Global sensitivity analysis for the boundary control of an open channel

Abstract

The goal of this paper is to solve the global sensitivity analysis for a particular control problem. More precisely, the boundary control problem of an open-water channel is considered, where the boundary conditions are defined by the position of a down stream overflow gate and an upper stream underflow gate. The dynamics of the water depth and of the water velocity are described by the Shallow-Water equations, taking into account the bottom and friction slopes. Since some physical parameters are unknown, a stabilizing boundary control is first computed for their nominal values, and then a sensitivity analysis is performed to measure the impact of the uncertainty in the parameters on a given to-be-controlled output. The unknown physical parameters are described by some probability distribution functions. Numerical simulations are performed to measure the first-order and total sensitivity indices.

Appendix: Technical proofs

Proof of Proposition 1

Let us first note that it follows from (2) and $Q=BHV$ that (by omitting the time variable)

$$\begin{aligned} B H(0) V(0) = U _0 B \mu _0\sqrt{2g(z_{\mathrm{up}}-H(0))} \end{aligned}$$

which is equivalent to

$$\begin{aligned} V(0) = U _0 \frac{1}{H(0)} \mu _0\sqrt{2g(z_{\mathrm{up}}-H(0))}. \end{aligned}$$

(37)

On the other side, the first line of (13) yields with (7) and the definitions of v and h, the following

$$\begin{aligned} V(0) - V ^\star + ( H (0) - H ^\star ) \sqrt{\frac{g}{H^\star }} = k_0 ( V(0)- V ^\star ) - k_0 (H(0) - H ^\star ) \sqrt{\frac{g}{H^\star }} , \end{aligned}$$

which may be rewritten as

$$\begin{aligned} V(0) =V ^{\star } -\frac{1+ k_0}{1- k_0} ( H (0) - H ^\star ) \sqrt{\frac{g}{H^{\star }}} . \end{aligned}$$

(38)

Therefore, with (37) and (38), (2) and the first line of (13) are equivalent as soon as the first control is defined by (11).

Let us compute the control $U_L$ in a similar way. To do that, we first deduce from (3) the following

$$\begin{aligned} (H(L)- h _ s - U _L)^3 = \frac{ H (L) ^2V (L) ^2}{2 g \mu _L ^2} \end{aligned}$$

and thus

$$\begin{aligned} V(L)=\frac{\sqrt{2 g \mu _L ^2(H(L)- h _ s - U _L)^3}}{H (L)}. \end{aligned}$$

(39)

Moreover, from the second line of (13), with (7) and the expressions of v and h, it holds

$$\begin{aligned} V(L) -V ^\star - ( H( L) - H ^\star ) \displaystyle \sqrt{\frac{g }{H^\star }} = k_L (V(L) -V ^\star ) + k_L (H(L) -H ^\star ) \displaystyle \sqrt{\frac{g }{H^\star }} \end{aligned}$$

and also

$$\begin{aligned} V(L) = V ^\star + \frac{1+ k_L}{1- k_l} ( H( L) - H ^\star ) \sqrt{\frac{g }{H^\star }}. \end{aligned}$$

(40)

From (39) and (40) we get that (3) and the second line (13) are equivalent as soon as the control $U_L$ is defined by

$$\begin{aligned} \sqrt{2 g \mu _L ^2 ( H(L) - h_s - U_L ) ^3} = H (L)\left( V ^\star + \frac{1+ k_L}{1- k_L} ( H( L) - H ^\star ) \sqrt{\frac{g }{H^\star }} \right) \end{aligned}$$

which is equivalent to (12). This concludes the proof of Proposition 1. $\square$

Proof of Proposition 3

For sake of conciseness, we omit the time variable: for instance H(0) stands for H(0, t) for each t. Let us define $\alpha =\frac{1+k_0}{1-k_0}, \quad \beta =\sqrt{\frac{g}{H^\star _\text {nom}}} .$

At $x=0$, we take, from the proof of Proposition 1:

$$\begin{aligned} H(0) V(0) = U_0 \mu _0 \sqrt{2g(z_{\mathrm{up}}-H(0))} \end{aligned}$$

with

$$\begin{aligned} U_0 = \frac{H(0)(V^\star _\text {nom}-\alpha (H(0)-H^\star _\text {nom})\beta )}{\mu _0\sqrt{2g(z_{\mathrm{up},\text {nom}}-H(0))}} \end{aligned}$$

that is:

$$\begin{aligned} v(0) = \sqrt{\frac{z_{\mathrm{up}}-H(0)}{z_{\mathrm{up},\text {nom}}-H(0)}} ( V^\star _\text {nom}- \alpha ( H(0)-H^\star _\text {nom}) \beta ) - V^\star . \end{aligned}$$

