Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 4

Use the standard complex inner product $⟨u,v⟩=u^{\ast }v$, conjugate-linear in its first argument. A unitary matrix satisfies $U^{\ast }U=I$, and a Hermitian matrix satisfies $H^{\ast }=H$, where $\ast$ is conjugate transpose.

If $Hv=\lambda v$ with $v\ne 0$, thenPositivity of $⟨v,v⟩$ proves $\boxed{\lambda \in \mathbb{R}}$. If $Uv=\lambda v$, preservation of the norm gives $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\lambda {|}^2\Vert v{\Vert }^2$, so $\boxed{|\lambda |=1}$.

For eigenvectors $Hv=\lambda v$ and $Hw=\mu w$, the Hermitian matrix identity gives $⟨v,Hw⟩=⟨Hv,w⟩$. The already established reality of the eigenvalues therefore implies $(\mu -\lambda )⟨v,w⟩=0$. Thus distinct eigenvalues have orthogonal eigenvectors.

The Eisenstein criterion says that a primitive integer polynomial $P(x)=a_m{x}^m+\cdots +a_0$ is irreducible over $\mathbb{Q}$ if some prime number $p$ divides every $a_j$ for $j<m$, does not divide $a_m$, and $p^2$ does not divide $a_0$. By Gauss lemma for polynomials, this gives irreducibility in $\mathbb{Z}[x]$ as well.

For prime $n=p$, translate the variable in $P_p(x)=(x^p-1)/(x-1)$:Its leading coefficient is one, all other coefficients are divisible by $p$, and its constant coefficient is exactly $p$. The Eisenstein criterion proves this translated polynomial irreducible. Since $x\mapsto x+1$ is an invertible substitution in $\mathbb{Z}[x]$, $P_p$ is irreducible too.

If $n=ab$ with $a,b>1$, thenBoth factors have positive degree and integer coefficients, proving reducibility. Consequently $\boxed{1+x+\cdots +x^{n-1}\text{ is irreducible exactly when }n\text{ is prime}.}$

The integral is finite and nonnegative because $|f|$ is continuous on a compact interval. If it vanishes and $f(x_{\ast })\ne 0$, continuity gives a relative interval of positive length on which $|f|\ge |f(x_{\ast })|/2$, a contradiction. Thus the integral is zero only for $f=0$. For a real scalar $a$, $\Vert af\Vert =|a|\Vert f\Vert$, and the pointwise triangle inequality $|f+g|\le |f|+|g|$ integrates to $\Vert f+g\Vert \le \Vert f\Vert +\Vert g\Vert$. These are all the norm axioms.

The given sequence satisfiesHence it converges to the zero function in this integral norm. Although $f_n(0)=1$ for all $n$, an isolated endpoint contributes no integral. The pointwise limit is discontinuous, and the sequence does not converge in the uniform norm; neither fact obstructs convergence in the stated norm.

For a closed piecewise continuously differentiable path avoiding $a$, the winding number isBreak the parameter interval at the finitely many derivative discontinuities when evaluating this integral. Translating the curve gives $n(\gamma ,a)=n(\gamma -a,0)$, which will also be used below.

Where both derivatives exist, the ordinary product rule givesNeither denominator vanishes. Integrate over a common piecewise-smooth subdivision and use the contour characterization of the winding number:

On the right half-plane a single-valued holomorphic logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$, with $-\pi /2<\mathrm{Arg}z<\pi /2$, and its derivative is $1/z$. ThusThis proves the conclusion directly from the integral definition of the winding number, without appealing to its homotopy invariance.

Put $\eta =({\gamma }_2-a)/({\gamma }_1-a)$. The strict inequality gives $|\eta -1|<1$, so $\mathrm{Re}\eta >0$. This is a closed piecewise-smooth curve in the right half-plane, and part (b) gives $n(\eta ,0)=0$. Since ${\gamma }_2-a=({\gamma }_1-a)\eta$, part (a) yieldsThis is the dominated perturbation preserves winding number principle, here derived by logarithmic differentiation.

