Write $\theta =T-T_e$ and use the Boussinesq approximation for air, ${\rho }_e-\rho \simeq \rho \theta /T_e$. A uniformly heated, well-mixed room has the buoyancy pressure difference $\Delta p_b=\rho gH\theta /T_e$ between its openings. Let $C_o$ denote the combined hydraulic coefficient of the two equal openings, so the effective opening area for pressure-driven ventilation givesFor example, if each orifice has conventional discharge coefficient $C_d$ in $V=C_{d}A\sqrt{2\Delta p/\rho }$, the two pressure losses add and $C_o=C_d$. The question's coefficient is the prefactor in the final heat-load law; define $c=(2C_o^2{)}^{1/3}$. This explicitly states the numerical loss-coefficient convention. Steady heat balance is $Q=\rho C_{p}V\theta$. Eliminating $\theta$ givesHere $Q$ is total heating power and $C_p$ is specific heat capacity at constant pressure. Writing $c=C_d$ without this conversion would incorrectly change the power of the discharge coefficient.

After the reduction, cooler newly heated air occupies the bottom and displaces the old warmer air upward. Use the ideal two-layer displacement ventilation model: each layer is well mixed, the upper layer remains at ${\theta }_0$, walls have negligible heat capacity, and the sharp interface is advected without exchange between the layers. Let the lower layer have depth $h$ and excess temperature $\theta$. Integrating the hydrostatic buoyancy over the full height givesThe room's excess thermal energy is $E_T=\rho C_p{A}_r[h\theta +(H-h){\theta }_0]$. The floor supplies $\lambda Q$, incoming air has zero excess temperature, and the ceiling removes $\rho C_{p}V{\theta }_0$. Thus the required heat balance isThis moving-interface energy term must include the loss of old upper-layer volume; balancing only the lower-layer temperature would miss it.

For a genuine reduction assume $0<\lambda <1$. SetThe integrated buoyancy factor is $H{\theta }_0(1-x)$, so $V/V_0=\sqrt{1-x}$. Since $Q=\rho C_p{V}_0{\theta }_0$, the thermal balance becomes the ventilation after a heating reduction equationThe material interface also moves with the upward displacement velocity, $A_r\dot{h}=V$. Therefore ${\eta }_s=\sqrt{1-x}$, and subtracting the two equations givesThe second relation is also the lower-layer energy balance: its total excess energy increases at rate $\lambda Q$. It does not assume its temperature stays constant. Since $x_s>0$ until $x=1-{\lambda }^2$, the solution remains $0\le x<1-{\lambda }^2$, and $V/V_0\ge \lambda$. Hence ${\eta }_s\ge \lambda$, so the lower layer reaches $\eta =1$ in finite time $s_H\le 1/\lambda$. Also ${\eta }_s\le 1$, so $s_H\ge 1$.

An explicit implicit solution is useful. Put $y=\sqrt{1-x}$, which decreases from $1$ toward $\lambda$. Separation givesFilling occurs when $1=1-y_H^2+\lambda s_H$, so $\lambda s_H=y_H^2$ and ${\theta }_H/{\theta }_0=y_H^2$. These relations determine the exit time and the continuous temperature at that time. For complete switch-off $\lambda =0$, the limiting solution is $y=1-s/2$, $x=\eta =s-s^2/4$ for $0\le s\le 2$, so the cold lower layer fills the room at $s_H=2$ even though its final ventilation rate is zero.

After the interface exits, the whole room is well mixed. For $z=\theta /{\theta }_0$, the natural ventilation rate is $V=V_0\sqrt{z}$, and the well-mixed ventilation temperature balance becomesEquivalently, $\rho C_p{A}_{r}H\dot{\theta }=\lambda Q-\rho C_p{C}_{o}A\sqrt{gH/T_e}{\theta }^{3/2}$. The heat-removal term is strictly increasing in $\theta \ge 0$, so every nonnegative solution tends to the unique stable equilibriumFor $\lambda =0$ the final temperature is $T_e$. The unchanged-load case $\lambda =1$ has no distinct new temperature layer and is outside the reduction scenario.

