homework and exercises - How do I recover the 1D wave equation from the Lagrangian?

Consider small displacements, $y(x,t)$, of an element of a string (circled in red and shown below) from equilibrium.

The force balance in the vertical direction yields: $$+\uparrow \Sigma F: -T\sin(\theta_1)+T\sin(\theta_2)=(mdx)\frac{\partial^2 y}{\partial t ^2}$$ Where we (well.. my prof) have equated the tension on either side of the element because we are considering the case in which there is no horizontal acceleration and because we are making a small angle approximation

$$T_{left}\cos(\theta_1)=T_{right}\cos(\theta_2).$$

After linearizing and simplifying, we end up with the 1D wave equation: $$c^2 \frac{\partial^2 y}{\partial x ^2} = \frac{\partial^2 y}{\partial t ^2}$$ where $c^2 = \frac{T}{m}$.

For the Lagrangian approach, I am able to write down the kinetic energy of the element and of the entire string (I think):

For the string: $$T=\frac{1}{2}\int_{_0}^{^L} \rho \dot y(x)^2 dx.$$

Where $y$ is a function of $x$ because we can have a different velocity in the $y$ direction at every point $x$.

For an element of the string: $$T=\frac{1}{2}m(dx)\dot y^2.$$

I am stumped at the potential energy for both the element and the string as a whole and I can't seem to figure out how to recover this wave equation from first principles. The potential energy should be proportional to the displacement $y$ from equilibrium but from what I'm seeing online it doesn't look anything like the potential energy due to a spring. What I've seen here is that "for a taut string, the stored energy of system described by the height function $u(x)$ is assumed to be proportional to the change in length": $$\int_a^b (\sqrt{1+u'^2}-1)dx$$ Which is a complete mystery to me.

So my question is how do I recover the EOMs for a taught string using the Lagrangian? I am also interested in how this particular case can be recovered from a more general solution but I may not have the background to understand that.

Answer

Say the density of the string is $\mu$ and the tension is $T$. It's clear that the kinetic energy of an infinitesimal piece of string is

$$dT = \frac{1}{2} (\mu \, dx) u_t(x)^2$$

The length of the infinitesimal piece of string from $(x, u(x))$ to $(x + dx, u(x + dx))$ is $$\begin{align} d\ell &= \sqrt{dx^2 + (u(x+dx) - u(x))^2} \\ &= \sqrt{dx^2 + (u_x(x) dx)^2} \\ &= dx \sqrt{1 + u_x(x)^2} \end{align}$$

For small displacements, $u_x(x) \ll 1$, and $$\begin{align} d\ell - dx &= \left(\sqrt{1 + u_x(x)^2} - 1\right) \, dx \\ &\approx \left(1 + \frac{1}{2}u_x(x)^2 - 1\right) \, dx \\ &= \frac{1}{2} u_x(x)^2 \, dx \end{align}$$ To extend the string by a small increment requires doing work against the tension force, $dW = T \, (d\ell - dx)$, so the potential energy of an infinitesimal piece of string from $x$ to $x + dx$ is $$\begin{equation} dV = \frac{1}{2} T u_x(x)^2 \, dx \end{equation}$$ and we can write down the action as $$\begin{equation} S = \iint \mathcal{L}(x, t) \, dx \, dt \end{equation}$$ with Lagrangian density $$\begin{equation} \mathcal{L}(x, t) = \frac{1}{2} \mu u_t(x, t)^2 - \frac{1}{2} T u_x(x, t)^2 \end{equation}$$ The Euler--Lagrange equation for an action of this form is $$\begin{equation} \frac{\partial \mathcal{L}}{\partial u} = \frac{\partial}{\partial x}\frac{\partial\mathcal{L}}{\partial u_x} + \frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial u_t} \end{equation}$$ so $$\begin{gather} 0 =\frac{\partial}{\partial x}(- T u_x) + \frac{\partial}{\partial t} (\mu u_t) \\ u_{tt} = \frac{T}{\mu} u_{xx} \end{gather}$$ and the speed is $c = \sqrt{\frac{T}{\mu}}$.