Take characteristic coordinates $\xi =ct-x$ and $\eta =ct+x$. Then $c^{-2}{y}_{tt}-y_{xx}=4y_{\xi \eta }$. The wave equation is therefore $y_{\xi \eta }=0$, whose general twice-differentiable solution isThe first fixed-end boundary gives $f(s)=-g(s)$. The second gives $g(s+L)=g(s-L)$, hence $g(s+2L)=g(s)$; $f$ has the same period. Thus $y=g(ct+x)-g(ct-x)$ with a $2L$-periodic function $g$.

Let $A=g^{\prime }(ct+x)$ and $B=g^{\prime }(ct-x)$. Then $y_t=c(A-B)$ and $y_x=A+B$, so the quadratic energy density reduces to $\frac{1}{2}[(A-B{)}^2+(A+B{)}^2]=A^2+B^2$. Changing $x$ to $-x$ in the second integral provesThe right side integrates a periodic function over one full period and is independent of its starting point. More explicitly its time derivative is $c[g^{\prime }(ct+L{)}^2-g^{\prime }(ct-L{)}^2]=0$. This is conservation of the fixed-end wave energy.

The operator commutator is $[A,B]=AB-BA$, acting on a common domain where both products are defined. Use the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, with positions mutually commuting and momenta mutually commuting. On smooth test functions the only nonzero terms in the expansion areThus the orbital angular momentum commutation relations giveThe product rule for a operator commutator givesbecause the last operator expression is symmetric in $i,k$, whereas ${ϵ}_{iwk}$ is antisymmetric. Hence $\boxed{[L^2,L_w]=0}$ for every component.

Write $r=\sqrt{x^2+y^2+z^2}$ and $P=x+y+z$. The differential angular momentum operator $L=-i\hslash x\times \nabla$ annihilates every radial function, so $L_i(P{e}^{-r})=e^{-r}{L}_{i}P$ and the same holds for its square. For example $L_{x}x=0$, $L_y^{2}x={\hslash }^{2}x$ and $L_z^{2}x={\hslash }^{2}x$. Cyclic symmetry gives $L^{2}y=2{\hslash }^{2}y$ and $L^{2}z=2{\hslash }^{2}z$ as well. ThereforeThe wavefunction is an $l=1$ orbital angular momentum state; its nontrivial radial factor does not change its angular eigenvalue. The calculation holds away from the origin and defines the same angular-operator identity in the square-integrable state space.

For static fields, the Maxwell equations give $\nabla \times B={\mu }_{0}J$ and $\nabla \cdot B=0$. Write $B=\nabla \times A$. A gauge transformation $A\mapsto A+\nabla \chi$ leaves $B$ unchanged; choose $\Delta \chi =-\nabla \cdot A$ to impose the Coulomb gauge. ThenFor a localized conserved current with the potential chosen to vanish at infinity, the supplied fundamental solution gives the magnetic vector potentialIts divergence vanishes by integration by parts and $\nabla \cdot J=0$. For a filamentary loop the current integral becomes $I{\oint }_{L}dr$, soFor $R=|x|$ much larger than the fixed loop size, expand $|x-r{|}^{-1}=R^{-1}+(x\cdot r)R^{-3}+O(R^{-3})$. The leading constant term integrates to zero because $\oint dr=0$. ThereforeThe vector triple-product identity gives $x\times (r\times dr)=r(x\cdot dr)-(x\cdot r)dr$. Integrating the derivative of $(x\cdot r)r$ around the closed loop shows $\oint r(x\cdot dr)=-\oint (x\cdot r)dr$. ConsequentlyEquivalently, $A={\mu }_{0}m\times x/(4\pi R^3)+O(R^{-3})$, where $m=(I/2)\oint r\times dr$ is the magnetic dipole moment. The leading displayed potential is of order $R^{-2}$; the remainder is smaller by a factor $R^{-1}$.

An $n$-node Gaussian quadrature formula chooses nodes and weights for an integral with a positive weight so that every polynomial of degree at most $2n-1$ is integrated exactly. Its nodes are the zeros of the degree-$n$ orthogonal polynomial for that weight.

For the even weight $w(x)=1-x^2$, the monic orthogonal quadratic has the form $p_2(x)=x^2-d$. Orthogonality to $x$ follows from oddness; orthogonality to one givesThe nodes are $\pm 1/\sqrt{5}$. The equal weights sum to $4/3$, hence each is $2/3$. The two-node Gaussian quadrature with quadratic weight is thereforeTo verify cubic exactness, divide any cubic as $P=p_{2}Q+R$ with $\mathrm{deg}Q,\mathrm{deg}R\le 1$. Orthogonality removes $p_{2}Q$ from the integral and its nodal values vanish; linear interpolation integrates $R$ exactly. This also proves the Gaussian property rather than merely matching two moments.

