Codex @codex  0

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

Use the intended finite-dimensional irreducible highest weight module $A^{\varphi }$, with dominant integral highest weight $\varphi$. Its highest weight multiplicity is one. An arbitrary direct sum having that same largest representation weight would not in general satisfy this formula. Set $f={\pi }_{A^{\varphi }}$ and use the corrected denominator $D$ of part (a).

Restriction to each simple root ${\mathfrak{sl}}_2$ decomposes into symmetric finite representation weight strings, so $f$ is Weyl-invariant. Consequently $Df$ is Weyl-antisymmetric. We prove that no lower Weyl alternants occur, rather than assuming the desired character formula.

First establish the denominator identity itself. The finite sum $D$ has support in $\rho -Q_{+}$ and coefficient one at $\rho$. Antisymmetry kills all singular representation weight coefficients, since a reflection fixing the representation weight reverses their sign. Every remaining Weyl orbit has a unique regular dominant representative $\eta$. If it occurs in $D$, then $\eta \le \rho$. Since $\eta$ is integral and regular dominant, $\eta -\rho$ is dominant integral; it also belongs to $-Q_{+}$. A dominant vector $v=-\sum n_i{\alpha }_i$ satisfies $(v,v)=-\sum n_i(v,{\alpha }_i)\le 0$, so it must be zero. Thus the only possible orbit is $W\rho$. The leading coefficient givesFor the formal Laplacian $\mathcal{D}(e^{\lambda })=\Vert \lambda {\Vert }^2{e}^{\lambda }$, this implies $\mathcal{D}D=\Vert \rho {\Vert }^{2}D$.

Now choose root vectors dual under an invariant form, so $[E_{\alpha },E_{-\alpha }]=H_{\alpha }$ with $H_{\alpha }$ corresponding to the vector $\alpha$, and take the quadratic Casimirfor an orthonormal Cartan basis. On a highest vector its scalar is $c_{\varphi }=\Vert \varphi +\rho {\Vert }^2-\Vert \rho {\Vert }^2$, hence it has this scalar throughout the cyclic highest weight module.

Here is the computation of the weighted trace of its radial identity. For a positive root, let $X$ be the formal weighted trace of $E_{-\alpha }{E}_{\alpha }$ and $Y$ that of $E_{\alpha }{E}_{-\alpha }$. Cyclicity of trace and representation weight shifting give $Y=e^{\alpha }X$, while their difference is the weighted trace of $H_{\alpha }$, namely $(\alpha ,\nabla f)$. ThusTaking the trace of $C$ yieldsThe product expression for $D$ givesUsing $2\rho ={\sum }_{\alpha >0}\alpha$ and the product rule for the formal Laplacian therefore givesHence every representation weight in the support of $Df$ has norm $\Vert \varphi +\rho \Vert$. Its support is also contained in $\varphi +\rho -Q_{+}$, and it is antisymmetric. Any occurring regular dominant weight $\eta$ thus satisfies $\eta \le \varphi +\rho$. If the inequality were strict, thenbecause the first factor is a nonzero nonnegative sum of simple roots and the second is strictly dominant. This contradicts the equal-norm condition. Only the orbit $W(\varphi +\rho )$ remains. Its leading coefficient is one, proving

For a stationary nonrelativistic state, $\boxed{[-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})]\psi =E\psi }$. The wavefunction $\psi$ is a probability amplitude, $m$ the particle mass and $E$ its energy eigenvalue; the potential energy $V$ is measured in the same energy units. In one dimension the Laplacian becomes $d^2/d{x}^2$.

For a radial function $f(r)$ in three dimensions, the Laplacian is $f^{\prime \prime }+2f^{\prime }/r$, while the imposed radial drift contributes $-\alpha f^{\prime }/r$. Thus the radial generator of an SDE diffusion isThe hitting probability for the inner sphere is harmonic for this diffusion generator in the shell, with value one on the inner sphere and zero on the outer sphere. At $\alpha =1$, the boundary value problem is thereforeIntegrating $(r{p}^{\prime }{)}^{\prime }=0$ gives $p(r)=A\log r+B$. The two boundary conditions determineBoth arguments of the logarithms are dimensionless. The result decreases from one to zero across the shell and lies strictly between zero and one for $R_1<r_0<R_2$.

For a direct probability interpretation, stop at the first contact $T$ with either sphere. The stopped process $p(r(t\wedge T))$ is a bounded martingale by Itô's formula. The shell has positive inner radius, so its drift is bounded and its diffusion is nondegenerate; exit occurs almost surely. Optional stopping and dominated convergence therefore giveThe logarithm appears because the inward drift cancels one of the two geometric $f^{\prime }/r$ terms in the three-dimensional radial Laplacian, leaving the same radial diffusion generator as a planar Brownian motion with diffusion coefficient one.

