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Substitution of $y=h$ gives

The steady parallel-flow Navier-Stokes equations reduce toThe boundary conditions areThe last condition balances the liquid's tangential traction with the opposing air stress. Hence $p=p_0$ and

Take $x$ vertically downward along the wall and $y$ normally outward, with the wall at $y=0$ and free surface at $y=h$. The expected steady velocity is $\mathbf{u}=u(y){\mathbf{e}}_x$: it vanishes at the wall and is opposed by an upward air shear $ku(h)$ at the surface. Gravity supplies the downward body force $\rho g{\mathbf{e}}_x$, while the surface normal stress is the atmospheric pressure $p_0$.

Integrate $z{e}^{i\pi z}/(z^2+a^2)$ over the upper semicircle. Jordan lemma removes the arc, and the pole at $z=ia$ has residue $e^{-\pi a}/2$. HenceThe cosine contribution is odd, while the sine contribution is even. Therefore

The integrandhas one pole inside the rectangle, at $z=0$, with residue $1$. The residue theorem therefore gives ${\int }_{C}F(z)dz=2\pi i$. The vertical-side integrals tend to zero as $R\to \infty$.

On the horizontal sides, direct substitution $z=x\pm \pi i$, with the upper side oppositely oriented, shows that their sum isTaking the limit and comparing with $2\pi i$ gives

HereThis matrix acts as multiplication by the complex number $(2x-1)+2iy$, so its operator norm is $\sqrt{(2x-1{)}^2+4y^2}$. The largest set is therefore

For distinct $x,y\in U$, convexity keeps the segment $x+t(y-x)$ in $U$. Continuity of $Df$ makesThe fundamental theorem of calculus along a line segment givesso $\Vert f(y)-f(x)\Vert \ge (1-q)\Vert y-x\Vert >0$. Thus $f$ is injective.

The bound $\Vert Df(a)-I\Vert <1$ makes $Df(a)$ invertible by the Neumann series. The inverse function theorem therefore makes $f$ locally open at every point, so $f(U)$ is open.

Surjectivity need not hold: the identity map on a proper convex open set has proper image. Even for $U={\mathbb{R}}^n$ it can fail. In one dimension, $f(x)=\arctan x$ satisfies $|f^{\prime }(x)-1|=x^2/(1+x^2)<1$, but its image is $(-\pi /2,\pi /2)$.

The map $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ when there is a linear map $Df(a)$ such thatIt is $C^1$ when it is differentiable everywhere and $a\mapsto Df(a)$ is continuous. The definition immediately gives $\Vert f(a+h)-f(a)\Vert \le (\Vert Df(a)\Vert +o(1))\Vert h\Vert \to 0$, so differentiability implies continuity.

The inverse function theorem says that if $f$ is $C^1$ near $a$ and $Df(a)$ is invertible, then $f$ restricts to a diffeomorphism between neighbourhoods of $a$ and $f(a)$.

For $|G|=12$, the third Sylow theorem gives $n_3\equiv 1(\mathrm{mod}3)$ and $n_3\mid 4$, so $n_3=1$ or $4$. In the first case the Sylow $3$-subgroup is unique.

If $n_3=4$, conjugation acts transitively on the four Sylow $3$-subgroups. Each normalizer has order $|G|/4=3$ and is the subgroup it normalizes. The action kernel lies in every such subgroup, whose intersection is trivial, so $G$ embeds in $S_4$. Its image has order $12$ and hence index two. The unique index-two subgroup of $S_4$ is $A_4$, so

Proceed by induction on $|G|$. If $p\mid |Z(G)|$, Cauchy's theorem for finite groups gives a central subgroup $C$ of order $p$. By induction $G/C$ has a subgroup of order $p^{n-1}$; its inverse image in $G$ has order $p^n$.

If $p\nmid |Z(G)|$, the class equation shows that some noncentral conjugacy class has size not divisible by $p$. For a representative $x$, its class size is $[G:C_G(x)]$, so $p^n\mid |C_G(x)|$. This centralizer is proper because $x$ is noncentral, and induction gives it, hence $G$, a subgroup of order $p^n$. This proves Sylow's first theorem.

Write $|G|=p^{n}m$ with $p\nmid m$. The Sylow theorems state:

Let $M$ be the mass of a complete uniform sphere of radius $a$. The intact marble has $I_1=2M{a}^2/5$. Each removed bubble has mass $M/8$, radius $a/2$, and centre a distance $a/2$ from the rolling axis. By the parallel axis theorem, each removesThus $M_2=3M/4$, $I_2=2M{a}^2/5-7M{a}^2/80=5M{a}^2/16$, and $I_2/(M_2{a}^2)=5/12$. The two accelerations are therefore $(5/7)g\sin \alpha$ and $(12/17)g\sin \alpha$. From rest, distance is proportional to acceleration, so

For rolling without slipping, $\dot{s}=a\Omega$. Conservation of energy givesand differentiation yields

For a sphere with density proportional to $r^c$,so $I/(M{a}^2)=2(c+3)/(3(c+5))$. A uniform disc has $I/(M{a}^2)=1/2$. Equal launch speeds give equal return times exactly when the accelerations, and hence these ratios, agree. Solving gives $c=3$.

The equation of motion isMultiplication by $\dot{x}$ givesso the mechanical energy $E=\frac{1}{2}m{\dot{x}}^2+V(x)$ is conserved. On a segment where the direction of motion is fixed,

The Principle of Galilean relativity states that the laws of mechanics have the same form in every inertial frame.

The basic Galilean transformations include spacetime translations ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{a}$, $t^{\prime }=t-b$; spatial rotations ${\mathbf{x}}^{\prime }=R\mathbf{x}$, $t^{\prime }=t$; and boosts ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$, where $R$ is a rotation matrix and $\mathbf{v}$ is constant.

An inertial frame is a reference frame in which a force-free particle moves with constant velocity, so Newton's first law holds.

Work in the rest frame of $B$. Two-body decay kinematics gives the massless-particle energiesIf $\theta$ is the angle between their momenta, thenSince $0\le 1-\cos \theta \le 2$,