= Solution
{c}
Let $f$ be continuous in its first variable and uniformly <Lipschitz continuous> in its second variable, with constant $L\geq0$. Retain the finite-interval smoothness assumptions and the bound $\|\tau_n\|\leq Kh^2$ from the preceding subpart. We first establish solvability of each implicit step. For a proposed starting state $u$, the map
$$
G_u(v)=u+h f\left(x_n+\frac h2,\frac{u+v}2\right)
$$
has <Lipschitz constant> $hL/2$. If $hL/2<1$, the <contraction mapping theorem> on the complete state space gives a unique next state $v=\Psi_{n,h}(u)$.
To compute a useful <Lipschitz constant> for the method, introduce its increment function by
$$
\Psi_{n,h}(u)=u+h\Phi_{n,h}(u),\qquad
\Phi_{n,h}(u)=f\left(x_n+\frac h2,u+\frac h2\Phi_{n,h}(u)\right).
$$
For two starting states $u,w$,
$$
\|\Phi_{n,h}(u)-\Phi_{n,h}(w)\|
\leq L\|u-w\|+\frac{hL}{2}
\|\Phi_{n,h}(u)-\Phi_{n,h}(w)\|.
$$
Rearranging gives the <implicit midpoint Lipschitz estimate>
$$
\boxed{\operatorname{Lip}(\Phi_{n,h})\leq\frac{L}{1-hL/2},
\qquad
\operatorname{Lip}(\Psi_{n,h})\leq
R_h:=\frac{1+hL/2}{1-hL/2}.}
$$
Thus, for example, $hL\leq1$ gives an increment <Lipschitz constant> at most $2L$, uniformly as $h\to0$.
For an explicit <global error> estimate, set $e_n=Y_n-y_n$. Subtract the numerical formula from the exact-value identity containing $h\tau_n$:
$$
e_{n+1}=e_n+h\left[
f\left(x_n+\frac h2,\frac{Y_n+Y_{n+1}}2\right)
-f\left(x_n+\frac h2,\frac{y_n+y_{n+1}}2\right)\right]+h\tau_n.
$$
The <Lipschitz condition> and the <triangle inequality> imply
$$
(1-hL/2)\|e_{n+1}\|
\leq(1+hL/2)\|e_n\|+h\|\tau_n\|,
$$
and therefore
$$
\|e_{n+1}\|\leq R_h\|e_n\|+
\frac{Kh^3}{1-hL/2}.
$$
Iterating this inequality is the elementary <discrete Gronwall inequality> argument. For $L>0$, summing the <geometric series> and using $R_h-1=hL/(1-hL/2)$ gives
$$
\|e_n\|\leq R_h^n\|e_0\|+
\frac{Kh^2}{L}(R_h^n-1).
$$
For $hL\leq1$, $R_h=1+hL/(1-hL/2)\leq1+2hL\leq e^{2hL}$. If $nh\leq T$, this proves
$$
\boxed{\max_{nh\leq T}\|e_n\|
\leq e^{2LT}\|e_0\|+
\frac{K}{L}(e^{2LT}-1)h^2.}
$$
With the exact initial value, $e_0=0$, the <global error> is $O(h^2)$ and tends uniformly to zero on every such finite interval. If $L=0$, the recurrence instead gives $\|e_n\|\leq\|e_0\|+KTh^2$, with the same conclusion. Starting values merely converging to the exact initial value also suffice for <numerical convergence>; second-order accuracy requires their errors to be $O(h^2)$. The result concerns exact implicit solves and convergence on fixed finite intervals, not uniform accuracy over an unbounded time interval.
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