Delay differential equations and history discretization — Examples

On the first block $0\le t\le \tau$, the delayed value is $1$, so $y(t)=1+at$. On the second block, writing $s=t-\tau \in [0,\tau ]$, the delayed value is $y(s)=1+as$, hence $y^{\mathrm{\prime }}(\tau +s)=a+a^{2}s$ and $y(2\tau )=1+2a\tau +a^2{\tau }^2\mathrm{/}2$.

The first Euler block gives $Y_j=1+ajh$ for $0\le j\le m$. On the second block, [F2] gives $Y_{m+j+1}=Y_{m+j}+ah{Y}_j$ for $0\le j<m$. Summing yields $Y_{2m}=1+a\tau +ah{\sum }_{j=0}^{m-1}(1+ajh)$. Pairing the terms $j$ and $m-1-j$ (and retaining the middle term when $m$ is odd) gives ${\sum }_{j=0}^{m-1}j=m(m-1)\mathrm{/}2$, so $Y_{2m}=1+2a\tau +a^2{h}^{2}m(m-1)\mathrm{/}2$.

Subtracting step 1.2 from step 1.1 and using $\tau =mh$ gives $y(2\tau )-Y_{2m}=a^2{h}^{2}m\mathrm{/}2=a^2\tau h\mathrm{/}2$. Since this is nonnegative, its absolute value is the claimed error, including $a=0$. On the first Euler block, $\mathrm{\mid }{Y}_j\mathrm{\mid }\le 1+\mathrm{\mid }a\mathrm{\mid }\tau$; on the second, $\mathrm{\mid }{Y}_{m+j}\mathrm{\mid }\le 1+\mathrm{\mid }a\mathrm{\mid }\tau +\mathrm{\mid }a\mathrm{\mid }\tau (1+\mathrm{\mid }a\mathrm{\mid }\tau )$ for $0\le j\le m$. The exact solution is polynomial on each of the first three delay blocks: besides step 1.1, on $[2\tau ,3\tau ]$ it is $y(2\tau +s)=y(2\tau )+(a+a^2\tau )s+(a^2\mathrm{/}2)s^2+(a^3\mathrm{/}6)s^3$. These formulas give a common bounded tube independent of $h$. The data therefore satisfy [F3] with $\eta =h_0=\tau$, because $f(t,x,z)=az$ is time-independent, the history has zero second derivative, and the tube is compact. Thus the exact error formula is consistent with the first-order guarantee.

For $0\le t\le \tau$, the delayed value is $\varphi (t-\tau )=t-\tau$, so $y^{\mathrm{\prime }}(t)=t-\tau$ and the solution matching $y(0)=0$ is $y(t)=t^2\mathrm{/}2-\tau t$. Earlier RK4 steps lie wholly in this interval, where $g(t)=t-\tau$ is affine and the RK4 rule in the Example integrates it exactly. Moreover $y^{\mathrm{\prime }\mathrm{\prime }}$ jumps at $\tau$: its left limit is $1$, while its right limit is $y^{\mathrm{\prime }}(0+)=-\tau$.

Write $x=t-\tau$. On the crossing step, the delayed right-hand side is $g(t)=x$ for $x\le 0$ and $g(t)=x^2\mathrm{/}2-\tau x$ for $x\ge 0$. The terminal stage queries $y(t_{+}-\tau )=y(h\mathrm{/}2)$, and $h\mathrm{/}2<t_{-}=mh$ for $m\ge 1$, so this value is already computed exactly by step 1.1; the midpoint uses $y(0)=0$. Therefore the RK4 stage values are $g(t_{-})=-h\mathrm{/}2$, $g(\tau )=0$, and $g(t_{+})=h^2\mathrm{/}8-\tau h\mathrm{/}2$.

The RK4 increment from the stated rule is $Q=h[-h\mathrm{/}2+(h^2\mathrm{/}8-\tau h\mathrm{/}2)]\mathrm{/}6=-(1+\tau )h^2\mathrm{/}12+h^3\mathrm{/}48$.

The exact increment is ${\int }_{-h\mathrm{/}2}^{0}xdx+{\int }_0^{h\mathrm{/}2}(x^2\mathrm{/}2-\tau x)dx=-(1+\tau )h^2\mathrm{/}8+h^3\mathrm{/}48$. Subtracting this from $Q$ gives the exact-start defect $(1+\tau )h^2\mathrm{/}24>0$.

This defect is not $O(h^5)$ as $h\to 0$, so the smooth-RHS fourth-order local-defect premise cannot be transferred across this breakpoint. Aligning or restarting at the breakpoint removes this particular crossing step but supplies no higher-order theorem without separate smoothness and error bounds.

Differentiating the displayed polynomial gives ${\varphi }^{\mathrm{\prime }}=q$, $q(-\tau )=1$, and $q(0)=a$. With $f=0$, the compatibility equation [F1] reads $a=a\cdot 1+0$, so it holds. The zero right-hand side is continuous and Lipschitz in the two state variables. Therefore [F2] supplies the unique solution on every finite horizon $K\tau$.

