Method of steps gives a unique compatible neutral-delay solution

Suppose the solution is known through $k\tau$. On $I_k=[k\tau ,(k+1)\tau ]$, both $z_k(t)=y(t-\tau )$ and $g_k(t)=y^{\mathrm{\prime }}(t-\tau )$ are known continuous functions, including the first interval where they come from $\varphi$ and ${\varphi }^{\mathrm{\prime }}$. Extend these known functions constantly just past the interval endpoints. The neutral equation on $I_k$ becomes $x^{\mathrm{\prime }}=a{g}_k(t)+f(t,x,z_k(t))$, a continuous ODE field that is globally $L$-Lipschitz in $x$; [F2] gives a unique local solution from the prescribed endpoint.

On a partial segment, [F7] gives the integral equation; [F4], [F6], and the Lipschitz bound then give $\mathrm{\mid }x(t)\mathrm{\mid }\le \mathrm{\mid }x(k\tau )\mathrm{\mid }+{\int }_{k\tau }^t(B_k+L\mathrm{\mid }x(s)\mathrm{\mid })ds$, where $B_k=\mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_{\mathrm{\infty }}+{\sup }_{I_k}\mathrm{\mid }f(s,0,0)\mathrm{\mid }+L\mathrm{\parallel }{z}_k{\mathrm{\parallel }}_{\mathrm{\infty }}$ is finite by [F5] and continuity of $f$. Since the interval length is at most $\tau$, this implies $\mathrm{\mid }x(t)\mathrm{\mid }\le \mathrm{\mid }x(k\tau )\mathrm{\mid }+B_k\tau +L{\int }_{k\tau }^t\mathrm{\mid }x(s)\mathrm{\mid }ds$. By [F3], $\mathrm{\mid }x(t)\mathrm{\mid }\le R_k:=(\mathrm{\mid }x(k\tau )\mathrm{\mid }+B_k\tau )e^{L\tau }$ throughout that partial segment. On the ball $\mathrm{\mid }x\mathrm{\mid }\le R_k+1$, the ODE field is bounded by $M_k:=B_k+L(R_k+1)$. Choose ${\delta }_k>0$ within the time-domain margin with ${\delta }_k{M}_k\le 1$ and $L{\delta }_k<1$; [F2] continues the solution by at least ${\delta }_k$ whenever it has not reached the right endpoint. Finitely many such continuations cover $I_k$.

The history derivative and the solution derivative agree at zero by the compatibility equation in [F1]. At a later breakpoint $(k+1)\tau$, the left and right limits of $a{y}^{\mathrm{\prime }}(t-\tau )+f(t,y(t),y(t-\tau ))$ agree because $y$ and $y^{\mathrm{\prime }}$ already agree continuously at $k\tau$ and $f$ is continuous. Thus the stepwise solution is $C^1$ across each breakpoint.

The first interval is uniquely determined by [F2]. If two solutions agree through $k\tau$, they have the same known functions $z_k,g_k$ on $I_k$, so ODE uniqueness in [F2] makes them agree throughout that interval. Induction over the finite $K$ intervals proves existence and uniqueness on $[-\tau ,T]$.