= Solution
A <Runge-Kutta method> advances $y'=f(t,y)$ through stages
$$
Y_i=y_n+h\sum_j a_{ij}F_j,\qquad
F_j=f(t_n+c_jh,Y_j),\qquad
y_{n+1}=y_n+h\sum_i b_iF_i.
$$
The <Butcher tableau> records $A,b,c$, normally with $c=Ae$. Linear <absolute stability> tests the amplification of a linear decay mode; nonlinear <B-stability> compares distances between two solutions of a <dissipative vector field>. Both describe propagation of perturbations, but they impose different conditions.
For the <Dahlquist test equation>, elimination of stages gives
$$
R(z)=1+zb^T(I-zA)^{-1}e,\qquad z=h\lambda.
$$
The <linear stability domain> consists of $z$ for which the step is defined and $|R(z)|\leq1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z)^ny_0$: the powers are bounded exactly when $|R(z)|\leq1$. For $y'=Ly$, the update is $R(hL)$. If $L$ is a <normal matrix>, <unitary diagonalization of a normal matrix> makes the <operator norm> equal to the largest scalar amplification <modulus>. For a diagonalizable <matrix>, powers are bounded by the <condition number> of the <eigenvector> matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus <eigenvalue> can instead produce a growing <Jordan block>; the scalar spectral test alone does not control a general <matrix>.
An <A-stable> method accepts the entire closed left half-plane. The <Explicit Euler method> has $R(z)=1+z$ and a disk of stability $|1+z|\leq1$, so it is not <A-stable>. No nonconstant <polynomial> <stability function> can be <A-stable>, since it is unbounded on the negative real axis; this excludes all consistent explicit <Runge Kutta methods>. The <Backward Euler method> has $R(z)=1/(1-z)$, and $|1-z|\geq1$ when $\operatorname{Re}z\leq0$, proving <A-stability>. It is also <L-stable>, because $R(z)\to0$ for large left-half-plane $z$.
The <implicit midpoint rule> and the <trapezoidal rule> have the same scalar <stability function>,
$$
R(z)=\frac{1+z/2}{1-z/2}.
$$
The identity $|1-z/2|^2-|1+z/2|^2=-2\operatorname{Re}z$ proves <A-stability>. On the imaginary axis their amplification has unit <modulus>, and along the negative real axis it tends to $-1$. Thus neither is <L-stable>; very stiff decay can persist as alternating numerical values. In contrast, the two-stage <Radau IIA method> has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is <L-stable>. This illustrates why <L-stability> matters beyond <A-stability> for a <stiff differential equation>.
For nonlinear problems, assume a real or complex Euclidean <inner product> and the dissipativity hypothesis
$$
\operatorname{Re}\langle f(t,u)-f(t,v),u-v\rangle\leq0
$$
for all relevant states at each common time. Exact solutions satisfy
$$
\frac{d}{dt}\|u-v\|^2
=2\operatorname{Re}\langle f(t,u)-f(t,v),u-v\rangle\leq0.
$$
A <B-stable> <Runge-Kutta method> reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-\widetilde y_n$, $D_i=Y_i-\widetilde Y_i$, and $G_i=f(t_n+c_ih,Y_i)-f(t_n+c_ih,\widetilde Y_i)$. Then $D_i=d+h\sum_j a_{ij}G_j$. Expansion of the output squared <norm>, followed by substituting $d=D_i-h\sum_j a_{ij}G_j$ in its linear terms, proves the <Runge-Kutta contractivity identity>
$$
\|d_{\mathrm{new}}\|^2-\|d\|^2
=2h\sum_i b_i\operatorname{Re}\langle D_i,G_i\rangle
-h^2\sum_{i,j}m_{ij}\operatorname{Re}\langle G_i,G_j\rangle,
\quad
m_{ij}=b_i a_{ij}+b_j a_{ji}-b_i b_j.
$$
If $b_i\geq0$ and $M=(m_{ij})$ is a <positive semidefinite matrix>, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage <vectors> to express it as a sum of nonnegative <quadratic forms>. These are exactly the conditions for <algebraic stability of a Runge-Kutta method>. Hence \b[<algebraic stability> implies <B-stability>]. Applying <B-stability> to the linear dissipative problem $y'=\lambda y$, or its two-dimensional real form, also proves <A-stability> whenever stages are well defined.
The <Backward Euler method> has $b=A=1$, hence $M=1$; the <implicit midpoint rule> has $b=1$, $A=1/2$, hence $M=0$. Both are <algebraically stable> and <B-stable>. For the two-stage <Radau IIA method>,
$$
M=\frac1{16}\begin{pmatrix}1&-1\\-1&1\end{pmatrix},
$$
a <positive semidefinite matrix>, while both weights are positive. It therefore combines third-order accuracy, <B-stability>, and <L-stability>. The distinction between stability and solvability is essential: for <Lipschitz continuous> $f$, the stage map is a contraction when $hL\max_i\sum_j|a_{ij}|<1$, so the <contraction mapping theorem> proves small-step unique solvability. The preceding <algebraic stability> proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.
Linear <A-stability> is insufficient for nonlinear <B-stability>. The <trapezoidal rule fails B-stability>, even though its scalar <stability function> is identical to that of the <B-stable> <implicit midpoint rule>. Its <Runge-Kutta method> coefficients are
$$
A=\begin{pmatrix}0&0\\1/2&1/2\end{pmatrix},\qquad
b=(1/2,1/2)^T,\qquad
M=\begin{pmatrix}-1/4&0\\0&1/4\end{pmatrix}.
$$
Failure of this sufficient <algebraic stability> test alone would not prove failure of <B-stability>. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows from
$$
(f(u)-f(v))(u-v)=-(u-v)^2(u^2+uv+v^2)\leq0.
$$
The unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit <differentiation> gives
$$
T'(1)=\frac{1-3(1)^2}{1+3T(1)^2}=-2.
$$
Nearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.
Finally, <stability of a numerical method> on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has <norm> at most $Ch^{p+1}$, then the error satisfies $\|e_{n+1}\|\leq\|e_n\|+Ch^{p+1}$, hence $\|e_n\|\leq\|e_0\|+CT h^p$ for $nh\leq T$. A Lipschitz step bound $1+Kh$ gives the same order using the <discrete Gronwall inequality>. For a <stiff differential equation>, the value of <B-stability> is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform <derivative> bounds; stability alone does not establish a mesh-uniform error order.
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