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In diffusion image processing, the observed grey-value image signal $g$ is the initial condition for an evolution $u(x,t)$; time controls the image smoothing scale. A useful model must suppress image noise while respecting the image signal's geometric boundaries. Unless exterior values are intended, use periodic boundaries or Neumann boundary conditions, so the filter does not lose intensity through the image signal border.

The basic linear model is the heat equation, $u_t=\Delta u$, $u(0)=g$. On the whole plane,It is Gaussian image smoothing with variance $2t$ in each coordinate. In Fourier variables, $\hat{u}(\xi ,t)=e^{-t|\xi {|}^2}\hat{g}(\xi )$: high spatial frequencies are damped most strongly. Under zero-flux or periodic boundaries the mean is conserved, the maximum principle keeps values within the initial range, andThese give stable image noise suppression, but a sharp step also contains high frequencies and is blurred across a width of order $\sqrt{t}$. Constant diffusivity has no mechanism for distinguishing image noise from an image edge.

Linear sharpening by the backward heat equation would multiply Fourier modes by $e^{t|\xi {|}^2}$ and amplify arbitrarily fine image noise without bound. It is an ill-posed initial-value problem. A controlled unsharp masking step instead uses $g+\lambda (g-G_t\ast g)$, with multiplier $1+\lambda (1-e^{-t|\xi {|}^2})$. This amplifies high frequencies by at most $1+\lambda$; it can improve apparent contrast but also amplifies image noise and cannot reliably restore information already lost by image smoothing.

Nonlinear models use image signal structure to select the diffusion. A gradient-based energy and its formal $L^2$ gradient flow areFor a smooth solution with the corresponding zero-flux condition, $dE/dt=-\int u_t^2\le 0$. The fidelity term prevents indefinite drift towards a constant reconstruction; pure diffusion is usually stopped at a selected finite time.

The local principal diffusion coefficients distinguish suppression of diffusion from backward diffusion. Where $s=|\nabla u|>0$, the flux derivative isIts coefficient along an image signal level curve is $c(s)$; across that curve it is $c(s)+s{c}^{\prime }(s)={\Psi }^{\prime \prime }(s)$. Forward parabolic behavior requires both coefficients nonnegative, with strict positive lower bounds giving uniform parabolicity. Merely choosing $c>0$ and decreasing does not establish well-posedness.

For a convex area-type penalty $\Psi (s)=\sqrt{{\epsilon }^2+s^2}$, the coefficients areDiffusion across steep image edges is weak but remains forward. As $\epsilon \to 0$, the total variation flow formally becomes $u_t=\mathrm{div}(\nabla u/|\nabla u|)$. Its convex subgradient formulation handles flat regions and discontinuities. It favors piecewise constant image signals and preserves sharp transitions better than Gaussian image smoothing, but can produce staircasing in total variation denoising, shrink small objects and move boundaries by curvature. image edge preservation does not mean exact preservation of all image edge locations or amplitudes.

The Perona-Malik equation takes a decreasing diffusivity such as $c(s)=1/(1+s^2/{\kappa }^2)$. Small gradients are smoothed strongly, while large gradients have weak flux. More precisely,For $s>\kappa$, diffusion in the gradient direction is backward, so a strong transition can steepen. This gives formal image edge enhancement but also causes instability and continuum ill-posedness. The exponential choice $c(s)=e^{-s^2/{\kappa }^2}$ similarly has a negative normal coefficient for $s>\kappa /\sqrt{2}$. Discrete results depend on the stencil, step size and implicit regularization; appealing visual results are not a proof of a well-posed PDE.

A regularized Perona-Malik diffusion computes the conductance from a smoothed image signal, for exampleThe flux still acts on $u$, but the image edge detector is less sensitive to raw image noise. With a smooth kernel and a positive conductance bounded away from zero on the attained range, the local principal diffusion is forward; appropriate regularity and boundary assumptions give a well-posed nonlocal parabolic model. The exact regularization and fixed image smoothing scale are part of the model.

