Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3

Take the integral domain $R=\mathbb{Z}[\sqrt{-5}]$, embedded in $\mathbb{C}$. It exhibits failure of unique factorization in the ring of integers generated by square root of minus five:Here is why these are genuinely distinct factorizations into irreducible elements. The multiplicative algebraic norm $N(a+b\sqrt{-5})=a^2+5b^2$ is a nonnegative integer and equals one only for $\pm 1$, which are therefore the only units. Neither two nor three is a value of $N$: if $b\ne 0$ the value is at least five, and if $b=0$ it is a square. A factorization of $2$ into nonunits would require norms two and two; one of $3$ would require norms three and three; one of $1\pm \sqrt{-5}$ would require norms two and three. All are impossible. Thus all four factors are irreducible. The factors of norms four and nine cannot be associate elements of those of norm six, and $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are not related by a sign. Therefore $R$ is an integral domain which is not a unique factorization domain.

Use the standard convention that an integral domain has $1\ne 0$. Since $(n)=(-n)$, the sign of a nonzero $n$ makes no difference. For $|n|=p$ prime, every nonzero residue has an inverse by Bézout's identity, so the quotient ring is a field and hence an integral domain. If $|n|=ab$ is composite with $1<a,b<|n|$, the nonzero classes of $a$ and $b$ multiply to zero, giving zero divisors. For $n=\pm 1$, the quotient is the zero ring, excluded by the convention above. Finally, $n=0$ gives $\mathbb{Z}/(0)\cong \mathbb{Z}$, an integral domain. ThusIf a convention restricts this notation to positive moduli, the answer is simply the positive primes.

Parametrize the surface by $r(x,y)=(x,y,xy)$. Its tangent vectors are $r_x=(1,0,y)$ and $r_y=(0,1,x)$. Choose the upward unit normalThe first fundamental form coefficients areSince $r_{xx}=r_{yy}=0$ and $r_{xy}=(0,0,1)$, the second fundamental form coefficients are $e=g=0$ and $f=(1+x^2+y^2{)}^{-1/2}$. The determinant of the shape operator gives the Gaussian curvature:Changing the normal's sign changes both second-form factors and leaves this determinant unchanged. Although the question calls the surface a hyperboloid, $z=xy$ is a hyperbolic paraboloid; the calculation uses the stated surface equation.

The function is Fréchet differentiable at $(a,b)$ when there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatIf it exists, $L(h,k)=f_x(a,b)h+f_y(a,b)k$, as is seen by taking increments along each coordinate axis.

To prove differentiability from continuous partial derivatives, take a small rectangle in the given neighbourhood and split the increment along its sides. The one-variable mean value theorem givesfor some $0<\theta ,\eta <1$ when the corresponding increment is nonzero. If an increment is zero, its term is simply omitted. The required one-variable continuity follows from existence of the corresponding partial derivative on the segment. Subtracting $L(h,k)$ gives a remainder whose absolute value, by continuity of both partial derivatives at $(a,b)$, is at mostfor sufficiently small increments. Since $ϵ$ is arbitrary, the normalized remainder tends to zero. Thus the function is differentiable at $(a,b)$, with the displayed derivative.

This does not define a topology on $\mathbb{Z}$. Every singleton $\{2k\}$ is declared open, but their union over $k\in \mathbb{Z}$ is $2\mathbb{Z}$, an infinite proper subset. It is therefore not declared open, violating closure under arbitrary unions. This is the example that finite subsets do not define a topology on an infinite set. The empty set, whole space and finite-intersection requirements do hold, but they do not compensate for the failed union axiom. This is different from the cofinite topology, which uses finite complements rather than finite sets.

If an open set contains an even integer, invariance under every even translation makes it contain all even integers. Similarly, containing an odd integer makes it contain all odd integers. Conversely each parity class is invariant under those translations. Thus the proposed open sets are exactlyThey form a partition topology. Both the empty and whole sets are included, and any union or finite intersection is again a union of parity classes. Hence this is a topological space. It is not the discrete topology: no singleton is open.

Write $z=x+iy$ and $w=u+iv$. Multiplying the numerator and denominator of the Möbius transformation by $1+\bar{z}$ givesPull back the given upper-half-plane solution to obtain the harmonic measure of a semicircle in a disk:The principal real arctangent is appropriate because $v>0$. By conformal invariance of harmonicity, composing a harmonic function with this analytic conformal map is harmonic. More explicitly, ${\Delta }_z(f\circ w)=|w^{\prime }(z){|}^2({\Delta }_{w}f)\circ w=0$. At a boundary point on the open upper semicircle the ratio tends to $+\infty$, yielding one; on the lower semicircle it tends to $-\infty$, yielding zero.