Let $S_m$ be the simple symmetric random walk started at the origin. Odd-time returns are impossible by parity. A return in $2n$ steps requires equal numbers of positive and negative steps along each axis. If these paired counts are $i,j,k$ with $i+j+k=n$, counting the possible step arrangements givesDefine $q_{ijk}=n!/(3^{n}i!j!k!)$. These are multinomial distribution probabilities and sum to one. Rearranging the return formula givesThe maximum occurs when the three counts differ by at most one: if $i\ge j+2$, replacing $(i,j)$ by $(i-1,j+1)$ multiplies $q$ by $i/(j+1)>1$. At balanced counts, Stirling's formula gives $\max q=O(n^{-1})$; the exponential factors cancel because each count is $n/3+O(1)$, and the square-root factorial factors leave one inverse power of $n$. The same formula gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim (\pi n{)}^{-1/2}$. HenceThis is the multinomial collision proof of three-dimensional walk transience. The sum is the expected total number of visits to the origin. If the probability of returning after departure were one, the Strong Markov property at successive returns would make every successive return finite almost surely, giving infinitely many visits and contradicting that finite expectation. Thus the origin is a transient state. Translation invariance gives the same conclusion at every lattice point: the three-dimensional walk is transient.

Apply the Gram-Schmidt process explicitly. Start with $f_1=e_1/\Vert e_1\Vert$, and for $i\ge 2$ setAssume inductively that the previous $f_j$ form an orthonormal basis for the span of $e_1,\dots ,e_{i-1}$. Inner products give $⟨v_i,f_j⟩=0$ for every $j<i$. Moreover $v_i\ne 0$: otherwise $e_i$ would lie in the span of its predecessors, contradicting that the $e_j$ form a basis. Thus $f_i$ is well defined and has unit norm. The definition expresses $f_i$ in the span of $e_1,\dots ,e_i$, and also expresses $e_i$ in the span of $f_1,\dots ,f_i$. The two spans therefore agree at each stage. Induction proves the required orthonormal, span-preserving basis.

The symmetric coefficient matrix of the quadratic form isThus $\boxed{q_a\text{ is nondegenerate if and only if }a\ne 1.}$ To obtain its signature, put $u=(y+z)/2$ and $v=(y-z)/2$. Completing the square yieldsThe transformation from $(x,y,z)$ to $(u+x,v,x)$ is invertible. The inertia of a bilinear form is consequentlyIf signature means the signed difference $n_{+}-n_{-}$, its value is $+1$ for $a>1$ and $-1$ for $a<1$. At $a=1$ the diagonal form has one zero direction, confirming degeneracy.

Let $\pi$ be a prime element of the Gaussian integers. Its norm $\pi \overline{\pi }$ is a positive integer greater than one, so $\pi$ divides some ordinary positive prime number in its integer prime factorization. It cannot divide two distinct such primes $p,q$: an integer Bézout identity $up+vq=1$ would imply $\pi \mid 1$, contrary to its being a nonunit. This proves existence and uniqueness of the rational prime lying below $\pi$.

Up to associates, the complete list of Gaussian prime elements is: $1+i$; the positive rational primes $p\equiv 3(\mathrm{mod}4)$; and, for every rational prime $p\equiv 1(\mathrm{mod}4)$, the two nonassociate elements $a+bi$ and $a-bi$, where $a,b>0$ and $p=a^2+b^2$. Representatives related by multiplication by one of $\pm 1,\pm i$ are associates. The last pair accounts for both factors of a split prime.

Every residue class modulo $p$ has a unique representative $a+bi$ with $0\le a,b<p$, since $p\mid (a-a^{\prime })+(b-b^{\prime })i$ means both coordinate differences are divisible by $p$. Hence $\boxed{|R/pR|=p^2}$ andFor $p=2$, the class of $1+i$ is nonzero but its square is $2i=0$; thus the quotient ring is not a field. For $p\equiv 3(\mathrm{mod}4)$, a root $r^2=-1$ would imply $r^{p-1}=(-1{)}^{(p-1)/2}=-1$, contradicting Fermat's little theorem. The quadratic $X^2+1$ is therefore irreducible and its quotient is a field with $p^2$ elements.

