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For a short channel slice, the fluid volume per unit length is $b(H+\eta )$ and the axial volume flux is $b(H+\eta )u$. The quasi-one-dimensional conservation of mass equation is thereforeNeglecting products of perturbations gives ${\eta }_t+(H/b)(bu{)}_x=0$. This uses the section-averaged approximation stipulated in the problem; small transverse velocities ensure impermeability of varying-width walls but are not leading axial terms.

The Euler equations are ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-{\rho }^{-1}\nabla p-g{\mathbf{e}}_z$. For irrotational flow, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$, soAt the zero-pressure free surface, linearization about rest gives ${\varphi }_t+g\eta =C(t)$; differentiate in $x$ to get $u_t+g{\eta }_x=0$. Differentiating the mass equation in time and substituting givesPut $c_0=\sqrt{gH}$. For the exponential width, a time-harmonic amplitude satisfiesWriting $N=e^{-\mu x/2}V$ removes the first derivative and gives $V^{\prime \prime }+k^{2}V=0$, where $k^2={\omega }^2/c_0^2-{\mu }^2/4$. A nonzero real wavenumber, and hence a spatially travelling phase, exists precisely whenFor this case an outward-going solution matching the oscillating boundary isSubstitution in both linear equations verifies it, and $\mathrm{Im}(U(0)e^{i\omega t})=u_0\sin \omega t$. Below the cutoff the spatial exponents are real, giving evanescent responses; at equality the repeated-root solution has no travelling spatial phase.

This is the cutoff frequency in an exponentially widening shallow-water channel.

Orient the body boundary counterclockwise. Its outward body normal satisfies $nds=(dy,-dx)$, so the pressure force has complex formBy steady Bernoulli's equation, $p=p_{\infty }+\rho (U_{\infty }^2-|q{|}^2)/2$, where $q=w^{\prime }=u-iv$. Constant terms integrate to zero around a closed curve. On an impermeable boundary, velocity is tangent, so $qdz=\overline{q}d\overline{z}$. Therefore $|q{|}^{2}d\overline{z}=q^{2}dz$, yielding the Blasius force theoremFor a slit body, the contour encloses both faces and the tips, interpreted by a limiting exterior contour.

Choose the analytic branch $s(z)=\sqrt{z^2-a^2}$ outside the slit with $s(z)\sim z$ at infinity. It may be written $\sqrt{z-a}\sqrt{z+a}$ using the principal individual square roots. For $-a<x<a$, approaching from above gives $s_{+}(x)=i\sqrt{a^2-x^2}$, and from below gives $s_{-}(x)=-i\sqrt{a^2-x^2}$.

NowThere is no $z^{-1}$ term in $q^2$. Since $q^2$ is analytic in the exterior, enlarging the contour to a large circle shows its integral is zero. ThusThis total-force argument retains the slit-tip contribution automatically.

On the upper face, $q_{+}(x)=\cos \alpha -x\sin \alpha /\sqrt{a^2-x^2}$; on the lower face, $q_{-}(x)=\cos \alpha +x\sin \alpha /\sqrt{a^2-x^2}$. For $0<\alpha \le \pi /2$, their zeros areThey are distinct face points even at $\alpha =\pi /2$, when both have coordinate zero. At $\alpha =0$, $w=z$ and $q=1$ everywhere in the exterior: there are no velocity-zero stagnation points. The limiting positions at the two tips must not be substituted into a singular expression and treated as zeros.

For the sketch at $\alpha =\pi /4$, the far-field velocity is $(1,1)/\sqrt{2}$, and streamlines are levels of $\mathrm{Im}w$. The plate is the zero-stream-function boundary, with the two stagnation points at $\pm a/\sqrt{2}$ on the indicated faces.

In SI units, Maxwell's equations areDotting the last equation with $\mathbf{E}/{\mu }_0$, and using the curl equation for $\mathbf{E}$, givesIntegrate over the fixed region and apply the divergence theorem. This proves Poynting's theorem: the rate of increase of electromagnetic energy stored inside, plus the outward Poynting flux, plus the rate of work on charges, is zero. The last term need not be positive for arbitrary externally driven currents; when positive it is power transferred from the field to matter.

For the travelling wave put $\phi =kz-\omega t$. A purely oscillatory magnetic field satisfying Faraday's law isBoth divergences vanish. The vacuum Ampère-Maxwell law then requires $k^2={\mu }_0{ϵ}_0{\omega }^2$, so with the positive constants in the question,A superposed static background is not part of the pure plane-wave solution. The electric and magnetic energy contributions are equal, givingConsequently $\boxed{|\mathbf{P}|=c\mathcal{U}}$ pointwise, including the nodes where both sides vanish.

The lower particle has position $(x+l\sin \theta ,-l\cos \theta )$, so its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. The bead contributes $m_1{\dot{x}}^2/2$ to the kinetic energy. The two equal springs have combined potential energy $k{x}^2$ up to a constant: their lengths are $a+x$ and $a-x$, and equal natural lengths cancel the linear terms in the sum of their squared extensions. The gravitational potential is $-m_{2}gl\cos \theta$. ThereforeThere is no explicit time dependence, so the total energyis conserved. The Euler-Lagrange equations areAt rest these require $x=0$ and $\sin \theta =0$, giving the downward and upward positions $\theta =0,\pi$ modulo $2\pi$. The downward position has positive-definite quadratic potential and is stable, whereas the upward position has a negative angular potential curvature.

Keeping terms of second order about the stable position and dropping a constant givesFor unit parameters the mass matrix and stiffness matrix are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. A normal mode with squared frequency $s$ requiresThus the positive angular frequencies are

For fixed endpoints and a regular integral constraint, Lagrange multipliers replace the constrained stationarity problem by stationarity of $I-\lambda J$, with constant $\lambda$. Writing $H=F-\lambda G$, the Euler-Lagrange equation is $H_y-(d/dx)H_{y^{\prime }}=0$. Regularity here means the first variation of $J$ is not identically zero; abnormal constraint extrema otherwise need separate consideration.

If $H$ has no explicit $x$ dependence, thenThis is the Beltrami identity, giving the required constant of integration.

For the chain, $ds=\sqrt{1+y^{\prime 2}}dx$, so its fixed arc length is exactly $J[y]=L$, and its gravitational energy is $\rho g{\int }_{-a}^{a}y\sqrt{1+y^{\prime 2}}dx$. The positive constant $\rho g$ can be suppressed in minimization. UseThe equation, with a consistent sign convention for the multiplier, isThe first integral becomes $(y-\lambda )/\sqrt{1+y^{\prime 2}}=c$. It implies $y^{\prime \prime }=(1+y^{\prime 2})/(y-\lambda )$; solving givesThe equal endpoint heights force $x_0=0$, by evenness and strict increase of $\cosh$ on the positive half-line. Thus the extrema have the asserted catenary form. The length and height conditions giveFor the hanging minimum choose $c>0$, the upward-convex branch lying below its endpoints. For $L>2a$ this positive $c$ is unique: writing $t=a/c$, the ratio $L/(2a)=\sinh (t)/t$ increases continuously from one to infinity. The negative-$c$ branch describes an upward arch, not the hanging minimum. If $L=2a$, only the horizontal straight chain is admissible and is the limiting degenerate case; $L<2a$ is impossible.