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An analytic semigroup is a strongly continuous semigroup whose bounded-operator-valued extension is holomorphic in a sector about the positive time axis. Its infinitesimal generator of a semigroup $L$ satisfies $S(t)H\subseteq D(L)$ for $t>0$ and, on each bounded time interval, $\Vert LS(t)\Vert \le C/t$. This estimate gives instantaneous spatial smoothing but creates an integrable-singularity question in an inhomogeneous variation-of-constants formula.

For a closed operator $L$ on a Banach space $H$, a positive-time classical solution belongs to $C([0,T],H)\cap C^1((0,T],H)$, takes values in $D(L)$ for positive time, and has $Lu\in C((0,T],H)$. It satisfies the differential equation and the initial condition in $H$. Classical regularity also at time zero requires the corresponding regularity of the initial datum, in particular $u_0\in D(L)$. For an analytic semigroup and Hölder continuous forcing, the variation-of-constants formula gives such a solution: subtracting the forcing at the current time turns the singular estimate $\Vert LS(r)\Vert \le C/r$ into the integrable bound $C{r}^{\theta -1}$.

The displayed integral is $(e^{t\beta }-e^{t\gamma })/(\beta -\gamma )$ for distinct exponents and $t{e}^{t\gamma }$ when they coincide. The latter is the continuous limit, so a zero denominator is a removable singularity. This quantity appears in variation-of-constants formula estimates of matrix evolution and avoids omitting an equal-exponent case.

For finite matrices, differentiate the ordered product to obtain ${\Phi }^{\prime }-(A+B)\Phi =[e^{tA},B]e^{tB}$. Multiplying the error by $e^{-t(A+B)}$ and integrating gives $\Phi (t)-e^{t(A+B)}={\int }_0^t{e}^{(t-s)(A+B)}[e^{sA},B]e^{sB}ds$. This exact variation-of-constants formula identifies noncommutativity as the source of Lie-Trotter splitting error.

For $u_t=\Delta u+\lambda u-u^3$ with zero Dirichlet boundary condition, choose $K\ge \max (\Vert u_0{\Vert }_{\infty },\sqrt{\max (\lambda ,0)})$. The constants $K$ and $-K$ are a supersolution and a subsolution. More explicitly, testing with $(u-K{)}_{+}$ gives $\frac{1}{2}\frac{d}{dt}\Vert (u-K{)}_{+}{\Vert }_2^2\le \lambda \Vert (u-K{)}_{+}{\Vert }_2^2$, since the reaction derivative is at most $\lambda$ and the reaction at $K$ is nonpositive. The Gronwall inequality and the corresponding argument for $-u$ prove that $[-K,K]$ is invariant. A local variation-of-constants formula in the supremum norm then has a lifespan depending only on $K$, which prevents finite-time loss of the classical solution.

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.

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 $K$.

Set $S=A+B$ and $E(t)=F(t)-e^{tS}$. Since $F(0)=I$, $E(0)=0$. Differentiate the ordered exponential products, using that each matrix commutes with its own exponential:Here the commutator convention is $[X,Y]=XY-YX$; this fixes both signs. The error satisfies $E^{\prime }=SE+D$, so the variation-of-constants formula givesThis is the symmetrized exponential-splitting defect identity. Symmetry of $A,B$ was not needed for the identity itself; it will be used to bound their exponentials in part (c).

Let $T$ be the $M\times M$ tridiagonal matrix with diagonal two and adjacent entries minus one, incorporating zero Dirichlet boundary conditions. The actual PDF scheme isAn orthonormal basis of eigenvectors of $T$ has components $\sqrt{2/(M+1)}\sin (jm\pi /(M+1))$, and substitution givesThey form an orthonormal basis by the spectral theorem for real symmetric matrices. Hence $I+\mu T$ is invertible for every $\mu >0$, and the amplification eigenvalues are $g_j=(1+\mu {\lambda }_j{)}^{-1}\in (0,1)$. For the discrete L2 norm $\Vert U{\Vert }_h^2=h{\sum }_m|U_m{|}^2$,This bound has constant one independent of time step and mesh, which is the required stability of a numerical method. Therefore every positive Courant number is stable. If a forcing or local-defect sequence is added, the discrete variation-of-constants formula gives a bound by the initial norm plus the sum of those perturbation norms; this also justifies the global error conclusion in part (a).

The variation-of-constants formula gives, when $a\ne 0$,Therefore, with $m=\min (s,t)$,If $a=0$, then $X_t=x+bt+\sigma B_t$ and

Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.

SetThen the forced equation is the abstract Cauchy problemFor $Z_0\in \mathcal{H}$ and $F\in C([0,T];\mathcal{H})$, a mild solution of an abstract Cauchy problem is a function $Z\in C([0,T];\mathcal{H})$ satisfying the variation-of-constants formula

Suppose $F(s)\in D(A)$ and both $F$ and $AF$ are continuous. If also $Z_0\in D(A)$, then the closedness of $A$ permits differentiation under the Bochner integral:Thus $Z\in C^1([0,\infty );\mathcal{H})$ and $Z^{\prime }=AZ+F$. The assumption $Z_0\in D(A)$ is necessary here: a unitary group has no smoothing, so the conditions on $F$ alone cannot make $U(t)Z_0$ differentiable for arbitrary $Z_0\in \mathcal{H}$.

For the resulting strong solution, skew symmetry of $A$ gives the energy estimateIntegration yields

For bounded matrices, differentiating the symmetrized product gives $F^{\prime }-(A+B)F=\frac{1}{2}([e^{tB},A]e^{tA}+[e^{tA},B]e^{tB})$. The variation-of-constants formula turns this into an integral expression for $F-e^{t(A+B)}$. This retains the noncommutativity of the factors explicitly; cancellation of their leading opposite commutators explains why the averaged product is more accurate than either ordering alone.