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The Dirichlet Laplacian eigenfunctions and positive Dirichlet Laplacian eigenvalues on the interval of length two areA Sturm-Liouville eigenfunction expansion gives $u={\sum }_j{q}_j(t){\varphi }_j$, with $q_j^{\prime \prime }+({\lambda }_j-\alpha )q_j=0$. Thus all normal modes oscillate precisely when $\alpha <{\lambda }_1$.

To justify a spatially uniform bound for arbitrary appropriate data, assume the usual wave equation energy class $u(0)\in H_0^1(-1,1)$ and $u_t(0)\in L^2(-1,1)$, or smoother compatible data. The energy method conservesThe sharp Poincare inequality is $\Vert u{\Vert }_{L^2}^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_{L^2}^2$. For $0\le \alpha <{\lambda }_1$, this makes $E$ coercive with constant $1-\alpha /{\lambda }_1$; for $\alpha <0$, coercivity is immediate. The one-dimensional Sobolev representative satisfies $|u(x,t)|\le \sqrt{2}\Vert u_x(t){\Vert }_{L^2}$ by the fundamental theorem of calculus and the Cauchy-Schwarz inequality. Hence the conserved energy bounds $u$ uniformly in both variables.

At $\alpha ={\lambda }_1$, the smooth solution $u=t{\varphi }_1$ is unbounded. At $\alpha >{\lambda }_1$, $u=e^{\sqrt{\alpha -{\lambda }_1}t}{\varphi }_1$ is unbounded. ThereforeThe bound depends on the initial data; it is not a single bound for all arbitrarily rescaled solutions. The energy-class hypothesis supplies pointwise meaning and excludes undefined rough-data interpretations.

The all-time conclusion in the original PDF is false. The expanding-flow transformation of a wave equation makes both the correct forward statement and a counterexample transparent. PutFor $D={\partial }_t+\rho {\partial }_{\rho }$, the expanded operator from part (i) is $D^2+D-{\partial }_{\rho }^2$. In the new variables $D=-S{\partial }_S$ and ${\partial }_{\rho }=S{\partial }_X$, henceThus it is the ordinary wave equation on the half-plane $S>0$. By the D'Alembert formula, any $C^2$ solution has the formAt $t=0$, differentiating the zero displacement with respect to $\rho$ and using the zero velocity gives, for $-1<\rho <1$,Subtracting $(\rho -1)$ times the first equation from the second gives $2F^{\prime }(\rho +1)=0$, and then $G^{\prime }(\rho -1)=0$. Thus $F=c$ on $(0,2)$ and $G=-c$ on $(-2,0)$. If $t\ge 0$ and $-1<\rho <1$, both arguments remain in those intervals. Consequently the valid forward result isFor negative time the arguments need not stay in the initial intervals. Take $F(s)=(s-2{)}_{+}^3$ and $G=0$, where $q_{+}=\max \{q,0\}$. Thenis a global $C^2$ solution, has the prescribed zero initial displacement and velocity on $(-1,1)$, but $u(0,-\log 3)=1$. This directly disproves the assertion for $t\in \mathbb{R}$.

The connection with part (i) is directional finite propagation speed. At a future point in the strip, the two characteristics traced back to $t=0$ meet it at ${\rho }_0=e^{-t}(\rho +1)-1$ and ${\rho }_0=e^{-t}(\rho -1)+1$, both in $(-1,1)$. For negative time these intersections can lie outside that interval. The lines $\rho =\pm 1$ are characteristic barriers for this forward domain of dependence, not barriers in both time directions.

In conformal gauge, ${\gamma }_{\mu \nu }=e^{2\omega }\mathrm{diag}(-1,1)$. The Weyl invariance of the Polyakov action removes $\omega$, leavingIts complete variation isFix the string embedding map on the initial and final time slices. The interior Euler-Lagrange field equation is the wave equationAt each spatial endpoint the remaining variation vanishes under either Dirichlet boundary conditions, $\delta X^m=0$, or Neumann boundary conditions, $X^{m\prime }=0$, independently for each target-space component. Thus a fixed endpoint or a free endpoint makes the boundary contribution vanish. Fixing a spatial point throughout time imposes Dirichlet boundary conditions on its spatial coordinates, while the time coordinate may retain Neumann boundary conditions.

