3.1. Mathematical formulation

In the limit $\textit {Re} \gg 1$, the problem is formulated based on a thin viscous boundary layer that governs the drag. The fluid outside this layer, to leading order in $\textit {Re}$, remains stationary as the plate moves. To model the flow in the boundary layer, we use a coordinate system attached to the leading edge of the plate, as shown in figure 1, and a reference frame attached to far field stationary fluid. The coordinates ${\tilde {{x}}}$, ${\tilde {{y}}}$ attached to the plate, and the boundary layer flow velocity $({\tilde {{u}}}({\tilde {{x}}},{\tilde {{y}}},{\tilde {{t}}}), {\tilde {{v}}}({\tilde {{x}}},{\tilde {{y}}},{\tilde {{t}}}))$ in a reference frame of the far field fluid, satisfy the Prandtl equations

(3.1a)$$\begin{gather} {\tilde{{u}}}_{\tilde{{t}}} + ({\tilde{{u}}}+V) {\tilde{{u}}}_{\tilde{{x}}} + {\tilde{{v}}} {\tilde{{u}}}_{\tilde{{y}}} - \nu {\tilde{{u}}}_{{\tilde{{y}}}{\tilde{{y}}}} = 0, \end{gather}$$

(3.1b)$$\begin{gather}{\tilde{{u}}}_{\tilde{{x}}} + {\tilde{{v}}}_{\tilde{{y}}} = 0, \end{gather}$$

where subscripts ${\tilde {{x}}}$, ${\tilde {{y}}}$ and ${\tilde {{t}}}$ denote partial differentiation with respect to that variable. The essential nonlinearity in fluid dynamics governing equations is due to the advection, which is retained in this formulation. Exploiting the reflection symmetry, we consider only the flow field for ${\tilde {{y}}}\geqslant 0$ and the drag on one face of the plate. These equations are subject to the boundary conditions

(3.2a)$$\begin{gather} {\tilde{{u}}}(0\leqslant {\tilde{{x}}}\leqslant L, {\tilde{{y}}}, {\tilde{{t}}}) + V (t) = 0, \end{gather}$$

(3.2b)$$\begin{gather}{\tilde{{u}}}_{\tilde{{y}}}({\tilde{{x}}}<0 \text{ or } {\tilde{{x}}}>L, {\tilde{{y}}}, {\tilde{{t}}}) = 0, \end{gather}$$

(3.2c)$$\begin{gather}{\tilde{{u}}}({\tilde{{x}}}, {\tilde{{y}}}\to \infty, {\tilde{{t}}}) = 0, \end{gather}$$

(3.2d)$$\begin{gather}{\tilde{{u}}}({\tilde{{x}}}\to -\infty, {\tilde{{y}}}, {\tilde{{t}}}) = 0, \end{gather}$$

(3.2e)$$\begin{gather}{\tilde{{u}}}({\tilde{{x}}},{\tilde{{y}}},{\tilde{{t}}}=0) = 0. \end{gather}$$

The drag force is given by ${\tilde {{f}}} = 2 \int _0^L\mu \, {\tilde {{u}}}_{\tilde {{y}}}({\tilde {{x}}},{\tilde {{y}}}=0,{\tilde {{t}}})\,{\rm d}\kern0.7pt {\tilde {{x}}}$, and the net work done for the motion is

(3.3)\begin{equation} {\tilde{{\mathcal{W}}}} = \int_0^t V(t) \int_0^l 2 \mu\, {\tilde{{u}}}_{\tilde{{y}}}({\tilde{{x}}},{\tilde{{y}}}=0,{\tilde{{t}}})\,{\rm d}{\tilde{{x}}}\,{\rm d}{\tilde{{t}}}. \end{equation}

