Put the broken dam at $x=0$ and let $c_0=\sqrt{gH/3}$. A dry bed is assumed downstream. The left-family rarefaction wave joins the undisturbed reservoir $h=H,u=0$ to the dry front. The other Riemann invariant is fixed at $u+6c=6c_0$. Within the fan its characteristic speed equals $\xi =x/t=u-c$, soThus the dry-bed dam break with cubic wetted area has the fieldsVelocity is not a meaningful fluid quantity in the dry region; its wet-side limiting value is $u_f=6c_0=2\sqrt{3gH}$. The upstream head travels at $-c_0$, while the front travels at the stated constant speed in this inviscid model.

Bottom drag slows the flow and changes the dry-tip structure, where the depth is small and bed stress per unit fluid mass becomes especially important. It can replace the inviscid zero-depth high-speed tail by a drag-controlled nose. An exact altered profile or front speed requires a specified drag law; the inviscid speed should not be extrapolated to a thin real irrigation-channel front without that qualification.

Use the explicit printed width law $b(z)=\alpha z^2$. Its wetted area and hydrostatic pressure integral areAssuming hydrostatic pressure, a horizontal prismatic bed and cross-sectionally uniform velocity, volume conservation and momentum conservation give the prismatic-channel shallow water equationsEquivalently,The coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/3\\ g& u\end{array}\right)$. Its eigenvalues and the corresponding gravity-wave speed areFor $h>0$ these eigenvalues are distinct and real, establishing strict hyperbolicity in the hyperbolic partial differential equation sense. Since $dc/dh=c/(2h)$, substitution in the two equations showsFor example $({\partial }_t+(u+c){\partial }_x)(u+6c)=0$. This provides both the slopes and the conserved characteristic quantities. At a dry bed $h=0$, strict hyperbolicity degenerates and the front is interpreted as the limit from positive depth.

The descriptive word “parabolic” alone would normally suggest $b\propto \sqrt{z}$ and different coefficients. The explicit $z^2$ width in the PDF determines the calculation here; it must not be silently replaced by that other geometry.

Bed and sidewall stress, turbulent mixing at the interface, wave radiation and hydraulic jumps convert or remove mechanical energy. In the specified local head-loss model, differentiate $B=u^2/2+g^{\prime }h$ while keeping $Q=bhu$ constant:At a regular hydraulic control the coefficient of $h^{\prime }$ vanishes. Therefore $B^{\prime }=-\alpha$ requiresThus the control moves to the widening downstream side, $x_c>0$. This is the exact local compatibility condition; it is the head-loss correction to hydraulic control with loss expressed as Bernoulli potential per unit length.

For a small loss and a control remaining near the throat, use the lossless depth $h_c=2H/3+O(\alpha )$ and $b^{\prime }/b=2x_c/L^2+O(x_c^3/L^4)$. The leading displacement isFor a finite loss, the local derivative alone does not specify the accumulated loss from the far-upstream reservoir, so it does not fix a unique revised discharge or numerical control depth. Given $Q$, the exact condition can instead be solved implicitly for $x_c$; if the inviscid discharge is imposed, it becomes $\alpha =(4g^{\prime }H/3L)(x_c/L)[1+(x_c/L{)}^2{]}^{-5/3}$. The small-displacement root is the one continuous from the throat. This distinction avoids treating the original upstream energy as unchanged after an unspecified finite upstream loss.

Assume the jump is stationary, short compared with the scale of width variation, hydrostatic on either side, and has negligible bed stress integrated over its short length. Treat its width $b_j=\sqrt{32/27}$ as constant. It must occur on the downstream supercritical branch, at $x_j=L\sqrt{b_j-1}>0$. The discharge from the control gives the local unit-width dischargeBefore the jump the energy is still $H$. Writing $y=h/H$, its equation is $y+1/(8y^2)=1$. Its positive roots are $1/2$ and $(1+\sqrt{5})/4$; the former is supercritical. ThusMass conservation gives $h_1{u}_1=h_2{u}_2=q_j$. The hydraulic jump momentum balance conserves $q_j^2/h+g^{\prime }{h}^2/2$. Factoring out the non-jump solution $h_2=h_1$ gives the conjugate-depth relationConsequently the conditions immediately after the jump areIts Froude number is below one. The specific-energy loss isSelecting the deeper root of the unchanged energy equation would be wrong: that would conserve the energy across a turbulent jump instead of the momentum.

