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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
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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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After I read some books dedicated to variational calculus, I came up with the following answer, which is perfectly consistent with the accepted answer by @user10354138. From the Lagrangian density $$\mathcal L(u,u_t,u_x) = T(u_t)-V(u_x)$$ defined in OP, we formulate the principle of stationary action $\delta \mathcal A = 0$, where $\mathcal A = \iint \mathcal L \,\text{d}x\text d t$ is the action, and $\delta \mathcal A$ denotes its first variation w.r.t. $u$, $u_t$, $u_x$. Using integration by parts and the commutativity of variation and differentiation, we find \begin{aligned} \delta A &= \iint_{\Bbb R^2} \left(\frac{\partial \mathcal L}{\partial u}\delta u + \frac{\partial \mathcal L}{\partial u_t}\delta u_t + \frac{\partial \mathcal L}{\partial u_x}\delta u_x\right) \text{d}x\text d t \\ &= \iint_{\Bbb R^2} \left(\frac{\partial \mathcal L}{\partial u} - \frac{\partial}{\partial t}\frac{\partial \mathcal L}{\partial u_t} - \frac{\partial}{\partial x}\frac{\partial \mathcal L}{\partial u_x}\right) \delta u\, \text{d}x\text d t \, , \end{aligned} where natural boundary conditions have been assumed. This gives us the Euler-Lagrange equation $$ \frac{\partial \mathcal L}{\partial u} - \frac{\partial}{\partial t}\frac{\partial \mathcal L}{\partial u_t} - \frac{\partial}{\partial x}\frac{\partial \mathcal L}{\partial u_x} = 0 $$ and therefore the wave equation $\rho u_{tt} - E u_{xx} = 0$.

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