The variables are rescaled as

(3.4a–e)\begin{equation} {\tilde{{t}}} = Tt,\quad {\tilde{{x}}} = Lx,\quad {\tilde{{y}}} = \delta y,\quad {\tilde{{u}}} = \dfrac{D}{T}\,u, \quad {\tilde{{v}}} = \dfrac{D\delta}{TL}\,v, \end{equation}

where $\delta = \sqrt {\nu T}$, to yield the dimensionless form of the governing boundary layer equations as

(3.5a)$$\begin{gather} \mathcal{M} \equiv u_t + \epsilon (u+f)u_x + \epsilon v u_y - u_{yy} = 0, \end{gather}$$

(3.5b)$$\begin{gather}\mathcal{C} \equiv u_x + v_y = 0, \end{gather}$$

in the semi-infinite half-space $y\geqslant 0$, and

(3.6a)$$\begin{gather} u|_{x\in P, y=0} + f(t) = u_y|_{x \not\in P, y=0} = 0, \end{gather}$$

(3.6b)$$\begin{gather}v|_{y=0} = u_y|_{y=\infty} = u|_{x={\pm}\infty} = u|_{t=0} = 0, \end{gather}$$

where $P = [0,1]$ specifies the extent of the flat plate, and $f(t) = (T/D)\,V({\tilde {{t}}})$ is the dimensionless plate speed. Subscripts $x$, $y$ and $t$ on $u$ and $v$ denote partial derivatives.

The following notation will be also used

(3.7)\begin{equation} \langle \phi \rangle_{\gamma_1 \dots \gamma_m} \equiv \iiint \phi\,\mathrm{d} \gamma_1 \dots \mathrm{d} \gamma_m, \end{equation}

where $\gamma _i$ could be any of $x$, $y$ or $t$. The limits of integration are $[0,1]$ on $t$, $[0,\infty ]$ on $y$, and $[-\infty, \infty ]$ on $x$. For example, $\langle \phi \rangle _{xy}$ is $\int _0^\infty \int _{-\infty }^\infty \phi \,\mathrm {d}\kern0.7pt x \,\mathrm {d} y$. For simplicity, $\langle \phi \rangle \equiv \langle \phi \rangle _{xyt}$.

The work done is ${\tilde {{\mathcal {W}}}} = 2D^2 L \sqrt {{\mu \rho }/{T^3}}\ \mathcal {W}[f]$, where $\mathcal {W}[f] = \langle f(t)\,u_y(x,y=0,t)\rangle _{xt}$. For $(u,v)$ satisfying (3.5) and (3.6), the work done must appear as an increase in the kinetic energy of the fluid or be viscously dissipated, i.e.

(3.8)\begin{equation} \mathcal{W}[f] = \hat{\mathcal{W}}[f] \equiv \left\langle\left. \tfrac{1}{2} u^2\right|_{t=1} \right\rangle_{xy} + \langle u_y^2 \rangle, \end{equation}

which is necessarily positive.

The objective of the optimization is to minimize $\mathcal {W}[f]$ subject to $\mathcal {M} = \mathcal {C}= 0$ in the fluid. The Lagrangians using $\mathcal {W}[f]$ and $\hat {\mathcal {W}}[f]$ are

(3.9a)\begin{equation} \mathcal{L} = \mathcal{W}[f] - \langle \alpha \mathcal{M} + \epsilon \beta \mathcal{C} \rangle - \lambda \mathcal{D} \end{equation}

and

(3.9b)\begin{equation} \hat{\mathcal{L}} = \hat{\mathcal{W}}[f] - \langle \hat{\alpha} \mathcal{M} + \epsilon \beta \mathcal{C} \rangle - \lambda \mathcal{D}, \end{equation}

respectively, where $\alpha (x,y,t)$, $\hat {\alpha }(x,y,t)$, $\beta (x,y,t)$ and $\lambda$ are Lagrange multipliers, and

