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Let $T$ be a self-map of a nonempty closed subset of a Banach space, with the displayed bounds. Then $T$ has a unique fixed point $p$ and every orbit $x_n=T^n{x}_0$ converges to it. Indeed, $\Vert x_{n+1}-x_n\Vert \le {\alpha }_n\Vert x_1-x_0\Vert$ makes the orbit Cauchy by summing successive increments. The first bound makes $T$ continuous, so the limit is fixed. Since ${\alpha }_n\to 0$, the bounds also imply uniqueness. The tail sum gives $\Vert p-x_n\Vert \le {\sum }_{k\ge n}{\alpha }_k\Vert x_1-x_0\Vert$ for $n\ge 1$. A Volterra integral equation supplies factorially decreasing bounds even when the first iterate is not a contraction mapping.

Consider the Volterra integral equationwhere $h$ is continuous, $|h|\le 1$, $h(0)=1$, and the continuous kernel satisfies $0\le K(t,s)\le e^{-c(t-s)}$ for $0\le s\le t$, with $c>0$. If $0<|b|<c/4$, there is exactly one continuous solution on the whole half-line. It is bounded, and Picard iteration beginning at $y_0=bh$ converges in supremum norm with contraction ratio at most $\gamma =4|b|/c<1$.

For a proof, the integral of the kernel is at most $1/c$. Thus $T$ maps the closed ball of radius $2|b|$ in the Banach space of bounded continuous functions into itself, since $\Vert Ty{\Vert }_{\infty }\le |b|+4b^2/c<2|b|$. Factoring $y^2-v^2$ gives $\Vert Ty-Tv{\Vert }_{\infty }\le \gamma \Vert y-v{\Vert }_{\infty }$ on this ball. The contraction mapping theorem gives its unique fixed point. A first-exit argument shows that every continuous solution, even one not initially assumed bounded, stays strictly inside the same ball: a first time with $|y|=2|b|$ would contradict $|y|\le |b|+4b^2/c$. Reapplying the estimate gives the strict global bound $\Vert y{\Vert }_{\infty }\le |b|(1+\gamma )<2|b|$, so uniqueness is unrestricted within the continuous solution class. If $b=0$, zero is the only solution by the Gronwall inequality on each finite interval.

A function weighting the influence of past values in a history-dependent equation. In a Volterra integral equation its convolution with the history is causal: only earlier times appear. For an integrable kernel, $\tilde{\kappa }(0)={\int }_0^{\infty }\kappa (u)du$ measures its response to a constant input. In a generalized Langevin equation, the friction memory kernel records delayed resistance to the velocity.

Integrating $e^{-pt}\sin (ct)$, or taking the imaginary part of ${\int }_0^{\infty }{e}^{-(p-ic)t}dt$, givesThe Laplace transform convolution theorem states $\mathcal{L}(f\ast g)=\bar{f}\bar{g}$ whenever the transforms and convolution are defined. Applying it to the Volterra integral equation gives $\bar{x}+\bar{x}/(p^2+1)=\bar{F}$, soSince ${\mathcal{L}}^{-1}[(p^2+2{)}^{-1}]=\sin (\sqrt{2}t)/\sqrt{2}$, a second application of the convolution theorem yields the sine-kernel Volterra resolventThe transform argument assumes suitable growth for the transforms to exist. The resulting causal resolvent formula also applies to locally integrable forcing on finite intervals, as can be checked by substitution and exchanging integrals over a finite triangle.

The fundamental theorem of calculus first gives $y^{\prime }(x)=1+{\int }_0^{x}f(y(s))ds$. Integrating again, or differentiating the proposed expression twice and checking its initial values, yields the Volterra integral equationFor $f(y)=-y$, the initial value problem is solved by $\boxed{y(x)=\sin x}$. Its Picard iteration constructs the sine series directly. The base approximation is $y_0(x)=x$. Ifthen elementary integration givesSubstituting this finite sum in the next update, then shifting its index, proves the same formula with $n+1$ in place of $n$. This is the required mathematical induction. The resulting partial sums converge to $\boxed{\sin x}$ for every real $x$, and uniformly on every bounded interval: the absolute terms on $|x|\le R$ are bounded by the convergent series ${\sum }_k{R}^{2k+1}/(2k+1)!$.

