High-resolution TVD, ENO and WENO reconstruction — Examples

Write $x=h\mathrm{/}2+y$ on $I_0$, so $y\in [-h\mathrm{/}2,h\mathrm{/}2]$ and $u=1-y^2$. Averaging gives $U_0=1-h^2\mathrm{/}12$. On either neighboring cell the centered coordinate $y$ has mean $\pm h$ and variance $h^2\mathrm{/}12$, hence $U_{-1}=U_1=1-(h^2+h^2\mathrm{/}12)=1-13h^2\mathrm{/}12$.

Therefore $U_0-U_{-1}=h^2$ and $U_1-U_0=-h^2$. Their signs differ, so [F1] gives $s_0=0$ and both reconstructed face states equal $U_0$.

The exact profile at either face $x=h\mathrm{/}2\pm h\mathrm{/}2$ is $1-h^2\mathrm{/}4$. The reconstructed face error is $\mathrm{\mid }{U}_0-(1-h^2\mathrm{/}4)\mathrm{\mid }=h^2\mathrm{/}6$. The reconstructed value equals the largest of the three cell averages, so this clipping introduces no new face value outside their range while retaining a second-order face error.

The initial sequence has one jump from $1$ to $0$, so $\mathrm{TV}(U^0)=1$. The centered increments are ${\sigma }_{-1}=(U_0-U_{-2})\mathrm{/}2=-1\mathrm{/}2$ and ${\sigma }_0=(U_1-U_{-1})\mathrm{/}2=-1\mathrm{/}2$; all other increments vanish.

The relevant upwind fluxes are $F_{-3\mathrm{/}2}=1$, $F_{-1\mathrm{/}2}=3\mathrm{/}4$, $F_{1\mathrm{/}2}=-1\mathrm{/}4$, and $F_{3\mathrm{/}2}=0$. With $\mathrm{\Delta }t\mathrm{/}h=1\mathrm{/}2$, the update gives $U_{-1}^1=9\mathrm{/}8$, $U_0^1=1\mathrm{/}2$, and $U_1^1=-1\mathrm{/}8$, with the adjacent exterior states still $1$ and $0$.

The new variation is $\mathrm{\mid }9\mathrm{/}8-1\mathrm{\mid }+\mathrm{\mid }1\mathrm{/}2-9\mathrm{/}8\mathrm{\mid }+\mathrm{\mid }-1\mathrm{/}8-1\mathrm{/}2\mathrm{\mid }+\mathrm{\mid }0-(-1\mathrm{/}8)\mathrm{\mid }=3\mathrm{/}2>1$. The method creates both an overshoot and an undershoot at a CFL number of one half, refuting the statement.

Since the average of an affine function over a cell equals its value at the cell center, $U_i^n=x_i-a{t}_n$. Hence $U_i^n-U_{i-1}^n=U_{i+1}^n-U_i^n=h$ and the minmod slope is $s_i=1$. The right MUSCL trace is $U_i^n+h\mathrm{/}2=(i+1)h-a{t}_n=u(x_{i+1\mathrm{/}2},t_n)$, with the reflected left trace giving the same exact value at the face.

The WENO candidates reproduce affine polynomials by [F2], so $p_0=p_1=p_2=u(x_{i+1\mathrm{/}2},t_n)$. Each normalized weight triple therefore returns this same exact value. The uniform smooth profile also satisfies [F3] and [F4], with zero error here because it is affine.

For $a>0$, the upwind flux is $a((i+1)h-a{t}_n)$ and the left face flux is $a(ih-a{t}_n)$; their difference is $ah$, so the update gives $U_i^{n+1}=x_i-a{t}_n-a\mathrm{\Delta }t$, the exact average at $t_{n+1}$. For $a<0$ the reflected right trace yields the same interface value and flux difference; for $a=0$ both the flux and time change vanish.

The five stencil averages, ordered from $i-2$ through $i+2$ with $i=-1$, are $(U_{i-2},U_{i-1},U_i,U_{i+1},U_{i+2})=h^2(19\mathrm{/}3,7\mathrm{/}3,1\mathrm{/}3,1\mathrm{/}3,7\mathrm{/}3)+h^3(-65\mathrm{/}4,-15\mathrm{/}4,-1\mathrm{/}4,1\mathrm{/}4,15\mathrm{/}4)$. Substitution in the three candidate formulas gives $(p_0,p_1,p_2)=h^3(-3\mathrm{/}2,1\mathrm{/}2,-1\mathrm{/}2)$; the quadratic contributions vanish by [F2].

Substitution of the same averages in the indicator formulas gives exactly ${\beta }_0=(16\mathrm{/}3)h^4-37h^5+(355\mathrm{/}4)h^6$, ${\beta }_1=(16\mathrm{/}3)h^4-17h^5+(55\mathrm{/}4)h^6$, and ${\beta }_2=(16\mathrm{/}3)h^4+15h^5+(43\mathrm{/}4)h^6$. Put $C=19\mathrm{/}3$ and $(c_0,c_1,c_2)=(-37,-17,15)$; then ${ϵ}_h+{\beta }_k=C{h}^4+c_k{h}^5+O(h^6)$.

