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The initially irrotational inviscid flow remains irrotational, so $u=\nabla \varphi$; incompressibility gives ${\nabla }^2\varphi =0$. At the plates, normal velocity matches their motion:With $\varphi =r^{2}f(\theta )$, Laplace's equation gives $f^{\prime \prime }+4f=0$. The symmetric solution isHenceA streamfunction isso streamlines are $r^2\sin 2\theta =\text{constant}$ and fluid is expelled radially as the plates close. The outward flux through $r=R$ is

With signature $(+---)$ one convenient convention iswhich gives the stated dual after index lowering. The invariants areFor the boost,with zero $x$ components.

In the final configuration, parallelism givesThe physical root isThus $v\sim c\theta /2$ for small angle. As $\theta \to \pi /2$, $v\to c$, so exactly crossed equal-magnitude null fields cannot be aligned by any finite inertial boost.

Sincethe divergence theorem proves Green's second identity.

If $x_0=(x_0,y_0,z_0)$ and $x_0^{\ast }=(x_0,y_0,-z_0)$, the image solution isIt vanishes on the plane and has the required unit delta source.

Green's identity yields the Poisson-kernel integralOn the axis this becomes

Cauchy's formula gives $|f^{\prime }(z_0)|\le M_R/R$ for an entire function bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to \infty$ gives $f^{\prime }=0$, proving Liouville's theorem.

Set $h(w)=f(1/w)$. Its limit at zero makes the singularity removable, soConsequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2{f}^{\prime }(z)\to -c_1=:b$.

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to 1$. Thus $m=n$. The functionhas removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence

For a closed surface, $\chi =V-E+F$ for any finite cell decomposition. Gauss–Bonnet for a geodesic polygon saysand for a closed smooth surface it says ${\int }_{S}KdA=2\pi \chi (S)$.

For a large radius, the graph surface bounded over that circle is flat near its boundary. The boundary geodesic-curvature integral is $2\pi$, so Gauss–Bonnet on the disc gives $\int K=0$. Since $K\ge 0$, continuity forces $K\equiv 0$.

For the torus,Thus $T_{+}$ has $\cos u>0$ and $T_{-}$ has $\cos u<0$. Cap the two boundary circles of the outer half $T_{+}$ by flat discs. The result is a sphere, and the caps contribute no Gaussian curvature, so Gauss–Bonnet gives ${\int }_{T_{+}}K=4\pi$. The full torus has Euler characteristic zero, hence total curvature zero and

A space is connected when it is not a union of two disjoint nonempty open subsets. If $f(X)$ had such a separation, its inverse images would separate connected $X$, so continuous images of connected spaces are connected. Equivalently, a nonconstant continuous map to discrete $\{0,1\}$ records a separation, and every separation defines such a map.

If $\mathbb{R}$ were separated, points in opposite pieces and the intermediate value theorem applied to the associated $\{0,1\}$-valued map would give a contradiction. Thus $\mathbb{R}$ is connected. Its quotient by $x\sim y$ is connected because the quotient map is continuous and surjective.

Finally let $A\subset B\subset \mathrm{Cl}(A)$ and suppose $B=U\cup V$ were a separation. Connectedness puts $A$ wholly in one side, say $U$. Every point of $V$ nevertheless lies in the closure of $A$, while $V$ is relatively open and disjoint from $A$, a contradiction. Hence $B$ is connected.

A euclidean domain is an integral domain with a function $d$ on nonzero elements such that for $a,b\ne 0$ there are $q,r$ with $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.

A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.