With the complex velocity convention $w^{\prime }(z)=u-iv$, differentiation gives $w^{\prime }=-i\Gamma /(2\pi z)$. Thus $u=-\Gamma y/(2\pi r^2)$ and $v=\Gamma x/(2\pi r^2)$. Resolving into polar components gives $\boxed{u_r=0,u_{\theta }=\Gamma /(2\pi r)}$.

Superposition for the opposite-signed pair givesup to a constant and branch choices. A vortex moves with the regular velocity generated by the other vortex, excluding its own singular field. At $z=ia$, the other contribution is $i\Gamma /[2\pi (2ia)]=\Gamma /(4\pi a)$; at $z=-ia$ it is the same real number. Hence the pair keeps its separation and translates parallel to the real axis withfor $a>0$.

For a vortex at $z_1=x_1+i{y}_1$ in the first quadrant, reflect it across either wall with opposite strength, then across both walls with its original strength. Thus the images are ${\bar{z}}_1$ and $-{\bar{z}}_1$ of strength $-\Gamma$, and $-z_1$ of strength $+\Gamma$. The potential isOn either real or imaginary axis, $z^2$ is real and the numerator and denominator of $(z^2-z_1^2)/(z^2-{\bar{z}}_1^2)$ are conjugates. Its modulus is one, making the stream function constant on both walls and enforcing no normal flow. A regular entire background flow could be added only with additional far-field data; the displayed image solution is the one generated by the specified vortex and its walls.

For a material volume, conservation of mass and the Reynolds transport theorem give ${\rho }_t+\nabla \cdot (\rho u)=0$. Constant material density, or equivalently incompressibility with the stated constant initial density, reduces this to $\nabla \cdot u=0$.

In an fluid undergoing inviscid flow the stress is the isotropic pressure stress $-pI$. Momentum conservation for a material volume then givesTransport and the divergence theorem localize this toFor a steady flow, $u_t=0$. For an irrotational flow the vorticity is zero, so the supplied identity gives $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Uniform downward gravity is $g=-\nabla (gz)$, and constant density permits pulling $1/\rho$ inside the pressure gradient. ThereforeThe Bernoulli invariant is constant throughout each connected fluid region. Without irrotationality the corresponding steady statement generally gives a constant only along each streamline.

The inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-x_i{x}_j)dm$. The mixed moments vanish by reflection in $x$ or $y$. Every diagonal integrand is even in $z$, so its integral over the upper hemisphere is half its integral over the corresponding full ball of mass $2M$. A uniform ball's coordinate moments are each $(2M)a^2/5$, obtained by spherical symmetry and $⟨r^2⟩=3a^2/5$. Hence the hemisphere has each coordinate second moment $M{a}^2/5$ andThe tensor parallel axis theorem gives $I_O=I_C+M(|d{|}^{2}I-d{d}^t)$ with $d=(0,0,3a/8)$. ThereforeTranslation along the symmetry axis changes the two transverse moments but leaves the axial one unchanged.

The total derivative is the unique linear map $D{f}_a:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfyingThis describes the best first-order linear approximation without choosing coordinates. The chain rule isThe evaluation point $f(a)$ in the outer derivative matters, particularly when composing a quadratic matrix map with the exponential.

The product topology is generated by the basis of a topology of open rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. For a fixed $y$,Both possibilities are open; checking this basis of a topology proves that $i_y$ is continuous. The analogous map $j_x:Y\to X\times Y$, $j_x(y)=(x,y)$, is continuous by the same argument.

If either factor is empty the product is empty and connected. Otherwise let $h:X\times Y\to \{0,1\}$ be continuous. For each $y$, $h\circ i_y$ is constant since $X$ is connected. Thus $h(x,y)=h(x_0,y)$ for any fixed $x_0\in X$. The right-hand side is $h\circ j_{x_0}$ and is constant in $y$ since $Y$ is connected. Therefore $h$ is constant on the whole product. The given characterization proves $X\times Y$ is connected.

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

The indicated measurable function is nonnegative. Tonelli's theorem applies without first assuming integrability and givesThus $h$ is integrable. Fubini's theorem now allows the reverse order. For $x>1$,ConsequentlyAt $x=1$ the quotient has the removable limit $b-a$; that endpoint does not create an additional divergence.

For $x>0$ the geometric ratio $e^{-2x}$ has modulus less than one, andEvery summand is nonnegative. The monotone convergence theorem therefore permits integration of the sum:The integral is finite because its integrand is bounded above by $e^{-x}$, whose integral is one. The displayed paired sums are the even partial sums of the alternating harmonic series; its odd partial sums differ by a term tending to zero. Thus the required equality holds with the usual series ordering. One can additionally evaluate it by $t=e^{-x}$:Grouping into nonnegative pairs supplies the justified interchange; absolute convergence of the ungrouped integrated alternating series is not needed and would be false.

The polynomial $x^2-2$ is an irreducible polynomial over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. This intermediate field is real, so it does not contain $i$. Hence $x^2+1$ has no root in it and adjoining $i$ has degree two. The tower law givesA basis is $1,\sqrt{2},i,i\sqrt{2}$. Square-bracket adjoining notation in this algebraic example gives the same field as round-bracket adjoining notation.