The printed update lists $n=1,2,\dots$, which would leave $y_1$ unspecified if read literally. The usual Picard iteration, and the requested induction from $y_0$, use the update starting at $n=0$; that intended convention is used here.

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.

The difference of two successive Picard iterates factors asThe preceding bound and the kernel estimate therefore giveMoreover $D_1:=\Vert y_1-y_0{\Vert }_{\infty }\le B^2/C_1$. Summing the resulting geometric series, for $m>n$,Thus there is a uniform limit $y$ on $[0,\infty )$, continuous by the uniform limit theorem, with $\Vert y{\Vert }_{\infty }\le 2B$. The integral passes to this limit becauseTherefore $y$ satisfies the Volterra integral equation, and hence the initial value problem. This proves global existence by Global Picard iteration with a decaying kernel. The same estimate gives the useful error boundIf $b=0$, every iterate is zero and the convergence assertion is immediate.

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

Put $\theta =kx-4k^{3}t$. Conjugation by $e^{i\theta {\sigma }_3}$ is the meaning of $e^{i\theta {\hat{\sigma }}_3}$. The normalized eigenfunctions have the Volterra integral equation representationChoose ${\gamma }_1$ from $(0,T)$ down the time boundary and then right, ${\gamma }_2$ from $(0,0)$ up the time boundary and then right, and ${\gamma }_3$ horizontally from infinity. Compatibility of the Lax pair makes the resulting construction path-independent.

Let $a=\mathrm{Im}k$ and $b=\mathrm{Im}{k}^3$. The nontrivial exponential in column one has modulus $e^{2a(x-x^{\prime })-8b(t-t^{\prime })}$; column two has its reciprocal. For ${\gamma }_1$, $x-x^{\prime }\ge 0$ and $t-t^{\prime }\le 0$, giving signs $a,b\le 0$ for column one and $a,b\ge 0$ for column two. For ${\gamma }_2$, both differences are nonnegative, giving $(a\le 0,b\ge 0)$ and $(a\ge 0,b\le 0)$. For ${\gamma }_3$, $x-x^{\prime }\le 0$ and the time difference is zero, giving upper and lower half-planes, respectively.

Define $D_1:(a>0,b>0)$, $D_2:(a>0,b<0)$, $D_3:(a<0,b>0)$ and $D_4:(a<0,b<0)$. The bounded analytic column domains are thereforeThese are the Volterra analyticity sectors for reverse-dispersion mKdV, shown in the right panel of the figure. $D_1$ is the pair of upper outer sectors, $D_2$ the upper middle sector, $D_3$ the lower middle sector and $D_4$ the lower outer pair. Their boundaries are the six rays $\arg k=m\pi /3$. The Volterra integral equation converges and differentiates analytically where these kernels are bounded, assuming the usual sufficiently integrable spatial decay and smoothness needed by these normalizations. If decay is interpreted as merely pointwise convergence to zero without an integrability condition, existence and the stated large-$k$ normalization need additional hypotheses. A finite-path eigenfunction may be entire for fixed coordinates, but boundedness for large $k$ holds in the displayed domains; entire continuation does not enlarge those bounded sectors.

Remove the commutator gauge by setting ${\Psi }_j={\mu }_j{e}^{i\theta {\sigma }_3}$. These matrices all satisfy the ordinary first-order systemsBoth coefficient matrices are traceless, so the normalization implies $\det {\Psi }_j=\det {\mu }_j=1$. For any two solutions, differentiating ${\Psi }_2^{-1}{\Psi }_j$ shows that it is independent of both $x$ and $t$. Calling these constant matrices $s$ and $S$ givesAt $(0,0)$, ${\mu }_2=I$, so $s={\mu }_3(0,0,k)$. More generally,At $(0,T)$ the other normalization is ${\mu }_1=I$, soThus $s$ is determined by the initial spatial Volterra integral equation, while $S$ is determined by the boundary time equation evaluated at $T$. The latter uses all boundary values occurring in $\tilde{Q}$, including the initially unknown second-derivative trace. It is the global relation, rather than a claim of independence of those traces, that supplies their constraint.

