At zero surface shear stress, the solution connected to part (a) isIndeed, it has $f_0^{\prime }=1$, $f_0^{\prime \prime }=f_0^{\prime \prime \prime }=0$ and $f_0(0)=0$. Insert the regular perturbation $f=\zeta +\gamma f_1+O({\gamma }^2)$ into the nonlinear ordinary differential equation. The terms areThe coefficient of $\gamma$ therefore givesExpanding the surface boundary conditions and the far-field matching condition givesThe printed $+\infty$ limit is a sign error. It is present in the native PDF on paper page 3, not introduced by transcription. The fluid occupies $z<0$, the inner variable is $\zeta =z/\delta <0$, and part (b)(iii) correctly matches at $-\infty$. We retain that physical domain throughout. A change to the positive coordinate $\eta =-\zeta$ would also change the differential equation and surface derivative signs; it cannot be made by changing only the far-field sign.

The inequality ${\delta }^2\ll \gamma$ ensures that effects discarded at order ${\delta }^2$ in the boundary-layer scaling are smaller than the retained order-$\gamma$ perturbation. The inequality $\gamma \ll 1$ justifies this regular perturbation. The two conditions control different approximations.

With $u=x{f}^{\prime }(\zeta )$ and $W=-f(\zeta )$, incompressibility is automatic because $u_x=f^{\prime }$ and $W_{\zeta }=-f^{\prime }$. In the streamwise boundary-layer equation,Equating coefficients of $x$ givesThis coefficient comparison determines the equation even though the streamwise equation vanishes identically at the stagnation point $x=0$.

The no-penetration boundary condition gives $f(0)=0$, the surface shear stress gives $x{f}^{\prime \prime }(0)=\gamma x$, and matching the tangential velocity to the outer solution gives $f^{\prime }\to 1$. ThusIf the correction $f^{\prime }-1$ is integrable at the far end, then $f=\zeta +O(1)$ there, which is consistent with $W\sim -\zeta$. The small-shear stress solution below has precisely this property.

In a thin boundary layer, take streamwise velocity and distance to be order one, set $z=\delta \zeta$, and use incompressibility to set $w=\delta W$. The streamwise convective acceleration is order one, whereas its dominant viscous contribution after division by the Reynolds number is ${\mathrm{Re}}^{-1}{\delta }^{-2}{u}_{\zeta \zeta }$. Their dominant balance selectsWith this choice, the exact rescaled Navier-Stokes equations areThus the leading pressure is independent of $\zeta$. Matching it to the outer solution from part (a), evaluated near $z=0$, gives $p=p_{\ast }-x^2/2$, hence $-p_x=x$. The leading boundary-layer equation is thereforeThe exact surface shear stress condition rescales to $u_{\zeta }+{\delta }^2{W}_x=\delta \beta x=\gamma x$. Since $W(x,0)=0$ for all $x$, its tangential derivative also vanishes there. The leading surface boundary conditions areMatching to the outer solution requiresThe second matching condition specifies the leading growth of $W$, allowing a constant displacement correction. Requiring $W+\zeta \to 0$ would impose an extra condition not supplied by the leading outer solution. This is the stress-driven stagnation-point boundary layer.

The held-flat interface imposes the no-penetration boundary condition, while its specified tangential traction gives a shear stress condition:A differentiable boundary value $w(x,0)\equiv 0$ implies $w_x(x,0)=0$, so the second boundary condition is also $u_z(x,0)=\beta x$. There is no imposed no-slip boundary condition at this interface; tangential velocity is determined by the shear stress and the fluid equations.

The solution in part (a) has zero tangential traction, and so remains valid for the whole interface if and only if $\beta =0$. At the single point $x=0$ the imposed shear stress vanishes for any $\beta$, but this does not satisfy a boundary condition that must hold for every $x$.

The volumetric flow rate is the integral of axial velocity over the cross-section:The plane Poiseuille flow contribution isFor each correction eigenfunction,Uniform convergence permits termwise integration. Multiplying these two integrals by $A_n$ givesConsequentlyEvery summand in the subtracted quantity is positive. Thus the side walls reduce the flux below the infinite-width plane Poiseuille flow prediction. Its dimensions are volume per unit time, as required for a volumetric flow rate.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

For a profile depending only on $y$, the Poisson equation becomes $w_1^{\prime \prime }=-G/\mu$. Integrating twice givesThe no-slip boundary conditions $w_1(b)=w_1(-b)=0$ imply $C_1=0$ and $C_2=G{b}^2/(2\mu )$. ThusThis is plane Poiseuille flow between the two horizontal walls. It is positive in the interior for $G>0$, so the velocity points down the pressure gradient. It does not satisfy the side-wall no-slip boundary conditions, which is precisely what the correction in part (iii) repairs.