(3.10)\begin{equation} \mathcal{D} = \langle f \rangle_t - 1 \end{equation}

is the constraint on the distance travelled by the plate. We also define $\hat {u} = u + f$, which is the $x$-component of the fluid velocity in the reference frame of the plate. Gradient descent is convenient using $\mathcal {L}$ because in these variables, the adjoint satisfies an equation similar to that for $u$. On the other hand, the spectral condition is derived using $\hat {\mathcal {L}}$, which when expressed in $\hat {u}$ reveals an eigenvalue problem with homogeneous boundary conditions. By virtue of (3.8), the two formulations in $\mathcal {L}$ and $\hat {\mathcal {L}}$ are equivalent, related by $\hat {\alpha } = \alpha + u$.

The multipliers $\alpha$ and $\beta$ satisfy the condition that first variations of $\mathcal {L}$ due to $u$ and $v$ vanish, i.e.

(3.11a)$$\begin{gather} \alpha_t + \epsilon (u+f) \alpha_x + \epsilon (v\alpha)_y + \epsilon \beta_x + \alpha_{yy} = 0, \end{gather}$$

(3.11b)$$\begin{gather}\beta_y = \alpha u_y, \end{gather}$$

in $y>0$, subject to the boundary conditions

(3.12a)$$\begin{gather} \alpha|_{x \in P, y=0} - f (t) = \alpha_y|_{x \not\in P, y=0} = 0, \end{gather}$$

(3.12b)$$\begin{gather}\beta|_{y=\infty} = \alpha_y|_{y=\infty} = \alpha|_{x=\infty} = \alpha|_{t=1} = 0. \end{gather}$$

The first variation with respect to $f$, given by

(3.13)\begin{equation} \dfrac{\delta \mathcal{L}}{\delta f} = \left\langle \left[u_y - \alpha_y\right]_{y=0} \right\rangle_x - \epsilon \langle \alpha u_x \rangle_{xy} - \lambda, \end{equation}

must also vanish.

The optimization is carried out using gradient descent in $f$ using (3.13) while solving (3.5), (3.6), (3.11) and (3.12) numerically, as described in Appendix A. We find that this procedure converges to $f=f_*$, $u=u_*$, $v=v_*$, $\alpha = \alpha _*$ and $\mathcal {W}[f]=\mathcal {W}_{min}$, as $\delta \mathcal {L}/\delta f$ approaches 0. The converged profiles for $f(t)$ are shown in figure 2, and the corresponding $\mathcal {W}_{min}$ in figure 3(a).

Figure 2.

Optimal $f(t)$ for different $\epsilon$ coded according to the colour in the adjoining colourbar. The dotted curve shows $f_0(t)$ from (3.19a), and the dashed curve shows unity.

Figure 3.

Minimum work for translating a flat plate and the verification of the corresponding spectral constraint. (a) Plot of $\mathcal {W}_{min}$ as a function of $\epsilon$. The dotted line shows $\mathcal {W}_{0,min}\approx 1.014$ from (3.19b), and the dashed line shows $0.664 \epsilon ^{1/2}$. (b) Minimum $\lambda _{min} = \min _{t\in (0,1)} \lambda (t)$ (blue circles) and maximum $\lambda _{max} = \max _{t\in (0,1)} \lambda (t)$ (red squares) of $\lambda (t)$. Thus the region shaded grey shows the values taken by $\lambda (t)$ for the corresponding $\epsilon$. By virtue of (3.18), the second variation (3.17) is given by $\lambda \langle \psi ^2 + \psi _x^2 + \psi _y^2 \rangle$. Since the quantity in the angled brackets is positive definite, $\lambda (t)$ being positive for all $t$ implies that the second variation in (3.17) is also positive definite, thus the spectral condition is satisfied.

