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Let $h_{\mu \nu }={\partial }_{\mu }{X}^a{\partial }_{\nu }{X}_a$ be the induced worldvolume metric and write $H={\gamma }^{\mu \nu }{h}_{\mu \nu }$. Varying the embedding in the auxiliary-metric brane action and integrating by parts givesThe boundary contribution, including its relative factor, isAt initial and final times one fixes the endpoint configurations or uses variations of compact support. At a spatial boundary, free target directions require Neumann boundary conditions $n_{\mu }{\gamma }^{\mu \nu }{\partial }_{\nu }{X}^a=0$; fixed directions require Dirichlet boundary conditions $\delta X^a=0$. Mixed conditions must make this boundary pairing vanish. A closed brane has no spatial boundary. Since the auxiliary worldvolume metric enters without derivatives, its variation produces no extra boundary term.

Using $\delta \sqrt{-\gamma }=-(1/2)\sqrt{-\gamma }{\gamma }_{\mu \nu }\delta {\gamma }^{\mu \nu }$, variation of the inverse worldvolume metric givesTaking the trace yields $(p-1)(H-p-1)=0$. For $p\ne 1$, this forces $H=p+1$, and substitution gives ${\gamma }_{\mu \nu }=h_{\mu \nu }$. For a nondegenerate timelike embedding, eliminating the auxiliary worldvolume metric therefore gives twice the worldvolume area, with the overall physical brane tension supplied by the action normalization.

For $p=1$, the traced metric equation is an identity, and the remaining equation only says $h_{\mu \nu }=(H/2){\gamma }_{\mu \nu }$. It fixes the worldsheet metric up to a Weyl transformation, rather than determining it uniquely. Indeed $\sqrt{-\gamma }{\gamma }^{\mu \nu }$ is Weyl invariant precisely in two worldvolume dimensions, and the constant term vanishes precisely at $p=1$. This is the Weyl-invariance exception for the string among branes. In a nondegenerate interior it gives the familiar classical equivalence of Polyakov and Nambu–Goto actions; degeneracies at a free string endpoint must be treated through the original equations.

For the open string, choose conformal gauge ${\gamma }_{\mu \nu }=\mathrm{diag}(-1,1)$ on $0\le \sigma \le \pi$. The embedding equation is the wave equation, and NN means ${\partial }_{\sigma }{X}^a=0$ at both spatial endpoints. For these operator formulas restore the conventional overall normalization $S=-TI/2$, with $T=(2\pi \alpha \prime {)}^{-1}$; the overall factor does not change the preceding classical equations. Its open-string mode expansion isReality requires $({\alpha }_n^a{)}^{†}={\alpha }_{-n}^a$. Canonical quantization givesThe metric equation is the vanishing of the worldsheet stress tensor: $({\partial }_{\tau }X+{\partial }_{\sigma }X{)}^2=({\partial }_{\tau }X-{\partial }_{\sigma }X{)}^2=0$. The two endpoint-compatible mode expansions contain the same string oscillator family. Their quadratic coefficients are the classical Virasoro constraints.

For the operators, use normal ordering with positive-index string oscillators as annihilators:Thus $({\partial }_{\tau }X\pm {\partial }_{\sigma }X{)}^2$ has modes $4{\alpha }^{\prime }{L}_m$ before the quantum ordering correction. The physical-state Virasoro conditions for an open string areOne imposes only the positive modes on kets, with the adjoint conditions on bras, as in Gupta-Bleuler quantization. Requiring every positive and negative mode to annihilate the same state would conflict with the Virasoro central extension. The intercept $a$ is the zero-mode ordering constant. The standard critical bosonic string theory has $d=26$ and $a=1$; in light-cone gauge in string theory the transverse zero-point energy gives $a=(d-2)/24$, while full anomaly-free Lorentz or BRST quantization fixes the critical values. The mass constraint is then ${\alpha }^{\prime }{M}^2=N-a$, with $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$.

