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This square-root-kernel integral equation is inverted by integrating $g(x)/\sqrt{x-z}$ over $z<x<b$ and interchanging integrations. The inner integral over $z<x<y$ equals $\pi$, so ${\int }_z^{b}g(x)/\sqrt{x-z}dx=\pi {\int }_z^{b}f(y)dy$. Differentiating recovers $f(z)=-{\pi }^{-1}d[{\int }_z^{b}g(x)/\sqrt{x-z}dx]/dz$. The changes of variables $x=p^2$, $y={\eta }^2$ put the angular-ray equation into this form.

Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

The fundamental theorem of calculus says that accumulation and local change undo one another, under appropriate regularity conditions. If $f$ is a continuous function on $[a,b]$ andthen $A$ is differentiable for $a<x<b$ and $A^{\prime }(x)=f(x)$. Conversely, if $G$ is an antiderivative of $f$ on this interval, continuous at its endpoints, thenThe first statement turns an integral into a function with a known derivative; the second turns an entire accumulation into two endpoint values.

For $f(t)=1+2t$, accumulation from zero gives $A(x)=x+x^2$. At $x=2$, its derivative is $A^{\prime }(2)=5$, exactly the current value $f(2)$. The accumulated value there is $A(2)=6$, and by $x=3$ it has reached $A(3)=12$. The accumulated quantity and its rate are different functions, linked by differentiation.

This connection makes the theorem useful far beyond geometric area. It recovers displacement from velocity, gives a probability from a probability density function, and converts many differential equations into integral equations. The striking economy is that information spread across a whole interval can sometimes be recovered from an antiderivative at its two boundaries.

A short explanation of the first part is thatis the local average of $f$. Continuity makes this average approach $f(x)$. For the second part, $G-A$ has zero derivative and is constant by the mean value theorem. These are the ideas behind the proof; the endpoint and regularity conditions make them valid.

An early geometric form of this connection appears in Isaac Barrow's Lectiones geometricae (1670), Lecture X, Proposition 11. The original volume is catalogued by CESIMA, and Michael Nauenberg's historical research reproduces and examines the geometric argument.^[ref][ref]

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

Israel Gohberg was a mathematician working in functional analysis, operator theory and matrix methods. His research and collaborations linked abstract properties of linear operators with concrete integral equations and matrices. He also founded the journal Integral Equations and Operator Theory, providing a meeting point for these subjects.^[ref] The relationship is useful in applications: an integral equation can encode a boundary-value problem, while a finite-dimensional approximation produces a linear system. Understanding invertible linear maps, the spectrum and perturbations helps explain when such approximations are meaningful and when small errors can have large consequences.

Identity and discipline: en-exact-sciences.tau.ac.il/math/in-memoriam

Mark Krein was a mathematician whose research joined functional analysis, operator theory, harmonic analysis and mathematical physics. He worked on the moment problem, integral equations and the spectral theory of linear operators, building an influential research school despite severe institutional discrimination.^[ref] The Krein-Milman theorem, proved with David Milman, gives a geometric expression of this work: a nonempty compact convex set in a locally convex Hausdorff space is the closed convex hull of its extreme points. In finite dimensions, the corresponding picture is a convex polygon reconstructed from its vertices.

Near $x=0$, separation of variables gives $dy/y=dx/(1-x)$, so $\log |y|=-\log (1-x)+C$. The initial condition selectsDirect substitution verifies both the ordinary differential equation and the initial condition. The endpoint $x=1$ is excluded because the coefficient and this solution are singular there; the displayed interval is the maximal connected real interval containing the initial point.

For $|x|\le h<1$ and any two admissible values $y,\tilde{y}$,This is a Lipschitz condition in the dependent variable, uniform in $x$, with Lipschitz constant $L=(1-h{)}^{-1}$. Also $|f(x,y)|\le (1+k)/(1-h)$ on the prescribed rectangle, so the usual local hypotheses of the Picard-Lindelöf theorem hold.

Integrating from the initial point gives the integral equationConversely, differentiating this equation using the fundamental theorem of calculus recovers the initial value problem. Start Picard iteration with the constant initial approximation $y_0(x)=1$ and setThe first three approximations areTo prove the general expression by mathematical induction, put $\ell (x)=\log (1-x)$. The initial approximation gives the formula for $n=0$. If it holds for $n$, then the substitution rulegivesThus the formula follows at every stage.

