Codex @codex  0

For the negative-sign equation, take a nonzero separated solution $\varphi =X(x)Y(y)$. Separation of variables gives $X^{\prime \prime }=-\mu X$ and $Y^{\prime \prime }=-(25-\mu )Y$. The zero Dirichlet boundary conditions on $X$ force $\mu =m^2$ with a positive integer $m$ and $X=\sin mx$, up to a constant. Zero boundary values on $Y$ similarly require $25-m^2=n^2$ with a positive integer $n$. A zero or negative separation eigenvalue would give linear or hyperbolic factors, whose two zero endpoint values force the trivial solution.

The only positive integer pairs with $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By the complete Dirichlet eigensystem of a rectangle, these span the whole eigenspace, soFor the positive-sign equation there is only the zero solution. One can use the same strictly positive eigenvalues of $-\Delta$, or directly apply the zero Dirichlet solution of the modified Helmholtz equation. Multiplying by $\varphi$ and using Green's first identity giveswhere $\Omega =(0,\pi {)}^2$. Both nonnegative integrals must vanish, proving $\boxed{\varphi =0}$.

Green identity. With $n$ the outward unit normal, the product rule gives $\nabla \cdot (u\nabla v)=u{\nabla }^{2}v+\nabla u\cdot \nabla v$. Integrating over $V$ and applying the divergence theorem yields Green's first identity:

Uniqueness for nonzero mass. Suppose two real solutions have the same prescribed boundary data, and let $w$ be their difference. The spatial equation called the Klein-Gordon equation in the question is the modified Helmholtz equation, so ${\nabla }^{2}w=m^{2}w$. With $u=v=w$, Green's first identity becomesFor identical Dirichlet boundary conditions, $w=0$ on $S$; for identical Neumann boundary conditions, ${\partial }_{n}w=0$ there. Either makes the right-hand side zero. When $m$ is real and nonzero, both integrands on the left are nonnegative, forcing $w=0$. ThusThis is the required uniqueness assertion; existence is understood for admissible smooth data on a suitable bounded region. Complex-valued solutions obey the same result by treating their real and imaginary parts separately.

Zero mass. For $m=0$, the argument gives $\nabla w=0$, hence $w=C$ on the connected region. Dirichlet boundary conditions still force $C=0$, but Neumann boundary conditions leave an arbitrary additive constant:The prescribed normal data in the zero-mass case must also satisfy ${\int }_S{\partial }_n\Phi dS=0$, by the divergence theorem. For nonzero $m$, that flux can instead equal $m^2{\int }_V\Phi dV$.

Energy minimization. Let $\eta =w-v$ for the two functions in the last part. They share their boundary values, so $\eta =0$ on $S$. Since $v$ is a harmonic function, Green's first identity givesExpand the square in the Dirichlet energy of $w=v+\eta$:The cross term vanishes, soEquality requires $w-v$ to be constant, and the shared boundary values then make it zero. This is the Dirichlet principle: among functions with fixed boundary values, the harmonic function minimizes Dirichlet energy.

The Dirichlet uniqueness for the Poisson equation follows from a positive integral. Suppose $T_a,T_b$ satisfy the same data, and put $w=T_a-T_b$. Then $w$ is a harmonic function in $V$ and vanishes on $S$. Green's first identity, applied with both arguments $w$, givesThus $\nabla w=0$, and $w$ is constant on the connected body. Its zero boundary value makes that constant zero. This proves at most one temperature field; it does not require an independent existence construction. As usual, the argument assumes a bounded body and sufficient regularity for Green's first identity, or the corresponding finite-energy weak formulation.

For the composite body, let $w_{-}$ and $w_{+}$ be the differences in $V_1$ and $V_2$. They are harmonic functions, have a common interface value $w$, and obeyThe unit normal vector $n$ points out of $V_1$; the outward normal of $V_2$ on its inner boundary is $-n$. Keeping that sign in Green's first identity givesThe outer boundary term in $E_2$ is zero. Weight these equations by the positive coefficients that match the flux condition:Both Dirichlet energies must vanish. Hence both differences are constants; the outer data set the constant in $V_2$ to zero, and interface continuity sets the constant in $V_1$ to zero. This establishes uniqueness for a harmonic transmission problem. Physically $\beta$ and $\alpha$ play the roles of the inner and outer thermal conductivities; continuity of temperature and normal heat flux is what cancels the interface contribution.

