $y(x)=\omega^{2}\int_{0}^{b}K(x,t)y(t)\,\mathrm{d}t$ with $K(x,t) = \begin{cases} t(b-x)/b, & t<x,\\[2pt] x(b-t)/b, & t>x,\end{cases}$ of which they list three properties: it is symmetric, it is continuous across $t=x$ , and $\partial K/\partial t$ jumps by $-1$ as $t$ increases through $t=x$ . Continuity, a unit jump in the $t$ -derivative across the diagonal and symmetry are the characteristic features of the Green's function of a self-adjoint second-order operator, and that is how they make the identification. How much of this survives for general $A$ and $B$ is the subject of the note below.
Read off the classification. Initial conditions at one end give a Volterra equation; boundary conditions at both ends give a Fredholm equation with a Green's-function kernel. Which of the four kinds appears is decided by the side conditions, not by the differential operator.
Going back. Differentiate the integral equation in $x$ as often as needed, using $\frac{\mathrm{d}}{\mathrm{d}x}\int_{a}^{x}F(x,t)\,\mathrm{d}t = F(x,x)+\int_{a}^{x}\partial_{x}F(x,t)\,\mathrm{d}t$ , and evaluate the equation and its derivatives at $x=a$ to recover the initial data. This direction is not always available: Arfken, Weber and Harris warn that some integral equations have no differential counterpart that anyone has found.

Riley, Hobson and Bence carry out step 1 for the more general $y''=F(x,y)$ , which produces the nonlinear Volterra equation $y(x)=\int_0^x (x-z)F(z,y(z))\,\mathrm{d}z + c_1x+c_2$ ; the conversion does not need linearity, only that $F$ is free of $y'$ .

Note(What the Fredholm kernel of step 2 looks like in general)Original

The three properties listed in step 2 belong to the kernel of the oscillator example, not to every equation the method starts from. Carrying out step 2 for general $A$ and $B$ on $[a,b]$ with $y(a)=y(b)=0$ gives the following, and the two computations below were checked numerically against the exact solutions of the corresponding boundary-value problems.

$A\equiv0$ . Write $\Gamma$ for the Green's function of $-\mathrm{d}^{2}/\mathrm{d}x^{2}$ on $[a,b]$ with Dirichlet data, $\Gamma(x,t) = \frac{1}{b-a}\begin{cases}(t-a)(b-x), & t<x,\\[2pt] (x-a)(b-t), & t>x,\end{cases}$ which is continuous and symmetric with $\partial_{t}\Gamma$ jumping by $-1$ across $t=x$ . Then step 2 produces the Fredholm equation $y=f+\int_{a}^{b}K(x,t)y(t)\,\mathrm{d}t$ with total kernel $K(x,t) = B(t)\,\Gamma(x,t),$ because the equation can be read as $y''=g-By$ with $y(a)=y(b)=0$ , whose solution is $y(x)=-\int_{a}^{b}\Gamma(x,t)\bigl[g(t)-B(t)y(t)\bigr]\mathrm{d}t$ , the Green's function of $\mathrm{d}^{2}/\mathrm{d}x^{2}$ with these boundary conditions being $-\Gamma$ ; the first term is $f$ and the second is the integral operator. This kernel is continuous, but it is symmetric only when $B$ is constant, and its $t$ -derivative jumps by $-B(x)$ , not by $-1$ . Taking $B\equiv\omega^{2}$ and dividing the constant out as the parameter leaves exactly the kernel $\Gamma$ , which is Arfken, Weber and Harris's $K$ of step 2 with $a=0$ : their three properties are the properties of $\Gamma$ . For a variable $B$ they fail: on $[0,1]$ with $B(x)=x$ the kernel is $t^{2}(1-x)$ for $t<x$ and $xt(1-t)$ for $t>x$ , continuous but not symmetric, with a jump of $-x$ in $\partial K/\partial t$ .
$A\not\equiv0$ . Then even continuity goes. Take $A\equiv1$ , $B\equiv0$ on $[0,1]$ with $y(0)=y(1)=0$ , so that step 1 gives the kernel $-1$ and $f(x)=\int_{0}^{x}(x-t)g(t)\,\mathrm{d}t+cx$ with $c=y'(0)$ unknown. Imposing $y(1)=0$ fixes $c=\int_{0}^{1}y-\int_{0}^{1}(1-t)g(t)\,\mathrm{d}t$ , and substituting it back gives the Fredholm equation with total kernel $K(x,t) = \begin{cases} x-1, & t<x,\\[2pt] x, & t>x,\end{cases}$ which jumps by $+1$ across the diagonal, is not symmetric, and has $\partial K/\partial t\equiv0$ on both sides.

What is true in general is the structural statement, and that is all step 2 claims: the two-point data turn the problem into a Fredholm equation of the second kind whose kernel is built from the Green's function of the boundary-value problem. Symmetry of the kernel is a property of the operator — it needs the operator to be formally self-adjoint, which $\mathrm{d}^{2}/\mathrm{d}x^{2}+A\,\mathrm{d}/\mathrm{d}x+B$ is only when $A\equiv0$ — and the clean jump of $-1$ belongs to the Green's function itself rather than to the kernel, which carries the extra factor $B(t)$ (Green's function for a boundary-value problem, Sturm-Liouville eigenvalue problems).

Idea of the proofOriginal

Why should such a conversion exist at all? A differential operator asks for local information and then needs side conditions to pin down the two constants of integration. Integrating twice replaces the derivatives by an operator that already knows the whole history of $y$ on $[a,x]$ , and the two constants of integration are precisely the two pieces of data; so the data can be built into the inhomogeneous term instead of being imposed afterwards. The only technical step is exchanging the order in the double integral, which replaces $\int_a^x\mathrm{d}u\int_a^u\mathrm{d}t$ by the single integral against the weight $(x-t)$ .

Proof(of the formulas in step 1)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.1, eqs. (21.7)–(21.10), pp. 1049–1050

Integrate $y''+Ay'+By=g$ from $a$ to $x$ and use $y'(a)=y_0'$ :

y'(x) = -\int_{a}^{x}A(t)y'(t)\,\mathrm{d}t - \int_{a}^{x}B(t)y(t)\,\mathrm{d}t + \int_{a}^{x}g(t)\,\mathrm{d}t + y_{0}' .

The first integral still contains $y'$ . Remove it by parts, which is legitimate because $A$ is continuously differentiable and $y$ is continuously differentiable, so both $Ay'$ and $A'y$ are continuous on $[a,x]$ and the by-parts formula applies:

\int_{a}^{x}A y'\,\mathrm{d}t = A(x)y(x)-A(a)y_{0}-\int_{a}^{x}A'(t)y(t)\,\mathrm{d}t .

Hence

y'(x) = -A(x)y(x) - \int_{a}^{x}\bigl[B(t)-A'(t)\bigr]y(t)\,\mathrm{d}t + \int_{a}^{x}g(t)\,\mathrm{d}t + A(a)y_{0}+y_{0}' .

Now integrate once more from $a$ to $x$ and use $y(a)=y_0$ . Each of the three $t$ -integrals becomes a double integral of the form $\int_{a}^{x}\mathrm{d}u\int_{a}^{u}\Phi(t)\,\mathrm{d}t$ . The region of integration in the $(u,t)$ plane is the triangle $a\le t\le u\le x$ ; integrating over $u$ first instead of $t$ gives

\int_{a}^{x}\mathrm{d}u\int_{a}^{u}\Phi(t)\,\mathrm{d}t =\int_{a}^{x}\Phi(t)\Bigl(\int_{t}^{x}\mathrm{d}u\Bigr)\mathrm{d}t =\int_{a}^{x}(x-t)\,\Phi(t)\,\mathrm{d}t ,

the exchange being permitted because the integrand is continuous on the closed triangle: for $\Phi=-[B-A']y$ this is where continuity of $A'$ is needed, alongside continuity of $B$ and of the solution $y$ , and for $\Phi=g$ it is continuity of $g$ . Applying this to $\Phi = -[B-A']y$ and to $\Phi=g$ , and integrating the constant term, we obtain

y(x) = -\int_{a}^{x}A(t)y(t)\,\mathrm{d}t - \int_{a}^{x}(x-t)\bigl[B(t)-A'(t)\bigr]y(t)\,\mathrm{d}t +\int_{a}^{x}(x-t)g(t)\,\mathrm{d}t + \bigl[A(a)y_{0}+y_{0}'\bigr](x-a)+y_{0}.

Collecting the two terms that carry the unknown gives exactly $\int_a^x K(x,t)y(t)\,\mathrm{d}t$ with $K(x,t)=(t-x)[B(t)-A'(t)]-A(t)$ , and the remaining terms are $f(x)$ . $\square$

Remark(Where the hypotheses are used)Original

$A$ differentiable is used in the integration by parts, and it is the reason $A'$ appears in the kernel. If $A$ is merely continuous, keep $\int_a^x A y'$ and work with the pair $(y,y')$ instead, or convert the ODE to a first-order system and apply step 1 componentwise.
$A'$ continuous — the extra strength in "continuously differentiable" — is used three times over, and each use is about $A'$ rather than $A$ . It makes $A'y$ continuous, hence integrable, in the by-parts step; it makes the integrand $-[B-A']y$ continuous on the closed triangle, which is the justification given for the exchange of the order of integration; and, since the kernel is $K(x,t)=(t-x)[B(t)-A'(t)]-A(t)$ , it is what makes $K$ continuous there. Mere differentiability of $A$ would leave $A'$ possibly unbounded and all three steps unjustified.
$F$ free of $y'$ is what Riley, Hobson and Bence assume in the nonlinear version; with $y'$ present in $F$ the double integration no longer closes on $y$ alone.
Continuity of $B$ and $g$ , together with that of $A'$ , is what makes $K$ continuous on the closed triangle $a\le t\le x\le b$ , which is the hypothesis under which the Volterra theory of Volterra equations applies.
The conversion is an equivalence only with the side conditions attached. The ODE alone has a two-parameter family of solutions; the integral equation of step 1 has exactly one, because the data are inside $f$ .

Example(An initial-value problem and its Volterra equation, both ways)Original

Set-up. Take

y''-2y'+y = 0, \qquad y(0)=1,\quad y'(0)=1 ,

so that $A(x)=-2$ , $B(x)=1$ , $g\equiv0$ , $a=0$ , $y_{0}=y_{0}'=1$ . The ODE has general solution $(c_{1}+c_{2}x)\mathrm{e}^{x}$ , and the data give $c_{1}=1$ , $c_{2}=0$ : the solution is $y(x)=\mathrm{e}^{x}$ .

Step 1: the kernel and the inhomogeneous term. Since $A'\equiv0$ ,

K(x,t) = (t-x)\bigl[B(t)-A'(t)\bigr]-A(t) = (t-x)\cdot 1 - (-2) = t-x+2,

f(x) = 0 + \bigl[A(0)y_{0}+y_{0}'\bigr](x-0)+y_{0} = (-2+1)x+1 = 1-x .

So the claim is that $y=\mathrm{e}^{x}$ is the unique solution of

y(x) = 1-x+\int_{0}^{x}(t-x+2)\,y(t)\,\mathrm{d}t .

Check it directly. With $y(t)=\mathrm{e}^{t}$ and $\int_{0}^{x}t\mathrm{e}^{t}\,\mathrm{d}t=(x-1)\mathrm{e}^{x}+1$ , $\int_{0}^{x}\mathrm{e}^{t}\,\mathrm{d}t=\mathrm{e}^{x}-1$ ,

\int_{0}^{x}(t-x+2)\mathrm{e}^{t}\,\mathrm{d}t = (x-1)\mathrm{e}^{x}+1+(2-x)(\mathrm{e}^{x}-1) = \mathrm{e}^{x}+x-1 ,

because the coefficient of $\mathrm{e}^x$ is $(x-1)+(2-x)=1$ . Adding $f(x)=1-x$ returns $\mathrm{e}^{x}$ , as required.

Step 4: recover the ODE. Differentiate the integral equation. With $F(x,t)=(t-x+2)y(t)$ we have $F(x,x)=2y(x)$ and $\partial_{x}F=-y(t)$ , so

y'(x) = -1+2y(x)-\int_{0}^{x}y(t)\,\mathrm{d}t .

Differentiating again, $y''(x)=2y'(x)-y(x)$ , which is the original ODE. Setting $x=0$ in the integral equation gives $y(0)=1$ , and in the first derivative gives $y'(0)=-1+2\cdot1=1$ : both initial conditions are reproduced. The two equations really are equivalent.

With a source term. Changing the right-hand side of the ODE to $g\equiv2$ changes nothing in $K$ and adds $\int_{0}^{x}(x-t)\cdot2\,\mathrm{d}t = x^{2}$ to $f$ , so

y''-2y'+y=2,\quad y(0)=1,\ y'(0)=1 \qquad\Longleftrightarrow\qquad y(x)=x^{2}-x+1+\int_{0}^{x}(t-x+2)y(t)\,\mathrm{d}t .

The ODE now has solution $y(x)=(2x-1)\mathrm{e}^{x}+2$ , which satisfies both the ODE and the data, and substituting it into the integral equation reproduces it exactly (checked symbolically).

A boundary-value problem, for contrast. Take

y''=-\mu y \quad\text{on }[0,1], \qquad y(0)=0,\quad y'(1)=0 .

The data here are not the two-point data $y(a)=y(b)=0$ of step 2 — the condition at the right end is on $y'$ — but the elimination is the same one: $y'(0)$ is the unknown constant, and the condition at $x=1$ fixes it. Here $\mu$ is a parameter rather than part of the operator, and the Green's function of $-\mathrm{d}^{2}/\mathrm{d}x^{2}$ with these boundary conditions is $\min(x,t)$ , so the bookkeeping of the note above, with the constant $B\equiv\mu$ taken outside the integral as the parameter, leaves the total kernel $K(x,t)=\min(x,t)$ and a homogeneous Fredholm equation of the second kind,

y(x) = \mu\int_{0}^{1}\min(x,t)\,y(t)\,\mathrm{d}t .

Verify it on one eigenfunction. For $y(t)=\sin(\pi t/2)$ and $\omega=\pi/2$ ,

\int_{0}^{1}\min(x,t)\sin(\omega t)\,\mathrm{d}t =\int_{0}^{x} t\sin(\omega t)\,\mathrm{d}t + x\int_{x}^{1}\sin(\omega t)\,\mathrm{d}t = \frac{\sin(\omega x)}{\omega^{2}} ,

since the two terms $-x\cos(\omega x)/\omega$ and $+x\cos(\omega x)/\omega$ cancel, using $\cos\omega = \cos(\pi/2)=0$ . So the integral equation holds exactly with $\mu=\omega^{2}=\pi^{2}/4$ , which is indeed the lowest eigenvalue of the boundary-value problem. The kernel $\min(x,t)$ is continuous and symmetric and $\partial_t\min(x,t)$ drops from $1$ to $0$ as $t$ crosses $x$ — the three Green's-function properties of step 2, which hold here because the operator is $\mathrm{d}^{2}/\mathrm{d}x^{2}$ with no first-derivative term and the coefficient $\mu$ has been taken out as the parameter.

ApplicationOriginal

Why a physicist meets these. Three reasons, all visible in the sources.

The boundary conditions come built in. This is the practical payoff of the method above, and Arfken, Weber and Harris make it the headline: in an integral equation the data are part of the equation, not a condition imposed at the end. The same observation drives Stakgold and Holst's list of advantages of the integral formulation — the integral operator is bounded and often compact where the differential operator is unbounded; the kernel is a Green's function and so carries the boundary conditions; and a boundary-value problem for a partial differential equation can sometimes be rewritten as an integral equation over the boundary alone, of lower dimension. That last point is the whole basis of boundary-element methods in potential theory, elasticity and acoustics: a three-dimensional Laplace problem becomes a two-dimensional integral equation on the surface.

Quantum scattering. Arfken, Weber and Harris begin with the momentum representation of the Schrödinger equation (their Eq. (21.6) (Arfken, Weber & Harris (2013), §21.1, Example 21.1.1, eq. (21.6), p. 1048)): with a Coulomb potential in hartree atomic units, $\psi$ in momentum space satisfies

\frac{k^{2}}{2m}\varphi(\mathbf{k}) + \frac{1}{(2\pi)^{3/2}}\int\frac{4\pi}{\lvert\mathbf{k}-\mathbf{k}'\rvert^{2}}\,\varphi(\mathbf{k}')\,\mathrm{d}^{3}k' = E\varphi(\mathbf{k}),

an inhomogeneous-looking eigenvalue problem that is in fact a homogeneous integral equation. The kernel depends on $\mathbf{k}-\mathbf{k}'$ — a displacement kernel, in the terminology of Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §23.4.2, p. 809) — because the potential acts by multiplication in position space; displacement kernels on the whole line are exactly the ones a Fourier transform diagonalises. The same reorganisation in position space turns the scattering problem into the Lippmann–Schwinger equation, a Fredholm equation of the second kind whose iteration is the Born series (Scattering and the Born approximation).

First-kind equations are the inverse problems. Every measurement that smears a quantity before recording it — a spectrometer with finite resolution, a tomograph, a seismic survey, a radiative-transfer sounding of an atmosphere — delivers data $v$ related to the quantity $u$ of interest by $\int K(x,t)u(t)\,\mathrm{d}t = v(x)$ , a Fredholm equation of the first kind in which $K$ is the instrument response. The remark above says that this is precisely the situation in which no bounded inverse exists: because $\mathcal{K}$ is compact, it maps the high-frequency content of $u$ almost to zero, so recovering that content from $v$ amplifies the noise in $v$ without limit. Practitioners therefore never solve the first-kind equation as written; they regularise it, for example by solving the nearby second-kind equation $\alpha u + \mathcal{K}^{*}\mathcal{K}u = \mathcal{K}^{*}v$ with $\alpha>0$ , or by truncating a singular-value expansion of $\mathcal{K}$ (Singular value decomposition). The classification is thus not bookkeeping: it tells you in advance whether you are facing a well-posed problem or an ill-posed one.

Theorem

Neumann series

Adapted

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For a second-kind integral equation with a bounded kernel and a small enough parameter, substituting the equation into itself converges: the solution is the Neumann series in the iterated kernels, summed by the resolvent kernel, with an explicit error bound from the contraction mapping theorem and three usable sufficient conditions for convergence.

Prerequisites: Classification of integral equations, The contraction mapping theorem, Uniform convergence

Definition(Iterated kernels, the Neumann series and the resolvent kernel)AdaptedAdapted from Riley, Hobson & Bence (2006), §23.5, eqs. (23.35)–(23.37), pp. 813–814; Kreyszig (1978), §5.4, Problems 4 and 7, pp. 324–325; Stakgold & Holst (2011), §6.1, eqs. (6.1.25)–(6.1.27), pp. 358–359

Let $K$ be a kernel on $[a,b]\times[a,b]$ and $\mathcal{K}$ its integral operator, as in Classification of integral equations. The iterated kernels are

K_{1}(x,z) = K(x,z), \qquad K_{m}(x,z) = \int_{a}^{b} K(x,s)\,K_{m-1}(s,z)\,\mathrm{d}s \quad (m\ge 2),

so that $K_m$ is the kernel of $\mathcal{K}^{m}$ .

Given the second-kind Fredholm equation $y = f + \lambda\mathcal{K}y$ , the iterates generated from a starting function $y_{0}$ are

y_{n+1} = f + \lambda\mathcal{K}y_{n}, \qquad n=0,1,2,\dots

With the particular choice $y_{0}=f$ , the $n$ -th iterate is the $n$ -th order approximation

y_{n}(x) = f(x) + \sum_{m=1}^{n}\lambda^{m}\int_{a}^{b}K_{m}(x,z)f(z)\,\mathrm{d}z ,

and its formal limit

y(x) = f(x) + \sum_{m=1}^{\infty}\lambda^{m}\int_{a}^{b}K_{m}(x,z)f(z)\,\mathrm{d}z

is the Neumann series for the solution. Summing the series under the integral sign writes it as

y(x) = f(x) + \lambda\int_{a}^{b} R(x,z;\lambda)\,f(z)\,\mathrm{d}z , \qquad R(x,z;\lambda) = \sum_{m=0}^{\infty}\lambda^{m}K_{m+1}(x,z),

and $R$ is the resolvent kernel. The same definitions apply verbatim to the Volterra equation, with the upper limit $b$ replaced by $x$ throughout.

