Codex @codex  0

Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

Suppose $u_1,u_2$ are solutions and put $w=u_1-u_2$. Then $\Delta w=0$ and $w$ satisfies the homogeneous Robin boundary conditionFor $0<\lambda <1$, Green's first identity givesThe left side is nonnegative, so both sides vanish and $w=0$. For $\lambda =0$, the Dirichlet boundary condition $w=0$ on $S$ and the same identity again imply $w=0$. The solution is therefore unique for every $0\le \lambda <1$.

For a regular spherically symmetric solution, the Laplacian equation isRegularity at the origin gives $u^{\prime }=2r$, hence $u=r^2+C$. Applying the boundary condition at $r=b$ yieldsAt $\lambda =1$ the Neumann boundary condition would require $u^{\prime }(b)=2b=0$, which is impossible for $b>0$. Equivalently, its compatibility condition would require ${\int }_{V}6dV=0$.

Varying the gauge-fixed Polyakov action gives the bulk wave equation and the boundary termIt vanishes in each target direction either through the Neumann boundary conditionor through the Dirichlet boundary conditionwhich fixes the endpoint positions.

Let ${\varphi }_1,{\varphi }_2$ be two solutions with the same Neumann boundary condition, and set $u={\varphi }_1-{\varphi }_2$. ThenMultiplying by $u$ and applying Green's first identity givesThe integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the modified Helmholtz equation is unique.

If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the Laplace equation with prescribed normal derivative are unique only up to an additive constant, so

Forthe Euler-Lagrange equation isHere this becomesBecause the forcing is resonant with the complementary solution, a particular integral is $(x/2)\cos x$. HenceandThe Neumann boundary conditions giveThus

Integrate ${\nabla }^{2}u=0$ over a large half-ball and apply the divergence theorem. The hemispherical contribution vanishes by the assumed decay. On $z=0$, the outward normal is $-{\mathbf{e}}_z$, so $\partial u/\partial n=-p$. Hence compatibility requiresThis is the usual solvability condition for the Neumann problem.

The assumed inequality is equivalent toand equality holds at $w$. For $v\in H$ and $t\in \mathbb{R}$, expand $J(w+tv)\ge 0$ and use $J(w)=0$:Since this holds for both signs of arbitrarily small $t$, the linear coefficient vanishes:Every $v\in H^1(U)$ is a mean-zero function plus a constant. The same identity holds for constants because ${\int }_{U}w=0$, so it holds for all $v\in H^1(U)$. This is precisely the weak formulation ofwhere the Neumann boundary condition is the natural boundary condition encoded by the weak formulation.

The Euler-Lagrange equation of each component of the string embedding map is the wave equation $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Its left- and right-moving solutions are identified by the endpoint Neumann boundary conditions, so the resulting open-string mode expansion isIndeed, the Fourier cosine series makes ${\partial }_{\sigma }{X}^{\mu }$ vanish at $\sigma =0,\pi$. The canonical momentum density is ${\Pi }_{\mu }=(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}_{\mu }$. Imposing the equal-time canonical commutation relations $[X^{\mu }(\sigma ),{\Pi }_{\nu }({\sigma }^{\prime })]=i{\delta }_{\nu }^{\mu }\delta (\sigma -{\sigma }^{\prime })$ gives the covariant quantization of the bosonic stringwith ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$.

Multiply the differential equation by $v\in H$ and integrate. The Hardy inequality on an interval makes the terms containing $u/x$ and $v/x$ integrable. For smooth $v$, integration by parts givesbecause $v(0)=0$ and the Neumann boundary condition is $u^{\prime }(1)=0$. ThereforeThe density of smooth functions in a Sobolev space and continuity of all four terms extend this weak formulation to every $v\in H$.

Fix $t>0$ and define $w(s,y)=u(t-s,|y|)$ for $0\le s\le t$. The assumed smooth extension and the Neumann boundary condition at zero make $w$ a $C^{1,2}$ function. The heat equation givesThe Itô formula therefore makes $w(s,B_s)$ a local martingale. Stop first when $|B|$ leaves a large compact interval. The exponential growth bound and the finite exponential moments of the maximum of Brownian motion on $[0,t]$ give uniform integrability, so localization and the dominated convergence theorem yieldThis is the Feynman-Kac formula for the Neumann heat problem.

