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A Butcher tableau records a Runge-Kutta method's stage matrix $A$, time nodes $c$, and output weights $b$ in the block arrayRead row $i$ above the line as the recipe for Runge-Kutta stage $i$: the left entry gives $t_n+c_{i}h$, and the right entries give the coefficients in $Y_i=y_n+h{\sum }_j{a}_{ij}{k}_j$. Below the line, $b_i$ weights slope $k_i$ in the final update. Zeros on and above the diagonal identify an Explicit Runge-Kutta method. For the internally consistent convention, row sums of $A$ equal $c$; first-order consistency of a numerical method also requires ${\sum }_i{b}_i=1$. An embedded Runge-Kutta pair adds another weight row while retaining $A,c$.

For a one-step approximation $y_{n+1}=y_n+h\varphi (t_n,y_n,h)$ to an ordinary differential equation, insert the exact solution and define the unscaled local truncation errorConsistency of a numerical method requires ${\tau }_{n+1}/h\to 0$ uniformly on the finite interval being approximated, under the stated smoothness assumptions. Merely asking ${\tau }_{n+1}\to 0$ is too weak: the useless update $y_{n+1}=y_n$ has defect $h{y}^{\prime }(t_n)+O(h^2)$, which tends to zero but generally fails the scaled test. Here $O$ denotes Big O notation. MIT uses this division-by-step convention.^[ref]

For the Euler method, Taylor's theorem gives ${\tau }_{n+1}=h^2{y}^{\prime \prime }({\xi }_n)/2$. A bounded second derivative therefore gives local truncation error $O(h^2)$ and a scaled defect $O(h)$. Under a uniform Lipschitz continuity assumption on $f$ in $y$, within a region containing the exact and numerical solutions, numerical stability controls accumulation over about $T/h$ steps. This yields global error $O(h)$ with exact starting data. Consistency of a numerical method checks each step; numerical stability controls their combination.

For an initial value problem, convergence means that the largest global error on each fixed finite time interval tends to zero as the step size tends to zero and the starting errors vanish. For linear multistep methods, the Dahlquist equivalence theorem separates this requirement into consistency of a numerical method and zero-stability.

For $u_t=u_x$, the recurrence $U_m^{n+1}-U_m^{n-1}={\sum }_{j=\pm 1,\pm 2}{a}_j{U}_{m+j}^n$ has fourth-order normalized consistency of a numerical method at fixed $\mu =k/h$ ifTo derive this, match moments $M_r={\sum }_j{a}_j{j}^r$ to the temporal Taylor series: $M_0=M_2=M_4=0$, $M_1=2\mu$, $M_3=2{\mu }^3$. The even conditions force antisymmetry; the two odd conditions then determine the displayed coefficients. The normalized residual begins with $h^4({\mu }^2-1)({\mu }^2-4)u_{xxxxx}/120$. Its amplification polynomial of a multilevel finite difference scheme is $G^2-2iA(\theta )G-1$, where $A=a_1\sin \theta +a_2\sin 2\theta$. At $\mu =1/2$, $|A|\le 3/4$, giving a uniform power bound. At $\mu =3/2$, $A(\pi /3)=19\sqrt{3}/32>1$, and wave packets on the growing polynomial root prove Cauchy linear instability.

For the square-grid nine-point finite-difference stencil for the Laplacian,If $\Delta u=0$, both displayed correction terms vanish: the mixed sixth derivatives sum to $\Delta u_{xxyy}=0$, and the pure sixth derivatives sum to the negative of that sum. Expanding two orders further yields $D_{9}u=h^6{u}_{xxxxxxxx}/3024+O(h^8)$ for a smooth harmonic function. Thus the compact harmonic scheme has sixth-order normalized consistency of a numerical method and, with the discrete maximum principle inverse bound, sixth-order nodal convergence on a square with exact smooth Dirichlet boundary data. For example $\mathrm{Re}(x+iy{)}^8$ shows that the sixth-order term need not vanish.