Theorem(Convergence of the Neumann series)AdaptedAdapted from Kreyszig (1978), §5.4, eqs. (3), (5), (6) and Thm 5.4-1, pp. 319–321; Arfken, Weber & Harris (2013), §21.3, eqs. (21.52)–(21.55), pp. 1065–1066; Riley, Hobson & Bence (2006), §23.5, eqs. (23.35)–(23.38), pp. 813–814; Hassani (2013), §18.2, eqs. (18.12)–(18.13), p. 549

Let $[a,b]$ be a finite interval, let $K$ be continuous on the square $[a,b]\times[a,b]$ with

\lvert K(x,t)\rvert \le c \quad\text{for all } (x,t)\in[a,b]\times[a,b],

let $f$ be continuous on $[a,b]$ , and suppose

\lvert\lambda\rvert\,c\,(b-a) \;<\; 1 . \tag{S}

Then:

Existence and uniqueness. The equation $y = f+\lambda\mathcal{K}y$ has exactly one continuous solution $y$ on $[a,b]$ .
Convergence of the iteration. For any continuous starting function $y_{0}$ , the iterates $y_{n+1}=f+\lambda\mathcal{K}y_{n}$ converge to $y$ uniformly on $[a,b]$ , with the error bound $\max_{x\in[a,b]}\lvert y_{n}(x)-y(x)\rvert \;\le\; \frac{\alpha^{n}}{1-\alpha}\,\max_{x\in[a,b]}\lvert y_{1}(x)-y_{0}(x)\rvert , \qquad \alpha = \lvert\lambda\rvert\,c\,(b-a) < 1 .$
The series. The iterated kernels satisfy $\lvert K_{m}(x,z)\rvert\le c^{m}(b-a)^{m-1}$ , the resolvent series for $R(x,z;\lambda)$ converges uniformly on the square, and $y(x) = f(x) + \sum_{m=1}^{\infty}\lambda^{m}\!\int_{a}^{b}\!K_{m}(x,z)f(z)\,\mathrm{d}z = f(x) + \lambda\int_{a}^{b} R(x,z;\lambda)f(z)\,\mathrm{d}z .$
No characteristic values in the disc. Under (S) the homogeneous equation $y=\lambda\mathcal{K}y$ has only the solution $y\equiv0$ , so no $\lambda$ obeying (S) is a characteristic value.

Two further sufficient conditions appear in the sources and are weaker than (S) in the cases that matter.

Square-integral condition (Riley, Hobson and Bence): the series converges in a domain of $\lvert\lambda\rvert$ provided $\lvert\lambda\rvert^{2}\int_{a}^{b}\!\!\int_{a}^{b}\lvert K(x,z)\rvert^{2}\,\mathrm{d}z\,\mathrm{d}x \;<\;1 . \tag{S$'$}$
Operator-norm condition (Hassani; Stakgold and Holst): if $K$ is a Hilbert–Schmidt kernel and $\lvert\lambda\rvert\,\lVert\mathcal{K}\rVert<1$ , then $1-\lambda\mathcal{K}$ is invertible on $L^{2}(a,b)$ and $y=\sum_{n\ge0}\lambda^{n}\mathcal{K}^{n}f$ converges in $L^{2}$ . $(\mathrm{S}')$ is this condition with the Hilbert–Schmidt bound $\lVert\mathcal{K}\rVert\le(\iint\lvert K\rvert^{2})^{1/2}$ put in for $\lVert\mathcal{K}\rVert$ .

Idea of the proofOriginal

Read $y=f+\lambda\mathcal{K}y$ as " $y$ is $f$ , corrected by $\lambda\mathcal{K}y$ ". If the correction is small, guess $y\approx f$ , put the guess back in to get a better one, and repeat; each round pushes the error down by a factor of $\lvert\lambda\rvert\lVert\mathcal{K}\rVert$ -ish, and the accumulated corrections are the terms $\lambda^{m}\mathcal{K}^{m}f$ . The whole content of the theorem is that the map $y\mapsto f+\lambda\mathcal{K}y$ is a contraction on the complete space $C[a,b]$ when $\lvert\lambda\rvert c(b-a)<1$ , so the contraction mapping theorem supplies the fixed point, the uniqueness, the convergence from any starting function and the error bound at once. The algebra of the series is then just the geometric series $(1-\lambda\mathcal{K})^{-1}=\sum\lambda^{m}\mathcal{K}^{m}$ , legitimised by the same estimate.

ProofAdaptedAdapted from Kreyszig (1978), §5.4, eqs. (2)–(6) and Thm 5.4-1, pp. 319–321; Stakgold & Holst (2011), §4.4, Example 3, pp. 250–251; Riley, Hobson & Bence (2006), §23.5, p. 814

Work in $X=C[a,b]$ , the continuous functions on the finite closed interval with the metric

d(u,w) = \max_{x\in[a,b]}\lvert u(x)-w(x)\rvert ,

which is complete. Define $T\colon X\to X$ by

(Tu)(x) = f(x)+\lambda\int_{a}^{b}K(x,t)u(t)\,\mathrm{d}t .

$T$ really does map $X$ into $X$ : for continuous $u$ the integrand is continuous on the square, so $x\mapsto\int_a^b K(x,t)u(t)\,\mathrm{d}t$ is continuous by uniform continuity of $K$ on the compact square, and $f$ is continuous. A fixed point of $T$ is exactly a continuous solution of the integral equation.

Step 1: $T$ is a contraction. For $u,w\in X$ and any $x$ ,

\lvert (Tu)(x)-(Tw)(x)\rvert = \Bigl\lvert\lambda\int_{a}^{b}K(x,t)\bigl[u(t)-w(t)\bigr]\mathrm{d}t\Bigr\rvert \le \lvert\lambda\rvert\,c\int_{a}^{b}\lvert u(t)-w(t)\rvert\,\mathrm{d}t \le \lvert\lambda\rvert\,c\,(b-a)\,d(u,w),

using the bound $\lvert K\rvert\le c$ — which exists because $K$ is continuous on a compact square — and then replacing $\lvert u-w\rvert$ by its maximum. Taking the maximum over $x$ on the left,

d(Tu,Tw)\le \alpha\,d(u,w), \qquad \alpha = \lvert\lambda\rvert\,c\,(b-a).

Hypothesis (S) says $\alpha<1$ , so $T$ is a contraction.

Step 2: parts 1 and 2. The contraction mapping theorem applied to $T$ on the complete metric space $X$ gives a unique fixed point $y\in X$ , convergence of $y_{n}=T^{n}y_{0}$ to $y$ for every $y_{0}\in X$ , and the a-priori bound $d(y_{n},y)\le\frac{\alpha^{n}}{1-\alpha}d(y_{1},y_{0})$ . Convergence in the metric $d$ is uniform convergence on $[a,b]$ . This is parts 1 and 2.

Step 3: the bound on the iterated kernels. By induction. For $m=1$ , $\lvert K_{1}\rvert = \lvert K\rvert\le c = c^{1}(b-a)^{0}$ . Assuming $\lvert K_{m-1}\rvert\le c^{m-1}(b-a)^{m-2}$ for some $m\ge2$ ,

\lvert K_{m}(x,z)\rvert \le \int_{a}^{b}\lvert K(x,s)\rvert\,\lvert K_{m-1}(s,z)\rvert\,\mathrm{d}s \le c\cdot c^{m-1}(b-a)^{m-2}\cdot(b-a) = c^{m}(b-a)^{m-1},

which closes the induction.

Step 4: the series, part 3. The $m$ -th term of the resolvent series is bounded by $\lvert\lambda\rvert^{m}c^{m+1}(b-a)^{m} = c\,\alpha^{m}$ , a constant times a convergent geometric series; by the Weierstrass criterion (Uniform convergence) $\sum_{m\ge0}\lambda^{m}K_{m+1}(x,z)$ converges uniformly on the square, so $R(\cdot,\cdot;\lambda)$ is continuous there.

Next identify the iterates with the partial sums. Take $y_{0}=f$ . Then $y_{1}=f+\lambda\mathcal{K}f$ , and if $y_{n}=f+\sum_{m=1}^{n}\lambda^{m}\mathcal{K}^{m}f$ then

y_{n+1} = f+\lambda\mathcal{K}y_{n} = f+\lambda\mathcal{K}f+\sum_{m=1}^{n}\lambda^{m+1}\mathcal{K}^{m+1}f = f+\sum_{m=1}^{n+1}\lambda^{m}\mathcal{K}^{m}f ,

so by induction $y_{n}$ is the $n$ -th partial sum. Because $\mathcal{K}^{m}$ is the integral operator of $K_{m}$ — which is the definition of $K_m$ , read off by iterating $\mathcal{K}$ and exchanging the order of the finitely many integrations, legitimate since all integrands are continuous on compact sets — the partial sum is the $n$ -th order approximation of the definition block. Part 2 says $y_{n}\to y$ uniformly, which is the first displayed formula of part 3. Finally, in

y(x) = f(x)+\sum_{m=1}^{\infty}\lambda^{m}\int_{a}^{b}K_{m}(x,z)f(z)\,\mathrm{d}z

the series of continuous functions of $x$ converges uniformly (same geometric bound, now with the extra factor $(b-a)\max\lvert f\rvert$ ), so the sum and the integral may be exchanged, which gives the resolvent form.

Step 5: part 4. If $y=\lambda\mathcal{K}y$ with $y$ continuous, then $y$ is a fixed point of $T$ with $f\equiv0$ ; so is $y\equiv0$ ; uniqueness in part 1 forces $y\equiv0$ . $\square$

Remark(Where the hypotheses are used, and how far they can be relaxed)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.3, eqs. (21.55) and the discussion after Example 21.3.1, pp. 1066–1067; Stakgold & Holst (2011), §6.4, eqs. (6.4.2)–(6.4.4) and the discussion of Example 1, pp. 379–381; Riley, Hobson & Bence (2006), §23.5, eq. (23.38), p. 814, and §23.6, pp. 815–816; Hassani (2013), §18.1, p. 548

Where (S) enters. Condition (S) is used once, in step 1, and only to make $\alpha<1$ . Continuity of $K$ is used twice: to produce the bound $c$ on a compact square, and to keep $T$ inside $C[a,b]$ . The finiteness of $b-a$ is used in the estimate $\int_a^b\lvert u-w\rvert\le(b-a)d(u,w)$ ; on an infinite interval the same argument needs $\int\lvert K(x,t)\rvert\mathrm{d}t$ bounded instead.
(S) is far from sharp. Arfken, Weber and Harris are explicit about this: their Example 21.3.1 has a Neumann series that converges comfortably although (S) fails, and they report that the necessary and sufficient condition is $\lvert\lambda\rvert<\lvert\lambda_{e}\rvert$ , where $\lambda_{e}$ is the characteristic value of smallest modulus of the corresponding homogeneous equation. Stakgold and Holst make the same point in their convention: the series they obtain is valid out to the spectral radius of $\mathcal{K}$ , not merely to $\lVert\mathcal{K}\rVert$ , and they verify it on a one-term separable kernel where the sum can be continued analytically to a rational function of the parameter. In this handbook's convention their statement reads: the Neumann series converges for $\lvert\lambda\rvert<\lvert\lambda_{e}\rvert$ , with $\lvert\lambda_{e}\rvert^{-1}$ the spectral radius of $\mathcal{K}$ . Neither book proves the sharp statement, and neither do we; it is a consequence of the spectral theory of compact operators (Spectrum and resolvent).
Beyond the radius. Convergence of the series is sufficient but not necessary for a solution to exist. Even when the series diverges, the resolvent usually continues as a meromorphic function of $\lambda$ whose poles are the characteristic values; Riley, Hobson and Bence describe Fredholm's construction of $R=D(x,z;\lambda)/d(\lambda)$ as a ratio of two power series that converge for all $\lambda$ , so that the solution is unique and finite whenever $d(\lambda)\neq0$ and the zeros of $d$ are the characteristic values. We do not reproduce that construction; see also Hilbert-Schmidt theory for the symmetric case, where the continuation is written down explicitly as an eigenfunction expansion.
Volterra is different in kind. For a Volterra equation the factor $(b-a)^{m}$ in step 3 is replaced by $(x-a)^{m}/m!$ , and the series then converges for every $\lambda$ . That is the subject of Volterra equations; Hassani states it as his Theorem 18.1.2, and it is the reason he treats Volterra equations first.
Which condition to use. (S) is the easiest to check and the crudest. $(\mathrm{S}')$ is better whenever $K$ is peaked, since it averages $\lvert K\rvert^{2}$ instead of taking its maximum; it is also the natural condition for an $L^{2}$ kernel that is unbounded. The sharp condition requires knowing the characteristic values, which is usually harder than solving the equation.

Example(Three convergence conditions, measured against the truth)Original

Take $[a,b]=[0,1]$ , $f\equiv1$ and the separable kernel

K(x,t) = \mathrm{e}^{\,x-t},

so the equation is

y(x) = 1 + \lambda\int_{0}^{1}\mathrm{e}^{\,x-t}\,y(t)\,\mathrm{d}t .

The iterated kernels. Because $K(x,s)K(s,t)=\mathrm{e}^{x-s}\mathrm{e}^{s-t}=\mathrm{e}^{x-t}$ has no $s$ left in it,

K_{2}(x,t) = \int_{0}^{1}\mathrm{e}^{\,x-t}\,\mathrm{d}s = \mathrm{e}^{\,x-t} = K(x,t),

and by induction every $K_{m}=K$ . So the resolvent kernel is a geometric series with constant ratio:

R(x,t;\lambda)=\sum_{m=0}^{\infty}\lambda^{m}\mathrm{e}^{\,x-t} = \frac{\mathrm{e}^{\,x-t}}{1-\lambda} \qquad (\lvert\lambda\rvert<1).

The iterates. With $y_{0}=f=1$ and $\int_{0}^{1}\mathrm{e}^{-t}\,\mathrm{d}t = 1-\mathrm{e}^{-1}$ ,

y_{1}(x) = 1+\lambda(1-\mathrm{e}^{-1})\mathrm{e}^{x}, \qquad y_{2}(x) = 1+\lambda(1-\mathrm{e}^{-1})\mathrm{e}^{x}+\lambda^{2}(1-\mathrm{e}^{-1})\mathrm{e}^{x},

and in general $y_{n}(x) = 1+(1-\mathrm{e}^{-1})\mathrm{e}^{x}\sum_{m=1}^{n}\lambda^{m}$ : the second step is $1+\lambda\int_0^1 \mathrm{e}^{x-t}[1+\lambda(1-\mathrm{e}^{-1})\mathrm{e}^{t}]\mathrm{d}t$ , and $\int_0^1\mathrm{e}^{-t}\mathrm{e}^{t}\mathrm{d}t=1$ reproduces the same shape. Hence for $\lvert\lambda\rvert<1$

y(x) = 1 + \frac{\lambda(1-\mathrm{e}^{-1})}{1-\lambda}\,\mathrm{e}^{x} .

Check. Substituting $y=1+\beta\mathrm{e}^{x}$ into the equation gives $1+\lambda\mathrm{e}^{x}[(1-\mathrm{e}^{-1})+\beta]$ , which equals $1+\beta\mathrm{e}^{x}$ exactly when $\beta(1-\lambda)=\lambda(1-\mathrm{e}^{-1})$ — the value above. At $\lambda=\tfrac12$ the solution is $y(x)=1+(1-\mathrm{e}^{-1})\mathrm{e}^{x}$ , and at $x=1$ this is $1+\mathrm{e}-1=\mathrm{e}=2.718281\ldots$ , a number worth remembering as the check on the whole calculation.

Now compare the three conditions. The exact solution has a single pole, at $\lambda=1$ , so the Neumann series converges exactly for $\lvert\lambda\rvert<1$ .

Condition (S): $c=\max_{[0,1]^{2}}\mathrm{e}^{x-t}=\mathrm{e}$ and $b-a=1$ , so (S) reads $\lvert\lambda\rvert<\mathrm{e}^{-1}=0.3679$ .
Condition $(\mathrm{S}')$ : $\int_{0}^{1}\!\!\int_{0}^{1}\mathrm{e}^{2(x-t)}\mathrm{d}x\,\mathrm{d}t = \frac{\mathrm{e}^{2}-1}{2}\cdot\frac{1-\mathrm{e}^{-2}}{2}=\frac{\mathrm{e}^{2}+\mathrm{e}^{-2}-2}{4}=1.38110$ , so $(\mathrm{S}')$ reads $\lvert\lambda\rvert<1/\sqrt{1.38110}=0.8509$ .
The sharp condition: $\mathcal{K}u = \mathrm{e}^{x}\int_{0}^{1}\mathrm{e}^{-t}u(t)\,\mathrm{d}t$ has rank one, and $\mathcal{K}\mathrm{e}^{x} = \mathrm{e}^{x}\int_{0}^{1}\mathrm{e}^{-t}\mathrm{e}^{t}\mathrm{d}t = \mathrm{e}^{x}$ . So $1$ is the only nonzero eigenvalue of $\mathcal{K}$ , the only characteristic value is $\lambda_{e}=1$ , and the sharp radius is $\lvert\lambda\rvert<1$ .

The three numbers $0.3679 < 0.8509 < 1$ are exactly the hierarchy the remark describes. At $\lambda=\tfrac12$ , which (S) rejects, the series converges perfectly well; its terms shrink by a factor $2$ each step, which the error bound of part 2 reports as $\alpha=\mathrm{e}/2>1$ — no information at all — while the true geometric ratio is $\tfrac12$ .

The homogeneous equation. At $\lambda=1$ the equation $y=1+\int_{0}^{1}\mathrm{e}^{x-t}y\,\mathrm{d}t$ has no solution: putting $y=1+\beta\mathrm{e}^{x}$ gives $\beta = (1-\mathrm{e}^{-1})+\beta$ , which is impossible. But the homogeneous equation $y=\int_{0}^{1}\mathrm{e}^{x-t}y\,\mathrm{d}t$ has the one-dimensional family $y=\beta\mathrm{e}^{x}$ . This is the Fredholm alternative at work (Separable kernels): $f\equiv1$ fails the solvability condition, since here $K$ is real with $K^{*}(x,t)=\mathrm{e}^{\,t-x}$ , the adjoint homogeneous equation $g=\overline{\lambda}\,\mathcal{K}^{*}g$ at $\lambda=1$ has the solutions $g(x)=\gamma\mathrm{e}^{-x}$ (indeed $\int_{0}^{1}\mathrm{e}^{\,t-x}\gamma\mathrm{e}^{-t}\mathrm{d}t=\gamma\mathrm{e}^{-x}$ ), and for $\gamma=1$ , $(f,g)=\int_{0}^{1}\mathrm{e}^{-x}\,\mathrm{d}x = 1-\mathrm{e}^{-1}\neq0$ .

ApplicationOriginal

Perturbation theory is a Neumann series. This is the reason the result matters far beyond integral equations, and both Arfken, Weber and Harris and Hassani point it out.

The Dyson series. In quantum mechanics the evolution operator in the interaction picture satisfies

U(t,t_{0}) = 1 - \frac{\mathrm{i}}{\hbar}\int_{t_{0}}^{t} V(t_{1})\,U(t_{1},t_{0})\,\mathrm{d}t_{1},

a Volterra equation of the second kind for the operator-valued unknown $U$ , with $f=1$ and $\lambda=-\mathrm{i}/\hbar$ . Its Neumann series

U(t,t_{0}) = 1 - \frac{\mathrm{i}}{\hbar}\int_{t_{0}}^{t}\!V(t_{1})\,\mathrm{d}t_{1} +\Bigl(\frac{-\mathrm{i}}{\hbar}\Bigr)^{2}\int_{t_{0}}^{t}\!\mathrm{d}t_{1}\int_{t_{0}}^{t_{1}}\!\mathrm{d}t_{2}\,V(t_{1})V(t_{2})+\cdots

is the Dyson series, and the iterated kernels are exactly the time-ordered products. Because the equation is of Volterra type, the series converges for every coupling strength when $V$ is bounded (Volterra equations) — the ordering of the times supplies the factorials $1/m!$ that a Fredholm equation does not have.

