The selected two-stage SDIRK method has order two and is scalar L-stable

Since $\sqrt{2}\ge 0$ and $(\sqrt{2}{)}^2=2$ by [F3], [F4] rules out $\sqrt{2}\le 1$ because that would imply $2\le 1$. Thus $\sqrt{2}>1$, so $0<1\mathrm{/}\sqrt{2}<1$ and $0<\gamma <1$. The row sums of $A$ are $c$, and $b^T$ is the last row of $A$, so the method is internally consistent and stiffly accurate by [F1] and [F2]. Also ${\sum }_i{b}_i=1$ and $b^{T}c=(1-\gamma )\gamma +\gamma =2\gamma -{\gamma }^2=1\mathrm{/}2$, since substituting $\gamma =1-1\mathrm{/}\sqrt{2}$ gives $2\gamma -{\gamma }^2=1\mathrm{/}2$.

To check order, augment the nonautonomous equation by the clock variable and write $X^{\mathrm{\prime }}=Q(X)$, $Q(t,y)=(1,f(t,y))$. The row-sum condition in step 1.1 makes its Runge–Kutta stage time coordinates $t+c_{i}h$. Writing $K_i=Q(X_i)$, the stage bound [F5] gives $X_i-X=O(h)$. The componentwise Taylor bound [F6] and the bounded derivatives [A1] then give $K_i=Q+h{c}_i{Q}^{\mathrm{\prime }}Q+O(h^2)$: indeed $X_i-X=h{\sum }_j{a}_{ij}{K}_j=h{c}_{i}Q+O(h^2)$. Thus the update is $X+h({\sum }_i{b}_i)Q+h^2({\sum }_i{b}_i{c}_i)Q^{\mathrm{\prime }}Q+O(h^3)=X+hQ+\frac{h^2}{2}{Q}^{\mathrm{\prime }}Q+O(h^3)$, which agrees with the exact Taylor expansion through degree two. Its local defect is therefore $O(h^3)$.

On the scalar equation, solving the first stage gives $K_1=\lambda y\mathrm{/}(1-\gamma z)$; substitution into the second gives $Y_2\mathrm{/}y=(1+(1-2\gamma )z)\mathrm{/}(1-\gamma z{)}^2$. Stiff accuracy identifies $Y_2$ with the update, proving the displayed rational formula wherever the stages are defined.

To rule out order three or higher, apply the method from $(t,y)=(0,0)$ to $y^{\mathrm{\prime }}=t^2$. The exact solution is $y(t)=t^3\mathrm{/}3$, whereas [F1] and [F2] give the update $h^3{\sum }_i{b}_i{c}_i^2=h^3((1-\gamma ){\gamma }^2+\gamma )=h^3\mathrm{/}(2\sqrt{2})$. Since $1\mathrm{/}(2\sqrt{2})\mathrm{\ne }1\mathrm{/}3$ (equality would imply $9=8$ after squaring), its local defect has a nonzero cubic coefficient. Together with step 2.1 this proves that the classical order is exactly two.

Put $w=-z=s+it$, $s\ge 0$, $v=\gamma s\ge 0$, and $u=\gamma t$. From [F3] and $\gamma =1-1\mathrm{/}\sqrt{2}$, one obtains $(1-2\gamma )\mathrm{/}\gamma =\sqrt{2}$. Direct expansion gives $\mathrm{\mid }1+\gamma w{\mathrm{\mid }}^4-\mathrm{\mid }1-\sqrt{2}\gamma w{\mathrm{\mid }}^2=(4+2\sqrt{2})v+4v^2+4v^3+v^4+4v{u}^2+2v^2{u}^2+u^4\ge 0.$ Also $\mathrm{Re}(1+\gamma w)=1+v>0$, so the denominator is nonzero for $s\ge 0$. The inequality proves $\mathrm{\mid }R(z)\mathrm{\mid }\le 1$ for $\mathrm{Re}z\le 0$, and the numerator degree is one while the denominator degree is two, so $R(z)\to 0$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ in that half-plane. By the scalar definition this is L-stability.