A Mehrstellen method increases the accuracy of a compact finite difference method discretization by using the differential equation to express its leading truncation terms in terms of the prescribed source. For $\Delta u=f$, the square-grid nine-point finite-difference stencil for the Laplacian satisfies $D_{9}u=\Delta u+h^2{\Delta }^{2}u/12+O(h^4)$. Replacing ${\Delta }^{2}u$ by $\Delta f$ and approximating this source derivative gives $D_{9}U=(I+h^2{D}_5/12)f$, with fourth-order normalized consistency of a numerical method while retaining a compact nine-point matrix. The uncorrected nine-point finite-difference stencil operator is only second-order accurate on general functions.

A numerical flux estimates the rate of transfer across a cell face from nearby states. In the finite volume method update, $F_{j+1/2}^n$ approximates the conservation law flux averaged over the time step; $k{F}_{j+1/2}^n$ is the transferred amount. For a two-state numerical flux $F(L,R)$ approximating a scalar conservation law with conservation law flux $f$, consistency of a numerical method requires $F(U,U)=f(U)$. Thus equal states recover the physical transfer rate. With constant equal-width cells, this also preserves a constant state because its incoming and outgoing numerical fluxes agree.^[ref]

Consistency and shared face flux cancellation answer different questions: matching $f$ at equal states approximates the intended equation, while sharing one oriented face value preserves the total. The example periodic upwind finite volume advection uses $F(L,R)=aL$ for $a>0$, so the arriving state comes from the left. The Godunov numerical flux uses an entropy solution of a local Riemann problem; the Engquist-Osher flux provides a different construction for a scalar conservation law. A monotone conservative scheme uses a numerical flux nondecreasing in its left state and nonincreasing in its right state, together with a time step that makes the complete update monotone. Neither two-state consistency nor conservation by itself supplies that time-step restriction or a general stability or order of a numerical method guarantee.

Use the coefficient $2-\alpha$ in the numerator, as verified in the PDF; the converted TeX lost its $-\alpha$. Multiplying the method by $2+\alpha$ gives the characteristic polynomials of a linear multistep methodFor every admissible real $\alpha$, $\rho (1)=0$ and ${\rho }^{\prime }(1)=2\alpha =\sigma (1)$, giving consistency of a numerical method. Factorization givesThe second root is $r=(2-\alpha )/(2+\alpha )$. The root condition for a multistep method requires $|r|\le 1$ and forbids a repeated root at one. For real $\alpha \ne -2$,At $\alpha =0$ both roots equal one, violating zero-stability. For $\alpha >0$ the second root lies strictly inside the unit disk. The Dahlquist equivalence theorem therefore givesTo find the order rather than only consistency, expand the exact-solution residual through the generating function:Thus its unnormalized local truncation error begins with $(\alpha /3)h^3{y}^{(3)}$ when $\alpha >0$, and the global method order is exactly two for every convergent parameter. For comparison, $\alpha =0$ has formal order three but is not convergent because of its double unit root:These orders assume a smooth solution, the usual local Lipschitz condition and starting approximations of the required accuracy.

For one stage, first-order consistency of a numerical method gives $b_1=1$. The condition $b^T{A}^{-1}1=1$ then gives $a_{11}=1$. Hence the scheme is the Backward Euler method:It is a one-node collocation Runge-Kutta method with $c_1=1$. Specifically, choose a linear polynomial $p(t)$ on $[t_n,t_n+h]$ with $p(t_n)=y_n$ and impose $p^{\prime }(t_n+h)=f(t_n+h,p(t_n+h))$. Its derivative is constant, so $p(t_n+h)-p(t_n)=h{p}^{\prime }(t_n+h)$, exactly the Backward Euler method update.

Expanding the symmetric spatial stencil by the Taylor theorem givesConsistency of a numerical method and cancellation of the $h^2$ term require $\alpha +2\beta +2\gamma =0$, $\beta +4\gamma =1$, and $\beta +16\gamma =0$. Solving this linear system givesThe resulting fourth-order centered second derivative hasThe nonzero next coefficient proves that the highest spatial order of a numerical method is four for this three-parameter stencil.