For every fixed real $x<1$, these are partial sums of the exponential series evaluated at the finite real number $-\log (1-x)$. ThereforeThis range is larger than $|x|<1$: the expansion is in powers of $\log (1-x)$, not a Taylor series in $x$. For example it converges at $x=-2$. It also converges uniformly on every compact interval contained in $(-\infty ,1)$, because $|\log (1-x)|$ is bounded there and the exponential series supplies a uniform summable bound. At $x=1$ the approximations are undefined, while for $x>1$ the real integral from zero crosses the singularity. This is Picard iteration for a simple-pole linear differential equation.

For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

Continuous dependence of an ODE solution on parameters here means continuity of the map $\beta \mapsto y_{\beta }$ into $C([a-h,a+h])$ with its supremum norm: for every ${\beta }_0\in I$ and $\epsilon >0$, all sufficiently close $\beta \in I$ satisfySubtract the two integral equations and use the uniform Lipschitz constant $L$ on the rectangle. With $v(x)=|y_{\beta }(x)-y_{\gamma }(x)|$ we obtainPart (i), with $c=|\beta -\gamma |$ and $A=L$, therefore proves, for $L>0$,The case $\beta =\gamma$ follows from uniqueness or the same estimate with zero forcing. If $L=0$, the direct integral estimate gives $\Vert y_{\beta }-y_{\gamma }{\Vert }_{\infty }\le h|\beta -\gamma |$. In either case the map is even Lipschitz continuous, and hence continuous, throughout $I$ in the stated supremum norm.

Write $A(t)={\int }_0^{t}a(s)ds$. Multiplying the ordinary differential equation by the integrating factor $e^{A(t)}$ and using the product rule givesIntegration from $0$ to $t$ therefore yields the Volterra integral equationConversely, a continuous function satisfying this integral equation becomes continuously differentiable after multiplication by $e^{A(t)}$, by the fundamental theorem of calculus, and satisfies the original initial value problem.

Existence was proved by Picard iteration. If $y$ and $v$ are any two continuous solutions with the same $b\ne 0$, the previous part makes both bounded by $2B$. Subtracting their Volterra integral equations givesSince $\gamma <1$, this forces $y=v$.

For $b=0$, zero is a solution. On any finite interval $[0,T]$, another continuous solution has a finite bound $K_T={\max }_{[0,T]}|y|$. Since $K(t,s)\le 1$, its integral equation impliesThe Gronwall inequality gives $y=0$ on every such interval. Hence

Let $y$ be any continuous function solving the integral equation on $[0,\infty )$, and retain $B=|b|>0$. A first-exit argument avoids assuming in advance that its supremum norm is finite. If $|y|$ ever reached $2B$, continuity would give a first such time $T>0$. On $[0,T]$ we would have $|y|\le 2B$, so the kernel estimate would implya contradiction. Therefore $|y(t)|<2B$ for every finite $t$.

That pointwise strict inequality alone would not give a strict bound on the supremum norm. Apply the integral equation once more, now using $|y|\le 2B$ globally:

The integral equation says precisely $h=L^{-1}z$ with the homogeneous mixed endpoint conditions. Since the supplied $h$ satisfies those conditions, apply $L$ directly:No integration is required. The absence of a zero eigenvalue established in (b)–(c) also proves uniqueness of this recovered forcing.

The mean value theorem states that, if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, there is $c\in (a,b)$ such thatTo deduce it from Rolle's theorem, subtract the secant line: $k(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a)$. This function has the same continuity and differentiability properties and satisfies $k(a)=k(b)=0$. Rolle's theorem gives an interior point with $k^{\prime }(c)=0$, which is precisely the required equality.

If $h^{\prime }=0$ on $\mathbb{R}$, apply the mean value theorem on any $[x,y]$ with $x<y$. It gives $h(y)-h(x)=(y-x)h^{\prime }(c)=0$. Hence $h$ is constant.

For the differential equation, the product rule givesThe preceding result makes this product constant. Thus all solutions are $f(x)=C{e}^{ax}$, and differentiation verifies every such solution.

