Variational crimes and Strang lemmas — Examples

The space $V$ has finite dimension, so [F4] makes it Banach and hence Hilbert for the stated inner product. By the power rule, ${\varphi }^{\mathrm{\prime }}(x)=1-2x$; the displayed polynomial is an antiderivative of $(1-2x{)}^2$ by the same rule, so [F3] gives $a(\varphi ,\varphi )={\int }_0^1(1-2x{)}^{2}dx=[x-2x^2+\frac{4}{3}{x}^3{]}_0^1=1\mathrm{/}3$. Bilinearity then gives $a(c\varphi ,d\varphi )=cd\mathrm{/}3=\frac{1}{3}(c\varphi ,d\varphi {)}_V$, verifying boundedness and coercivity with constant $1\mathrm{/}3$.

The midpoint is $1\mathrm{/}2$, where ${\varphi }^{\mathrm{\prime }}(1\mathrm{/}2)=0$. Thus [F1] gives $a_h(\varphi ,\varphi )=Q_M(({\varphi }^{\mathrm{\prime }}{)}^2)=0$, although $\varphi \mathrm{\ne }0$.

The stated norm gives $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_V=1$, so a coercivity constant ${\alpha }_h>0$ would require $0=a_h(\varphi ,\varphi )\ge {\alpha }_h$, contradicting step 2.1. Therefore the first-Strang stability hypothesis [F2] is absent; the calculation asserts no error rate for the unstable discrete problem.

The functional $L(g):={\int }_0^{1}g$ is linear by [F12] and bounded by [F2], since $\mathrm{\parallel }1{\mathrm{\parallel }}_2=1$ by [F10]. Thus $V=\ker L$ is a closed linear subspace of $L^2$. If $(g_n)$ is Cauchy in $V$, [F1] gives an $L^2$ limit $g$; continuity of $L$ gives $L(g)=0$, so $g\in V$. Hence $V$ is a Hilbert space.

For ${\omega }_h=c_L{\omega }_L+c_R{\omega }_R$, [F10, F11, F13] give $\mathrm{\parallel }{\omega }_h{\mathrm{\parallel }}_h^2={\int }_0^1\mathrm{\mid }{\omega }_h{\mathrm{\mid }}^2=2(c_L^2+c_R^2)$ and $\mathrm{\parallel }{U}_{{\omega }_h}{\mathrm{\parallel }}_2^2=(c_L^2+c_R^2)\mathrm{/}6=\mathrm{\parallel }{\omega }_h{\mathrm{\parallel }}_h^2\mathrm{/}12$, by integrating the bounded piecewise-polynomial representatives. They also give $\mathrm{\parallel }6x{\mathrm{\parallel }}_2^2={\int }_0^{1}36x^{2}dx=12$. A Cauchy sequence in $V_h$ therefore has Cauchy coefficient sequences; [F8] gives their limits and the same norm formula gives convergence in $V_h$. Thus $V_h$ is a two-dimensional Hilbert space. Moreover, ${\omega }_L+{\omega }_R=2$ is constant, and every $g\in L^2(0,1)$ decomposes as $g=(g-{\int }_0^{1}g)+\frac{1}{2}({\int }_0^{1}g)({\omega }_L+{\omega }_R)$; the first summand lies in $V$ and the second in $V_h$. Thus $X_h=V+V_h=L^2(0,1)$, with the common $L^2$ norm restricting to both spaces. The discrete form has coercivity constant $1$, and the extension ${\tilde{a}}_h(g,{\omega }_h)=\int g{\omega }_h$ is bounded by [F2]. Holder [F2] and the displayed finite $L^2$ norms show $F_h({\omega }_h)={\int }_0^{1}6x{U}_{{\omega }_h}(x)dx$ is bounded. Hence the stability and bounded-extension hypotheses of [F6] hold.

