Semi-Lagrangian characteristic transport — Examples

Since $b=1$, the explicit Euler foot at $x_0=0$ is $-3\mathrm{/}10$, which wraps to $7\mathrm{/}10$ on the unit torus. On the lift, $(-3\mathrm{/}10)\mathrm{/}h=-24\mathrm{/}5=-5+1\mathrm{/}5$, so the lower index is $-5\equiv 11(\mathrm{m}\mathrm{o}\mathrm{d}16)$, the upper index is $12$, and $\theta =1\mathrm{/}5$.

The update is therefore $(1-\theta )2+\theta 10=(4\mathrm{/}5)2+(1\mathrm{/}5)10=18\mathrm{/}5$. By [F2] this value lies in $[2,10]$, despite the 4.8-cell displacement. The example uses no displacement CFL restriction.

Each denominator in $u_0$ is nonzero because $j\mathrm{/}N\mathrm{\notin }\mathbb{Z}$, each factor is smooth and $1$-periodic, and at node $x_0=0$ every factor equals $1$. At node $x_m=m\mathrm{/}N$ with $1\le m<N$, the factor with $j=m$ vanishes. Thus the grid samples are exactly the stated unit vector. The exact characteristic is $X(t;s,x)=x+(t-s)$, so at $T=h$ the exact grid vector has value $1$ at node $1$ and $0$ at node $2$.

For $K\ge 2$, put $\alpha =\mathrm{\Delta }t\mathrm{/}h=1\mathrm{/}K$. Repeated application of [F1] is convolution with the two coefficients $(1-1\mathrm{/}K,1\mathrm{/}K)$; induction using Pascal's identity shows that after $K$ steps the contribution from exactly two one-node shifts at node $2$ is $\left(\genfrac{}{}{0px}{}{K}{2}\right)K^{-2}(1-1\mathrm{/}K{)}^{K-2}$. Periodic wrap can add other nonnegative contributions, so $U_2^K\ge \left(\genfrac{}{}{0px}{}{K}{2}\right)K^{-2}(1-1\mathrm{/}K{)}^{K-2}$.

Since $\left(\genfrac{}{}{0px}{}{K}{2}\right)K^{-2}\to 1\mathrm{/}2$ and $(1-1\mathrm{/}K{)}^{K-2}=(1-1\mathrm{/}K{)}^K\mathrm{/}(1-1\mathrm{/}K{)}^2\to e^{-1}$ by [F2], we have ${\mathrm{lim\hspace{0.17em}inf}}_{K\to \mathrm{\infty }}{U}_2^K\ge e^{-1}\mathrm{/}2>0$, whereas the exact value at node $2$ is zero. Thus reducing $\mathrm{\Delta }t$ at fixed $h$ does not remove this interpolation error. The sufficient condition $h^2\mathrm{/}\mathrm{\Delta }t\to 0$ in the global error theorem cannot be dropped from that bound: here $h^2\mathrm{/}\mathrm{\Delta }t=hK\to \mathrm{\infty }$.

The squared-sine factors are smooth and period $3$; the value is zero at nodes $0$ and $1$ and equals $1$ at node $2$. Direct substitution into $b$ gives $(b(0),b(1),b(2))=(1\mathrm{/}4,1\mathrm{/}4,3\mathrm{/}4)$. Therefore the three lifted feet are $-1\mathrm{/}4,3\mathrm{/}4,5\mathrm{/}4$. Their interpolation rows, in node order $0,1,2$, are $(3\mathrm{/}4,0,1\mathrm{/}4)$, $(1\mathrm{/}4,3\mathrm{/}4,0)$, and $(0,3\mathrm{/}4,1\mathrm{/}4)$.

Applying those rows to $(0,0,1)$ gives the new grid values $(1\mathrm{/}4,0,1\mathrm{/}4)$. The grid sum changes from $1$ to $1\mathrm{/}2$. This is a pointwise interpolation update, so the grid sum is a diagnostic only; the result shows why pointwise interpolation is not itself a conservative remap.

Since $b(x)>0$ on $(0,1\mathrm{/}2)$, the forward trajectory from $1\mathrm{/}4$ strictly increases and cannot reach the equilibrium $1\mathrm{/}2$ at finite time by uniqueness in [F1]. Thus $\mathrm{\Phi }(1\mathrm{/}4)>1\mathrm{/}4$, while $\mathrm{\Phi }(0)=0$ and $\mathrm{\Phi }(1\mathrm{/}2)=1\mathrm{/}2$. Monotonicity in [F2] gives $0<r<1\mathrm{/}4$. With representatives in the open fundamental interval $(0,1)$, $D_1=(0,r)$ and $D_2=[r,1)$, so $\mathrm{\mid }{D}_1\mathrm{\mid }=r$, $\mathrm{\mid }{D}_2\mathrm{\mid }=1-r$, and the nonzero overlaps with the occupied source cell $C_1\cap (0,1)$ are $\mathrm{\mid }{D}_1\cap C_1\mathrm{\mid }=r$ and $\mathrm{\mid }{D}_2\cap C_1\mathrm{\mid }=1\mathrm{/}4-r$.

Formula [F3] gives ${\bar{\rho }}_1=1$ and ${\bar{\rho }}_2=(1\mathrm{/}4-r)\mathrm{/}(1-r)$. Since $\mathrm{\mid }{A}_1\mathrm{\mid }=1\mathrm{/}4$ and $\mathrm{\mid }{A}_2\mathrm{\mid }=3\mathrm{/}4$, the arrival averages are ${\rho }_1^1=4r>0$ and ${\rho }_2^1=\frac{4}{3}(1\mathrm{/}4-r)>0$. Their total mass is $\frac{1}{4}{\rho }_1^1+\frac{3}{4}{\rho }_2^1=r+(1\mathrm{/}4-r)=1\mathrm{/}4$, equal to the initial mass $\frac{1}{4}\cdot 1+\frac{3}{4}\cdot 0$.