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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 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 $b>0$ and $-b\le y\le b$, the even quadratic polynomial $b^2-y^2$ has the displayed expansion in even Dirichlet eigenfunctions. Put $k_n=(2n+1)\pi /(2b)$. Their squared norm on $[-b,b]$ is $b$. Two integrations by parts, using $\cos (k_{n}b)=0$, giveDivision by $b$ gives the displayed Fourier coefficients. The cubic decay gives uniform convergence. The half-integer frequencies arise from requiring both evenness and zero values at the endpoints, rather than from a periodic boundary condition. This expansion is useful in rectangular duct flow.

Rectangular duct flow is steady flow of a Newtonian fluid down a straight rectangular pipe under a constant axial pressure gradient $-G$. For stationary walls and unidirectional flow, incompressibility eliminates axial dependence and the Navier-Stokes equation becomes a two-dimensional Poisson equation with zero Dirichlet boundary conditions. Subtracting a plane Poiseuille flow profile leaves a harmonic function that repairs the side-wall values using separation of variables. Uniqueness follows from Green's first identity: the difference of two solutions has zero boundary values and zero integral of the squared gradient, and hence vanishes.