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Let the spatial spacing be $d$, the time step $k$, and $\mu =k/d$. Abbreviate the three new-time coefficients by $a=\mu (1+\mu )/2$, $b=(1+\mu )(2-\mu )$, $c=(1-\mu )(2-\mu )/2$, and the two old-time coefficients by $d_0=2-\mu$, $e_0=1+\mu$. These are the coefficients of the compact third-order advection stencil.

For a smooth solution of the advection equation, $u_t=u_x$ implies $u(x,t+k)=u(x+k,t)$. Substituting the exact solution into the stencil therefore evaluates all five terms on the same characteristic profile, at spatial offsets $(\mu -1)d$, $\mu d$, $(\mu +1)d$, $0$, and $d$. Its Taylor moments areDirect polynomial expansion givesConsequently the unscaled local truncation error isAt a fixed nonzero Courant number, dividing by the time step gives an $O(d^3)$ consistency defect. ThusWhen $K(\mu )=0$, the order is higher because the method simplifies to an exact integer shift: at $\mu =1$ the two sides have the common factor $2+e^{-i\theta }$ and the amplification is $e^{i\theta }$; at $\mu =2$ it is $e^{2i\theta }$. The signed value $\mu =-1$ gives the shift $e^{-i\theta }$, and $\mu =0$ gives the identity. These special cases do not invalidate the third-order statement for the family. Under the stability conditions established next, the usual accumulation of the fourth-order local defect over $O(1/k)$ steps gives third-order convergence for smooth data.