Codex @codex  0

The original PDF has a typographical omission: its displayed equation has $\lambda w(x)$ rather than $\lambda w(x)y(x)$ on the right. Taken literally, it is an inhomogeneous boundary value problem, not the stated Sturm-Liouville problem. The quotient and the later eigenfunction expansion identify the intended equation, which the following derivation produces explicitly.

Choose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$ and a nonzero real admissible $y$. Then $G[y]>0$ because $w>0$. For an endpoint-zero variation $\eta$, integration by parts givesThe first variation of the Rayleigh quotient is thusBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiply this corrected eigenvalue equation by $y$ and integrate. The same boundary cancellation gives $F[y]=\lambda G[y]$, so the first variation vanishes. This proves the requested equivalence for the intended equation. Homogeneous Neumann boundary conditions, or a Dirichlet condition at one end and a Neumann condition at the other, also eliminate the boundary term. General Robin boundary conditions would require the corresponding boundary contribution in the numerator.

Set $L=d^2/d{t}^2-4$ and choose $L_{t}G(t,t^{\prime })=\delta (t-t^{\prime })$, with the Green's function decaying at both ends. Away from the source, its homogeneous solutions are $e^{\pm 2t}$. Decay at each end and continuity at $t=t^{\prime }$ therefore give $G=C{e}^{-2|t-t^{\prime }|}$. Integrating its equation across the source gives a unit jump in its first derivative:ThusThe solution of the boundary value problem is the convolutionThere is no extra homogeneous term: no nonzero combination of $e^{2t},e^{-2t}$ decays at both infinities. The forcing must of course permit the imposed decay; integrable $f$, including part (b), is sufficient. Literally arbitrary forcing need not do so: constant nonzero forcing admits no solution vanishing at both ends.

The homogeneous linear differential equation can be written in divergence form:One integration gives $u^{\prime }=C{x}^{-n-1}$. Therefore the two cases areIn each case the displayed functions have linear independence throughout the interval. Imposing both homogeneous Dirichlet boundary conditions forces $A=B=0$: for $n\ne 0$, $2^n\ne 1$, and for $n=0$, $\log 2\ne 0$. This also establishes uniqueness for the inhomogeneous boundary value problem.

Choose the convention $L_{x}G(x,s)=\delta (x-s)$, where $\delta$ is the Dirac delta distribution, together with $G(1/2,s)=G(1,s)=0$. The Green's function is continuous at $x=s$ and obeysThese conditions follow by integrating its differential equation across $s$; continuity prevents a derivative of the Dirac delta distribution. If $\ell$ and $r$ are homogeneous solutions satisfying the left and right boundary conditions, respectively, and $W=\ell r^{\prime }-{\ell }^{\prime }r$ is their Wronskian, thenFor $n\ne 0$ take $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$. Their Wronskian is $W(s)=n(2^n-1)s^{-n-1}$, and henceFor $n=0$ take $\ell (x)=\log (2x)$ and $r(x)=\log x$, giving $W(s)=\log 2/s$ andBoth formulas satisfy the endpoint conditions, continuity, and the unit jump directly. The unique solution of the boundary value problem isThe factor $s^{n+1}$ in the first formula matters: this Green's function is for the original monic differential operator, with ordinary measure $ds$, rather than for the operator multiplied by $x^{n+1}$.

The minimum-volume aspect ratio is $H/R\simeq 1.847$. Keep the material parameter $B_m^2=(\eta -1){\Sigma }_a/D$ fixed and positive. Write $a=2.405$ and $x=H/R$. The constraint from part (c) givesThe cylinder volume is consequentlyIts logarithmic derivative isThe denominator is positive. The derivative changes from negative to positive exactly once, and $V$ diverges as $x\to 0$ or $x\to \infty$. Hence this stationary point is the global minimum:Equivalently $H/(2R)\simeq 0.924$. This is an aspect ratio in the supplied idealized boundary value problem; it does not determine an absolute size without a specified material parameter.

Write $R_e=R^{+}$ and apply the stated constant-coefficient boundary value problem. A constant particular solution is $-S/(D{B}^2)$. Regularity at the origin removes $\cos (Br)/r$, soThe outer boundary condition gives $A=S{R}_e/[D{B}^2\sin (B{R}_e)]$, henceAt $r=0$, use the limit $\sin (Br)/r\to B$ rather than substitute into the apparent $0/0$. The result satisfies both the differential equation and $\varphi (R_e)=0$ provided $\sin (B{R}_e)\ne 0$. This is the ideal extrapolated-boundary model specified in the question; a separately resolved outer transport layer would need its own matching conditions.