Take the boundary to be horizontal, put $z=0$ on it, and let $z=h(x,t)$ be the interface. Write $\Delta \rho =\rho -{\rho }_2>0$. Under the long-thin hydrostatic approximation, the deep ambient fluid has pressure $p_2=p_{\ast }-{\rho }_{2}gz$ to leading order, with negligible horizontal pressure gradient. Matching pressure at the interface gives, inside the dense porous gravity current,The Darcy velocity is superficial volume flux per total area, not pore speed. The horizontal Darcy law and depth-integrated flux therefore giveThe porosity $\varphi$ multiplies storage, so mass conservation is $\varphi h_t+q_x=0$. ConsequentlyThe deep-ambient and long-thin assumptions are what permit neglecting the ambient return-flow pressure variation, vertical inertia, and nonhydrostatic pressure corrections.

For a localized finite release, first use the reflecting half-line $x\ge 0$, so $h_x(0,t)=0$. Let $S=V/\varphi$ be the conserved cross-sectional area of the current; $V$ here is actual pore-fluid volume per unit cross-flow distance. A similarity solution of the porous medium equation has thickness proportional to $t^{-\alpha }$ and extent proportional to $t^{\beta }$. Conservation gives $\alpha =\beta$, and balancing $h_t$ against $(h{h}_x{)}_x$ gives $\alpha +2\beta =1$. Thus $\alpha =\beta =1/3$. PutSubstitution gives $-(F+\xi F^{\prime })/3=(F{F}^{\prime }{)}^{\prime }$. Zero flux at $\xi =0$ allows one integration: $F{F}^{\prime }=-\xi F/3$. In the positive-current region, $F^{\prime }=-\xi /3$, so $F=({\xi }_f^2-{\xi }^2)/6$. The normalization ${\int }_0^{{\xi }_f}Fd\xi =1$ gives ${\xi }_f^3=9$. The one-sided constant-volume porous gravity current is thereforeHere $[a{]}_{+}=\max (a,0)$. Its Darcy flux vanishes at the moving front because $h=0$ there, and its physical volume is $\varphi {\int }_0^{L}hdx=V$. This is the exact concentrated-release Barenblatt solution of the depth equation, physically used after the short initial adjustment to a long-thin current; an arbitrary finite initial patch requires its initial profile as well as its volume, although the same profile describes the usual late spreading regime. If a line release spreads symmetrically in both directions and $V$ denotes the total volume, use the same parabola on $-L<x<L$ with $L^3=9\lambda Vt/(2\varphi )$. If instead volume has already been divided by porosity, replace $V/\varphi$ by that geometric volume.

Drainage at superficial rate $\varphi \Gamma h$ changes mass conservation to $\varphi h_t+q_x=-\varphi \Gamma h$, soFor constant $\Gamma >0$, use the exponential drainage transform for porous-medium diffusionIndeed, $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $(h{h}_x{)}_x=e^{-2\Gamma t}(H{H}_x{)}_x$. Choosing ${\tau }^{\prime }=e^{-\Gamma t}$ leaves $H_{\tau }=\lambda (H{H}_x{)}_x$. Substituting the undrained Barenblatt solution givesUse the factor $1/2$ in $L_d^3$ for a total-volume symmetric release, as above. For general initial data the same transform is exact, but $H$ is the corresponding undrained solution rather than automatically a source parabola.

The remaining physical volume is $V(t)=V{e}^{-\Gamma t}$. Early times have $\tau \sim t$, recovering undrained spreading. Late times have $\tau \to 1/\Gamma$, so the front approaches a finite runout while the current drains exponentially:The front slows because drainage reduces the hydrostatic driving head. The finite transformed time also explains why drainage does not guarantee that an arbitrary initial profile reaches its undrained universal large-time shape.

The natural expectation is orbital stability, since a spatial translation of a travelling wave is another travelling wave. Thus convergence for ordinary perturbations, when it holds, is to a suitably translated front. It also matters which perturbations are allowed in the leading edge: the zero state ahead of this front is linearly unstable, and a sufficiently shallow positive tail can select a different speed. One should not assert asymptotic stability for arbitrary disturbances without a norm and a class of initial data. The requested essential spectrum calculation is a necessary part of such an analysis, not by itself a complete nonlinear stability proof.