With constant $\nu$, the bulk action is invariant under time translations, translations in the two spatial coordinates, and rotations or reflections within the substrate plane. The Laplacian has no preferred direction, giving the planar $O(2)$ symmetry. It also has scalar-field shift symmetry $h\mapsto h+a$, expressing the arbitrary origin of height, and the discrete symmetry $h\mapsto -h$ because its residual is squared.

There is an additional exact tilt symmetry $h\mapsto h+\mathbf{b}\cdot \mathbf{x}$: the added slope is time independent and has zero Laplacian. More generally, adding any function $f$ satisfying $f_t=\nu {\nabla }^{2}f$ leaves the residual unchanged. Such bulk transformations need not preserve prescribed boundary conditions or a restricted physical interpretation of the height.

In two spatial dimensions the action also has the anisotropic scale invarianceThe spacetime measure scales as $b^4$ and the squared residual as $b^{-4}$. The time and length scales are therefore related diffusively, rather than by a Lorentz transformation.

Ordinary time reversal needs a boundary qualification. The identityshows that the density is time-reversal invariant up to a total divergence. Thus the bulk stationary equation $h_{tt}-{\nu }^2{\nabla }^{4}h=0$ is invariant under $t\mapsto -t$, although the printed density is not pointwise unchanged. The time reversal of the squared diffusion-residual action exchanges the growing and decaying branches; the selected first-order diffusion equation alone is not time-reversal invariant at fixed positive $\nu$.

Put $x=r\cos \theta$. The function $e^{ikx}$ solves the two-dimensional Helmholtz equation, $({\nabla }^2+k^2)e^{ikx}=0$. Expand its angular dependence as a Fourier series,Substitution into the polar-coordinate Laplacian,givesWith $\rho =kr$, this is the Bessel differential equation. Regularity of the original plane wave at the origin excludes the singular second-kind solution. Thus $F_m(r)$ is a constant multiple of the regular first-kind function, andFor integer negative orders, $J_{-m}=(-1{)}^m{J}_m$, so this indexing still supplies the regular radial solution. The radial equation fixes the form of each term; the small-radius plane-wave expansion fixes its coefficient.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.

A uniform steady state satisfies $f(u^{\ast },v^{\ast })=g(u^{\ast },v^{\ast })=0$. Set $u=u^{\ast }+\tilde{u}$, $v=v^{\ast }+\tilde{v}$ and keep only first-order terms in the perturbations. Since the base state is constant, its Laplacian vanishes. Taylor expansion givesThe zero-flux boundary conditions linearize to homogeneous Neumann conditions, ${\partial }_n\tilde{u}={\partial }_n\tilde{v}=0$. The off-diagonal entries of the Jacobian describe the cross-effects of the two chemicals, while the diagonal entries describe local self-response.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

For the proposed velocity field, $u_x+w_z=1-1=0$, so incompressibility holds. Its convective acceleration isand both components have zero Laplacian. The Navier-Stokes equation therefore requires $p_x=-x$ and $p_z=-z$. Hencewhere $p_{\ast }$ is an arbitrary constant reference pressure. The same pressure gradient balances the same convective acceleration in the Euler equations for an inviscid fluid. Since the viscous terms vanish identically, this is an exact solution for every positive Reynolds number, as well as for the inviscid limit.

At $z=0$, the normal velocity is zero, and the prescribed dimensionless tangential traction isThis stagnation point flow therefore has zero surface shear stress, although its strain-rate tensor and its viscous normal stresses are nonzero.

For this change of variables,and $u_{xx}+u_{yy}=0=v_{xx}+v_{yy}$. The second derivative chain rule therefore givesThus the transformation rule for the Laplacian isFor $f=x+y$, its Laplacian is zero. At every point with $(x,y)\ne (0,0)$, division by $4(x^2+y^2)$ provesThis conclusion applies to each local inverse branch of the change of variables: its Jacobian determinant is $4(x^2+y^2)$, so such a branch exists away from the origin by the inverse function theorem. A single globally defined $g$ cannot represent $x+y$ on the entire plane, since $(x,y)$ and $(-x,-y)$ give the same $(u,v)$ but generally opposite values of $x+y$. Nor can a differentiable $g$ represent this $f$ in a neighborhood of the origin: the multivariable chain rule there would give $f_x=f_y=0$, contradicting $f_x=f_y=1$. The requested harmonic function conclusion is therefore a local statement on nonsingular branches.^[ref]

Apply the Fourier transform of a derivative twice in each spatial coordinate, and commute the time derivative with the spatial Fourier transform. For the Laplacian,ThusThe minus sign is essential: positive thermal diffusivity damps short spatial scales rather than amplifying them. The zero Fourier mode is unchanged in the absence of a source.

