Reduced-order LL local ODE and shell — Examples

Define $I(t):={\int }_t^{\mathrm{\infty }}{e}^{(t-s)\mathrm{/}{\tau }_e}{F}_{\mathrm{e}\mathrm{x}\mathrm{t}}(s)ds$, interpreting the integral as a finite Riemann integral after any upper truncation beyond $\max \{t,T\}$; compact support makes the value independent of that truncation. For $t<T$, write $I(t)=e^{t\mathrm{/}{\tau }_e}{\int }_t^T{e}^{-s\mathrm{/}{\tau }_e}{F}_{\mathrm{e}\mathrm{x}\mathrm{t}}(s)ds$. By [F1] the integrand is integrable on this compact interval, and [F2] together with the fundamental theorem gives $I^{\mathrm{\prime }}(t)=-F_{\mathrm{e}\mathrm{x}\mathrm{t}}(t)+\frac{1}{{\tau }_e}I(t)(t<T)\mathrm{.}$ For $a=I\mathrm{/}(m{\tau }_e)$ this identity becomes ${\tau }_e{a}^{\mathrm{\prime }}=-F_{\mathrm{e}\mathrm{x}\mathrm{t}}\mathrm{/}m+a$, which rearranges to the displayed Abraham–Lorentz equation. For $t>T$, both $I$ and $F_{\mathrm{e}\mathrm{x}\mathrm{t}}$ vanish, so the same equation holds. At $t=T$, smoothness and support imply $F_{\mathrm{e}\mathrm{x}\mathrm{t}}(T)=0$, and the one-sided derivatives agree with the displayed identity.

For $t\ge T$, the integration interval contains no point where $F_{\mathrm{e}\mathrm{x}\mathrm{t}}$ is nonzero, so $a(t)=0$. For $t<0$, continuity and nonnegativity of the nonzero pulse give a closed subinterval of $(0,T)$ on which $F_{\mathrm{e}\mathrm{x}\mathrm{t}}\ge d>0$. The exponential factor is positive and continuous there by [F2, F4]; its minimum is positive by [F5]. The integral over that subinterval is strictly positive by the lower bound [F3], and all remaining contributions are nonnegative. Thus $a(t)>0$. Since $a(t)=0$ for $t\ge T$, it satisfies the future boundary condition.

If $a_1,a_2$ are two solutions, their difference $d=a_1-a_2$ satisfies $d^{\mathrm{\prime }}=d\mathrm{/}{\tau }_e$. The product rule gives $(e^{-t\mathrm{/}{\tau }_e}d(t){)}^{\mathrm{\prime }}=0$, so [F6] implies this product is a constant $C$. By [F4], $d(t)=C{e}^{t\mathrm{/}{\tau }_e}$; if $C\mathrm{\ne }0$, then [F7] makes $\mathrm{\mid }d(t)\mathrm{\mid }\to \mathrm{\infty }$ as $t\to +\mathrm{\infty }$, contradicting $d(t)\to 0$. Hence $C=0$ and $d=0$, proving uniqueness. Together with step 2.1, the unique future-decaying solution is strictly positive for all $t<0$: this pre-acceleration is a property of the stated model equation, not an empirical claim.

The integral defining $x$ is well-defined on every compact $\tau$ interval because $U$ is smooth by [F2]. The fundamental theorem [F2] gives $\dot{x}=U$. By [F1], $\eta (U,U)=c^2(-{\cosh }^2\theta +{\sinh }^2\theta )=-c^2$. Also $U^0=c\cosh \theta >0$, so $U$ is future directed in the chosen inertial chart. Thus $\tau$ is proper time and $x$ is future timelike.

Differentiating with [F1, F2] gives $a=c{\theta }^{\mathrm{\prime }}(\sinh \theta ,\cosh \theta ,0,0),\dot{a}=c{\theta }^{\mathrm{\prime }\mathrm{\prime }}(\sinh \theta ,\cosh \theta ,0,0)+c({\theta }^{\mathrm{\prime }}{)}^2(\cosh \theta ,\sinh \theta ,0,0),\eta (a,a)=c^2({\theta }^{\mathrm{\prime }}{)}^2\mathrm{.}$ Consequently $\dot{a}-\eta (a,a)U\mathrm{/}{c}^2=c{\theta }^{\mathrm{\prime }\mathrm{\prime }}(\sinh \theta ,\cosh \theta ,0,0)$, and the force-free equation [F3] is equivalent to ${\theta }^{\mathrm{\prime }}={\tau }_e{\theta }^{\mathrm{\prime }\mathrm{\prime }}$. The chosen $\theta$ has ${\theta }^{\mathrm{\prime }}=A{e}^{(\tau -{\tau }_0)\mathrm{/}{\tau }_e}$ and ${\theta }^{\mathrm{\prime }\mathrm{\prime }}={\theta }^{\mathrm{\prime }}\mathrm{/}{\tau }_e$, so the equation holds identically. This proves that the curve solves the adopted LAD equation for every real proper time.

The invariant $\eta (a,a)=c^2({\theta }^{\mathrm{\prime }}{)}^2$ gives $\sqrt{\eta (a,a)}=c\mathrm{\mid }A\mathrm{\mid }{e}^{(\tau -{\tau }_0)\mathrm{/}{\tau }_e}$, which is unbounded toward the future. For $F=0$, [F4] gives $a_{\mathrm{L}\mathrm{L}}=0$, hence its local solutions have constant $U$. The displayed LAD solution is non-inertial since $A\mathrm{\ne }0$, so it demonstrates that the force-free LAD equation admits runaway solutions even though the reduced LL force-free field does not. This is an ODE comparison, not an empirical conclusion.