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For the sharp energy estimate, $C_0=0$ stays uniform while the available bound $C_1=M^2/\epsilon$ diverges as $\epsilon \downarrow 0$ when $M>0$. If the coarser estimate is used for the first part, both displayed bounds $M^2/\epsilon$ diverge, but the first divergence is only an artifact of discarding an exact cancellation.

The method of characteristics explains why uniform control of the gradient cannot generally persist for the inviscid scalar conservation law. Before characteristic crossing, with initial data $u_0$,If $F^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )<0$ somewhere, the denominator reaches zero in finite positive time. The solution steepens and the smooth description breaks down; an entropy solution can subsequently contain shocks. Positive viscosity replaces such a discontinuity by a thin smooth layer, which can have large gradient even while its $L^2$ norm remains controlled. The divergent bound does not assert that every flux and every initial datum form a shock; a linear flux, for example, has no such steepening.

A one-dimensional Sobolev representative is absolutely continuous. For $x<y$, the fundamental theorem of calculus and Holder inequality giveThus $u$ has a representative in $C^{0,1-1/q}([0,1])$. The qualification about representatives matters because a Sobolev space element is an almost-everywhere equivalence class.

In two dimensions, the Sobolev fundamental theorem of calculus on lines and Fubini's theorem imply that almost every horizontal and vertical slice belongs to $W^{1,q}(0,1)$ and has this one-dimensional Hölder continuity. The slice seminorm depends on the slice; this does not give one uniform pointwise estimate on the square. For $q>2$, Morrey's inequality additionally gives a globally Hölder continuous representative of exponent $1-2/q$. For $1<q\le 2$, global continuity need not hold. For example, with a smooth cutoff around an interior point, $u(x)=|x-x_0{|}^{-a}$ lies in $W^{1,q}$ when $0<a<2/q-1$, but is unbounded. At $q=2$, the cutoff version of $\log \log (e/|x-x_0|)$ is unbounded while its gradient has finite squared integral, sinceThese examples distinguish Sobolev slicing and planar continuity from a false two-dimensional application of the interval exponent.

Put $\Omega =(0,1{)}^2$. A BV space is a function $u\in L^1(\Omega )$ whose distributional derivative $Du$ is a finite vector-valued Radon measure. Equivalently its total variation seminorm is finite:The BV space has norm $\Vert u{\Vert }_{L^1}+|Du|(\Omega )$. For $u\in W^{1,1}(\Omega )$, integration by parts against the compactly supported field gives $\int u\mathrm{div}\phi =-\int \nabla u\cdot \phi \le \int |\nabla u|$. Conversely the measurable choice $\phi =-\nabla u/|\nabla u|$ on nonzero gradients attains the pointwise bound. Approximating this bounded field by smooth fields, using interior cutoffs and the finite measure $|\nabla u|dx$, justifies the supremum and givesIt is a norm of the derivative measure, rather than a pointwise derivative at jump discontinuities.

There is a genuine mismatch in the printed definition of the next functional. Its constraints on ${\phi }_0$ and $\phi$ are independent. Hence its stated supremum, denoted $A_{\mathrm{box}}$, separates asThe scalar supremum is $|\Omega |$, by cutoffs approaching one, and the vector supremum is the variation. For an affine image signal with $|\nabla u|=1$, this gives $2|\Omega |$, whereas the displayed square-root area would give $\sqrt{2}|\Omega |$. The intended relaxed graph-area functional instead uses the coupled pointwise constraint ${\phi }_0^2+|\phi {|}^2\le 1$, givingwhere $D^{s}u$ is the singular part of $Du$. Both readings have a minimizer, but their equations are different.

Here is the direct method in the calculus of variations for either reading. Let $A_{\bullet }$ be the literal $A_{\mathrm{box}}$ or the corrected $A$, and define the energy on $BV(\Omega )\cap L^2(\Omega )$, assigning infinity elsewhere. A minimizing sequence has bounded energy by comparison with $u=0$. Both $A_{\bullet }\ge |Du|(\Omega )$, so its variation is bounded, and the fidelity bounds $\Vert u-g{\Vert }_2$, hence also $\Vert u{\Vert }_2$ and $\Vert u{\Vert }_1$. By bounded-variation compactness, a subsequence converges strongly in $L^1$ to $u\in BV$, and after another subsequence almost everywhere. Fatou's lemma provesThe regularizer is a supremum of affine functionals continuous in $L^1$, since the test-field divergence is bounded. It is therefore lower semicontinuous. Combining the two lower bounds proves existence of a minimizer. In fact the convex regularizer and the strictly convex squared fidelity make the minimizer unique up to null sets. This does not assert that the minimizer must belong to $W^{1,1}$.