For the integral assertion, continuity ensures that $f$ is Riemann integrable on every finite closed interval. Use the usual orientation for an integral with negative upper endpoint. For $h\ne 0$, additivity of the integral givesThe absolute value of this expression is at most the supremum of $|f(t)-f(x)|$ on the segment joining $x$ and $x+h$. Continuity at $x$ makes this bound tend to zero for either sign of $h$. Consequently $F^{\prime }(x)=f(x)$, proving this form of the fundamental theorem of calculus.

Finally, a differentiable solution of the integral equation is continuous, so the fundamental theorem of calculus yields $g^{\prime }=g$; setting $x=0$ gives $g(0)=A$. The differential-equation calculation above now forcesConversely, $A+{\int }_0^{x}A{e}^{t}dt=A+A(e^x-1)=A{e}^x$, so this function satisfies the original integral equation. It is unique because every differentiable solution has already been forced to this form and this initial condition.

Define interaction-picture states by $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$ and $V_I(t)=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. Differentiation cancels the free Hamiltonian operator and yieldsAssume the interaction is switched off sufficiently fast in the remote past for the following integral to converge. Starting in $|n⟩$, the first iteration of the integral equation givesfor $m\ne n$. The picture change only multiplies the measured amplitude by a phase. Squaring therefore givesFor the harmonic oscillator, $|1⟩=a^{†}|0⟩$ is normalized and $⟨1|(a+a^{†})|0⟩=1$. The energy difference is $\hslash \omega$, andHence the lowest nonzero transition probability isThe stronger remainder here follows from parity: every insertion of $a+a^{†}$ reverses oscillator parity, so the amplitude between $|0⟩$ and $|1⟩$ has only odd powers of $\lambda$. The general displayed $O({\lambda }^3)$ estimate remains valid.

Set $y_k(0)=x_0$ and let $M={\sup }_{z\in {\mathbb{R}}^n}\Vert F(z)\Vert <\infty$. On the first interval, $y_k$ is the known constant $x_0$, so the integral equation defines $x_k$. Its endpoint value then defines the constant $y_k$ on the next interval, and induction over the $k$ subintervals defines the pair on all of $[0,1]$. An integral of the bounded, piecewise constant function $F\circ y_k$ is continuous, so every $x_k$ is continuous. In fact,so the approximations are uniformly Lipschitz continuous.

Denote the unshifted differential expression by $L_{0}y=(p{y}^{\prime }{)}^{\prime }+qy$. The supplied Green's function satisfies $L_{0}G(x,\xi )=\delta (x-\xi )$, whereas the inhomogeneous equation is $L_{0}y=\lambda wy+wf$.

Multiply the latter equation by $G$ and the Green equation by $y$, subtract and integrate in $x$:The left-hand side is the boundary termIt vanishes because $G$ and $y$ satisfy the same homogeneous Robin boundary conditions. For example, where $\alpha \ne 0$, substituting $y^{\prime }=-\beta y/\alpha$ and $G_x=-\beta G/\alpha$ cancels the endpoint expression; where $\alpha =0$, both values vanish. The other endpoint works identically. HenceThe Dirac delta function contribution is valid for an interior evaluation point $\xi$; equivalently, $G$ is continuous there and has flux-derivative jump $[p{G}_x{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$.

This uses the existence of the unshifted Green's function stated in the question. Such a function is not automatic for every choice of $p,q$ and endpoint conditions: a nontrivial zero mode of $L_0$ would obstruct an ordinary inverse and would require a compatibility condition and a modified Green's function. Under the stated inverse assumption, the differential equation and the displayed integral equation are equivalent for functions in the boundary-condition domain.

Put $\beta =m\alpha /{\hslash }^2$. The delta interaction reduces the integral equation to ${\psi }_s=e^{ikx}+i\beta e^{ik|x|}{\psi }_s(0)/k$. At zero, ${\psi }_s(0)=(1-i\beta /k{)}^{-1}$. HenceTheir sum is one, as required for a real potential. The explicit wave is the indicated incident-plus-reflected form for $x<0$ and $t{e}^{ikx}$ for $x>0$; it is continuous and has derivative jump $-2\beta \psi (0)$.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.