In the moving coordinate $z=x-ct$, write $u(x,t)=w_c(z)+v(z,t)$. Expansion of the reaction term givesThe first equation to study is its linearization,and the associated eigenvalue equation $\mathcal{L}\varphi =\lambda \varphi$. Differentiating the front equation gives $\mathcal{L}{w}_c^{\prime }=0$, the neutral translation mode whenever it belongs to the chosen function space. Ahead of the front the limiting coefficient is $k$, so in unweighted L2 space the limiting dispersion relation is $\lambda =-{\xi }^2+ic\xi +k$: its rightmost real part is positive. An exponential weight penalizes precisely these disturbances ahead of the front.

There is a small functional-analytic correction to the printed definition. Use the complete exponentially weighted L2 spaceThe additional requirement of unweighted L2 space membership in the printed set makes that set incomplete under its stated weighted norm. Indeed $v_N=1_{[-N,0]}$ belongs to the intersection and is a weighted Cauchy sequence, but its weighted limit $1_{(-\infty ,0]}$ does not belong to unweighted L2 space. The usual Hilbert space for this spectral argument is the completion $X_{\gamma }$. We use that completion explicitly, rather than treating the intersection as a complete space.

Multiplication $T_{\gamma }v=e^{\gamma z}v$ is a unitary map from $X_{\gamma }$ onto L2 space. Take the domain $T_{\gamma }^{-1}{H}^2(\mathbb{R})$ for $\mathcal{L}$. The transformed closed operator isThis follows from $T_{\gamma }{\partial }_z{T}_{\gamma }^{-1}={\partial }_z-\gamma$. The two limiting constant-coefficient dispersion relations, obtained by the Fourier transform, areIn particular the necessary strict leading-edge condition is ${\gamma }^2-c\gamma +k<0$. For $c>2\sqrt{k}$ this meansA particularly convenient choice isWith this choice there is no first derivative, and ${\mathcal{L}}_{\gamma }={\partial }_z^2+k-c^2/4-2w_c(z)$ is a self-adjoint differential operator on $H^2(\mathbb{R})$. This also avoids possible ambiguities between definitions of essential spectrum for nonnormal operators.

The relevant perturbation theorem is essential-spectrum invariance under relatively compact perturbations: if $A_0$ is a closed operator, $V:D(A_0)\to H$ is compact with respect to the graph norm, and $A_0+V$ is closed on the same domain, their Fredholm essential spectrum is the same. For self-adjoint operators this is the usual essential spectrum after removing isolated finite-multiplicity eigenvalues. To see why the theorem applies, regard $A_0-\lambda$ and $A_0+V-\lambda$ as bounded maps from the graph norm domain to $H$, and use compact perturbation invariance of Fredholm operators for each $\lambda$.

Define a step-potential comparison operatorBoth $A_0$ and ${\mathcal{L}}_{\gamma }$ are self-adjoint on $H^2(\mathbb{R})$. Their difference is multiplication byIt is relatively compact, not necessarily compact as a multiplication map on all of L2 space. To prove the needed compactness, take a sequence bounded in $H^2(\mathbb{R})$. On each bounded interval the Rellich-Kondrachov compactness theorem gives a subsequence converging in L2 space. A diagonal subsequence makes the products by $V$ converge locally. Outside a sufficiently large interval, $\Vert V{\Vert }_{\infty }$ is arbitrarily small and the sequence has a uniform L2 norm bound, so those tails are uniformly small. Thus the products converge globally in L2 space, proving graph-domain compactness.