The Born series. The time-independent scattering problem in position space is the Lippmann–Schwinger equation, a Fredholm equation of the second kind with the free outgoing Green's function as kernel and the incident wave as $f$ . Truncating its Neumann series after the first correction is the first Born approximation; the whole series is the Born series (Scattering and the Born approximation). Here the Fredholm character is real: for a strong enough potential $\lambda$ passes a characteristic value, the series diverges, and the characteristic values that spoil it are the bound states and resonances. Convergence of the Born series is thus not a technicality but a statement about the physics — it fails precisely when the scatterer can hold the particle.

Iterative numerics. On the computational side, discretising $y=f+\lambda\mathcal{K}y$ by a quadrature rule turns it into a linear system $(\mathrm{I}-\lambda A)c=b$ , and the Neumann series is then the Richardson iteration $c_{n+1}=b+\lambda Ac_{n}$ for that system. Part 2 of the statement is the familiar fact that such an iteration converges at a geometric rate whenever the iteration matrix has norm less than one, and the error bound is the usual one; the remark about the sharp radius is the usual statement that convergence is governed by the spectral radius rather than the norm (Solving linear systems numerically). What the integral-equation setting adds is that the $m$ -th term has a meaning — the contribution of paths that scatter $m$ times — so truncating the series is a physical approximation and not only a numerical one.

Method

Separable kernels

Adapted

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A kernel that is a finite sum of products of a function of x with a function of t turns a Fredholm equation of the second kind into an n-by-n linear system, so the characteristic values are the roots of a polynomial and the Fredholm alternative is exactly the solvability trichotomy for that system.

Prerequisites: Classification of integral equations, Linear systems and rank

Definition(Separable kernel)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.2, eq. (21.33), p. 1057; Riley, Hobson & Bence (2006), §23.4.1, eq. (23.8), p. 807; Stakgold & Holst (2011), §6.1, p. 356, and §6.2, eq. (6.2.4), p. 361

A kernel $K$ on $[a,b]\times[a,b]$ is separable, or degenerate, if it is a finite sum of products of a function of the first variable with a function of the second:

K(x,t) = \sum_{j=1}^{n}\phi_{j}(x)\,\psi_{j}(t),

with $n<\infty$ and all $\phi_j,\psi_j$ in $L^{2}(a,b)$ . Stakgold and Holst write the second factor conjugated, $\sum_{i}p_{i}(x)\overline{q_{i}(t)}$ , which describes the same class of kernels, and they add the normalisation that the family $\{\phi_{j}\}$ and the family $\{\psi_{j}\}$ are each linearly independent; this costs nothing, since a dependent family can be thinned and the kernel rewritten with a smaller $n$ .

The rank of a separable kernel is defined to be $\dim\operatorname{ran}\mathcal{K}$ , the dimension of the range of the integral operator

(\mathcal{K}u)(x) = \sum_{j=1}^{n}\phi_{j}(x)\int_{a}^{b}\psi_{j}(t)u(t)\,\mathrm{d}t .

It is finite, and for a representation normalised as above it is exactly the number of terms. The display puts $\operatorname{ran}\mathcal{K}$ inside $\operatorname{span}\{\phi_{1},\dots,\phi_{n}\}$ , so the rank is at most $n$ ; and it is at least $n$ because the $\psi_{j}$ are independent, so the $n$ linear functionals $u\mapsto\int_{a}^{b}\psi_{j}u$ are independent as well and some $u_{1},\dots,u_{n}$ make the matrix $\bigl[\int\psi_{j}u_{i}\bigr]$ invertible, whence the images $\mathcal{K}u_{1},\dots,\mathcal{K}u_{n}$ span the whole of $\operatorname{span}\{\phi_{1},\dots,\phi_{n}\}$ , which is $n$ -dimensional because the $\phi_{j}$ are independent. Since the same containment applies to any representation of $K$ , no representation can have fewer than $\dim\operatorname{ran}\mathcal{K}$ terms: the normalised $n$ is also the smallest possible number of terms, which is why one number deserves both descriptions.

Separable kernels are Hilbert–Schmidt, and the corresponding operator is compact: it is bounded and its range is finite-dimensional, which is Stakgold and Holst's first step in proving that every Hilbert–Schmidt operator is compact. Arfken, Weber and Harris note that the class is larger than it looks: it contains all polynomials in $x$ and $t$ and many of the elementary transcendental functions, their example being

\cos(t-x) = \cos x\cos t + \sin x\sin t .

Their "many" should be read strictly: being an elementary function of $x-t$ is not enough. On $[0,2\pi]$ the kernel $K(x,t)=\bigl(2+\cos(x-t)\bigr)^{-1}$ is smooth and bounded, yet its Fourier expansion in the difference,

\frac{1}{2+\cos(x-t)} = \frac{1}{\sqrt3}\sum_{k\in\mathbb{Z}}(-1)^{k}\bigl(2-\sqrt3\bigr)^{\lvert k\rvert}\mathrm{e}^{\mathrm{i}k(x-t)},

has no vanishing coefficient. That expansion is the Poisson kernel in disguise, and is worth deriving rather than quoting. Put $r=2-\sqrt3$ , the root in $(0,1)$ of $1+r^{2}=4r$ ; then

1+2r\cos\theta+r^{2} = 4r+2r\cos\theta = 2r\bigl(2+\cos\theta\bigr).

Writing the Poisson identity $\sum_{k\in\mathbb{Z}}r^{\lvert k\rvert}\mathrm{e}^{\mathrm{i}k\vartheta}=(1-r^{2})/(1-2r\cos\vartheta+r^{2})$ at $\vartheta=\theta-\pi$ , where $\cos\vartheta=-\cos\theta$ and $\mathrm{e}^{\mathrm{i}k\vartheta}=(-1)^{k}\mathrm{e}^{\mathrm{i}k\theta}$ , and dividing by the displayed factor $2r$ ,

\frac{1}{2+\cos\theta} = \frac{2r}{1-r^{2}}\sum_{k\in\mathbb{Z}}(-1)^{k}r^{\lvert k\rvert}\mathrm{e}^{\mathrm{i}k\theta}, \qquad \frac{2r}{1-r^{2}} = \frac{2-\sqrt3}{2\sqrt3-3} = \frac{1}{\sqrt3},

the last step because $2\sqrt3-3=\sqrt3\,(2-\sqrt3)$ . Taking $\theta=x-t$ is the display above; we also checked it numerically. Write $c_{k}=(-1)^{k}(2-\sqrt3)^{\lvert k\rvert}/\sqrt3$ , so that $\lvert c_{k}\rvert$ decays geometrically with ratio $2-\sqrt3=0.2679<1$ and $\sum_{k}\lvert c_{k}\rvert<\infty$ ; the series therefore converges absolutely and uniformly on the square and may be integrated term by term against the bounded function $\mathrm{e}^{\mathrm{i}kt}$ . Doing so and using $\int_{0}^{2\pi}\mathrm{e}^{\mathrm{i}(k-m)t}\,\mathrm{d}t=2\pi\delta_{km}$ gives $\mathcal{K}\mathrm{e}^{\mathrm{i}kt}=2\pi c_{k}\mathrm{e}^{\mathrm{i}kx}$ for every $k\in\mathbb{Z}$ , so the range contains the infinitely many independent functions $\mathrm{e}^{\mathrm{i}kx}$ . This kernel therefore has infinite rank and is not separable, although it is a bounded, infinitely differentiable function of $x-t$ .

Method(Reduction to a linear system, and the Fredholm alternative)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.2, eqs. (21.34)–(21.42), pp. 1058–1059, and Exercise 21.2.16(b), p. 1063; Riley, Hobson & Bence (2006), §23.4.1, eqs. (23.9)–(23.11), p. 807, and §23.3, eq. (23.7), p. 806; Stakgold & Holst (2011), §6.2, Thm 6.2.1, p. 360, and Example 1, eqs. (6.2.7) and (6.2.11), pp. 361–362

Let $K(x,t)=\sum_{j=1}^{n}\phi_{j}(x)\psi_{j}(t)$ be separable with $\{\phi_{j}\}$ independent and $\{\psi_{j}\}$ independent, let $f\in L^{2}(a,b)$ and let $\lambda\neq0$ . Define the $n\times n$ matrix $A$ and the vector $b\in\mathbb{C}^{n}$ by

A_{ij} = \int_{a}^{b}\psi_{i}(x)\,\phi_{j}(x)\,\mathrm{d}x , \qquad b_{i} = \int_{a}^{b}\psi_{i}(x)\,f(x)\,\mathrm{d}x .

The correspondence. A function $y\in L^{2}(a,b)$ solves $y(x) = f(x)+\lambda\int_{a}^{b}K(x,t)y(t)\,\mathrm{d}t \tag{$\ast$}$ if and only if $y = f+\lambda\sum_{j=1}^{n}c_{j}\phi_{j} \qquad\text{for some } c\in\mathbb{C}^{n} \text{ with } (\mathrm{I}-\lambda A)\,c = b .$ The correspondence $y\leftrightarrow c$ is a bijection between the solution set of $(\ast)$ and the solution set of the linear system, with $c_{j}=\int_{a}^{b}\psi_{j}y$ .
Characteristic values. $\lambda\neq0$ is a characteristic value of the kernel — that is, $y=\lambda\mathcal{K}y$ has a nonzero solution — if and only if $\det(\mathrm{I}-\lambda A)=0 .$ This is a polynomial equation in $\lambda$ of degree at most $n$ , so a separable kernel of rank $n$ has at most $n$ characteristic values. At such a $\lambda$ , the homogeneous equation has exactly $\dim\ker(\mathrm{I}-\lambda A)$ independent solutions. In addition, $0$ is always an eigenvalue of $\mathcal{K}$ itself, of infinite multiplicity when $L^{2}(a,b)$ is infinite-dimensional: every $u$ with $(u,\overline{\psi_{j}})=0$ for all $j$ satisfies $\mathcal{K}u=0$ , and those $u$ form a subspace of codimension $n$ . (This null space is what Arfken, Weber and Harris isolate in their Exercise 21.2.16(b), where the equation is of the first kind, $\int_{a}^{b}K(x,t)\varphi(t)\,\mathrm{d}t=f(x)$ : there a function orthogonal to every $\psi_{j}$ may indeed be added to any solution without disturbing the equation, because the operator does not see it. For the second-kind equation $(\ast)$ that is false: $y+u$ solves $(\ast)$ only if $u=\lambda\mathcal{K}u$ , which for $u\in\ker\mathcal{K}$ forces $u=0$ . The non-uniqueness in $(\ast)$ comes from $\ker(\mathrm{I}-\lambda\mathcal{K})$ , part 3 below, not from $\ker\mathcal{K}$ .)
The alternative. Exactly one of the following holds.
- $\det(\mathrm{I}-\lambda A)\neq0$ . Then $(\ast)$ has exactly one solution for every $f$ , namely $y=f+\lambda\sum_j c_j\phi_j$ with $c=(\mathrm{I}-\lambda A)^{-1}b$ .
- $\det(\mathrm{I}-\lambda A)=0$ , say with $k=\dim\ker(\mathrm{I}-\lambda A)\ge1$ . Then the homogeneous equation $y=\lambda\mathcal{K}y$ has a $k$ -dimensional solution space; the adjoint homogeneous equation $g(x) = \overline{\lambda}\int_{a}^{b}\overline{K(t,x)}\,g(t)\,\mathrm{d}t$ has a solution space of the same dimension $k$ ; and $(\ast)$ has a solution exactly when the data satisfy the $k$ orthogonality conditions $(f,g) = \int_{a}^{b} f(x)\overline{g(x)}\,\mathrm{d}x = 0 \qquad\text{for each adjoint homogeneous solution } g .$ Where those conditions hold, the solutions of $(\ast)$ form a $k$ -parameter family: any one of them, plus an arbitrary solution of the homogeneous equation.

This is Theorem 6.2.1 of Stakgold and Holst, which holds for every compact $\mathcal{K}$ , specialised to the separable case where it can be proved outright; and it is the statement that Riley, Hobson and Bence quote in their §23.3 without proof. One warning before turning to their p. 360: they normalise the equation differently, writing it as $\mathcal{K}u-\lambda u=f$ in their (6.2.1), so their $\lambda$ is an eigenvalue of $\mathcal{K}$ and is the reciprocal of ours, and their Remarks 1–3 — on $\lambda=0$ , and on eigenvalues of infinite multiplicity — are phrased in that parametrisation. Dividing their equation through by $-\lambda$ turns it into our $(\ast)$ , with our parameter the reciprocal of theirs and our $f$ equal to $-f/\lambda$ ; the exclusion $\lambda\neq0$ is common to both, but it excludes opposite degenerate cases — for them the first-kind equation, for us the trivial one $y=f$ .

Idea of the proofOriginal

A separable kernel can only ever produce functions in the span of $\phi_{1},\dots,\phi_{n}$ . So whatever the solution of $(\ast)$ is, it must be $f$ plus something in that $n$ -dimensional span, and the only unknowns are the $n$ coefficients. The integral equation can be nothing but $n$ linear equations in those $n$ numbers — and the way to extract them is to hit the equation with each $\psi_{i}$ and integrate, which collapses every integral to a number. Once that translation is made, nothing is left to prove: every assertion about $(\ast)$ is the corresponding assertion about an $n\times n$ system, read off from Linear systems and rank. The only point needing care is that the orthogonality conditions in the alternative, which live in the function space, match the consistency conditions of the matrix system, which live in $\mathbb{C}^{n}$ ; that is a short computation identifying the adjoint integral equation with the adjoint matrix equation.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §21.2, eqs. (21.34)–(21.42), pp. 1058–1059; Stakgold & Holst (2011), §6.2, Example 1, eqs. (6.2.7)–(6.2.11), pp. 361–362; Riley, Hobson & Bence (2006), §23.4.1, eqs. (23.9)–(23.11), p. 807

Part 1, necessity. Suppose $y$ solves $(\ast)$ . Insert the separable form of $K$ and pull the $x$ -dependent factors out of the integral, which is legitimate because the sum is finite:

y(x) = f(x)+\lambda\sum_{j=1}^{n}\phi_{j}(x)\int_{a}^{b}\psi_{j}(t)y(t)\,\mathrm{d}t .

The $n$ integrals are numbers; call them $c_{j}=\int_{a}^{b}\psi_{j}y$ , each finite by the Cauchy–Schwarz inequality since $\psi_j,y\in L^2$ . Then $y = f+\lambda\sum_{j}c_{j}\phi_{j}$ , which is the stated shape. To get the system, multiply this identity by $\psi_{i}$ and integrate:

c_{i} = \int_{a}^{b}\psi_{i}y = \int_{a}^{b}\psi_{i}f+\lambda\sum_{j}c_{j}\int_{a}^{b}\psi_{i}\phi_{j} = b_{i}+\lambda\sum_{j}A_{ij}c_{j},

that is $(\mathrm{I}-\lambda A)c=b$ .

Part 1, sufficiency. Conversely let $c$ satisfy $(\mathrm{I}-\lambda A)c=b$ and set $y=f+\lambda\sum_{j}c_{j}\phi_{j}$ . Then

\int_{a}^{b}\psi_{i}y = b_{i}+\lambda\sum_{j}A_{ij}c_{j} = c_{i}

by the system, so

f(x)+\lambda\int_{a}^{b}K(x,t)y(t)\,\mathrm{d}t = f(x)+\lambda\sum_{j}\phi_{j}(x)\int_{a}^{b}\psi_{j}y = f(x)+\lambda\sum_{j}c_{j}\phi_{j}(x) = y(x),

so $y$ solves $(\ast)$ .

Part 1, bijectivity. The map $c\mapsto y=f+\lambda\sum c_{j}\phi_{j}$ is injective because $\lambda\neq0$ and the $\phi_{j}$ are independent, and the two displays above show it maps solutions onto solutions in both directions, with inverse $y\mapsto\bigl(\int\psi_{j}y\bigr)_{j}$ .

Part 2. Take $f\equiv0$ , so $b=0$ . By part 1 the nonzero solutions of $y=\lambda\mathcal{K}y$ correspond bijectively to the nonzero solutions of $(\mathrm{I}-\lambda A)c=0$ , and the correspondence is linear, so it matches the solution spaces dimension for dimension. By part 4 of the solvability statement in Linear systems and rank, a square homogeneous system has a nonzero solution exactly when its determinant vanishes; expanding $\det(\mathrm{I}-\lambda A)$ shows it is a polynomial in $\lambda$ of degree at most $n$ with constant term $1$ , hence not identically zero, so it has at most $n$ roots. For the last clause, $\mathcal{K}u = \sum_j\phi_j\int\psi_j u$ , and because the $\phi_{j}$ are independent this vanishes exactly when $\int\psi_{j}u=0$ for every $j$ — that is, $u$ lies in the orthogonal complement of $\operatorname{span}\{\overline{\psi_{1}},\dots,\overline{\psi_{n}}\}$ , a subspace of codimension $n$ .

Part 3, the two branches. By part 1 the solution sets of $(\ast)$ and of $(\mathrm{I}-\lambda A)c=b$ are in bijection, so $(\ast)$ has exactly one solution for every $f$ exactly when the square system does, which by Linear systems and rank is exactly when $\det(\mathrm{I}-\lambda A)\neq0$ ; and then $c=(\mathrm{I}-\lambda A)^{-1}b$ . In the other branch put $M=\mathrm{I}-\lambda A$ with $k=\dim\ker M\ge1$ . The finite-dimensional Fredholm alternative (Linear systems and rank) says that $Mc=b$ has a solution exactly when $(b,d)=0$ for each $d\in\ker M^{*}$ , that $\dim\ker M^{*}=k$ as well, and that the solution set is then a coset of $\ker M$ , of dimension $k$ . Transporting the last two facts through the bijection of part 1 gives the stated dimensions.

Part 3, matching the orthogonality conditions. It remains to show that the $k$ conditions $(b,d)=0$ , $d\in\ker M^{*}$ , are the conditions $(f,g)=0$ , $g$ an adjoint eigenfunction. The adjoint kernel is

\overline{K(t,x)} = \sum_{j}\overline{\psi_{j}(x)}\;\overline{\phi_{j}(t)},

again separable, with the roles of the two families exchanged and everything conjugated. So part 1 applied to the adjoint homogeneous equation $g=\overline{\lambda}\,\mathcal{K}^{*}g$ shows that its solutions are exactly

g = \overline{\lambda}\sum_{j}d_{j}\,\overline{\psi_{j}}, \qquad d_{j}=\int_{a}^{b}\overline{\phi_{j}}\,g ,

with $d$ running over the kernel of $\mathrm{I}-\overline{\lambda}\,A^{\sharp}$ , where

A^{\sharp}_{ij} = \int_{a}^{b}\overline{\phi_{i}(x)}\,\overline{\psi_{j}(x)}\,\mathrm{d}x = \overline{A_{ji}} = (A^{*})_{ij}.

Hence $d\in\ker(\mathrm{I}-\overline{\lambda}A^{*}) = \ker\bigl((\mathrm{I}-\lambda A)^{*}\bigr)=\ker M^{*}$ , exactly the vectors appearing in the matrix conditions. Finally compute the inner product in the function space:

(f,g) = \int_{a}^{b}f\,\overline{g} = \int_{a}^{b}f\;\overline{\overline{\lambda}\sum_{j}d_{j}\overline{\psi_{j}}} = \lambda\sum_{j}\overline{d_{j}}\int_{a}^{b}f\,\psi_{j} = \lambda\sum_{j}\overline{d_{j}}\,b_{j} = \lambda\,(b,d).