A genuinely directional filter uses a diffusion tensor for image filtering:Here $e_n$ estimates the image edge normal and $e_t$ its tangent. A structure tensor $J_{\rho }=G_{\rho }\ast (\nabla u_{\sigma }\nabla u_{\sigma }^T)$ provides robust directions at a second averaging scale. Strong tangent diffusion smooths image noise along an image edge, while weak normal diffusion reduces mixing across it. Related coherence-enhancing choices connect elongated features along their dominant orientation. Positive tensor eigenvalues preserve forward parabolicity; this type of enhancement must be distinguished from Perona–Malik's backward normal diffusion.

For more direct sharpening, a shock filter for image enhancement formally evolves $u_t=-\mathrm{sign}(\Delta u)|\nabla u|$. It is a Hamilton–Jacobi-type transport mechanism, not a positive diffusion operator, and steepens transitions around inflection boundaries. In practice it can be combined with regularized forward diffusion to control image noise. Any enhancement method needs a scale or stopping rule and an honest treatment of image noise amplification.

Linear heat flow offers predictable image smoothing but blurs image edges; convex nonlinear diffusion can preserve them; backward or shock mechanisms sharpen them at a greater stability cost. Gradient thresholds, conductance regularization and positive tensor directions determine which of these behaviors a proposed filter actually has.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

We use the homogeneous version of the Sobolev embedding theorem in three dimensions:Here homogeneous Sobolev space ${\dot{H}}^1$ is the completion of compactly supported smooth functions in the L2 norm of the gradient, identified with its $L^6$ representative. Apply this Sobolev inequality both to $\varphi$ and to its spatial derivatives.

LetThe Plancherel theorem identifies the L2 norm of the Hessian matrix with $\Vert \varphi {\Vert }_{{\dot{H}}^2}$, because ${\sum }_{i,j}{\xi }_i^2{\xi }_j^2=|\xi {|}^4$. In particular,Differentiating the defocusing semilinear wave equation gives $\square ({\partial }_i\varphi )=3{\varphi }^2{\partial }_i\varphi$. By the Holder inequality with exponents $3$ and $6$, the preceding Sobolev inequality, and conservation of the positive wave energy,Use the inhomogeneous wave energy estimate simultaneously for the three spatial derivatives. It givesThe Gronwall inequality therefore yields $B(t)\le D\exp (CEt)$. We may take the continuous, locally bounded functionThe constant is universal; the dependence on the initial data is only through $D$ and $E$. The H2 bound for the defocusing cubic wave equation holds on every existing smooth interval, without assuming the global conclusion. If $D=0$ or $E=0$, the zero solution satisfies the same estimate.

The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

The MHD induction equation can be written in material form asIts $y$ component immediately givesFor the Cartesian magnetic flux function, the $x,z$ components of the MHD induction equation, or the $y$ component of the magnetic vector potential equation, giveThe additive function of time in $\psi$ has no effect on $\mathbf{B}$. Choose this gauge so that $C=0$. Then the flux label is materially conserved:This is magnetic flux freezing in the two-dimensional geometry.

To prove the absence of axial motion, construct an invariant solution with $u_y=0$. Along each Lagrangian trajectory, $\psi$ is fixed, and the assumed divergence givesThus $B_y$ remains a function of $\psi$ and time alone. Its gradient stays parallel to $\nabla \psi$, so the $y$ component of the Lorentz force remains zero. The axial momentum equation isbecause $(B_x{\partial }_x+B_z{\partial }_z)\psi =0$. With $u_y=0$ initially, it remains zero. Hence neither axial force nor axial motion is generated. This flux-surface preservation during magnetic-tube expansion argument assumes the smooth ideal evolution for which the initial-value problem is unique; the particular in-plane rising motion need not be calculated.

There is a genuine conflict in the printed coefficient assumptions: no uniformly positive coefficient in a zero-boundary Sobolev space exists. A function $a\in H_0^1(\Omega )$ cannot also obey $a\ge a_{-}>0$ almost everywhere. Indeed, the Lipschitz truncation $g(s)=\min (\max (s,0),a_{-})$ has $g(0)=0$, so Lipschitz truncation preserves zero-boundary Sobolev spaces, giving $g(a)\in H_0^1(\Omega )$. But $g(a)$ would be the nonzero constant $a_{-}$. Its zero gradient contradicts the Poincare inequality in the zero-boundary Sobolev space. Thus the literal coefficient class is empty.