The polynomial $g_u=1-3x^2$ has Lebesgue integral zero by [F9, F11], so $g_u\in V$, and its primitive is $U_{g_u}=x-x^3$. By [F2], $\mathrm{\mid }{U}_h(x)\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_2$, hence $\mathrm{\parallel }{U}_h{\mathrm{\parallel }}_2\le \mathrm{\parallel }h{\mathrm{\parallel }}_2$ using [F10]. Since $\mathrm{\parallel }6x{\mathrm{\parallel }}_2^2=12$ by step 1.2, Holder [F2] bounds $F(h)={\int }_0^{1}6x{U}_h(x)dx$ for $h\in V$; [F3] makes the primitive representative-independent. The function $(x,t)\mapsto 6xh(t){\mathbf{1}}_{0\le t\le x\le 1}$ is absolutely integrable, with absolute integral at most $6\mathrm{\parallel }h{\mathrm{\parallel }}_1\le 6\mathrm{\parallel }h{\mathrm{\parallel }}_2$. Fubini [F4] gives $F(h)={\int }_0^{1}h(t)3(1-t^2)dt$. Linearity [F12] and $\int h=0$ yield $F(h)={\int }_0^{1}h(t)(1-3t^2)dt=a(g_u,h)$. The form is the $L^2$ inner product, hence bounded and coercive, so [F5] confirms the unique exact solution is $g_u$.

The associated broken function for ${\omega }_L$ is ${\varphi }_L$. By [F9, F11, F13], $F_h({\omega }_L)={\int }_0^{1\mathrm{/}2}12x^{2}dx=1\mathrm{/}2$, whereas ${\tilde{a}}_h(g_u,{\omega }_L)={\int }_0^{1\mathrm{/}2}2(1-3x^2)dx=2[x-x^3{]}_0^{1\mathrm{/}2}=3\mathrm{/}4$. The residual is $-1\mathrm{/}4$ and $\mathrm{\parallel }{\omega }_L{\mathrm{\parallel }}_h=\sqrt{2}$, so ${\delta }_h^{\mathrm{n}\mathrm{c}}(g_u)\ge 1\mathrm{/}(4\sqrt{2})>0$.

The slope field ${\omega }_L$ has integral $1$, so ${\omega }_L\mathrm{\notin }V$ and $V_h⊈V$. Thus the conforming hypothesis in [F7] fails; independently, step 2.2 shows that the exact solution does not satisfy the discrete equation on $V_h$. This is the consistency term retained by [F6].

By [F7], $\mu ([-1,1])=2$. Every relative neighborhood of a point in $[-1,1]$ contains a nondegenerate relative interval, so it has positive measure by [F7]; hence the relative support is all of $[-1,1]$ and contains at least three points. Each $\mathrm{\mid }x{\mathrm{\mid }}^k$, $0\le k\le 3$, is continuous and bounded on $[-1,1]$, hence Riemann integrable and Lebesgue integrable by [F8]. Thus the moment datum conditions [F1] hold.

Write $p_2(x)=x^2+bx+c$. Orthogonality [F2] to $1$ and $x$, using [F8] to read the Lebesgue moments as Riemann integrals, and [F5, F6] to evaluate ${\int }_{-1}^{1}1=2$, $\int x=0$, $\int x^2=2\mathrm{/}3$, and $\int x^3=0$, gives $2\mathrm{/}3+2c=0$ and $2b\mathrm{/}3=0$. Hence $p_2=x^2-1\mathrm{/}3$, and [F3] gives nodes $\pm 1\mathrm{/}\sqrt{3}$.

Using the nodes from step 1.2, the corresponding cardinal polynomials are ${\mathrm{\ell }}_{+}(x)=(x+1\mathrm{/}\sqrt{3})\mathrm{/}(2\mathrm{/}\sqrt{3})=(1+\sqrt{3}x)\mathrm{/}2$ and ${\mathrm{\ell }}_{-}(x)=(1\mathrm{/}\sqrt{3}-x)\mathrm{/}(2\mathrm{/}\sqrt{3})=(1-\sqrt{3}x)\mathrm{/}2$. By [F8], [F5], and [F6], their odd terms integrate to zero and their constant terms integrate to one, so [F3] gives both weights equal to $1$.

By [F4], this rule is exact for every polynomial of degree at most three. Using the nodes and weights from steps 1.2 and 2.1, its value on $x^4$ is $(1\mathrm{/}\sqrt{3}{)}^4+(-1\mathrm{/}\sqrt{3}{)}^4=2\mathrm{/}9$, whereas [F8], [F5], and [F6] give ${\int }_{-1}^1{x}^{4}d\mu =2\mathrm{/}5$. The absolute error is $2\mathrm{/}5-2\mathrm{/}9=8\mathrm{/}45$.