To compute the Virasoro algebra, commute a quadratic generator with one string oscillator:These identities and the Jacobi identity imply that $[L_m,L_n]-(m-n)L_{m+n}$ commutes with every string oscillator and with the center-of-mass coordinates and momenta. Mode number permits a scalar term only for $m+n=0$. Its coefficient follows from the formal zero-momentum Fock vacuum, on which $L_0=0$. For $m>0$,The two possible string oscillator contractions giveThe timelike target coordinate still contributes one to this central charge: its two metric signs cancel in ${\eta }_{ab}{\eta }^{ab}=d$. ConsequentlyThis is the free-boson Virasoro central term. It vanishes for the three global conformal modes $m=0,\pm 1$.

To keep the intercept convention separate, define ${\mathcal{L}}_m=L_m-a{\delta }_{m,0}$. The Virasoro zero-mode shift changes the displayed central term toThe matter central charge here is $d$, not $d-2$. Covariant worldsheet ghost fields contribute $-26$; their inclusion cancels the anomaly at $d=26$, while the intercept is handled by the appropriate zero-mode and physical-state convention.

The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of smooth amplitudes with estimatesThere is no loss of symbol order under an $x$ derivative. A phase function is real and smooth for $\theta \ne 0$, positively homogeneous of degree one in $\theta$, with its full differential $(d_x\Phi ,d_{\theta }\Phi )$ nowhere zero there, at least on the amplitude's conic support. Homogeneous phases need not be smooth at $\theta =0$; that bounded-frequency region is handled separately or the amplitude is cut off near it.

Choose $\rho \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The oscillatory integral is defined by regularization in distributions:This is an ordinary integral when decay makes it absolutely convergent. For general order, the definition is independent of the regularizer. To see the mechanism, at large $|\theta |$ setHomogeneity and the nonvanishing full differential give $A\gtrsim |\theta {|}^2$ on compact $x$ sets, and $L{e}^{i\Phi }=e^{i\Phi }$. Its $x$-derivative coefficients have order $-1$ and its $\theta$-derivative coefficients order zero, so each formal transpose $L^t$ lowers the amplitude order by one. Iterating more than $N+k$ times makes the test-function pairing integrable, and the same estimates control regularizer derivatives. This justifies the cutoff function limit and its independence. A fixed bounded-frequency integral is smooth in $x$.

The singular support of a distribution is the complement of the largest open set on which it is represented by a smooth function. For the support-sensitive bound one must use conic support of an oscillatory amplitude: a closed set of limiting high-frequency directions, locally in the base variable. One sufficient precise choice iswith closure taken in this product. If $x_0$ is outside the projection of $C_a\cap \{d_{\theta }\Phi =0\}$, compactness of the unit sphere gives a neighborhood of $x_0$ and a positive uniform lower bound on $|d_{\theta }\Phi |$ wherever the large-frequency amplitude is supported. On that region useIts coefficients have order zero and its transpose lowers order by one. To prove $C^j$ regularity, first differentiate $j$ times in $x$, raising the amplitude order by at most $j$, and then integrate by parts more than $N+j+k$ times. The resulting integrals and their derivatives converge uniformly on compact neighborhoods. Since $j$ is arbitrary, the stationary-direction bound for singular support follows:This is the standard conic interpretation of the support restriction in the question; directions at infinity, rather than finite-frequency stationary points, determine possible singularities.