For the intended graph area, conditionally assume that the minimizer is in $W^{1,1}$. For $\eta \in C_c^{\infty }(\Omega )$, differentiate at $u+t\eta$. The derivative of the integrand is bounded by $|\nabla \eta |$, so dominated convergence applies. The weak equation isthat is,This is the graph-area Euler-Lagrange equation. Compactly supported variations impose no boundary condition in this statement.

For the literal printed supremum, the constant $|\Omega |$ drops out and one obtains total variation denoising. Its total variation calibration form isThe distributional equation means $\alpha \int z\cdot \nabla \eta +\int (u-g)\eta =0$. In particular $z=\nabla u/|\nabla u|$ wherever the gradient is nonzero; writing this quotient without handling zero gradients would be incomplete. Formally the one-sided derivative of $\int |\nabla u|$ isMinimality in the directions $\eta$ and $-\eta$ bounds the remaining linear functional by the second integral. The Hahn-Banach theorem extends it on that zero-gradient set to a bounded vector field of magnitude at most one, furnishing $z$ and the displayed weak equation. Thus the literal definition has a nonsmooth subgradient equation, not the square-root equation above.

Interpret the printed Euclidean norm literally on the full vector array $X^2$. It is a single global norm, not the sum of pixelwise gradient lengths in discrete isotropic total variation. Set $A=\nabla$ and let $A^{\ast }$ be its exact adjoint operator for the chosen difference and boundary conventions. The closed dual ball and its image signal areThe set $C$ is nonempty, convex and compact, hence closed, because it is a linear image signal of a compact ball in finite dimensions. Euclidean norm duality givesthe support function of $C$.

The metric projection onto a closed convex set $w=P_{C}g$ uniquely minimizes $\Vert g-w{\Vert }^2$ over $C$. Its variational characterization of convex projection saysPut $u_{\ast }=g-w$. This inequality says precisely that ${\sigma }_C(u_{\ast })=⟨u_{\ast },w⟩$. For any $v\in X$, the primal energy satisfiesCompleting the square shows that the right side is uniquely minimized by $v=g-w$. At that point the inequality is equality, soThis also follows from the proximal operator of a support function and Moreau decomposition: the convex conjugate of ${\sigma }_C$ is the indicator functional of a constraint set for $C$. The squared fidelity is strictly convex and coercive, so the primal minimizer exists and is unique.

The convex projection in this formula is the removed component $g-u_{\ast }$, not generally $u_{\ast }$ itself. A metric convex projection onto one fixed closed convex set is idempotent. On a right singular vector direction with positive singular value of any nonzero $A$, this denoising map reduces to soft thresholding with a positive threshold, and applying it twice shrinks again. It therefore cannot be such a convex projection for all data. This qualifies the printed convex projection wording while giving the required global gradient-norm projection residual.

Compute $P_{C}g$ by solving the convex dual least-squares problemIts gradient is $\nabla f(p)=A(A^{\ast }p-g)$ and has Lipschitz constant $L=\Vert A{\Vert }^2$. Projected gradient descent givesFor fixed $0<\tau <2/L$, the finite-dimensional projected-gradient convergence theorem ensures that $p^n$ converges to a dual minimizer. Reconstruct $u^n=g-A^{\ast }{p}^n$; its limit is the unique primal solution even if dual minimizers are nonunique. If $A=0$, the primal solution is simply $g$ and no iteration is needed. For unit-grid forward differences with periodic or zero-difference boundaries, $\Vert A{\Vert }^2\le 8$, so $0<\tau <1/4$ is sufficient. Grid-spacing factors or other boundary stencils change this bound.

The normalization here is global. If a pixelwise sum of gradient lengths had instead been intended, the dual feasible set would be a product of pixelwise balls and convex projection would normalize each block separately; it is a different regularizer and should not be silently substituted.

For the discrete Hessian-norm denoising variant, write $H={\nabla }^2:X\to X^4$ and useThis is again a compact convex image signal of a Euclidean ball, now under the second-difference adjoint. In suitable conventions $H^{\ast }$ is a discrete double divergence, but its exact boundary adjoint is what defines the set. It lies in $(\ker H{)}^{\perp }$, so components in $\ker H$ are preserved by denoising. Interior second derivatives annihilate affine image signals; whether all such image signals remain in the kernel depends on the boundary convention. With a pixelwise Hessian norm, the corresponding balls would instead be four-component blocks.

