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In elasticity, the kinetic energy $T$ and the potential energy $V$ are $$ T(u_t) = \frac{1}{2}\rho{u_t}^2 \qquad\text{and}\qquad V(u_x) = \frac{1}{2} E {u_x}^2 , $$ where $u$ is the displacement, $u_t$ the velocity, and $u_x$ the strain. The symbols $\rho>0$ and $E>0$ denote respectively the mass density and Young's modulus. The corresponding Lagrangian is $\mathcal{L} = T-V$. Using the principle of least action, how do we show that the 1D wave equation $$\rho u_{tt} = E u_{xx}$$ is the corresponding Euler-Lagrange equation?

If this wave equation comes from $\frac{\text d}{\text d t} \frac{\partial \mathcal{L}}{\partial \dot q} = \frac{\partial \mathcal{L}}{\partial q}$, what is $q$? On the left-hand side, one would like $\dot{q} = u_t$, while on the right-hand side, ${q} = u$ gives zero. How to resolve this inconsistency? I am aware of this exercise from Evans, but would like to tackle the particular case of the wave equation in elasticity.

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  • $\begingroup$ What do you mean by $\simeq$? Is the domain of $u$ of the form $\mathbb{R}^d\times[0,T]$? $\endgroup$ – Uskebasi Sep 17 '18 at 20:50
  • $\begingroup$ @Uskebasi For clarity, I removed $\simeq$, which was a Taylor approximation. The displacement $u (x,t)$ belongs to $\Bbb R$, as usual for 1D wave equation. $\endgroup$ – Harry49 Sep 17 '18 at 22:09
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In this case you have $q=u$ is a single function of two variables $(x_1,x_2)=(t,x)$, so Euler-Lagrange takes the form $$ \dfrac{\partial\mathcal{L}}{\partial q}=\sum_{j=1}^{2}\frac{\partial}{\partial x_j}\left(\frac{\partial\mathcal{L}}{\partial q_{,j}}\right) $$ where $q_{,j}=\dfrac{\partial q}{\partial x_j}$.

In other words, it is $$ \dfrac{\partial\mathcal{L}}{\partial q}=\frac{\partial}{\partial t}\left(\frac{\partial\mathcal{L}}{\partial u_t}\right)+\frac{\partial}{\partial x}\left(\frac{\partial\mathcal{L}}{\partial u_x}\right). $$

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