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For a uniform taut sheet in the small-displacement approximation, the two-dimensional wave equation isSeek a harmonic solution $Y(y)e^{i\omega t-i{k}_{x}x}$. The clamped edges impose Dirichlet boundary conditions $Y(0)=Y(b)=0$. Hence $Y^{\prime \prime }+k_y^{2}Y=0$ has nontrivial solutions only forTaking the real part gives the stated sinusoidal transverse profile. There is no nonzero $n=0$ mode satisfying both clamped boundaries.

Substitution gives the dispersion relationThe three branches have cutoffs ${\omega }_{c,n}=n\pi v/b$ at $k_x=0$ and approach $\omega =v|k_x|$ at large $|k_x|$. Only frequencies above a branch's cutoff give traveling waves along an infinitely long guide; below it, the axial dependence is evanescent. For a forward propagating mode,The phase velocity exceeds $v$, while the group velocity is smaller: the carrier phase speed is not the signal speed.

Here the first two cutoff frequencies are $50$ and $100\mathrm{Hz}$, so both modes propagate at $200\mathrm{Hz}$. With $\omega =400\pi {\mathrm{s}}^{-1}$,Each transverse mode independently acquires its axial phase. The second mode first changes sign relative to the first when $(k_1-k_2)x_1=\pi$, givingAt that distance a common temporal phase is $\varphi =-k_1{x}_1$ modulo $2\pi$; the second term is negative relative to the first for every $y$ and every time, not just at one selected instant.

In an optical fibre, different guided modes have different group velocities, so an input pulse becomes a sum of differently delayed pulses. This intermodal dispersion broadens pulses and can make adjacent digital symbols overlap. A single-mode optical fiber removes this particular mode-delay mechanism; a graded-index optical fiber can reduce it in a multimode design. Single-mode propagation still permits wavelength-dependent group delay, so it does not remove every kind of optical dispersion.

The fixed edge at $y=0$ is automatically satisfied by $\sin (qy)$. At $y=a$, the displacement must vanish at every $x,t$, so $\sin (qa)=0$. Thus the allowed transverse normal modes haveThe case $n=0$ gives zero displacement for this sine solution and is not a physical propagating mode. Negative $n$ only reverses the amplitude sign and does not create a distinct transverse mode. This discrete transverse spectrum is a consequence of the Dirichlet boundary conditions, rather than quantization of the longitudinal wavenumber in an indefinitely long guide.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

Specifying potential on the complete boundary gives a Dirichlet boundary condition and a unique harmonic solution. Specifying its normal derivative gives a Neumann boundary condition, with the zero-net-flux compatibility condition and an undetermined additive constant. Mixed conditions on complementary portions also give uniqueness when a potential reference is fixed. In an exterior domain, conditions at infinity are needed. These statements follow by applying Green's first identity to the difference of two solutions. A metal in electrostatic equilibrium is an equipotential: $\Phi =V_0$, tangential electric field zero, and normal field $E_n=\sigma /{ϵ}_0$ just outside a vacuum interface.

For the conducting-wedge potential expansion, put $u=\Phi -V_0$ in the empty sector $0<\theta <\alpha$. For $u=r^{\nu }\Psi (\theta )$, Laplace's equation gives ${\Psi }^{\prime \prime }+{\nu }^2\Psi =0$. Vanishing values on both faces require $\Psi \propto \sin (n\pi \theta /\alpha )$, henceBounded potential at the corner sets $b_n=0$. The coefficients still require outer boundary data; the metal potential alone cannot select them. For a generic nonzero $a_1$, put $p=\pi /\alpha$. Differentiation givesIf $a_1=0$, the first nonzero higher harmonic replaces this leading term. For $\alpha =\pi /2$, $u\simeq 2a_{1}xy$, so $\mathbf{E}=-2a_1(y,x)$ and field lines satisfy $x^2-y^2=\mathrm{constant}$, meeting the metal faces normally and tending to zero field at the corner. For $\alpha =3\pi /2$, the electrostatic corner-field exponent is $-1/3$. A real tip's rounding cuts off this divergence; breakdown or field emission can limit the attainable field.

