Delay differential equations and history discretization

Let $d\ge 1$, $\tau >0$, and $T>0$. The history space for delay $\tau$ is ${\mathcal{C}}_{\tau }:=C([-\tau ,0],{\mathbb{R}}^d),\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}:={\sup }_{-\tau \le s\le 0}\mathrm{\parallel }\varphi (s){\mathrm{\parallel }}_2,$ where continuity of vector-valued functions and the Euclidean norm are as in Vector-valued functions $f:A\to {\mathbb{R}}^m$, their limits and continuity, with the dictionary to the metric notions and The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$. Its completeness, so that it is a Banach space with this norm, is established in The continuous delay-history space is Banach ↗.

A retarded delay initial-value problem on $[0,T]$ consists of a history $\varphi \in {\mathcal{C}}_{\tau }$ and a vector field $f:U\times {\mathbb{R}}^d\times {\mathbb{R}}^d\to {\mathbb{R}}^d$, where $U$ is an open interval containing $[0,T]$, such that $f$ is continuous and there is an $L\ge 0$ giving the following global Lipschitz bound (in the convention of Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction): $\mathrm{\parallel }f(t,u,v)-f(t,\tilde{u},\tilde{v}){\mathrm{\parallel }}_2\le L(\mathrm{\parallel }u-\tilde{u}{\mathrm{\parallel }}_2+\mathrm{\parallel }v-\tilde{v}{\mathrm{\parallel }}_2)$ for all $t\in U$ and all $u,v,\tilde{u},\tilde{v}\in {\mathbb{R}}^d$. A solution is a continuous function $y:[-\tau ,T]\to {\mathbb{R}}^d$ whose restriction to $[0,T]$ is continuously differentiable, for which $y(s)=\varphi (s)$ on $[-\tau ,0]$, and which satisfies the compatibility condition $y(0)=\varphi (0)$ and the retarded equation $y^{\mathrm{\prime }}(t)=f(t,y(t),y(t-\tau )),0\le t\le T\mathrm{.}$ At $t=0$ the derivative is the right derivative. The equation is an ordinary first-order IVP on each method-of-steps interval once the delayed history on that interval is known.

The definition concerns a fixed positive constant delay. It does not include state-dependent delays, neutral equations containing a delayed derivative, or integrals over a delay window; those are distinct interfaces.

For every integer $d\ge 1$ and delay $\tau >0$, the history space ${\mathcal{C}}_{\tau }=C([-\tau ,0],{\mathbb{R}}^d)$ from Retarded delay equations and their history space, equipped with $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{s\in [-\tau ,0]}\mathrm{\parallel }\varphi (s){\mathrm{\parallel }}_2$, is a Banach space.

Let $\tau ,T>0$, $L,H\ge 0$, let $q:[0,T]\to [0,\mathrm{\infty })$ be continuous, and let $u:[-\tau ,T]\to [0,\mathrm{\infty })$ be continuous with $u(s)\le H$ for $s\in [-\tau ,0]$. Suppose that $u(t)\le H+{\int }_0^t(q(s)+L(u(s)+u(s-\tau )))ds(0\le t\le T)\mathrm{.}$ Then, writing $U(t)=\max \{H,{\sup }_{0\le r\le t}u(r)\}$, one has $U(t)\le (H+{\int }_0^{t}q(s)ds)e^{2Lt}(0\le t\le T)\mathrm{.}$

Let $d\ge 1$, $T,\tau >0$, and let $\varphi \in {\mathcal{C}}_{\tau }$. Suppose $f:U\times {\mathbb{R}}^d\times {\mathbb{R}}^d\to {\mathbb{R}}^d$ is continuous on an open interval $U\supseteq [0,T]$ and satisfies the global state bound in Retarded delay equations and their history space with constant $L\ge 0$. Then the retarded-delay initial-value problem has exactly one solution on $[-\tau ,T]$. The construction is the method of steps: on each interval $[k\tau ,(k+1)\tau ]\cap [0,T]$, the delayed path is already known, so the current equation is an ordinary differential equation.

