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For an annulus $a<r<1$ with $0<a<1$, these factors interpolate the two boundary coefficients of a Fourier harmonic of positive integer order $n$ in a harmonic function. They obey $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$, $K_n(a)=1$. Thus a mode with outer coefficient $v$ and inner coefficient $w$ has radial factor $v{H}_n+w{K}_n$.

To derive them, separation of variables in Laplace's equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, whose solutions are $r^n,r^{-n}$. Imposing the two Dirichlet boundary conditions on $R=\alpha r^n+\beta r^{-n}$ gives $\alpha +\beta =v$ and $a^n\alpha +a^{-n}\beta =w$. Solving yields $\alpha =(v-a^{n}w)/(1-a^{2n})$ and $\beta =a^n(w-a^{n}v)/(1-a^{2n})$, proving the formula. For the zero angular mode, the analogous factor is $v+(w-v)\log r/\log a$, since its radial solutions are $1$ and the logarithm $\log r$.

For boundary data $f(u)$ on the sphere $r=a$, with $u=\cos \theta$, define its Legendre polynomial coefficients by $f_{\ell }=(2\ell +1){\int }_{-1}^{1}f(u)P_{\ell }(u)du/2$. Separation of variables in the Laplace equation gives radial factors $r^{\ell }$ and $r^{-\ell -1}$. Regularity at the origin selects the first for the interior Dirichlet problem; decay at infinity selects the second for the exterior problem. Matching each boundary coefficient gives the displayed series. For continuous boundary data the resulting harmonic functions attain those data, with convergence interpreted through their harmonic extension rather than requiring uniform convergence of the unweighted boundary series. The maximum principle for harmonic functions proves uniqueness in the bounded interior and decaying exterior classes.

The capacitor gap aspect ratio compares the gap $d$ with a lateral plate width $W$. For rectangular plates of widths $W_x,W_z$, require both $d/W_x\ll 1$ and $d/W_z\ll 1$ for the leading parallel-plate capacitor approximation. Large area alone is insufficient: a long narrow strip can have large area and still have $d$ comparable with its smaller width. In addition, the observation point must be many gap widths from an edge. Reducing $d/W$ improves the interior approximation without making the edge itself disappear.

In a two-dimensional straight gap, write the electric potential as the infinite-plate solution plus a perturbation:The perturbation satisfies the Laplace equation and vanishes on the two ideal plates. Separation of variables in a semi-infinite strip, with $s$ the distance inward from its entrance, gives transverse modesThus edge disturbances have a natural penetration scale set by $d$. Modes forbidden by a particular symmetry have zero coefficients; this expression is an explanation of localization in an ideal strip, not a universal numerical correction to finite-plate capacitance. The outside boundary conditions determine the coefficients.

There is no single aspect-ratio cutoff guaranteeing an assigned accuracy: the requested observable, plate shape and thickness, surroundings, dielectric distribution, and guard geometry all matter. The small-gap approximation is local and is not uniform across an edge. It also does not justify extrapolating $C\simeq ϵA/d$ to arbitrarily large $d$, where finite electrical conductors and their external electric field dominate. At fixed $V$, formally shrinking an ideal gap makes the interior electric field grow like $1/d$; real electric currents across an imperfectly insulating gap, surface roughness and contact eventually invalidate that ideal limit.

A body ejecting mass at constant rate $Q>0$ and relative exhaust speed $V>0$, against quadratic drag $\alpha u^2$, has remaining mass $m=m_0-Qt$. A short-interval momentum balance gives $m\dot{u}=QV-\alpha u^2$. From rest, separation of variables givesFor final mass $m_f>0$, this formula applies until $t=(m_0-m_f)/Q$. The zero-drag limit is the rocket equation $u_f=V\log (m_0/m_f)$. Two dimensionless variables are $\alpha V/Q$ and $m_0/m_f$.