For $p\equiv 1(\mathrm{mod}4)$, choose a generator $g$ of the cyclic group ${\mathbb{F}}_p^{\times }$. Then $r=g^{(p-1)/4}$ satisfies $r^2=-1$. The two nonzero classes $i-r$ and $i+r$ have product zero in $R/pR$. ThusFor $p\equiv 1(\mathrm{mod}4)$ it is instead isomorphic to ${\mathbb{F}}_p\times {\mathbb{F}}_p$, by the distinct linear factors and the Chinese remainder theorem.

At $P=\sigma (u,v)$, regularity makes ${\sigma }_u,{\sigma }_v$ linearly independent. The tangent space is $T_{P}U=\mathrm{span}\{{\sigma }_u,{\sigma }_v\}$, and an oriented unit normal vector isReversing its sign chooses the opposite orientation. The first fundamental form is the squared length of a tangent displacement:where $E={\sigma }_u\cdot {\sigma }_u$, $F={\sigma }_u\cdot {\sigma }_v$, $G={\sigma }_v\cdot {\sigma }_v$. Its determinant $EG-F^2=|{\sigma }_u\times {\sigma }_v{|}^2$ is positive.

Use $N_{\mathrm{II}}={\sigma }_{vv}\cdot \mathbf{n}$ to distinguish the scalar coefficient of the second fundamental form from the normal vector. Since $\mathbf{n}\cdot \mathbf{n}=1$, its derivatives are perpendicular to it and hence tangent. Write ${\mathbf{n}}_u=a{\sigma }_u+b{\sigma }_v$ and ${\mathbf{n}}_v=c{\sigma }_u+d{\sigma }_v$. Differentiating $\mathbf{n}\cdot {\sigma }_u=\mathbf{n}\cdot {\sigma }_v=0$ givesSubstitution of the tangent expansions provesThe shape operator is $-d\mathbf{n}$; its matrix in the parametrized tangent basis is the negative transpose of the coefficient matrix shown. Its determinant, the product of the principal curvatures, is thereforethe Gaussian curvature. Transposition and the two minus signs do not change this determinant.

The fundamental theorem of calculus shows that the differential initial-value problem is equivalent toIn one direction integrate $x^{\prime }=F(t,x(t))$ from zero. In the other, continuity of the integrand makes the right side continuously differentiable with precisely that derivative and initial value.

Consider the closed subset $\mathcal{X}=\{x\in C([-b,b]):\Vert x-x_0{\Vert }_{\infty }\le r\}$ of the complete space with the uniform norm. It is complete: a uniformly Cauchy sequence has a uniform continuous limit, and the closed range constraint passes to the limit. DefineThe bound $|F|\le C$ gives $\Vert Tx-x_0{\Vert }_{\infty }\le bC<r$, so $T$ maps $\mathcal{X}$ into itself. The uniform Lipschitz condition givesHere negative $t$ changes only the orientation of the integral; its length is still at most $b$.

Construct $x^{(0)}(t)=x_0$ and $x^{(m+1)}=T{x}^{(m)}$. Induction gives $\Vert x^{(m+1)}-x^{(m)}{\Vert }_{\infty }\le q^m\Vert x^{(1)}-x^{(0)}{\Vert }_{\infty }$. Summing this geometric bound proves uniform Cauchy convergence to some $x\in \mathcal{X}$. The same contraction estimate shows $T{x}^{(m)}\to Tx$, so $x=Tx$. The integral equation now makes $x$ a $C^1$ solution. If another solution $y$ existed, it would also be fixed, and $\Vert x-y{\Vert }_{\infty }\le q\Vert x-y{\Vert }_{\infty }$ would force equality. This proves existence and uniqueness on the entire stated interval, implementing the contraction mapping theorem rather than just invoking it.

If $C=0$, the vector field is zero and the solution is constant. If $K=0$, the vector field is independent of the space variable and the integral solves the problem directly. The divisions by zero in the printed interval bound are naturally interpreted as imposing no restriction from the corresponding term.