Finally,Here is a direct verification. Its quadratic form is at most $(k-c^2/4)\Vert h{\Vert }_2^2$, so no spectrum lies above this endpoint. For every $\lambda \le k-c^2/4$, choose $\xi$ with $\lambda =k-c^2/4-{\xi }^2$. Normalized, increasingly long cutoffs of $e^{i\xi z}$, translated far to the right, have disjoint supports and satisfy $\Vert (A_0-\lambda )h_N{\Vert }_2\to 0$: differentiating the cutoff gives errors of orders $N^{-1}$ and $N^{-2}$. They converge weakly to zero. This is a singular sequence in the Weyl criterion, proving that every such $\lambda$ belongs to the essential spectrum. The relatively compact perturbation theorem therefore givesThis proves the weighted essential spectrum of a Fisher travelling front is strictly stable when $c>2\sqrt{k}$. In fact here the quadratic form gives more:Since $w_c\ge 0$, the full transformed spectrum lies below $-(c^2/4-k)$, and the evolution given by its linearization decays at least as $e^{-(c^2/4-k)t}$ in this weighted norm. This does not contradict translation invariance: the nonzero translation mode cannot lie in this weighted operator domain, since otherwise it would be a zero eigenfunction, contradicting the strictly negative quadratic-form bound. The weight restricts the permitted perturbations and excludes these neutral translations. Nor does a weighted L2 norm alone control the nonlinear remainder at $z\to -\infty$; a nonlinear theorem needs further boundedness or regularity assumptions.

At the minimal speed $c_{\ast }=2\sqrt{k}$, the best choice is $\gamma =\sqrt{k}$. Now ${\gamma }^2-c\gamma +k=(\gamma -\sqrt{k}{)}^2$, andThe essential spectrum has no positive part but touches zero; no positive exponential weight makes it strictly negative. Thus the critical front has no spectral gap and this argument gives no exponential decay rate there. Strict weighted spectral stability holds above the minimal speed; at the minimal speed it is marginal.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

Use the usual fractional Dirichlet domain scale: $A=-{\Delta }_D$ on $H=L^2(\Omega )$, and $H_{\rho }=D(A^{\rho })$. Its Sobolev embedding theorem givesSuch a positive $\beta$ exists precisely because the assumed $\delta$ is greater than $n/4$. Consequently $v=u_t$ is continuous in time with values in this Hölder space on every compact positive-time interval. This embedding alone gives no two spatial derivatives for $v$; those must come from the equation.

First establish a bound that will also justify the nonlinear regularity steps. Put $F(s)=\lambda s-s^3$ andThe constants $K$ and $-K$ enclose both the initial data and the zero Dirichlet boundary data. Moreover $F(K)\le 0$ and $F(-K)\ge 0$, so they are a supersolution and a subsolution for the parabolic comparison principle. A direct energy proof avoids assuming pointwise continuity at time zero. Let $h=(u-K{)}_{+}$. This truncation has zero trace under the Sobolev trace operator; testing the evolution with $h$ and using integration by parts givesFor $u>K$, the mean value theorem and $F^{\prime }(s)=\lambda -3s^2\le \lambda$ imply $F(u)\le F(K)+\lambda (u-K)$. ThereforeThe Gronwall inequality gives $h=0$, since its initial L2 norm vanishes. For the initially rough datum, apply the inequality from $\epsilon >0$ and let $\epsilon \downarrow 0$, using $u(\epsilon )\to u_0$ in L2 space. The same argument for $-u$, which satisfies the same equation because $F$ is odd, provesThis is the invariant interval for a cubic reaction-diffusion equation; the energy calculation is valid by the Sobolev chain rule and approximation even before the stronger pointwise regularity is established.

Now fix $0<\tau <t_0$ and, if necessary, reduce the positive Hölder exponent $\beta$. Rearranging the evolution gives the elliptic identityIts right side is bounded on $[\tau ,t_0]\times \Omega$, because $v\in C([\tau ,t_0],C^{0,\beta })$ and $|u|\le K$. Boundary elliptic regularity first gives $u(t)\in W^{2,p}(\Omega )$ for every finite $p$, uniformly on this time interval. For $p>n$, the Sobolev embedding theorem makes $u(t)$ spatially continuously differentiable, with a positive Hölder exponent. Thus $F(u(t))$ lies in $C^{0,\beta }$ after reducing $\beta$ if necessary. The global Schauder estimate then gives $u(t)\in C^{2,\beta }(\overline{\Omega })$.

These are continuous-in-time, not just separate-time, regularity statements. Since $v\in C(C^{0,\beta })$, the identity $u(t)-u(s)={\int }_s^{t}v(r)dr$, initially an identity in L2 space, holds in $C^{0,\beta }$ once $u(s)$ is known to belong to that space. Hence $u$ is locally Lipschitz in time into $C^{0,\beta }$. Apply the global Schauder estimate to the difference of the two elliptic identities. Continuity of $v$ and of $F(u)$ into $C^{0,\beta }$ proves $u\in C([\tau ,t_0],C^{2,\beta })$.