For a square of side $\ell$ initially at constant temperature $T_0$, held at zero boundary temperature for $t>0$, the heat equation $T_t=\kappa \Delta T$ with $\kappa >0$ has solutionThe double Fourier sine series of the initial constant gives the coefficients. The spatial modes are Dirichlet Laplacian eigenfunctions, so separation of variables gives their decay rates. At late times the $(1,1)$ mode dominates, giving the displayed profile; the next nonzero decay exponent is $10\kappa {\pi }^2/{\ell }^2$. At early times the series approaches the constant at each interior point and in L2 space, while the boundary remains at zero. An instantaneous change to the boundary condition therefore need not produce a solution continuous at the initial boundary in space-time.

For unit channel width, incompressible flow towards a corner withdrawal point has $\psi =0$ on the lower wall and $\psi =-m$ on the upper and left walls, with $\psi \to -my$ far downstream. For irrotational flow, separation of variables gives the displayed harmonic correction: the Fourier sine coefficients of $-m(1-y)$ are $-2m/(n\pi )$. Near the corner, $\psi \sim -2m\theta /\pi$ and radial velocity is $-2m/(\pi r)$, whose integrated inward flux across a quarter-circle is $m$.

On a scalar, the covariant wave operator is $\square \varphi =|g{|}^{-1/2}{\partial }_a(|g{|}^{1/2}{g}^{ab}{\partial }_b\varphi )$. For the metric signature $(-+++)$ it has the flat-space form $-{\partial }_t^2+{\nabla }^2$. This divergence expression is convenient for separation of variables and for integrating the Klein-Gordon equation by parts.

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.

On $0<x<L_x$, $0<y<L_y$ with positive side lengths and zero Dirichlet boundary conditions, separation of variables for $-\Delta u=\lambda u$ giveswith the displayed eigenvalues. The one-dimensional endpoint conditions exclude constant and hyperbolic factors and force the positive integer sine frequencies. The product Fourier sine series basis is complete in the square-integrable functions on the rectangle. Consequently an eigenfunction with prescribed $\lambda$ is a linear combination of exactly those modes with ${\lambda }_{mn}=\lambda$; if none match, only the zero solution exists. Distinct pairs with the same sum give an eigenvalue with geometric multiplicity greater than one. On a square of side $\pi$, for instance, $\lambda =25$ has precisely the pairs $(3,4)$ and $(4,3)$, so its eigenspace has dimension two.

For a harmonic function between $\tau =a$ and $\tau =b$, $0<a<b$, in bipolar coordinates, suppose $\Phi (\sigma ,a)=0$ and $\Phi (\sigma ,b)=F(\sigma )$. Write the periodic boundary values as a Fourier seriesSeparation of variables givesEach radial factor solves $M^{\prime \prime }-n^{2}M=0$ and has endpoint values zero and one; the mean mode solves $M^{\prime \prime }=0$. Sufficiently regular boundary data give the required Fourier series solution. Uniqueness follows from the maximum principle for a harmonic function with zero Dirichlet boundary conditions on both circles. The formula follows the coordinate values; $\tau =b$ is the geometrically smaller circle.

For $n_t=\mu \Delta n+\gamma n$ in a rectangular box with zero Dirichlet boundary conditions, separation of variables gives spatial eigenfunctions $\sin (p\pi x/L_x)\sin (q\pi y/L_y)\sin (r\pi z/L_z)$ and temporal rates $\gamma -\mu {\pi }^2(p^2/L_x^2+q^2/L_y^2+r^2/L_z^2)$. The largest rate is the displayed one. Every nonzero nonnegative initial density has positive projection onto this first eigenfunction, so its sign determines eventual exponential growth or decay; equality leaves a stationary first mode. At fixed volume, write $a_j=L_j^{-2}$: their product is fixed, and the arithmetic-geometric mean inequality minimizes their sum when all three are equal. Thus the cube maximizes the growth rate among rectangular boxes of fixed volume; making a box sufficiently thin suppresses growth.