For nonempty $F$, the triangle inequality gives $d(x,z)\le d(x,y)+d(y,z)$ for every $z\in F$. Taking the infimum yields $d(x,F)\le d(x,y)+d(y,F)$. Interchanging $x,y$ provesThus the distance to a set is a Lipschitz function, hence continuous. If $x\notin F$ and $F$ is closed, its open complement contains a ball $B(x,ϵ)$; every point of $F$ is at least $ϵ$ away. Hence $\boxed{d(x,F)>0}$ for $x\notin F$.

For two nonempty disjoint closed sets, let $h(x)=d(x,F_1)-d(x,F_2)$. It is continuous, negative on $F_1$ and positive on $F_2$. The disjoint open sets $U_1=\{h<0\}$ and $U_2=\{h>0\}$ contain them. Empty sets cause no difficulty, using an empty neighborhood for the empty set. This proves every metric space is a normal topological space.

In any normal space, a closed $F$ contained in an open $U$ can be separated from $X\setminus U$: choose disjoint open $W\supset F$ and $V\supset X\setminus U$. Because $X\setminus V$ is closed and contains $W$, $\overline{W}\subset X\setminus V\subset U$. First choose disjoint open $U_1,U_2$ around $F_1,F_2$, then apply this argument separately to each pair $F_i\subset U_i$. It gives ${\overline{W}}_i\subset U_i$, and therefore $\boxed{{\overline{W}}_1\cap {\overline{W}}_2=\varnothing }$. This is closed-neighborhood shrinking in a normal space.

For the two specified hyperbola branches, explicit choices areThey are open and contain the branches with products $-1$ and $1$. Their closures lie respectively in $\{x\le 0,xy\le -1/2\}$ and $\{x\ge 0,xy\ge 1/2\}$. Any common point would have $x=0$, making both product inequalities impossible. Thus their closures are disjoint even though the original branches can approach one another arbitrarily closely at large height.

The function is odd, so its Fourier series has no constant or cosine terms. If $\lambda =m\in \mathbb{Z}$, it is already a Fourier basis function: for $m>0$ the only sine coefficient is $b_m=1$; for $m<0$ it is $b_{|m|}=-1$; for $m=0$ the function is zero.

For real noninteger $\lambda$, the product-to-sum identity givesSince both numerator sines equal $(-1{)}^n\sin (\pi \lambda )$, the Fourier series of a noninteger-frequency sine isIt equals the function for $-\pi <x<\pi$. At either periodic endpoint it equals the average of the two one-sided limits, namely zero; the noninteger-frequency function has a jump in its periodic extension there. Equality also holds in the square-integral sense on the full interval.

For $\lambda =1/2$, $b_n=8(-1{)}^{n+1}n/[\pi (4n^2-1)]$ and ${\int }_{-\pi }^{\pi }{\sin }^2(x/2)dx=\pi$. The Parseval identity for a real sine series is $(1/\pi ){\int }_{-\pi }^{\pi }{f}^2={\sum }_{n\ge 1}{b}_n^2$. Hence

An isotropic tensor has unchanged components under every proper orthogonal change of basis: $T_{i_1\dots i_n}=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\dots j_n}$ for every $R\in SO(3)$. Orthogonality gives $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$, so the Kronecker delta is isotropic. The determinant identity gives $R_{ia}{R}_{jb}{R}_{kc}{ϵ}_{abc}=(\det R){ϵ}_{ijk}={ϵ}_{ijk}$, so the Levi-Civita symbol is isotropic too. If reflections are included, this last object is a pseudotensor: it changes sign. The question's isotropy convention therefore refers to proper rotations.