The independent metric tensor equation must also be retained after gauge fixing. Its Virasoro constraints areTogether with the wave equation, these ensure equivalence to the Nambu–Goto action; varying only the already fixed metric would lose these Virasoro constraints.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

Write the density and pressure as ${\rho }_0+{\rho }^{\prime }$ and $p_0+p^{\prime }$, and let the small fluid velocity be the gradient of an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$. The linear homentropic acoustic equations arewhere the adiabatic sound speed of a perfect gas is $c_0^2=\gamma p_0/{\rho }_0$. Absorbing a function of time into $\varphi$, the momentum equation givesEliminating $p^{\prime }$ and ${\rho }^{\prime }$ gives the acoustic wave equation

Let the shell displacement be $a(t)-a_0=\mathrm{Re}(\eta e^{-i\omega t})$ and put $k=\omega /c_0$, $\theta =k{a}_0$. The spherically symmetric solution regular at the centre isThe kinematic boundary condition ${\varphi }_r(a_0,t)=\dot{a}(t)$ givesSince $p^{\prime }=-{\rho }_0{\varphi }_t$, the pressure amplitude on the shell is thereforeThe radial Newton's second law per unit area isso its complex amplitude satisfiesFor a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(m{c}_0^2)$ yield the frequency equation for spherically symmetric vibration of a gas-filled elastic shell:The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta \cot \theta -1$ runs from $+\infty$ to $-\infty$ on $(n\pi ,(n+1)\pi )$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

Write the density and pressure as $\rho ={\rho }_0+{\rho }^{\prime }$ and $p=p_0+p^{\prime }$, and let $\mathbf{u}$ be the small fluid velocity. The linear homentropic acoustic equations arewhere the adiabatic sound speed isIf the flow is irrotational, write $\mathbf{u}=\nabla \varphi$. The momentum equation integrates spatially toa purely time-dependent integration term having been absorbed into $\varphi$. Substitution into continuity gives the wave equationso the wave speed is $c_0$.

To obtain energy conservation, dot the momentum equation with $\mathbf{u}$ and use the product rule:The continuity and pressure-density relations giveTherefore the acoustic energy conservation law iswith acoustic energy density and intensity

Takeand define the positive normal decay rateThe decaying solutions of the acoustic wave equation that satisfy the kinematic conditions areTheir pressures at the membrane areSubstitution into the dynamic boundary condition givesand hence the acoustic wave on a tensioned massive membrane dispersion relation

Writing the phase speed as $c=\omega /k$, the positive added-inertia term givesPhysically, each evanescent acoustic surface wave accelerates a layer of fluid on both sides of the membrane, so the inertia per unit area exceeds $m$. Moreover,and the added mass of an evanescent fluid layer $2{\rho }_0/q$ diverges as $k\to 0$. In that limit $c\ll c_0$, so $q\sim k$ and

The string is initially at rest, so $y(x,0)=y_t(x,0)=0$. Taking the Laplace transform in $t$ converts the wave equation intoFor $\mathrm{Re}p>0$, the solution that decays as $x\to \infty$ is $\hat{y}=A(p)e^{-px/c}$. The transformed boundary condition $\hat{y}(0,p)=\hat{f}(p)$ fixes $A=\hat{f}$, and therefore

With $\eta =x-t$ and $\xi =x+t$, the chain rule givesHenceso the wave equation is equivalent to $U_{\eta \xi }=0$. Integrating in each variable givesThe initial conditions implySolving and integrating yields the D'Alembert formula with initial velocity

If $|x-x_0|>r+t$, then both $x-t$ and $x+t$ lie outside the support interval $[x_0-r,x_0+r]$, and every point between them does as well. All three terms therefore vanish. Thuswhich is finite propagation speed.