To gain the spatial regularity of the time derivative itself, take time difference quotients of the evolution. If $d_h(t)=[u(t+h)-u(t)]/h$, their equations arewhereThe coefficients have uniformly bounded spatial Hölder norms and temporal Hölder norms on compact positive-time cylinders: this follows from the spatial regularity just proved and the local time Lipschitz bound in $C^{0,\beta }$. Also $d_h={\int }_0^{1}v(t+sh)ds$ is uniformly bounded and tends to $v$ in $C^{0,\beta }$. Positive-time parabolic Schauder estimates, applied away from the lower time edge, give uniform $C^{1+\beta /2,2+\beta }$ bounds for $d_h$, with a slightly smaller exponent if needed. Passing to the limit proves that $v$ is continuous into $C^2(\overline{\Omega })$ and satisfiesThe Dirichlet boundary condition for $v$ follows also by differentiating the identically zero boundary value in $C^0$. Since $\tau$ was arbitrary, this proves the requested, genuinely stronger regularity:This is positive-time smoothing for a semilinear heat equation; it is the extra parabolic step for $v$, not just the Sobolev embedding theorem, that gives the $C^1$ topology asserted here.

Finally, a uniform bound excludes a finite maximal existence time for this semilinear partial differential equation. Restart at any positive time $s$, using the heat semigroup on the Banach space $X=\{g\in C(\overline{\Omega }):g{|}_{\partial \Omega }=0\}$ with the supremum norm. Its variation-of-constants formula isThe heat semigroup is a contraction semigroup on $X$. On the ball of radius $K+1$, $F$ is Lipschitz with constant at most $|\lambda |+3(K+1{)}^2$, and bounded by $|\lambda |(K+1)+(K+1{)}^3$. Choose a time length $\eta >0$ so that the integral is at most $1$ and its Lipschitz constant is less than $1$. The Banach fixed-point theorem then gives a local solution on $[s,s+\eta ]$, with $\eta$ independent of $s$, because $\Vert u(s){\Vert }_{\infty }\le K$. If a finite maximal time existed, choosing $s$ less than $\eta$ before it would extend the same solution past it, a contradiction. Uniqueness identifies this continuation with the original solution, and positive-time smoothing restores the classical solution on each extended interval. The solution is globally defined and remains between $-K$ and $K$.

A sufficient set of assumptions is that $L$ is a densely defined closed operator on the Hilbert space $H$, generating an analytic semigroup $S(t)$, and that $f\in C^{0,\theta }([0,T],H)$ for some $0<\theta \le 1$. For an operator criterion, one may assume that $A=\omega I-L$ is a sectorial operator of angle less than $\pi /2$, with dense domain. The analytic semigroup then satisfies, on a fixed finite time interval,The Hölder continuity assumption means $\Vert f(t)-f(s){\Vert }_H\le K|t-s{|}^{\theta }$. It supplies time regularity of the forcing, not spatial membership of $f(t)$ in $D(L)$.

With initial datum $u_0\in H$, a classical solution of an abstract Cauchy problem meanswith the evolution identity holding in $H$ at positive times and $u(t)\to u_0$ in $H$. Equivalently, the positive-time domain continuity is in the graph norm. If the definition requires differentiability also at zero, assume additionally $u_0\in D(L)$; the proof below then gives $u\in C^1([0,T],H)\cap C([0,T],D(L))$.

The variation-of-constants formula gives the candidateThe integral is a Bochner integral. Boundedness of $S$ and continuity of $f$ make the expression continuous into $H$, and its integral term tends to zero with $t$. Thus it is a mild solution of an abstract Cauchy problem with the correct initial value. The issue is showing that its integral term lies in the domain of the unbounded linear operator $L$.

