Conditional fixed-grid forward-Euler error bound

The formula [F6] is smooth on $\mathcal{G}$: $u=m\mathrm{/}\rho$, $\epsilon =E-\mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{/}(2\rho )$, $s=c_v\log ((\gamma -1)\epsilon \mathrm{/}{\rho }^{\gamma })+s_0$, and $\beta =c_v\rho \mathrm{/}\epsilon$ have positive denominators and logarithm arguments there. Thus $v$ is locally Lipschitz. Together with [F2] and [F3], this makes every term in the finite sum [F1] locally Lipschitz on the open finite-dimensional domain ${\mathcal{G}}^{\mathrm{\mid }\mathcal{I}\mathrm{\mid }}$, so $L_h$ is locally Lipschitz and continuous there. Compactness of ${\mathcal{C}}^{\mathrm{\mid }\mathcal{I}\mathrm{\mid }}$ gives a bound $M_h={\sup }_{Y\in {\mathcal{C}}^{\mathrm{\mid }\mathcal{I}\mathrm{\mid }}}\mathrm{\parallel }{L}_h(Y)\mathrm{\parallel }<\mathrm{\infty }$. Cover this compact set by finitely many smaller balls whose doubled balls lie in local Lipschitz neighborhoods, and let ${\delta }_0>0$ be the smallest smaller-ball radius. Any pair at distance less than ${\delta }_0$ lies together in one of those larger balls, so its Lipschitz quotient is bounded by the largest local constant. For pairs at distance at least ${\delta }_0$, boundedness gives quotient at most $2M_h\mathrm{/}{\delta }_0$. Thus a finite Lipschitz constant $L_h^{\ast }$ holds on ${\mathcal{C}}^{\mathrm{\mid }\mathcal{I}\mathrm{\mid }}$.

For an exact time $t_n=n\mathrm{\Delta }t$ and $t_{n+1}\le T$, the integral equation and the bound in step 1.1 give the one-step defect $r_n=\bar{U}(t_{n+1})-\bar{U}(t_n)-\mathrm{\Delta }t{L}_h(\bar{U}(t_n))$, with $\mathrm{\parallel }{r}_n\mathrm{\parallel }\le \frac{1}{2}{L}_h^{\ast }{M}_h(\mathrm{\Delta }t{)}^2$. Indeed, $r_n={\int }_0^{\mathrm{\Delta }t}(L_h(\bar{U}(t_n+s))-L_h(\bar{U}(t_n)))ds$, while $\mathrm{\parallel }\bar{U}(t_n+s)-\bar{U}(t_n)\mathrm{\parallel }\le M_{h}s$.

The unguarded Euler recurrence, as long as each preceding candidate has been accepted, therefore satisfies $e_{n+1}\le (1+L_h^{\ast }\mathrm{\Delta }t)e_n+\frac{1}{2}{L}_h^{\ast }{M}_h(\mathrm{\Delta }t{)}^2$, where $e_n=\mathrm{\parallel }{U}^n-\bar{U}(t_n)\mathrm{\parallel }$ and $e_0=0$. Summing the geometric bound and using $(1+L_h^{\ast }\mathrm{\Delta }t{)}^n\le e^{L_h^{\ast }T}$ yields $e_n\le C_{h,T}\mathrm{\Delta }t$ with, for example, $C_{h,T}=\frac{1}{2}{M}_h(e^{L_h^{\ast }T}-1)$ when $L_h^{\ast }>0$ and $C_{h,T}=0$ when $L_h^{\ast }=0$.

The continuous exact trajectory is compactly contained in $\mathrm{int}\mathcal{C}$ over the compact time interval, so there is a positive minimum distance $\delta$ from every nodal exact state to ${\mathbb{R}}^{d+2}\setminus \mathcal{C}$. Choose $\mathrm{\Delta }t$ so small that $C_{h,T}\mathrm{\Delta }t<\delta$. Inductively, after an accepted stage the recurrence in step 3.1 bounds the next candidate's distance from the exact next state by less than $\delta$, so every candidate node remains in $\mathcal{C}$ and [F4] accepts it. This proves all $N$ steps are accepted; [F5] confirms finite return within the stated budget. The estimate in step 3.1 then holds for every returned stage.