For the meaningful uniformly elliptic problem, take $a\in L^{\infty }(\Omega )$ with $0<a_{-}\le a\le a_{+}$, and impose the Dirichlet boundary condition on the unknown and test functions: $V=H_0^1((0,1{)}^2)$. Additional $H^1$ regularity of $a$ is harmless, but a zero trace for $a$ must be removed. The divergence-form elliptic operator is $L=-\mathrm{div}(a\nabla )$. Integration by parts defines the symmetric bounded bilinear formIn particular,Here $2{\pi }^2$ is the first Dirichlet Laplacian eigenvalue on the unit square; the Poincare inequality follows, for example, by applying the one-dimensional inequality in each coordinate and adding. Thus $B$ is coercive in the gradient norm on $V$, and the Dirichlet realization of $L$ is a positive definite symmetric operator. On its operator domain, $⟨Lv,v⟩=B(v,v)$.

The required functional and weak equation areBecause $f\in L^2$, the right side is a bounded linear functional by the Cauchy-Schwarz inequality and Poincare inequality. The Lax-Milgram theorem gives a unique weak solution, and part (b) proves that it uniquely minimizes $I$. These formulas prove the intended conclusion under the repaired coefficient hypothesis; under the literal hypothesis there is no coefficient to which the conclusion can be applied.

For a bounded domain, the nondivergence-form elliptic operator obeys the weak maximum principle: $Lu\ge 0$ and $c\le 0$ implyHere $u^{+}=\max \{u,0\}$. Boundedness of $\Omega$ is needed for this formulation: it is not explicitly imposed in this question. On an unbounded domain one needs an appropriate condition at infinity instead. Without either condition the claim is false, as $u(x)=x_n$ and $L=\Delta$ on the half-space $\{x_n>0\}$ show.

For the bounded-domain proof choose $\gamma >0$ so large that $\lambda {\gamma }^2-\Vert b^1{\Vert }_{\infty }\gamma -\Vert c{\Vert }_{\infty }>0$. The positive exponential $h(x)=e^{\gamma x_1}$ then satisfiesLet $u_{\epsilon }=u+\epsilon h$, $M={\sup }_{\partial \Omega }{u}^{+}$ and $H={\sup }_{\overline{\Omega }}h<\infty$. If $u_{\epsilon }$ had a value greater than $M+\epsilon H$, it would attain a positive maximum at an interior point. There its gradient vanishes and its Hessian matrix is negative semidefinite. The symmetric part of $(a^{ij})$ is positive-definite by uniform ellipticity, so $a^{ij}{D}_{ij}{u}_{\epsilon }\le 0$; also $c{u}_{\epsilon }\le 0$. Hence $L{u}_{\epsilon }\le 0$, contradicting $L{u}_{\epsilon }=Lu+\epsilon Lh>0$. Thus $u_{\epsilon }\le M+\epsilon H$. Letting $\epsilon \downarrow 0$ proves the weak maximum principle, without requiring continuity of the coefficients.

For example, the unbounded-domain version follows by exhausting the domain with bounded truncations if ${\mathrm{limsup}}_{|x|\to \infty ,x\in \Omega }u(x)\le M$: the added spherical boundaries then have supremum at most $M+o(1)$.

Use a Lipschitz continuous cutoff function $\eta$ which equals one on $B_R(x_0)$, vanishes outside $B_{2R}(x_0)$, takes values in $[0,1]$, and has $|\nabla \eta |\le 2/R$. Its gradient is supported in the annulus $A_R=B_{2R}(x_0)\setminus B_R(x_0)$. The function ${\eta }^2(u-c)$ is an admissible test function in the zero-boundary Sobolev space for the weak solution, by approximation with smooth compactly supported test functions. If ${\Lambda }_0$ bounds the operator norm of $(a^{ij})$, we can take ${\Lambda }_0=n\Lambda$. The weak equation and uniform ellipticity giveYoung inequality bounds the right side byAfter absorption, this annular Caccioppoli inequality isThe dimension is fixed in the notation $C(\lambda ,\Lambda )$ of the question; with the entrywise coefficient bound its dependence on $n$ is absorbed there. If the larger ball merely lies in $B$ without its closure being compactly contained, approximation from smaller cutoff functions gives the same admissible test function and estimate. Crucially, the right side uses only the annulus, where the cutoff function varies. The constant $c$ is arbitrary because constants have zero gradient.