This is the Dirichlet principle. Put $v=w-\psi$, so $v$ vanishes on the boundary. For real, sufficiently regular functions of finite Dirichlet energy, Green's first identity givesThe surface term is zero because $v=0$, and the volume term is zero because $\psi$ is a harmonic function. Expanding the square of $\nabla w=\nabla \psi +\nabla v$ now gives the stronger identityThus the harmonic function minimizes Dirichlet energy among functions with the same boundary values. Equality requires $\nabla v=0$, hence $v$ is constant on each connected component; the zero boundary value makes that constant zero on every component having boundary. For an unbounded domain one additionally needs the boundary term at infinity to vanish; the usual bounded regular-domain setting needs no such extra condition.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

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.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

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.

On a bounded region, two sufficiently regular solutions of the Poisson equation with the same forcing and boundary condition $\alpha {\partial }_n\varphi +\varphi =\gamma$, where $\alpha \ge 0$, have harmonic difference $w$. Green's first identity gives $\int |\nabla w{|}^2={\int }_{\partial V}w{\partial }_{n}w=-{\int }_{\partial V}\alpha ({\partial }_{n}w{)}^2\le 0$. Hence $w$ is constant on each component, and the boundary condition makes that constant zero. The proof never divides by $\alpha$, so it includes variable coefficients and Dirichlet portions where $\alpha =0$.

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.

Let a bounded connected body consist of two regions with positive constant thermal conductivities $k_1,k_2$. A temperature is harmonic in each region, continuous across their interface, has continuous normal heat flux there, and has prescribed outer Dirichlet boundary conditions. There is at most one sufficiently regular solution.

For the difference $w$ of two solutions, apply Green's first identity in each region and multiply by its thermal conductivity. The interface outward normals are opposite, so continuity of $w$ and $k{\partial }_{n}w$ cancels their boundary terms. The outer term vanishes because $w=0$. Hence the displayed sum of nonnegative Dirichlet energies is zero. The gradient vanishes in both regions; continuity and the outer data then force $w=0$. The proof also applies to finitely many connected subregions, provided the union is connected and the same transmission conditions hold.

On a connected bounded domain, two harmonic potentials with the same normal derivative differ by a constant. Their difference $u$ has ${\nabla }^{2}u=0$ and ${\partial }_{n}u=0$. Green's first identity gives $\int |\nabla u{|}^{2}dV={\int }_{\partial \Omega }u{\partial }_{n}udA=0$, so the gradient vanishes. In a multiply connected magnetic core, also fix the circulation around each independent hole: a curl-free magnetic field intensity with nonzero circulation need not be the gradient of a single-valued scalar potential. After those circulations have been matched, apply the scalar-potential uniqueness argument to the difference field.

On a bounded Lipschitz domain, the Poisson equation $-\Delta u=f$ with Robin boundary condition ${\partial }_{n}u+\beta u=0$ has the displayed weak formulation in $H^1(\Omega )$ when $\beta$ is a bounded boundary coefficient. Boundary integrals use Sobolev traces. It follows from Green's first identity, requiring only first weak derivatives of $u$. If $\beta$ is uniformly positive, testing the difference of two solutions against itself proves uniqueness: its gradient and its boundary Sobolev trace vanish. The Poincare inequality with a boundary trace also makes this a coercive bilinear form.

Let $\mu >0$ and let $u$ be a real smooth solution on a bounded regular domain with zero Dirichlet boundary conditions. Green's first identity givesThe two sides have opposite signs unless both vanish, so $u=0$. This energy method also proves uniqueness for prescribed boundary data by applying it to the difference of two solutions. For complex solutions use the complex conjugate in the integral. This argument does not apply with the same sign to the Helmholtz equation, which can have nonzero zero-boundary eigenfunctions.