If $\mathrm{supp}a(x,\cdot )$ literally means the support of the restricted function, the printed inclusion can fail. There is a slice-support obstruction for oscillatory singularities. In one dimension, take $\Phi =x\theta$ and $a=x\eta (\theta )/|\theta |$, where $\eta$ is smooth and even, zero for $|\theta |\le 1$, and one for $|\theta |\ge 2$. This is a symbol of order $-1$, and the full phase differential is nonzero for $\theta \ne 0$. Its regularized integral near $x=0$ isIndeed substitute $u=|x|\theta$ in the tail, split at $u=1$, and write $\cos u=1+(\cos u-1)$ below one. The first term gives $-\log |x|$, while the remainder has an even convergent power series in $x$ plus a constant; the bounded-frequency correction is smooth. Thus $0$ is in the singular support, but $a(0,\cdot )=0$ has empty slice support. The joint closed conic support retains this limiting base point and makes the proved theorem valid. No nonstationarity conclusion is inferred just from the amplitude vanishing at one base point.

For the wave equation, Fourier transformation in $x$ reduces the initial-value problem to ${\partial }_t^2\hat{E}+c^2|\xi {|}^2\hat{E}=0$, with $\hat{E}(\xi ,0)=0$ and ${\partial }_t\hat{E}(\xi ,0)=1$. For $c>0$ its solution iswith value $t$ at zero. Choose $\chi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one near zero. The low-frequency decomposition of the wave propagator iswhereThe low-frequency term is an ordinary smooth function: the sine quotient extends smoothly in $\xi$ at zero and all derivatives are integrable on the fixed compact frequency set. The high-frequency amplitudes belong to $\mathrm{Sym}({\mathbb{R}}^{n+1},{\mathbb{R}}^n;-1)$, vanish near zero, and the phases have nonzero full differential since $d_x{\Phi }_{\pm }=\xi \ne 0$. Their stationary-direction equations are $x\pm ct\xi /|\xi |=0$. Such a direction can occur only if $|x|=c|t|$. The smooth low-frequency term contributes no singular support, soThis is a statement about singular support: some dimensions have a smooth nonzero tail inside the cone, so it does not claim that the distribution is concentrated only on the cone.

Put $u=ct-x$ and $v=ct+x$. The chain rule turns the wave equation into $4y_{uv}=0$. Integrating first in $v$ and then in $u$ gives the D'Alembert formulaAt the left endpoint, $g(u)=-f(u)$ for $u>0$. At the right endpoint, $f(ct-L)+g(ct+L)=0$, henceThus the travelling waves can be represented by $g=-f$, with $f$ and $g$ both $2L$-periodic. More precisely, the positive-time boundary conditions constrain only $f$ on $(-L,\infty )$ and $g$ on $(0,\infty )$; their unused arguments may be filled in by this periodic extension. They do not constrain arbitrary pre-existing choices outside those ranges. This is periodic reflection for a fixed-end string.

Using $g=-f$, differentiation gives $y_t=c(f^{\prime }(ct-x)-f^{\prime }(ct+x))$ and $y_x=-f^{\prime }(ct-x)-f^{\prime }(ct+x)$. The cross terms in the wave energy cancel, soThis integrates a $2L$-periodic function over exactly one period. $E(t)$ is constant. Equivalently, the wave equation gives $E^{\prime }(t)=[y_t{y}_x{]}_0^L=0$, because the fixed endpoint values imply $y_t=0$ there.

For this geometry $\nabla \cdot \mathbf{u}=0$, $\nabla \cdot \mathbf{B}=0$ and $\mathbf{u}\cdot \nabla =0$. Consequently constant mass density and pressure satisfy the full nonlinear continuity and adiabatic pressure equations. Write ${\mathbf{B}}_{\perp }=(B_x,B_y)$ and ${\mathbf{u}}_{\perp }=(u_x,u_y)$. The ideal magnetohydrodynamic induction equation and the transverse ideal magnetohydrodynamic momentum equation becomeThe longitudinal momentum equation isThus constant transverse magnetic magnitude eliminates the otherwise unavoidable magnetic pressure acceleration. Differentiating the ideal magnetohydrodynamic induction equation in time gives, for either transverse component,A constant-magnitude nonlinear Alfvén wave is obtained by taking $B_z\ne 0$ andwhere $A$ and ${\mathbf{U}}_0$ are constants and $F$ is any sufficiently differentiable real function. Both first-order equations hold, and $B_x^2+B_y^2=A^2$. Taking $F(\zeta )=k\zeta$ gives an exact finite-amplitude circularly polarized wave. More general traveling rotations also work; an arbitrary superposition of oppositely traveling solutions of the wave equation need not preserve transverse magnetic magnitude and hence need not solve the full nonlinear system. If $B_z=0$, the coupled first-order equations instead require time-independent transverse fields and velocities, with spatially constant transverse magnetic magnitude; there is no propagating Alfvén wave in the $z$ direction.