For the usual hollow rectangular electromagnetic waveguide, let the propagation axis be $z$ and the interior be $0<x<a$, $0<y<b$. A transverse magnetic polarization has $B_z=0$ and $E_z\ne 0$. At a perfect conductor wall the tangential electric field vanishes. Since $E_z$ is tangential to all four side walls, it satisfies a homogeneous Dirichlet boundary condition around the rectangle.

Separating the transverse Helmholtz equation therefore gives the transverse magnetic mode of a rectangular waveguideA zero value of either index makes $E_z$ identically zero, rather than producing a nonzero field uniform along that direction. The transverse electric and magnetic fields of this TM family are obtained from transverse derivatives of $E_z$, so they also vanish. Thus both TM indices must be nonzero: neither ${\mathrm{TM}}_{n0}$ nor ${\mathrm{TM}}_{0m}$ exists in this rectangular hollow-guide model. In contrast, transverse electric modes obey a Neumann boundary condition on $B_z$ and may have one zero index. The claim is about this waveguide geometry, not all possible guides.

Inside the infinite square well, the time-independent Schrödinger equation and Dirichlet boundary conditions areand the wavefunction vanishes outside the interval. For a nonzero state, integration by parts givesThe strict inequality follows because a function with zero derivative and zero endpoint values is zero. Put $k=\sqrt{2mE}/\hslash$. The general solution is $A\sin kx+B\cos kx$; the first boundary condition gives $B=0$, and the second gives $ka=n\pi$, $n=1,2,\dots$. Since ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$, the normalized energy eigenstates and eigenvalues areThe normalization fixes the magnitude of the coefficient; its arbitrary global phase has been chosen real and positive.

At the midpoint the second sine vanishes, whereas the derivative of the first sine vanishes. Thus the wavefunction and its spatial derivative simplify toSubstitution in the probability current givesThe continuity equation, integrated over the left half, gives $P_{\mathrm{L}}^{\prime }(t)=j(0,t)-j(a/2,t)$. The Dirichlet boundary condition makes $j(0,t)=0$, and the preceding probability formula givesPositive midpoint current removes probability from the left half. This is precisely the midpoint current of a two-mode box state.

The operator $R$ is the midpoint parity operator about $a/2$. The angle-subtraction identity givesConsequently $HR{\psi }_n=RH{\psi }_n=E_n(-1{)}^{n+1}{\psi }_n$. The sine functions form a complete orthonormal basis of $L^2(0,a)$, so the simultaneous diagonal action implies commutation on the Hamiltonian operator's operator domain. More directly, $R$ preserves the Dirichlet boundary conditions and $(R\psi {)}^{\prime \prime }(x)={\psi }^{\prime \prime }(a-x)$; hence $HR\psi =RH\psi$ whenever $H\psi$ is defined. This checks the domain as well as the differential expressions.

For arbitrary $f,g\in L^2(0,a)$, use $y=a-x$ in the inner product:Also $\Vert Rf\Vert =\Vert f\Vert$, so $R$ is bounded and defined everywhere. The identity therefore proves $\boxed{R^{\ast }=R}$, with no extra domain restriction. In particular $R$ is a self-adjoint operator; $R^2=I$ also makes it a unitary operator.

Inside the infinite square well, the Schrödinger equation and Dirichlet boundary conditions areThe wavefunction vanishes outside the box. These walls constrain its values, not its first derivatives: requiring both to vanish at an endpoint would eliminate all the nonzero stationary solutions.

For a nonzero separated solution, separation of variables givesThe last equation is the time-independent Schrödinger equation. Its energy eigenvalue is real. Multiplication by ${\psi }^{\ast }$ and integration by parts, with the endpoint values zero, givesEquality would force $\psi$ to be constant and hence zero, so a nonzero energy eigenstate has $E>0$. Writing $k=\sqrt{2mE}/\hslash$, the general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$; the right gives $\sin ka=0$. Negative integers reproduce the same functions up to a constant, and $n=0$ gives the zero function, leaving $k=n\pi /a$ for $n=1,2,\dots$.