Write $v(t)={\int }_0^{t}S(r)f(t-r)dr$. For $\epsilon >0$, the truncated integral $v_{\epsilon }(t)={\int }_{\epsilon }^{t}S(r)f(t-r)dr$ lies in $D(L)$. Subtract the current forcing before applying $L$:Here the last term follows by integrating $LS(r)f(t)=\frac{d}{dr}S(r)f(t)$ away from zero; it does not assume that $f(t)\in D(L)$. The first integrand satisfiesSince $v_{\epsilon }(t)\to v(t)$ in $H$ and the displayed images also converge, the closed operator property givesThis is the reason for the Hölder continuity assumption: the unsplit estimate would have the nonintegrable factor $r^{-1}$. Merely knowing that $f$ is continuous does not provide the integrable bound needed here. More generally, a modulus of continuity $\eta$ satisfying ${\int }_0^1\eta (r)dr/r<\infty$ suffices for this argument; a positive Hölder exponent is a convenient sufficient hypothesis.

The formula also proves continuity of $Lv$ at positive times. On compact positive-time intervals, split the integral at a small $r=\epsilon$: its short part is uniformly bounded by $C{\epsilon }^{\theta }$, while the remaining part is continuous by the strongly continuous semigroup property and ordinary dominated convergence. Together with analytic smoothing of $S(t)u_0$, this makes $Lu$ continuous at positive times. The semigroup property givesBecause $u(t)\in D(L)$, division of this identity by $h$ after subtracting $u(t)$ gives the continuous right derivative $Lu(t)+f(t)$. The left derivative agrees by the same identity and continuity of $Lu+f$. Thus $u^{\prime }=Lu+f$, proving existence of the classical solution of an abstract Cauchy problem on the entire given finite interval, and in particular local existence.

For uniqueness, the difference $h$ of two such classical solutions of an abstract Cauchy problem satisfies $h^{\prime }=Lh$ and $h(0)=0$. For $0<\epsilon <t$, differentiation of $S(t-s)h(s)$ on $[\epsilon ,t]$ gives zero, so $h(t)=S(t-\epsilon )h(\epsilon )$. Letting $\epsilon \downarrow 0$ yields $h(t)=0$. Finally, if $u_0\in D(L)$, then $LS(t)u_0=S(t)L{u}_0\to L{u}_0$, and the expression for $Lv(t)$ tends to zero. Hence $u^{\prime }(t)\to L{u}_0+f(0)$ and the stronger time-zero regularity follows. The solution is unique; the forcing regularity controls the singularity in the differentiated integral.

Eigenvalue analysis proves mesh-uniform stability when it is accompanied by control of eigenvectors or an energy norm; the signs of eigenvalues alone are insufficient. For an evolution equation, the method of lines first produces $u_h^{\prime }=L_h{u}_h$. A time integrator then gives an amplification matrix $S_{h,\delta }$, for example $S_{h,\delta }=R(\delta L_h)$ for a Runge-Kutta method. The useful stability requirement on a fixed physical time interval iswhere $C_T$ is independent of the refining spatial and temporal meshes. The semidiscrete analogue is $\Vert e^{t{L}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. These are bounds on propagation of initial errors and residuals, not merely statements that each finite matrix has a bounded solution over its own finite time interval.

If $L_h$ is a normal matrix in the chosen discrete inner product, it has a unitary eigenbasis. ThenFor dissipative problems it is sufficient that every $\delta {\lambda }_j$ lie in the integrator's linear stability domain. A diagonalizable family also suffices if its eigenvector matrices have uniformly bounded condition numbers: the same maximum is multiplied by $\Vert V_h\Vert \Vert V_h^{-1}\Vert$. Without that uniform bound the argument does not establish mesh stability. For finite-time estimates, an amplification modulus $1+O(\delta )$ can also be acceptable; strict contraction at every step is a stronger property.

For a constant-coefficient stencil on the full lattice or a periodic grid, von Neumann stability analysis provides the eigenvectors explicitly. A Fourier mode diagonalizes each translation-invariant spatial difference, giving a scalar symbol ${\lambda }_h(\theta )$ or a small matrix symbol for a system. The discrete Fourier transform and Parseval identity turn a uniform modal bound into an actual discrete L2 norm bound. The two-dimensional stencil in question 3 has $\mathrm{Re}{\lambda }_h=-(1-\cos \eta {)}^2/h\le 0$, so its continuous-time amplification is contractive; a later time integrator must still be checked. With boundaries one must analyze the boundary rows as well: a whole-line Fourier symbol cannot detect an unstable boundary closure.