3.2. Spectral condition

To prove that the computed minimum is global, we choose $\hat {\alpha }=\hat {\alpha }_* = \alpha _* + u_*$ and the Lagrange dual

(3.14)\begin{equation} \mathcal{D}[\hat{\alpha}] \equiv \min_{u,v,\beta,f} \hat{\mathcal{L}}. \end{equation}

Because $\hat {\mathcal {L}}$ is quadratic in $(u,v,\beta,f)$, by virtue of (3.11) and (3.12) $(u_*,v_*, \beta _*,f_*)$ is a stationary point of $\hat {\mathcal {L}}$ in those variables, for which $\hat {\mathcal {L}} = \mathcal {W}_{min}$. It is the second variation

(3.15)\begin{equation} \mathcal{H}[\hat{\alpha}_*; u,v,f] = \hat{\mathcal{W}}[f] - \epsilon \left\langle \hat{\alpha}_* ((u+f)u_x + vu_y) \right\rangle \end{equation}

in $(u,v,f)$ for $(u,v)$ satisfying (3.5b) that determines whether $(u_*,v_*, \beta _*,f_*)$ is a minimum of $\hat {\mathcal {L}}$. This leads to

(3.16)\begin{equation} \mathcal{W}[f] \geqslant \mathcal{D}[\hat{\alpha}_*] = \begin{cases} \mathcal{W}_{min}, & \text{if } \mathcal{H} \geqslant 0 \text{ for all } (u,v,f), \\ -\infty, & \text{otherwise}. \end{cases} \end{equation}

The condition $\mathcal {H} \geqslant 0$ is satisfied if for each $t\in (0,1)$, we have

(3.17)\begin{equation} \varDelta[u,v] = \langle \hat u_y^2 -\hat{\alpha}_* \epsilon (\hat u \hat u_x + v \hat u_y) \rangle_{xy} \geqslant 0, \end{equation}

for all $(\hat u,v)$ satisfying (3.5b) and (3.6).

Equation (3.5b) implies a stream function $\psi$ such that $\hat {u} = \psi _y$ and $v = -\psi _x$. Substituting this in (3.17) and integrating by parts yields the related optimization problem

(3.18)\begin{equation} \left.\begin{gathered} \lambda(t) = \min_\psi \varDelta = \min_\psi \langle \psi_{yy}^2 +\epsilon \hat{\alpha}_{*x} \psi_y^2 - \epsilon \hat{\alpha}_{*y} \psi_x\psi_y \rangle_{xy}, \\ \text{subject to }\ \langle \psi^2 + \psi_x^2 + \psi_y^2 \rangle_{xy} = 1 \end{gathered}\right\} \end{equation}

for each $t\in (0,1)$. The solution of (3.18) implies $\varDelta \geqslant \lambda (t)\,\langle \psi ^2 + \psi _x^2 + \psi _y^2 \rangle _{xy}$ and hence the spectral constraint is satisfied if $\lambda (t)\geqslant 0$ for $t\in (0,1)$. Here, $\lambda (t)$ is determined numerically by constructing the generalized eigenvalue problem from the linear operators underlying the quadratic forms in (3.18), which is described in Appendix A. As shown in figure 3(b), $\lambda (t)$ is positive for each $t$, thus proving that the local minimum found in § 3.1 is a global one.

3.3. Interpretation of the results

As seen in figures 2 and 3(a), the analytical solution by Mandre (Reference Mandre2020) derived for $\epsilon \ll 1$,

(3.19a)\begin{equation} f(t) = f_0(t) = C {t^{1/4} (1-t)^{1/4}} \end{equation}

and

(3.19b)\begin{equation} \mathcal{W}_{min} = \mathcal{W}_{0,min} \approx 1.014, \end{equation}

where $C \approx 1.62$, approximates the solution for finite $\epsilon \leqslant 0.5$. As $\epsilon$ increases beyond, the optimal $f(t)$ departs from $f_0(t)$, while $\mathcal {W}_{min}$ rises above $\mathcal {W}_{0,min}$. In particular, $f(t)$ starts from $f(0) = 0$ and ends at $f(1) = 0$ but flattens out in the middle. For $\epsilon \gtrsim 5$, $f(t)$ approaches unity, except at the start and the end. In other words, the optimum kinematics to cover a distance $D\gg L$ in time $T$ is to cruise at the average speed $U\approx D/T$, except to start and stop. For $\epsilon \gg 1$, $\mathcal {W}_{min}$ is also observed to rise $\propto \epsilon ^{1/2}$.