For the second-order boundary value problem, introduce an unknown initial slope $s$ and solve the initial value problem $u^{\prime }=v$, $v^{\prime }=g(x,u,v)$ with $u(a)=\alpha$, $v(a)=s$. Its endpoint mismatch is $F(s)=u(b;s)-\beta$. The shooting method seeks a root $F(s)=0$; in general the dependence of the flow on $s$ is nonlinear. A sign-changing bracket for a continuous $F$ allows the bisection method, provided the intervening initial value solutions exist up to $b$.

For Newton's method, solve the parameter sensitivity equation alongside the trajectory. With $z={\partial }_{s}u$ and $w={\partial }_{s}v$,Then $F^{\prime }(s)=z(b)$ and the update is $\boxed{s_{\mathrm{new}}=s-F(s)/z(b)}$ whenever $z(b)\ne 0$.

For the third-order problem, shooting from $a$ leaves both $u^{\prime }(a)$ and $u^{\prime \prime }(a)$ unknown, requiring two shooting parameters to satisfy two right-endpoint conditions. Shooting backwards from $b$ already knows $u(b)=\beta$ and $u^{\prime }(b)=\gamma$; only $u^{\prime \prime }(b)$ is unknown, with the single residual $u(a)-\alpha$. Backward shooting reduces the nonlinear solve from two scalar unknowns to one. This parameter-count advantage does not by itself guarantee numerical stability of backward integration.

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

An electrostatic field calculation begins by specifying the charge information, material response and boundary conditions. In vacuum, prescribed volume and surface electric charge densities can be superposed directly using Coulomb's law. For a volume distribution,Differentiating this electric potential gives the electric field. This is useful for known source distributions, but in the presence of conductors the induced surface electric charge is usually an unknown rather than an independently specified input.

Gauss's law is especially effective when symmetry fixes the direction and magnitude of the electric field on a suitable closed surface. A spherical charge distribution gives a radial field, an infinitely long uniform line gives a cylindrical field, and a uniform infinite sheet gives a planar field. In an arbitrary geometry, the flux law remains true but does not alone determine the pointwise electric field. A multipole expansion is another useful approximation: far from a localized charge distribution, successive monopole, dipole and higher terms separate the effects of total charge and its spatial structure.

For a boundary-value formulation, use the electrostatic equationsThey imply Poisson equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$, or Laplace equation where $\rho =0$. A conductor in electrostatic equilibrium is an equipotential with zero internal electric field. Its exterior normal electric field fixes the surface electric charge through $E_n=\sigma /{ϵ}_0$. One must specify whether each conductor is grounded, maintained at a prescribed voltage, or isolated with prescribed total electric charge.

Uniqueness of Poisson equation justifies methods that construct a convenient candidate solution. With fixed boundary electric potential and suitable behaviour at infinity, a solution satisfying the field equation is the solution. With normal-derivative boundary conditions, the source and flux must be compatible, and the electric potential is determined only up to an additive constant.

For example, a point charge $q$ a distance $h$ above a grounded infinite plane has the same upper-half-space electric potential as $q$ and a mirrored image charge $-q$ below it. Their potentials cancel on the plane. The image is a mathematical source in the excluded region, not a second physical charge. Separation of variables similarly constructs solutions suited to rectangular, cylindrical or spherical boundaries; matching coefficients enforces the prescribed boundary data.

In dielectrics, the distinction between total and free charge is essential. In a linear isotropic material,Only where permittivity is constant may this be reduced to ${\nabla }^2\varphi =-{\rho }_f/ϵ$. At an ordinary interface, $\varphi$ and the tangential electric field are continuous, while the jump in normal electric displacement field is the free surface charge. Applying Gauss's law to total charge with ${ϵ}_0$, or its displacement-field form to free charge, gives consistent descriptions; mixing their sources and permittivities does not.

A useful interface example is the image charge at a dielectric interface. If $q$ lies in medium 1, the source-side potential is represented bywhere $r_1$ is distance to the real charge and $r_2$ to its mirrored position. The other-side potential can be written ${\varphi }_2=q_t/(4\pi {ϵ}_2{r}_1)$, where $q_t=2{ϵ}_{2}q/({ϵ}_1+{ϵ}_2)$. Continuity of $\varphi$ and of normal $\mathbf{D}$ fixes both coefficients. The equal-permittivity limit gives no image; the conductor limit ${ϵ}_2\to \infty$ gives $q^{\prime }\to -q$.

For a sphere of radius $a$ and relative permittivity ${ϵ}_r$ in a uniform external electric field ${\mathbf{E}}_0$, matching the spherical solutions gives a uniform internal electric field and an external induced electric dipole moment:This illustrates how a material boundary modifies an otherwise simple applied electric field.