Finally ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$. Thus wavefunction normalization with positive coefficient givesAn arbitrary constant global phase has no effect on the quantum state.

Write $k=\pi /a$, $A=\sqrt{2/a}$ and ${\Psi }_0=A{e}^{ikx}\sin kx$. Its derivative isIntegration by parts with the Dirichlet boundary conditions gives the energy expectationIt is constant in time, since the time-independent Hamiltonian operator has fixed energy measurement probabilities: $⟨H{⟩}_t={\sum }_n{E}_n|c_n{|}^2$.

For the midpoint parity operator, $R{\Psi }_0=-A{e}^{-ikx}\sin kx$. Hence, again putting $u=kx$,Indeed the imaginary part integrates to zero by reflection and ${\int }_0^{\pi }\cos 2u{\sin }^{2}udu=-\pi /4$. This expectation value is also constant in time, because the bounded self-adjoint operator $R$ commutes with $H$ and hence with its time-evolution operator. Equivalently, $⟨R{⟩}_t={\sum }_n(-1{)}^{n+1}|c_n{|}^2$.

For the momentum operator, direct evaluation givesThe real sine-cosine integral vanishes. This expectation value is not constant in time: the infinite square well walls reverse motion. One can prove the change exactly without approximating the wavefunction. Atthe energy eigenstate expansion and $R{\psi }_n=(-1{)}^{n+1}{\psi }_n$ give $\Psi (t_{\ast })=-R{\Psi }_0$. Differentiation shows $PR=-RP$, soThis half revival in an infinite square well is a concrete counterexample to momentum conservation. The sign from $-R$ is only a global phase.

The apparent interior identity $[H,P]=0$ cannot justify conservation here: the momentum operator does not preserve the Dirichlet boundary conditions of the Hamiltonian operator's operator domain. For sufficiently regular states, retaining the boundary terms in Ehrenfest theorem instead gives the boundary force in an infinite square well,In the strictly confined interval, $-i\hslash {\partial }_x$ with both endpoint values zero is symmetric but not itself a self-adjoint operator. The requested integral is the conventional mean momentum; it also equals the full-line momentum mean of the zero-extended state, as measured upon releasing the walls. This domain distinction explains why the formal interior commutator misses the wall force.

Multiply the equation for $y_n$ by $y_m$ and that for $y_m$ by $y_n$, then subtract. Integrating the resulting identity yieldsAll endpoint terms vanish because both functions satisfy the zero Dirichlet boundary conditions. Since the multipliers differ,This is the weighted orthogonality of Sturm-Liouville eigenfunctions.

Inside the infinite square well, $V=0$, and the time-independent Schrödinger equation isFor $E>0$, write $k^2=2mE/{\hslash }^2$ and $\psi =A\sin kx+B\cos kx$. The first Dirichlet boundary condition gives $B=0$, and the second requires $kL=n\pi$ for a nonzero state. The case $n=0$ is identically zero; negative $n$ only changes an overall quantum phase. For $E=0$ a linear solution satisfying both walls is zero, and for $E<0$ the hyperbolic solutions likewise cannot satisfy both walls nontrivially.

Wavefunction normalization, ${\int }_0^L|\psi {|}^{2}dx=1$, fixes $|A|=\sqrt{2/L}$. The orthonormal energy eigenstates and energy eigenvalues are thereforeEach state vanishes outside the well. Its stationary time-dependent form is ${\Psi }_n(x,t)={\psi }_n(x)e^{-i{E}_{n}t/\hslash }$. General states are normalized superpositions of these modes, rather than additional single energy eigenstates.

Multiply each time-sliced path weight by $\exp [-{\sum }_{j}V(x_j)\Delta t]$. A nonnegative finite $V$ is a local killing rate: the Feynman-Kac weight attenuates paths in proportion to the time spent in regions where $V>0$. Let that rate tend to infinity outside the allowed interval. Paths that exit are removed, leaving Brownian motion killed at its first boundary hit.

Consequently the required kernel is the Gaussian path measure restricted to surviving paths, equivalentlywith the infinite-barrier interpretation stated above. It solves ${\partial }_{t}K=D{\partial }_x^{2}K$ in the interior with absorbing Dirichlet boundary conditions at $\pm L$. A reflecting boundary condition for a diffusion would describe a different process.