At $z=\pm 1$ the two boundary prescriptions jump, so no continuous boundary value can be imposed there and the interior limit depends on the approach. The bounded harmonic solution is interpreted with the specified values away from those endpoints. Boundedness also supplies the usual uniqueness for this Dirichlet problem; without a boundedness or comparable regularity restriction, singular harmonic functions at the omitted endpoints would allow extra solutions.

Choose the coordinate vectors $b=e_i$ and $c=e_j$ in the assumed bilinear identity. It gives $T_{ij}=0$ for each pair of indices, hence the tensor is zero.

Let $\Omega$ be the enclosed region. Apply the divergence theorem to the vector field $(b\cdot x)c$. Since $b,c$ are constant,ThereforeWriting the left side as $b_i{c}_j{\int }_S{x}_i{n}_{j}dS$ and using the initial tensor argument gives the boundary integral of position and normal componentsContract with the Kronecker delta to obtain ${\int }_{S}x\cdot ndS=3V$. Contract with the Levi-Civita symbol to obtainThus the requested integrals are $3V$ and the zero vector, respectively. These applications require the usual sufficiently regular closed boundary and outward-normal orientation of the divergence theorem.

In the position representation, $\hat{p}=-i\hslash \nabla$. The orbital angular momentum operator and its squared magnitude areFor example, with $D=x\cdot \nabla$, the latter is $-{\hslash }^2[r^2\Delta -D^2-D]$. Operator products here act on smooth wavefunctions in a common invariant domain.

To calculate the angular momentum commutation relations, use $[x_a,p_b]=i\hslash {\delta }_{ab}$ and the vanishing position-position and momentum-momentum commutators. Expanding the products givesMultiplication by ${ϵ}_{iab}{ϵ}_{jcd}$ then givesFinally, the product commutator rule yieldsThe final cancellation pairs the antisymmetric Levi-Civita symbol with an expression symmetric in $j,k$. Thus

Under the null hypothesis the three preference probabilities for girls are each $1/3$. Conditional on their total of sixty, the counts have a multinomial distribution, giving expected counts $(20,20,20)$. The Pearson chi-squared goodness-of-fit test statistic isNo probabilities have been fitted, so the asymptotic null distribution is chi-squared distribution with $3-1=2$ degrees of freedom. All expected counts are comfortably large for this approximation. For two degrees of freedom its upper-tail probability is $e^{-X^2/2}$, soAt the conventional five-percent or one-percent significance level, reject equal attractiveness: there is substantial evidence against the null distribution. No level was specified, so reporting the p-value makes the decision rule explicit. The test assumes independent responses and a sampling interpretation appropriate to the children surveyed.

Use the Pearson chi-squared test of homogeneity for two multinomial samples. Under equal preference distributions, estimate the common probabilities from pooled counts, giving $(1/4,1/4,1/2)$. With row totals forty and sixty, the expected table isThe Pearson chi-squared statistic for contingency tables isThere are $(2-1)(3-1)=2$ degrees of freedom, after fitting the common category probabilities. All expected counts exceed five. ThusAt five percent, do not reject the common preference distribution. These data are consistent with that hypothesis; failure to reject does not prove equality. This conclusion differs from part a because a common nonuniform distribution, especially an overall preference for chocolate, is allowed here.

Order the states as sandwich $N$, Hall $H$, and own cooking $C$. The transition matrix, with current state indexing the row, isThe two probabilities $1/4$ in the first row are unconditional alternatives to the probability $1/2$ of returning. Let $a_n=(P^n{)}_{HH}$, so $n$ counts days after the initial Hall lunch. Expanding the determinant gives the characteristic polynomialThe three distinct eigenvalues imply that the sequence has the formusing the specified $0^0=1$. Directly, $a_0=1$, $a_1=1/3$ and $a_2=7/36$. The last two equations give $A=3/14$ and $B=-5/7$; then $C=3/2$. ConsequentlyThe limiting value is also the Hall entry of the stationary distribution $(4/7,3/14,3/14)$. The tiny finite-time correction is retained in the boxed exact probability. Sixty days later means sixty transitions, rather than the sixtieth day including the initial day.

The matrix trace is the sum of diagonal entries, $\mathrm{tr}A={\sum }_{i=1}^n{A}_{ii}$. For any two square matrices,Apply this cyclic property of the trace with $B=T$ and $C=A{T}^{-1}$:Thus matrix similarity preserves the matrix trace; no assumption that either matrix is diagonalizable is needed.