Since $\lambda\neq0$ , $(f,g)=0$ for all adjoint eigenfunctions $g$ if and only if $(b,d)=0$ for all $d\in\ker M^{*}$ . That closes part 3. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Stakgold & Holst (2011), §6.2, Thm 6.2.1 and Remark 1, p. 360; Arfken, Weber & Harris (2013), §21.2, Exercise 21.2.16(b), p. 1063

$\lambda\neq0$ . It is what makes $(\ast)$ a second-kind equation, and it is used three times: to write the system in the form $(\mathrm{I}-\lambda A)c=b$ at all, in the bijectivity argument of part 1, and in the final identity $(f,g)=\lambda\,(b,d)$ that matches the function-space orthogonality conditions with the matrix ones. At $\lambda=0$ the equation degenerates to $y=f$ : still solvable, but the correspondence of part 1 collapses, since every $c$ now gives the same $y$ , and $\det(\mathrm{I}-0\cdot A)=1$ makes part 2 vacuous. The genuinely different problem at $\lambda=0$ is the first-kind equation $\mathcal{K}y=f$ , which this unit does not treat; Stakgold and Holst exclude it from their theorem for the same reason and warn that orthogonality to $\ker\mathcal{K}^{*}$ is not sufficient there, because the range of a compact operator need not be closed. Arfken, Weber and Harris's Exercise 21.2.16(b), discussed in part 2, is a first-kind equation and behaves accordingly.

Each of the two families independent. This is a hypothesis about the representation, not about the kernel, and it buys two things. In the definition it makes the rank exactly $n$ : independence of the $\phi_{j}$ makes $\operatorname{span}\{\phi_{j}\}$ $n$ -dimensional, and independence of the $\psi_{j}$ makes the $n$ functionals $u\mapsto\int\psi_{j}u$ independent, so the range fills that span. In part 2 it is what identifies $\ker\mathcal{K}$ : $\mathcal{K}u=\sum_{j}\phi_{j}\int\psi_{j}u$ vanishes only if every coefficient $\int\psi_{j}u$ does (independence of the $\phi_{j}$ ), and the resulting space has codimension exactly $n$ , not less (independence of the $\psi_{j}$ ). Parts 1 and 3 are not disturbed by a padded representation — one simply thins it first, which is step 1 of the method, replacing $A$ and $b$ accordingly — but every statement in which the number $n$ appears is about the thinned representation and would be false for a padded one.

$f\in L^{2}(a,b)$ . Together with $\psi_{i}\in L^{2}$ this is what makes each $b_{i}=\int\psi_{i}f$ a finite number, by Cauchy–Schwarz, so that there is a linear system to write down at all; it also keeps $y=f+\lambda\sum_{j}c_{j}\phi_{j}$ in $L^{2}$ , which is where the solution is being sought, and it is needed for the inner products $(f,g)$ of part 3 to converge. Drop it and the method does not so much fail as lose its meaning: with $b$ undefined there is nothing finite-dimensional to solve. One can still pose $(\ast)$ in a larger space, but then the adjoint, the inner product and hence the alternative all have to be set up again in that space.

What is not assumed. Nothing here requires $K$ to be symmetric, real, continuous, or even bounded as a function; $\phi_{j},\psi_{j}\in L^{2}$ is the whole regularity hypothesis. Nor is any smallness of $\lvert\lambda\rvert$ needed, which is the method's advantage over a Neumann series (Neumann series). And finiteness of $n$ is not a hypothesis one can relax by a limit: the example at the end of the definition block is a smooth bounded kernel of infinite rank, for which $A$ would be an infinite matrix and the determinant of part 2 would have to be replaced by a Fredholm determinant.

Method(Solving an equation with a separable kernel)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.2, the generating-function method, eqs. (21.30)–(21.32), pp. 1056–1057, and eqs. (21.33)–(21.42), pp. 1057–1059; Riley, Hobson & Bence (2006), §23.4.1, eqs. (23.8)–(23.11), p. 807; Hassani (2013), §18.2, Example 18.2.3, eq. (18.16) and the closing remark, pp. 553–554

Given $y(x)=f(x)+\lambda\int_{a}^{b}K(x,t)y(t)\,\mathrm{d}t$ with $\lambda\neq0$ :

Separate the kernel. Write $K(x,t)=\sum_{j=1}^{n}\phi_{j}(x)\psi_{j}(t)$ with as few terms as possible. Expand sums and differences of arguments ( $\cos(x-t)$ , $\sin(x+t)$ , $(x-t)^{2}$ , $\mathrm{e}^{x-t}$ ) and collect powers. Discard any $\phi_{j}$ that is a combination of the others, folding its $\psi_{j}$ into theirs, and likewise on the other side, so that both families end up independent.
Build the small system. Compute the $n^{2}$ numbers $A_{ij}=\int_{a}^{b}\psi_{i}\phi_{j}$ and the $n$ numbers $b_{i}=\int_{a}^{b}\psi_{i}f$ . Note the order of the indices: the row index is the $\psi$ , the column index the $\phi$ .
Find the characteristic values by solving the polynomial equation $\det(\mathrm{I}-\lambda A)=0$ , which Arfken, Weber and Harris call the secular equation. Dividing each row by $\lambda$ turns it into the ordinary eigenvalue problem for $A$ with eigenvalue $1/\lambda$ , which is often the easier form.
Branch on $\lambda$ .
- If $\det(\mathrm{I}-\lambda A)\neq0$ , solve $(\mathrm{I}-\lambda A)c=b$ and return $y=f+\lambda\sum_{j}c_{j}\phi_{j}$ . This is the unique solution.
- If $\det(\mathrm{I}-\lambda A)=0$ , first test solvability: find a basis $d^{(1)},\dots,d^{(k)}$ of $\ker\bigl((\mathrm{I}-\lambda A)^{*}\bigr)$ and check $(b,d^{(i)})=0$ for each. If some test fails there is no solution and you are done. If all pass, find one solution $c_{\mathrm{p}}$ of the system (by elimination, which reports the consistency as it goes) and a basis $e^{(1)},\dots,e^{(k)}$ of $\ker(\mathrm{I}-\lambda A)$ ; the general solution is $y = f+\lambda\sum_{j}\bigl(c_{\mathrm{p}}+\textstyle\sum_{i}t_{i}e^{(i)}\bigr)_{j}\phi_{j}, \qquad t_{1},\dots,t_{k}\in\mathbb{C}\ \text{arbitrary.}$
For the homogeneous equation only ( $f\equiv0$ ), skip to step 3, then for each root $\lambda$ solve $(\mathrm{I}-\lambda A)c=0$ ; each independent $c$ gives the eigenfunction $\sum_{j}c_{j}\phi_{j}$ , whose normalisation is arbitrary.

Notes on using it. When $n=1$ the system is one scalar equation and no linear algebra is needed; Hassani does exactly this for $K(x,t)=xt$ , introducing the single constant $A=\int t\,u(t)\,\mathrm{d}t$ , and gets the closed-form answer $u(x)=x/(1-\lambda A_{11})$ , where in our notation $\phi_{1}(x)=x$ , $\psi_{1}(t)=t$ and $A_{11}=\int_{a}^{b}x^{2}\,\mathrm{d}x=(b^{3}-a^{3})/3$ , the quantity he writes as the norm of the kernel. The scope of that is exactly what he claims for it: no smallness hypothesis on $\lambda$ is needed, which is the virtue of the method over a Neumann series (Neumann series), whose convergence in his example needs $\lvert\lambda\rvert(b^{3}-a^{3})<3$ . The one value of $\lambda$ that must still be excluded is the characteristic value $\lambda=1/A_{11}=3/(b^{3}-a^{3})$ , visible as the zero of his own denominator, where part 2 of the statement above takes over. When $K$ is not separable, the move Arfken, Weber and Harris make is to expand it in a series of special functions: their worked case is the kernel $(1-2xt+x^{2})^{-1/2}$ on $[-1,1]$ , whose generating-function expansion $\sum_{n}P_{n}(t)x^{n}$ is a series of products of a function of $x$ with a function of $t$ , and they recommend the device whenever such an expansion is available and the interval suits. Their equation is of the first kind, with no term outside the integral, so the recipe above — which needs $\lambda\neq0$ and the unknown outside as well as inside — does not apply to it; what they do instead is expand the unknown in the same polynomials and match coefficients. What the device does hand to this unit is the shape of the kernel: a kernel written as a series of products of a function of $x$ with a function of $t$ becomes separable as soon as the series is truncated, and for a second-kind equation the recipe above then solves the truncated problem exactly. That is the Galerkin idea, and the error analysis becomes a question of how well the truncation approximates $K$ in the Hilbert–Schmidt norm.

Example(A rank-two Legendre kernel: both branches of the alternative)Original

Take $[a,b]=[-1,1]$ and

K(x,t) = \frac{1+3xt}{2},

which is separable of rank $2$ : with $\phi_{1}=1$ , $\psi_{1}(t)=\tfrac12$ , $\phi_{2}(x)=3x$ , $\psi_{2}(t)=\tfrac{t}{2}$ , indeed $\phi_{1}\psi_{1}+\phi_{2}\psi_{2} = \tfrac12+\tfrac{3xt}{2}$ . Both families are independent.

Step 2: the matrix. Using $\int_{-1}^{1}1\,\mathrm{d}x=2$ , $\int_{-1}^{1}x\,\mathrm{d}x=0$ and $\int_{-1}^{1}x^{2}\,\mathrm{d}x=\tfrac23$ ,

A_{11}=\int_{-1}^{1}\tfrac12\cdot1 = 1,\quad A_{12}=\int_{-1}^{1}\tfrac12\cdot3x = 0,\quad A_{21}=\int_{-1}^{1}\tfrac{x}{2}\cdot1 = 0,\quad A_{22}=\int_{-1}^{1}\tfrac{x}{2}\cdot3x = \tfrac32\cdot\tfrac23 = 1,

so $A=\mathrm{I}$ . (This is no accident: $1$ and $x$ are the first two Legendre polynomials, and $K$ is the truncation after two terms of the Legendre expansion of the delta function on $[-1,1]$ .)

Step 3: the characteristic values. $\det(\mathrm{I}-\lambda\mathrm{I}) = (1-\lambda)^{2}$ , so the only characteristic value is $\lambda=1$ , and $\dim\ker(\mathrm{I}-A)=2$ : it is doubly degenerate. The eigenfunctions are $\sum_{j}c_{j}\phi_{j}$ for arbitrary $c$ , that is the whole span of $1$ and $x$ . Check directly:

\int_{-1}^{1}\frac{1+3xt}{2}\cdot 1\,\mathrm{d}t = \frac12\cdot 2 + \frac{3x}{2}\cdot 0 = 1, \qquad \int_{-1}^{1}\frac{1+3xt}{2}\,t\,\mathrm{d}t = 0+\frac{3x}{2}\cdot\frac23 = x,

so $\mathcal{K}$ fixes both $1$ and $x$ : $\lambda=1$ with a two-dimensional eigenspace, as predicted.

First branch: $\lambda=\tfrac12$ , $f(x)=x^{2}$ . Here $\det(\mathrm{I}-\tfrac12\mathrm{I})=\tfrac14\neq0$ , so there is exactly one solution. The right-hand side is

b_{1}=\int_{-1}^{1}\tfrac12 t^{2}\,\mathrm{d}t = \tfrac13 , \qquad b_{2}=\int_{-1}^{1}\tfrac{t}{2}\,t^{2}\,\mathrm{d}t = 0 ,

and $(\mathrm{I}-\tfrac12\mathrm{I})c = \tfrac12 c = b$ gives $c=(\tfrac23,0)$ . Hence

y(x) = x^{2}+\tfrac12\bigl(\tfrac23\cdot 1 + 0\cdot 3x\bigr) = x^{2}+\tfrac13 .

Verify from scratch:

\tfrac12\int_{-1}^{1}\frac{1+3xt}{2}\Bigl(t^{2}+\tfrac13\Bigr)\mathrm{d}t =\tfrac14\int_{-1}^{1}\Bigl(t^{2}+\tfrac13\Bigr)\mathrm{d}t + \tfrac{3x}{4}\int_{-1}^{1}t\Bigl(t^{2}+\tfrac13\Bigr)\mathrm{d}t =\tfrac14\Bigl(\tfrac23+\tfrac23\Bigr)+0 = \tfrac13 ,

and adding $f=x^{2}$ returns $x^{2}+\tfrac13$ , as required.

Second branch, no solution: $\lambda=1$ , $f(x)=x^{2}$ . Now $M=\mathrm{I}-A=0$ , so $M^{*}=0$ and $\ker M^{*}=\mathbb{C}^{2}$ : there are $k=2$ solvability conditions, $b_{1}=b_{2}=0$ . But $b_{1}=\tfrac13\neq0$ , so the equation

y(x) = x^{2}+\int_{-1}^{1}\frac{1+3xt}{2}\,y(t)\,\mathrm{d}t

has no solution. The conditions are the orthogonality statements $(f,1)=0$ and $(f,x)=0$ in disguise: the adjoint eigenfunctions are again $\operatorname{span}\{1,x\}$ , because $K$ is real and symmetric and $\lambda=1$ is real, and $(x^{2},1)=\int_{-1}^{1}x^{2}\,\mathrm{d}x=\tfrac23\neq0$ .

Second branch, infinitely many solutions: $\lambda=1$ , $f(x)=x^{2}-\tfrac13$ . This $f$ is the Legendre polynomial $P_{2}$ up to a factor, so $(f,1)=\tfrac23-\tfrac23=0$ and $(f,x)=0$ : both conditions hold, and $b=0$ . Then $Mc=0$ is satisfied by every $c\in\mathbb{C}^{2}$ , and the solution set is

y(x) = x^{2}-\tfrac13+\alpha+\beta x, \qquad \alpha,\beta\in\mathbb{C}\ \text{arbitrary},

a two-parameter family, as part 3 requires. Check it: writing $y(t)=t^{2}-\tfrac13+\alpha+\beta t$ , the orthogonality of $t^{2}-\tfrac13$ to both $1$ and $t$ leaves

\int_{-1}^{1}\frac{1+3xt}{2}y(t)\,\mathrm{d}t = \frac12\int_{-1}^{1}\alpha\,\mathrm{d}t + \frac{3x}{2}\int_{-1}^{1}\beta t^{2}\,\mathrm{d}t = \alpha+\beta x ,

so $f+\mathcal{K}y = x^{2}-\tfrac13+\alpha+\beta x = y$ , for every $\alpha$ and $\beta$ . Every number in this example was checked symbolically.

ApplicationOriginal

Separable potentials in nuclear and few-body physics. The Lippmann–Schwinger equation for scattering off a potential $V$ is a Fredholm equation of the second kind whose kernel involves $V$ ; it is solvable in closed form if $V$ is separable in momentum space, $\langle k\lvert V\rvert k'\rangle = \sum_{j}g_{j}(k)\,\lambda_{j}\,g_{j}(k')$ . This is exactly the situation of the method above with $n$ small, so the $T$ -matrix comes out as the inverse of an $n\times n$ matrix instead of as the solution of an integral equation. The Yamaguchi potential ( $n=1$ , $g(k)=1/(k^{2}+\beta^{2})$ ) is the textbook case, and separable expansions of realistic nucleon–nucleon potentials are what make three-body Faddeev calculations tractable. A rank- $n$ separable potential also supports at most $n$ bound states in each partial wave — but that is the finite rank at work, not part 2 of the statement applied to the Lippmann–Schwinger equation: its kernel contains the free resolvent and so depends on the energy, which means the parameter is not held fixed and the characteristic values move as the energy moves. What does transfer is the shape of the argument: because $V$ has rank $n$ , a bound state at energy $E$ forces the wavefunction into the $n$ -dimensional span of the functions $(H_{0}-E)^{-1}g_{j}$ , and the bound-state condition is the vanishing of one $n\times n$ determinant built from the form factors and the free resolvent. The count does not follow from the size of that determinant: its entries depend on $E$ , and a determinant of $E$ -dependent entries can vanish at more than $n$ energies. It follows from the rank itself, by the Birman–Schwinger argument. Take first the attractive case $V\le 0$ , in which the counting is usually stated: a bound state at $E<0$ makes $1$ an eigenvalue of the Birman–Schwinger operator $\lvert V\rvert^{1/2}(H_{0}-E)^{-1}\lvert V\rvert^{1/2}$ , which inherits rank at most $n$ from $V$ and so has at most $n$ nonzero eigenvalues at any $E$ ; monotonicity of those eigenvalues in $E$ then allows at most $n$ bound states in all. The hypothesis $V\le 0$ is needed for that operator to be the right one. For a potential of either sign, write $V=U\lvert V\rvert$ with $U=\operatorname{sgn}V$ and $\varphi=\lvert V\rvert^{1/2}\psi$ ; then $H\psi=E\psi$ rearranges to $\varphi=-\lvert V\rvert^{1/2}(H_{0}-E)^{-1}U\lvert V\rvert^{1/2}\varphi$ , so it is $-\lvert V\rvert^{1/2}(H_{0}-E)^{-1}U\lvert V\rvert^{1/2}$ that has eigenvalue $1$ , and $U$ collapses to $-\mathrm{I}$ only when $V\le 0$ . The rank bound is unaffected by the extra factor. But the rank bound alone caps only the multiplicity at one fixed energy, which is the weaker conclusion dismissed above when the determinant was set aside; what converts it into a bound over all energies is the monotonicity in $E$ , and that is exactly the step the attractive case was needed for. So the general-sign count has to be reached another way, and it is. Split $V$ into its positive and negative parts, $V=V_{+}+V_{-}$ with $V_{-}\le 0$ . A self-adjoint operator of rank at most $n$ has at most $n$ nonzero eigenvalues, and $V_{-}$ is carried by the negative ones, so $\operatorname{rank}V_{-}\le n$ as well. Since $V\ge V_{-}$ as quadratic forms, the min–max principle gives $H_{0}+V$ no more negative eigenvalues, counted with multiplicity, than $H_{0}+V_{-}$ ; and $V_{-}$ is attractive of rank at most $n$ , so the attractive argument above applies to it and returns at most $n$ . The count of at most $n$ bound states therefore does hold for either sign — by running the attractive case on $V_{-}$ , not by running Birman–Schwinger on $V$ itself.

Galerkin and boundary-element discretisation. Approximating a general kernel by a separable one is not merely a trick for exercises: it is the standard numerical method. Choose orthonormal basis functions $\chi_{1},\dots,\chi_{N}$ for the unknown — orthonormality is what makes the next step a projection — replace $K$ by $\sum_{i,j}K_{ij}\chi_{i}(x)\overline{\chi_{j}(t)}$ with $K_{ij}=(\mathcal{K}\chi_{j},\chi_{i})$ , which is the orthogonal projection of $\mathcal{K}$ onto $\operatorname{span}\{\chi_{i}\}$ on both sides. The recipe above then gives $A=K$ and $b=K\hat f$ , where $\hat f_{i}=(f,\chi_{i})$ , so the integral equation becomes the $N\times N$ system $(\mathrm{I}-\lambda K)c=K\hat f$ with answer $y=f+\lambda\sum_{j}c_{j}\chi_{j}$ . A Galerkin boundary-element code assembles and factorises the same matrix $\mathrm{I}-\lambda K$ , but in a different unknown: projecting the equation onto each $\chi_{i}$ gives $(\mathrm{I}-\lambda K)\hat y=\hat f$ for the coefficients $\hat y_{i}=(y,\chi_{i})$ of the solution itself. The two are one system seen through the change of unknown $\hat y=\hat f+\lambda c$ , which is what $y=f+\lambda\sum_{j}c_{j}\chi_{j}$ says after projection. (A Nyström code reaches a system of the same shape by replacing the integral with a quadrature rule instead of projecting; the matrix is then $K(x_{i},x_{j})$ times the weights, and no orthonormal family is involved.) Step 3 becomes the numerical eigenvalue problem for $K$ (Computing eigenvalues), and the characteristic values of the discretised problem converge to those of the original as $N$ grows.