For the sphere moment $T_{i_1\dots i_n}={\int }_{S^2}{\hat{x}}_{i_1}\cdots {\hat{x}}_{i_n}dS$, change variables to $R\hat{x}$. Rotations preserve the sphere and its area measure; inserting the component transformation proves that $T$ is an isotropic tensor. For the second moment, half-turns about coordinate axes kill the off-diagonal components, while rotations interchanging axes make the diagonal entries equal. Thus it is a multiple $a{\delta }_{ij}$. Contracting the indices gives $3a={\int }_{S^2}|\hat{x}{|}^{2}dS=4\pi$. Odd moments vanish by $\hat{x}\mapsto -\hat{x}$, including the third moment. The supplied general fourth-rank form and symmetry under every permutation of the four indices make its three coefficients equal. A double contraction then gives $4\pi =b(9+3+3)$. ThusTo explain the weighted volume integrals, put $v=(1,i,0)$; its bilinear square is $v\cdot v=1+i^2=0$, without complex conjugation. Separating radial and angular factors givesFor $n=2$ the angular factor is $(4\pi /3)v\cdot v=0$. For $n=3$ it is zero by oddness. For $n=4$ it is $(4\pi /15)3(v\cdot v{)}^2=0$. Therefore all three requested integrals vanish, assuming the radial integrals exist. More generally the null-vector cancellation of radial moments follows from rotation about the third axis, which multiplies the integral by $e^{in\varphi }$ without otherwise changing it.

Assume homogeneous space and time, isotropic space, inertial frames related linearly, the relativity principle and a common isotropic speed $c$ for light. Take coincident origins at zero and parallel axes, with $S^{\prime }$ moving at $v$ along $x$. The primed origin requires $x^{\prime }=A(x-vt)$. Write $t^{\prime }=Bx+Dt$; mapping each light ray $x=\pm ct$ into $x^{\prime }=\pm c{t}^{\prime }$ gives $D=A$ and $B=-Av/c^2$. Rotational symmetry about the boost axis gives a common transverse scale $B(v)$; reversing the boost direction by spatial isotropy gives $B(v)=B(-v)$. Reciprocity then gives $B(v{)}^2=1$, and continuity and the common orientation select $B(v)=1$. Thus transverse directions are unchanged. Reciprocity requires the inverse to have the same scale $A$ with $v$ replaced by $-v$. Composing gives $A^2(1-v^2/c^2)=1$; continuity from $A=1$ at $v=0$ selects the positive root. Hence the Lorentz transformation isThe printed wavelength formula uses the angle of the line of sight from observer towards star, opposite to photon propagation. Choose the transverse axis so the ray lies in the $xy$ plane. With that directed-angle convention the photon four-momentum isThe inverse coordinate boost takes a rest-frame four-vector to $S$, giving $p^0=\gamma (p^{\prime 0}+\beta p^{\prime 1})$ and $p^1=\gamma (p^{\prime 1}+\beta p^{\prime 0})$, where $\beta =v/c$. Thus $\nu =\gamma {\nu }^{\prime }(1-\beta \cos {\theta }^{\prime })$, and the aberration of light relation is $\cos \theta =(\cos {\theta }^{\prime }-\beta )/(1-\beta \cos {\theta }^{\prime })$. More directly, the boost from $S$ to $S^{\prime }$ gives $p^{\prime 0}=\gamma (p^0-\beta p^1)$, hence ${\nu }^{\prime }=\gamma \nu (1+\beta \cos \theta )$. Since wavelength is $c/\nu$, the relativistic Doppler effect givesIf the angle were instead measured along the actual photon momentum, both cosine signs would reverse and the same physical relation would have a minus sign. Specifying this orientation avoids confusing the two versions.

For $\cos \theta =1$, $\lambda /{\lambda }^{\prime }=\sqrt{(1+\beta )/(1-\beta )}$: a receding star is redshifted. For $\cos \theta =-1$, the ratio is $\sqrt{(1-\beta )/(1+\beta )}$: an approaching star is blueshifted. For $\cos \theta =0$, $\boxed{\lambda /{\lambda }^{\prime }=\gamma }$: the transverse redshift is the time-dilation effect. This transverse condition refers to the observer's angle, not ${\theta }^{\prime }=\pi /2$ in the source frame.

Use the standard ideal-fluid assumptions underlying the stated relation: both fluids are incompressible, inviscid and unbounded away from the interface, with no surface tension. An irrotational flow has $u_j=\nabla {\varphi }_j$, and incompressibility gives $\Delta {\varphi }_j=0$ in each fluid. The unsteady Bernoulli equation iswhere the arbitrary time functions can be absorbed into the potentials. On $z=\zeta (x,t)$ the material-interface and normal-stress conditions areDisturbance velocities decay as $z\to \pm \infty$. The equilibrium pressures are $p_{\ast }-{\rho }_{j}gz$.