Take the linear limit and write $g_j(t)={\partial }_x^{j}q(0,t)$. The spatial and temporal Volterra integral equations give, to first order,while $a=A=1$ to first order and $c(k,T)=-\hat{q}(2k,T)$. The minus sign in $b$ comes from the spatial path running from infinity to zero. For $B$, the inverse in $S=e^{4i{k}^{3}T{\hat{\sigma }}_3}{\mu }_2(0,T{)}^{-1}$ supplies the minus sign; its conjugation cancels the final-time exponential in the temporal kernel. Substitution in the global relation therefore yieldsThis is the backward-sign Airy half-line global relation, because the linear limit is $q_t-q_{xxx}=0$, independently of $\lambda$.

Its constraint is especially transparent after a time Laplace transform. Let $\mathrm{Re}p>0$, choose the principal root $r=p^{1/3}$ with positive real part, and set $k=-ir/2$ in the lower transform half-plane. Then $8i{k}^3=-p$. Taking the infinite-horizon Laplace limit for a solution of sufficient exponential order givesThis proves that the second-derivative trace is fixed by the initial data and two boundary traces; it is not a third freely prescribable datum. Finite-horizon versions retain the terminal transform and express the same relation on $0<t<T$.

To see that two are necessary, rather than merely sufficient to eliminate a transform, the homogeneous transformed spatial equation is ${\partial }_x^{3}u-pu=0$. Its roots are $r$, $e^{2\pi i/3}r$, and $e^{4\pi i/3}r$. Since $-\pi /6<\arg r<\pi /6$, exactly two roots have negative real part. Decay at infinity retains two independent spatial modes. Their value/first-derivative boundary matrix has determinant equal to the difference of those two roots, which is nonzero. Thus prescribed $g_0,g_1$ fix both modes; one condition leaves a free mode, while three arbitrary traces overdetermine them.

Two independent admissible boundary conditions are needed at $x=0$, for example $q(0,t)$ and $q_x(0,t)$. This is the two boundary traces for reverse-dispersion Airy flow count and hence the perturbative count for the nonlinear modified Korteweg-De Vries equation. The argument determines the small-norm boundary count; it does not claim a global nonlinear existence theorem for arbitrary large boundary data or arbitrary incompatible choices of traces.

Let $D_1,D_2,D_3,D_4$ denote the open quadrants, counterclockwise from the first. Write $\theta =kx+2k^{2}t$. The Lax pair has the Volterra integral equation kernel $e^{-i[k(x-x^{\prime })+2k^2(t-t^{\prime })]{\hat{\sigma }}_3}$, obtained by integrating its closed matrix differential form from a normalization point.

In column one, the off-diagonal factor is $e^{2i[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$, whose modulus is $e^{-2\mathrm{Im}[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$. Column two has the reciprocal factor. For ${\mu }_2$, choose the path $(0,0)\to (0,t)\to (x,t)$, so both differences are nonnegative. Column one requires $\mathrm{Im}k\ge 0$ and $\mathrm{Im}{k}^2\ge 0$, giving $D_1$; column two requires the opposite inequalities, giving $D_4$.

For ${\mu }_1$, the time path starts at $T$ and runs backward, while the spatial path runs rightward. Its column-one conditions are $\mathrm{Im}k\ge 0$, $\mathrm{Im}{k}^2\le 0$, giving $D_2$; column two is bounded in $D_3$.