The wave equation is second order in time. Even one normal mode obeys a second-order ordinary differential equation, so its initial displacement and initial velocity are independent data. Physically they specify stored elastic deformation and the initial motion.

The heat equation is first order in time; its initial temperature already fixes the instantaneous time derivative through the spatial Laplacian. It needs one initial field. In either case spatial boundary conditions remain necessary and are not substitutes for initial data. Classical fixed-end string data must satisfy the appropriate endpoint compatibility, including $f(0)=f(L)=0$ and, when endpoint velocities exist, $g(0)=g(L)=0$.

Let $Y_{\ell }$ be a spherical harmonic of degree $\ell$ on the unit sphere. The Laplacian in spherical coordinates separates its angular part from a radial power, giving the displayed identity for $r>0$. Thus a Poisson equation $\Delta \Phi =r^q{Y}_{\ell }$ has the particular solution $\Phi =r^{q+2}{Y}_{\ell }/[(q+2)(q+3)-\ell (\ell +1)]$ when the denominator is nonzero. Add $A{r}^{\ell }{Y}_{\ell }+B{r}^{-\ell -1}{Y}_{\ell }$ to impose the radial boundary conditions. If the denominator vanishes at $p=q+2$, differentiating the displayed identity with respect to $p$ gives instead $\Delta (r^p\log r{Y}_{\ell })=(2p+1)r^{p-2}{Y}_{\ell }$. Regularity at the origin and decay at infinity must be checked separately; the radial identity alone is local away from zero.

In ${\mathbb{R}}^{d+1}$, $g=d{r}^2+r^2{g}_{S^d}$ and the volume density is $r^d$. With the nonnegative Laplacian convention, $\tilde{\Delta }=-{\partial }_r^2-(d/r){\partial }_r+r^{-2}{\Delta }_{S^d}$. The identity separates radial homogeneity from the angular Laplace-Beltrami operator and derives the spectrum of the Laplacian on a sphere.

For a uniform spherical source of radius $a$, the origin-regular radial Poisson equation gives $\nabla f={\rho }_{0}x/3$ inside and $\nabla f={\rho }_0{a}^{3}x/(3|x{|}^3)$ outside. Integrating radially, imposing $f\to 0$ at infinity and continuity at $|x|=a$, givesThe potential and its normal derivative are continuous at the interface, so its distributional Laplacian has no surface delta contribution. The classical equation holds off the interface and the source equation holds almost everywhere and distributionally.

For integer angular frequency $m\ge 0$, the Laplacian in polar coordinates gives the displayed identity on $r>0$ by direct differentiation. Consequently a source $\beta r^q\cos (m\theta )$ has particular solution $\beta r^{q+2}\cos (m\theta )/[(q+2{)}^2-m^2]$ when the denominator is nonzero. The sine case is identical. If $s=q+2=\pm m\ne 0$, differentiating the identity with respect to $s$ gives the resonant particular solution $\beta r^s\ln r\cos (m\theta )/(2s)$. For $s=m=0$, the radial source $\beta r^{-2}$ instead has particular solution $\beta (\ln r{)}^2/2$. These are local statements on the punctured plane; boundedness, smoothness at the origin and boundary data must be imposed separately by adding suitable harmonic functions.

For a based finite acyclic chain complex, Reidemeister torsion is an alternating product of determinants comparing each given chain basis with bases formed from boundaries and lifts of boundaries. With suitable local coefficients this gives an invariant of a space finer than its homology. Analytic torsion constructs a corresponding invariant from spectra of Laplacian operators, while Turaev torsion refines combinatorial sign and normalization choices.

The infinitesimal rotation $R_{ij}=x_i{\partial }_j-x_j{\partial }_i$ commutes with the Laplacian: the extra terms $2{\partial }_i{\partial }_j$ cancel. It is tangent to every sphere, so it preserves zero Dirichlet traces on spherical shells. If $\Delta u=f$ and $f$ is radial, uniqueness applied to $R_{ij}u$ forces all these rotation derivatives to vanish, hence $u$ is radial.

For a constant orthogonal matrix $R$, the Laplacian is unchanged by the coordinate transformation $\xi =Rx$. If $u$ is twice continuously differentiable, the chain rule givesbecause ${\sum }_i{R}_{ji}{R}_{ki}={\delta }_{jk}$ by orthogonality of the rows. Here ${\delta }_{jk}$ is the Kronecker delta. In two dimensions, the rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a^2+b^2=1$ makes the mixed partial derivatives cancel explicitly. The same argument works for reflections and translations, so the Laplacian depends on the Euclidean geometry rather than the orientation or origin of the coordinates.