Differentiate the specific energy at fixed discharge:In the lossless flat channel $E$ is constant. A smooth hydraulic control has $F=1$, so regularity requires $b^{\prime }=0$. For the prescribed width this occurs at the minimum width $x=0$. The required minimum specific energy is largest at the narrowest section; a controlled flow supplies exactly that head.

Far upstream $b\to \infty$, so the velocity of a finite-discharge flow with depth tending to $H$ tends to zero. Thus the Bernoulli equation gives $E=H$. At the control $u_c^2=g^{\prime }{h}_c$ and $E=3h_c/2$, henceThe printed width normalization has $b(0)=1$. Downstream on the accelerating controlled branch the flow becomes shallower and supercritical.

The shallow-water approximation requires the depth to be small compared with the horizontal scale of variation, small vertical accelerations and nearly hydrostatic pressure. Take cross-sectionally uniform lower-layer velocity and a sufficiently deep, effectively stationary upper layer. Under the Boussinesq approximation, $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0$ is the reduced gravity. Conservation of lower-layer volume gives constant discharge $Q=bhu$.

Dividing the Bernoulli equation by $g^{\prime }$ gives the shallow-water specific energyAt fixed $Q,b$, the energy curve has its minimum at the critical depth $h_c=(Q^2/(g^{\prime }{b}^2){)}^{1/3}$, where $E_c=3h_c/2$. Above this minimum there are two depth branches: deeper subcritical flow has $F<1$, and shallower supercritical flow has $F>1$.

The two long-wave speeds relative to the bed are $u\pm \sqrt{g^{\prime }h}$. In subcritical flow one family can propagate upstream and one downstream; in supercritical flow both travel downstream for $u>0$. At hydraulic control, $u-\sqrt{g^{\prime }h}=0$, and a smooth transcritical solution must also satisfy the geometric regularity condition. The flow may remain entirely on either branch, cross smoothly through the control from subcritical to supercritical, or take the reverse smooth branch under appropriate boundary conditions. A downstream hydraulic jump can return supercritical flow to the subcritical branch while losing energy.

The sketch distinguishes smooth branch changes at a critical section from a jump: a jump conserves mass and momentum, not the original Bernoulli head.

Cooling from above makes dense cold fluid overlie lighter warm fluid, so the upper boundary layer is unstable and can mix the interior. Assume high Rayleigh number, rapid interior mixing, constant properties, negligible wall heat capacity and a quasi-steady cooling flux obeying the four-thirds convective heat-transfer law. Write $\theta (t)=T_{\mathrm{bulk}}(t)-T_0$ and use a coefficient $C$ appropriate to this single cooling boundary. The volumetric heat capacity is $\rho c_p=k/\kappa$, so the area-wise heat balance isWith $K=C\kappa [g\beta /(\nu \kappa ){]}^{1/3}/h$, integration gives convective cooling under the four-thirds heat-transfer law:The power-law decay is a bulk approximation between the initial boundary transient and the later stage when the decreasing temperature contrast no longer maintains strong convection. It is not an exact spatially uniform solution satisfying the cold boundary itself.

Cooling from below is different: dense cold fluid now lies below warm fluid and stabilizes the layer. With no other imposed motion, cooling is by thermal conduction, and temperature becomes depth dependent. Let $z=0$ be the cooled bottom and $z=h$ the insulating top. Solve ${\theta }_t=\kappa {\theta }_{zz}$ with $\theta (0,t)=0$, ${\theta }_z(h,t)=0$ and $\theta (z,0)=\Delta T$. The mixed-boundary eigenfunctions have ${\lambda }_n=(n+1/2)\pi /h$, givingIts volume-averaged temperature isThe late conductive decay is exponential on the timescale $4h^2/({\pi }^2\kappa )$, rather than the convective $t^{-3}$ bulk law. This assumes the usual positive thermal expansion coefficient and no destabilizing compositional stratification.

Continue the expansion established in part (b), with $p={ϵ}^2{p}_2+\cdots$ after removal of the background pressure. The radial momentum equation at order ${ϵ}^2$ givesIntegrating across the core, then differentiating in $x$, givesThis is the dispersive contribution from transverse inertia.

At second order the axial momentum equation isFor the displacement fields just obtained,Use $u_0=0$ on each wall. The lower and upper regular core limits respectively givewhere $v_2(1)=0$ at the fixed outer wall.

