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The original PDF puts both neighbouring values at time $n+1$. Thus the actual scheme is Backward Euler method in time with a centered three-point approximation to the second spatial derivative. The old-time neighbours in the TeX transcription describe a different method.

Let $h=\Delta x$ and $k=\Delta t$. Insert a sufficiently smooth exact solution and divide by $k$. Expanding at $(x_m,t_{n+1})$ givesThe PDE cancels the $u_t-u_{xx}$ term. Therefore the normalized local truncation error is $O(k+h^2)$ andFor compatible smooth data with exact nodal initialization, the stability of a numerical method established below gives global error $O(k+h^2)$ on fixed time intervals, by summing the one-step defects with the contraction bound. Thus no relation between $k$ and $h$ is needed for stability; choosing $k=O(h^2)$ makes both consistency contributions $O(h^2)$.

Let ${\delta }_x^2{u}_m=u_{m-1}-2u_m+u_{m+1}$. For an exact smooth solution of the diffusion equation, the second-order central difference givesSince the Courant number $\mu =k/h^2$ is fixed, $h^2=O(k)$. Taylor expansion in time givesUsing $u_t=u_{xx}$ and $u_{tt}=u_{xxt}$ cancels the leading terms. Because $k{h}^2=O(k^2)$, the residual, and hence the local truncation error in the convention of the question, is

A one-step or linear multistep method has order $p$ if, when the exact sufficiently smooth solution is substituted for the numerical values, its unscaled local truncation error is $O(h^{p+1})$, and this estimate is not generally $O(h^{p+2})$. For a stable method this corresponds to a global error of order $O(h^p)$ on a fixed time interval.

The local truncation error is the error made by one numerical step started from the exact solution:A one-step method has local order $p+1$ when ${\tau }_{n+1}=O(h^{p+1})$ uniformly for $t_n$ in each fixed bounded time interval.

Let $A_h$ be the symmetric Dirichlet second-difference matrix. Its eigenvalues are nonpositive, so every eigenvalue ofhas modulus at most one. Thus $\Vert R_h{\Vert }_{2,h}\le 1$.

Insert the exact smooth solution into the scheme. The local truncation error per step has discrete norm $O(k^3+k{h}^2)$, from the order-three trapezoidal time residual and the order-two centred spatial residual. If $e^n$ is the grid error, stability and iteration giveThere are at most $1/k$ steps and $e^0=0$, henceThis proves convergence for every $\mu =k/h^2>0$ without invoking the Lax equivalence theorem.

Substitute the exact solution and expand every value about $t_n$. The coefficient of $h^q{y}^{(q)}(t_n)$ in the defect is zero for $q=0,\dots ,p$ exactly whenThese are the order conditions for a linear multistep method, specialized to the two nonzero derivative coefficients. The first failed identity determines the leading local truncation error.

Let ${\Phi }_h$ be the one-step map of a numerical method. It is a time-symmetric numerical method when reversing the step exactly reverses the update:

Let ${\phi }_h$ be the exact flow map and suppose the method has order $p$ with a nonzero leading local truncation error:Inverting this expansion changes the sign of its leading perturbation, sowhere transport by the exact flow only changes $d$ by $O(h)$ and hence does not affect the leading parity. On the other hand, replacing $h$ by $-h$ in the first expansion givesTime symmetry equates these expressions, so $(-1{)}^{p+1}=-1$. Hence $p+1$ is odd and

Insert the exact solution and expand about $t=t_{n+2}$. Since $f(y)=y^{\prime }$ and $f^{\prime }(y)f(y)=y^{\prime \prime }$, the left side isThe coefficient of $h^j{y}^{(j)}$ on the right side isfor $j\ge 1$, with the last term absent for $j=0$. These coefficients agree through $j=3$; at $j=4$ the left side minus the right side is $1/10$. Thus the local truncation error is $h^4{y}^{(4)}/10+O(h^5)$ and the method has order three.