# Variational derivation of wave equation (Euler-Lagrange equation)

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.

• What do you mean by $\simeq$? Is the domain of $u$ of the form $\mathbb{R}^d\times[0,T]$? Sep 17, 2018 at 20:50

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 $$\dfrac{\partial\mathcal{L}}{\partial q}=\frac{\partial}{\partial t}\left(\frac{\partial\mathcal{L}}{\partial u_t}\right)+\frac{\partial}{\partial x}\left(\frac{\partial\mathcal{L}}{\partial u_x}\right).$$
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 $$\rho u_{tt} - E u_{xx} = 0$$.