Take a reservoir with constant horizontal cross-sectional area $A$, one small orifice of area $a$, no inflow, and constant stipulated discharge coefficient $C_d>0$. Let the initial head be $H>0$. For constant mass density $\rho$, the stored mass above the outlet level is $\rho Ah$ and outgoing mass flow rate is $\rho Q$. Mass conservation cancels $\rho$ and gives the initial value problemThe downward speed of the free surface is $Q/A$, smaller than the ideal jet speed by the factor $C_{d}a/A$. Thus a small opening relative to the reservoir supports the negligible-surface-velocity assumption.

For $h>0$, separation of variables or the chain rule gives $d\sqrt{h}/dt=-k/2$. Integrating yieldsThe model reaches zero head in finite time even though volumetric flow rate tends to zero. Extending the squared expression beyond $T$ would make head grow again and violate the original ordinary differential equation. The zero-head continuation enforces no further outflow. For an elevated side opening, zero head means the surface has reached that opening, leaving fluid below it; $T$ is not the time to remove every drop from such a tank.

For an original synthetic example, stipulateThe initial volume above outlet level is $AH=0.0250{\mathrm{m}}^3=25.0\mathrm{L}$. The initial ideal jet speed is $2.215\mathrm{m}{\mathrm{s}}^{-1}$, the initial volumetric flow rate is $Q_0=0.1373\mathrm{L}{\mathrm{s}}^{-1}$, and the predicted zero-head time is $T=364.1\mathrm{s}$. These are consequences of stipulated parameters, not experimental measurements. At $t=T/2=182.1\mathrm{s}$, head is $H/4=0.0625\mathrm{m}$, stored volume is $6.25\mathrm{L}$, and volumetric flow rate is $Q_0/2=0.06866\mathrm{L}{\mathrm{s}}^{-1}$. Half the initial volume has already left at $t=T(1-1/\sqrt{2})=106.7\mathrm{s}$.

An independent mass conservation check integrates the outgoing volumetric flow rate:Multiply both sides by mass density to obtain discharged mass. A dimensional analysis check gives $[k]={\mathrm{m}}^{1/2}{\mathrm{s}}^{-1}$, $[Q]={\mathrm{m}}^3{\mathrm{s}}^{-1}$ and $[T]=\mathrm{s}$. The triangular area under $Q(t)$ explains why the initial-rate estimate $AH/Q_0$ is only half the actual model time. Doubling $A$ doubles $T$; doubling $a$ or $C_d$ halves $T$, within the same assumptions.

This finite-time result is an idealized prediction. The constant $C_d$ and quasi-steady approximation need not survive very small head or the initial startup. Appreciable viscosity in a narrow outlet tube requires a different pressure balance, as in Poiseuille flow; appreciable surface tension can prevent zero head altogether.

A constant function $y(x)=c$ is an equilibrium solution of $y^{\prime }=F(x,y)$ on an interval when $F(x,c)=0$ throughout that interval. Dividing by a factor such as $y-c$ during separation of variables excludes this solution, so it must be checked separately. For an autonomous differential equation, the constant state is an equilibrium point.

For a unidirectional incompressible flow between stationary plates a distance $H>0$ apart, removing the driving pressure gradient leaves $u_t=\nu u_{yy}$ when the kinematic viscosity $\nu >0$ is constant. The no-slip boundary condition gives $u(0,t)=u(H,t)=0$. For an initial profile $g$ in L2 space, separation of variables and the Fourier sine series give the displayed solution, withEach normal mode decays at a rate proportional to $n^2$. The lowest mode present dominates at late times; when $b_1\ne 0$, its time constant is $H^2/(\nu {\pi }^2)$, consistent with the viscous diffusion time. The solution approaches the initial data in L2 space; compatible smooth profiles also give the familiar pointwise initial condition. This is a heat equation for momentum diffusion, rather than temperature.

A real harmonic function on the unit disk has the displayed expansion, convergent locally uniformly with its derivatives in the interior. If its continuous boundary data have Fourier coefficients $A_0,A_n,B_n$ in the convention $f(\theta )=A_0/2+{\sum }_{n\ge 1}(A_n\cos (n\theta )+B_n\sin (n\theta ))$, its harmonic extension has $a_0=A_0/2$, $a_n=A_n$, $b_n=B_n$.

