Choose a standard mollifier $\rho \in C_0^{\infty }({\mathbb{R}}^n)$ with $\rho \ge 0$, $\int \rho =1$ and support in the unit ball, and set ${\rho }_{\epsilon }(z)={\epsilon }^{-n}\rho (z/\epsilon )$. Define the convolutionThe smoothing properties of a mollifier give $u_{\epsilon }\in C^{\infty }$ and ${\partial }_j{u}_{\epsilon }={\rho }_{\epsilon }\ast D_{j}u$. Apply the chain rule to $t\mapsto u_{\epsilon }(y+t(x-y))$ and integrate its derivative by the fundamental theorem of calculus:

An integrable example on the whole real line is the tent functionIt is not differentiable at $-1,0,1$. For a test function $\phi$, split the integral into $(-1,0)$ and $(0,1)$ and use integration by parts. The endpoint terms at $-1,1$ vanish because $u=0$, and those at zero cancel because $u$ is continuous. The result is $\int u{\phi }^{\prime }=-\int (Du)\phi$. Thus the corners obstruct classical differentiability but not this weak derivative.

The fundamental lemma of the calculus of variations says that if $h\in L_{\mathrm{loc}}^1(\Omega )$ and $\int h\phi =0$ for every $\phi \in C_0^{\infty }(\Omega )$, then $h=0$ almost everywhere.

If $v_1,v_2$ both represent $D^{\alpha }u$, subtracting their defining identities gives $\int (v_1-v_2)\phi =0$ for every test function. The lemma gives $v_1=v_2$ almost everywhere, proving uniqueness. If $u\in C^1$, part (i) gives exactly the defining identity with $v={\partial }_{j}u$. Since this classical derivative is locally integrable, it is the unique weak derivative $D_{j}u$, up to equality almost everywhere.

The function $v$ is the weak derivative $D^{\alpha }u$ ifThe integrals exist by local integrability and the compact support of the test function. This is the definition of a distributional derivative represented by a locally integrable function.

For one coordinate, integration by parts gives ${\int }_{\Omega }({\partial }_{j}u)\phi =-{\int }_{\Omega }u{\partial }_j\phi$: the test function vanishes outside a compact subset of $\Omega$, so there is no boundary contribution. Repeat this operation once for each of the $|\alpha |$ coordinate differentiations in the multi-index $\alpha$. Each operation contributes a factor $-1$, givingAll intermediate functions are continuous on the relevant compact support, since $u\in C^k$ and $|\alpha |\le k$.

We prove existence directly through quadratic energy minimization by the parallelogram law. Let $m={\inf }_{v\in H_g^1}A[v]$. The set contains $g$, and uniform ellipticity gives $0\le m\le A[g]<\infty$. Choose a minimizing sequence $v_j$ with $A[v_j]\to m$. Since $(v_j+v_k)/2\in H_g^1$, the parallelogram law givesThe differences belong to $H_0^1$, so uniform ellipticity and the Poincaré inequality implyThus $(v_j)$ is a Cauchy sequence. Completeness of $H_0^1$, applied to $v_j-g$, produces $u\in H_g^1$ with $v_j\to u$ in $H^1$. The bounded coefficient matrix makes $a$ a bounded bilinear form, so $A[v_j]\to A[u]$. Consequently $A[u]=m$, and part (i) makes $u$ a weak solution.

For uniqueness, the difference $w$ of two weak solutions lies in $H_0^1$ and satisfies $a(w,w)=0$. Uniform ellipticity gives $\nabla w=0$ almost everywhere, and the Poincaré inequality gives $w=0$. The argument proves existence and uniqueness without assuming the Lax-Milgram theorem.

Write $a(v,w)={\int }_{\Omega }a\nabla v\cdot \nabla w$ and $A[v]=a(v,v)$. The admissible set is the closed affine subspace $H_g^1=g+H_0^1$. Symmetry of the coefficient matrix givesIf $u$ minimizes the energy, its first variation vanishes: differentiating at $t=0$ gives $a(u,\phi )=0$. Hence $u$ is a weak solution.