Put ${\rho }^{\prime }=\rho -{\rho }_0$, and take the retarded acoustic Green function for ${\partial }_t^2-c_0^2{\nabla }^2$:This selects the causal, outgoing acoustic density perturbation, with no additional incoming homogeneous wave equation solution. Convolving the acoustic dipole forcing with this Green function and moving its spatial derivative outside the integral givesThe surface delta distribution converts the spatial integral to the moving surface. At fixed surface labels $(p,q)$, setThe radial Mach number enters the moving-surface retarded Jacobian, because $dr/d\tau =-\hat{r}\cdot v$ and henceLet ${\tau }^{\ast }$ be a root of the retarded time equationThe delta change-of-variable rule now givesHere $h_p{h}_q$ is the surface-area factor from the orthogonal surface coordinates. For a subsonic surface, $|v|<c_0$, the retarded equation is monotone in $\tau$ and has one root when the motion is defined for the required past times. For more general motion, sum the displayed contribution over all simple retarded roots. A root with $1-M_r=0$ requires a separate limiting treatment; the simple-root formula does not apply there.

With zero initial time derivative, the integral term in the D'Alembert formula vanishes. The Heaviside step function representation givesThe rectangular pulse splitting under the wave equation consists of two width-two pulses of height $1/2$, moving left and right at speed $c$. At $t=0$ they coincide and add to the initial height-one pulse. Before $ct=1$ the overlapping interval has height one; after $ct=1$ the two pulses are separated. Values exactly at the fronts depend on the chosen $H(0)$ convention. Away from the fronts the equation holds classically, and across the fronts it holds as a weak solution of the wave equation.

For $\mathrm{Re}s>0$, integration by parts recursively in the Laplace transform givesThe initial exponential boundary term vanishes, and $\mathcal{L}[1]=1/s$ starts the recursion. Here $H$ is the Heaviside step function.

Transform the displacement in time, writing $Y(x,s)={\mathcal{L}}_t[y](x,s)$. Zero initial displacement and velocity give $s^{2}Y=c^2{Y}_{xx}+g/s$. A particular solution is $g/s^3$, and the spatial homogeneous solutions are $e^{\pm sx/c}$. The prescribed far-field behavior excludes the growing exponential. The fixed boundary sets the remaining coefficient to $-g/s^3$, henceSince $\mathcal{L}[t^2/2]=1/s^3$, the Laplace transform time-shift rule givesEquivalently, $y=g{t}^2/2$ for $t\le x/c$, and $y=gxt/c-g{x}^2/(2c^2)$ for $t\ge x/c$. This solves the forced wave equation in each region and has continuous displacement and first derivatives across the characteristic $t=x/c$, so it also solves the equation weakly there. It satisfies the fixed-end and initial conditions and equals $g{t}^2/2$ sufficiently far away at each fixed time. The end's influence travels at speed $c$.

For a defining function $\varphi$ with $d\varphi \ne 0$, the characteristic hypersurface test is that the principal symbol vanish at $d\varphi$.

For the wave equation with speed $c>0$,Thus its characteristic hypersurfaces satisfy ${\varphi }_t^2=c^2|{\nabla }_x\varphi {|}^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.

For the free Schrodinger equation, in normalized units,Its total-order characteristic hypersurfaces satisfy ${\nabla }_x\varphi =0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.