Suppose $v=(v_1,\dots ,v_r{)}^t$ belongs to the kernel of the displayed matrix $\Lambda$. Define the polynomial $q(z)={\sum }_{k=1}^r{v}_k{z}^k$. Then $q({\lambda }_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a polynomial of degree at most $r$. Repeated application of the factor theorem therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional rank-nullity theorem, invertible. Its transpose is invertible with inverse $({\Lambda }^{-1}{)}^t$, proving both kernels are trivial.

For the second matrix, membership of the all-ones vector in the kernel meansGroup the distinct nonzero values as ${\mu }_1,\dots ,{\mu }_r$, with multiplicities $m_1,\dots ,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations areor $\Lambda (\mu {)}^{t}m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ andConversely, all zero values plainly give a zero matrix. This proves the zero power sums force a finite complex multiset to vanish criterion, including repetitions and zero values rather than assuming all eigenvalues distinct.

Use the permitted triangularization to write $A=TU{T}^{-1}$ with $U$ upper triangular and diagonal ${\lambda }_1,\dots ,{\lambda }_n$. Products of upper-triangular matrices are upper triangular, and their diagonal entries multiply, soThe result of part b makes every ${\lambda }_j$ zero. Thus $U$ is strictly upper triangular. To see directly that $U^n=0$, a potentially nonzero entry in a product of $n$ factors requires a chainof $n+1$ indices in $\{1,\dots ,n\}$, which is impossible. HenceThis is nilpotence from vanishing traces of powers. It proves nilpotence without substituting an unproved diagonalization for the allowed triangularization.

Use the usual commutative unital-ring convention for Noetherian rings. A ring is Noetherian exactly when every ideal is finitely generated, equivalently when ascending chains of ideals stabilize. Indeed, the union of an ascending chain is an ideal; finitely many generators all lie in a single chain member. Conversely, if an ideal cannot be finitely generated, adjoining a new element at each stage constructs a strictly ascending chain.

First assume $R[X]$ is Noetherian. Evaluation at zero gives a surjective ring homomorphism $R[X]\to R$ with kernel $(X)$. A quotient of a Noetherian ring is Noetherian: lift an ideal to the source, take finitely many generators there, and map those generators down. Thus $R\cong R[X]/(X)$ is Noetherian.

For the converse, assume $R$ is Noetherian and take any ideal $I\subseteq R[X]$. For $d\ge 0$, let $J_d$ be the set of coefficients of $X^d$ in elements of $I$ of degree at most $d$. It is an ideal of $R$, since sums and scalar multiples of such polynomials stay in $I$. Multiplication by $X$ shows $J_d\subseteq J_{d+1}$. The ascending chain stabilizes, say at $N$.

For each $0\le d\le N$, choose finitely many generators $a_{d,j}$ of $J_d$ and corresponding $f_{d,j}\in I$ of degree at most $d$ having coefficient $a_{d,j}$ at $X^d$. Discard zero generators. We claim these finitely many polynomials generate $I$. If a nonzero $f\in I$ has degree $m\le N$, express its leading coefficient as ${\sum }_j{r}_j{a}_{m,j}$ and subtract ${\sum }_j{r}_j{f}_{m,j}$. The degree decreases. If $m>N$, use $J_m=J_N$, express the leading coefficient using the $a_{N,j}$, and subtractAgain the leading term cancels and the degree decreases. Induction on degree, including degree zero, expresses every $f$ in the claimed ideal. Therefore every ideal of $R[X]$ is finitely generated, proving the Hilbert basis theorem andThe argument is the bounded-degree coefficient proof of Hilbert basis theorem; it never divides by a leading coefficient and therefore does not require $R$ to be an integral domain. With the analogous left-ideal convention the same proof applies to left-Noetherian rings with a central indeterminate.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.

The inverse function theorem says: if $U\subseteq {\mathbb{R}}^2$ is open, $F:U\to {\mathbb{R}}^2$ is smooth, $a\in U$, and $DF(a)$ is invertible, then there are open neighbourhoods $V\subseteq U$ of $a$ and $W$ of $F(a)$ for which $F:V\to W$ is a bijection with a smooth inverse. Its derivative is $D(F^{-1})(F(x))=[DF(x){]}^{-1}$ after shrinking $V$ if necessary. A smooth function here has derivatives of every order.

For the given map,Thus its local smooth-invertibility set isA differentiable local inverse is impossible at either excluded line by the chain rule. Even a merely continuous local inverse is impossible there: at fixed $y$, the cubic first component has a strict local extremum at these $x$ values, giving equal values at two nearby $x$ coordinates and preventing local injectivity.