Where the alternative bites: the Neumann problem and its discrete twin. The degenerate branch of part 3 is not a curiosity. The potential problem $\nabla^{2}u=0$ in a region with prescribed normal derivative on the boundary becomes, by the single-layer representation $u=\mathcal{S}\sigma$ , the second-kind boundary integral equation $(\pm\tfrac12\mathrm{I}+\mathcal{K}^{*})\sigma=f$ , in which $\mathcal{K}$ is the double-layer operator, $\mathcal{K}^{*}$ its adjoint, and the sign is fixed by the side from which the normal derivative is taken. It is the adjoint homogeneous equation $(\pm\tfrac12\mathrm{I}+\mathcal{K})v=0$ that the constants solve, because $\mathcal{K}1=\mp\tfrac12$ on a smooth closed boundary. The equation's own homogeneous version does not have them: its null space is spanned by the equilibrium (capacity) density of the surface, which is constant only for a disc or a ball. Part 3 does not say otherwise — it gives the two null spaces the same dimension, not the same elements — and the dimension is all that is wanted here. So $k=1$ and the adjoint null space is spanned by a nonzero constant $g$ . The condition is on the data, not on $g$ : with $g$ constant, $(f,g)=\overline{g}\oint f\,\mathrm{d}S$ , so the solvability condition $(f,g)=0$ reads

\oint f\,\mathrm{d}S = 0 ,

the vanishing of the integral of the prescribed normal derivative over the boundary — the physical statement that the total flux through the closed surface must vanish, as it must for a source-free region. ( $\oint g\,\mathrm{d}S$ , by contrast, is the constant times the area and never vanishes; reading the condition off the wrong slot would produce a requirement that can never be met.) The solution is then pinned down only up to an added constant — one free parameter, matching $k=1$ (Green's identities, The Fredholm alternative and modified Green's functions). A code that ignores this gets a singular matrix at step 4 and, worse, a nearly singular one after discretisation; the cure is to impose the flux condition on the data and pin the constant by an extra equation, which is precisely the $k$ -parameter freedom made explicit.

Theorem

Hilbert-Schmidt theory

Adapted

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For a Hermitian Hilbert-Schmidt kernel the characteristic values are real, the eigenfunctions are orthogonal and span the range of the operator, the kernel itself expands bilinearly in them, and the inhomogeneous second-kind equation is solved term by term with a resolvent whose poles are exactly the characteristic values.

Prerequisites: Classification of integral equations, Spectral theorem for compact self-adjoint operators, Orthonormal bases in Hilbert space

Definition(Hermitian kernels, characteristic values, eigenfunctions)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.4, eqs. (21.61), (21.64)–(21.67), pp. 1069–1070; Riley, Hobson & Bence (2006), §23.7, eq. (23.49), p. 817; Stakgold & Holst (2011), §6.3, p. 371

Let $[a,b]$ be a finite interval and let $K$ be a Hilbert–Schmidt kernel on $[a,b]\times[a,b]$ (Classification of integral equations) that is Hermitian,

K(x,t) = \overline{K(t,x)} ,

the case $K(x,t)=K(t,x)$ with $K$ real being the symmetric one that Arfken, Weber and Harris and Riley, Hobson and Bence treat. With the inner product $(u,w)=\int_{a}^{b}u\overline{w}$ , the operator $\mathcal{K}$ is then self-adjoint and compact.

A scalar $\lambda\neq0$ is a characteristic value of the equation, with eigenfunction $y\neq0$ , if

y(x) = \lambda\int_{a}^{b}K(x,t)\,y(t)\,\mathrm{d}t , \qquad\text{equivalently}\qquad \mathcal{K}y = \lambda^{-1}y .

So the characteristic values are the reciprocals of the nonzero eigenvalues of $\mathcal{K}$ . We write $\lambda_{1},\lambda_{2},\dots$ for the characteristic values, repeated according to multiplicity and ordered so that $\lvert\lambda_{1}\rvert\le\lvert\lambda_{2}\rvert\le\cdots$ , and $e_{1},e_{2},\dots$ for a corresponding orthonormal family of eigenfunctions, $(e_{m},e_{n})=\delta_{mn}$ .

Symmetrising a weighted kernel. Arfken, Weber and Harris note that an equation

\varphi(x) = f(x)+\lambda\int_{a}^{b}K(x,t)\,\rho(t)\,\varphi(t)\,\mathrm{d}t

with $K$ symmetric has the non-symmetric total kernel $K(x,t)\rho(t)$ ; multiplying by $\sqrt{\rho(x)}$ and substituting $\psi=\sqrt{\rho}\,\varphi$ gives

\psi(x) = \sqrt{\rho(x)}f(x)+\lambda\int_{a}^{b}\Bigl[K(x,t)\sqrt{\rho(x)\rho(t)}\Bigr]\psi(t)\,\mathrm{d}t ,

whose total kernel is symmetric. Their equations (21.62)–(21.63) state no condition on the weight $\rho$ ; the hypothesis $\rho>0$ on $[a,b]$ is added here, since it is what makes $\sqrt{\rho}$ real and the substitution $\psi=\sqrt{\rho}\,\varphi$ reversible, and it also keeps the new kernel Hermitian rather than merely formally symmetric. Many kernels that arise with a weight can be brought into the scope of this unit this way.

Theorem(Hilbert–Schmidt theory)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.4, eqs. (21.74)–(21.80), pp. 1071–1072; Riley, Hobson & Bence (2006), §23.7, eqs. (23.49), (23.54), (23.55), pp. 817–818; Stakgold & Holst (2011), §6.3, Thms 6.3.4, 6.3.5, 6.3.7 and 6.3.8, pp. 373–377, and §6.4, eqs. (6.4.6)–(6.4.13a), pp. 381–383; Hassani (2013), §18.2, eqs. (18.20)–(18.22), pp. 554–555

Let $K$ be a Hermitian Hilbert–Schmidt kernel on the finite square $[a,b]\times[a,b]$ , not identically zero, and let $\mathcal{K}$ , $\lambda_{n}$ , $e_{n}$ be as in the definition.

Reality and orthogonality. Every characteristic value is real, and eigenfunctions belonging to different characteristic values are orthogonal. Each characteristic value has finite multiplicity, and within one eigenspace an orthonormal basis may be chosen by Gram–Schmidt (Orthonormal bases and Gram-Schmidt), so the family $\{e_{n}\}$ above exists. There is at least one characteristic value, $\lvert\lambda_{1}\rvert^{-1}=\lVert\mathcal{K}\rVert$ , and if there are infinitely many then $\lvert\lambda_{n}\rvert\to\infty$ .
Expansion theorem. For every $u\in L^{2}(a,b)$ , $\mathcal{K}u \;=\; \sum_{n}\lambda_{n}^{-1}(u,e_{n})\,e_{n}$ in $L^{2}$ . Hence $\{e_{n}\}$ is an orthonormal basis of $\overline{\operatorname{ran}\mathcal{K}}$ . It is an orthonormal basis of all of $L^{2}(a,b)$ exactly when $\mathcal{K}$ has trivial null space; in general $u$ splits uniquely as $u = h+\sum_{n}(u,e_{n})e_{n}$ with $\mathcal{K}h=0$ .
Bilinear expansions. For $m\ge2$ the iterated kernel $K_{m}$ has the expansion $K_{m}(x,t) = \sum_{n}\lambda_{n}^{-m}\,e_{n}(x)\,\overline{e_{n}(t)} ,$ convergent in mean square in either variable for each fixed value of the other. (Which values: when $K$ is merely square-integrable, the section $K_{m}(\cdot,t)$ is itself defined only for almost every $t$ , and the expansion is asserted at exactly those $t$ . The qualification is about the set of $t$ , not about the mode of convergence, which is mean square throughout.) If $K_{2}$ is continuous on the closed square, the convergence is uniform there. For $m=1$ the same expansion holds under the extra hypotheses of Mercer's theorem: if $K$ is continuous and all but finitely many characteristic values have the same sign, then $K(x,t) = \sum_{n}\lambda_{n}^{-1}\,e_{n}(x)\,\overline{e_{n}(t)} \qquad\text{uniformly on } [a,b]\times[a,b].$
The inhomogeneous equation. Consider $y(x) = f(x)+\lambda\int_{a}^{b}K(x,t)\,y(t)\,\mathrm{d}t , \qquad f\in L^{2}(a,b),\ \lambda\neq0 . \tag{$\ast$}$
- If $\lambda$ is not a characteristic value, $(\ast)$ has exactly one solution, $y = f + \lambda\sum_{n}\frac{(f,e_{n})}{\lambda_{n}-\lambda}\,e_{n} ,$ the series converging in $L^{2}$ . Equivalently $y = f+\lambda\int_{a}^{b}R(x,t;\lambda)f(t)\,\mathrm{d}t$ with the resolvent kernel $R(x,t;\lambda) = \sum_{n}\frac{e_{n}(x)\,\overline{e_{n}(t)}}{\lambda_{n}-\lambda} .$
- If $\lambda=\lambda_{p}$ for some $p$ , let $e_{m_{1}},\dots,e_{m_{k}}$ be the orthonormal eigenfunctions belonging to $\lambda_{p}$ . Then $(\ast)$ has a solution precisely when the $k$ conditions $(f,e_{m_{i}})=0$ all hold, and in that case the solutions are $y = f + \lambda_{p}\sum_{n\,:\,\lambda_{n}\neq\lambda_{p}}\frac{(f,e_{n})}{\lambda_{n}-\lambda_{p}}\,e_{n} \;+\;\sum_{i=1}^{k}c_{i}\,e_{m_{i}} , \qquad c_{1},\dots,c_{k}\in\mathbb{C}\ \text{arbitrary} ,$ a $k$ -parameter family, of which the one with all $c_{i}=0$ has the smallest norm.

Idea of the proofOriginal

Everything follows from one move: a Hermitian kernel makes $\mathcal{K}$ self-adjoint, and a Hilbert–Schmidt kernel makes it compact, so the spectral theorem for compact self-adjoint operators applies and $\mathcal{K}$ is diagonal in the orthonormal family $\{e_{n}\}$ . In that basis the integral equation $(1-\lambda\mathcal{K})y=f$ splits into one scalar equation for each coefficient, $(1-\lambda/\lambda_{n})(y,e_{n})=(f,e_{n})$ , which one solves by dividing — unless the factor on the left vanishes, that is unless $\lambda$ is a characteristic value, and then the equation reads $0=(f,e_{n})$ , which is either a condition on the data or a coefficient left free. The bilinear expansion of the kernel is the same diagonalisation read as a statement about $K$ rather than about $\mathcal{K}$ : the coefficients of $K(\cdot,t)$ in the basis are $\lambda_{n}^{-1}\overline{e_{n}(t)}$ . The only delicate points are convergence: of the coefficient series (easy, because $\lambda_{n}^{-1}\to0$ makes the coefficients smaller than those of $f$ ) and of the bilinear expansion for $m=1$ (genuinely delicate, which is why Mercer's theorem needs hypotheses).

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §21.4, eqs. (21.69)–(21.76), pp. 1071–1072, and eqs. (21.85)–(21.89), pp. 1073–1075; Stakgold & Holst (2011), §6.3, Thm 6.3.4 and its proof, p. 373, Thm 6.3.5, p. 374, and §6.4, eqs. (6.4.6)–(6.4.12), pp. 381–382; Riley, Hobson & Bence (2006), §23.7, eqs. (23.51)–(23.54), pp. 817–818

Part 1. That $\mathcal{K}$ is bounded and compact is the statement quoted in Classification of integral equations (Reed and Simon's characterisation of Hilbert–Schmidt operators, with Stakgold and Holst's proof that they are compact). Self-adjointness is a direct computation: for $u,w\in L^{2}$ ,

(\mathcal{K}u,w) = \int_{a}^{b}\!\!\int_{a}^{b}K(x,t)u(t)\overline{w(x)}\,\mathrm{d}t\,\mathrm{d}x = \int_{a}^{b}u(t)\,\overline{\int_{a}^{b}\overline{K(x,t)}\,w(x)\,\mathrm{d}x}\,\mathrm{d}t = (u,\mathcal{K}w),

where the order of integration may be exchanged because the double integral converges absolutely: the function $(x,t)\mapsto u(t)\overline{w(x)}$ lies in $L^{2}$ of the square, with $\iint\lvert u(t)\rvert^{2}\lvert w(x)\rvert^{2}=\lVert u\rVert^{2}\lVert w\rVert^{2}$ , so Cauchy–Schwarz on the square bounds $\iint\lvert K(x,t)u(t)\overline{w(x)}\rvert$ by $\bigl(\iint\lvert K\rvert^{2}\bigr)^{1/2}\lVert u\rVert\lVert w\rVert<\infty$ . The last step uses the Hermitian symmetry $\overline{K(x,t)}=K(t,x)$ .

Reality and orthogonality can now be quoted from the spectral theorem, but it is worth seeing them directly, following Arfken, Weber and Harris. Let $\mathcal{K}e=\mu e$ and $\mathcal{K}e'=\mu'e'$ with $e,e'\neq0$ . Then

\mu(e,e') = (\mathcal{K}e,e') = (e,\mathcal{K}e') = \overline{\mu'}\,(e,e') .

Taking $e'=e$ gives $(\mu-\overline{\mu})\|e\|^{2}=0$ , and $\|e\|\neq0$ , so $\mu$ is real; hence the characteristic values $\lambda_{n}=\mu_{n}^{-1}$ are real too. Taking $\mu\neq\mu'$ , both now real, gives $(\mu-\mu')(e,e')=0$ and therefore $(e,e')=0$ .

The remaining assertions of part 1 are the content of the spectral theorem for compact self-adjoint operators (Spectral theorem for compact self-adjoint operators): each nonzero eigenvalue has finite multiplicity, the nonzero eigenvalues can only accumulate at $0$ , and — this is Stakgold and Holst's Theorem 6.3.1, their extremal principle — the supremum of $\lvert(\mathcal{K}v,v)\rvert$ over unit vectors is attained at a normalised eigenfunction whose eigenvalue has modulus $\lVert\mathcal{K}\rVert$ and is the largest in modulus. Since $\mathcal{K}\neq0$ there is at least one such eigenvalue, giving $\lvert\lambda_{1}\rvert^{-1}=\lVert\mathcal{K}\rVert$ ; and $\mu_{n}\to0$ translates into $\lvert\lambda_{n}\rvert\to\infty$ .

Part 2. This is Stakgold and Holst's expansion theorem 6.3.4. Their argument, in our notation: for $u\in L^{2}$ put $v_{N}=u-\sum_{n\le N}(u,e_{n})e_{n}$ , which is orthogonal to $e_{1},\dots,e_{N}$ by construction. The extremal principle applied on the orthogonal complement of $\operatorname{span}\{e_{1},\dots,e_{N}\}$ gives $\lVert\mathcal{K}v\rVert\le\lvert\mu_{N+1}\rvert\lVert v\rVert$ for every $v$ there, so

\Bigl\lVert \mathcal{K}u-\sum_{n\le N}\mu_{n}(u,e_{n})e_{n}\Bigr\rVert = \lVert\mathcal{K}v_{N}\rVert \le \lvert\mu_{N+1}\rvert\,\lVert v_{N}\rVert \le \lvert\mu_{N+1}\rvert\,\lVert u\rVert ,

using $\mathcal{K}e_{n}=\mu_{n}e_{n}$ and Bessel's inequality $\lVert v_{N}\rVert\le\lVert u\rVert$ . Since $\mu_{N+1}\to0$ , the right-hand side tends to $0$ and the series converges to $\mathcal{K}u$ ; with $\mu_{n}=\lambda_{n}^{-1}$ this is the displayed formula. (Should the nonzero eigenvalues be finite in number, the extremal principle gives $\mathcal{K}v_{N}=0$ outright for $N$ past the last one, and the finite sum is exact.) Consequently every element of $\operatorname{ran}\mathcal{K}$ is an $L^{2}$ -limit of combinations of the $e_{n}$ , so $\{e_{n}\}$ is an orthonormal basis of $\overline{\operatorname{ran}\mathcal{K}}$ . (Theorem 6.3.4 is printed as a basis for $\operatorname{ran}\mathcal{K}$ itself; the closure is the reading we state, since the closed span of an orthonormal family is closed while $\operatorname{ran}\mathcal{K}$ in general is not — see the fourth bullet of the remark.) For the splitting, set $h=u-\sum_{n}(u,e_{n})e_{n}$ ; the series converges by Bessel, and $\mathcal{K}h = \mathcal{K}u-\sum_{n}\mu_{n}(u,e_{n})e_{n}=0$ by the formula just proved, so $h\in\ker\mathcal{K}$ ; uniqueness holds because $\ker\mathcal{K}$ is orthogonal to every $e_{n}$ (if $\mathcal{K}h=0$ then $\mu_{n}(h,e_{n})=(h,\mathcal{K}e_{n})=(\mathcal{K}h,e_{n})=0$ and $\mu_{n}\neq0$ ). In particular $\{e_{n}\}$ is a basis of $L^{2}$ exactly when $\ker\mathcal{K}=\{0\}$ , which is Stakgold and Holst's Theorem 6.3.5.

Part 3. Fix $t$ and expand $K_{m}(\cdot,t)$ . Because $K$ is Hermitian, so is every $K_{m}$ , and

\int_{a}^{b}K_{m}(x,t)\,\overline{e_{n}(x)}\,\mathrm{d}x = \overline{\int_{a}^{b}\overline{K_{m}(x,t)}\,e_{n}(x)\,\mathrm{d}x} = \overline{\int_{a}^{b}K_{m}(t,x)e_{n}(x)\,\mathrm{d}x} = \overline{(\mathcal{K}^{m}e_{n})(t)} = \mu_{n}^{m}\,\overline{e_{n}(t)} ,

using $\mu_{n}$ real. So the Fourier coefficients of $K_{m}(\cdot,t)$ in the basis $\{e_{n}\}$ are $\mu_{n}^{m}\overline{e_{n}(t)}$ , and the Fourier series is $\sum_{n}\mu_{n}^{m}e_{n}(x)\overline{e_{n}(t)}$ as claimed. For $m\ge2$ the relation $K_{m}(\cdot,t)=\mathcal{K}K_{m-1}(\cdot,t)$ puts $K_{m}(\cdot,t)$ in $\operatorname{ran}\mathcal{K}$ , where by part 2 the Fourier series does converge to the function in mean square; by symmetry the same holds in the other variable. (Stakgold and Holst say this for each fixed value of the second variable; for a kernel that is only square-integrable the function $K_{m}(\cdot,t)$ is defined for almost every $t$ , which is the qualification carried in part 3.) The uniform-convergence refinement when $K_{2}$ is continuous is Stakgold and Holst's Theorem 6.3.7, and the $m=1$ case under continuity of $K$ plus a sign condition on all but finitely many characteristic values is their Theorem 6.3.8, Mercer's theorem. We take both from them without proof; the delicacy is exactly that $K(\cdot,t)$ need not lie in $\operatorname{ran}\mathcal{K}$ , so part 2 does not apply.

Part 4, the coefficients. Let $y\in L^{2}$ solve $(\ast)$ , that is $y=f+\lambda\mathcal{K}y$ . Take the inner product with $e_{n}$ and use self-adjointness with $\mu_{n}=\lambda_{n}^{-1}$ real:

(y,e_{n}) = (f,e_{n})+\lambda(\mathcal{K}y,e_{n}) = (f,e_{n})+\lambda(y,\mathcal{K}e_{n}) = (f,e_{n})+\frac{\lambda}{\lambda_{n}}(y,e_{n}),

hence

\Bigl(1-\frac{\lambda}{\lambda_{n}}\Bigr)(y,e_{n}) = (f,e_{n}), \qquad\text{that is}\qquad (\lambda_{n}-\lambda)\,(y,e_{n}) = \lambda_{n}\,(f,e_{n}). \tag{C}

This single relation carries both cases, exactly as the corresponding relation (6.4.6) does for Stakgold and Holst.

