To start with, we consider a two-dimensional diffuser where the inflow, given by (
1), is almost uniform. We study this situation because it illustrates the limiting case of the earlier developing shear layer examples in Fig.
4, except where the inner and outer plug regions are of similar speeds. In such situations, the shear rate
\(\varepsilon _y\) is small, such that the shear layer develops slowly between the plug flow regions (see Eq. (
2)). Furthermore, we restrict our attention to situations where the channel is sufficiently short that the shear layer never reaches across the channel (see Fig.
1b). Therefore, we consider an inflow given by (
1) with small
\(U_1-U_2\), which is a perturbation to a uniform velocity profile. Then, we asymptotically expand our governing equations (
1)–(
6) about the leading order uniform flow solution, resulting in a reduced set of equations which are amenable to analysis using Pontryagin’s maximum principle.
Hence, let us introduce the small parameter
\(\epsilon \ll 1\), which is defined by the difference in speed between the two plug regions at the inflow
$$\begin{aligned} \epsilon V=U_1(0)-U_2(0)=U_0-U_2(0), \end{aligned}$$
(30)
where
V is a dimensional velocity scale. If the plug regions always exist, with positive width
\(h_1>0\),
\(h_2>0\), and we assume that the slower flow never stagnates
\(U_2>0\), then we need not consider the complementarity format for Bernoulli’s equation. Therefore, we replace Eqs. (
5) and (
6) with
$$\begin{aligned} p+\frac{1}{2}\rho U_i^2=\frac{1}{2}\rho U_{i}(0)^2 \quad \mathrm {for}\quad i=1,2. \end{aligned}$$
(31)
We consider the distinguished limit where the friction factor
f is small such that
\(f=\epsilon S F \), where
\(F=O(1)\). Note that, whilst
f is small, over the long diffuser length scales we consider, friction has a significant effect on the pressure variations. For a Reynolds number of
\(Re=10^6\) and hydraulically smooth walls, the friction factor is calculated as
\(f=0.01\), using the Blasius relationship [
30,
31]. Therefore, if
\(\epsilon = 0.1\) and
\(S=0.11\), then
\(F=0.91\). Choosing these parameter values and setting
\(V/U_0=2\), we achieve the small shear limit example in Fig.
6a, c, e.
By considering perturbations to uniform flow, we expand variables in powers of the small parameter
\(\epsilon \),
$$\begin{aligned}&U_1=U_{1_0}+\epsilon \hat{U}_1+\cdots , \end{aligned}$$
(32)
$$\begin{aligned}&U_2=U_{2_0}+\epsilon \hat{U}_2+\cdots ,\end{aligned}$$
(33)
$$\begin{aligned}&h_1=h_{1_0}+\epsilon \hat{h}_1+\cdots ,\end{aligned}$$
(34)
$$\begin{aligned}&h_2=h_{2_0}+\epsilon \hat{h}_2+\cdots ,\end{aligned}$$
(35)
$$\begin{aligned}&p=p_0+\epsilon \hat{p}+\cdots ,\end{aligned}$$
(36)
$$\begin{aligned}&\delta =\delta _0+\epsilon \hat{\delta }+\cdots ,\end{aligned}$$
(37)
$$\begin{aligned}&\varepsilon _y=\varepsilon _{y_0} +\epsilon \hat{\varepsilon }_y+\cdots . \end{aligned}$$
(38)
In the limit
\(\epsilon \rightarrow 0\), Eqs. (
2)–(
4) and (
31) are satisfied by
$$\begin{aligned}&U_{1_0}=U_{2_0}=\frac{U_0h_0}{h},\end{aligned}$$
(39)
$$\begin{aligned}&h_{2_0}=h-h_{1_0},\end{aligned}$$
(40)
$$\begin{aligned}&p_0=\frac{1}{2}\rho U_0^2\left( 1-\frac{h_0^2}{h^2}\right) ,\end{aligned}$$
(41)
$$\begin{aligned}&\delta _0=0,\end{aligned}$$
(42)