Conversely, if $u$ is a weak solution, every admissible $v$ has $v-u\in H_0^1$ and thereforeby uniform ellipticity. This is the Dirichlet principle for the stated symmetric energy.

A weak solution is a function $u\in H_0^1(\Omega )$ satisfying the weak formulationMultiplication by a test function and integration by parts give this identity from the classical equation. Conversely it is the equation in the sense of distributions, with the zero boundary condition encoded by the zero-boundary Sobolev space. Testing only against $C_0^{\infty }(\Omega )$ is equivalent, by its density in $H_0^1$ and continuity of both sides.

The space of test functions is $\boxed{C_0^{\infty }(\Omega )=\{\phi \in C^{\infty }(\Omega ):\mathrm{supp}\phi \text{ is a compact subset of }\Omega \}}$. The Sobolev space iswhere $D_{j}u$ denotes a weak derivative, and its Sobolev norm is $\Vert u{\Vert }_{H^1}^2=\Vert u{\Vert }_2^2+{\sum }_j\Vert D_{j}u{\Vert }_2^2$. The zero-boundary Sobolev space is $\boxed{H_0^1(\Omega )={\overline{C_0^{\infty }(\Omega )}}^{H^1}}$. This closure defines the boundary condition even when the boundary has insufficient regularity for a classical boundary value.

Here uniform ellipticity means that there is a constant $\lambda >0$, independent of $x$, such thatThe assumption that the entries are essentially bounded functions also gives an upper bound $|a(x)\xi |\le \Lambda |\xi |$ with a finite constant $\Lambda$. For this symmetric divergence-form operator, the lower bound makes its energy control the squared Euclidean norm of the gradient.

Let $c_{\uparrow }(w)$ denote the increasing-front speed from the preceding solution. Reversing that heteroclinic reverses its speed, so a decreasing back has speed $c_{\downarrow }(w)=-c_{\uparrow }(w)$. A rigidly travelling pulse has one speed throughout, requiringThis is the correct use of the integral formula: compare the oppositely oriented transitions, including their signs, rather than equating two increasing-front speeds.

For completeness, write $f(v)-w=-(v-r_1)(v-r_2)(v-r_3)$, $r_1<r_2<r_3$. The explicit increasing frozen-recovery cubic front isSubstitution, using $v_X=(v-r_1)(r_3-v)/\sqrt{2}$, verifies the equation. Since $r_1+r_2+r_3=1+a$, speed matching givesFor $0<a<1/2$ this selects the appropriate positive back level within the cubic's three-root range. The high branch connects $B$ to $C$ slowly, and the low branch supplies recovery after $D$.

The stated positive-speed, rising-front orientation imposes $a<1/2$ at leading order. The printed range $0<a<1$ by itself does not guarantee that pulse construction: at $a=1/2$ the leading front is stationary, while for $a>1/2$ the calculated increasing-front speed is negative. Existence and stability of a full finite-$ϵ$ pulse also require the slow connections; speed matching is its leading necessary construction, rather than a proof for every parameter in the printed range.

Put $X=z/ϵ$. The travelling equations become $c{v}_X=f(v)-w+v_{XX}$ and $c{w}_X=ϵ(\gamma v-w)$. Hence, for nonzero $O(1)$ speed and $ϵ\gamma \ll 1$, $w$ is constant to leading order across the front.

Writing $p=v_X$ gives the planar system $v_X=p$, $p_X=cp-f(v)+w$. At the outer cubic roots $A=(V_{-},0)$ and $B=(V_{+},0)$, $f^{\prime }(V_{\pm })<0$. The Jacobian determinant is $f^{\prime }(V_{\pm })<0$, so both are saddle points. The increasing front is a heteroclinic orbit from the unstable manifold of $A$ to the stable manifold of $B$. Varying $c$ supplies the shooting parameter required to match these manifolds. For this cubic one can also construct it explicitly, as in the next solution.