On the curve, $y^2=x^3-x=x(x-1)(x+1)\ge 0$. Its sign intervals give exactly $x\in [-1,0]\cup [1,\infty )$, so the two prescribed subsets cover the curve and are disjoint. For fixed $y$, the function $x^3-x$ is strictly increasing on $[1,\infty )$, starts at zero and tends to infinity. The intermediate value theorem therefore gives a unique $p(y)\ge 1$ with $p(y{)}^3-p(y)=y^2$.

At every point of this second component, $3x^2-1\ge 2$, so the inverse function theorem applies. Locally its smooth inverse evaluated at $(0,y)$ has the form $(p(y),y)$, by the proved uniqueness. These local descriptions agree on overlaps, proving $p$ is smooth on all of $\mathbb{R}$. For example,whose nonzero denominator also makes the smoothness mechanism transparent.

Let $p$ first be an arbitrary degree-$n$ polynomial with leading coefficient $a_n\ne 0$. Choose $R$ so large that on $|z|=R$,The homotopy $a_n{z}^n+s[p(z)-a_n{z}^n]$, $0\le s\le 1$, never meets zero on that circle. Hence the image curve $p(R{e}^{i\theta })$ has winding number $n$, the same as $a_n{R}^n{e}^{in\theta }$. The argument principle counts precisely $n$ zeros inside, with multiplicity, since there are no poles. Choose $R$ large enough that the same leading-term domination holds for every $|z|\ge R$; the lower-term ratios decrease with radius, so this also excludes all exterior zeros. This proves the fundamental theorem of algebra with its multiplicity count. A degree-zero nonzero polynomial has no zeros.

Now use the specified bound for the monic quartic. Compare $\Gamma$ with the closed curveIts first piece travels once counterclockwise around a circle, and its second stays at the endpoint, so its winding number is one. On the first piece, $|\Gamma -{\gamma }_0|<R^4$. On the second, both $p(R)$ and $p(iR)$ differ from $R^4$ by less than $R^4$; every convex combination of those errors also has modulus less than $R^4$. Thus the quoted winding number stability result gives

For real coefficients define $q(y)=y^4-b{y}^2+d$. It has no real zeros and positive leading coefficient, so $q(y)>0$ for every real $y$, including $d=q(0)>0$. Thereforelies strictly in the right half-plane for every real $y$. Also $p(x)>0$ for $x\ge 0$: there are no zeros there, and the polynomial is positive for sufficiently large positive $x$. These facts exclude roots on the positive real and imaginary axes.

Consider the positively oriented quarter-disk boundary: first the arc from $R$ to $iR$, then the imaginary axis to zero, then the real axis back to $R$. Under $p$, its last two pieces form a path from $p(iR)$ to $p(R)$ entirely within the right half-plane. The chord closing $\Gamma$ also lies there. Since this half-plane is convex and excludes zero, deforming one closing path into the other changes no winding number. Thus the full image of the quarter-disk boundary has winding number one. By the argument principle there is exactly one root inside that quarter-disk, counted with multiplicity.

Finally, all polynomial roots lie in $|z|<R$ under the given strict coefficient bound: for $|z|\ge R$, the sum of lower terms divided by $|z{|}^4$ is smaller than one. HenceIts multiplicity is one. This is the quarter-sector winding test for a real quartic; it accounts for the axis exclusions and does not replace the sector boundary by a chord without justifying the deformation.

For the Fourier sine series on $(0,1)$, orthogonality givesThereforeThe series is that of the odd, two-periodic extension. At $x=0$ its value is zero, agreeing with $f(0)$; at $x=1$ every summand is zero and the series converges to zero, the average of the two one-sided extension limits. Thus it represents the function in the interior and in the square-integrable sense, not the prescribed value one at the endpoint $x=1$. This endpoint issue matters when using sine expansions with homogeneous boundary conditions.

Set $y=e^{-x/2}v$, the Dirichlet gauge transform for constant drift. Direct differentiation givesThe homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v^{\prime \prime }=(\lambda +1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar{v}$ and integrate by parts:for a nonzero eigenfunction. Therefore $\lambda +1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin (kx)+B\cos (kx)$ give $B=0$ and $k=n\pi$, $n\ge 1$. HenceEquivalently, $L[y]=\lambda y$ is the Sturm-Liouville form $(e^x{y}^{\prime }{)}^{\prime }=\lambda e^{x}y$, with positive weight $e^x$. Its orthogonality relation isIt is the weighted inner product, not the unweighted one, that makes these eigenfunctions orthogonal.