A constant-step linear multistep method has the formwith ${\alpha }_k\ne 0$. The characteristic polynomials of a linear multistep method encode both accuracy and error propagation. The method has order of a numerical method $p$ when insertion of a sufficiently smooth exact solution leaves an unnormalized local truncation error $O(h^{p+1})$. The exponential-symbol order criterion for a multistep method expresses this as $\rho (e^z)-z\sigma (e^z)=O(z^{p+1})$. In particular, first-order consistency of a numerical method requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Accuracy alone is insufficient. For $f=0$, perturbations obey the recurrence $\rho (E)e_n=0$. The root condition for a multistep method requires every root of a polynomial $\rho$ to lie in the closed unit disk, with each unit-circle root simple. This is zero-stability. A root outside the unit disk causes geometric amplification as $n\sim T/h$ increases; a repeated unit-circle root permits polynomial growth in $n$. Either can destroy convergence even for starting perturbations tending to zero.

The Dahlquist equivalence theorem states that, for a fixed-coefficient method and a well-posed ordinary differential equation with a suitably Lipschitz continuous right-hand side, consistency of a numerical method and zero-stability are equivalent to convergence for all convergent starting data. If the starting errors are $O(h^p)$ and the exact solution is sufficiently smooth, the global error is $O(h^p)$ on a fixed time interval. A stability estimate bounds the propagated sum of $O(h^{p+1})$ local defects over $O(1/h)$ steps; the discrete Gronwall inequality controls the dependence of $f$ on the accumulated error. Implicit methods also require the locally consistent, uniquely solvable update branch for sufficiently small $h$.

For example, the explicit Euler method has order one and $\rho (\zeta )=\zeta -1$, satisfying the root condition for a multistep method. The two-step Adams-Bashforth method has order two and $\rho (\zeta )=\zeta (\zeta -1)$, again satisfying it. The trapezoidal rule has order two with one step. The two-step Simpson multistep method, $Y_{n+2}-Y_n=h(f_{n+2}+4f_{n+1}+f_n)/3$, has order four and the simple roots $1,-1$: it is zero-stable, although its parasitic amplification root can spoil long-time decay. These examples distinguish finite-time numerical convergence from absolute stability.

The first Dahlquist barrier gives the maximum possible order of a convergent ordinary real $k$-step method:These bounds assume only first-derivative evaluations, fixed coefficients and zero-stability; the sixth-order multiderivative multistep method in solution is outside that class.

To obtain arbitrarily high convergent order within the ordinary class, increase the number of steps. The Adams-Bashforth method integrates a polynomial interpolation of $k$ past derivative values and has order $k$. The Adams–Moulton method includes the new endpoint and has order $k+1$ for $k$ steps. In both families $\rho (\zeta )={\zeta }^{k-1}(\zeta -1)$, so the root condition for a multistep method holds for every fixed $k$. Starting values can be generated by a Runge-Kutta method of sufficient accuracy. The backward differentiation formula instead differentiates an interpolating polynomial; its order-$k$ members are zero-stable only for $1\le k\le 6$.

Finally, high order and favorable stiff stability are separate design goals. By the Second Dahlquist barrier, an ordinary A-stable linear multistep method has order at most two. One may accept a smaller linear stability domain, use suitable backward differentiation formula members for stiff problems, or switch to high-order implicit Runge-Kutta methods or multiderivative multistep methods. A convergent high-order formula still needs time steps appropriate to its stability properties and the problem's eigenvalues.

A spatial finite difference method and time discretization for a linear evolution problem give a recurrence $U^{n+1}=B_{h,k}{U}^n$. To obtain numerical stability on $0\le nk\le T$, one needs a bound $\Vert B_{h,k}^n\Vert \le C_T$ uniform in the permitted grid sizes and time steps. More generally a growth estimate $\Vert B_{h,k}^n\Vert \le C{e}^{cnk}$ is sufficient. The finite-time stability versus power boundedness distinction matters: growth per step of $1+O(k)$ can be compatible with the latter finite-time estimate, whereas unrestricted geometric growth at a fixed Courant number is not.

The eigenvalue method diagonalizes $B_{h,k}$ into modes. If $B_{h,k}$ is a normal matrix, the spectral theorem gives $\Vert B_{h,k}^n{\Vert }_2={\max }_j|{\mu }_j{|}^n$. Thus $|{\mu }_j|\le 1$ gives contraction, and $|{\mu }_j|\le e^{ck}$ gives the preceding growth estimate. A diagonalizable matrix $B=V\Lambda V^{-1}$ has $\Vert B^n{\Vert }_2\le \Vert V{\Vert }_2\Vert V^{-1}{\Vert }_2{\max }_j|{\mu }_j{|}^n$, so its condition number must remain bounded uniformly as the mesh changes. A stable-looking list of eigenvalues alone is insufficient.

