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For a material volume, mass conservation is $d{\int }_{V(t)}\rho dV/dt=0$. The transport theorem and arbitrariness of that volume give $\boxed{{\rho }_t+\nabla \cdot (\rho u)=0}$. Balance of linear momentum includes the pressure traction and body force:Apply transport, the divergence theorem and the mass equation to obtainFor small perturbations of the uniform resting barotropic state, write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=p_0+c_0^2{\rho }^{\prime }+\cdots$. Linearized mass and momentum areTaking the curl shows that initially vanishing vorticity remains zero, so the acoustic velocity has a potential $u=\nabla \varphi$. Integrating momentum, using the additive freedom of a velocity potential to remove the spatially constant integration term, gives ${\rho }^{\prime }=-({\rho }_0/c_0^2){\varphi }_t$. Continuity now yields $\boxed{{\varphi }_{tt}=c_0^2{\nabla }^2\varphi }$.

For spherical symmetry, set $\varphi =e^{-i\omega t}F(r)$ and $k=\omega /c_0$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }+k^2{r}^{2}F=0$, or $(rF{)}^{\prime \prime }+k^2(rF)=0$. HenceRigid boundaries require radial velocity $F^{\prime }=0$ at both radii. A form automatically satisfying the condition at $a$ isAt $b$, the condition $b(rF{)}^{\prime }-(rF)=0$ becomesTherefore the nonzero oscillatory eigenfrequencies obeyThe undivided sine-cosine form avoids artificial issues from dividing by a cosine. The formal $k=0$ root is the spatially constant zero-frequency potential additive freedom of a velocity potential, not a propagating acoustic oscillation. This is the radial acoustic spectrum between rigid concentric spheres.

In an unbounded incompressible flow with spherical symmetry, the kinematic boundary condition on a sphere of radius $R(t)$ gives $u_r=R^2\dot{R}/r^2$ and velocity potential $\varphi =G(t)-R^2\dot{R}/r$. The Unsteady Bernoulli equation at infinity has $F(t)=\dot{G}+p_{\infty }/\rho$. Subtracting it from the finite-radius equation gives the displayed pressure, independently of $G$. Time differentiation is at fixed $r$. Evaluating at $r=R(t)$ yields the Rayleigh equation for an inviscid spherical bubble when the boundary pressure is prescribed.

Let $\kappa >0$ be an inverse length. In three spatial dimensions, seek the Green's function that decays at infinity and satisfieswhere ${\nabla }^2$ is the Laplacian and ${\delta }^{(3)}$ the three-dimensional Dirac delta function. The point source and the differential operator have spherical symmetry. For $r>0$, the radial Laplacian givesWriting $u(r)=r{G}_{\kappa }(r)$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$. Decay at infinity removes the growing exponential function, leaving $G_{\kappa }(r)=C{e}^{-\kappa r}/r$.

The Dirac delta function fixes the coefficient. Integrating over a ball $B_{ϵ}$ and using the divergence theorem givesThe first term tends to $4\pi C$, while the second is $O({ϵ}^2)$. Hence $C=1/(4\pi )$ andThis solves the source equation in the sense of distributions, including at the origin. Merely verifying the equation for $r>0$ would leave the normalization undetermined. The limit $\kappa \to 0$ at fixed $r>0$ is $1/(4\pi r)$, the Green function of the Laplacian for the convention $-{\nabla }^2$.

For an attractive scalar toy interaction, absorb constants into $A>0$ and setHere $A$ has units of energy times length, and negative $F_r$ points inward. At equal $A$, comparison with the unscreened model $V_0=-A/r$ givesThus the potential energy and force have the same exponential scale but different suppression factors. At $x=1$, these ratios are approximately $0.368$ and $0.736$; at $x=3$, they are $0.0498$ and $0.199$. Finite range means rapid decay, not an abrupt cutoff.

The Tresca yield criterion bounds the largest resolved shear stress, expressed through the principal stresses. In spherical symmetry, the two tangential principal stresses coincide, so $f=|{\tau }_{rr}-{\tau }_{\theta \theta }|/2$. Here ${\tau }_Y$ is a shear yield stress; the corresponding uniaxial yield strength is $2{\tau }_Y$.