For the unidirectional flow $\mathbf{u}=(0,0,w)$, incompressibility gives ${\partial }_{z}w=0$. Hence $w=w(x,y,t)$, and the convective acceleration $w{\partial }_{z}w$ vanishes. The transverse components of the Navier-Stokes equation give $p_x=p_y=0$, while the axial component givesA steady-flow assumption is implicit in the question. Neither unidirectional flow nor a time-independent pressure gradient alone eliminates ${\partial }_{t}w$: a decaying transient can be superposed on the final pipe profile. The native PDF does not explicitly say steady flow in this part. For the intended steady flow, $w_t=0$, and thereforeTaking the pipe walls to be stationary, the no-slip boundary condition isThe no-penetration boundary condition is already satisfied by the assumed velocity field. This is the Poisson equation defining rectangular duct flow.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

For an arbitrary material volume $V(t)$, Newton's second law equates the rate of change of linear momentum to the resultant contact force, since there is no body force:Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leavesOn the right, the divergence theorem applied to each row of the Cauchy stress tensor givesConsequently the volume integral of $\rho D{u}_i/Dt-{\partial }_j{\sigma }_{ij}$ vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:The left side is mass density times material acceleration; the right side is the divergence of the Cauchy stress tensor.

At an oriented internal surface with unit normal vector $\mathbf{n}$, the traction $\mathbf{t}(\mathbf{x},t;\mathbf{n})$ is the limiting contact force per unit surface area exerted by the material on the positive-$\mathbf{n}$ side on the material on the negative-$\mathbf{n}$ side:For a boundary of a chosen material volume, this convention gives the external traction when $\mathbf{n}$ is outward. The Cauchy stress tensor has componentsso the second index specifies the surface orientation and the first specifies the force component. The vectors ${\mathbf{e}}_i$ form the standard Cartesian coordinate system basis of unit vectors.

Cauchy's stress theorem states that the dependence of traction on the normal vector is linear:Repeated indices are summed according to Einstein notation. In particular, reversing the normal vector reverses the traction, and knowing the tractions on three mutually perpendicular coordinate planes determines the traction on every plane. Symmetry of the Cauchy stress tensor, for an ordinary fluid without body couples, follows separately from angular momentum conservation.

Let $V(t)$ have a boundary moving with velocity ${\mathbf{v}}_b$, and let $\mathbf{n}$ be its outward unit normal vector. For a smooth scalar field $q$, the Reynolds transport theorem isIn the last expression, extend ${\mathbf{v}}_b$ smoothly into $V(t)$ and use the divergence theorem. For a material volume, its boundary moves with the fluid velocity $\mathbf{u}$, so ${\mathbf{v}}_b=\mathbf{u}$. Equivalently, the Reynolds transport theorem for a material volume readswhere $D/Dt$ is the material derivative.

For an arbitrary moving region, mass conservation includes the relative mass flow rate across its boundary:Substitution of the Reynolds transport theorem and the divergence theorem givesSince the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an integral of the same sign. This gives the continuity equation. With constant positive mass density $\rho$, it reduces toThus the flow is incompressible. For a material volume, the same argument starts directly from conservation of its total mass, with zero relative boundary flux.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

The classification theorem for closed combinatorial surfaces states that the geometric realization of every closed combinatorial surface is homeomorphic to exactly one of the following:

With ${\Sigma }_0=S^2$, the complete distinguishing data are orientability and the Euler characteristic:Here $\chi (K)=V-E+F$, where $V,E,F$ count the vertices of a simplicial complex, edges, and triangles of a finite triangulation. Thus two such connected closed surfaces are homeomorphic if and only if they have the same orientability and Euler characteristic. In particular $N_1$ is the real projective plane and $N_2$ is the Klein bottle. Every listed type admits a finite triangulation. For disconnected finite closed combinatorial surfaces, apply the theorem to each of the finitely many connected components; the resulting unordered collection of types determines the homeomorphism type.

