= Solution
A <Runge-Kutta method> with stage <matrix> $A$, weights $b$ and <step size> $h$ is
$$
Y_i=y_n+h\sum_j a_{ij}f(Y_j),\qquad y_{n+1}=y_n+h\sum_i b_i f(Y_i).
$$
Its order measures accuracy as $h\to0$, while stability concerns how errors or stiff components propagate. Linear <A-stability>, strong stiff damping through <L-stability>, and nonlinear <B-stability> address different questions.
For the <Dahlquist test equation> $y'=\lambda y$, elimination of the stages gives the <stability function>
$$
R(z)=1+zb^T(I-zA)^{-1}\mathbf1,\qquad z=h\lambda.
$$
The <linear stability domain> is $\{z:|R(z)|\leq1\}$. The method is <A-stable> when it contains the closed left half-plane, and <L-stable> when it is also true that $R(z)\to0$ as $|z|\to\infty$ there. These conditions describe, respectively, stability for every non-growing scalar linear mode and damping of very rapidly decaying modes.
The <Forward Euler method> has $R(z)=1+z$, so it fails <A-stability>, for example at $z=-3$. In fact no nontrivial explicit <Runge-Kutta method> is <A-stable>: its <stability function> is a nonconstant <polynomial>, which is unbounded along the negative real axis. The <Backward Euler method> has $R(z)=1/(1-z)$. For $\operatorname{Re}z\leq0$, $|1-z|\geq1$, and the function tends to zero at infinity, proving both <A-stability> and <L-stability>. The <implicit midpoint rule> has $R(z)=(1+z/2)/(1-z/2)$. The identity $|1-z/2|^2-|1+z/2|^2=-2\operatorname{Re}z$ proves <A-stability>, but its limit $-1$ rules out <L-stability>.
For nonlinear equations, a <dissipative vector field> satisfies $\langle f(x)-f(y),x-y\rangle\leq0$. The corresponding exact solutions contract because the time derivative of their squared distance is twice that <inner product>. A <B-stable> method preserves this contraction for every $h>0$, whenever its stages are well defined.
Let $d=y_n-\widetilde y_n$, $D_i=Y_i-\widetilde Y_i$, and $F_i=f(Y_i)-f(\widetilde Y_i)$. Expanding the update's squared <norm>, substituting $d=D_i-h\sum_j a_{ij}F_j$, and symmetrizing the double sum proves the <Runge-Kutta contractivity identity>
$$
\|d_{n+1}\|^2-\|d\|^2
=2h\sum_i b_i\langle D_i,F_i\rangle-h^2\sum_{i,j}m_{ij}\langle F_i,F_j\rangle,
\quad m_{ij}=b_i a_{ij}+b_j a_{ji}-b_i b_j.
$$
If $b_i\geq0$ and the <matrix> $M=(m_{ij})$ is a <positive semidefinite matrix>, the first sum is nonpositive by dissipativity. The second double sum is nonnegative: sum the <quadratic form> of $M$ over each coordinate of the $F_i$. Thus \b[<algebraic stability> implies <B-stability>]. Applied to complex scalar linear equations with the real part of the Hermitian <inner product>, the same argument also implies <A-stability>, where the stages are defined.
The <Backward Euler method> has $b_1=a_{11}=1$, hence $M=(1)$, and is <algebraically stable>. One can prove its nonlinear contraction directly: $d_{n+1}=d+h(f(y_{n+1})-f(\widetilde y_{n+1}))$ gives $\|d_{n+1}\|^2\leq\langle d,d_{n+1}\rangle\leq\|d\|\|d_{n+1}\|$. The <implicit midpoint rule> has $a_{11}=1/2$, $b_1=1$ and $M=0$, so it is <B-stable> even though it is not <L-stable>.
Linear <A-stability> does not by itself imply nonlinear <B-stability>. The ODE <trapezoidal rule> has the same <stability function> as the <implicit midpoint rule>, but take the dissipative scalar field $f(y)=-\max(y,0)$ and $h=6$. For any positive starting value, its trapezoidal update is $y_{n+1}=y_n+3(f(y_n)+f(y_{n+1}))=-2y_n<0$. Thus two positive starting values have their distance doubled, violating <B-stability>. Its <algebraic stability> <matrix> also has a negative first diagonal entry. The field is globally <Lipschitz continuous>, so this is a well-defined nonlinear counterexample.
These examples distinguish accuracy, scalar stiff-mode stability, nonlinear contraction, and rapid stiff-mode damping. For implicit methods, solvability of the stage equations remains an additional requirement; a formal <stability function> alone does not establish it for arbitrary nonlinear problems.
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