Part 4, $\lambda$ not a characteristic value. Then $\lambda_{n}-\lambda\neq0$ for every $n$ , so (C) determines $(y,e_{n}) = \lambda_{n}(f,e_{n})/(\lambda_{n}-\lambda)$ . Since $y-f=\lambda\mathcal{K}y$ lies in $\operatorname{ran}\mathcal{K}$ , part 2 gives

y-f = \lambda\mathcal{K}y = \lambda\sum_{n}\lambda_{n}^{-1}(y,e_{n})\,e_{n} = \lambda\sum_{n}\frac{(f,e_{n})}{\lambda_{n}-\lambda}\,e_{n},

which is the stated formula; so a solution, if it exists, is unique. The series converges in $L^{2}$ : because $\lambda_{n}\to\infty$ (or the family is finite), the quantity $\lvert\lambda_{n}/(\lambda_{n}-\lambda)\rvert$ is bounded by some $C$ over all $n$ , so $\sum_{n}\lvert(y,e_{n})\rvert^{2}\le C^{2}\sum_{n}\lvert(f,e_{n})\rvert^{2}\le C^{2}\lVert f\rVert^{2}<\infty$ by Bessel, and the Riesz–Fischer theorem (Orthonormal bases in Hilbert space) makes the series converge. Conversely, define $y$ by the formula and check it solves $(\ast)$ : applying $\mathcal{K}$ term by term (legitimate since $\mathcal{K}$ is bounded and the series converges in $L^{2}$ ),

f+\lambda\mathcal{K}y = f+\lambda\mathcal{K}f+\lambda^{2}\sum_{n}\frac{(f,e_{n})}{\lambda_{n}-\lambda}\lambda_{n}^{-1}e_{n} = f+\lambda\sum_{n}\Bigl[\frac{1}{\lambda_{n}}+\frac{\lambda}{\lambda_{n}(\lambda_{n}-\lambda)}\Bigr](f,e_{n})\,e_{n},

where $\mathcal{K}f=\sum_{n}\lambda_{n}^{-1}(f,e_{n})e_{n}$ by part 2. The bracket is

\frac{(\lambda_{n}-\lambda)+\lambda}{\lambda_{n}(\lambda_{n}-\lambda)} = \frac{1}{\lambda_{n}-\lambda},

so the right-hand side is $y$ . For the resolvent form, one first has to know that $R(\cdot,\cdot;\lambda)$ is a function at all, which the $L^{2}$ convergence of the solution series does not say. It is: the eigenvalues of a Hilbert–Schmidt operator are square-summable, $\sum_{n}\lambda_{n}^{-2}=\sum_{n}\mu_{n}^{2}\le\iint\lvert K\rvert^{2}<\infty$ , and $\lvert\lambda_{n}\rvert\to\infty$ makes $\lvert\lambda_{n}-\lambda\rvert^{-1}$ comparable to $\lvert\lambda_{n}\rvert^{-1}$ for large $n$ , so $\sum_{n}\lvert\lambda_{n}-\lambda\rvert^{-2}<\infty$ ; since the products $e_{n}(x)\overline{e_{n}(t)}$ are orthonormal in $L^{2}([a,b]\times[a,b])$ , the series for $R$ converges there and $R$ is itself a Hilbert–Schmidt kernel. Now insert $(f,e_{n})=\int_{a}^{b}f(t)\overline{e_{n}(t)}\,\mathrm{d}t$ and exchange sum and integral, which that $L^{2}$ convergence on the square permits.

Part 4, $\lambda=\lambda_{p}$ . Group the indices into the resonant ones $m_{1},\dots,m_{k}$ , for which $\lambda_{n}=\lambda_{p}$ , and the rest. For a resonant index, (C) reads $0 = \lambda_{p}(f,e_{m_{i}})$ , and $\lambda_{p}\neq0$ ; so if some $(f,e_{m_{i}})\neq0$ no solution exists. Suppose all $k$ conditions hold. Then (C) leaves $(y,e_{m_{i}})$ completely undetermined — call it $c_{i}$ — and fixes $(y,e_{n})=\lambda_{n}(f,e_{n})/(\lambda_{n}-\lambda_{p})$ for every non-resonant $n$ . Running the computation of the previous paragraph again, with the resonant terms now contributing $c_{i}e_{m_{i}}$ to $y-f$ and $\lambda_{p}\cdot\lambda_{p}^{-1}c_{i}e_{m_{i}}=c_{i}e_{m_{i}}$ to $\lambda_{p}\mathcal{K}y$ , shows that every choice of the $c_{i}$ gives a solution and that these are all of them. Since the $e_{m_{i}}$ are orthonormal and orthogonal to the rest, $\lVert y\rVert^{2}$ is the norm of the $c$ -free part plus $\sum_{i}\lvert c_{i}\rvert^{2}$ , minimised by $c=0$ . $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.4, p. 1072, p. 1073, Exercises 21.4.1 and 21.4.4, pp. 1077–1078; Stakgold & Holst (2011), §6.3, Thm 6.3.5, p. 374, Thms 6.3.7–6.3.8 and the trace inequality (6.3.19), p. 377, Exercise 6.3.2, p. 378, the opening paragraph of §6.3, p. 371, and §6.2, Thm 6.2.1 with its remarks, p. 360, and Example 3, pp. 365–366; Riley, Hobson & Bence (2006), §23.7, p. 817

Hermitian symmetry is not a local convenience but the hypothesis every part rests on. It enters at the very first step of the proof, where it makes $\mathcal{K}$ self-adjoint, and the rest of part 1 is then the spectral theorem for compact self-adjoint operators. Part 2 uses it again through that theorem: the extremal bound $\lVert\mathcal{K}v\rVert\le\lvert\mu_{N+1}\rvert\lVert v\rVert$ is a statement about self-adjoint operators, and $\mathcal{K}$ carries the orthogonal complement of $\operatorname{span}\{e_{1},\dots,e_{N}\}$ into itself only because it is self-adjoint. Part 4 uses it a third time, to move $\mathcal{K}$ across the inner product in deriving (C). Beyond these uses through self-adjointness, symmetry is used twice at the level of the kernel: in the direct reality-and-orthogonality computation of part 1, and in the identification of the Fourier coefficients of $K_{m}$ in part 3 (where the $\mu_{n}$ being real is also needed). Drop the hypothesis and there is nothing left: a real non-symmetric kernel can have complex characteristic values and need not have any eigenfunctions at all. Stakgold and Holst give the Volterra operator as such an example, and point out what survives — one replaces $\mathcal{K}$ by $\mathcal{K}^{*}\mathcal{K}$ and $\mathcal{K}\mathcal{K}^{*}$ and gets a singular-value decomposition instead of an eigenfunction expansion (Singular value decomposition). Arfken, Weber and Harris mostly assume $K$ real and symmetric and then drop the complex conjugates; their Exercise 21.4.1 asks for the Hermitian case, which is what is stated here.
The Hilbert–Schmidt condition is what makes $\mathcal{K}$ compact, and compactness is what gives the discrete spectrum, the finite multiplicities and the accumulation of $\mu_{n}$ at $0$ . It does this by itself, with no help from the interval: the characterisation quoted in Classification of integral equations asks only that $K$ be square-integrable on the product, and a Gaussian kernel $\mathrm{e}^{-x^{2}-t^{2}}$ is square-integrable on the whole plane and so compact there. What the finite interval buys is the rest of the unit — continuity on a compact square, which is what Mercer's theorem and the uniform expansion of part 3 ask for, the convergence of the trace integral $\int K_{m}(x,x)$ , and the Sturm–Liouville reading in the application. Both books warn by example, and in both examples it is the square-integrability that fails. Arfken, Weber and Harris invite the reader to apply the theory to $\varphi(x)=\lambda\int_{0}^{\infty}\mathrm{e}^{-xt}\varphi(t)\,\mathrm{d}t ,$ of which they say that it has an infinite number of solutions in all, one of them being the non-normalisable $\varphi(x)=x^{-1/2}$ at $\lambda=\pi^{-1/2}$ . (They say nothing about how many solutions belong to that one value of $\lambda$ .) Their Note puts the anomaly down to the infinite range of integration, which makes the equation singular; and on p. 1073 they warn that for a singular equation the bilinear expansion need not exist. Stakgold and Holst give $K(x,t)=\mathrm{e}^{-\lvert x-t\rvert}/2$ on the whole line (their §6.2, Example 3, pp. 365–366), which is symmetric and bounded — a Holmgren kernel, with $\lVert\mathcal{K}\rVert=1$ — but not Hilbert–Schmidt, since $\iint\lvert K\rvert^{2}$ diverges; it is not compact, and it has no eigenvalues whatever, the interval $0\le\lambda\le1$ lying instead in its continuous spectrum. They recall the example at the head of §6.3, p. 371, as the standing warning that boundedness and self-adjointness alone buy nothing.
Completeness is of the range, not of the space. Part 2 says $\{e_{n}\}$ spans $\overline{\operatorname{ran}\mathcal{K}}$ . Arfken, Weber and Harris state the completeness they need in exactly this guarded form — a function that can be written as $\mathcal{K}h$ has an eigenfunction expansion — and Riley, Hobson and Bence assert completeness without the qualification. The qualification matters: if $\mathcal{K}$ has a nontrivial null space, there are functions with no expansion in the $e_{n}$ , and the solution formula of part 4 still holds because $y-f$ , not $y$ , is the object being expanded.
$\lambda=0$ is a separate question. Part 4 assumes $\lambda\neq0$ ; the case $\lambda=0$ of $(\ast)$ is the first-kind equation $\mathcal{K}y=f$ , and Stakgold and Holst remark that orthogonality of $f$ to $\ker\mathcal{K}^{*}$ is then necessary but usually not sufficient, because $\operatorname{ran}\mathcal{K}$ need not be closed. Formally the solution would be $y=\sum\lambda_{n}(f,e_{n})e_{n}$ , and the factors $\lambda_{n}\to\infty$ show what goes wrong: the series diverges unless the coefficients of $f$ decay fast enough. That divergence is the ill-posedness of first-kind equations, quantified.
Mercer's hypotheses. Continuity alone is not enough for the $m=1$ bilinear expansion to converge uniformly; Stakgold and Holst say so explicitly before stating Mercer's theorem, whose extra hypothesis is that all but finitely many characteristic values share a sign — automatic for a positive kernel such as a Green's function of a definite operator. For $m\ge2$ the hypotheses are weaker because $K_{m}$ lies in the range of $\mathcal{K}$ .
A bonus from part 3. Setting $x=t$ in the expansion of $K_{m}$ and integrating gives $\sum_{n}\lambda_{n}^{-m}=\int_{a}^{b}K_{m}(x,x)\,\mathrm{d}x$ — valid for $m\ge2$ , where part 3 supplies the expansion unconditionally, and at $m=1$ only under Mercer's hypotheses, which is exactly the caveat Stakgold and Holst attach to their (6.3.19). From it they derive the trace inequality $\lvert\lambda_{1}\rvert^{-1}\le\bigl(\int_{a}^{b}K_{m}(x,x)\,\mathrm{d}x\bigr)^{1/m}$ for a positive operator — a computable upper bound for $\lVert\mathcal{K}\rVert$ , sharpening as $m$ grows.

Example(The kernel min(x, t): eigenfunctions, Mercer's expansion and a solved equation)Original

Take $[a,b]=[0,1]$ and

K(x,t)=\min(x,t),

which is real, symmetric, continuous and bounded, hence Hilbert–Schmidt on the unit square. It is the Green's function of $-\mathrm{d}^{2}/\mathrm{d}x^{2}$ with $y(0)=0$ , $y'(1)=0$ (Green's function for a boundary-value problem). (The same kernel reflected in $x\mapsto1-x$ , namely $1-\max(x,\xi)$ , is the running example of Stakgold and Holst (Stakgold & Holst (2011), §6.2, Example 2, pp. 363–364), with the same characteristic values and the cosines in place of the sines; the Mercer check, the closed form and the degenerate case below are ours.)

The eigenfunctions. Set $\omega_{n}=(n-\tfrac12)\pi$ and $e_{n}(x)=\sqrt2\,\sin(\omega_{n}x)$ for $n=1,2,\dots$ . These are orthonormal on $[0,1]$ : $2\int_{0}^{1}\sin(\omega_{m}x)\sin(\omega_{n}x)\mathrm{d}x=\delta_{mn}$ . To see that they are the eigenfunctions, compute for any $\omega$ with $\cos\omega=0$ :

\int_{0}^{1}\min(x,t)\sin(\omega t)\,\mathrm{d}t = \int_{0}^{x}t\sin(\omega t)\,\mathrm{d}t + x\int_{x}^{1}\sin(\omega t)\,\mathrm{d}t .

The first integral is $-\dfrac{x\cos(\omega x)}{\omega}+\dfrac{\sin(\omega x)}{\omega^{2}}$ by parts; the second is $\dfrac{x\bigl(\cos(\omega x)-\cos\omega\bigr)}{\omega}=\dfrac{x\cos(\omega x)}{\omega}$ because $\cos\omega=0$ . The two $x\cos(\omega x)/\omega$ terms cancel, leaving

\mathcal{K}e_{n} = \omega_{n}^{-2}\,e_{n} .

So the characteristic values are $\lambda_{n}=\omega_{n}^{2}=(n-\tfrac12)^{2}\pi^{2}$ , that is $\pi^{2}/4,\ 9\pi^{2}/4,\ 25\pi^{2}/4,\dots$ , all positive and tending to infinity, with $\lvert\lambda_{1}\rvert^{-1}=4/\pi^{2}=0.4053=\lVert\mathcal{K}\rVert$ . There is no null space here ( $\mathcal{K}u=0$ forces $u=-(\mathcal{K}u)''=0$ ), so by part 2 the $e_{n}$ form an orthonormal basis of $L^{2}(0,1)$ — they are the familiar quarter-wave sine basis.

Mercer's expansion, checked at a point. All the characteristic values are positive and $K$ is continuous, so Mercer's theorem applies:

\min(x,t) = \sum_{n=1}^{\infty}\frac{2\sin(\omega_{n}x)\sin(\omega_{n}t)}{\omega_{n}^{2}} \qquad\text{uniformly on }[0,1]^{2}.

Test it at $x=t=1$ , where $\sin\omega_{n}=(-1)^{n+1}$ so every term is positive and the left side is $1$ :

\sum_{n=1}^{\infty}\frac{2}{\omega_{n}^{2}} = \frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{(n-\tfrac12)^{2}} = \frac{2}{\pi^{2}}\cdot\frac{\pi^{2}}{2} = 1 ,

using $\sum_{n\ge1}(n-\tfrac12)^{-2}=\pi^{2}/2$ (which is $\sum_{m\ \mathrm{odd}}4/m^{2}=4\cdot\pi^{2}/8$ ). Numerically the first $20{,}000$ terms give $0.99999$ , consistent.

Solving an inhomogeneous equation. Take $f(x)=x$ and $\lambda=1$ :

y(x) = x + \int_{0}^{1}\min(x,t)\,y(t)\,\mathrm{d}t .

Since $\lambda=1$ is not among the $\lambda_{n}$ (the smallest is $\pi^{2}/4=2.467$ ), part 4 gives a unique solution. The coefficients of $f$ are

(f,e_{n}) = \sqrt2\int_{0}^{1}x\sin(\omega_{n}x)\,\mathrm{d}x = \sqrt2\Bigl[-\frac{\cos\omega_{n}}{\omega_{n}}+\frac{\sin\omega_{n}}{\omega_{n}^{2}}\Bigr] = \frac{\sqrt2\,(-1)^{n+1}}{\omega_{n}^{2}} ,

y(x) = x + 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin(\omega_{n}x)}{\omega_{n}^{2}\bigl(\omega_{n}^{2}-1\bigr)} .

The same answer in closed form. Differentiate the integral equation twice. Writing the integral as $\int_{0}^{x}ty\,\mathrm{d}t+x\int_{x}^{1}y\,\mathrm{d}t$ , the terms $\pm xy(x)$ cancel on differentiating, leaving $y'(x)=1+\int_{x}^{1}y\,\mathrm{d}t$ and then $y''=-y$ . The equation itself gives $y(0)=0$ and the first derivative gives $y'(1)=1$ . So $y=\sin x/\cos 1$ , and at $x=1$ this is $\tan 1 = 1.5574077$ . The series at $x=1$ is $1+2\sum_{n}\bigl[\omega_{n}^{2}(\omega_{n}^{2}-1)\bigr]^{-1}$ , whose first four terms are $0.27619$ , $0.0021235$ , $0.00026714$ , $0.00006898$ ; summing eleven terms and doubling gives $0.5574026$ , so the series value is $1.5574026$ against $\tan1=1.5574077$ — agreement to the expected truncation error.

Where the poles are. The closed form $y=\sin(\sqrt\lambda\,x)/(\sqrt\lambda\cos\sqrt\lambda)$ , obtained the same way for general $\lambda$ , blows up exactly when $\cos\sqrt\lambda=0$ , that is at $\lambda=\omega_{n}^{2}=\lambda_{n}$ — the poles of the resolvent, exactly as part 4 predicts. And the degenerate case behaves as advertised: at $\lambda=\lambda_{1}=\pi^{2}/4$ the data $f=x$ has $(f,e_{1})=4\sqrt2/\pi^{2}\neq0$ , so there is no solution; replacing $f$ by $\sin(3\pi x/2)=e_{2}/\sqrt2$ , which is orthogonal to $e_{1}$ , makes the equation solvable, with the one-parameter family

y = \frac{e_{2}}{\sqrt2}+\frac{\lambda_{1}}{\sqrt2\,(\lambda_{2}-\lambda_{1})}\,e_{2}+c\,e_{1} = \Bigl(1+\frac{\pi^{2}/4}{2\pi^{2}}\Bigr)\sin\frac{3\pi x}{2}+c\sqrt2\sin\frac{\pi x}{2},

since $\lambda_{2}-\lambda_{1} = (9-1)\pi^{2}/4=2\pi^{2}$ .

ApplicationOriginal

Sturm–Liouville problems are this theorem in disguise. Stakgold and Holst make the connection the point of their §6.3: take a formally self-adjoint differential operator $L$ of order $p$ with real coefficients and $p$ boundary functionals making the problem self-adjoint, build the Green's function $g(x,\xi)$ of $L$ with those boundary conditions, and the eigenvalue problem $Lu=\theta u$ becomes $u=\theta\,\mathcal{G}u$ with $\mathcal{G}$ the integral operator of $g$ . That $g$ is real and symmetric, and $\mathcal{G}$ compact, so parts 1 and 2 apply and deliver the standard facts of Sturm-Liouville eigenvalue problems: real eigenvalues, orthogonal eigenfunctions, and completeness. Their Theorem 6.3.6 states the conclusion in one line: for a self-adjoint boundary-value problem, the eigenfunctions are an orthonormal basis of the whole space. The proof is the integral-equation argument above together with the observation that $\mathcal{G}$ has no null space, because $Lz=u$ with homogeneous boundary data has the unique solution $z=\mathcal{G}u$ . The example worked above is the smallest instance: $\min(x,t)$ is the Green's function, and the quarter-wave sines are the eigenfunctions. The direction of the implication is worth noticing. The discreteness and completeness of the normal modes of a vibrating string, a quantum well or a resonant cavity are not facts about differential equations; they are facts about compact self-adjoint integral operators, inherited by the differential problem through its Green's function.

