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Initial data whose normal derivative is nonzero at a wall cannot satisfy homogeneous Neumann boundary conditions continuously all the way to the initial time. For example, $-\mathrm{erf}(ay)$ has wall derivative $-2a{e}^{-a^2}/\sqrt{\pi }$ at $y=\pm 1$, which is nonzero for finite $a>0$. A parabolic mild solution of an abstract Cauchy problem can still take those initial data in an integral or square-integrable sense and satisfy the boundary condition for every positive time. An exactly compatible smooth profile is needed for a classical solution on the closed time interval.

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.

On each finite interval use the Banach space $X_T=C([0,T];E)$ with the supremum norm. The free term $g(t)=U_t{f}_0$ belongs to $X_T$, and the time-integral operator $\tau$ maps $X_T$ to itself with norm at most $2T$. The strongly continuous semigroup property and boundedness of $B$ justify continuity of the Bochner integral.

DefineThe factorial bound for a Volterra iterate gives $\Vert {\tau }^{j}g{\Vert }_{X_T}\le (2T{)}^j\Vert f_0{\Vert }_1/j!$. The series therefore converges absolutely in $X_T$. Since $\tau$ is a bounded linear operator on $X_T$, it can be passed through the convergent sum, givingThus $f=g+\tau f$, the required integral formulation, and $f(0)=f_0$. Each term on a larger interval restricts to the identical term on a smaller interval, so these constructions define a single global solution without having to restart at successive times. This is an integrable Volterra solution for normalized velocity relaxation. It is a mild solution of an abstract Cauchy problem in $L^1$ and hence an $L^1$ weak solution in the paper's integral-formulation sense. No smallness condition such as $2T<1$ is needed.

The temporal integration by parts terms areThe terminal states are free and the objective has no explicit terminal-velocity dependence. ThereforeThe factor two follows from the displayed objective without a $1/2$ prefactor. The initial scalar field perturbation is fixed, so $\delta \theta (0)=0$ and there is no independently prescribed adjoint-scalar condition at $t=0$. Its value there is obtained by backward integration. Independent variation of the initial velocity state gives ${\mathbf{u}}^{†}(0)={\mathbf{u}}_0^{†}$.

Variation of the candidate ${\mathbf{u}}_0$ in the energy-augmented functional gives the fixed-energy initial-condition optimality condition. For the usual divergence-free no-slip control space $H$, write $P_H$ for its orthogonal projection, the Leray-Helmholtz projection in the usual incompressible velocity space. This is the standard formal control condition when pressure is determined by incompressible flow and normal momentum. If the independently imposed direct pressure flux is retained as an additional constraint, the admissible control variations must also satisfy its compatibility conditions along the trajectory; they need not fill the usual space $H$. The standard projection formula alone does not prove stationarity for that more restricted problem. ThenFor the natural solenoidal adjoint space, $P_H{\mathbf{u}}^{†}(0)={\mathbf{u}}^{†}(0)$. Equivalently, its component tangent to the sphere in a normed vector space of fixed kinetic energy vanishes:Here $E_0>0$ is needed for the sphere in a normed vector space to be regular and for division by $2E_0$. The multiplier is real and can have either sign; normalizing the initial adjoint with a prescribed positive sign is not a general necessary condition for minimization. If $E_0=0$, the only feasible initial velocity is zero and this tangent formula for the sphere in a normed vector space is inapplicable. First-order stationarity alone also allows maxima or saddles; a local minimum requires the appropriate nonnegative constrained second variation.

The spatial boundary terms vanish with periodic adjoints in $x,z$, homogeneous no-slip boundary conditions ${\mathbf{u}}^{†}=0$ at $y=\pm 1$, and ${\partial }_y{\theta }^{†}=0$ there. To see this, velocity variations vanish at the wall while their normal derivatives need not, so the viscous boundary term forces the adjoint velocity to vanish. Scalar variations have zero normal derivative but free values, forcing the adjoint scalar's normal derivative to vanish. Pressure variation gives adjoint incompressible flow; no independent terminal or initial datum is assigned to $p^{†}$. Its additive time-dependent constant may be fixed by a zero spatial mean.

In particular, a homogeneous Neumann condition for the direct pressure does not by variational transposition require ${\partial }_y{p}^{†}=0$. For a smooth adjoint solution, the wall-normal adjoint momentum equation instead suppliesat the flat walls, because the direct scalar normal derivative is zero there. A zero adjoint-pressure derivative would be an additional compatibility choice, valid only when this right-hand side vanishes.

There are also literal direct-data compatibility qualifications in the printed setup. The initial total scalar field is $\overline{\theta }=-\mathrm{erf}(30y)$, whose wall derivative is $-60e^{-900}/\sqrt{\pi }\ne 0$. It is extraordinarily small but mathematically not zero. Thus exact initial scalar data and exact zero wall flux are incompatible for a classical solution smooth at $t=0$. One may use a parabolic mild solution of an abstract Cauchy problem with the boundary condition enforced for $t>0$, or replace the initial profile by an exactly compatible smooth one; these are conventions, not an equality $e^{-900}=0$. The reference scalar field is also not a stationary profile of the diffusion equation at finite $Pe$: the full $\Delta \Theta$ term must be retained.

The pressure compatibility at a no-slip wall is similarly important. Direct normal momentum gives ${\partial }_{y}p=\nu \Delta u_y-R{i}_B\theta$. The separately imposed zero pressure derivative requires this right side to vanish. It is not automatic for arbitrary fixed-energy initial velocities. For example, the divergence-free no-slip field generated by the periodic stream function $S=A(1-y^2{)}^2\sin (\alpha x)$ with $\alpha =\pi /L_x>0$ has $u_x={\partial }_{y}S$, $u_y=-{\partial }_{x}S$ and $\Delta u_y{|}_{y=\pm 1}=-8A\alpha \cos (\alpha x)\ne 0$, while $\theta (0)=0$. With $A\ne 0$ adjusted to any positive energy, no classical solution smooth up to the initial wall can satisfy that extra pressure condition. The displayed adjoints are the formal necessary equations along admissible smooth trajectories; a well-posed physical formulation normally determines pressure from incompressible flow and normal momentum, rather than imposing independent homogeneous pressure flux for every control.

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