For the Laplace equation,There are no real characteristic hypersurfaces for the Laplace equation, since their normal cannot be zero. These statements concern the ordinary total-order principal symbol, not a weighted space-time grading.

For $u\in {\mathcal{E}}^{\prime }$, choose a cutoff function $\chi =1$ on a neighborhood of its support. Its extension to smooth functions makesindependent of the choice of cutoff. For real frequency, the definition of the distributional Fourier transform gives $F(\lambda )=\hat{u}(\lambda )$: interchange the pairing with the integral of a Schwartz function, using the finite-order estimate on the cutoff's compact support. On each compact set of complex frequencies the exponential and all its $x$ derivatives have uniformly convergent power series. Continuity of the distribution allows termwise differentiation, withConsequently $F$ is entire on ${\mathbb{C}}^n$.

For the precise exponential type, a fixed enlarged support would give an unnecessarily enlarged radius. Instead use the shrinking-cutoff exponential-type estimate. On one fixed compact neighborhood of the closed radius-$\delta$ ball, $u$ has finite order of a distribution $m$. Choose ${\chi }_{\epsilon }=1$ near that ball, supported in the radius-$(\delta +\epsilon )$ ball, with $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1/2$. The finite-order estimate givesTake $\epsilon =[2(1+|z|){]}^{-1}$. Its extra exponential is bounded by $e^{1/2}$, while the derivative cost is polynomial. ThusThe entire function was defined with a fixed cutoff; only its bound uses a frequency-dependent cutoff, so no holomorphic dependence is lost. This proves the required estimate with some $N\ge 0$ without enlarging $\delta$.

For the converse, set $V(z)=e^{iz\cdot y}U(z)$. Its restriction to real frequencies has polynomial growth. Define the inverse tempered distribution $v$ byRapid decay of the Schwartz function transform proves convergence and continuity, and Fourier inversion gives $\hat{v}=V$ on real frequencies.

We prove its support directly by contour shifting, rather than assuming the support conclusion of the Paley–Wiener–Schwartz theorem. Let $\omega$ be a real unit vector and take a test function $\varphi$ supported in $x\cdot \omega \ge \delta +ϵ$ for some $ϵ>0$. For every integer $M\ge 0$, integration by parts in the real-frequency Fourier integral givesIndeed, move $(1-{\Delta }_x{)}^M$ from the oscillatory exponential to $e^{-t\omega \cdot x}\varphi$; its derivatives supply at most $2M$ powers of $t$.

The product $V(z)\hat{\varphi }(-z)$ is entire. Rotate coordinates so that $\omega$ is the first coordinate vector and apply Cauchy integral theorem to a rectangle in that one complex coordinate, integrating the remaining real coordinates afterwards. For fixed $t$, choose $2M>N+n+2$; the displayed decay estimate, uniformly on the intervening imaginary segment, makes the vertical faces vanish as the real rectangle width tends to infinity. The horizontal integrals are absolutely convergent. HenceUsing $(1+|\xi +it\omega |{)}^N\le C(1+t{)}^N(1+|\xi |{)}^N$ and the growth hypothesis bounds this pairing byIt tends to zero as $t\to \infty$, so the pairing vanishes. Every point outside the closed radius-$\delta$ ball lies in such a separating half-space; a finite partition of unity for the support of a test function outside the ball proves $\mathrm{supp}v\subset \overline{B}(0,\delta )$.

Translate this compactly supported distribution by $y$: define $⟨u,\varphi ⟩=⟨v,\varphi (\cdot +y)⟩$. The Translation property of the Fourier transform givesThe compact-support entire extension agrees with $U$ everywhere by the identity theorem, applied successively in the complex coordinates. Fourier inversion also gives uniqueness. This completes both directions of the ball version of the Paley–Wiener–Schwartz theorem.

