A joint probability density function must be measurable, nonnegative almost everywhere, and normalized:It determines the joint law by $\mathrm{Pr}((X,Y)\in B)={\int }_{B}f(x,y)dxdy$ for measurable sets $B$. Values on sets of planar measure zero may be changed without changing that law.

The sequence is a Markov chain when the conditional law of its next state depends on the past only through the current state. Precisely, for each history of positive probability and every $j\in S$,A time-homogeneous chain additionally has transition probabilities independent of $n$, written $P_{ij}=\mathrm{Pr}(X_{n+1}=j\mid X_n=i)$. The displayed single transition matrix in the subsequent parts specifies this time-homogeneous case.

For a point vortex in stagnation flow beside a wall, a freely moving point vortex is transported by the regular velocity due to the background and its image, excluding its own singular velocity. Differentiate the potential and evaluate that regular part at $z_0=x+iy$:Thus the equations of motion areWriting $c=\Gamma /(4\pi )$, differentiate the proposed path function:HenceFor initial position $(x_0,y_0)$, an explicit check is $y(t)=y_0{e}^{-t}$ and $x(t)=e^t(x_0+ct/y_0)$. Their substitution makes the linear terms in $t$ cancel in the path function and also shows that the vortex stays above the wall at every finite time.

A point vortex of circulation $\Gamma$ at $z_0$ contributes the complex potential of a two-dimensional potential flow $-i\Gamma \log (z-z_0)/(2\pi )$ with the stated convention. The method of images places a vortex of the opposite circulation at ${\overline{z}}_0$ below the wall, outside the fluid domain. Superposition with the stagnation flow givesFor real $z=x$, the distances $|x-z_0|$ and $|x-{\overline{z}}_0|$ agree. The two vortex contributions to the stream function are $-\Gamma \log |x-z_0|/(2\pi )$ and $+\Gamma \log |x-{\overline{z}}_0|/(2\pi )$, which cancel. The background stream function is also zero there, so $y=0$ remains an impermeable streamline. The image changes the boundary condition without introducing another singularity in the upper fluid region.

Use the convention $w=\varphi +i\psi$, with $dw/dz=u_x-i{u}_y$. ThenIndeed $w^{\prime }(z)=z=x+iy$, giving velocity components $(x,-y)$. The streamlines are the level sets of the stream function:For $c\ne 0$ these are rectangular hyperbolas, restricted to $y>0$. The level $c=0$ consists of the vertical axis and the wall $y=0$. Directly on the wall the normal velocity is $-y=0$, so it is a streamline. The arrows point down toward the wall and then away from the stagnation point along the wall.

The Free surface of a Rankine vortex follows from setting its pressure constant. At a free surface with constant pressure $p_s$, solve the two pressure expressions from part (b) for $z$. With $C=(B-p_s)/(\rho g)$, this givesSubstituting $r^2=x^2+y^2$ recovers the required Cartesian form. The two values coincide at $r=a$, and both radial slopes there equal ${\Gamma }^2/(g{a}^3)$. The surface is parabolic in the core and approaches height $C$ as $r\to \infty$; at the centre its depression below $C$ is ${\Gamma }^2/(g{a}^2)$.

This is a Rankine vortex. Write $r=\sqrt{x^2+y^2}$. Inside the core the velocity is $u=(\Gamma y/a^2,-\Gamma x/a^2,0)$, soThis is the core value obtained from the inner formula; outside, the given irrotationality makes the vorticity zero. For nonzero $\Gamma$, the sharp boundary $r=a$ has different one-sided velocity gradients, so a single classical vorticity is not defined exactly on that edge; the inner trace is the boxed value and the outer trace is zero.

Inside the core, direct differentiation gives $(u\cdot \nabla )u=-{\Gamma }^2(x,y,0)/a^4$. The steady Euler equations therefore give $p_r=\rho {\Gamma }^{2}r/a^4$ and $p_z=-\rho g$. Integrating,Outside, $|u{|}^2={\Gamma }^2/r^2$, so part (a) givesMatching pressure at the core edge $r=a$ yields $A=B-\rho {\Gamma }^2/a^2$. Hence the complete pressure field, up to one additive constant $B$, isBoth the velocity and pressure are continuous at $r=a$. The inside expression has been derived from the momentum equation because the core is rotational.

For a steady flow the local time derivative in the Euler equations vanishes. The vector identityreduces to a gradient when the vorticity $\nabla \times u$ is zero. With constant density and gravity $-gk$, the momentum equation becomesThus on each connected flow region,This is Bernoulli's theorem for a steady irrotational flow: irrotationality makes the same constant apply throughout a connected region, rather than only along an individual streamline.