Finally, ${\mu }_3$ integrates spatially from infinity to $x$, where $x-x^{\prime }\le 0$. Column one is bounded in the lower half-plane and column two in the upper half-plane. Normalization at $(\infty ,T)$ is equivalent to normalization at $(\infty ,t)$ because the decaying potential vanishes on the vertical path at infinity. ThusThe Volterra integral equation Neumann series converges locally uniformly under the usual integrable smooth scattering assumptions, proving holomorphic dependence in the open domains and continuous boundary values. These are the Volterra analyticity sectors for a half-line NLS Lax pair; at $x=0$, absence of a spatial segment allows the larger time-only domains used for $A,B$.

Undo the commutator gauge by putting ${\psi }_j={\mu }_j{e}^{-i\theta {\sigma }_3}$. Then ${\psi }_{j,x}=(-ik{\sigma }_3+Q){\psi }_j$ and ${\psi }_{j,t}=(-2i{k}^2{\sigma }_3+V){\psi }_j$, with $V=2kQ-i{Q}_x{\sigma }_3-i|q{|}^2{\sigma }_3$. The trace of each coefficient matrix is zero, so all normalized fundamental matrices have determinant one. Their ratios ${\psi }_2^{-1}{\psi }_j$ are independent of both $x$ and $t$, because differentiation cancels the common left-hand coefficient.

Call these constant matrices $s$ for $j=3$ and $S$ for $j=1$. Returning to the commutator gauge givesHere $e^{\alpha {\hat{\sigma }}_3}C=e^{\alpha {\sigma }_3}C{e}^{-\alpha {\sigma }_3}$. Evaluation at $(0,0)$ gives $s={\mu }_3(0,0,k)$, computed solely by the spatial Volterra integral equation for the initial potential. At $(0,T)$ the normalization of ${\mu }_1$ givesThus $S$ is computed by integrating the time Lax pair for ${\mu }_2(0,t,k)$ from zero to $T$. It uses both boundary traces $q(0,t)$ and $q_x(0,t)$ before their compatibility is enforced.

Eliminate ${\mu }_2$ between the two connection relations and evaluate at $(0,T)$, where ${\mu }_1=I$. This yields the matrix global relationIts upper off-diagonal entry is especially useful. Since $S^{-1}=\left(\begin{array}{cc}A& -B\\ -B^{♯}& A^{♯}\end{array}\right)$, it isThe function $c_T$ is analytic in the upper half-plane and is $O(1/k)$ there by the spatial Volterra integral equation. This is the half-line NLS global relation. It expresses compatibility between initial and boundary scattering and the final state. At finite $T$ the right-hand side is generally nonzero; setting it to zero would silently impose an extra assumption.

To first order in the potential, $a=A=1+O(q^2)$, while the spatial and temporal Volterra integral equations giveThus the linear limit of the global relation is exactlyfor the linear equation $i{q}_t+q_{xx}=0$. This ties the two boundary traces together, rather than allowing them to be independent data.

The count is made explicit by a temporal Laplace transform. For $\mathrm{Re}p>0$, the transformed equation is ${\tilde{q}}_{xx}+ip\tilde{q}=i{q}_0$. Put $\alpha =\sqrt{-ip}$ with positive real part. Of the two homogeneous solutions $e^{\pm \alpha x}$, only $e^{-\alpha x}$ decays at infinity. There is therefore exactly one free complex coefficient to be fixed at the endpoint. In particular, prescribed Dirichlet trace ${\tilde{g}}_0$ gives the definite Neumann traceThis follows by differentiating the decaying Dirichlet Green kernel $-(e^{-\alpha |x-y|}-e^{-\alpha (x+y)})/(2\alpha )$ at $x=0$. Thus one complex boundary condition is needed, such as Dirichlet, Neumann or a suitable Robin combination. Prescribing both traces arbitrarily would overdetermine the problem. In a Sobolev space with the required trace regularity, the cubic term is a locally Lipschitz perturbation, so the small-data perturbative problem retains the same boundary-condition count; corner compatibility still must be imposed for a classical solution.