For the wave equation, take a real constant $c\ne 0$. The Fourier transform method for the wave equation givesThis inverse tempered distribution is twice differentiable in $t$: time derivatives introduce polynomial frequency factors, still integrable against every Schwartz function. It has the specified initial displacement and zero initial velocity and satisfies the equation distributionally.

To apply the support theorem, replace the real norm by the entire wave cosine multiplierThis series is entire and equals $\cos (ct\sqrt{z\cdot z})$ regardless of the square-root choice. Write $z=a+ib$ and $q=\sqrt{z\cdot z}$. Since $|z\cdot z|\le |a{|}^2+|b{|}^2$,Together with $|\cos w|\le e^{|\mathrm{Im}w|}$, this yields $|C_t(z)|\le e^{|c|t|\mathrm{Im}z|}$. The forward estimate for the translated initial support now givesThe converse therefore proves finite propagation with the stated speed:In particular, $\sqrt{z\cdot z}$ itself need not be entire; it is the even cosine series that provides the required entire extension.

For completeness this construction identifies the distributional Cauchy solution even without initially assuming spatial temperedness. The zero-displacement wave multiplier is $S_{\tau }(z)={\int }_0^{\tau }{C}_s(z)ds$, entire with bound $|S_{\tau }(z)|\le \tau e^{|c|\tau |\mathrm{Im}z|}$. Given a compactly supported smooth $\varphi$ and final time $T$, the backward solution $w(s)={\mathcal{F}}^{-1}[S_{T-s}\hat{\varphi }]$ is smooth, has support in one compact ball for $0\le s\le T$ by the same support theorem, and satisfies $w(T)=0$, $w_s(T)=-\varphi$. For the difference $h$ of two distributional solutions with zero initial data,All pairings use compactly supported test functions. The expression vanishes initially and equals $⟨h(T),\varphi ⟩$ finally. Hence $h(T)=0$, establishing uniqueness in the usual time-differentiable distributional solution class.

An acoustic analogy is an exact rearrangement of the fluid equations into a chosen linear propagation operator acting on an acoustic variable, with everything left over placed on the right as effective forcing. The rearrangement becomes a sound-prediction method only after a reference medium, boundary conditions and approximations to the forcing are specified. In particular, a right-hand-side term need not represent independently generated sound.

Use Einstein summation convention and write ${\partial }_i=\partial /\partial x_i$. Differentiating the continuity equation in time and taking the divergence of the momentum equation eliminates ${\partial }_t(\rho u_i)$:Set $q=p-{\hat{p}}_0$ and $W_{ij}=\rho u_i{u}_j-{\sigma }_{ij}$. Since $p+\chi =q+P$, the constant $P$ has zero Laplacian. The prescribed reference mass density and reference speed of sound are independent of time, soCombining the two identities givesOnly conservation of mass and conservation of momentum have been used; no equation of state or energy equation was needed. Spatial variation of ${\hat{c}}_0$ creates no omitted derivative in this identity, because it multiplies a time derivative. The double divergence of the momentum flux tensor has the structure of an acoustic quadrupole.

For a localized flow, choose the reference fields to match the stationary surrounding medium: ${\hat{\rho }}_0={\rho }_0$ and ${\hat{c}}_0^2=c_0^2$ outside the flow, with $P$ chosen so ${\hat{p}}_0=p_0$ there. A uniform surrounding fluid permits ambient constant mass density and adiabatic sound speed, giving the familiar homogeneous wave equation. A nonuniform surrounding fluid calls for its actual stationary reference profiles, extended sensibly through the flow region. This makes the acoustic variable vanish in the unperturbed exterior and minimizes artificial contrast terms. In a uniform isentropic exterior the leading acoustic relation $p^{\prime }=c_0^2{\rho }^{\prime }$ also makes $Q$ vanish to first order. In a stratified exterior, propagation and entropy-advection effects can remain in $Q$, as the next part demonstrates.