A topological surface is a Hausdorff space with a countable basis of a topology in which every point has an open neighbourhood homeomorphic to an open subset of the Euclidean plane. We use the convention that a surface has no boundary unless a surface with boundary is explicitly specified. A closed surface is compact and has no boundary. Some conventions also include connectedness in the definition; we impose it on the closed combinatorial surface below.

A closed combinatorial surface is a finite connected simplicial complex $K$ with the following properties: every simplex is a face of a simplex of dimension two, and the link of a simplex at every vertex is a cycle graph. Explicitly, for a vertex $v$ its link is the simplicial subcomplexTo say this is a cycle graph means that there are distinct vertices $w_1,\dots ,w_m$, with $m\ge 3$, and exactly the link edges $\{w_i,w_{i+1}\}$, with indices modulo $m$. In particular, every edge of $K$ belongs to exactly two triangles. The vertex-link condition also requires that the triangles around each vertex form one cyclic fan; edge incidence alone would allow singular examples.

These conditions make $|K|$ a closed surface. A point inside a triangle has a disc neighbourhood, the two triangles at an edge give a disc neighbourhood, and the cyclic fan at a vertex gives a disc neighbourhood. Finiteness gives compactness, the Hausdorff property, and a countable basis of a topology from a realization in a finite-dimensional Euclidean space. If disconnected closed combinatorial surfaces are allowed, each connected component satisfies this definition separately.

Fix a vertex of a simplicial complex $v$, and let $R$ be the set of vertices reachable from $v$ by an edge path. If a simplex contains a vertex of $R$, every vertex of that simplex lies in $R$: append its edge from the reachable vertex to the existing edge path. Thus every simplex has either all its vertices in $R$ or all outside $R$.

Partition the geometric realization as $|K|=A\bigsqcup B$, according to whether the support in the barycentric coordinate system lies in $R$ or outside $R$. On each geometric simplex, the intersection with $A$ is either the whole simplex or the empty set, and the same holds for $B$. The defining realization topology therefore makes both $A$ and $B$ closed, hence also open because they are complements.

The vertex $v$ belongs to $A$. If any vertex were unreachable, it would belong to the nonempty set $B$, giving a separation of a topological space of $|K|$. This contradicts connectedness. Therefore every vertex is reachable from $v$, and any two vertices can be joined by an edge path.

An abstract simplicial complex consists of a set $V$ of vertices of a simplicial complex and a family $K$ of finite nonempty subsets of $V$, called abstract simplexes, such that every singleton $\{v\}$ belongs to $K$ and every nonempty subset of a member of $K$ also belongs to $K$. One may include the empty set as an additional face by convention. A member with $k+1$ vertices is a $k$-simplex; its nonempty subsets are its faces of a simplex.

The geometric realization has the following explicit description using a barycentric coordinate system:Here the support is $\mathrm{supp}(t)=\{v:t(v)>0\}$. Each $\rho \in K$ gives a standard geometric simplexwith its usual Euclidean topology. Common faces of a simplex are identified using the same barycentric coordinate system.

Precisely, the topology on $|K|$ is defined by declaring $A\subseteq |K|$ closed if and only if $A\cap |\rho |$ is closed in $|\rho |$ for every $\rho \in K$. Equivalently, use the quotient topology on the disjoint union of all the standard geometric simplexes, identifying identical points of common faces of a simplex. For a finite simplicial complex, this agrees with the subspace topology in ${\mathbb{R}}^V$. For an infinite simplicial complex, the stated rule specifies the realization topology without presuming a finite-dimensional ambient space.

False. Take the metric space $X=\mathbb{R}\times \{0,1\}$ with the Euclidean distance and subspace topology inherited from ${\mathbb{R}}^2$. For every $t\ne 0$, identify $(t,0)$ with $(t,1)$; leave the two points $(0,0)$ and $(0,1)$ in separate equivalence classes. All equivalence classes have one or two points, so all are closed in this metric space.

Let $o_0,o_1$ be the two distinct images of the origins. If $U_0,U_1$ are any open neighbourhoods of $o_0,o_1$ in the quotient space, their preimages contain respectivelyfor some $\epsilon ,\delta >0$. Choose $0<t<\min (\epsilon ,\delta )$. The common image $p(t,0)=p(t,1)$ belongs to both $U_0$ and $U_1$. Therefore the origins have no disjoint open neighbourhoods and the quotient space is not Hausdorff. This is the line with two origins.