For the finite string, differentiate its energy and use $y_{tt}=y_{xx}$:The fixed-end conditions hold for every $t$, so differentiation gives $y_t(0,t)=y_t(L,t)=0$. The boundary term vanishes and

With $\xi =x+ct$ and $\eta =x-ct$,so the wave equation becomesThus $u=F(\xi )+G(\eta )$. At $t=0$ the initial data giveSolving for $F^{\prime }$ and $G^{\prime }$ and integrating gives d'Alembert's formula>

For the wave operator ${\partial }_x^2-{\partial }_y^2$, use the symboland complexify $\eta$. Its roots $\eta =\pm \xi$ are real for every real $\xi$, so the single horizontal stepnever meets a root and is an explicit Hörmander staircase.

Write $\rho ={\rho }_0+\tilde{\rho }$, $p=p_0+\tilde{p}$, and let $\tilde{\mathbf{u}}$ be small. Linearized mass and momentum conservation areThe linearized equation of state is
$\tilde{p}=c_0^2\tilde{\rho }$. Differentiate the first equation in time and eliminate the velocity with the second to obtain the wave equation

Take the spatial Fourier transform of the wave equation. The transformed initial-value problem isand hence

Part (c) and the convolution theorem turn the first term intoFor the second term, choose $p$ with $p^{\prime }=g$. Part (b) says $\tilde{g}=ik\tilde{p}$, soPart (c) now makes its inverse transformCombining the two terms gives the D'Alembert formula

Let $A=2V$ and $B=W-V$. Part (c) makes $A$ and $B$ independent identically distributed exponential variables of rate $\lambda$. The change of variables $(A,B)=(rs,(1-r)s)$ has Jacobian $s$, and integrating ${\lambda }^{2}s{e}^{-\lambda s}$ over $s>0$ gives density one for $0<r<1$. Therefore the uniform ratio of independent exponential variables gives

Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula

For a smooth function $g$ define its spherical meanThe Kirchhoff formuladefines a smooth solution of the three-dimensional wave equation for all positive and negative $t$ and has the prescribed data at $t=0$.

For uniqueness, apply the local energy estimate to the difference of two solutions on a backward light cone. Its energy at the cone tip is bounded by the zero initial energy on the cone base, so the difference and its derivatives vanish. Covering spacetime by such cones proves uniqueness among $C^2$ solutions.

Compact support is unnecessary for existence or uniqueness: the sphere in the Kirchhoff formula is compact for each $(t,x)$, so arbitrary smooth data suffice, and the cone-energy proof is local.

The Euler-Lagrange equation of each component of the string embedding map is the wave equation $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Its left- and right-moving solutions are identified by the endpoint Neumann boundary conditions, so the resulting open-string mode expansion isIndeed, the Fourier cosine series makes ${\partial }_{\sigma }{X}^{\mu }$ vanish at $\sigma =0,\pi$. The canonical momentum density is ${\Pi }_{\mu }=(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}_{\mu }$. Imposing the equal-time canonical commutation relations $[X^{\mu }(\sigma ),{\Pi }_{\nu }({\sigma }^{\prime })]=i{\delta }_{\nu }^{\mu }\delta (\sigma -{\sigma }^{\prime })$ gives the covariant quantization of the bosonic stringwith ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$.

Hereso the Euler-Lagrange field equation is the variable-coefficient wave equationAt leading order, $\Phi =A(X,T)\cos \theta$ hasand henceThe leading part of the modulated-wave first integral isA nonconstant periodic wave therefore requires the local dispersion relationThis is usually called the eikonal equation, or equivalently the leading geometric-optics dispersion relation.

Let $ϵ$ be the group parameter. The infinitesimal generator of a Lie point symmetryhas flow equationsTherefore it generates the scaling group

Here ${\xi }^t=t$, ${\xi }^x=x$, and $\eta =0$. The total-derivative formulas for the second prolongation of a Lie point symmetry giveandThusFor the wave equation expression $F=u_{tt}-u_{xx}$,It therefore vanishes whenever $F=0$, proving that $V$ generates the simultaneous spacetime scaling symmetry of the wave equation.

The independent invariant of the scaling orbits is the similarity variablewhile $u$ itself is invariant. Seek a group-invariant solution $u(t,x)=f(z)$. Direct partial differentiation givesThe wave equation reduces toHence, on any interval avoiding $z=\pm 1$,