Expanding ${\alpha }_k^J=d_k(C{h}^4+c_k{h}^5+O(h^6){)}^{-2}$ and normalizing gives ${\omega }_k^J=d_k[1-2(c_k-\bar{c})h\mathrm{/}C+O(h^2)]$, where $\bar{c}=\sum d_k{c}_k=-47\mathrm{/}5$. Since $\sum d_k{p}_k=0$ by [F2], and ${\sum }_k{d}_k(c_k-\bar{c})(-3\mathrm{/}2,1\mathrm{/}2,-1\mathrm{/}2{)}_k=-9\mathrm{/}5$, the reconstructed state is ${\sum }_k{\omega }_k^J{p}_k=-\frac{2h^4}{C}(-9\mathrm{/}5)+O(h^5)=\frac{54}{95}{h}^4+O(h^5)$. Thus the face error is not $O(h^5)$.

By step 3.1 the scaled-epsilon JS state has a nonzero $h^4$ leading error. For WENO-Z, ${\tau }_5=\mathrm{\mid }-52h^5+78h^6\mathrm{\mid }=52h^5+O(h^6)$, so ${\tau }_5\mathrm{/}({ϵ}_h+{\beta }_k)=(52\mathrm{/}C)h+O(h^2)$ and its square has the same leading $O(h^2)$ term for each $k$. That common term cancels on normalization, giving ${\omega }_k^Z-d_k=O(h^3)$. Since the ideal combination is exactly zero on this cubic, [F2] gives ${\sum }_k{\omega }_k^Z{p}_k=O(h^6)$. This establishes the stated scaled-epsilon contrast; the fixed-${ϵ}_0$ theorem in [F3] remains a separate branch.

The jump data are bounded and locally integrable. For any translation length $r>0$, $\mathrm{\mid }{u}_0(x+r)-u_0(x)\mathrm{\mid }$ equals $1$ on an interval of length $r$ and vanishes elsewhere, so $\mathrm{TV}(u_0)=1$ by [F1]. The data are not in global $L^1(\mathbb{R})$, so [F3] uses only its ${\mathrm{A}\mathrm{C}}_{\omega }$ branch and imposes no DC assumption.

For $0\le a\le b\le 1$, monotonicity of $f$ gives ${\min }_{v\in [a,b]}f(v)=f(a)$; for $0\le b<a\le 1$, it gives ${\max }_{v\in [b,a]}f(v)=f(a)$. Thus the flux defined in the Example is $F_G(a,b)=a^2\mathrm{/}2$ on $[0,1{]}^2$. Use the same polynomial $F(a,b)=a^2\mathrm{/}2$ on all of ${\mathbb{R}}^2$ for the theorem's globally defined input. Its restriction is consistent, nondecreasing in $a$, independent of $b$, and satisfies $\mathrm{\mid }F(a,b)-F(a^{\mathrm{\prime }},b)\mathrm{\mid }\le \mathrm{\mid }a-a^{\mathrm{\prime }}\mathrm{\mid }$ on $[0,1{]}^2$; hence $L_1=1,L_2=0$ are valid in [F2]. For the particular jump $(1,0)$, [F5] gives speed $s=(f(1)-f(0))\mathrm{/}(1-0)=1\mathrm{/}2>0$, so the interface ray lies on the left state and the flux is the physical Riemann flux.

Here ${\lambda }_m={\tau }_m\mathrm{/}{h}_m=1\mathrm{/}2$, $h_m,{\tau }_m\to 0$, and ${\lambda }_m(L_1+L_2)=1\mathrm{/}2\le 1$ with ${\lambda }_m\ge 1\mathrm{/}2$. Therefore [F3] applies to the cell-average recurrence and gives convergence to the unique bounded entropy solution, provided the explicit shock below is identified as that solution.

Define $u(t,x)=1$ for $x<t\mathrm{/}2$ and $u(t,x)=0$ for $x>t\mathrm{/}2$. The jump speed is $s=1\mathrm{/}2=(f(1)-f(0))\mathrm{/}(1-0)$, so [F5] gives the weak jump law. For $0\le k\le 1$, the Kruzhkov entropy-flux jump is $q_k(0)-q_k(1)-\frac{1}{2}(\mathrm{\mid }0-k\mathrm{\mid }-\mathrm{\mid }1-k\mathrm{\mid })=k^2-k\le 0$; for $k\le 0$ and $k\ge 1$ direct substitution gives zero. The initial trace is strong locally because the shock moves a distance $t\mathrm{/}2\to 0$. Hence [F4] identifies this profile as a bounded Kruzhkov solution; uniqueness in [F3] makes it the convergence target from step 2.1.

All cell averages are constant on each side of the initial interface, and both neighboring differences change sign or vanish at the two cells adjacent to it; [F6] therefore gives zero minmod slope in every cell. At the first step $F(1,0)=F(1,1)=1\mathrm{/}2$ and $F(0,0)=0$, so the cell to the right of the interface becomes $0-(1\mathrm{/}2)(0-1\mathrm{/}2)=1\mathrm{/}4$, while the cell to the left remains $1$. The exact shock at time $\tau =h\mathrm{/}2$ has advanced by $h\mathrm{/}4$, so its average on the first right cell is also $1\mathrm{/}4$.