Multiply the second-order equation by $v_X$ and integrate across the front. Because $v_X\to 0$ at both ends, the term $\int v_{XX}{v}_{X}dX=[v_X^2/2{]}_{-\infty }^{\infty }$ vanishes. The remaining identity isThe denominator is positive for a nonconstant front, so its direction is determined by the signed area between the cubic and the horizontal recovery line: Maxwell balance for a scalar reaction-diffusion front gives zero speed when that area vanishes.

The derivative in the denominator must use the inner coordinate $X$. If one instead uses the source's original $z$ and denotes $J_z=\int v_z^{2}dz$, then $J_z={ϵ}^{-1}\int v_X^{2}dX$ and the identity is $c=\int (f-w)dv/(ϵJ_z)$. Thus the printed $v^{\prime }(z)$ is valid only if $z$ has silently been reused for the rescaled inner coordinate; with the original coordinate it misses a factor $ϵ$.

For $z=ct-x$ the chain rule gives ${\partial }_t=cd/dz$, ${\partial }_x=-d/dz$, ${\partial }_{xx}=d^2/d{z}^2$. ThusAs $z$ increases from the material ahead of the wave towards that behind it, the singular pulse construction is $A\to B\to C\to D\to A$. The front $AB$ raises $v$ at approximately fixed $w=0$ from the low root to the high root. On the plateau $BC$, $f(v)\simeq w$ on the right attracting branch while $w$ rises slowly. The back $CD$ decreases $v$ at approximately fixed $w=w_0$ to the left attracting branch. The recovery tail then lowers $w$ back to zero on that branch, returning to $A=(0,0)$. Both sharp transitions have thickness $O(ϵ)$ in $z$; the other segments have the slower recovery scale.

The figure labels all four points and the front/back orientations. This frozen-recovery construction needs $ϵ\gamma \ll 1$ as well as $ϵ\ll 1$ if $\gamma$ is large, so that the recovery variable remains nearly fixed in an inner transition. It also needs an admissible positive front speed for the chosen orientation, addressed below.

For the first Hopf boundary of scalar delayed negative feedback, first, if a root has nonnegative real part then $|\lambda +\beta |=\gamma e^{-\mathrm{Re}\lambda }\le \gamma$. For $\beta \ge \gamma$ this is impossible, apart from the potential equality case $\lambda =0$, already excluded. Thus $\beta <\gamma$ is necessary for instability, although not sufficient by itself.

At an imaginary root $\lambda =i\omega$ with $\omega >0$, separate real and imaginary parts:The first crossing uses ${\omega }_c\in (\pi /2,\pi )$ for $\beta >0$, and satisfiesThe period is in scaled time; its physical value is $\tau {\mathcal{T}}_c$. Higher-frequency branches have larger critical gains. For fixed $\beta$, stability at zero delayed gain and absence of any earlier imaginary or zero-root crossing identify this lowest branch as the first stability boundary. Its crossing is transverse: implicit differentiation of $\lambda +\beta +\gamma e^{-\lambda }=0$ gives at the imaginary rootFor very small $\beta$, ${\omega }_c=\pi /2+2\beta /\pi +O({\beta }^2)$, so ${\mathcal{T}}_c\to 4$ and ${\gamma }_c=\pi /2+2\beta /\pi +O({\beta }^2)$. Therefore ${\gamma }_c=\pi /2$ is exact in the limit $\beta =0$ and is the leading approximation for small positive $\beta$. At that limit $\lambda e^{\lambda }=-\gamma$ first has an imaginary pair at $\lambda =\pm i\pi /2$; subsequent pairs occur at $\pi /2+2n\pi$ and cannot be the first loss of stability. A nonlinear Hopf theorem, including its nondegeneracy conditions, is needed to turn the linear imaginary mode into a claim about a particular finite-amplitude periodic branch.

