Boundary-incompatible forcing reduces Strang order to five quarters in a Dirichlet modal model

For each mode, solving $u_k^{\mathrm{\prime }}=-k^2{u}_k+g_k$, $u_k(0)=0$, gives $u_k(t)=\frac{g_k}{k^2}(1-\exp (-k^{2}t))\mathrm{.}$ Differentiating verifies the equation and initial value. Also $\mathrm{\mid }{u}_k(t)\mathrm{\mid }\le \mathrm{\mid }{g}_k\mathrm{\mid }\mathrm{/}{k}^2\le k^{-3}$, so [F4] shows the reference state belongs to ${\mathrm{\ell }}^2$ for every $t\ge 0$.

Put $x=k^{2}h$. By [F1], each Strang coordinate obeys ${\tilde{u}}_{n+1,k}=\exp (-x){\tilde{u}}_{n,k}+h\exp (-x\mathrm{/}2)g_k\mathrm{.}$ Starting from zero and summing this scalar recurrence gives ${\tilde{u}}_{N,k}=h\exp (-x\mathrm{/}2)\frac{1-\exp (-Nx)}{1-\exp (-x)}{g}_k,$ where the finite geometric identity follows by multiplying the sum by $1-\exp (-x)$; the denominator is positive because $x>0$ and the exponential is strictly increasing.

Define $\sigma (x):=\exp (x\mathrm{/}2)-\exp (-x\mathrm{/}2)$ for real $x$, and $q(x):=1-x\mathrm{/}\sigma (x)$ for $x>0$. For $x>0$, ${\sigma }^{\mathrm{\prime }}(x)=\frac{1}{2}(\exp (x\mathrm{/}2)+\exp (-x\mathrm{/}2))>1$, because $\exp (x\mathrm{/}2)+\exp (-x\mathrm{/}2)-2=(\exp (x\mathrm{/}4)-\exp (-x\mathrm{/}4){)}^2>0.$ Thus [F6] gives $\sigma (x)>x$, so $0<q(x)<1$. Direct differentiation gives $\sigma (0)=0$, ${\sigma }^{\mathrm{\prime }}(0)=1$, and ${\sigma }^{\mathrm{\prime }\mathrm{\prime }}(0)=0$; [F7] therefore gives $\sigma (x)=x+O(x^3)$ on $0<x\le 1$. Hence $q(x)=(\sigma (x)-x)\mathrm{/}\sigma (x)\le C{x}^2$ there.

By [F2] and [F3], this constant Dirichlet forcing is outside the generator domain and has nonzero boundary value while the sine modes vanish at both endpoints. The modal comparison below therefore concerns the boundary-incompatible data identified in the Statement.

Since $Nx=k^{2}T$, the identity $\frac{h\exp (-x\mathrm{/}2)}{1-\exp (-x)}=\frac{1}{k^2}\frac{x}{\exp (x\mathrm{/}2)-\exp (-x\mathrm{/}2)}$ turns the numerical value into the exact modal value times $x\mathrm{/}\sigma (x)$. Therefore for odd $k$, $\mathrm{\mid }{\tilde{u}}_{N,k}-u_k(T)\mathrm{\mid }=k^{-3}(1-\exp (-k^{2}T))q(k^{2}h),$ and the error is zero for even $k$.

The quotient $\sigma (x)\mathrm{/}x$ is increasing for $x>0$: its derivative has numerator $p(x)=x{\sigma }^{\mathrm{\prime }}(x)-\sigma (x)$, with $p(0)=0$ and $p^{\mathrm{\prime }}(x)=x{\sigma }^{\mathrm{\prime }\mathrm{\prime }}(x)=x\sigma (x)\mathrm{/}4>0$. Therefore $x\mathrm{/}\sigma (x)$ decreases and $q$ increases. In particular $q(x)\ge q(1)>0$ on $1\le x\le 4$.

For sufficiently small $h\le 1$, set $H:=1\mathrm{/}\sqrt{h}$; [F8] gives $H\ge 1$. For odd $k\le H$, steps 2.2 and 1.3 give $\mathrm{\mid }{\tilde{u}}_{N,k}-u_k(T)\mathrm{\mid }\le C{h}^{2}k$. The sum of their squared errors is at most $C{h}^4{H}^3=C{h}^{5\mathrm{/}2}$. For $k>H$, use $q<1$ and $1-\exp (-k^{2}T)<1$; dyadic blocks $[2^{j}H,2^{j+1}H)$ contain at most $2^{j+1}H$ integers and contribute at most $2^{j+1}H(2^{j}H{)}^{-6}$. Summing the geometric series gives a tail at most $C{H}^{-5}=C{h}^{5\mathrm{/}2}$. Thus the squared ${\mathrm{\ell }}^2$ error is at most $C_T{h}^{5\mathrm{/}2}$.

Choose $h$ sufficiently small that $h\le 1\mathrm{/}64$ and $h\le T$. Then $\sqrt{h}\le 1\mathrm{/}8$ by [F8], so $H\ge 8$. The interval $[H,2H]$ contains at least $H-1$ integers, and among any $m$ consecutive integers at least $(m-1)\mathrm{/}2$ are odd; hence it contains at least $H\mathrm{/}4$ odd integers. For each such $k$, $1\le k^{2}h\le 4$, $k^{-6}\ge (2H{)}^{-6}$, $q(k^{2}h)\ge q(1)$ by step 2.3, and $1-\exp (-k^{2}T)\ge 1-\exp (-T\mathrm{/}h)>1\mathrm{/}2$ since $T\mathrm{/}h\ge 1$ and $e=1+1+{\sum }_{n\ge 2}1\mathrm{/}n!>2$. Therefore the sum of squared errors over these modes is at least $\frac{H}{4}(2H{)}^{-6}\frac{q(1{)}^2}{4}=c{H}^{-5}=c{h}^{5\mathrm{/}2}$ for a constant $c>0$. Taking square roots of the upper bound in step 3.1 and this lower bound, then using [F9], proves the claimed two-sided $h^{5\mathrm{/}4}$ error rate.