Indeed,Both eigenvalues are one, but the Jordan block gives growth proportional to $n\sim T/k$, violating uniform finite-time numerical stability. A nonnormal Fourier amplification matrix can similarly exhibit large amplification even with its eigenvalues inside the unit disk. Eigenvector conditioning and the full powers of the amplification operator are essential.

For constant-coefficient schemes on a periodic or infinite grid, a Fourier transform diagonalizes scalar translation-invariant stencils. This yields von Neumann stability analysis using the Fourier amplification symbol $G(\theta )$. The discrete Parseval identity turns a bound on $G(\theta {)}^n$ into a discrete L2 norm estimate. For systems, the symbol is a matrix; a uniform power bound for matrix Fourier symbols is required, rather than a bound on its spectral radius alone.

For the heat equation, a central finite difference in space and the explicit Euler method in time giveThe scalar condition $|G|\le 1$ for every frequency is exactly $0\le r\le 1/2$. With zero Dirichlet boundary conditions, the spatial matrix is symmetric and diagonalized by sine modes, leading to the same uniform sufficient restriction. By contrast, the Backward Euler diffusion scheme has $G(\theta )=[1+4r{\sin }^2(\theta /2){]}^{-1}$ and is stable for every $r\ge 0$.

For the advection equation $u_t+a{u}_x=0$, $a>0$, centered space with the explicit Euler method gives $G=1-i\nu \sin \theta$, $\nu =ak/h$, and $|G{|}^2=1+{\nu }^2{\sin }^2\theta$. At fixed positive Courant number, repeated powers are unbounded as the mesh is refined. The upwind finite difference scheme instead haswhich gives contraction precisely for $0\le \nu \le 1$. This comparison shows why the spatial stencil and time method must be analyzed together.

Boundary treatment can destroy the normality or mode decomposition available on a periodic grid, so a finite-interval scheme needs its own uniform operator estimate. Multilevel time schemes likewise require the root condition for a multistep method and uniform bounds for the associated amplification matrices; simple scalar root tests do not establish these bounds for arbitrary systems. Once consistency of a numerical method, a well-posed linear evolution problem and the required numerical stability estimate hold, the Lax equivalence theorem gives numerical convergence. Eigenvalues provide a powerful route to that estimate when the accompanying eigenvector and boundary analysis is justified.

The first characteristic polynomial factors asAlso $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the order conditions for a linear multistep method hold for every $\alpha$. Convergence of the full linear multistep method recurrence requires, in addition, the root condition for a multistep method: all polynomial roots of $\rho$ lie in the closed unit disk and its polynomial roots on the unit circle are simple.

For $-1<\alpha <1$, write $\alpha =\cos \theta$ with $0<\theta <\pi$. The other polynomial roots are $e^{\pm i\theta }$, distinct from each other and from 1, so the root condition for a multistep method holds. For $|\alpha |>1$, the two quadratic polynomial roots are real and reciprocal, and one has modulus greater than one. At $\alpha =1$, $\rho =(w-1{)}^3$; at $\alpha =-1$, $\rho =(w-1)(w+1{)}^2$. Both endpoints violate simplicity. Consistency of a numerical method plus zero-stability therefore givesThe conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha =1$ the polynomials contain $(w-1{)}^2$, and at $\alpha =-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the common-factor cancellation defect in a multistep recurrence, so those parameter values are not added to the interval.

Let $h=\Delta x$, $k=\mu h$ and $M_r={\sum }_{j\in \{-2,-1,1,2\}}{a}_j{j}^r$. Along a smooth solution, ${\partial }_t^{r}u={\partial }_x^{r}u$. The exact difference $u(x,t+k)-u(x,t-k)$ contains only odd derivatives. Matching terms through degree four therefore requiresThe first two conditions give $a_{-1}+a_1=a_{-2}+a_2=0$, which also implies $M_4=0$. With $a_{-j}=-a_j$, the remaining equations are $a_1+2a_2=\mu$ and $a_1+8a_2={\mu }^3$. HenceThe fifth spatial moment is $M_5=10{\mu }^3-8\mu$. After dividing the exact-minus-scheme residual by $2k$, its first possible nonzero term isThis is the fourth-order two-step advection stencil. Thus the normalized local truncation error is $O(h^4)$ at fixed positive Courant number, as required. At $\mu =1$ or $2$ the formula translates the physical advection branch exactly, but consistency of a numerical method alone does not decide stability of all two-level perturbations.

