Sufficient explicit Runge–Kutta order conditions through order four

Write $f_0=f(y)$, $J=Df(y)$, $H=D^{2}f(y)$, and $G=D^{3}f(y)$ at the exact starting state. The hypothesis $f\in C^4(U)$ is in the coordinate sense of [F3]. To justify the Banach chain rules used below, regard each finite array of order-$r$ coordinate partials as a map into a finite-dimensional coordinate space. For $r=0,1,2,3$, its first coordinate partials exist and are continuous by $f\in C^4(U)$, so the continuous-partials theorem in [F3], applied successively, gives continuous Fréchet derivatives through order four, with entries the corresponding coordinate partials. Thus [F5] applies to $f$ and the exact solution. The exact solution is differentiable, hence continuous from the Fréchet remainder in its derivative definition; since $y^{\mathrm{\prime }}=f\circ y$ and $f$ is continuous, $y^{\mathrm{\prime }}$ is continuous and $y$ is $C^1$. If $y$ is $C^r$ for $1\le r\le 4$, the iterated chain and product rules in [F5] and continuity of the derivative arrays show $f\circ y$ is $C^r$, so $y^{\mathrm{\prime }}=f\circ y$ makes $y$ $C^{r+1}$. Thus $y$ is $C^5$ on the admitted interval. A further differentiation expresses $y^{(5)}$ using $f$ and its derivatives through order four, which are bounded on $K$; hence the order-four Taylor remainder of $y$ is uniform. Differentiating gives $y^{\mathrm{\prime }\mathrm{\prime }}=J{f}_0$, $y^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=H(f_0,f_0)+J^2{f}_0$, and $y^{(4)}=G(f_0,f_0,f_0)+3H(f_0,J{f}_0)+JH(f_0,f_0)+J^3{f}_0$; the factor $3$ uses the symmetry in [F3].

Prove the stage expansions simultaneously by induction on $i$. The empty first row gives $Y_1=y$ and $K_1=f_0$, the displayed formulas below with their zero row coefficients. Suppose the formulas hold for $j<i$. Substituting them into $Y_i=y+h{\sum }_{j<i}{a}_{ij}{K}_j$ gives $Y_i-y=h{c}_i{f}_0+h^2(Ac{)}_{i}J{f}_0+h^3((A^{2}c{)}_i{J}^2{f}_0+\frac{1}{2}(A{c}^{[2]}{)}_{i}H(f_0,f_0))+O(h^4)$. Taylor-expand $f(Y_i)$ at $y$ through cubic order by [F4]: $f(y+\delta )=f_0+J\delta +\frac{1}{2}H(\delta ,\delta )+\frac{1}{6}G(\delta ,\delta ,\delta )+O(\mathrm{\parallel }\delta {\mathrm{\parallel }}_2^4)\mathrm{.}$ Substituting the preceding stage-state expansion and collecting powers of $h$ gives, uniformly in $i$, $K_i=f_0+h{c}_{i}J{f}_0+h^2((Ac{)}_i{J}^2{f}_0+\frac{1}{2}{c}_i^{2}H(f_0,f_0))+h^3((A^{2}c{)}_i{J}^3{f}_0+\frac{1}{2}(A{c}^{[2]}{)}_{i}JH(f_0,f_0)+c_i(Ac{)}_{i}H(f_0,J{f}_0)+\frac{1}{6}{c}_i^{3}G(f_0,f_0,f_0))+O(h^4)$. The cross term uses [F3] to exchange the two arguments of $H$. Compactness, bounded derivatives, and finiteness of $s$ make every remainder uniform.

Substituting step 2.1 into $y^{+}=y+h{\sum }_i{b}_i{K}_i$ gives coefficients of the four orders: $h{f}_0$ has coefficient ${\sum }_i{b}_i$; $h^{2}J{f}_0$ has coefficient ${\sum }_i{b}_i{c}_i$; $h^3{J}^2{f}_0$ and $h^{3}H(f_0,f_0)$ have coefficients ${\sum }_i{b}_i(Ac{)}_i$ and $\frac{1}{2}{\sum }_i{b}_i{c}_i^2$; and the four $h^4$ terms $J^3{f}_0$, $JH(f_0,f_0)$, $H(f_0,J{f}_0)$, and $G(f_0,f_0,f_0)$ have coefficients ${\sum }_i{b}_i(A^{2}c{)}_i$, $\frac{1}{2}{\sum }_i{b}_i(A{c}^{[2]}{)}_i$, ${\sum }_i{b}_i{c}_i(Ac{)}_i$, and $\frac{1}{6}{\sum }_i{b}_i{c}_i^3$.

Taylor's formula [F4] and the derivatives in step 1.1 give the exact solution coefficients $1$, $1\mathrm{/}2$, $1\mathrm{/}6$ for each order-three term, and $1\mathrm{/}24$, $1\mathrm{/}24$, $3\mathrm{/}24$, $1\mathrm{/}24$ for the four order-four terms in the same order as step 3.1.

The displayed coefficient equations make the method and exact expansions agree through $h^p$ for each requested $p\le 4$. The stage remainder in step 2.1 is $O(h^4)$ and is multiplied by $h$, while the exact Taylor remainder is uniform on the compact tube, so matching through $h^p$ leaves a uniform defect $O(h^{p+1})$. By [F2] this is uniform local order at least $p$.