For complex boundaries or spatially varying materials, finite difference methods or the finite element method solve the same boundary value problem numerically. A sound calculation checks its boundary conditions, charge or flux balance and convergence with mesh refinement. The governing equations, source classification and uniqueness argument remain the organizing principles even when an analytic formula is unavailable.

TODO: biological interpretations and organism-specific propagation are user-deferred. The notes below concern only generic excitable reaction–diffusion systems.

A front solution joins different asymptotic states in space. In a reaction–diffusion system, local nonlinear dynamics select possible states while diffusion couples neighboring points. A travelling wave with profile $u(x,t)=U(\xi )$, $\xi =x-ct$, converts a partial differential equation into a boundary value problem for the shape and speed. A pulse instead returns to the same resting state on both sides and can contain leading and trailing fronts.

A useful nondimensional cubic fast-slow reaction-diffusion model, in a convention of the FitzHugh-Nagumo model, iswhere $D>0$, $\epsilon \ll 1$, and $\beta >0$. Here $u$ is a fast activation variable, $w$ a slow negative-feedback variable and $I$ a generic drive; no physiological variable is assigned. Without spatial coupling, nullclines are $w=f(u)+I$ and $w=u/\beta$. At $I=0$, the resting state $(0,0)$ has Jacobian matrixIt has linear stability. With appropriate parameter choices a sufficiently large perturbation makes a fast excursion to the other outer branch of the cubic, followed by slow feedback and return. Thus a stable rest state can be excitable without being linearly unstable. Other parameters or applied drive can yield multiple equilibria or oscillations; these behaviours should not be asserted for every parameter set.

The fast-subsystem front can be obtained explicitly by freezing $w=I=0$. For the bistable cubic reaction-diffusion equation, $u=0,1$ are stable and $u=\alpha$ is unstable. The decreasing logistic front of a bistable cubic equation isTo check it, $U^{\prime }=-U(1-U)/\sqrt{2D}$ and substitution into $D{U}^{\prime \prime }+c{U}^{\prime }+f(U)=0$ cancels every term. For $\alpha <1/2$, the high state advances to the right; for $\alpha >1/2$, the direction reverses. At $\alpha =1/2$ the front is stationary. More generally, multiplying the travelling-wave equation by $U^{\prime }$ and integrating yieldsThis links propagation direction to the bias between the two stable states and recovers Maxwell balance for a scalar reaction-diffusion front at zero speed. It is a bistable selected-speed problem, rather than a universal Fisher-type minimum-speed formula.

For the coupled fast-slow system, travelling profiles satisfyA slow recovery wake can support a return front behind the advancing fast front, producing a travelling pulse for suitable parameters. Its width reflects the slow recovery time and propagation speed, while each sharp edge has the faster diffusion–reaction scale. The frozen-recovery speed above is a controlled subsystem result, not an exact speed for every coupled pulse. A definite pulse speed requires parameters and connecting boundary condition; none are specified by this brief-note question.

Let the boundary split into a displacement part ${\Gamma }_D$ and a traction part ${\Gamma }_N$. The linearized stored elastic energy functional isIts corresponding total potential subtracts the work of fixed body and surface loads:The problem in the calculus of variations is to minimize $\Pi$ among admissible displacements with $u=u_D$ on ${\Gamma }_D$. For a stable quadratic elastic energy this is the usual energy minimization formulation. Its stationarity condition suffices to derive equilibrium.

Vary by $u+sv$, with $v=0$ on ${\Gamma }_D$. Major symmetry combines the two differentiated quadratic terms, givingIntegration by parts, first with test functions compactly supported in the interior, gives $-\mathrm{Div}{S}^{\mathrm{lin}}=f$. Then arbitrary traces on ${\Gamma }_N$ give the natural traction condition. The complete boundary value problem isFor a homogeneous hyperelastic material the elasticity tensor is constant. In isotropic linear elasticity the constitutive equation is $S^{\mathrm{lin}}=\lambda \mathrm{div}uI+\mu (Du+D{u}^T)$, giving the familiar Navier-Cauchy equation $-\mu \Delta u-(\lambda +\mu )\nabla \mathrm{div}u=f$.

At $\alpha =0$ the transverse coordinates are independent Brownian motions with diffusion coefficient $D=1$. The cylinder is infinitely long, so the axial coordinate cannot affect its first contact time. By rotational invariance of Brownian motion, the mean first-passage time $u$ depends only on the transverse radius $\rho =\sqrt{x^2+y^2}$.