We derive the Robin spectral elimination while retaining the finite-time global relation. This distinguishes the actual ratio $B/A$ from the equivalent, time-independent boundary coefficient needed by the Riemann-Hilbert problem.

At $x=0$ the time Lax pair coefficient has off-diagonal entries $(2k+ic)q$ and $(2k-ic)\bar{q}$. LetDirect multiplication shows $V(0,t,k)=K(k)V(0,t,-k)K(k{)}^{-1}$. Because the commutator term depends on $k^2$ and $K$ commutes with ${\sigma }_3$, uniqueness of the time Volterra integral equation gives $S(k)=K(k)S(-k)K(k{)}^{-1}$. ThereforeThe exceptional parameter points of $K$ are treated by analytic continuation. These identities express linearizable Robin scattering for defocusing NLS.

For finite horizon $T$, the exact upper-half-plane ratio isConsequently the often used answer $B/A=b/a$ is not a literal finite-$T$ identity. Its effective time-independent replacement, in the two bounded time sectors, isThe second branch follows from the Robin symmetry; its argument $-k$ lies in the upper half-plane where the initial scattering functions are known. We now justify this replacement for the inverse problem without assuming that boundary values decay as $t\to \infty$.

Only $\Gamma =B^{♯}/(ad)$ is needed for the connection across the upper imaginary ray. The conjugate global relation at $-k$ and the Robin symmetry giveFor upper-half-plane $k$, the notation $b^{♯}(-k)=\overline{b(-\bar{k})}$ uses only values of $b$ in its legitimate upper-half-plane domain. PutThe effective boundary coefficient isThis Robin spectral determinant for defocusing NLS is explicitly computed from the initial data and $c$.

For completeness, the finite-horizon defect can be removed algebraically, not by an unproved limit. If $r=B^{♯}/A^{♯}$, then $\Gamma =r/[a(a-br)]$. Subtracting the same expression with $r_0=-h{b}^{♯}(-k)/a^{♯}(-k)$ givesIts jump multiplier isThe exponential decays in $D_2$ for $x>0$ and $t<T$, since $\mathrm{Im}k>0$ and $\mathrm{Im}{k}^2<0$. The coefficient is holomorphic away from the stated determinant zeros and is appropriately decaying at infinity. From the explicit matrices in part (c), $C_2=C_{1}L(\Gamma e^{2i\theta })$, where $L(z)=\left(\begin{array}{cc}1& 0\\ z& 1\end{array}\right)$. Hence replace the sector-two solution byThe reflected upper-triangular change is made in $D_3$. These analytic changes remove the finite-time defect, leave $M_1,M_4$ unchanged, and leave the reconstruction from the second column in $D_2$ unchanged. At apparent poles, the old $d$-dependent singularities cancel by the connection formula; genuine poles of the new determinant remain as explicit residue data. Equivalently, small circles can first isolate all zeros and the same algebra can be performed on the regularized contours. This is finite-horizon Robin spectral elimination.

Every modified jump is now explicitly known: keep $C_1,C_4$ from part (c) and replaceAll jumps are again $C_{-}^{-1}{C}_{+}$, now involving only $a,b,c$. If $\mathcal{D}$ has a simple zero $k_j$ giving a genuine pole, its coefficient iswith the reflected second-column condition at ${\bar{k}}_j$. Common numerator/denominator zeros are checked for removability, and zeros on the cross require contour indentation or local circle jumps. Thus no unknown boundary norming data remain. One must not assume away such zeros for every real $c$; attractive Robin parameters can support localized modes.

It follows that the equivalent normalized Riemann-Hilbert problem is specified entirely by $q_0$ and $c$, including its pole data. When two solutions exist, their quotient has no jump, its prescribed pole singularities cancel, and it tends to $I$ at infinity; Liouville theorem makes the quotient $I$, proving uniqueness. Smooth corner evolution additionally requires $q_0^{\prime }(0)=c{q}_0(0)$ and higher compatibility conditions at higher regularity. Without that compatibility, the same initial data are interpreted in the weak evolution sense, rather than as a smooth solution through the corner.