The Engquist-Osher method is an upwind finite volume method. For cell values $U_j^n$, mesh width $h$ and time step $k$, put $r=k/h$ and use the conservative updatewhere the numerical flux isConsistency of a numerical method follows from $F(u,u)=f(u)$. Positive characteristic speeds use the left state, and negative speeds use the right state. The flux difference telescopes when summed over cells, giving a discrete conservation law. The method is first order in space and time for smooth solutions; its conservation law form is also appropriate when shock waves develop.

Let $s_{\ast }$ be the unique sonic point of a scalar flux, $f^{\prime }(s_{\ast })=0$. Convexity implies $f^{\prime }<0$ to its left and $f^{\prime }>0$ to its right. Put $f_{\ast }=f(s_{\ast })$ and splitThen $f=f_{\ast }+f^{+}+f^{-}$, $(f^{+}{)}^{\prime }\ge 0$, $(f^{-}{)}^{\prime }\le 0$, andThis is the sonic-point splitting of a convex Engquist-Osher flux. If both states are above $s_{\ast }$, $F=f(a)$; if both are below it, $F=f(b)$. Across a transsonic rarefaction wave, $a\le s_{\ast }\le b$, the flux is the minimum $f_{\ast }$. In the opposite crossing it is $f(a)+f(b)-f_{\ast }$, which in general differs from the exact Godunov numerical flux. This distinction does not affect the monotonicity proof.

Here stability requires the usual Courant–Friedrichs–Lewy condition. Let the initial values lie in $[m,M]$, and chooseA differentiable convex function has a continuous derivative on this interval, so the maximum is finite. Write the three-input update as $H(a,b,c)$. Differentiating givesThus the update $T$ preserves componentwise order. Constants are fixed, since $H(c,c,c)=c$, so comparison with the constants $m,M$ proves $m\le U_j^n\le M$ for every time level. In particular the same CFL condition remains applicable. This is a mesh-independent maximum-principle stability bound.

A stronger stability statement is contraction in discrete $L^1$. Consider two solutions $U,V$ in the same invariant interval, on a periodic mesh or on the infinite lattice with summable differences. Let $Q=U\vee V$ be their componentwise maximum. Monotonicity gives $TQ\ge TU,TV$, henceThe discrete conservation law for the difference gives the equality: on a finite periodic mesh the flux sum cancels exactly, and on the infinite lattice it follows by truncation and the fact that the flux is a Lipschitz function on $[m,M]$. Interchanging $U,V$ and adding provesThis is L1 contraction of a monotone conservative scheme. Since $T$ commutes with the cell shift, applying this contraction to $U$ and its shift gives ${\sum }_j|U_{j+1}^{n+1}-U_j^{n+1}|\le {\sum }_j|U_{j+1}^n-U_j^n|$. The scheme is therefore a total variation diminishing scheme as well.

The same order argument proves a discrete entropy inequality, explaining why this stable discretization selects entropy solutions rather than nonphysical expansion shock waves. For a constant $c\in [m,M]$, setBecause $T(U\vee c)\ge TU,c$ and $T(U\wedge c)\le TU,c$, their componentwise difference is at least $|TU-c|$. Subtracting their updates givesFor constants outside $[m,M]$, the entropy inequality is an equality by the invariant bound and the conservative update. At equal states $Q_c(u,u)=\mathrm{sgn}(u-c)[f(u)-f(c)]$, the Kruzhkov entropy flux. The flux is a Lipschitz function on the invariant interval: writing $L=\max |f^{\prime }|$, the conservative update gives $h{\sum }_j|U_j^{n+1}-U_j^n|\le 2kL{\sum }_j|U_{j+1}^n-U_j^n|$. This time-translation bound, the invariant bound and the total variation estimate give local space-time compactness for bounded-variation initial conditions as the mesh is refined; the conservative update passes to the weak conservation law, and the displayed inequality passes to its Kruzhkov entropy inequality. This identifies the limit as the entropy solution.