Fix positive integers $d,m$, positive $T,\tau$, $h=\tau \mathrm{/}m$, and $N=\lceil T\mathrm{/}h\rceil$. Consider the following finite storage schedule: initialize and retain $d$-vectors $Y_j=\varphi (jh)$ for $j=-m,\dots ,0$; at each index $n=0,\dots ,N-1$, first test an optional nonnegative integer budget $B$, stopping if $n\ge B$, and otherwise evaluate one vector $f(nh,Y_n,Y_{n-m})$, form $Y_{n+1}=Y_n+hf(nh,Y_n,Y_{n-m})$, and append it. Assume every reached evaluation is defined and returns in finite time. A complete schedule uses exactly $N$ calls to $f$ and retains $N+m+1$ state vectors. If $B<N$, it stops after exactly $B$ calls with $m+B+1$ vectors. Initialization uses $m+1$ history evaluations in either case. Full history storage is $O(d(N+m))=O(d(T+\tau )\mathrm{/}h)$ real entries. With direct array indexing and a constant-cost scalar floor operation, a query of the piecewise-linear interpolant of the stored vectors costs $O(d)$ arithmetic. Counts exclude internal evaluation costs.

For a complete stored history, fix a positive integer $q$ and $n\in \{0,\dots ,N\}$, put $r=qm$, $\mathrm{\Delta }=h\mathrm{/}q$, and ${\theta }_j=(j+1\mathrm{/}2)\mathrm{\Delta }$. Consider the schedule that initializes $A=0\in {\mathbb{R}}^d$ and, for $j=0,\dots ,r-1$, adds $\mathrm{\Delta }K({\theta }_j){\widehat{y}}_h(nh-{\theta }_j)$ to $A$. With returning kernel evaluations, this makes $r$ kernel evaluations and $r$ history queries, using $O(rd)$ arithmetic besides kernel evaluations. Streaming the scalar weights uses $O(d)$ additional storage; caching all $r$ weights uses $O(r+d)$, excluding the input history. Every stored-grid knot, including a delay breakpoint $k\tau$, encountered by $\theta \mapsto {\widehat{y}}_h(nh-\theta )$ on $[0,\tau ]$ is a quadrature-cell endpoint. No constant-memory claim is made for the complete stored history.

Inputs and admissibility. Take a retarded-delay problem from Retarded delay equations and their history space with $\tau >0$, a target $T>0$, a positive integer $m$, its history $\varphi$ and vector field $f$, and an optional evaluation budget $B\in {\mathbb{N}}_0$. The budget counts calls to $f$; if omitted, it imposes no limit. These conditions are the algorithm's admissible inputs.

Output. On completion, return the nodes and their piecewise-linear interpolant through $T_h=Nh$, together with its restriction to $[0,T]$. If the budget expires first, return budget-exhausted, the number of completed updates, all values stored so far, and the piecewise-linear interpolant through the last accepted node.

1. Set $h=\tau \mathrm{/}m$, $N=\lceil T\mathrm{/}h\rceil$ (the least integer with $Nh\ge T$), and $T_h=Nh$, so $0\le T_h-T<h$. Initialize and store $Y_j=\varphi (jh)$ for $j=-m,-m+1,\dots ,0$.

2. For $n=0,1,\dots ,N-1$, if a budget was supplied and $n\ge B$, return budget-exhausted with $n$ completed updates, the stored values $Y_{-m},\dots ,Y_n$, and their interpolant on $[-\tau ,t_n]$. Otherwise evaluate $F_n=f(t_n,Y_n,Y_{n-m})$ and set $Y_{n+1}=Y_n+h{F}_n$. The delayed query is a stored node because $t_n-\tau =t_{n-m}$. Every required evaluation is defined: $n<N$ implies $t_n<T$, and the vector field is defined for every state on the time domain containing $[0,T]$.

3. On each grid cell in $[-\tau ,T_h]$, define ${\widehat{y}}_h$ by linear interpolation between the stored endpoints. Return completed, the nodes $Y_{-m},\dots ,Y_N$, and ${\widehat{y}}_h$ on $[-\tau ,T_h]$ and its restriction to $[0,T]$.

