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For $-\Delta \Phi =\lambda \Phi$ on the square, separation of variables gives Dirichlet eigenfunctions $\sin (jx)\sin (ky)$ with $j,k\ge 1$, and Neumann eigenfunctions $\cos (jx)\cos (ky)$ with $j,k\ge 0$. The eigenvalues in both cases are $j^2+k^2$, with multiplicity equal to the number of ordered admissible pairs. In particular, an arbitrary expression $\lambda =m^2+p^2$ with integers $m,p$ need not be a Dirichlet eigenvalue when one index is zero. The Neumann zero eigenvalue has the constant eigenfunction, whereas the Dirichlet zero problem has only the zero solution by Dirichlet uniqueness for the Poisson equation.

A Dirichlet eigenfunction is an eigenfunction of a differential operator satisfying a homogeneous Dirichlet boundary condition. The term applies to more general boundary-value operators as well as the Dirichlet Laplacian.

Put $t=\log x$ and $Y(t)=y(e^t)$. Then $x{y}^{\prime \prime }+y^{\prime }=Y^{\prime \prime }/x$, so the eigenvalue problem reduces toA nonzero Dirichlet eigenfunction therefore has $4+\lambda =n^2$ for an integer $n\ge 1$. The complete family isThe alternatives $4+\lambda \le 0$ give only the zero solution under both endpoint conditions.

The operator itself is self-adjoint with the given unweighted inner product, on the usual regular Dirichlet domain. Indeed $My=(x{y}^{\prime }{)}^{\prime }+4y/x$, and integration by parts givesfor functions satisfying the endpoint conditions. This is a regular Sturm-Liouville operator with positive leading coefficient $x$, whose Dirichlet realization is self-adjoint. The displayed generalized eigenproblem contains the additional weight $1/x$; its eigenfunctions are accordingly orthogonal with respect to $\int fg/x$, which is separate from the self-adjointness assertion about $M$.

Use the real weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}wuvdx$ and the energy bilinear form $B(u,v)={\int }_a^b(p{u}^{\prime }{v}^{\prime }+quv)dx$. Integration by parts, the endpoint conditions, and the Sturm-Liouville problem giveThus orthonormality in the stated expansion givesand hence the Rayleigh quotientThe expansion is understood for nonzero admissible functions with finite norm and energy.

For a general endpoint-preserving variation $y+\epsilon \eta$, explicitlysoThe fundamental lemma of the calculus of variations shows that stationarity is exactly the eigenvalue equation with $\lambda =\Lambda$. Conversely an eigenfunction has vanishing first variation and quotient equal to its eigenvalue. Completeness gives the same weighted-average formula for any nonzero admissible expansion, even one with zero ground-state coefficient. Since every ${\lambda }_i\ge {\lambda }_0$,This is the Rayleigh-Ritz variational principle.

For the interval problem, the normalized Dirichlet eigenfunctions and their eigenvalues, with the paper's zero-based indexing, areFor $u=x(1-x)$, direct integration gives $F={\int }_0^1(1-2x{)}^{2}dx=1/3$ and $G={\int }_0^1{x}^2(1-x{)}^{2}dx=1/30$. ThereforeThe sine expansion and Rayleigh quotient of a parabolic Dirichlet trial follow from the coefficient integralConsequently the requested expansion hasEquivalently $u=(8/{\pi }^3){\sum }_{n\ge 1,n\text{ odd}}\sin (n\pi x)/n^3$. In particular $c_1=0$: the first excited eigenfunction $\sqrt{2}\sin (2\pi x)$ is antisymmetric about $1/2$, whereas the trial function is symmetric. Since $10-{\pi }^2>0$, the final ratio is exactlyThe first nonzero correction is $c_2=1/27$, not $c_1$. Reindexing only the nonzero terms would change the meaning of the requested ratio and is not used here.

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 $b>0$ 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.

On $(0,1)$, integration by parts gives the displayed Fourier sine series. For normalized Dirichlet eigenfunctions $y_i=\sqrt{2}\sin [(i+1)\pi x]$, indexed from $i=0$, it has the form $A[y_0+{\sum }_{i\ge 1}{c}_i{y}_i]$, with $A=4\sqrt{2}/{\pi }^3$ and $c_i=[1-(-1{)}^{i+1}]/[2(i+1{)}^3]$. Its Rayleigh quotient for $-d^2/d{x}^2$ is $10$, since ${\int }_0^1(u^{\prime }{)}^{2}dx=1/3$ and ${\int }_0^1{u}^{2}dx=1/30$. Reflection symmetry forces $c_1=0$, because that first excited eigenfunction is antisymmetric about $1/2$. Thus the first excited state's contribution to the error is exactly zero; the first nonzero correction is $c_2=1/27$.