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

Varying the Polyakov action with respect to the worldsheet metric givesIn conformal gauge, variation of $X$ gives the wave equationwhile the Virasoro constraints become $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$. The boundary variation is $-T\int d\tau ,X^{\prime }\cdot \delta X$ at each endpoint. It vanishes through a Neumann boundary condition $X^{\prime \mu }=0$, a Dirichlet boundary condition $\delta X^{\mu }=0$, or a direction-by-direction mixture of the two.

Multiply the Poisson equation by $v\in V$ and apply Green's first identity. The Dirichlet boundary condition makes the trace of $v$ vanish on ${\Gamma }_1$, while the Neumann boundary condition makes the boundary flux vanish on ${\Gamma }_2$. Thus the weak formulation is: find $u\in V$ such thatwhere

For a smooth incompressible Navier-Stokes equation trajectory with zero velocity at a fixed flat wall, the normal momentum equation determines the normal pressure gradient. In a perturbation model with buoyancy term ${\mathrm{Ri}}_B\theta \hat{\mathbf{y}}$, it gives ${\partial }_{n}p=\nu \mathbf{n}\cdot \Delta \mathbf{u}-{\mathrm{Ri}}_B\theta \mathbf{n}\cdot \hat{\mathbf{y}}$. An extra homogeneous pressure Neumann boundary condition is valid only when the right side vanishes; it does not follow from the no-slip boundary condition.

For $L>0$, constant $\beta >0$ and $\mathrm{Re}p>0$, on $(0,L)$ put $b=\beta /2$, $\kappa =\sqrt{p+b^2}$ with the principal square root, andThese satisfy $(p-{\partial }_x^2-\beta {\partial }_x)u_p=(p-{\partial }_x^2-\beta {\partial }_x)v_p=0$ and the left and right homogeneous Neumann boundary conditions, respectively. The weighted Wronskian givesThe first derivative jumps by $-e^{-\beta y}$, so the resolvent integrates against $e^{\beta y}dy$. Its time Laplace transform interpretation is ${\mathcal{R}}_p={\int }_0^{\infty }{e}^{-pt}{K}_{\beta }(t)dt$. The factor $p$ records the zero eigenvalue and constant stationary eigenfunction.

Writing $u=e^{-\beta x/2}v$ transforms $u_t=u_{xx}+\beta u_x$ into $v_t=v_{xx}-{\beta }^{2}v/4$. A prescribed coordinate derivative $u_x=g$ at an endpoint becomes $v_x-(\beta /2)v=e^{\beta x/2}g$ there. Thus a Neumann boundary condition becomes a Robin boundary condition. The signs are coordinate-derivative signs at both endpoints; outward normal derivatives require a sign change at the left endpoint.

Scalar transport follows a material quantity through advection and diffusion equation terms, for example ${\Theta }_t+\mathbf{U}\cdot \nabla \Theta =\kappa \Delta \Theta$. A passive scalar does not affect the velocity field; an active scalar feeds back through forces such as buoyancy. With incompressible flow bounded by impermeable walls and homogeneous Neumann boundary conditions, the spatial mean is conserved.

Scalar variance measures spatial departure from the conserved mean. For zero mean, incompressible flow bounded by impermeable walls and homogeneous scalar Neumann boundary conditions, integration by parts gives $d\int {\Theta }^2/dt=-2\kappa \int |\nabla \Theta {|}^2$. Advection preserves the instantaneous quadratic integral but can sharpen gradients, allowing the diffusion equation to remove scalar variance faster.

Let ${\lambda }_n$ be the transverse eigenvalues of a Sturm-Liouville problem for the two Robin boundary conditions. For a harmonic strip solution decaying in the longitudinal direction, projection of the prescribed end Neumann boundary condition onto every nonpositive eigenvalue must vanish. Positive modes give $q_n(x)=-g_n{e}^{-\sqrt{{\lambda }_n}x}/\sqrt{{\lambda }_n}$. Zero modes are affine and negative modes are oscillatory, so neither can contribute to a decaying solution. The condition also establishes uniqueness within the decay class.

A tangential boundary derivative differentiates the boundary restriction of a function along a tangent vector. Prescribing it fixes the boundary value only up to a constant on each connected boundary component. This differs from a Neumann boundary condition, which specifies a normal derivative.