> [!NOTE] > The [[Wave Equation in 1 Spatial Dimension]]: $u_{tt}-c^2u_{xx}=0$describes a standing wave field in one spatial dimension $x$ over time $t$ where $u$ is related to acoustic pressure and $c$ to the speed of the sound wave. **Derivation**: Consider a fluid in arbitrary section $a \leq x \leq b$ with mass density $\rho=\rho(x,t)$; velocity $v=v(x,t)$ and pressure $p=p(x,t).$ The total mass of the section is obtained by integrating $\rho(x,t)$ between $a$ and $b$. **Conservation of mass** (rate of change in this mass is the rate of mass entering at $a$ minus the mass leaving at $b$) yields the equation $\frac{d}{dt} \int_{a}^{b} \rho(x,t) \, dx = \rho(a,t)v(a,t)-\rho(b,t)v(b,t).$ Assuming the integrand above is $C^1,$ we get $\frac{d}{dt}\int_{a}^{b} \rho(x,t) \; dx=\int_{a}^{b}\partial_{t}\rho(x,t)\;dx.$ Applying the [[Fundamental theorem of calculus|second FTC]] to the RHS expression yields $\int_{a}^{b} \partial_{t} \rho(x,t)+\partial_{x}(\rho(x,t),v(x,t)) \; dx = 0.$Since this integrand is continuous, we obtain the **first Euler equation**: $\partial_{t} \rho + \partial_{x}(\rho v)=0.\tag{1}$ Now, the total momentum of the section is obtained by integrating $\rho v$ between $a$ and $b$. By Newton's second law, this is equal to the total force exerted on either side of the section: $\frac{d}{dt}\int_{a}^{b} \rho(x,t)v(x,t)\; dx = p(a,t)+\rho(a,t)v(a,t)^{2} - p(b,t)-\rho(b,t)v(b,t)^{2}.$Once again, we can take time derivative into the integral and use the FTC on the RHS to obtain **second Euler equation**: $\partial_{t}(\rho v)+\partial_{x}(\rho v^2)=-\partial_{x}p.\tag{2}$ To linearise, write $\rho=\bar{\rho}+u$where $u$ is a sufficiently small perturbation of a constant density $\bar{\rho}.$ Likewise, assume that velocity $v$ is sufficiently small too.