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Put $h=\Delta x$, $k=\Delta t$ and $\mu =k/h$, consistent with $u_t=u_x$ and the exact transport $u(x,t)=F(x+t)$. The PDF's middle implicit value is $u_m^{n+1}$; the converted TeX incorrectly places it at $m+1$.

Write the new-level coefficients at offsets $-1,0,1$ as $a=\mu (1+\mu )/2$, $b=(1+\mu )(2-\mu )$ and $c=(1-\mu )(2-\mu )/2$, and the old-level coefficients as $d=2-\mu$, $e=1+\mu$. Their sums are $a+b+c=d+e=3$. To compute the local truncation error, expand the exact values $F(x_m+t_n+h(\mu +j))$ at the new level and $F(x_m+t_n+jh)$ at the old level. Equivalently the residual symbol isThe coefficients through $z^3$ vanish, whileFor example, the first three moment identities are ${\sum }_j{a}_j(\mu +j{)}^r=e$ for $r=1,2,3$, where $a_{-1}=a,a_0=b,a_1=c$; the fourth-moment difference is $\mu (\mu -2)(\mu -1)(\mu +1)$. Thus the exact residual isNormalizing the equation by $3k$ givesFor fixed nonzero $\mu \notin \{-1,1,2\}$, the method has order three. Its one-step residual is fourth degree, but the differential-equation truncation error is third order because of division by the time step.

There are exact transport exceptions. At $\mu =-1$ the equation reduces to $u_m^{n+1}=u_{m-1}^n$. At $\mu =2$ it gives $u_m^{n+1}=u_{m+2}^n$. At $\mu =1$, the exact shift $u_m^{n+1}=u_{m+1}^n$ satisfies the implicit equation, whose symbol is invertible as shown below. These are exact for every grid profile, not merely fourth-order exceptions. At $\mu =0$ the equation is the zero-time-step identity. These properties are summarized by the rational implicit advection stencil with exact shift exceptions.