$$
\begin{align}ed
& d z=a_0, z=n \cdot a_0 \\
& x_n=x\left(n a_0, t\right)=x(z, t) \\
& x_n \pm 1=x\left(n a_0 \pm a_0, t\right)=x(z \pm d z, t) \\
& m \fracd^2 x_n(t)d t^2=-k\left(x_n(t)-x_n-1(t)\right)+k\left(x_n+1(t)-x_n(t)\right) \text devient: \\
& m \frac\partial^2 x(z, t)\partial t^2=-k(x(z, t)-x(z-d z, t))+k(x(z+d z, t)-x(z, t))
\end{align}ed
$$

On introduit dz :
$$
M \frac\partial^2 x(z, t)\partial t^2=-k d z\left(\fracx(z, t)-x(z-d z, t)d z-\fracx(z+d z, t)-x(z, t)d z\right)
$$
\(\mathrm{dz}\) est petit, on peut le faire tendre vers 0 :
$$
\fracx(z, t)-x(z-d z, t)d z \cong \frac\partial x(z, t)\partial z \quad \fracx(z+d z, t)-x(z, t)d z \cong \frac\partial x(z+d z, t)\partial z
$$
On a alors :
$$
M \frac\partial^2 x(z, t)\partial t^2=-k d z\left(\frac\partial x(z, t)\partial z-\frac\partial x(z+d z, t)\partial z\right)
$$

$$
M \frac\partial^2 x(z, t)\partial t^2=-k d z\left(\frac\partial x(z, t)\partial z-\frac\partial x(z+d z, t)\partial z\right)
$$
On pose : \(f(z, t)=\frac{\partial x(z, t)}{\partial z}\)
$$
M \frac\partial^2 x(z, t)\partial t^2=-k d z(f(z, t)-(f(z+d z, t))
$$

En divisant à nouveau par \(\mathrm{dz}\), on obtient :
$$
\begin{align}ed
& m \frac\partial^2 x(z, t)\partial t^2=-k d z^2\left(\fracf(z, t)-(f(z+d z, t)d z\right) \\
\Rightarrow \quad & m \frac\partial^2 x(z, t)\partial t^2=-k d z^2\left(-\frac\partial f(z, t)\partial z\right)=+k d z^2\left(\frac\partial^2 x(z, t)\partial z^2\right) \\
\Rightarrow & \frac\partial^2 x(z, t)\partial z^2=\fracmk d z^2 \frac\partial^2 x(z, t)\partial t^2
\end{align}ed
$$