Past exam of the theoretical mechanics course of the University of Oxford / 2009 / B / 4

The inviscid equations are ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$ and $\rho [{\mathbf{u}}_t+(\mathbf{u}\cdot \nabla )\mathbf{u}]=-\nabla p$. Write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=P({\rho }_0)+p^{\prime }$ and retain only linear terms. With $c_0^2=P^{\prime }({\rho }_0)>0$, barotropy gives $p^{\prime }=c_0^2{\rho }^{\prime }$. Irrotational perturbations allow ${\mathbf{u}}^{\prime }=\nabla \varphi$. The momentum equation integrates to $p^{\prime }=-{\rho }_0{\varphi }_t$, after using the time-dependent gauge of $\varphi$ to remove the spatially constant integration term. ThusSubstitution into ${\rho }_t^{\prime }+{\rho }_0\Delta \varphi =0$ gives the acoustic wave equation $\boxed{{\varphi }_{tt}=c_0^2\Delta \varphi }$.

Impermeability imposes homogeneous Neumann data: ${\varphi }_x=0$ at $x=0,a$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables therefore givesOne may take $n,m\ge 0$ without duplication. For a nontrivial oscillation they are not both zero. A time-independent spatially constant potential is a gauge mode at zero frequency, rather than an additional sound oscillation.

At leading order the two wall conditions may be evaluated on the undeformed line $y=b$: shifting the evaluation point multiplies small perturbations and creates only second-order errors. The dynamic relation and the linear Bernoulli relation give $\eta =p^{\prime }/({\rho }_{0}k)=-{\varphi }_t/k$. Differentiate this in time and use ${\varphi }_y={\eta }_t$:The sign follows from the convention that positive pressure displaces the upper wall upward.

Retain the rigid-wall cosine in $x$ and the lower-wall cosine in $y$:The wave equation gives ${\omega }^2=c_0^2[(n\pi /a{)}^2+(\beta /b{)}^2]$. The upper-wall condition gives $-k\beta \sin \beta /b-{\omega }^2\cos \beta =0$, henceFor nonzero real $\beta$, eliminating $\omega$ gives $-\tan \beta =M(\beta +{\kappa }_n^2/\beta )$ with $M=c_0^2/(kb)>0$, ${\kappa }_n=n\pi b/a$.

On every interval $((j+1/2)\pi ,(j+1)\pi )$, the left side decreases from positive infinity to zero, while the right side is finite and positive. Continuity gives at least one intersection per interval for each $n$. The functions are analytic between their poles, so roots are isolated there; their union over countably many intervals is a countably infinite set. Reflection gives the redundant negative-$\beta$ solutions. For $n=0$, the limiting formal root $\beta =0$ is the zero-frequency gauge mode, not one of these positive-frequency branches.

The graph uses representative dimensionless values $M=1$ and $b/a=1$; the interval argument applies for all positive values. For fixed $n$ and fixed branch $j$, take $M\to \infty$. The positive right side becomes large, forcing the intersection towards the tangent pole from the right:More precisely, if ${\beta }_j^0=(j+1/2)\pi$, then ${\beta }_j-{\beta }_j^0\sim {\beta }_j^0/[M(({\beta }_j^0{)}^2+{\kappa }_n^2)]$. Therefore the requested real-branch frequencies satisfyThe limiting upper condition is effectively pressure release for these acoustic branches. No claim is made here about imaginary or other complex $\beta$ branches, which were excluded from the requested analysis.