= Solution
Put $h=\Delta x$ and suppose $L=(n+1)h$, so there are $n$ interior grid points $x_j=jh$, $j=1,\ldots,n$. Write $u_j(t)\simeq u(x_j,t)$. The fixed ends impose homogeneous <Dirichlet boundary conditions>,
$$
u_0(t)=u_{n+1}(t)=0.
$$
For a sufficiently smooth <displacement> field, the two <Taylor series> at fixed time give
$$
\begin{aligned}
u(x_j+h,t)&=u(x_j,t)+hu_x+\frac{h^2}{2}u_{xx}+\frac{h^3}{6}u_{xxx}+O(h^4),\\
u(x_j-h,t)&=u(x_j,t)-hu_x+\frac{h^2}{2}u_{xx}-\frac{h^3}{6}u_{xxx}+O(h^4),
\end{aligned}
$$
where all displayed <derivatives> on the right are evaluated at $(x_j,t)$. Adding cancels the odd terms, leaving the <second-order central difference>
$$
u_{xx}(x_j,t)=\frac{u(x_j-h,t)-2u(x_j,t)+u(x_j+h,t)}{h^2}+O(h^2).
$$
Substitute this <finite difference> into the <wave equation> and retain continuous time. The resulting <method of lines> gives coupled second-order <ordinary differential equations>,
$$
\ddot u_j=\frac{c^2}{h^2}(u_{j-1}-2u_j+u_{j+1}),\qquad j=1,\ldots,n.
$$
In particular, the first and last equations are $\ddot u_1=(c^2/h^2)(-2u_1+u_2)$ and $\ddot u_n=(c^2/h^2)(u_{n-1}-2u_n)$ for $n\ge2$. With the interior <displacement> <vector> only,
$$
\boxed{\ddot{\mathbf u}=A\mathbf u,\qquad
\mathbf u=(u_1,\ldots,u_n)^T,\qquad
A=\frac{c^2}{h^2}\begin{pmatrix}
-2&1&0&\cdots&0\\
1&-2&1&\ddots&\vdots\\
0&\ddots&\ddots&\ddots&0\\
\vdots&\ddots&1&-2&1\\
0&\cdots&0&1&-2
\end{pmatrix}.}
$$
Equivalently, $A_{jj}=-2c^2/h^2$, $A_{j,j+1}=A_{j+1,j}=c^2/h^2$, and all other entries vanish. This entrywise definition also covers small $n$, including $A=(-2c^2/h^2)$ when $n=1$. The <matrix> is a <symmetric matrix> and a <tridiagonal matrix>. The spatial approximation has error $O(h^2)$ in <Big O notation>; time has not yet been discretized.
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