Finally the time-step qualification is essential. For the convex Inviscid Burgers equation flux $f(u)=u^2/2$, whose unique sonic point of a scalar flux is zero, linearize about a positive constant $U$. The perturbation update becomes $v_j^{n+1}=(1-rU)v_j^n+rU{v}_{j-1}^n$. At Fourier frequency $\pi$ its multiplier is $1-2rU$, with modulus greater than one if $rU>1$. Wave packets near this phase give linear instability even for summable Cauchy perturbations. Thus the assumptions on $f$ imply the stability results above under the CFL condition, not for arbitrary $k/h$. This is CFL necessity for explicit Engquist-Osher stability.

A Mehrstellen method gains accuracy by exploiting the differential equation in the local truncation error of a compact finite difference method stencil. The extra nearby values alone do not guarantee higher order: one must also modify the source term. Consider $\Delta u=f$ on a square with prescribed Dirichlet boundary data, using the same spacing $h$ in both directions.

For comparison, the five-point Laplacian has expansionIts leading error is not a multiple of ${\Delta }^{2}u$, because the mixed fourth derivative is missing. To obtain an isotropic leading error, consider a symmetric nine-point finite-difference stencil with weight $a$ on each axial neighbor, $b$ on each diagonal neighbor, and $c$ on the center, all divided by $h^2$. Vanishing on constants and consistency of a numerical method require $4a+4b+c=0$ and $a+2b=1$. Taylor expansion then gives the fourth-derivative contributionFor this to equal $h^2{\Delta }^{2}u/12$ we must choose $b=1/6$, hence $a=2/3$, $c=-10/3$. The resulting nine-point finite-difference stencil operator isIts expansion, retaining the next term for later use, isThus $D_9$ by itself is second order on a general function. On a Poisson equation solution, however, ${\Delta }^{2}u=\Delta f$. Approximating this known source derivative by $D_{5}f$ gives the fourth-order Mehrstellen method equationEquivalently, its two stencils areThe normalized local truncation error is $O(h^4)$, since $D_{5}f-\Delta f=O(h^2)$. If the equation is multiplied by $6h^2$, the row residual is $O(h^6)$; this change of normalization does not make the solution sixth order. The fourth-order correction of the nine-point Poisson stencil retains only the closest axial and diagonal unknowns. A fourth-order one-dimensional second derivative instead uses $(-U_{i+2}+16U_{i+1}-30U_i+16U_{i-1}-U_{i-2})/(12h^2)$, widening the unknown stencil and introducing negative outer weights. Mehrstellen method accuracy is obtained by correcting known source values rather than widening that unknown stencil.

The compact scheme has useful stability and solvability properties. With homogeneous Dirichlet boundary data, $-D_9$ is symmetric, has positive diagonal and nonpositive off-diagonal entries, and is a positive definite symmetric operator. Indeed its quadratic form is a sum of positive edge weights times squared differences, with boundary values set to zero. Vanishing of this sum would force a constant on the connected grid and then zero by connection to the boundary. Consequently the Dirichlet boundary condition linear system has a unique solution. The same weights give a discrete maximum principle: if $D_{9}v\ge 0$ at every interior node and $v\le 0$ on the boundary, a positive maximum inside would make every weighted difference to its neighbors nonpositive. Equality forces the maximum to propagate to the boundary, which is impossible. Thus $v\le 0$ everywhere.

This also turns consistency of a numerical method into a quantitative convergence result, rather than merely suggesting an order. Let $B_{h}f=f+h^2{D}_{5}f/12$, let $\tau =D_{9}u-B_{h}f$ be the normalized residual of the exact solution, and set $e=U-u$. Exact Dirichlet boundary data imply $e=0$ on the boundary and $D_{9}e=-\tau$. On the unit square the quadratic barrierhas $D_{9}q=-1$ exactly, is nonnegative on the boundary, and has maximum $1/8$. If $\Vert \tau {\Vert }_{\infty }\le \epsilon$, then $D_9(e-\epsilon q)\ge 0$ and $D_9(-e-\epsilon q)\ge 0$, while both comparison functions are nonpositive on the boundary. The discrete maximum principle provesThis is maximum-norm convergence of the corrected nine-point Poisson scheme. It requires sufficient smoothness of the solution up to the boundary for the uniform truncation bound. On other domains the source and boundary closures must retain the required accuracy; boundary singularities can invalidate a smooth-solution order estimate.

