Bootstrap closure and nonlinear stability boundaries — Examples

On any finite $[0,T]$, set $u=E$, $a(t)=E_0$ and $b(t)=Cϵ(1+t{)}^{-1-\delta }$. These functions satisfy [F1]'s continuity and nonnegativity hypotheses, and $a$ is nondecreasing. Therefore $E(t)\le E_0\exp \text{\hspace{0.17em}⁣}(Cϵ{\int }_0^t(1+s{)}^{-1-\delta }ds)$.

For $t=0$ the integral is zero. For $t>0$, the function $G(s)=-{\delta }^{-1}(1+s{)}^{-\delta }$ has derivative $G^{\mathrm{\prime }}(s)=(1+s{)}^{-1-\delta }$ by [F3] and [F4]; $G^{\mathrm{\prime }}$ is continuous on $[0,T]$. Applying [F5] gives ${\int }_0^t(1+s{)}^{-1-\delta }ds={\delta }^{-1}[1-(1+t{)}^{-\delta }]$. By [F2] and [F6], $(1+t{)}^{-\delta }>0$, so the exponent $q(t):=\frac{Cϵ}{\delta }[1-(1+t{)}^{-\delta }]$ satisfies $q(t)\le Cϵ\mathrm{/}\delta$. By [F7], $\exp (q(t))\le \exp (Cϵ\mathrm{/}\delta )$; since $Cϵ\mathrm{/}\delta >0$, [F7] and [F8] give $\exp (Cϵ\mathrm{/}\delta )>1$. As $E_0\ge 0$, these inequalities prove both displayed bounds and $E_0\le E_0\exp (Cϵ\mathrm{/}\delta )$.

If additionally $E_0{e}^{Cϵ\mathrm{/}\delta }<A{ϵ}^2$, the bound in step 2.1 implies $E(t)<A{ϵ}^2$ on every finite $[0,T]$ with $T<T_{\max }$ and $E(0)\le E_0<A{ϵ}^2$. Thus the strict improvement required by [F9] holds whenever its hypothesis $E\le 2A{ϵ}^2$ holds. For a continuation application, assume in addition that this $E$ is the normalized energy of a maximal solution of a fixed evolution problem in the solution class of [F9], that the problem satisfies the stated finite-endpoint continuation criterion, and that $m_j(t)\ge {\mu }_j$ for every listed $j$ and every $t<T_{\max }$. All hypotheses of [F10] now hold, so $T_{\max }=\mathrm{\infty }$. These solution, continuation and margin hypotheses are separate inputs, not consequences of the scalar inequality.

The case $t=0$ is immediate. For finite $t>0$, set $u=E$, $a(s)=E_0$ and $b(s)=Cϵ(1+s{)}^{-1}$ on $[0,t]$. The given inequality holds there; $u$ is continuous and nonnegative, $a$ is a nonnegative constant and hence nondecreasing, and $b$ is continuous and nonnegative, so [F1] applies and gives $E(t)\le E_0\exp \text{\hspace{0.17em}⁣}(Cϵ{\int }_0^t(1+s{)}^{-1}ds)$. Since $1+s>0$, [F2] and [F4] give $G^{\mathrm{\prime }}(s)=(1+s{)}^{-1}$ for $G(s)=\log (1+s)$; this derivative is continuous. By [F5], ${\int }_0^t(1+s{)}^{-1}ds=G(t)-G(0)=\log (1+t)$.

For the stated boundary example take $T_{\max }=\mathrm{\infty }$, set $Cϵ=1$, $E_0=1$ and $E(t)=1+t$. Then $Cϵ(1+s{)}^{-1}E(s)=1$. The function $H(s)=s$ has continuous, hence integrable, derivative $1$, so [F5] gives ${\int }_0^{t}1ds=H(t)-H(0)=t$ for $t>0$; for $t=0$, both sides equal $1$. Thus $E_0+Cϵ{\int }_0^t(1+s{)}^{-1}E(s)ds=1+t=E(t)$.

By [F3], $\exp (Cϵ\log (1+t))=(1+t{)}^{Cϵ}$, which proves the stated polynomial upper bound on every finite subinterval.

Put $T=B$. Because $B>1$, for $0\le t\le B$ we have $E(0)=1<B$ and $E(t)\le 1+B\le 2B$, but $E(B)=1+B>B$. Thus the contract's input condition holds on $[0,B]$ while its claimed improvement fails there; [F7] cannot be applied from this integral inequality alone.