To obtain the radial factors, separation of variables in the Laplace equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-\lambda R=0$ and ${\Theta }^{\prime \prime }+\lambda \Theta =0$. Angular periodicity restricts nonconstant modes to $\lambda =n^2$ for positive integers $n$. The radial factors are $r^n,r^{-n}$, with $1,\log r$ for the constant angular mode. Boundedness at the origin removes $r^{-n}$ and $\log r$. Each surviving factor $r^n\cos (n\theta )$ or $r^n\sin (n\theta )$ is a harmonic polynomial, the real or imaginary part of $(x+iy{)}^n$. The expansion can also be derived from the Poisson integral on the unit disk for the disk.

For $0<\theta <\gamma$ and $0<r<R$, a bounded harmonic function that vanishes on the straight edges and has outer boundary data $\phi$ has coefficientsThis follows by separation of variables and Fourier orthogonality. Negative radial powers are excluded by boundedness at the vertex. Reflection-symmetric data $\phi (\theta )=\phi (\gamma -\theta )$ have only odd Fourier modes. Knowing the resulting solution on the bisector determines all these odd coefficients by its radial power series. Without reflection symmetry, the even angular modes vanish on the bisector and cannot be recovered from that restriction alone.

On $0<x<a$, $0<y<b$, the bounded harmonic function with value $F$ on the open upper edge and zero on the other three open edges has the sine-hyperbolic series in the title. Separation of variables gives the modes, and the Fourier sine series of a constant selects only odd $n$. The series converges exponentially on compact subsets of the interior, justifying its derivatives in Laplace's equation.

For $F\ne 0$, the upper corners have incompatible one-sided boundary limits, so the solution cannot extend continuously to them. Assigning isolated corner values does not alter the bounded interior Dirichlet problem. For numerical evaluation, the stable mode ratio isThis avoids overflow from separately evaluating large hyperbolic sines.

On $0<x<a$, $0<y<b$, impose zero Dirichlet boundary conditions at $x=0,a$ and zero Neumann boundary conditions at $y=0,b$. Separation of variables for $T_t=\lambda \Delta T$ gives the displayed sine-cosine expansion. The zero cosine mode must be included: its norm is $b$, whereas the positive cosine modes have norm $b/2$. Thus $A_{n0}=2(ab{)}^{-1}\int \int T_0\sin (n\pi x/a)$ and $A_{nm}=4(ab{)}^{-1}\int \int T_0\sin (n\pi x/a)\cos (m\pi y/b)$ for $m\ge 1$. Expanding $x(a-x)$ gives sine coefficient $4a^2[1-(-1{)}^n]/(n^3{\pi }^3)$. Initial data $x(a-x){\sin }^2(2\pi y/b)$ therefore involve only $m=0,4$ and odd $n$. Their leading long-time mode is $4a^2\sin (\pi x/a)e^{-\lambda {\pi }^{2}t/a^2}/{\pi }^3$.

On $0<x<L$, let $u_t=\nu u_{xx}+G\sin (n\pi x/L)$, with $\nu >0$, zero Dirichlet boundary conditions, and zero initial data. Write $k_n=n\pi /L$. Its stationary solution is $u_s=G\sin (k_{n}x)/(\nu k_n^2)$. Subtracting it removes the forcing from the partial differential equation and leaves a heat equation with initial data $-u_s$.

Because the forcing is a single spatial eigenfunction, separation of variables gives the entire solution without mixing modes:Its integral is $G[1-(-1{)}^n](1-e^{-\nu k_n^{2}t})/(\nu k_n^3)$; even modes have zero signed net integral. Removing forcing from the evolution equation does not remove its amplitude from the initial data for the transient.

In plane polar coordinates, Laplace's equation isSeparation of variables gives radial powers $r^n,r^{-n}$ for positive angular modes and $1,\log r$ for the zero mode.