The evaluation counts, history storage, and direct-index interpolation cost are proved in Stored retarded Euler and distributed midpoint work, storage, and grid alignment.

Let $h=\tau \mathrm{/}m$ with integer $m\ge 1$, and let the mesh nodes cover a finite interval containing $[-\tau ,T]$, with every delay breakpoint $k\tau$ a mesh node. Suppose $y$ is continuous and each restriction to a mesh cell is $C^2$ (with one-sided derivatives at its endpoints), with $M\ge 0$ and $\mathrm{\parallel }{y}^{\mathrm{\prime }\mathrm{\prime }}(t){\mathrm{\parallel }}_2\le M$ on every cell interior. Let the stored nodal values $Y_j$ satisfy $\mathrm{\parallel }{Y}_j-y(t_j){\mathrm{\parallel }}_2\le E$ and let ${\widehat{y}}_h$ be their piecewise-linear interpolant as in Explicit Euler with a stored piecewise-linear delay history. Then ${\sup }_t\mathrm{\parallel }{\widehat{y}}_h(t)-y(t){\mathrm{\parallel }}_2\le E+\frac{M{h}^2}{8}\mathrm{.}$

Fix $T,\tau ,\eta ,\rho >0$ and $L_t\ge 0$. Write $B_r=\{x\in {\mathbb{R}}^d:\mathrm{\parallel }x{\mathrm{\parallel }}_2\le r\}$. Let $f$ and a history $\varphi :[-\tau ,0]\to {\mathbb{R}}^d$ define the retarded problem of Retarded delay equations and their history space on $[0,T+\eta ]$, with global state Lipschitz constant $L_s$, and assume $[-\rho ,T+\eta +\rho ]\subset U$. Let $y$ be the unique solution there by Method of steps gives a unique retarded-delay solution. Suppose $\varphi$ is twice continuously differentiable on the interval interior, with continuous one-sided extensions of ${\varphi }^{\mathrm{\prime }}$ and ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}$ to the endpoints, and $\mathrm{\parallel }{\varphi }^{\mathrm{\prime }\mathrm{\prime }}(s){\mathrm{\parallel }}_2\le M_{\varphi }$. For $0<h=\tau \mathrm{/}m\le h_0\le \eta$, where $m$ is a positive integer, assume the exact path and all Euler nodes generated for target $T$ lie in a fixed closed ball $B_R$. Choose $Q\ge 0$ such that $\mathrm{\parallel }f(t,0,0){\mathrm{\parallel }}_2\le Q$ for $-\rho \le t\le T+\eta +\rho$, and set $R_z=(\mathrm{\parallel }\varphi (0){\mathrm{\parallel }}_2+(Q+L_{s}R)(T+\eta ))e^{L_s(T+\eta )},R_{\ast }=1+\max \{R,R_z\}\mathrm{.}$ Assume that on $[-\rho ,T+\eta +\rho ]\times B_{R_{\ast }}\times B_{R_{\ast }}$, $f$ is bounded in norm by $M_f$, and on $[0,T+\eta ]\times B_{R_{\ast }}\times B_{R_{\ast }}$ it satisfies $\mathrm{\parallel }f(t,x,z)-f(s,x,z){\mathrm{\parallel }}_2\le L_t\mathrm{\mid }t-s\mathrm{\mid }$ for fixed $x,z$ and all $s,t$ in this time interval. Put $S=\max \{{\sup }_{[-\tau ,0]}\mathrm{\parallel }{\varphi }^{\mathrm{\prime }}(s){\mathrm{\parallel }}_2,M_f\}$ and $C_{\mathrm{l}\mathrm{o}\mathrm{c}}=\frac{L_t+L_s(S+M_f)}{2},C_{\mathrm{o}\mathrm{d}\mathrm{e}}=\{\begin{array}{ll}{C}_{\mathrm{l}\mathrm{o}\mathrm{c}}(e^{L_s(T+\eta )}-1)\mathrm{/}{L}_s,& L_s>0,\\ C_{\mathrm{l}\mathrm{o}\mathrm{c}}(T+\eta ),& L_s=0,\end{array}$ $C_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}=C_{\mathrm{o}\mathrm{d}\mathrm{e}}+M_f\mathrm{/}2+M_{\varphi }{h}_0\mathrm{/}8,C_{\mathrm{d}\mathrm{d}\mathrm{e}}=C_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}(1+L_s(T+\eta )e^{2L_s(T+\eta )})\mathrm{.}$ Then, when Explicit Euler with a stored piecewise-linear delay history completes all $N$ updates (no evaluation budget or an integer budget $B\ge N$), its returned piecewise-linear interpolant through $T_h=Nh$, where $N$ is the least integer with $Nh\ge T$, satisfies ${\sup }_{-\tau \le t\le T}\mathrm{\parallel }{\widehat{y}}_h(t)-y(t){\mathrm{\parallel }}_2\le C_{\mathrm{d}\mathrm{d}\mathrm{e}}h\mathrm{.}$ All constants are independent of $m$. The claim uses a fixed positive delay, grid alignment $h=\tau \mathrm{/}m$, time-Lipschitz $f$, and a common compact tube; it asserts no order for state-dependent delays or a higher-order method crossing a delay breakpoint.