$$\begin{aligned}&\varepsilon _{y_0}=\frac{h_0U_0}{S\int _0^x h(\hat{x})\,\mathrm{d}\hat{x}}. \end{aligned}$$
(43)
The function
\(h_{1_0}\), which represents the location of the centre of the shear layer to leading order, is determined at order
\(O(\epsilon )\). Bernoulli’s equation (
31) for each plug region, at order
\(O(\epsilon )\), is
$$\begin{aligned} \hat{p}+\rho \hat{U}_1 U_{1_0}&=0, \end{aligned}$$
(44)
$$\begin{aligned} \hat{p}+\rho \hat{U}_2 U_{1_0}&=-\rho U_0 V. \end{aligned}$$
(45)
From the relationship
\(h_1+h_2+\delta =h\) at order
\(O(\epsilon )\), we find that
$$\begin{aligned} \hat{h}_1 +\hat{h}_2 + \hat{\delta }=0. \end{aligned}$$
(46)
Thus, the conservation of mass equation (
3) is
$$\begin{aligned} \hat{U}_1h_{1_0} + \hat{U}_2 (h - h_{1_0}) = -V h_2(0), \end{aligned}$$
(47)
and the momentum equation (
4) is
$$\begin{aligned} h\frac{d\hat{p}}{\mathrm{d}x}+\rho \frac{\mathrm{d}}{\mathrm{d}x}\left( 2 U_{1_0}\hat{U}_1 h_{1_0} +2 U_{1_0}\hat{U}_2 (h-h_{1_0}) \right) =-\frac{1}{8}SF\rho U_{1_0}^2. \end{aligned}$$
(48)
Thus, using Eqs. (
44)–(
46), we can simplify Eq. (
48) to an equation purely involving
\(h_{1_0}\) and
h$$\begin{aligned} V h^2 \frac{\mathrm{d}h_{1_0}}{\mathrm{d}x} -V h h_{1_0} \frac{\mathrm{d}h}{\mathrm{d}x} =-\frac{1}{8}F Sh_0^2 U_0. \end{aligned}$$
(49)
Equation (
49) has solution
$$\begin{aligned} h_{1_0}=h\left( \frac{h_1(0)}{h_0} -\int _0^x \frac{SFh_0^2U_0}{8V h(\hat{x})^3}\,\mathrm{d}\hat{x} \right) . \end{aligned}$$
(50)
Combining (
47) and (
48), we obtain a differential equation for the pressure correction
\(\hat{p}\) in terms of the channel shape and its derivative
\(\alpha =h'(x)\),
$$\begin{aligned} \frac{\mathrm{d}\hat{p}}{\mathrm{d}x}=-\frac{\rho U_0^2 h_0^2 }{h^{3}} \left( \frac{2 V}{U_0} \left( 1- \frac{h_1(0)}{h_0}\right) \alpha + \frac{S F}{8} \right) . \end{aligned}$$
(51)
We now solve the optimal control problem outlined in Sect.
2.2, using the approach outlined in Sect.
2.3.2. Since the inflow conditions are fixed and we take
\(p(0)=0\), maximising
\(C_p\) is equivalent to maximising pressure at the outlet
p(
L). Furthermore, the constraint equations have been reduced to (
51). Therefore the optimal control problem, including terms up to and including order
\(O(\epsilon )\), and written as a system of first order differential equations, is as follows:
$$\begin{aligned} \max _{\alpha _\mathrm{min} \le \alpha (x)\le \alpha _\mathrm{max}}\quad \varPhi : = p(L), \end{aligned}$$
(52)
such that
$$\begin{aligned}&\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{\rho U_0^2 h_0^2 }{h^{3}} \left( \left( 1- \epsilon \frac{2 V}{U_0} \left( 1- \frac{h_1(0)}{h_0}\right) \right) \alpha -\epsilon \frac{S F}{8} \right) , \end{aligned}$$
(53)
$$\begin{aligned}&\frac{\mathrm{d}h}{\mathrm{d}x}=\alpha ,\end{aligned}$$
(54)
$$\begin{aligned}&h(0)=h_0,\end{aligned}$$
(55)
$$\begin{aligned}&p(0)=0, \end{aligned}$$
(56)
$$\begin{aligned}&h(L)=h_L. \end{aligned}$$
(57)
We now solve this reduced problem using Pontryagin’s maximum principle [
25], as outlined earlier. The Hamiltonian for this system is
$$\begin{aligned} H=\lambda _p \frac{\mathrm{d}p}{\mathrm{d}x}+\lambda _h \frac{\mathrm{d}h}{\mathrm{d}x}, \end{aligned}$$
(58)
where
\(\lambda _p\) and
\(\lambda _h\) are the adjoint variables which satisfy the adjoint equations