On the first segment, the equation gives $y^{\mathrm{\prime }}(s)=a{\varphi }^{\mathrm{\prime }}(s-\tau )=aq(s-\tau )$. If on segment $k-1$ the derivative is $a^{k}q(s-\tau )$, then on segment $k$ the same equation gives $y^{\mathrm{\prime }}(k\tau +s)=a{y}^{\mathrm{\prime }}((k-1)\tau +s)=a^{k+1}q(s-\tau )$. Compatibility makes the left and right endpoint values agree at every breakpoint, so induction proves the formula for $0\le k<K$. Since $q(-\tau )=1$, each segment derivative has supremum at least $\mathrm{\mid }a{\mathrm{\mid }}^{k+1}$; for $\mathrm{\mid }a\mathrm{\mid }>1$ this is geometric amplification.

For $a=1$, on the $K$-segment space from [F1], let $r_k(s)=1$ and $u_k(s)=k+1$ for $0\le k<K$. The defined shift has $(S_{K}u{)}_0=0$ and $(S_{K}u{)}_k=u_{k-1}$ for $k\ge 1$, so $(I-S_K)u=r$. Also $S_K^K=0$, hence $(I-S_K{)}^{-1}=I+S_K+\cdots +S_K^{K-1}$ by multiplying the finite sum. Since $\mathrm{\parallel }r{\mathrm{\parallel }}_{{\mathcal{X}}_K}=1$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{{\mathcal{X}}_K}=K$, the inverse norm is at least $K$. Thus no horizon-independent bound follows at $a=1$, consistent with the strict hypothesis $\mathrm{\mid }a\mathrm{\mid }<1$ in [F3].

At $t=0$, [F1] gives $I={\int }_0^{\tau }{\theta }^{2}d\theta ={\tau }^3\mathrm{/}3$. Put $Q_{\mathrm{e}\mathrm{x}}=\mathrm{\Delta }{\sum }_{j=0}^{r-1}{\theta }_j^2$, the midpoint sum using the exact history. On a cell centered at $c_j$, write $\theta =c_j+u$ with $-\mathrm{\Delta }\mathrm{/}2\le u\le \mathrm{\Delta }\mathrm{/}2$. Then ${\int }_{-\mathrm{\Delta }\mathrm{/}2}^{\mathrm{\Delta }\mathrm{/}2}(c_j+u{)}^{2}du=c_j^2\mathrm{\Delta }+{\mathrm{\Delta }}^3\mathrm{/}12$. Summing the cells gives $I-Q_{\mathrm{e}\mathrm{x}}=r{\mathrm{\Delta }}^3\mathrm{/}12=\tau {\mathrm{\Delta }}^2\mathrm{/}12$.

On the history cell $[\mathrm{\ell }h,(\mathrm{\ell }+1)h]$ in the variable $\theta =-s$, the stored chord of $\varphi$ satisfies ${\widehat{\varphi }}_h(-\theta )-{\theta }^2=(\theta -\mathrm{\ell }h)((\mathrm{\ell }+1)h-\theta )\mathrm{.}$ Its maximum is $h^2\mathrm{/}4$, so the uniform interpolation error on $[-\tau ,0]$ is $h^2\mathrm{/}4$. Since $\mathrm{\Delta }=h\mathrm{/}q$, every midpoint subcell is contained in one such chord cell.

The stored chord is affine on each width-$h$ cell, and the midpoint rule is exact for an affine function on each of its $q$ equal subcells: the centered linear term integrates to zero. Therefore the stored-history midpoint sum $Q_{\mathrm{s}\mathrm{t}}$ equals the integral of the chord. The area between the chord and ${\theta }^2$ on one width-$h$ cell is ${\int }_0^{h}v(h-v)dv=h^3\mathrm{/}6$. Summing $m$ cells gives $Q_{\mathrm{s}\mathrm{t}}-I=m{h}^3\mathrm{/}6=\tau h^2\mathrm{/}6$. Combining with step 1.1 yields $Q_{\mathrm{s}\mathrm{t}}-Q_{\mathrm{e}\mathrm{x}}=\tau h^2\mathrm{/}6+\tau {\mathrm{\Delta }}^2\mathrm{/}12$.

At $t=0$ every sampled time $-{\theta }_j$ lies in $[-\tau ,0]$, so [F2] and step 1.2 give $\mathrm{\mid }{Q}_{\mathrm{s}\mathrm{t}}-Q_{\mathrm{e}\mathrm{x}}\mathrm{\mid }\le \mathrm{\Delta }{\sum }_{j=0}^{r-1}\mathrm{\mid }{\widehat{\varphi }}_h(-{\theta }_j)-{\theta }_j^2\mathrm{\mid }\le \tau h^2\mathrm{/}4$. Combining this with step 1.1 directly gives $\mathrm{\mid }I-Q_{\mathrm{s}\mathrm{t}}\mathrm{\mid }\le \tau h^2\mathrm{/}4+\tau {\mathrm{\Delta }}^2\mathrm{/}12$, the same history/quadrature split as [F3]; this calculation needs no error bound for positive-time stored values. The exact node-history term from step 2.1 is $\tau h^2\mathrm{/}6+\tau {\mathrm{\Delta }}^2\mathrm{/}12\le \tau h^2\mathrm{/}4$ because $\mathrm{\Delta }\le h$. Its positive sign cancels the midpoint underestimate $Q_{\mathrm{e}\mathrm{x}}-I=-\tau {\mathrm{\Delta }}^2\mathrm{/}12$, leaving $\mathrm{\mid }{Q}_{\mathrm{s}\mathrm{t}}-I\mathrm{\mid }=\tau h^2\mathrm{/}6$.