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The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

Set $u(r,\theta )=R(r)\Theta (\theta )$. Periodicity and separation of variables give the angular modes $1$, $\cos n\theta$, and $\sin n\theta$. The radial equation iswhose solutions are $r^{\pm n}$ for $n\ge 1$; the zero mode has solutions $1$ and $\log r$. Thus a general real harmonic function on the annulus is

The boundary data contain only the $n=2$ cosine mode, so writeThe condition at $r=a$ gives $B=-A{a}^4$, and the condition at $r=b$ fixes $A$. The Dirichlet problem on an annulus for one Fourier mode therefore has solution

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain, then its integral around every closed piecewise smooth contour in that domain is zero.

Put $q=b-k$. Completing the square givesAfter the change of variable $z=x-iq/(2a)$, the integral runs along the horizontal line $\mathrm{Im}z=-q/(2a)$. The integrand $e^{-a{z}^2}$ is entire. Apply Cauchy's theorem to a rectangle joining this line to the real axis; the two vertical integrals tend to zero as their real parts tend to $\pm \infty$, because $a>0$. The contour may therefore be shifted to the real line. Hence the Fourier transform of a Gaussian gives

Use the global parametrizationIts tangent vectors areandThus $X$ is a regular parametrization and the graph is a smooth surface.

The first fundamental form isChoosing the upward orientation, the Gauss map is

If the bound is to hold with finite $c$, the error functional must vanish on every polynomial whose fourth derivative is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ givesSolving,This is the four-point one-sided second-derivative formula.

The Peano kernel theorem says that if a linear functional $L$ annihilates all polynomials of degree below $r$, then for $f\in C^r[a,b]$,

Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,The integral equals $e(q)$ for $q(x)=(x+1{)}^4/24$, since $q^{(4)}=1$. Now $q^{\prime \prime }(-1)=0$ andThereforeEquality is attained by $q$, so the sharp Peano-kernel constant for the four-point endpoint second derivative is

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are

The fluid kinetic energy iswhere $V=4\pi a^3/3$. Thus the added mass of a sphere is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density ${\rho }_s$ lowers the gravitational potential energy at rateThe total kinetic energy of sphere and fluid isConservation of total energy givesFor $U\ne 0$, the acceleration of a freely falling sphere with added mass isThe same formula holds at release by continuity.

In electrostatic equilibrium, a nonzero electric field inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential electric field is zero. A Gaussian pillbox across the surface givesSince $E_{\mathrm{in}}=0$, the electrostatic boundary conditions at a conductor give

For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requiresThus the charge sharing between distant connected spheres is

For the charge on connected concentric spherical shells, the potentials at the two radii areEquality forces $Q_1=0$. Henceso all charge lies on the exterior of the outer shell.

For the neutral sphere in the uniform field, the far-field condition gives $\alpha =-E$. Constancy of the potential on $r=R$ gives $\beta =E{R}^3$. ThereforeThe outward normal field at the surface isThe induced charge on a conducting sphere in a uniform electric field is consequentlyFinally,confirming neutrality.