The following dimensional argument rationalizes these observations. The drag according to Blasius (Reference Blasius1907) for a flat plate moving at steady speed $U$ is given by $0.664 \sqrt {\mu \rho U^3 L}$, and consequently, the work done to move the plate is ${\tilde {{\mathcal {W}}}}_\infty = 0.664 T \sqrt {\mu \rho U^5 L}$. Consider covering the distance $D=UT$ in two stages, of durations $aT$ and $(1-a)T$, moving with speeds $U_1 = bU/a$ and $U_2 = (1-b)U/(1-a)$, respectively, for constants $a$ and $b$ between 0 and 1. When $\epsilon \gg 1$, steady state is reached much faster than the duration of each segment, and the modified kinematics approximately incurs the work

(3.20)\begin{equation} 0.664 T \sqrt{\mu \rho U^5 L} \left(\dfrac{b^{5/2}}{a^{3/2}} + \dfrac{(1-b)^{5/2}}{(1-a)^{3/2}} \right). \end{equation}

This work is minimized when $b=a$, or $U_1 = U_2 = U$. In other words, the penalty incurred in the work done when traveling fast outweighs the benefits accrued when traveling slower, explaining why the optimum avoids modulation of the speed. Converting ${\tilde {{\mathcal {W}}}}_\infty$ to a dimensionless form yields $\mathcal {W}_{min} \approx 0.664 \epsilon ^{1/2}$ for $\epsilon \gg 1$, agreeing up to leading order with the results of the computations as shown in figure 3(a). Accounting for the two sides of the plate leads to (1.1a,b).

It is seen readily why the optimal profile avoids an impulsive start and stop. For $t\ll 1$ (and $1-t\ll 1$), owing to the development of the viscous boundary layer, the unsteady inertia $u_t$ and shear viscosity $u_{yy}$ in (3.5) dominate, while advection $\epsilon (u+f)u_x + \epsilon vu_y$ is negligible. Therefore, in the first variation condition (3.13), the dominant balance is between $u_y$ and $\alpha _y$. For an impulsive start, the initial shear stress profile on the plate is $u_y|_{y=0}\approx f(0)/\sqrt {{\rm \pi} t}$ due to the growth of the boundary layer thickness proportional to $\sqrt {t}$. The adjoint dynamics, due to their backwards evolution in time, does not ‘know’ about the impulsive start, and therefore cannot generate an $\alpha _y$ that matches this asymptotic behaviour. This behaviour can be observed explicitly in the analytical solution for $\epsilon \ll 1$ (Mandre Reference Mandre2020) and carries over to finite $\epsilon$. Therefore, for small $t$, one can always reduce the work done by eliminating any impulsive start (and analogously an impulsive stop).

We also find that for finite $\epsilon$, the optimal starting and stopping dynamics behave proportionally to $t^{1/4}$ and $(1-t)^{1/4}$, respectively. This is observed in the numerical solution for over four orders of magnitude in $t$, as shown in figure 4. The reason is analogous to that in the analytical solution for vanishing $\epsilon$ in (3.19). Near the starting and stopping times, the advection is negligible and the optimal kinematics is governed by the viscous diffusion of momentum within the fluid, just as is the case when $\epsilon \ll 1$ (Mandre Reference Mandre2020), which causes this behaviour. To see this more clearly, consider the approximations of (3.5a) and (3.11a) for small $t$,