Bounding a fundamental frequency. The trace identity $\sum_{n}\lambda_{n}^{-m}=\int_{a}^{b}K_{m}(x,x)\,\mathrm{d}x$ — free for $m\ge2$ , and available at $m=1$ here because $\min(x,t)$ is continuous with all characteristic values positive, so Mercer's theorem applies — is a practical tool: it converts a spectral quantity into an integral of the kernel, computable without solving the eigenvalue problem. For the example above, $m=1$ gives $\sum_{n}\omega_{n}^{-2}=\int_{0}^{1}x\,\mathrm{d}x=\tfrac12$ — a closed-form evaluation of $\sum(n-\tfrac12)^{-2}\pi^{-2}$ , matching the Mercer check — and the trace inequality then bounds the lowest eigenvalue of the boundary-value problem by $\lambda_{1}\ge 2$ , against the true value $\pi^{2}/4=2.467$ . Taking $m=2$ gives $\sum_{n}\omega_{n}^{-4}=\int_{0}^{1}K_{2}(x,x)\mathrm{d}x$ and a sharper bound, and this is how a lowest resonant frequency can be estimated from the response kernel alone. It is the integral-equation twin of the Rayleigh–Ritz bound (The Rayleigh-Ritz method).

Resolvent poles, response functions and resonances. Part 4 says the solution of the driven problem is $\sum_{n}\frac{(f,e_{n})}{\lambda_{n}-\lambda}e_{n}$ up to the term $f$ : a sum over modes, each weighted by the overlap of the drive with that mode and by a factor that blows up as the drive parameter approaches the mode's characteristic value. This is the universal shape of a linear response function, and the resolvent kernel $R(x,t;\lambda)$ is the response at $x$ to a unit drive at $t$ . Reading it as a function of $\lambda$ gives a meromorphic function whose poles sit at the eigenvalues and whose residues are the projections onto the eigenspaces — the same object as the eigenfunction expansion of a Green's function (Eigenfunction expansion of Green's functions). The degenerate case of part 4 is resonance: when the drive is tuned to a mode, a bounded response exists only if the drive is orthogonal to that mode, and then the mode's amplitude is undetermined by the equation and must be fixed by something else — initial conditions, or damping that the model left out.

Truncating the expansion: compression and regularisation. Because $\lambda_{n}^{-1}\to0$ , the bilinear expansion of part 3 is dominated by a few terms, and discarding the rest is both the standard way to compress a smooth kernel and the standard way to regularise a first-kind equation: keeping only the terms with $\lvert\lambda_{n}\rvert$ below a cutoff replaces $\mathcal{K}$ by a separable kernel of small rank (Separable kernels) and makes the inversion stable at the cost of resolution. For a non-Hermitian kernel the same construction with singular values in place of eigenvalues is the truncated singular-value decomposition (Singular value decomposition), which is what image deblurring and tomographic reconstruction actually use.

Theorem

Volterra equations

Adapted

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A second-kind Volterra equation with a continuous kernel has exactly one continuous solution for every value of the parameter, so it has no characteristic values at all; the proof is that some power of the iteration map is a contraction, with a factorial in the bound. Abel's weakly singular equation of the first kind is then inverted in closed form.

Prerequisites: Neumann series, Classification of integral equations, The contraction mapping theorem, The Beta function, The Gamma function

Definition(Volterra equations and Abel's equation)AdaptedAdapted from Kreyszig (1978), §5.4, eq. (7), p. 321; Arfken, Weber & Harris (2013), §21.2, Example 21.2.2, eq. (21.27), p. 1055; Stakgold & Holst (2011), §6.1, Example 1, eq. (6.1.1), p. 352

Fix a finite interval $[a,b]$ and let

R = \{(x,t) : a\le t\le x\le b\}

be the closed triangle below the diagonal. A Volterra equation of the second kind is

y(x) = f(x)+\lambda\int_{a}^{x}K(x,t)\,y(t)\,\mathrm{d}t , \qquad a\le x\le b ,

with $f$ and $K$ given, $K$ defined on $R$ , and $\lambda$ a scalar parameter; a Volterra equation of the first kind is

\int_{a}^{x}K(x,t)\,y(t)\,\mathrm{d}t = f(x) .

Only the values of $K$ on $R$ enter, so a Volterra equation is the Fredholm equation on $[a,b]$ whose kernel has been extended by $K(x,t)=0$ for $t>x$ — a kernel that need not be continuous across the diagonal.

Abel's equation is the first-kind Volterra equation with the weakly singular kernel $(x-t)^{-\alpha}$ ,

\int_{0}^{x}\frac{y(t)}{(x-t)^{\alpha}}\,\mathrm{d}t = f(x), \qquad 0<\alpha<1 ,

which Arfken, Weber and Harris call the generalized Abel equation; the classical case is $\alpha=\tfrac12$ . The kernel blows up as $t\uparrow x$ , so this is a singular integral equation. What puts it outside the theorem below, however, is that it is of the first kind: the theorem is about second-kind equations, and the remark after it shows that Abel's kernel in a second-kind equation would cause no trouble at all. Abel's equation is therefore stated and solved separately.

Theorem(Second-kind Volterra equations are always uniquely solvable)AdaptedAdapted from Kreyszig (1978), §5.4, Thm 5.4-2 and Lemma 5.4-3, pp. 321–323; Hassani (2013), §18.1, Thm 18.1.2 and eq. (18.5), pp. 544–545; Stakgold & Holst (2011), §4.4, Example 4, eqs. (4.4.13)–(4.4.15), pp. 251–252

Let $f$ be continuous on $[a,b]$ and let $K$ be continuous on the triangular region $R=\{a\le t\le x\le b\}$ . Then:

Unique solvability, for every $\lambda$ . For every scalar $\lambda$ the equation $y(x) = f(x)+\lambda\int_{a}^{x}K(x,t)\,y(t)\,\mathrm{d}t$ has exactly one continuous solution $y$ on $[a,b]$ . No restriction on $\lambda$ is needed, and the conclusion holds however long the interval is.
No characteristic values. Taking $f\equiv0$ , the only solution of the homogeneous equation is $y\equiv0$ , whatever $\lambda$ may be: a second-kind Volterra equation has no characteristic value, equivalently the operator $1-\lambda\mathcal{K}$ is a bijection of $C[a,b]$ for every $\lambda$ , with bounded inverse $\sum_{m\ge0}\lambda^{m}\mathcal{K}^{m}$ supplied by part 3, so the spectrum of the Volterra operator $\mathcal{K}$ contains no nonzero point.
The Neumann series converges everywhere. With $c=\max_{R}\lvert K\rvert$ , the iterated kernels obey $\lvert K_{m}(x,t)\rvert \le c^{m}\,\frac{(x-t)^{m-1}}{(m-1)!} \qquad\text{on } R ,$ so the Neumann series of Neumann series converges uniformly on $[a,b]$ for every $\lambda$ , and the unique solution is $y(x) = f(x)+\sum_{m=1}^{\infty}\lambda^{m}\int_{a}^{x}K_{m}(x,t)\,f(t)\,\mathrm{d}t , \qquad K_{m}(x,t)=\int_{t}^{x}K(x,s)K_{m-1}(s,t)\,\mathrm{d}s .$ The iterated kernels of Classification of integral equations are defined with the full range $\int_{a}^{b}$ ; the limits $t$ to $x$ written here are not a new definition but the same one evaluated, since for the Volterra kernel extended by $K=0$ above the diagonal the integrand vanishes for $s<t$ and for $s>x$ . Step 5 of the proof carries out that reconciliation.

Hassani states parts 1 and 2 as his Theorem 18.1.2. He derives them from the pointwise bound

\bigl\lvert(\lambda\mathcal{K})^{m}[u](x)\bigr\rvert \le \lvert\lambda\rvert^{m}c^{m}\lVert u\rVert_{\infty}\frac{(x-a)^{m}}{m!},

with $c=\max_{R}\lvert K\rvert$ (his $M$ ), which he then replaces by $x=b$ and calls an operator-norm inequality; as printed, his right-hand side still carries the factor $\lVert u\rVert_{\infty}$ , which for an operator norm has to be divided out, giving $\lVert(\lambda\mathcal{K})^{m}\rVert\le\lvert\lambda\rvert^{m}c^{m}(b-a)^{m}/m!$ . Either way $\sum_{m}\lVert(\lambda\mathcal{K})^{m}\rVert$ converges, by comparison with $\mathrm{e}^{c\lvert\lambda\rvert(b-a)}$ , for every $\lambda$ , and he identifies the sum of the series with the inverse of $1-\lambda\mathcal{K}$ . Kreyszig states part 1 as his Theorem 5.4-2, with the same factorial estimate; Stakgold and Holst prove the corresponding statement for the nonlinear Volterra equation $u(x)=f(x)+\int_{a}^{x}k(x,y,u(y))\,\mathrm{d}y$ under a Lipschitz condition $\lvert k(x,y,z_{2})-k(x,y,z_{1})\rvert\le M\lvert z_{1}-z_{2}\rvert$ on $k$ , of which the linear case is the choice $k(x,y,z)=\lambda K(x,y)z$ .

Idea of the proofOriginal

For a Fredholm equation the iteration map $T u = f+\lambda\mathcal{K}u$ shrinks distances by at most $\lvert\lambda\rvert c(b-a)$ per step, and that is a contraction only if $\lambda$ is small. For a Volterra equation the integral runs only up to $x$ , so the first step loses a factor $(x-a)$ rather than $(b-a)$ ; the second step integrates that and loses $(x-a)^{2}/2$ ; after $m$ steps the factor is $(x-a)^{m}/m!$ . A factorial beats any geometric growth, so however large $\lvert\lambda\rvert$ is, some power $T^{m}$ is a contraction. A map whose $m$ -th power is a contraction has a unique fixed point too — that is a short lemma — and we are done. The triangular support of the kernel is thus not a technical convenience: it is the whole reason the Volterra theory is unconditional, and it is the same mechanism that makes the Dyson series converge where the Born series need not.

ProofAdaptedAdapted from Kreyszig (1978), §5.4, eqs. (8)–(10) and Lemma 5.4-3 with its proof, pp. 321–323; Stakgold & Holst (2011), §4.4, Example 4, eqs. (4.4.12)–(4.4.14), pp. 251–252

Work in the complete metric space $X=C[a,b]$ with $d(u,w)=\max_{[a,b]}\lvert u-w\rvert$ , and define

(Tu)(x) = f(x)+\lambda\int_{a}^{x}K(x,t)u(t)\,\mathrm{d}t .

$T$ maps $X$ into $X$ : the integrand is continuous on $R$ , which is closed and bounded, so $x\mapsto\int_{a}^{x}K(x,t)u(t)\,\mathrm{d}t$ is continuous. Fixed points of $T$ are exactly the continuous solutions. Since $K$ is continuous on the compact set $R$ , it is bounded there; put $c=\max_{R}\lvert K\rvert$ and $\kappa=\lvert\lambda\rvert c$ .

Step 1: the factorial estimate. We claim that for all $u,w\in X$ , all $m\ge1$ and all $x\in[a,b]$ ,

\bigl\lvert (T^{m}u)(x)-(T^{m}w)(x)\bigr\rvert \;\le\; \kappa^{m}\,\frac{(x-a)^{m}}{m!}\,d(u,w). \tag{E}

For $m=1$ ,

\lvert (Tu)(x)-(Tw)(x)\rvert = \Bigl\lvert\lambda\int_{a}^{x}K(x,t)[u(t)-w(t)]\,\mathrm{d}t\Bigr\rvert \le \kappa\int_{a}^{x}\lvert u-w\rvert\,\mathrm{d}t \le \kappa\,(x-a)\,d(u,w),

using $\lvert K\rvert\le c$ on $R$ and bounding $\lvert u-w\rvert$ by its maximum. Suppose (E) holds for some $m\ge1$ . Applying $T$ once more and inserting the inductive bound inside the integral,

\bigl\lvert (T^{m+1}u)(x)-(T^{m+1}w)(x)\bigr\rvert \le \kappa\int_{a}^{x}\bigl\lvert (T^{m}u)(t)-(T^{m}w)(t)\bigr\rvert\,\mathrm{d}t \le \kappa^{m+1}\,\frac{d(u,w)}{m!}\int_{a}^{x}(t-a)^{m}\,\mathrm{d}t ,

and $\int_{a}^{x}(t-a)^{m}\mathrm{d}t = (x-a)^{m+1}/(m+1)$ , which turns $1/m!$ into $1/(m+1)!$ . That closes the induction. Taking the maximum over $x$ and using $x-a\le b-a$ ,

d(T^{m}u,T^{m}w) \le \alpha_{m}\,d(u,w), \qquad \alpha_{m} = \frac{\bigl[\lvert\lambda\rvert c\,(b-a)\bigr]^{m}}{m!} .

Step 2: some power is a contraction. For fixed $\lambda$ the numbers $\alpha_{m}$ are the terms of the convergent series for $\mathrm{e}^{\lvert\lambda\rvert c(b-a)}$ , so $\alpha_{m}\to0$ ; choose and fix an $m$ with $\alpha_{m}<1$ . Then $B=T^{m}$ is a contraction on $X$ . This is where the Volterra structure is used, and it is the only place: nothing is assumed about the size of $\lambda$ .

Step 3: a map one of whose powers is a contraction has a unique fixed point. This is Kreyszig's Lemma 5.4-3, which we prove. Let $B=T^{m}$ be a contraction with constant $\alpha<1$ on the complete space $X$ , and note that $T$ is continuous, being Lipschitz by (E) with $m=1$ . By the contraction mapping theorem, $B$ has a unique fixed point $y$ , and $B^{n}u_{0}\to y$ for every $u_{0}\in X$ .

$y$ is a fixed point of $T$ . Apply the convergence statement to the starting point $Tu_{0}$ : since $B$ and $T$ are both powers of $T$ they commute, so

B^{n}(Tu_{0}) = T\bigl(B^{n}u_{0}\bigr) \longrightarrow Ty

by continuity of $T$ , while $B^{n}(Tu_{0})\to y$ because $y$ attracts every starting point. A sequence has one limit, so $Ty=y$ .

It is the only one. If $Tz=z$ then $Bz=T^{m}z=z$ , so $z$ is a fixed point of $B$ and therefore $z=y$ .

Every iterate of $T$ converges to $y$ , not only the iterates of $B$ . Given $u_{0}\in X$ and $n\ge0$ , write $n=qm+j$ with $0\le j<m$ ; then $T^{n}u_{0}=B^{q}\bigl(T^{j}u_{0}\bigr)$ , because both are $T$ applied $n$ times. The contraction mapping theorem gives $B^{q}v\to y$ as $q\to\infty$ for each of the $m$ starting points $v=T^{j}u_{0}$ , $j=0,\dots,m-1$ , so each of the $m$ arithmetic subsequences of $(T^{n}u_{0})_{n}$ tends to $y$ ; since they cover all $n$ , $T^{n}u_{0}\to y$ . This is the form in which step 5 uses the conclusion.

Step 4: parts 1 and 2. Step 3 applied to our $T$ gives exactly one continuous $y$ with $Ty=y$ , which is part 1, valid for every $\lambda$ by step 2. For part 2 take $f\equiv0$ : then $y\equiv0$ is a fixed point, so by uniqueness it is the only one, and no $\lambda$ admits a nonzero solution of $y=\lambda\mathcal{K}y$ . In operator terms, part 1 says that $(1-\lambda\mathcal{K})y=f$ has a solution for every $f\in C[a,b]$ , so $1-\lambda\mathcal{K}$ is onto $C[a,b]$ , and the uniqueness says it is injective; it is therefore a bijection of $C[a,b]$ for every $\lambda$ . Bijectivity alone would need the open mapping theorem to make the inverse bounded, and we do not appeal to it: part 3 exhibits the inverse outright as the sum of the Neumann series $\sum_{m\ge0}\lambda^{m}\mathcal{K}^{m}$ , whose terms are bounded in operator norm by $\lVert(\lambda\mathcal{K})^{m}\rVert\le\bigl[\lvert\lambda\rvert c(b-a)\bigr]^{m}/m!$ — this is estimate (E) read with $w=0$ and $f\equiv0$ , where $T$ is just $\lambda\mathcal{K}$ — so the series converges in operator norm and $\lVert(1-\lambda\mathcal{K})^{-1}\rVert\le\mathrm{e}^{\lvert\lambda\rvert c(b-a)}$ .

Step 5: part 3. The bound on $K_{m}$ is the same induction as (E), run on the kernels. The base case is $m=1$ : $\lvert K_{1}\rvert=\lvert K\rvert\le c$ , which is the claim with $(x-t)^{0}/0!=1$ . For the inductive step take $m\ge2$ , so that the previous index $m-1$ is at least $1$ and the factorial $(m-2)!$ in the inductive hypothesis is defined. Suppose, then, that $\lvert K_{m-1}(s,t)\rvert\le c^{m-1}(s-t)^{m-2}/(m-2)!$ for $t\le s$ . Then for $t\le x$

\lvert K_{m}(x,t)\rvert \le \int_{t}^{x}\lvert K(x,s)\rvert\,\lvert K_{m-1}(s,t)\rvert\,\mathrm{d}s \le \frac{c^{m}}{(m-2)!}\int_{t}^{x}(s-t)^{m-2}\,\mathrm{d}s = c^{m}\frac{(x-t)^{m-1}}{(m-1)!},

where the limits of the inner integral are $t$ to $x$ because $K_{m-1}(s,t)=0$ for $s<t$ and $K(x,s)=0$ for $s>x$ . Integrating that bound over $t$ gives, for every $x\in[a,b]$ ,

\int_{a}^{x}\lvert K_{m}(x,t)\rvert\,\mathrm{d}t \le \frac{c^{m}}{(m-1)!}\int_{a}^{x}(x-t)^{m-1}\,\mathrm{d}t = c^{m}\,\frac{(x-a)^{m}}{m!} \le c^{m}\,\frac{(b-a)^{m}}{m!} ,

so the $m$ -th term of the Neumann series is bounded on $[a,b]$ by

\lvert\lambda\rvert^{m}\max\lvert f\rvert\int_{a}^{x}\lvert K_{m}(x,t)\rvert\,\mathrm{d}t \;\le\; \max\lvert f\rvert\;\frac{\bigl[\lvert\lambda\rvert c\,(b-a)\bigr]^{m}}{m!} ,

the factor $\lvert\lambda\rvert^{m}$ having been absorbed into the bracket. These are the terms of a convergent exponential series, so by the Weierstrass criterion the Neumann series converges uniformly for every $\lambda$ . Its sum is the limit of the iterates $T^{n}f$ (the identification of iterates with partial sums is proved in Neumann series and uses nothing about the limits of integration), and those converge to $y$ by the last paragraph of step 3. So the series represents the unique solution. $\square$

Theorem(Abel's equation)AdaptedAdapted from Arfken, Weber & Harris (2013), §21.2, Example 21.2.2, eqs. (21.27)–(21.29), pp. 1055–1056, and Exercise 21.2.6, p. 1061

Fix $0<\alpha<1$ and consider the generalized Abel equation

\int_{0}^{x}\frac{y(t)}{(x-t)^{\alpha}}\,\mathrm{d}t = f(x), \qquad x>0 .

Arfken, Weber and Harris obtain the solution

y(x) = \frac{\sin\pi\alpha}{\pi}\,\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x}\frac{f(t)}{(x-t)^{1-\alpha}}\,\mathrm{d}t .

The two steps behind it, each of independent use, are:

Integrated form. If $y$ is a solution, then for every $z>0$ $\int_{0}^{z} y(t)\,\mathrm{d}t \;=\; \frac{\sin\pi\alpha}{\pi}\int_{0}^{z}\frac{f(x)}{(z-x)^{1-\alpha}}\,\mathrm{d}x .$ This holds whenever the double integral converges absolutely, and the proof below gets that from $\int_{0}^{z}\lvert y\rvert<\infty$ alone: integrability of $y$ on $[0,z]$ is all that is used, so continuity of $y$ — which any classical solution has anyway — is far more than the step needs.
Differentiation. Differentiating step 1 in $z$ gives the formula, at every point where the right-hand side of step 1 is differentiable.