At the moving inner wall $y=ϵF$, no penetration requires $v=ϵ\lambda St{F}_t+ϵu{F}_x$. Expanding about $y=0$, the axial edge velocity is $ϵ{\gamma }_0(F+A/R)+O({ϵ}^2)$ and $v_{1y}(0)=-{\gamma }_0{A}_x/R$. ThusorThe thin wall layers do not change this normal-flux condition at the retained order, because their flux correction is $o({ϵ}^2)$.

Subtract the two axial pressure-gradient limits and use the radial-pressure difference. With ${\alpha }_1={\gamma }_0/R+{\gamma }_1/(R+1)$ and ${\alpha }_2={\gamma }_0^2/R^2-{\gamma }_1^2/(R+1{)}^2$, the result is the forced annular displacement equationEvery coefficient follows from the given profile and wall slopes; no explicit base-flow solution has been used. The derivation is conditional on the regular core limits and wall layers remaining asymptotically thin.

The positive drift places the boundary layer at the left endpoint. Set $y_{\mathrm{out}}=y_0+ϵy_1+\cdots$ and impose the right boundary condition order by order. The leading ordinary differential equation is $[(1+x)y_0{]}^{\prime }=0$, so $y_0=2/(1+x)$. At the next order,which integrates toThe outer limit at the left endpoint is two, so it cannot satisfy the left boundary value without a rapidly varying correction.

For the inner expansion use $X=x/ϵ$ and $y=Y_0(X)+ϵY_1(X)+\cdots$. Its equations and boundary values areMatching the leading constant gives $Y_0=2-e^{-X}$. The first-order forcing is then $-2+(1-X)e^{-X}$. A particular solution is $-2X+\frac{1}{2}{X}^2{e}^{-X}$, and matching determines the constant to be $3/2$. Enforcing $Y_1(0)=0$ givesIn the overlap region the common expansion is $2+ϵ(3/2-2X)$. Adding the inner and outer approximations and subtracting that common part gives the uniform asymptotic approximationIt has error $O({ϵ}^2)$ uniformly for $0\le x\le 1$, apart from an exponentially small endpoint adjustment of the same asymptotic irrelevance. At $x=0$ it equals one exactly, and at $x=1$ it differs from one only exponentially. The polynomial multiplying the exponential remains bounded on the inner scale.

There is an independent check from a first integral. The equation is $[ϵy^{\prime }+(1+x)y{]}^{\prime }=0$, so its exact solution has the formwhere $C$ is selected by the right boundary value. Endpoint expansion of this integral gives $C=2-ϵ/2+O({ϵ}^2)$, the same outer coefficients and the same inner expansion. This check also establishes the uniform order of the composite approximation without treating an inner residual as an outer estimate.

For two distinct symmetric fibre families use reference directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0{)}^T$, with $0<\theta <\pi /2$. The second sign is needed for the orthogonality conclusion: the PDF prints only one direction. If both families literally had that same direction, their images under an invertible deformation would remain parallel, so part (ii) would be false. The following derivation states the intended two-family geometry explicitly.

Write $c=\cos \theta$, $s=\sin \theta$ and take positive $\lambda ,\alpha$, as required for a positive stretch tensor. Applying the inextensible fibre constraint to either family givesWith $b={\alpha }^2$, multiplication by $b$ yields $c^2{b}^2-\lambda b+s^2=0$. Its discriminant is ${\lambda }^2-4c^2{s}^2={\lambda }^2-{\sin }^{2}2\theta$, so the two-family inextensible-fibre pure stretch hasThis also shows why the angle endpoints need separate treatment: at $\theta =0$ the admissible root is ${\alpha }^2=\lambda$, and at $\theta =\pi /2$ it is ${\alpha }^2=1/\lambda$. The nominal two families then represent the same unoriented fibre direction, and the two-branch/orthogonality assertions do not apply.

Write $N_{i,k}$ for the order-$k$ B-spline. At order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, and the finite sum is one on its basic knot interval. For larger order, the Cox-de Boor recurrence iswith a zero-denominator term defined as zero. Sum over $i=1,\dots ,n$. For each interior lower-order index $j=2,\dots ,n$, the combined coefficient isThe two unmatched lower-order terms have supports $[t_1,t_k]$ and $[t_{n+1},t_{n+k}]$, respectively, and vanish on the open basic interval $(t_k,t_{n+1})$. Induction therefore givesFor simple knots and $k\ge 2$, continuity extends the identity to both endpoints. In the order-one case and at admissible repeated endpoint knots, use the one-sided value from the basic interval. Extending the knot sequence in both directions also proves the subpartition of unity for B-splines, $0\le {\sum }_i{N}_i\le 1$ everywhere: the finite sum is a subset of a nonnegative full partition. That weaker global inequality will be needed in Question 7(a).