Seek a power series $y={\sum }_{j\ge 0}{a}_j{x}^j$. The initial values give $a_0=0$ and $a_1=1$. Matching the coefficient of $x^j$ for $j\ge 1$ gives $j(j+1)a_{j+1}+a_j=0$. Induction yields $a_j=(-1{)}^{j-1}/[(j-1)!j!]$, henceThe ratio of successive absolute terms is $|x|/[(k+1)(k+2)]$, so the series and its derivatives converge on every compact interval and satisfy the equation and initial values.

The first recursive approximations are $y_1=x-x^2/2$, $y_2=x-x^2/2+x^3/12$ and $y_3=x-x^2/2+x^3/12-x^4/144$. The general formula isIt holds for $n=0$. Assuming it for $n-1$, twice integrating $y_n^{\prime \prime }=-y_{n-1}/x$ with the required initial values gives the Volterra integral equation iterationEach monomial integrates by ${\int }_0^x(x-s)s^{k}ds=x^{k+2}/[(k+1)(k+2)]$, producing precisely the next truncated series and proving the induction. The integrand is continuous at zero by its limiting value, since each previous iterate has zero constant term. Thus the Volterra iteration for a singular power-series problem produces successive partial sums, converging with derivatives on compact sets to the analytic solution already found.

Introduce the null coordinatesThen $\square =-4{\partial }_{\xi }{\partial }_{\eta }$, so the equation becomesWrite the compatible boundary values asTwice integrating the equation gives the equivalent Volterra integral equation

Let $T$ denote the double-integral operator including the factor $-1/4$, and put $g=a+b-c$. Successive approximation gives the Neumann seriesOn a rectangle $|\xi |\le R$, $|\eta |\le R$,The series and its differentiated series converge locally uniformly. Since $a$ and $b$ are analytic, the sum is analytic and solves the equation and data near the origin.

If two solutions have the same data, their difference $w=Tw$. Iterating and using the same factorial estimate gives $\Vert w{\Vert }_{\infty }=0$ on every sufficiently small rectangle. This proves uniqueness. The argument is the Analytic Goursat problem for a Klein--Gordon equation.

The space $C([a,b])$ with the supremum norm is a Banach space. Continuity of the kernel makes $K$ a bounded linear operator. Induction on $n$ givesso $A^{n}y-A^{n}z=K^n(y-z)$. The ordered integration region gives the factorial bound for a Volterra iterate:For completeness, if this holds at $n$, then integrating $M{M}^n(s-a{)}^n/n!$ from $a$ to $t$ gives the bound at $n+1$; the initial case is the defining integral. Thusand ${\sum }_{n\ge 1}{\alpha }_n=e^{M(b-a)}-1<\infty$. Part (b) now gives uniform convergence of every iteration sequence to the unique fixed point. Since $y=Ay$ is precisely the Volterra integral equation, this is its unique continuous solution. No restriction $M(b-a)<1$ is needed: the factorial controls all iterates on the whole finite interval.

For $c_1=2$, $c_2=4$, $c_3=5$, the friction memory kernel and its Laplace transform areUsing the generalized fluctuation-dissipation relation with equilibrium velocity variance $5$, the velocity autocorrelation function has Laplace transformAll three poles are negative, so the stationary relaxation is stable. A partial fraction decomposition givesTaking the inverse Laplace transform givesThe negative middle coefficient is compatible with an autocorrelation function; individual terms in a partial fraction decomposition need not themselves be separate physical modes with positive weights. Several useful checks follow directly:The Volterra integral equation for $\chi$ predicts exactly ${\chi }^{\prime }(0)=0$ and ${\chi }^{\prime \prime }(0)=-\kappa (0)\chi (0)=-10$. Its integral is also $8-10/9+1/18=125/18=5/\tilde{\kappa }(0)$, agreeing with the Green-Kubo formula for diffusion.