Strictly, this is a transition probability density in the final position, jointly with survival; a probability for an interval of final positions includes an integral over $x_f$. Its integral over the whole allowed interval is at most one. It is not the normalized endpoint density conditional on survival, which would require division by the survival probability. In these diffusion units $V$ and the eigenvalues of $H$ have units of inverse time.

Separate the Klein-Gordon equation as $\varphi (t,x)=e^{\pm i\omega t}X(x)$. The spatial equation is $X^{\prime \prime }+k^{2}X=0$, with ${\omega }^2=k^2+m^2$. The first Dirichlet boundary condition removes the cosine solution, and the second requires $\sin (kL)=0$. ThusThere is no nonzero $n=0$ Dirichlet mode. The real field is built from both complex-conjugate time dependences, with mode functionsThe sine functions satisfy ${\int }_0^L\sin (k_{n}x)\sin (k_{m}x)dx=(L/2){\delta }_{nm}$. Consequently the factor $1/\sqrt{{\omega }_{n}L}$ gives unit Klein-Gordon inner product for each positive-frequency mode and the canonical equal-time field commutator after quantization. A general solution is a superposition of these modes, rather than a claim that a single mode exhausts all solutions.

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.

The heat equation and its homogeneous Dirichlet boundary conditions are linear, so superposition of the separated modes givesThe Fourier sine basis represents the initial profile on the interior of the interval. At $t=0$ its Fourier sine series must equal $x(1-x)$. Multiplying by $\sin (m\pi x/L)$, integrating from $0$ to $L$, and using orthogonality givesTherefore the required Fourier sine coefficients areFor every $t>0$, the exponential factors ensure convergence of the differentiated series, and every term satisfies the partial differential equation and the boundary conditions. The Fourier sine series recovers the smooth initial profile at every $0<x<L$ as $t\downarrow 0$.

The scan explicitly prints $x(1-x)$ on $(0,L)$; it must not silently become $x(L-x)$. For $L\ne 1$, the interior profile tends to $L(1-L)\ne 0$ at the right endpoint, whereas the prescribed boundary condition is zero for $t>0$. The solution is therefore not continuous at the corner $(L,0)$, although it satisfies every condition on the printed open ranges. A solution continuous on the closed space-time domain with these data would require $L=1$.^[ref]

For a nonzero separable solution $u=XT$, separation of variables givesThese Dirichlet boundary conditions follow because $T$ is not identically zero. If $\kappa =0$, the affine function $X=A+Bx$ obeys both conditions only when $A=B=0$. If $\kappa =-{\rho }^2<0$, $X=A\cosh (\rho x)+B\sinh (\rho x)$ also has only the zero solution under both conditions.

For $\kappa ={\omega }^2>0$, $X=A\cos (\omega x)+B\sin (\omega x)$. The first boundary condition forces $A=0$, and the second forces $\omega L=n\pi$ for a nonzero solution. Hence all nonzero spatial modes areThe identically zero function is also a solution. Allowing $B_n=0$ includes it in the displayed family. Here $L>0$ is the interval length and $n$ labels the eigenvalues of the operator $-d^2/d{x}^2$ with these boundary conditions.

For $t>0$, fixed zero endpoint temperatures are homogeneous Dirichlet boundary conditions:An initial condition is also needed to determine a particular solution of the heat equation.

Put $k_n=n\pi /L$ and ${\lambda }_n=\kappa k_n^2$. Direct differentiation of $T_n=A\sin (k_{n}x)e^{-{\lambda }_{n}t}$ givesso ${\partial }_t{T}_n=\kappa {\partial }_x^2{T}_n$, the heat equation. At $x=0$, the sine factor vanishes; at $x=L$, it is $\sin (n\pi )=0$ for integer $n$. Thus the homogeneous Dirichlet boundary conditions hold. The nonzero, nonredundant spatial modes use $n=1,2,\dots$; $n=0$ is the zero function and negative $n$ only change the amplitude sign.