Let the equilibrium be $A_{\ast }=R{e}^{i\theta }$ and write $\delta A=e^{i\theta }(u+iv)$. The linearization of the cubic term gives $\delta (|A{|}^{2}A)=2s\delta A+A_{\ast }^2\delta \overline{A}$. Therefore the real stability matrix isIts eigenvalues obeyThe square root may be imaginary; stability uses real parts. The determinant $D$ is exactly $F_s$, so a zero eigenvalue occurs only at the turning points derived above. Away from the cusp those are ordinary saddle-node bifurcations; the parameterization joining point has $D>0$.

At a candidate Hopf bifurcation, the trace vanishes: ${\mu }_H=2s_H$. Substitution into the steady cubic gives $s_H(s_H^2+{\Lambda }^2)=1$, which has one positive root. The pair is genuinely imaginary only ifThus $s_H<2^{-1/3}$ and, equivalently,This direct calculation proves the threshold without needing the additional monotonicity assumption. The equality gives a double-zero eigenvalue, not an ordinary Hopf bifurcation. Along a fixed-$\Lambda$ branch, implicit differentiation gives $ds/d\mu =-2s(\mu -s)/D$; at this crossing $d\mathrm{Re}\sigma /d\mu =1+4s_H^2/D_H>0$. A nonzero nonlinear Hopf coefficient then gives the usual Hopf bifurcation.

The fold and Hopf thresholds of a directly forced cubic amplitude divide the diagrams into three regimes. At noncritical points the two eigenvalues, counted with multiplicity by the signs of their real parts, are classified as follows:

At a fold one eigenvalue is zero; at a Hopf point the pair has zero real part. These nonhyperbolic points are marked separately in the sketches.

Choose the charge conjugation matrix so that $C^T=-C$ and $({\gamma }^m{)}^T=-C{\gamma }^m{C}^{-1}$, and write $\bar{\lambda }={\lambda }^{T}C$. Since the components are odd Grassmann variables,Thus a symmetric matrix gives zero, whereas a nonzero antisymmetric matrix can give a nonzero bilinear. The four-dimensional Majorana bilinear vanishing criterion follows by transposing an antisymmetrized product and reversing its $n$ factors:For $n=0,1,2,3$, the symmetry types are respectively antisymmetric, symmetric, symmetric and antisymmetric. These are the same-spinor selection rules of the Majorana spinor; they would be reversed for a commuting spinor. The vanishing choices are (a) and (c).

In phi cubed theory, a two-point bubble diagram has two cubic vertices, two internal scalar propagators and symmetry factor $1/2$. Its connected amputated insertion is $+{\mu }^{ϵ}{g}^2{B}_d(p)/2$. The inverse scalar propagator, or quadratic one-particle-irreducible vertex, therefore receives the negative insertion:Equivalently, the connected two-point correction is $G_0(p{)}^2{\mu }^{ϵ}{g}^2{B}_d(p)/2$, and expansion of its inverse changes the sign. The same sign follows from the second term of $\frac{1}{2}\mathrm{Tr}\log (K+{\mu }^{ϵ/2}g\varphi )$.

Using the six-dimensional residue and ${\int }_0^{1}x(1-x)dx=1/6$, the divergent inverse-propagator correction isIts momentum term requires wave-function renormalization, unlike the massive quartic tadpole. The minimal-subtraction two-point counterterms in cubic scalar theory areThese are additive coefficients in a Lagrangian expressed in terms of the renormalized field. If ${\varphi }_0=Z^{1/2}\varphi$, $Z=1+\delta Z$, and the bare mass is separately written $m_0^2=m^2+\Delta m^2$, then $\delta m^2=m^2\delta Z+\Delta m^2$ at this order, so $\Delta m^2=-5g^2{m}^2/[6(4\pi {)}^3ϵ]$. Distinguishing these conventions prevents a spurious disagreement between mass counterterms. The local Minkowski counterterms are $\delta Z(\partial \varphi {)}^2/2-\delta m^2{\varphi }^2/2$. A complete cubic theory also has one-point and three-point subtractions; they are not the specified two-point bubble poles.