The product rule for the divergence givesSubtracting cancels the two gradient products. ThereforeTo obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is $\hat{\mathbf{r}}$; on the inner sphere it is $-\hat{\mathbf{r}}$. Write ${\partial }_r=\hat{\mathbf{r}}\cdot \nabla$ on both spheres. Then the resulting Green second identity isThe minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

Away from the origin, the given vector field is the gradient of the globally single-valued radius:The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the chain rule gives $\mathbf{F}\cdot d\mathbf{x}=dr$. Consequently, independently of orientation,The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a potential of a conservative vector field throughout the punctured space.

Integrating the oscillatory term together with the constant components gives a potential of a conservative vector fieldDirect partial derivatives recover both components of $\mathbf{V}$, proving it is a conservative vector field. Its line integral is therefore path independent. At an integer endpoint $(m,n)$, the sine term vanishes, so for any of the specified pathsAny other potential has the same gradient, so differs from $F$ by a constant on the connected plane. In particular every potential changes by $1$ under $x\mapsto x+1$ and by $2$ under $y\mapsto y+1$. Equivalently the integral from $(0,0)$ to $(1,0)$ is $1$, whereas a unit-periodic potential would make that difference zero. There is no potential periodic with period one in both coordinate directions. This is an example of how periods obstruct a periodic potential: a periodic vector field need not have a periodic potential.

Fix nested relatively compact subsets of $\Omega$ and let $M$ bound the gradient on the larger one. The fundamental theorem of calculus along a line segment gives $|w_h|\le M$ on the smaller one. Part (i) gives coefficient bounds independent of small $h$, so the De Giorgi-Nash-Moser theorem gives uniform interior Hölder continuity estimates for these weak solutions. Since $Du$ is continuous, the difference quotients $w_h$ converge locally uniformly to $D_{\ell }u$. Passing to the limit in the Hölder seminorm estimate, and repeating for every coordinate, yieldsThe first derivatives are locally Hölder continuous. Here the initially obtained exponent depends on $n$ and the local gradient bound. This is the interior interpretation of the printed regularity conclusion; a uniform global estimate does not follow from $u\in C^1(\Omega )$ alone.

Write the minimal surface equation for a graph as $\mathrm{div}F(Du)=0$ in its weak formulation, where $F(p)=p/\sqrt{1+|p{|}^2}$. For negative increments use the same difference quotient convention, soFor $\zeta \in C_c^1({\Omega }^{\prime \prime })$, the function ${\delta }_{\ell ,h}\zeta$ is an admissible test function in $\Omega$. Commuting the first partial derivative with translation, then applying discrete integration by parts, givesThe fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equationThus the backward difference quotient solves a linear equation in divergence form:To verify the ellipticity of the minimal surface flux, computeand henceThe eigenvalue parallel to $p$ is $(1+|p{|}^2{)}^{-3/2}$; every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. The same lower and upper bounds pass to the averaged symmetric matrix $A_h$ whenever both endpoint gradients have norm at most $M$:The printed $C^1(\Omega )$ hypothesis supplies such an $M$ on each relatively compact subset, uniformly for sufficiently small $h$. It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of ${\Omega }^{\prime \prime }$ additionally requires bounded gradients there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.

For a harmonic function $u$ on $\mathbb{C}\cong {\mathbb{R}}^2$, the Itô formula givesStop on leaving increasing discs and at increasing deterministic times. On each such interval the gradient is bounded, so the stochastic integral is a martingale. This proves that $u(B)$ is a continuous local martingale.

If $|u|\le K$, this is a bounded local martingale and hence a true martingale. Start Brownian motion at an arbitrary $z$. Its Gaussian heat kernel givesFor a unit vector $e$, ${\partial }_e{p}_t(v)=-(e\cdot v)p_t(v)/t$, so $\int |{\partial }_e{p}_t(v)|dv=\sqrt{2/\pi }/\sqrt{t}$. Integrating the directional derivative along the segment from $z$ to $w$ yieldsLetting $t\to \infty$ gives the harmonic Liouville theorem:This Gaussian heat-kernel proof of the harmonic Liouville theorem uses the bounded martingale identity established above.