The same expansion reveals how to add still more accuracy. Differentiating $\Delta u=f$ givesThus the sixth derivatives appearing in the $h^4$ term can also be expressed solely through the source. With exact source derivatives,is the sixth-order source correction of the nine-point Poisson stencil, with normalized residual $O(h^6)$. If source derivatives are replaced by differences, $\Delta f$ must be approximated through fourth order and the fourth derivatives through second order to retain that accuracy. The unknowns still occupy the same nine-point finite-difference stencil; only evaluation of known source data becomes more elaborate. The same inverse bound yields sixth-order convergence when these approximations and the boundary data have matching accuracy.

A particularly instructive case is the Laplace equation, $f=0$. Both displayed error corrections automatically vanish: ${\Delta }^{2}u=0$ and $u_{xxxxyy}+u_{xxyyyy}=\Delta u_{xxyy}=0$, which also makes the pure sixth derivatives sum to zero. Expanding the stencil two orders further gives, for a sufficiently smooth harmonic function,For verification, the three eighth-derivative contributions are $(u_{xxxxxxxx}+u_{yyyyyyyy})/20160$, $(u_{xxxxxxyy}+u_{xxyyyyyy})/2160$ and $u_{xxxxyyyy}/864$. Because $u$ is a harmonic function, their sum is $u_{xxxxxxxx}(1/10080-1/1080+1/864)=u_{xxxxxxxx}/3024$. The example $u=\mathrm{Re}(x+iy{)}^8$ has $D_{9}u(0,0)=40h^6/3$, so this term is genuinely present. The unmodified harmonic nine-point scheme therefore has sixth-order nodal convergence under the same smoothness and boundary assumptions. This harmonic superconvergence of the nine-point stencil is special to harmonic functions; it should not be confused with the fourth-order source-corrected scheme for general Poisson equation data.

The appeal of Mehrstellenverfahren is this combination of a compact sparse matrix, favorable discrete maximum principle properties, and high accuracy obtained by using the governing equation. The derivation also shows precisely which accuracy belongs to the operator on arbitrary functions, which belongs to its action on solutions, and which survives in the computed boundary-value solution.

For this two-step family with a third-order member, use the exponential-symbol order criterion for a multistep method. Expanding the exact-solution residual symbol givesThe constant, linear and quadratic coefficients vanish for every parameter. ThusAt the exceptional value the fourth-degree coefficient is $-1/30$, so there is no further increase. This means an unscaled exact-solution residual of order $h^{p+1}$, or a residual divided by the step of order $h^p$.

The first characteristic polynomial factors as $\rho (w)=(w-1)(w-a)$. The root condition for a multistep method requires every root to have modulus at most one and each unit-modulus root to be simple. The root $a=-1$ is allowed because it is distinct from the simple root one; $a=1$ is not, since it makes one a double root. Hence zero-stability holds exactly forOn this interval, $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=1-a\ne 0$ establish consistency of a numerical method. The Dahlquist equivalence theorem, for a fixed-step consistent linear multistep method applied to an ordinary differential equation with a Lipschitz continuous right-hand side with convergent starting values and well-defined small-step updates, says that convergence of a numerical method is equivalent to zero-stability. Therefore the same interval is exactly the convergence of a numerical method range. With sufficiently accurate starting values the convergent members have the orders just computed.

At $a=1$ the Taylor expansion still gives a formal order-two residual, but ${\rho }^{\prime }(1)=\sigma (1)=0$ and the double root prevents a convergent method. Canceling the common factor $w-1$ produces the common-factor cancellation defect in a multistep recurrence and changes the starting relations and cannot repair convergence of a numerical method of the original recurrence. At $a=0$, cancellation of a zero root instead reveals the ordinary trapezoidal rule; it is consistent with the order-two result.