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The wave equation in one space dimension is given as $$ u_t + Au_x = 0 $$ where $$ u := \begin{bmatrix} v(x,\, t) \\ w(x,\, t) \end{bmatrix}, \quad A = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}. $$ We are also given the following finite difference scheme, where $V_j^k$ is the approximation to $v(x_j,\, t_k)$: $$ \begin{align} V_j^{k+1} &= V_j^k + \frac{1}{2} p \left( W_{j+1}^k - W_{j-1}^k \right), \\ W_j^{k+1} &= W_j^k + \frac{1}{2} p \left( V_{j+1}^{k+1} - V_{j-1}^{k+1} \right), \end{align} $$ where $p = \Delta t / \Delta x$. My task is to find the leading error term in the local truncation error. I am given the answer, but no thurrow explanation. The answer is as follows:

Using the Taylor expansion and the fact that $v_t=w_x$, $w_t = v_x$, we find $$ \begin{align} \tau_j^k(v) &= \frac{1}{2}t \partial_x^2 v_j^k - \frac{1}{6} \Delta x^2 \partial_x^3 w_j^k + \mathcal{O}(\Delta t^2 + \Delta x^3), \\ \tau_j^k(w) &= -\frac{1}{2}t \partial_x^2 w_j^k - \frac{1}{6} \Delta x^2 \partial_x^3 v_j^k + \mathcal{O}(\Delta t^2 + \Delta x^3). \end{align} $$

However, I do not understand which Taylor expansions have been applied; obviously $v$ and $w$ must be approximated, but at which points are we to approximate them? And where do we center the approximations?

Thiank you in advance for your help and your time.

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The clue is in the answer: Only terms $v_j^k=v(x_j,t_k)$ and $w_j^k=w(x_j,t_k)$ appear in the right-hand sides so the expansions must have been made around the space-time point $(x_j, t_k)$.

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