Now, given the change of variables between (V, H), (v, h), this equation could be rewritten as a nonlinear relation between v(0) and h(0), hence, as a nonlinear relation between $\xi _1(0)$ and $\xi _2(0)$. However, to keep the resolution simple, we have to linearize this equation. This linearization is made in accordance with the linearization of the Shallow-Water equation for (H, V) near $(H^\star , V^\star )$, hence for (h, v) near the origin. We do the same here; hence by Taylor expansion around $h=0$, we get $v(0)= \mathcal {A}+ \mathcal {B}h(0) + o (h(0))$ with

$$\begin{aligned} \mathcal {A}= \frac{\mu _0}{\mu _{0,\text {nom}}}\sqrt{ \frac{H^\star -z_{\mathrm{up}} }{H^\star -z_{\mathrm{up},\text {nom}}}}( V^\star _\text {nom}-\alpha \beta (H^\star -H^\star _\text {nom}) ) - V^\star , \end{aligned}$$

and

$$\begin{aligned} \mathcal {B}= \frac{\mu _0}{\mu _{0,\text {nom}}}\sqrt{\frac{H^\star -z_{\mathrm{up}} }{H^\star -z_{\mathrm{up},\text {nom}}}}\left( - \alpha \beta + \frac{(z_{\mathrm{up}}-z_{\mathrm{up},\text {nom}})( V^\star _\text {nom}- \alpha \beta (H^\star -H^\star _\text {nom}) ) }{2(H^\star -z_{\mathrm{up}})(H^\star -z_{\mathrm{up},\text {nom}})}\right) . \end{aligned}$$

Similarly, at $x=L$, we have

$$\begin{aligned} V(L) = \frac{\sqrt{2g } \mu _L}{H(L)} \left( e_h + \left( \frac{H(L)(V^\star _\text {nom}+\alpha _L \beta e_H)}{\sqrt{2g} \mu _{L}} \right) ^{2/3} \right) ^{3/2} \end{aligned}$$

where $\alpha _L=\frac{1+k_L}{1-k_L}$, and

$$\begin{aligned} e_h = h_{s,\text {nom}} - h_s, \quad e_H = H(L)- H^\star _\text {nom}. \end{aligned}$$

Therefore,

$$\begin{aligned} v(L) = \mathcal {C}+ \mathcal {D}h(L) + o(h(L)), \end{aligned}$$

with

$$\begin{aligned} \mathcal {C}= & {} {\frac{\sqrt{2g}}{4}}\mu _L \left( 2 h_{s,\text {nom}}-2h_s\right. \\&+\left. 2^{\frac{2}{3}} \left( -H^\star \frac{\left( -V^\star _{\text {nom}}+\alpha _L \beta _{\text {nom}} \times H^\star _{\text {nom}}\right) }{\left( \sqrt{g}\mu _{L,\text {nom}}\right) }\right) ^{\frac{2}{3}}\right) \\&\times \frac{\sqrt{\left( 4 h_{s,\text {nom}}-4 h_s+2\times 2^{\frac{2}{3}} \left( -H^\star \times \frac{\left( -V^\star _{\text {nom}}+\alpha _L \beta _{\text {nom}} H^\star _{\text {nom}}\right) }{\left( \sqrt{g} \mu _{L,\text {nom}}\right) }\right) ^{\frac{2}{3}}\right) }}{H^\star }\\ \mathcal {D}= & {} -{\frac{1}{2}} \left( 2^{\frac{1}{6}} \left( -H^\star \frac{(-V^\star _{\text {nom}}+\alpha _L \beta _{\text {nom}} H^\star _{\text {nom}})}{\left( \sqrt{g} \mu _{L,\text {nom}}\right) }\right) ^{\frac{2}{3}} \alpha _L \beta _{\text {nom}} H^\star \right. \\&-\left. \sqrt{2} V^\star _{\text {nom}} h_{s,\text {nom}}+\sqrt{2} V^\star _{\text {nom}} h_s+\sqrt{2} \alpha _L \beta _{\text {nom}} H^\star _{\text {nom}} h_{s,\text {nom}}\right. \\&\left. -\sqrt{2}\alpha _L \beta _{\text {nom}} H^\star _{\text {nom}} h_s\right) \\&\times \sqrt{\left( 4 h_{s,\text {nom}}-4h_s+2 \times 2^{\frac{2}{3}} \left( -H^\star \frac{(-V^\star _{\text {nom}}+\alpha _L \beta _{\text {nom}} \times H^\star _{\text {nom}})}{\left( \sqrt{g}\mu _{L,\text {nom}}\right) }\right) ^{\frac{2}{3}}\right) }\\&\times \mu _L \frac{\sqrt{g}}{((-V^\star _{\text {nom}}+\alpha _L \beta _{\text {nom}} H^\star _{\text {nom}})H^{\star 2} )}. \end{aligned}$$

Thus, the linearized boundary relations satisfied by $\xi _1$ and $\xi _2$ in the real-life model are given by (18) and (19). $\square$