The regular level set theorem says that for a $C^1$ map $g:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^r$, if $Dg$ has rank $r$ on $g^{-1}(c)$, then $g^{-1}(c)$ is an $(n-r)$-dimensional submanifold andHere $r=1$ and the given rank condition makes the zero set a two-dimensional submanifold. It is nonempty because $g(0)=f(0)=0$. Part (c) givesConsequentlyThere is also a direct curve proof: $g(p+t(1,1,1))=g(p)=0$, so the entire line through $p$ in this direction lies in $M$. Its velocity at zero is exactly the required tangent vector.

An embedded $k$-dimensional submanifold $M\subseteq {\mathbb{R}}^n$ is locally a coordinate $k$-plane: for every $p\in M$, there are open neighbourhoods $U\ni p$, $V\subseteq {\mathbb{R}}^n$ and a diffeomorphism $\Phi :U\to V$ withFor the present $C^1$ setting, $\Phi$ and its inverse are continuously differentiable; in the smooth convention both are smooth.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a continuously differentiable curve $\gamma :(-ϵ,ϵ)\to M$ with $\gamma (0)=p$. The tangent space is the collection of all these vectors:In the local coordinates above, it is $D\Phi (p{)}^{-1}({\mathbb{R}}^k\times \{0{\}}^{n-k})$, so it is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$.

Let $L(x,y,z)=(x-y,y-z,z-x)$. This is a linear map, hence smooth; composing it with the given $C^1$ function proves that $g=f\circ L$ is continuously differentiable by the chain rule. Write $f_1,f_2,f_3$ for the first partial derivatives, evaluated at $L(p)$. ThenAdding these identities givesGeometrically, $L(p+t(1,1,1))=L(p)$, so $g$ is constant along every line in the diagonal direction; its derivative in that direction must vanish.

For differentiable maps with compatible domains, the multivariable chain rule isIn Jacobian matrix notation this is matrix multiplication. In particular, if both maps are continuously differentiable, the composition is continuously differentiable. Componentwise,

The function is continuously differentiable when all its first partial derivatives exist on $U$ and are continuous there. Equivalently, $h$ has a Fréchet derivative at every point and the map $x\mapsto Dh(x)$ is continuous. For a scalar-valued function,On the open domain, continuous first partial derivatives ensure that this linear expression is the total derivative.

Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The familyis countable, since the Cartesian product of two countable sets is countable. It is a basis of a topology for the product topology: given a product-open neighbourhood of $(x,y)$, choose a contained open rectangle $U\times V$, then basis members with $x\in B_i\subseteq U$ and $y\in C_j\subseteq V$. Their product contains $(x,y)$ and is contained in the original neighbourhood. ThereforeIn standard terminology, a finite product of second-countable spaces is second-countable.

The product topology on $X\times Y$ has as a basis all open rectanglesThus a subset $W\subseteq X\times Y$ is open precisely when, around every $(x,y)\in W$, there are open $U\ni x$ and $V\ni y$ with $U\times V\subseteq W$. These rectangles cover the product, and their pairwise intersections are rectangles because $(U\times V)\cap (U^{\prime }\times V^{\prime })=(U\cap U^{\prime })\times (V\cap V^{\prime })$. Hence they define a topology by part (b).

This is equivalently the coarsest topology making the two coordinate projection maps continuous: the preimages of factor-open sets are the open strips $U\times Y$ and $X\times V$, whose finite intersections are the basis rectangles.

Define $\mathcal{T}$ to be all unions of members of $\mathcal{B}$, including the empty union. Then $\varnothing \in \mathcal{T}$ and the covering assumption gives $X\in \mathcal{T}$. Arbitrary unions of such sets are again unions of basis members.

For two members $U={\bigcup }_i{B}_i$ and $V={\bigcup }_j{B}_j^{\prime }$, distribute intersection over union:Each intersection is a union of members of $\mathcal{B}$ by the second assumption, so $U\cap V\in \mathcal{T}$. Repeating gives closure under all finite intersections. These are precisely the topology axioms. Every $B\in \mathcal{B}$ belongs to $\mathcal{T}$, and every set in $\mathcal{T}$ is a union of such members by construction. Thus

A basis of a topology $\mathcal{B}$ is a family of open subsets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every $U\in \mathcal{T}$ and $x\in U$, some $B\in \mathcal{B}$ satisfiesThe empty open set is represented by the empty union.