Use the fixed positive-delay/history convention of Retarded delay equations and their history space in dimension $d=1$. The canonical isometric coordinate map between $\mathbb{R}$ and ${\mathbb{R}}^1$ is as specified in Vector-valued functions $f:A\to {\mathbb{R}}^m$, their limits and continuity, with the dictionary to the metric notions; the formulas below write the single coordinate as a real scalar.

Let $\tau >0$, $T=K\tau$ with integer $K\ge 1$, $a\in \mathbb{R}$, and let $U$ be an open interval containing $[0,T]$. Let $f:U\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ be continuous and globally Lipschitz in its last two variables: for some $L\ge 0$, $\mathrm{\mid }f(t,x,z)-f(t,\tilde{x},\tilde{z})\mathrm{\mid }\le L(\mathrm{\mid }x-\tilde{x}\mathrm{\mid }+\mathrm{\mid }z-\tilde{z}\mathrm{\mid })(t\in U)\mathrm{.}$ Let $\varphi :[-\tau ,0]\to \mathbb{R}$ be continuously differentiable, with continuity and derivative as in Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point and The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set. A scalar neutral delay initial-value problem is $y^{\mathrm{\prime }}(t)-a{y}^{\mathrm{\prime }}(t-\tau )=f(t,y(t),y(t-\tau )),0\le t\le T,$ with history $y(t)=\varphi (t)$ for $-\tau \le t\le 0$. A solution is continuous on $[-\tau ,T]$, continuously differentiable on $[0,T]$ with domain-relative endpoint derivatives, and satisfies the equation. The history is compatible at zero when ${\varphi }^{\mathrm{\prime }}(0)=a{\varphi }^{\mathrm{\prime }}(-\tau )+f(0,\varphi (0),\varphi (-\tau ))\mathrm{.}$ The parameter restriction $\mathrm{\mid }a\mathrm{\mid }<1$ is the stable derivative-shift branch used below; the equation itself is defined for every real $a$. No state-dependent delay is included.

For the finite horizon $T=K\tau$, write a derivative path as segments $u_k(s)=y^{\mathrm{\prime }}(k\tau +s)$ for $0\le k<K$ and $0\le s\le \tau$. Let ${\mathcal{X}}_K$ be the set of $K$-tuples of continuous real functions on $[0,\tau ]$, with continuity as in Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point, and define $\mathrm{\parallel }u{\mathrm{\parallel }}_{{\mathcal{X}}_K}={\max }_{0\le k<K}{\sup }_{0\le s\le \tau }\mathrm{\mid }{u}_k(s)\mathrm{\mid },$ and the delay shift $S_K$ by $(S_{K}u{)}_0(s)=0$ and $(S_{K}u{)}_k(s)=u_{k-1}(s)$ for $1\le k<K$. The known derivative history on the first segment is a forcing term; on each later segment the delayed derivative is exactly the preceding segment $S_{K}u$. Thus the neutral derivative relation has the operator form $(I-a{S}_K)u=r$, with $r$ containing the right-hand side and the known initial derivative history.