$$\begin{aligned} \frac{\mathrm{d}\lambda _p}{\mathrm{d}x}&=-\frac{\partial H}{\partial p},\end{aligned}$$
(59)
$$\begin{aligned} \frac{\mathrm{d}\lambda _h}{\mathrm{d}x}&=-\frac{\partial H}{\partial h}. \end{aligned}$$
(60)
From (
24), since the objective function only depends on pressure at the outlet
\(\varPhi =p(L)\), we have the natural boundary condition
$$\begin{aligned} \lambda _p(L)=\frac{\partial \varPhi }{\partial p}=1. \end{aligned}$$
(61)
Considering Eq. (
61) and the fact that there is no dependance of the Hamiltonian (
58) on the pressure
p, Eq. (
59) tells us that
\(\lambda _p=1\) for all values of
x. There is no natural boundary condition for
\(\lambda _h\) since we are enforcing a condition on
h at the outlet
\(x=L\) (see (
24)). The last condition from Pontryagin’s maximum principle is the optimality condition, and since the Hamiltonian is linear in the control
\(\alpha \), this takes the form of Eq. (
26).
Next we investigate whether the optimal control is bang-bang, or whether any singular arcs exist (see Sect.
2.3.2). Using (
58) with (
53), (
54) and (
60), and noting that
\(\lambda _p=1\), we see that
$$\begin{aligned} \frac{\mathrm{d}H_\alpha }{\mathrm{d}x}=-\frac{3 \epsilon \rho S F U_0^2 h_0^2 }{8h^{4}}, \end{aligned}$$
(62)
which is negative for all values of
x. Hence, it is impossible for singular arcs to exist in this case. Therefore the control is bang-bang with
$$\begin{aligned} \alpha (x)={\left\{ \begin{array}{ll} \alpha _\mathrm{max} &{} \mathrm {for} \quad x\in [0,\gamma ], \\ \alpha _\mathrm{min} &{} \mathrm {for} \quad x\in [\gamma ,L], \\ \end{array}\right. } \end{aligned}$$
(63)
where the switching point
\(\gamma \) is given by
$$\begin{aligned} \gamma =\frac{h_L-h_0-\alpha _\mathrm{min} L}{\alpha _\mathrm{max} - \alpha _\mathrm{min} }. \end{aligned}$$
(64)
In Fig.
6c, e, we plot the solution to the optimal control problem found using the Hamiltonian approach on top of the solution found using the numerical optimisation routine outlined in Sect.
2.3.1. It is clear that the numerical optimisation routine has correctly found the bang-bang control which we have derived here, with
\(\gamma /h_0=10.59\) (the small discrepancy is probably due to the finite value of
\(\epsilon \)).
It should be noted that the adjoint variable
\(\lambda _h\) is only solved for up to a constant of integration
C (from integrating Eq. (
60)) since it has no boundary condition. Instead,
C is determined by the condition that
$$\begin{aligned} H_\alpha (x=\gamma )=0. \end{aligned}$$
(65)
In the case where we also allow the channel length
L to be a control as well as
\(\alpha \), as described in Sect.
2.3.2, we have the additional constraint on the Hamiltonian at the final point (
25). By calculating
H (
58), it is straightforward to show that (
65) and (
25) are inconsistent unless
\(\gamma =0\) or
\(\gamma =L\). For the case in Fig.
6a, c, e, it is clear that
\(\gamma =0\) is impossible, so we conclude that the optimal diffuser length is
\(L=\gamma \). Therefore, including
L as a control and taking
\(\alpha _\mathrm{min}=0^\circ \), the optimal diffuser shape for the small shear limit is one which expands at the maximum angle until
h reaches
\(h_L\), at which point the channel terminates (i.e. conventional diffuser design for uniform flow).