For the heat equation $u_t=\kappa u_{xx}$ on a unit interval with zero Dirichlet data, the centred second difference has a sine eigenbasis and eigenvaluesExplicit Euler method multiplies mode $j$ by $1+\delta {\lambda }_j$. Requiring all factors to have modulus at most one yields $\kappa \delta /h^2\le 1/[2{\sin }^2((N-1)\pi /(2N))]$ on a given grid, and the sharp grid-independent sufficient limit isThese parabolic time-step restrictions express the growth of the largest-magnitude diffusion eigenvalue like $h^{-2}$. Backward Euler method instead has factor $(1-\delta {\lambda }_j{)}^{-1}$ and the Crank-Nicolson method factor $(1+\delta {\lambda }_j/2)/(1-\delta {\lambda }_j/2)$, both bounded by one for every $\delta \ge 0$. Their unconditional stability follows from A-stability, with much stronger damping of the shortest scales for backward Euler.

For $u_t=-a{u}_x$, $a>0$, explicit time stepping with a backward spatial difference givesThus the upwind finite difference scheme is contractive exactly for $0\le \nu \le 1$, a Courant–Friedrichs–Lewy condition matching the direction and speed of information propagation. A centred spatial difference has purely imaginary eigenvalues, and forward Euler gives $|S{|}^2=1+{\nu }^2{\sin }^2\theta >1$. Under a fixed positive hyperbolic Courant number, high-frequency data grow by a fixed factor in each of $O(1/h)$ steps, proving instability as the mesh is refined. This is not a claim that no exceptionally small coupled step can ever give a finite-time bound: if $\delta =O(h^2)$, the excess per step is only $O(\delta )$, but the usual hyperbolic scaling is unstable. Higher-order explicit integrators can possess nontrivial imaginary-axis stability intervals and are then suitable for skew-adjoint advection or the energy-form first-order wave equation.

Variable coefficients usually prevent Fourier diagonalization, but may preserve a self-adjoint spatial operator. The positive-face diffusion discretization in question 5 is an example. Its discrete energy identity makes $A+B$ symmetric negative definite, so a unitary eigenbasis still exists even without a closed eigenvalue formula. More generally, a dissipative operator $L_h$ in a discrete energy inner product obeyswhich directly proves a uniform semigroup estimate. The dissipative Cayley-transform contraction also follows without normality: for $\alpha \ge 0$,Taking $v=(I-\alpha L_h{)}^{-1}u$ proves contraction of the Crank–Nicolson factor. The contractivity of split Crank-Nicolson diffusion then follows from multiplying two contraction bounds; simultaneous eigenvectors of the two directional matrices are unnecessary.

The negative spectra do not imply uniform semidiscrete stability example shows what can go wrong. LetBoth eigenvalues are $-1$ for every mesh, but applying the propagator to the second unit vector at $t=1$ gives norm at least $1/(eh)$. The evolution is therefore not mesh-uniformly stable. For a discrete matrix, a unit-modulus eigenvalue with a nontrivial Jordan block gives powers growing polynomially in the step number. Even strictly interior eigenvalues can allow large transient growth when eigenvectors are badly conditioned. A pseudospectrum or numerical range of an operator, and especially a direct energy estimate, can reveal behavior missed by eigenvalues.

A resolvent estimate provides another way to see the needed uniformity. If $\Vert S_h^n\Vert \le C$ for every $n$, then for $|z|>1$,This shows directly that a uniform power estimate imposes more than the location of the spectrum. Conversely, resolvent-to-power estimates in varying dimension must themselves be uniform; a bound with constants depending on the number of grid points is not enough.

For a linear multistep method, each spatial eigenvalue gives an amplification polynomial, and all roots must satisfy the root condition for a multistep method, including simplicity of any unit-modulus root. Merely following the root approximating $e^{\delta \lambda }$ misses parasitic modes. The temporal and spatial analyses must therefore be combined with the appropriate zero-stability and uniform modal bounds.