The beta integral that drives it (The Beta function) is

\int_{t}^{z}\frac{\mathrm{d}x}{(z-x)^{1-\alpha}(x-t)^{\alpha}} \;=\; B(1-\alpha,\alpha) \;=\; \Gamma(\alpha)\Gamma(1-\alpha) \;=\; \frac{\pi}{\sin\pi\alpha},

the same value for every $0\le t<z$ ; the middle equality is $B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)$ at $p+q=1$ , and the last is Euler's reflection formula (The Gamma function), which is where the constant $\sin\pi\alpha$ in the inversion formula comes from.

Idea of the proofOriginal

The kernel $(x-t)^{-\alpha}$ depends only on $x-t$ , so the equation is a convolution and the natural moves are the ones that diagonalise convolution. Arfken, Weber and Harris take the Laplace transform, where the equation becomes multiplication by $\Gamma(1-\alpha)s^{\alpha-1}$ and inverting means multiplying by $s^{1-\alpha}$ — a transform that cannot be inverted directly, so they first divide by $s$ (which integrates) and then differentiate at the end. The route taken below does the same thing without transforms: convolving with the complementary power $(z-x)^{\alpha-1}$ turns the composite kernel into a beta integral whose value is a constant, so the two fractional powers annihilate each other and leave the plain integral of $y$ . The final differentiation is the price of the division by $s$ , in either route.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §21.2, Exercise 21.2.6 and its note, p. 1061, and eq. (21.29), p. 1056

The beta integral. Fix $0\le t<z$ and substitute $x = t+(z-t)s$ , $\mathrm{d}x=(z-t)\,\mathrm{d}s$ , so that $z-x=(z-t)(1-s)$ and $x-t=(z-t)s$ :

\int_{t}^{z}\frac{\mathrm{d}x}{(z-x)^{1-\alpha}(x-t)^{\alpha}} = (z-t)^{\alpha-1+(-\alpha)+1}\int_{0}^{1}(1-s)^{\alpha-1}s^{-\alpha}\,\mathrm{d}s = B(1-\alpha,\alpha).

The powers of $(z-t)$ cancel exactly, which is the point of choosing the exponent $\alpha-1$ for the multiplier. The remaining integral is the beta function $B(p,q)=\int_{0}^{1}s^{p-1}(1-s)^{q-1}\mathrm{d}s$ (The Beta function) at $p=1-\alpha$ , $q=\alpha$ , finite because both exponents $-\alpha$ and $\alpha-1$ exceed $-1$ for $0<\alpha<1$ ; and $B(1-\alpha,\alpha)=\Gamma(1-\alpha)\Gamma(\alpha)/\Gamma(1)=\Gamma(\alpha)\Gamma(1-\alpha)$ , which is $\pi/\sin\pi\alpha$ by Euler's reflection formula $\Gamma(\alpha)\Gamma(1-\alpha)=\pi/\sin\pi\alpha$ (The Gamma function). (Numerically, at $\alpha=\tfrac14$ both sides are $4.44288$ ; at $\alpha=\tfrac12$ both are $\pi$ .)

Step 1. Multiply the equation, written at the point $x$ , by $(z-x)^{\alpha-1}$ and integrate over $0\le x\le z$ :

\int_{0}^{z}\frac{f(x)}{(z-x)^{1-\alpha}}\,\mathrm{d}x = \int_{0}^{z}\frac{1}{(z-x)^{1-\alpha}}\int_{0}^{x}\frac{y(t)}{(x-t)^{\alpha}}\,\mathrm{d}t\,\mathrm{d}x .

The double integral is over the triangle $0\le t\le x\le z$ , on which the integrand is $y(t)$ times a positive function; it converges absolutely, because integrating $\lvert y(t)\rvert$ last and using the beta integral gives the finite value $\frac{\pi}{\sin\pi\alpha}\int_{0}^{z}\lvert y\rvert$ . So the order may be exchanged (Fubini). Doing the $x$ -integral first, with $t$ fixed and $x$ ranging over $[t,z]$ :

\int_{0}^{z}\frac{f(x)}{(z-x)^{1-\alpha}}\,\mathrm{d}x = \int_{0}^{z} y(t)\Bigl[\int_{t}^{z}\frac{\mathrm{d}x}{(z-x)^{1-\alpha}(x-t)^{\alpha}}\Bigr]\mathrm{d}t = \frac{\pi}{\sin\pi\alpha}\int_{0}^{z}y(t)\,\mathrm{d}t ,

the bracket being the constant just computed. Rearranging gives step 1 of the statement.

Step 2. The left-hand side of step 1 is $\int_{0}^{z}y$ . Wherever the right-hand side is differentiable in $z$ , differentiating both sides gives $y(z)$ on the left, which is the formula. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Kreyszig (1978), §5.4, pp. 321–323; Arfken, Weber & Harris (2013), §21.2, p. 1056 and p. 1061, and §21.3, p. 1067 (the Volterra Neumann series converges at every parameter value for a square-integrable kernel); Stakgold & Holst (2011), §4.4, Example 4, p. 252, and §6.1, Example 1, p. 352; Hassani (2013), §18.1, p. 545

Continuity of $K$ on the closed triangle is used twice in the Volterra proof: to keep $T$ inside $C[a,b]$ , and to produce the bound $c$ . The bound is all the estimate (E) needs, so continuity can be weakened to boundedness plus enough measurability for the integrals to exist — but then the solution need not be continuous and one works in $L^{\infty}$ or $L^{2}$ instead. Boundedness itself is a convenience of this proof rather than a limit on the conclusion: Arfken, Weber and Harris record (p. 1067) that square-integrability of the kernel already suffices, in the Volterra case, for the Neumann series to converge at every value of $\lambda$ . Abel's kernel shows that even square-integrability is not where the conclusion stops — but it is not an instance of Arfken's criterion, because it fails that criterion too: at $\alpha=\tfrac12$ the kernel $(x-t)^{-1/2}$ is unbounded on the triangle and not square-integrable there, since $\iint_{R}(x-t)^{-1}\mathrm{d}t\,\mathrm{d}x$ diverges logarithmically at the diagonal. It is a case lying past both hypotheses, for which the conclusion happens to be true all the same, and the way to see that is to compute rather than to invoke a criterion. The beta integral computed below gives, at $\alpha=\tfrac12$ , the constant second iterate $K_{2}(x,t)=B(\tfrac12,\tfrac12)=\pi$ , and the same computation gives $K_{m}(x,t)=\pi^{m/2}(x-t)^{m/2-1}/\Gamma(m/2)$ , whose integral $\pi^{m/2}x^{m/2}/\Gamma(\tfrac{m}{2}+1)$ is summable for every $\lambda$ ; so the second-kind equation with Abel's kernel is uniquely solvable for every $\lambda$ after all. For $f\equiv1$ the sum is $\sum_{m\ge0}(\lambda\sqrt{\pi x})^{m}/\Gamma(\tfrac{m}{2}+1)$ , which we checked numerically against the equation at $\lambda=3$ . What takes Abel's equation outside the theorem is therefore not the singularity of its kernel but that it is of the first kind, with no term outside the integral, and so has a completely different solvability theory — which is why it is stated and proved separately below.
The triangular support is what produces the factorials. For the Fredholm equation the same induction gives only $(b-a)^{m}$ , and the convergence condition of Neumann series is then genuinely restrictive. Kreyszig and Riley–Hobson–Bence both note that a Volterra equation is a Fredholm equation with a kernel vanishing above the diagonal; the content of this unit is that this particular family of Fredholm kernels has empty nonzero spectrum.
Arbitrarily long intervals. Stakgold and Holst stress that their conclusion survives however large $b$ is taken, provided their Lipschitz bound continues to hold there. This is because the hypothesis never involves the size of $\lvert\lambda\rvert c(b-a)$ , only that it is finite; the price is that the chosen $m$ in step 2 grows with it, so the rate of convergence degrades even though convergence does not fail.
Abel: hypotheses on $f$ . Arfken, Weber and Harris state the inversion formula without spelling out conditions on $f$ , and the proof above shows why some are needed: step 1 is an identity about any solution, but step 2 requires the function $z\mapsto\int_{0}^{z}f(x)(z-x)^{\alpha-1}\mathrm{d}x$ to be differentiable, and the formula then gives $y$ only where that holds. The argument also shows the uniqueness and the obstruction at once: step 1 determines $\int_{0}^{z}y$ for every $z$ , hence $y$ as a distribution, so a solution is unique if it exists; but it exists only if that primitive is differentiable, which is a smoothness condition on $f$ . Stakgold and Holst, discussing Abel's equation as their opening example, make the same point — read purely mathematically, $f$ may have to satisfy conditions for a solution to exist — and add that the physical interpretation imposes a further restriction, namely that the solution generate a feasible arclength, which forces $f(y)\ge(2y/g)^{1/2}$ : the prescribed time of descent cannot beat free fall.
$0<\alpha<1$ is used in the convergence of the beta integral (both exponents must exceed $-1$ ) and in $\sin\pi\alpha\neq0$ . At $\alpha=0$ the equation is $\int_{0}^{x}y=f$ , inverted by one differentiation; at $\alpha\ge1$ the kernel is not locally integrable and the left-hand side is generally infinite.

Example(A Volterra equation summed by iteration, and an Abel inversion)Original

Part 1: iteration for every $\lambda$ . Solve

y(x) = x + 2\int_{0}^{x}\mathrm{e}^{\,x-t}\,y(t)\,\mathrm{d}t

on $[0,b]$ for any $b>0$ . Here $f(x)=x$ , $K(x,t)=\mathrm{e}^{x-t}$ is continuous on the triangle, and $\lambda=2$ — well outside anything the Fredholm condition would allow on a long interval, yet part 1 guarantees a unique continuous solution.

By differentiation. Write the equation as $y(x)=x+2\mathrm{e}^{x}\int_{0}^{x}\mathrm{e}^{-t}y(t)\,\mathrm{d}t$ and differentiate:

y'(x) = 1+2\mathrm{e}^{x}\!\int_{0}^{x}\!\mathrm{e}^{-t}y\,\mathrm{d}t + 2y(x) = 1+\bigl(y(x)-x\bigr)+2y(x) = 3y(x)-x+1 ,

where the middle step used the integral equation itself to replace the integral term. With $y(0)=0$ (set $x=0$ in the equation), the linear ODE $y'-3y=1-x$ has particular solution $\tfrac{x}{3}-\tfrac29$ and hence

y(x) = \tfrac29\,\mathrm{e}^{3x}+\tfrac{x}{3}-\tfrac29 .

Checked at $x=1$ : $y(1)=\tfrac29\mathrm{e}^{3}+\tfrac13-\tfrac29 = 4.5745638$ , and substituting this $y$ back into the right-hand side of the integral equation reproduces it to $17$ significant digits (numerical quadrature at $x=0.3,0.7,1.0,1.5$ ).

By iteration. Start from $y_{0}=f=x$ . Using $\int_{0}^{x}t\mathrm{e}^{-t}\mathrm{d}t = 1-(x+1)\mathrm{e}^{-x}$ ,

y_{1}(x) = x+2\mathrm{e}^{x}\bigl[1-(x+1)\mathrm{e}^{-x}\bigr] = 2\mathrm{e}^{x}-x-2 ,

and continuing,

y_{2}(x)=4x\mathrm{e}^{x}-6\mathrm{e}^{x}+3x+6, \qquad y_{3}(x)=4x^{2}\mathrm{e}^{x}-12x\mathrm{e}^{x}+18\mathrm{e}^{x}-5x-18 .

At $x=1$ these give $1.000000$ , $2.436564$ , $3.563436$ , $4.182818$ , and the next iterate $4.446551$ , climbing to $4.574564$ : convergence, as promised, with no smallness condition on $\lambda$ . The factorial in part 3 is visible in the rate — the gaps to the limit are $3.575$ , $2.138$ , $1.011$ , $0.392$ , $0.128$ , shrinking by factors $0.60$ , $0.47$ , $0.39$ , $0.33$ — a ratio that keeps falling, as a factorial bound requires, rather than settling at a fixed geometric value. The crude estimate of step 2 does not see this: with $b=1$ the constant is $\kappa=\lvert\lambda\rvert c=2\mathrm{e}=5.437$ , and $\alpha_{m}=\kappa^{m}/m!$ rises to $39.6$ at $m=5$ and first drops below $1$ only at $m=13$ ( $\alpha_{13}=0.582$ ). Convergence from the start is therefore much better than the proof promises; the proof only has to establish that it happens eventually.

Part 2: an Abel inversion. Solve

\int_{0}^{x}\frac{y(t)}{\sqrt{x-t}}\,\mathrm{d}t = x ,

that is $\alpha=\tfrac12$ , $f(x)=x$ . By the formula, with $\sin(\pi/2)=1$ ,

y(x) = \frac{1}{\pi}\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x}\frac{t}{\sqrt{x-t}}\,\mathrm{d}t .

Substituting $t=xs$ gives $\int_{0}^{x}t(x-t)^{-1/2}\mathrm{d}t = x^{3/2}\int_{0}^{1}s(1-s)^{-1/2}\mathrm{d}s = x^{3/2}B(2,\tfrac12)$ , and

B(2,\tfrac12) = \frac{\Gamma(2)\Gamma(\tfrac12)}{\Gamma(\tfrac52)} = \frac{\sqrt\pi}{\tfrac34\sqrt\pi} = \frac43 ,

so $y(x)=\frac{1}{\pi}\cdot\frac43\cdot\frac32\sqrt{x} = \frac{2}{\pi}\sqrt{x}$ .

Check, independently of the formula. Substitute $y(t)=\frac{2}{\pi}\sqrt t$ and put $t=xs$ :

\int_{0}^{x}\frac{2\sqrt t/\pi}{\sqrt{x-t}}\,\mathrm{d}t = \frac{2x}{\pi}\int_{0}^{1}\frac{\sqrt s}{\sqrt{1-s}}\,\mathrm{d}s = \frac{2x}{\pi}B\bigl(\tfrac32,\tfrac12\bigr) = \frac{2x}{\pi}\cdot\frac{\Gamma(\tfrac32)\Gamma(\tfrac12)}{\Gamma(2)} = \frac{2x}{\pi}\cdot\frac{\tfrac12\sqrt\pi\cdot\sqrt\pi}{1} = x ,

as required. Both parts of the example were verified symbolically.

ApplicationOriginal

The tautochrone, which is where Abel's equation came from. A bead slides without friction down a wire in a vertical plane, starting from rest at height $y$ and ending at the origin. Conservation of energy gives speed $\sqrt{2g(y-\eta)}$ at height $\eta$ , so the time of descent is

T(y) = \int_{0}^{y}\frac{\mathrm{d}s}{\sqrt{2g(y-\eta)}} = \int_{0}^{y}\frac{u(\eta)}{\sqrt{y-\eta}}\,\mathrm{d}\eta , \qquad u(\eta) = \frac{1}{\sqrt{2g}}\,\frac{\mathrm{d}s}{\mathrm{d}\eta} ,

with $s(\eta)$ the arclength from the bottom: Abel's equation with $\alpha=\tfrac12$ , as Stakgold and Holst present it in their opening example. Abel's question was the inverse one — prescribe the descent time and find the wire — and the inversion formula answers it. Ask for a tautochrone, a curve whose descent time is the same from every height, so $T(y)\equiv T$ . Then

u(y) = \frac{1}{\pi}\frac{\mathrm{d}}{\mathrm{d}y}\int_{0}^{y}\frac{T\,\mathrm{d}\eta}{\sqrt{y-\eta}} = \frac{1}{\pi}\frac{\mathrm{d}}{\mathrm{d}y}\bigl(2T\sqrt{y}\bigr) = \frac{T}{\pi\sqrt{y}} ,

and therefore

\Bigl(\frac{\mathrm{d}s}{\mathrm{d}y}\Bigr)^{2} = 1+\Bigl(\frac{\mathrm{d}x}{\mathrm{d}y}\Bigr)^{2} = \frac{2gT^{2}}{\pi^{2}y} = \frac{a}{y}, \qquad a=\frac{2gT^{2}}{\pi^{2}} ,

so $\mathrm{d}x/\mathrm{d}y=\sqrt{a/y-1}$ . That is the differential equation of a cycloid generated by a circle of diameter $a$ , and reading the relation backwards, $T=\pi\sqrt{a/(2g)}$ is the descent time from every starting height. A bead released on the arc swings to the lowest point in the time $T$ and back out to the other side in the same time, so $T$ is a quarter of the full period: the cycloidal pendulum has period

4T = 4\pi\sqrt{\frac{a}{2g}} = 2\pi\sqrt{\frac{2a}{g}} ,

which is the period $2\pi\sqrt{L/g}$ of a simple pendulum of length $L=2a$ , four times the radius $a/2$ of the generating circle. With $a=1\ \mathrm{m}$ and $g=9.80665\ \mathrm{m\,s^{-2}}$ that is $T=0.70937\ \mathrm{s}$ and $4T=2.83749\ \mathrm{s}$ , independently of the amplitude — Huygens's isochronous pendulum, recovered from an integral equation. The same inversion, with $T(y)$ measured rather than prescribed, is how a profile is reconstructed from timing data.

Abel inversion in diagnostics and astronomy. The $\alpha=\tfrac12$ kernel is not confined to mechanics. Any measurement that integrates a spherically or cylindrically symmetric quantity along chords produces it: the line-of-sight emission of an axisymmetric plasma column at impact parameter $r$ is $\int_{r}^{R}\varepsilon(\rho)\,2\rho\,\mathrm{d}\rho/\sqrt{\rho^{2}-r^{2}}$ , which is Abel's equation in the variable $\rho^{2}$ . Inverting it to get the local emissivity $\varepsilon(\rho)$ from the measured profile is called Abel inversion, and it is standard in plasma spectroscopy, flame diagnostics, shock-tube interferometry and the deprojection of the surface brightness of a spherical star cluster or galaxy. The remark above is the practical warning attached to all of these: the inversion involves a derivative of the data, so measurement noise is differentiated and amplified, and real codes smooth or fit the data before applying the formula.

Why Volterra structure makes evolution problems safe. Part 1 is the reason causal problems are better behaved than boundary-value problems. An initial-value problem for a linear ODE converts into a linear second-kind Volterra equation (Classification of integral equations), and part 1 applies to it as it stands: exactly one solution, on the whole interval, with no smallness condition. For a general initial-value problem $y'=F(x,y)$ , $y(a)=y_{0}$ , the equivalent Volterra equation $y(x)=y_{0}+\int_{a}^{x}F(t,y(t))\,\mathrm{d}t$ is nonlinear, so part 1 — a theorem about linear equations — does not reach it; what does is the Lipschitz version of the same contraction argument, which is the one Stakgold and Holst carry out and which is the Picard–Lindelöf theorem (Existence and uniqueness of solutions). The mechanism is identical in the two cases, and it is the triangular range of integration in both. The same mechanism is at work in the Dyson series of time-dependent perturbation theory and in the retarded response kernels of linear-response theory: time ordering restricts the integral to the past, the factorials appear, and the series converges for every coupling. Fredholm structure, by contrast, means the data at one end are coupled to the data at the other, characteristic values exist, and solvability becomes conditional — the Fredholm alternative (Separable kernels, The Fredholm alternative and modified Green's functions). Hassani organises his chapter around exactly this contrast, and says he takes up the Volterra case first for being the easier one; part 2 of the statement above is the precise sense in which it is easier.

Volterra equations of the second kind in viscoelasticity and renewal theory. The constitutive law of a linear viscoelastic material, $\sigma(x) = \int_{0}^{x}G(x-t)\dot\varepsilon(t)\,\mathrm{d}t$ with a relaxation modulus $G$ , is a Volterra equation whenever one of the histories is to be recovered from the other; likewise the renewal equation $m(x)=F(x)+\int_{0}^{x}m(x-t)\,\mathrm{d}F(t)$ of reliability theory, and the age-structured population balances of epidemiology. In every case part 1 says the model is well posed for any loading and any kernel strength, and part 3 says that marching forward in small steps — which is what a numerical Volterra solver does — converges, because each step only ever uses values already computed.