A rigorous Mumford–Shah functional permits nonsmooth image signals and free discontinuities. One classical admissible class consists of relatively closed countably rectifiable sets $K\subset \Omega$ with finite Hausdorff measure ${\mathcal{H}}^1(K)$, and $u\in W^{1,2}(\Omega \setminus K)\cap L^{\infty }(\Omega )$ with finite energy. No exterior boundary values are prescribed. For an existence argument, use the equivalent relaxed classA special bounded-variation space excludes the Cantor part of a bounded-variation derivative of the derivative: $Du=\nabla udx+[u]{\nu }_u{\mathcal{H}}^1\lfloor J_u$. The jump set of a bounded-variation function $J_u$ is the relaxed image edge set. Clipping to $[-M,M]$ decreases squared fidelity, does not increase the gradient term, and does not create jumps, so this bound loses no minimizers.

Take a minimizing sequence and compare with a constant image signal. Its $L^2$ gradient norms and jump lengths are bounded. Alsoso the sequence is bounded in $BV$. The SBV compactness theorem for bounded values, superlinear gradient growth and bounded jump measure yields an $L^1$ limit in $SBV$, weak convergence of gradients in $L^2$, and lower semicontinuity of both the Dirichlet term and the jump measure. The uniform value bound upgrades convergence to $L^2$, so fidelity converges. This proves existence of a relaxed minimizer. Essential closedness of Mumford–Shah jump sets then supplies a relatively closed representative $K=\overline{J_u}\cap \Omega$, without added length, and $u\in W^{1,2}(\Omega \setminus K)$. This completes the outline for the classical pair problem. Arbitrary Hausdorff convergence of image edge sets alone is not an adequate substitute for these compactness and regularity results. No uniqueness is claimed for segmentation.

As $\alpha \to \infty$ with $\beta$ fixed, bounded energy forces $\nabla u\to 0$ in $L^2$. The reduced piecewise-constant Mumford–Shah problem isEquivalently, use a Caccioppoli partition $\{E_i\}$ of the image signal domain and constants $c_i$:The relative perimeter counts only interior boundaries, and the factor one half counts each interface once. Adjacent equal-valued regions can be merged, removing unnecessary boundaries.

For fixed $K$, let $E_i$ be its positive-area regions. Minimization over $u$ reduces to independent scalar least-squares fits:The minimized fidelity is ${\int }_{\Omega }{g}^2-{\sum }_i({\int }_{E_i}g{)}^2/|E_i|$. Thus region means in piecewise-constant segmentation give the optimal grey values for a fixed segmentation.

For a fixed full spatial function $u$, the image edge set must contain its jumps; any extra curve only adds length. The optimal choice is its essential jump set, with a relatively closed representative when appropriate. There is no independent relocation of boundaries while that full function is held fixed. A different common alternating step fixes the values $c_i$ but allows the labels $E_i$ to move. It minimizes the fidelity-plus-perimeter partition functional above. Without the perimeter term each point takes its nearest grey value; with it, interface length is penalized. At a smooth interface between two labels, outward normal displacement of $E_i$ has first variationwhere $\kappa ={\mathrm{div}}_{\Gamma }{\nu }_i$ is positive for an outward normal to a circle. The stationary segmentation interface curvature balance isThis is the geometric interpretation of optimizing boundaries with fixed grey levels, and distinguishes it from fixing the whole spatial image signal.

As $\beta \to \infty$ with $\alpha$ fixed, a constant competitor bounds the minimum independently of $\beta$, forcing ${\mathcal{H}}^1(J_u)\to 0$. compactness in the relaxed formulation leaves no jump or Cantor part of a bounded-variation derivative, so the limit is in $W^{1,2}(\Omega )$ on the connected rectangle. The reduced edge-free Mumford–Shah limit isA set of zero length can be omitted; this does not impose a zero image signal or a Dirichlet boundary value. Comparison with any fixed $H^1$ competitor and lower semicontinuity justify the limit minimization.

For completeness, the bilinear form $B(u,v)=\int uv+\alpha \int \nabla u\cdot \nabla v$ on $H^1(\Omega )$ is continuous and coercive, with $B(u,u)\ge \min (1,\alpha )\Vert u{\Vert }_{H^1}^2$. The right-hand side $\int gv$ is bounded because $g\in L^2$ on the bounded rectangle. The Lax-Milgram theorem gives a unique $u$ satisfyingIt is the unique minimizer by strict convexity. Formally,with the Neumann condition understood through this weak formulation. Equivalently, subtracting the weak equation shows that the energy increase at $u+v$ is $\Vert v{\Vert }_2^2+\alpha \Vert \nabla v{\Vert }_2^2>0$ for nonzero $v$.

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 Poincaré 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 Poincaré 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 Poincaré 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.