(3.21a)$$\begin{gather} u_t - u_{yy} \approx 0, \end{gather}$$

(3.21b)$$\begin{gather}\alpha_t + \alpha_{yy} \approx 0, \end{gather}$$

driven by the boundary conditions $u|_{y=0} = -f(t)$ and $\alpha |_{y=0} = f(t)$. The approximate gradient for small $t$ is dominated by

(3.22)\begin{equation} \dfrac{\delta \mathcal{L}}{\delta f} \approx \left\langle \left[u_y - \alpha_y\right]_{y=0} \right\rangle_x. \end{equation}

The first term in this expression is the part of the variation arising from a change in $f(t)$, while the second term arises from the resulting variation in the shear force on the plate. Because $x$ derivatives drop out of this approximation, the solution to (3.21) is uniform in $x$, and the integral in $x$ can be dropped. Elimination of an impulsive start implies a power-law behaviour for $f$ at small $t$, say $f(t) \approx a t^n$ for $0< t<\hat {t}$, where $\hat {t} \gg t$ is a time until which the approximation remains valid. The Dirichlet-to-Neumann map for the heat equation (3.21) then implies

(3.23a)$$\begin{gather} u_y|_{y=0} \approx{-}\int_0^t \dfrac{a n s^{n-1}}{\sqrt{{\rm \pi}(t-s)}}\,\mathrm{d} s ={-} \dfrac{a n t^{n-1/2}}{\sqrt{\rm \pi}} \int_0^1 \dfrac{w^{n-1}}{\sqrt{1-w}}\,\mathrm{d} w, \end{gather}$$

(3.23b)$$\begin{gather}\alpha_y|_{y=0} \approx{-}\int_t^{\hat{t}} \dfrac{a n s^{n-1}}{\sqrt{{\rm \pi}(s-t)}}\, \mathrm{d} s ={-}\dfrac{a n t^{n-1/2}}{\sqrt{\rm \pi}} \int_{t/\hat{t}}^1 \dfrac{1}{w^{n+1/2} \sqrt{1-w}}\,\mathrm{d} w, \end{gather}$$

where the second half of each equation is derived by using the transformation $w=s/t$ for (3.23a) and $w=t/s$ for (3.23b). Substituting in (3.22) and using $t/\hat {t} \approx 0$ yields

(3.24)\begin{equation} \dfrac{\delta \mathcal{L}}{\delta f} \approx \dfrac{a n t^{n-1/2}}{\sqrt{\rm \pi}} \int_0^1 \dfrac{1}{\sqrt{1-w}} \left(\dfrac{1}{w^{n+1/2}} - w^{n-1} \right)\mathrm{d} w. \end{equation}

For this leading-order estimate of the gradient to vanish, the integrand must vanish, which yields $n=1/4$. In other words, the variations arising from perturbing the plate speed and those arising from the resulting perturbation in shear stress both scale as $t^{n-1/2}$ for plate speed $\propto t^n$. The coefficient of proportionality is calculated as the integral in (3.24). It is necessary that the coefficient vanishes for the variation to be zero, which happens to leading order for $n=1/4$.

Figure 4.

Optimal starting and stopping kinematics plotted on a logarithmic scale. (a) Starting kinematics; colour code identical to that in figure 2; dashed line shows $t^{1/4}$. (b) Stopping kinematics; dashed line shows $(T-t)^{1/4}$.

Equations (3.21)–(3.22) are invariant under the transformation where $u$ and $\alpha$ are exchanged, $f \to -f$, and $t \to 1-t$. Thus the transformed version of the derivation above also predicts the dependence proportional to $(1-t)^{1/4}$ for $f(t)$ near $t=1$. Furthermore, the invariance described above and asymptotic independence to departures from the power law (which enable setting $t/\hat {t} \to 0$ in (3.23b)) implies that even the constant of proportionality, $a$, is the same for the starting and stopping kinematics. In other words, $f(t) \sim at^{1/4}$ near $t=0$ and $f(t) \sim a (1-t)^{1/4}$ near $t=1$, with the same value for the coefficient $a$. This is observed readily in the analytical solution for the case $\epsilon \ll 1$ from (3.19), and is verified numerically in figure 5 for all $\epsilon$. Thus the optimal starting and stopping kinematics are asymptotically identical for all $\epsilon$.

