Background
$\def\d{\mathop{}\!\mathrm{d}}$Consider the following first-order system
\begin{equation}\label{1}\tag{1}
\frac{\partial V_i}{\partial t} + \lambda_i(x,t) \frac{\partial V_i}{\partial x}
= \sum_{j=1}^n \alpha_{ij}(x,t)V_j + \beta_i(x,t) \quad i=1,\ldots,N
\end{equation}
with the initial value condition
\begin{equation}\label{2}\tag{2}
V_i(x,0) = \varphi_i(x) \quad i=1,\ldots,N.
\end{equation}
Suppose that the eigenvalues are mutually distinct, i.e.,
$$ \lambda_1(x,t) < \lambda_2(x,t) < \cdots < \lambda_N(x,t). $$
Then through every point there are $N$ characteristic curves $L_i$ ($i=1,\ldots,N$) passing and
they are given by
$$ \frac{\d x}{\d t} = \lambda_i(x,t). $$
Suppose the IVC \eqref{2} is given on the interval $[a,b]$.
Draw two characteristic curves $L_N$ and $L_1$ passing through $(a,0)$
and $(b,0)$ respectively. And let $G$ be the domain enclosed by $L_N$,
$L_1$, $x$-axis and $t=T$.
We suppose that in the closed domain $\overline{G}$, functions
$\alpha_{ij}$, $\beta_i$ and $\varphi_i$ are all $C^1$.
Fix any point $(x,t)\in\overline{G}$, draw a characteristic curve $L_i$
which intersects $x$-axis at $(x_i,0)$. See the following figure.
Note that the left-hand side of \eqref{1} is the derivative of $V_i$
with respect to $t$ along the curve $L_i$. Integrate \eqref{1} from $(x_i,0)$
to $(x,t)$ with respect to $t$ along $l_i$, we have
\begin{equation}\label{3}\tag{3}
V_i(x,t) - V_i(x_i,0) = \int_{l_i} \biggl(\sum_{j=1}^N \alpha_{ij} V_j + \beta_i\biggr) \d\tau,
\end{equation}
that is
\begin{equation}\label{4}\tag{4}
V_i(x,t) = \varphi_i(x_i) + \int_{l_i} \biggl(\sum_{j=1}^N \alpha_{ij} V_j + \beta_i\biggr) \d\tau.
\end{equation}
Now we use \eqref{4} to construct the iteration sequence. Define for $n\geq 0$
\begin{equation}\label{5}\tag{5}
V_i^{(n+1)}(x,t) = \varphi_i(x_i) + \int_{l_i} \biggl[\sum_{j=1}^N \alpha_{ij} V_j^{(n)}+\beta_i\biggr] \d\tau \quad (i=1,\ldots,N),
\end{equation}
or more precisely,
\begin{equation}\label{6}\tag{6}
\begin{aligned}
V_i^{(n+1)}(x,t) & = \varphi_i(x_i(0;x,t)) + \int_0^t \biggl[\sum_{j=1}^N
\alpha_{ij}(x_i(\tau;x,t),\tau) V_j^{(n)}(x_i(\tau;x,t),\tau) \\
& \quad + \beta_i(x_i(\tau;x,t),\tau)\biggr] \d\tau,
\end{aligned}
\end{equation}
where $x_i(\tau;x,t)$ is the $i$th characteristic curve passing through $(x,t)$.
Let
$$M := \max_{(x,t)\in\overline{G},\,i=1,\ldots,N} \bigl\{|V_i^{(0)}|, |V_i^{(1)}|\bigr\},$$
$$A := \max_{(x,t)\in\overline{G},\,i,j=1,\ldots,N} |\alpha_{ij}|.$$
By a simple induction argument, we can prove that
\begin{equation}\label{7}\tag{7}
|V_i^{(n+1)}(x,t) - V_i^{(n)}(x,t)| \leq 2M \frac{(ANt)^n}{n!} \quad i=1,\ldots,N,
\end{equation}
which means that $(V_i^{(n)})$ converges uniformly to some function $V_i$ and it is the solution
to \eqref{1}.
Problem
I want to use the iteration method to solve the following first-order system $$ \begin{cases} u_t + u_x = v, \\ v_t - v_x = u \end{cases} $$ with the initial condition $u|_{t=0}=\sin x$, $v|_{t=0}=\cos x$. Obviously, $u(x,t)=\sin x$ and $v(x,t)=\cos x$ is the only solution to the system.
For fixed $(x,t)$, the first characteristic curve (straight line in fact) $l_1$ passes through $(x,t)$ and $(x-t,0)$ while the second $l_2$ passes through $(x,t)$ and $(x+t,0)$. So the system is equivalent to
\begin{align*} u(x,t) & = \sin(x-t) + \int_{l_1} v(x_1(\tau;x,t),\tau) \d\tau, \\ v(x,t) & = \cos(x+t) + \int_{l_2} u(x_2(\tau;x,t),\tau) \d\tau, \end{align*} where $x_1(\tau;x,t) = x-t+\tau$ and $x_2(\tau;x,t) = x+t-\tau$.
Now we go to the iteration procedure. Let $u^{(0)} = \sin(x-t)$ and $v^{(0)} = \cos(x+t)$, then \begin{align*} u^{(1)}(x,t) & = \sin(x-t) + \int_0^t v^{(0)} (x-t+\tau,\tau) \d\tau \\ & = \frac{1}{2}[\sin(x+t) - \sin(x-t)], \\ v^{(1)}(x,t) & = \cos(x+t) + \int_0^t u^{(0)} (x+t-\tau,\tau) \d\tau \\ & = \frac{1}{2} [\cos(x+t) + \cos(x-t)], \end{align*} Similarly, we calculate that \begin{align*} u^{(2)}(x,t) & = \frac34 \sin(x-t) + \frac14 \sin(x+t) + \frac12 t\cos(x-t), \\ v^{(2)}(x,t) & = \frac34 \cos(x+t) + \frac14 \cos(x-t) + \frac12 t\sin(x+t), \\ u^{(3)}(x,t) & = \frac12 [\sin(x+t)+\sin(x-t)] + \frac14 t[\cos(x-t)-\cos(x+t)], \\ v^{(3)}(x,t) & = \frac12 [\cos(x+t)+\cos(x-t)] + \frac14 t[\sin(x+t)-\sin(x-t)], \\ u^{(4)}(x,t) & = \frac{5}{16}\sin(x+t) + \frac{11}{16} \sin(x-t) + \frac{t}{2}\cos(x-t) \\ & \quad -\frac{t}{8}\cos(x+t) + \frac{t^2}{8}\sin(x-t), \\ v^{(4)}(x,t) & = \frac{11}{16} \cos(x+t) + \frac{5}{16}\cos(x-t) + \frac{t}{2}\sin(x+t) \\ & \quad -\frac{t}{8}\sin(x-t) - \frac{t^2}{8}\cos(x+t). \end{align*}
However, it's not obvious at all to see that $u^{(n)}\to \sin x$, and $v^{(n)}\to \cos x$.
