Second-order Crank–Nicolson nodal error for exact initial data under H2 time regularity

Fix a finite basis of $V_h$ and express one CN step as a square linear system. Its bilinear form is $b_j(x,v)={\tau }_j^{-1}(x,v{)}_H+\frac{1}{2}a(x,v)$. If the homogeneous system has solution $x$, testing with $x$ gives ${\tau }_j^{-1}\mathrm{\parallel }x{\mathrm{\parallel }}_H^2+\frac{1}{2}a(x,x)=0$. Coercivity [F4] implies $x=0$; the square matrix is therefore invertible. Thus the step has exactly one solution by finite-dimensional linear algebra, without a choice axiom.

Integrating the semidiscrete equation from [F2] over $I_j$ gives the exact step with $u_h$ replaced in the form by its time average ${\bar{u}}_h^j={\tau }_j^{-1}{\int }_{I_j}{u}_h(t)dt$. Subtract it from the CN equation. With $e^j=U^j-u_h(t_j)$ and $r_j={\bar{u}}_h^j-(u_h(t_j)+u_h(t_{j-1}))\mathrm{/}2$, the result is ${\tau }_j^{-1}(e^j-e^{j-1},v_h{)}_H+a\text{\hspace{0.17em}⁣}(\frac{e^j+e^{j-1}}{2},v_h)=a(r_j,v_h)\mathrm{.}$

Test the CN equation by $U^{\mathrm{m}\mathrm{i}\mathrm{d},j}$ and multiply by $2{\tau }_j$. The midpoint identity follows from [F3], since $2(U^j-U^{j-1},U^{\mathrm{m}\mathrm{i}\mathrm{d},j}{)}_H=2(U^j-U^{j-1},U^j{)}_H-\mathrm{\parallel }{U}^j-U^{j-1}{\mathrm{\parallel }}_H^2$. Coercivity and Young's inequality give $\mathrm{\parallel }{U}^j{\mathrm{\parallel }}_H^2-\mathrm{\parallel }{U}^{j-1}{\mathrm{\parallel }}_H^2+\alpha {\tau }_j\mathrm{\parallel }{U}^{\mathrm{m}\mathrm{i}\mathrm{d},j}{\mathrm{\parallel }}_V^2\le {\alpha }^{-1}{\tau }_j\mathrm{\parallel }{\bar{f}}_j{\mathrm{\parallel }}_{V_h^{\ast }}^2\mathrm{.}$ Sum through any $m$ and apply Jensen to obtain $\mathrm{\parallel }{U}^m{\mathrm{\parallel }}_H^2+\alpha {\sum }_{j=1}^m{\tau }_j\mathrm{\parallel }{U}^{\mathrm{m}\mathrm{i}\mathrm{d},j}{\mathrm{\parallel }}_V^2\le C_U\mathrm{.}$ Taking the maximum of the endpoint term and setting $m=N$ for the sum prove the separate stability bounds.

On $I_j=[a,b]$ of length ${\tau }_j$, let $L_j$ be the affine interpolant of $u_h(a),u_h(b)$. Since $r_j={\tau }_j^{-1}{\int }_{I_j}(u_h-L_j)$, solve $w^{\mathrm{\prime }\mathrm{\prime }}=u_h^{\mathrm{\prime }\mathrm{\prime }}$ with $w(a)=w(b)=0$ by two integrations. The Green kernel for $-d^2\mathrm{/}d{t}^2$ with zero endpoint values is $G_j(t,s)=\frac{(\min (t,s)-a)(b-\max (t,s))}{{\tau }_j}\mathrm{.}$ It has zero endpoint values and derivative jump $-1$ at $t=s$, so $w(t)=-{\int }_a^b{G}_j(t,s)u_h^{\mathrm{\prime }\mathrm{\prime }}(s)ds$; the difference between this $w$ and $u_h-L_j$ has second derivative zero and zero endpoint values, hence is zero. Therefore $u_h(t)-L_j(t)=-{\int }_a^b{G}_j(t,s)u_h^{\mathrm{\prime }\mathrm{\prime }}(s)ds\mathrm{.}$ The kernel satisfies $0\le G_j(t,s)\le {\tau }_j\mathrm{/}4$. Therefore $\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_V\le \frac{1}{{\tau }_j}{\int }_a^b\mathrm{\parallel }{u}_h^{\mathrm{\prime }\mathrm{\prime }}(s){\mathrm{\parallel }}_V({\int }_a^b{G}_j(t,s)dt)ds\le \frac{{\tau }_j}{4}{\int }_a^b\mathrm{\parallel }{u}_h^{\mathrm{\prime }\mathrm{\prime }}(s){\mathrm{\parallel }}_{V}ds\le \frac{{\tau }_j^{3\mathrm{/}2}}{4}\mathrm{\parallel }{u}_h^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(I_j;V)}$ by Tonelli's theorem for the nonnegative integrand and Cauchy–Schwarz. Consequently ${\sum }_j{\tau }_j\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_V^2\le \frac{{\tau }_{\max }^{}}{16}\mathrm{\parallel }{u}_h^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(0,T;V)}^2\mathrm{.}$

Test the error equation from step 1.2 by $e^{\mathrm{m}\mathrm{i}\mathrm{d},j}=(e^j+e^{j-1})\mathrm{/}2$, multiply by $2{\tau }_j$, and use the midpoint consequence of [F3]. Coercivity and Young's inequality give $\mathrm{\parallel }{e}^j{\mathrm{\parallel }}_H^2-\mathrm{\parallel }{e}^{j-1}{\mathrm{\parallel }}_H^2+\alpha {\tau }_j\mathrm{\parallel }{e}^{\mathrm{m}\mathrm{i}\mathrm{d},j}{\mathrm{\parallel }}_V^2\le \frac{M^2}{\alpha }{\tau }_j\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_V^2\mathrm{.}$ Sum through any $m$ and apply step 2.2 to obtain $\mathrm{\parallel }{e}^m{\mathrm{\parallel }}_H^2+\alpha {\sum }_{j=1}^m{\tau }_j\mathrm{\parallel }{e}^{\mathrm{m}\mathrm{i}\mathrm{d},j}{\mathrm{\parallel }}_V^2\le C_e\mathrm{.}$ Taking the maximum of the endpoint term and setting $m=N$ for the sum prove the separate error bounds.