Figure 5.

The coefficient $a$ in the asymptotic power-law dependence $f(t)\sim at^{1/4}$ for $t \ll 1$ (blue circles), and $f(t) \sim a(1-t)^{1/4}$ for $1-t \ll 1$ (red crosses). The coefficients are obtained by determining the best-fit coefficient in the range $0.5< f(t)<0.3$. The dotted line shows $C=1.62$, which is the asymptotic value for $\epsilon \ll 1$, according to (3.19). The dashed curve shows the result for the asymptotic limit $\epsilon \gg 1$ from (3.29a,b).

Finally, we observe that for $\epsilon \gg 1$, the starting and stopping dynamics as a function of $\epsilon t$ are independent of $\epsilon$, as seen in figure 6. (Here, $\epsilon t$ is the ratio of the dimensional distance covered when travelling at speed $D/T$ to the dimensional plate length.) For $\epsilon \geqslant 5$, these optimal starting and stopping dynamics approach successively closer to limiting curves. These limiting curves denote the work-minimizing kinematics for the flat plate to attain a constant cruising speed from rest, and to stop from the cruising motion, respectively. These starting and stopping kinematics do not depend separately on the total distance travelled $D$ or the time taken $T$, but depend only on the cruising speed. For the purpose of determining the optimal startup and stopping kinematics, we non-dimensionalize the target cruising speed to unity. The governing equations to determine this kinematics are identical to the ones developed in § 3.1, with the following exceptions. The variables $t$ and $y$ are rescaled as $\epsilon t = t'$ and $y\sqrt {\epsilon } = y'$, and formally, $t'$ ranges from 0 to $\tau \to \infty$ (all the integrals in (3.9) are now over the semi-infinite time interval). The distance travelled condition $\mathcal {D}=0$ is replaced by the unit target cruising speed

(3.25)\begin{equation} \mathcal{D}' = \lim_{\tau \to \infty} \dfrac{1}{\tau} \int_0^\tau f_{start}(t') \,\mathrm{d} t' - 1 = 0. \end{equation}

An additional constraint that the final velocity profile approaches the Blasius steady-state profile $u_s(x,y,t)$ with $u_s(x\in P, y=0, t) = -1$ is added. Imposing a target final state for $u$ implies that the condition for starting the backwards time-integration of $\alpha$ must be determined as part of the solution. This condition is determined trivially to be the steady solution of (3.11) with $u = u_s$, $u(x\in P,y=0,t)=-1$ and $\alpha (x\in P, y=0,t) = 1$. The numerical procedure described in Appendix A then yields the optimum startup kinematics.

Figure 6.

Optimal starting and stopping kinematics. (a) Startup dynamics of optimal kinematics for $\epsilon =5$, 10, 20 and 50 (same colour code as in figure 2) plotted against $\epsilon t$. The solid black curve shows the optimal starting kinematics, and the overlapping dotted green curve shows the empirical fit in (3.27a,b). (b) Same as (a), but for stopping kinematics.