Expand $x{e}^{-x/2}={\sum }_n{b}_n{y}_n$ with the coefficients from part a. Completeness and the nonzero eigenvalues give the Sturm-Liouville eigenfunction expansionThe coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v^{\prime \prime }-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.

For a normalized state and an observable $A$, define $⟨A⟩=⟨\psi ,A\psi ⟩$ and ${\Delta }_{\psi }A=\Vert (A-⟨A⟩)\psi \Vert$. Suppose first that the position and momentum variances are finite and the state has the regularity needed for the usual position-momentum quadratic forms. Put $a=⟨x⟩$, $b=⟨p⟩$, $u=(x-a)\psi$ and $v=(-i\hslash {\partial }_x-b)\psi$. The Cauchy-Schwarz inequality givesThe term involving $b$ has zero imaginary part. Integration by parts givesConsequently $\Im ⟨u,v⟩=\hslash /2$, andThis proves the Heisenberg uncertainty relation. For smooth rapidly decaying states the boundary terms vanish directly. The same proof extends to $x\psi \in L^2$ and weak derivative ${\psi }^{\prime }\in L^2$ using smooth cutoffs and taking their radius to infinity; the necessary mixed products are integrable by Cauchy-Schwarz. Normalization alone need not give finite variances or an operator-domain momentum expectation. A state with infinite spread is interpreted through the corresponding extended bound, rather than assigning a finite uncertainty to an undefined expression.

For the specified Gaussian wave packet, write $s=\hslash t$ and $a_t=1+is$. Its probability density isIt is a normalized centered normal distribution with variance $1+s^2$. Symmetry gives $⟨x⟩=0$, and henceAlso ${\partial }_x\psi =-x\psi /(2a_t)$, so $p\psi =i\hslash x\psi /(2a_t)$ and $⟨p⟩=0$ by oddness. Integration by parts, or directly taking the squared norm, givesThusThe packet saturates the bound at $t=0$ and has a larger position-momentum uncertainty product at every nonzero time. Its spatial spreading does not change the momentum distribution.

In SI units, the source-free Maxwell equations in vacuum areUse complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, soThe magnetic divergence equation follows because $k\cdot (k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-{\mu }_0{ϵ}_0\omega E_0$. Since $k\times (k\times E_0)=-|k{|}^2{E}_0$, a nonzero wave satisfiesThus both fields are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.

For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_{+}-E_{-}$, similarly for $B$. Applying the integral Maxwell equations to a thin pillbox and a thin loop gives the electromagnetic boundary conditionsHere $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.

For the perfect conductor, take the time-varying interior fields to vanish. Choose positive $k$ with $\omega =ck$. The incident electric field has complex form $E_0{e}_x{e}^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected field $-E_0{e}_x{e}^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum fields are thereforeThe sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the electric field is zero and the magnetic field is tangential. The jump conditions yieldFor a real incident amplitude this is $j=(2E_0/{\mu }_{0}c)\cos (\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior field is separate from this oscillatory response.

For an inviscid flow of constant density $\rho$, driven by a conservative body force $-\nabla \Phi$, the Euler equation isThe vector identity $(u\cdot \nabla )u=\nabla (|u{|}^2/2)-u\times (\nabla \times u)$ and irrotationality give the Unsteady Bernoulli equation. Specifically, wherever a single-valued potential $u=\nabla \varphi$ is available,On a connected potential-flow region this meansThe integration function is spatially constant but need not be constant in time; it can be removed by changing the potential's time-dependent gauge. A simply connected irrotational region provides the requisite global potential. For a barotropic fluid the pressure term is replaced by $\int dp/\rho (p)$; here the liquid density is constant.

For the tube, let $U(t)$ be its approximately uniform axial velocity. The large reservoir's surface velocity is negligible and gravity is neglected as stipulated. Entrance-region inertia contributes an effective length of order $a$, negligible against $L$. The potential difference between the outlet and the reservoir is consequently $LU$ to leading order. Subtract Bernoulli at the reservoir surface, where pressure is $P+p_a$, from its value at the outlet, where pressure is $p_a$ and speed is $U$. This givesThe inertial term cannot be dropped during startup. With $U_{\infty }=\sqrt{2P/\rho }$, separate variables:Hence the inviscid startup in a pressure-driven tube givesInitially $U\sim Pt/(\rho L)$, the acceleration of the stationary liquid column. At long times $U\to \sqrt{2P/\rho }$, the inviscid pressure-to-kinetic-energy limit. The result applies while the tube remains full and reservoir dimensions are large enough that surface motion and the neglected entrance inertia remain small.