For real $\lambda \ge 0$, draw $y=\lambda +\beta$ and $y=-\gamma e^{-\lambda }$. The first is nonnegative and increasing, while the second is negative and approaches zero from below. They cannot intersect on the nonnegative real axis. At zero the characteristic function equals $\beta +\gamma >0$, so a stationary zero-root crossing is excluded as well.

Consequently a loss of stability from a stable regime cannot occur by a real eigenvalue crossing through zero. It must involve a conjugate pair crossing the imaginary axis at nonzero frequency, yielding oscillatory instability. This does not mean that every characteristic root is nonreal: negative real roots may exist and are harmless for stability. The graph and imaginary-axis conditions are shown together below.

Since the steady state is strictly positive, the positive-part function is differentiable in a neighbourhood of it. Write $p=p_{\ast }+q$. Retaining first-order terms givesSubstitute the exponential mode $q=e^{\lambda T}$ to obtainHere the last equality uses the exact steady-state equation. For small $\mu$, $\beta =\alpha \mu +O(\alpha {\mu }^2)$ and $\gamma =\alpha (1+\mu )+O(\alpha {\mu }^2)$. Linearizing at the exact root avoids carrying a spurious constant residual from the printed approximate equilibrium.

Distinguish physical time $t$ from scaled time $T=t/\tau$. PutThen $K(P_s/\tau )p^{\prime }=M-[P_0+P_{s}p](M/P_0)v$, givingA steady state must be positive, since otherwise ventilation vanishes while production remains positive. It solves $\mu p_{\ast }^2+p_{\ast }-1=0$, soThe value $1/(1+\mu )=1-\mu +{\mu }^2+O({\mu }^3)$ printed in the question agrees only through first order. Thus it is a small-$\mu$ approximation with error of order ${\mu }^2$, not an exact equilibrium for nonzero $\mu$. The other quadratic root is negative and inadmissible for this steady ventilation balance.

The loaded-carrier balance gives $J=k(p_e-p_i)$. Substitute the preceding two complex concentrations and collect the current terms:Next carrier conservation givesEliminate $c_e$ between these relations and set $K_d=k/k_{+}$. Simplification yieldsHere $c_0$ must be the total carrier protein, as in the conservation equation. Calling it total substrate in the final line of the scan is a typographical inconsistency; total substrate is the different invariant $s_0$.

For carrier-turnover saturation of passive transport at fixed $s_i$, write $A=s_i+K+K_d$, $D=A(K+K_d)-K_d^2>0$ and $C=K_{d}K{k}_{+}{c}_0/2>0$. Then $J=C(s_e-s_i)/(A{s}_e+D)$, soThe flux is outward when the substrate gradient is reversed and inward when $s_e>s_i$. Saturation occurs because there is a finite carrier population and finite binding-release/conformational turnover: arbitrarily abundant exterior substrate cannot make carriers cycle arbitrarily quickly. This is passive facilitated transport, so no maintained flux remains when the two free concentrations agree. Varying $s_e$ at fixed $s_i$ compares reservoir-controlled steady states or systems with different total substrate; it is not a trajectory preserving one fixed $s_0$.

The exterior free-substrate equation gives $B_e=J$ at steady state. Thus $k_{+}{s}_e{c}_e-k_{-}{p}_e=J$, andThe sign of the current term is opposite to the interior relation because substrate is supplied to the exterior and removed from the interior. Nonnegative occupancies are selected by the complete steady transport solution and compatible substrate totals.

The relation $B_i=-J$ means $k_{-}{p}_i=k_{+}{s}_i{c}_i+J$. Put $K=k_{-}/k_{+}$ and substitute the preceding expression for $c_i$:The first term is the equilibrium binding contribution from the outward free-carrier pool; the additional terms account for the maintained current and the excess inward-facing free carriers. This is a steady throughput relation, not a detailed-balance assumption.