The Kolmogorov backward equation for a mean first-passage time is $Lu=-1$. Here $L={\partial }_x^2+{\partial }_y^2$, so the radial boundary value problem isThe absorbing boundary condition for a diffusion is imposed on the circular cross-section. At $\rho =0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating givesSmoothness at the axis excludes the logarithm term, and $u(R)=0$ fixes $C_0=R^2/4$. ConsequentlyThis also verifies the finiteness assumption behind the Kolmogorov backward equation. Let $T$ be the first contact time and apply the stopped Itô formula to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped stochastic integral a martingale, givingBy monotone convergence, $T$ has finite expected value; continuity and dominated convergence then make the final term vanish as $t\to \infty$. Hence $u({\hat{r}}_0)=\mathbb{E}T$. In particular, the mean first-passage time is $R^2/4$ on the axis and tends to zero at the adsorbing wall.

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.

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.

Set $g=f_1^{\prime }$. The boundary value problem becomesFor $y_1=1+{\zeta }^2$, direct substitution gives $y_1^{\prime \prime }+\zeta y_1^{\prime }-2y_1=2+2{\zeta }^2-2(1+{\zeta }^2)=0$. This growing polynomial solution is useful for reduction of order.

Write $g=y_{1}v$. Substitution into the linear ordinary differential equation cancels the terms proportional to $v$ and givesAn integrating factor for this first-order equation for $v^{\prime }$ yieldsThe far-field boundary condition selects the convergent integral from $-\infty$:Any additional multiple of $1+{\zeta }^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta \ge 1$,Here $(1+t^2{)}^{-2}\le (1+r^2{)}^{-2}$ and ${\int }_r^{\infty }{e}^{-t^2/2}dt\le r^{-1}{\int }_r^{\infty }t{e}^{-t^2/2}dt$. Thus the far-field boundary condition is verified without a cancellation of large terms.

At $\zeta =0$, differentiating gives $g^{\prime }(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answerHere $\mathrm{erf}$ is the error function. Finally,recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal velocity matching. This is the weak surface-shear perturbation of stagnation flow.

If the printed $+\infty$ condition is instead imposed literally on a positive half-line, reduction of order givesIt satisfies $g_{+}^{\prime }(0)=1$ but grows as $\zeta \to -\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.

For $y=1+x^2$, the derivatives are $y^{\prime }=2x$ and $y^{\prime \prime }=2$. Its endpoint values are $y(0)=1$ and $y(2)=5$, so it satisfies both the ordinary differential equation and the boundary conditions. Moreover any solution of $y^{\prime \prime }=2$ has the form $x^2+Ax+B$; the boundary conditions force $B=1$ and $A=0$. Thus the unique solution of the boundary value problem is

Put $a=\pi /2$. Substituting $x=-a$ and $x=a$ into part (ii), using $\cos (\pi /4)=\sin (\pi /4)=1/\sqrt{2}$, givesThus $A-B=\sqrt{2}(1+e^{-a})$ and $A+B=\sqrt{2}(1-e^a)$. Solving these two linear equations givesConsequently the solution of the boundary value problem isThe two independent boundary conditions determine $A,B$ uniquely.

Suppose the mixed boundary value problem $\Delta w=g$, with zero Dirichlet boundary condition on a nonempty boundary portion and zero Neumann boundary condition on the remainder, has an $H^2$ solution with $\Vert w{\Vert }_{H^2}\le C\Vert g{\Vert }_{L^2}$. Assume the domain is connected and that zero Sobolev trace on the specified portion excludes nonzero constants. Then the same boundary conditions for $-\Delta z+z=e$ give $\Vert z{\Vert }_{H^2}\le 2C\Vert e{\Vert }_{L^2}$.

Indeed, the Lax-Milgram theorem first constructs $z$ in the Sobolev space with a partial Dirichlet condition; testing with $z$ shows $\Vert z{\Vert }_{H^1}\le \Vert e{\Vert }_{L^2}$. Apply the assumed elliptic regularity result to $g=z-e$. Its solution $w$ and $z$ satisfy the same weak Laplace equation, so testing their difference shows $\nabla (w-z)=0$. The Sobolev function with zero weak gradient property and zero Sobolev trace give $w=z$. Finally $\Vert z-e{\Vert }_{L^2}\le 2\Vert e{\Vert }_{L^2}$. This argument supplies the dual regularity needed by the Aubin–Nitsche duality argument for the equation with a positive reaction term.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

A stationary solution of an evolution differential equation is independent of time. Substituting ${\partial }_{t}u=0$ leaves the spatial boundary value problem that determines its profile. For an ordinary differential equation this is an equilibrium; for a partial differential equation it may vary in space. A stationary profile need not have zero flux, although zero flux is forced in a one-dimensional conserved system with reflecting endpoints.