For the optimum stopping kinematics, the time variable $t'$ is shifted so that the final time is zero. The initial state for $u$ is $u_s$, and the final state is unknown. Therefore, $\alpha =0$ at $t'=0$ holds. The distance travelled condition is replaced by

(3.26)\begin{equation} \mathcal{D}'' = \lim_{\tau \to \infty} \dfrac{1}{(-\tau)} \int_{-\tau}^0 f_{stop}(t') \,\mathrm{d} t' - 1 = 0. \end{equation}

Following the numerical procedure in Appendix A then yields the optimal stopping kinematics. Figure 6 shows that the profiles for finite but large $\epsilon$ converge to the optimal starting and stopping kinematics. An empirical but convenient fit

(3.27a,b) \begin{equation} f_{start}(t') = (1 - {\rm e}^{{-}At'} )^{1/4}, \quad f_{stop}(t') = (1 - {\rm e}^{Bt'^m} )^{1/(4m)}, \end{equation}

with $A\approx 2.62$, $B \approx 2.06$ and $m\approx 0.70$, approximates the computed starting and stopping dynamics with correct asymptotic behaviour and an error with an $L^1$ norm ${<}1\,\%$.

A composite expression for the optimal kinematics for $\epsilon \gg 1$ can now be constructed using $f_{start}$ and $f_{stop}$. Because $f_{start} \leqslant 1$, the distance travelled by the plate always lags behind one moving with cruising speed. Indeed, $\int _0^\infty (f(t')-1)\,\mathrm {d} t' \approx -0.14$, implying in dimensional terms that by the time steady cruising at a speed $U_c$ is reached, the optimal kinematics lags a distance $0.14 L\times (U_c T/D)$ behind. Similarly, an additional distance $0.43 L\times (U_c T/D)$ is lost when stopping. Thus the total distance travelled in time $T$ is $U_c T (1 - 0.57/\epsilon )$. Equating this distance to $D$ yields $U_c = (D/T)/(1 - 0.57/\epsilon )$. The corresponding composite expression for $f(t)$ is

(3.28)\begin{equation} f(t) \approx \dfrac{f_{start}(\epsilon t) + f_{stop}(\epsilon (t-1)) - 1 }{1 - ({0.57}/{\epsilon})}. \end{equation}

Near $t=0$ and $t=1$, respectively, this composite expression approaches

(3.29a,b)\begin{equation} f(t) \sim \dfrac{(A\epsilon t)^{1/4}}{1-(0.57/\epsilon)} \quad \text{and} \quad f(t) \sim [\epsilon(1-t)]^{1/4}\,\dfrac{B^{1/(4m)}}{1-(0.57/\epsilon)}, \end{equation}

which agrees with the computed solutions for large but finite $\epsilon$, as shown in figure 5. The equality of the asymptotic startup and stopping kinematics in (3.29a,b) implies $A\approx B^{1/m}$, which is satisfied approximately to about 8 %.

Finally, we test the validity of the Prandtl boundary layer approximation, which neglects the $\nu {\tilde {{u}}}_{{\tilde {{x}}}{\tilde {{x}}}}$ term from the Navier–Stokes equations. The size of this term depends on the boundary layer thickness $\delta$, which in turn depends on the asymptotic regime in $\epsilon$. So long as the boundary layer is thinner than $L$, the boundary layer approximation applies far from the edges of the plate. In a region of length $\ell \propto \sqrt {\nu t}$ from the edges, the boundary layer approximation fails. The contribution of this region to the drag force on the plate scales as $\mu U$ (where $U=D/T$), whereas the drag on the rest of the plate scales as $\mu U L/\delta$, which is a factor $O(L/\delta )$ larger. Thus from both of these sources, the boundary layer approximation commits an error of $O(\delta /L)$.

For $\epsilon \ll 1$, the boundary layer thickness is $\delta \propto \sqrt {\nu t}$, which grows no thicker than $\sqrt {\nu T}$. Thus in this regime, the condition for validity of the approximation is $\delta \ll L$, which is independent of the Reynolds number. An identical conclusion was reached by Mandre (Reference Mandre2020) in a more general case. For $\epsilon \gg 1$, the boundary layer grows to a maximum thickness that scales as $\delta \propto \sqrt {\nu L/U}$, and the error in the boundary layer approximation scales as $\delta /L = O(\textit {Re}^{-1/2})$. Thus in this regime, the condition for validity of the approximation is $\textit {Re} \gg 1$.