Vibration Control of a Flexible Aerial Refueling Hose with Input Saturation

Abstract

In Chap. 5, we developed a PDE model for a flexible aerial refueling hose system during coupling. As we know, a hose-drogue aerial refueling system consists of a flexible hose and an active drogue control actuator, which are the most universal refueling equipments of probe-drogue refueling (PDR). The probe and drogue systems are comparatively simpler and more compact than the flying boom, and their arrangement on the tanker enables multiple aircraft to be refueled simultaneously. The significant drawback is that PDR requires a skillful piloting technique of maneuvering a probe into the center of a moving drogue with an acceptable closure rate. However, due to the flexible property of the hose, the deflection of the flexible hose has a significant influence on the dynamics and control performance of the AAR, which brings difficulties to the coupling. In order to solve this problem, we will establish a model for a flexible aerial refueling hose system before coupling, and develop an efficient boundary control scheme to suppress vibrations. Moreover, considering that common active drogue control actuators which consists of a set of aerodynamic control surface [5, 7] or other kinds of controllable drogues [6] cannot provide enough control input, which degrades the performance of the control system or leads to the instability.

Appendices

Appendix 1: Proof of Lemma 6.1

Proof

To motivate the followings, we first focus our attention on the term ${{V}_{2}}(t)$. It satisfies the following inequality

$$ \left| {{V}_{2}}(t)\right| \le \alpha \rho L\int _{0}^{L}{{w_{t}^{2}}(x,t)dx}+\alpha \rho L\int _{0}^{L}{{{\left[ {w_{x}}(x,t)\right] }^{2}}dx}\le {{\beta }_{1}}{{V}_{1}}(t)$$

where ${{\beta }_{1}}=\frac{2\alpha \rho L}{\beta \min \left( \rho ,{{P}_{\min }}\right) }$. We then obtain

$$ -{{\beta }_{1}}{{V}_{1}}(t)\le {{V}_{2}}(t)\le {{\beta }_{1}}{{V}_{1}}(t) $$

Assuming that $\alpha$ is a small positive weighting constant satisfying $0<\alpha <\frac{\beta \min \left( \rho ,{{P}_{\min }}\right) }{2\rho L}$, we can obtain $0<{{\beta }_{1}}<1$, and

$$\begin{aligned} {{\beta }_{2}}\left( {{V}_{1}}(t)+{{V}_{b}}(t)\right) \le V(t)\le {{\beta }_{3}}\left( {{V}_{1}}(t)+{{V}_{b}}(t)\right) \end{aligned}$$

(6.34)

where ${{\beta }_{2}}=\min \left( 1-{{\beta }_{1}},1\right) =1-{{\beta }_{1}}$ and ${{\beta }_{3}}=\max \left( 1+{{\beta }_{1}},1\right) =1+{{\beta }_{1}}$.

Differentiating Eq. (6.31) with respect to time, we have

$$\begin{aligned} \dot{V}(t)={{\dot{V}}_{1}}(t)+{{\dot{V}}_{2}}(t)+{{\dot{V}}_{b}}(t) \end{aligned}$$

(6.35)

The term ${{\dot{V}}_{1}}(t)$ is rewritten as

$$ {{\dot{V}}_{1}}(t)={{\dot{V}}_{11}}+{{\dot{V}}_{12}}+{{\dot{V}}_{13}}$$

where

$$\begin{aligned} {{\dot{V}}_{11}}&=\beta \rho \int _{0}^{L}{{w_{t}}(x,t){w_{tt}}(x,t)dx}\end{aligned}$$

(6.36)

$$\begin{aligned} {{\dot{V}}_{12}}&=\beta \int _{0}^{L}{P(x,t){w_{x}}\left( x,t\right) {{w_{xt}}}\left( x,t\right) dx}\end{aligned}$$

(6.37)

$$\begin{aligned} {{\dot{V}}_{13}}&=\frac{\beta }{2}\int _{0}^{L}{{P_{t}}(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx} \end{aligned}$$

(6.38)

Substituting Eq. (6.8) into (6.36), we get

$$\begin{aligned} {{{\dot{V}}}_{11}}&=\beta \rho \int _{0}^{L}{{w_{t}}(x,t){w_{tt}}(x,t)dx}\\&=\beta \int _{0}^{L}{{w_{t}}(x,t)\left( {P_{x}}(x,t){w_{t}}\left( x,t\right) +P(x,t){w_{xx}}\left( x,t\right) +Q(x,t)\right) dx}\end{aligned}$$

and integrating Eq. (6.37) by parts with the boundary conditions, we obtain

$$\begin{aligned} {{{\dot{V}}}_{12}}&=\beta \left[ P(L,t){w_{x}}\left( L,t\right) {w_{t}}\left( L,t\right) -P(0,t){w_{x}}\left( 0,t\right) {w_{t}}\left( 0,t\right) \right] \\&-\beta \int _{0}^{L}{{w_{t}}\left( x,t\right) {P_{x}}(x,t){w_{x}}\left( x,t\right) dx}\\&-\beta \int _{0}^{L}{{w_{t}}\left( x,t\right) P(x,t){w_{xx}}\left( x,t\right) dx}\end{aligned}$$

Then, we have

$$\begin{aligned} {{{\dot{V}}}_{1}}(t)&=\beta \int _{0}^{L}{{w_{t}}(x,t)Q(x,t)dx}+\frac{\beta }{2}\int _{0}^{L}{{P_{t}}(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\nonumber \\&+\frac{\beta P(L,t)}{2}\left[ z_{2}^{2}-{{\left( {w_{x}}\left( L,t\right) \right) }^{2}}-{{\left( {w_{t}}\left( L,t\right) \right) }^{2}-c}_{1}^{2}w^{2}(L,t)\right] \nonumber \\&-\beta c_{1}P(L,t){w_{t}}\left( L,t\right) w\left( L,t\right) -\beta c_{1}P(L,t){w_{x}}\left( L,t\right) w\left( L,t\right) \end{aligned}$$

(6.39)

According to Lemma 2.4, we obtain

$$\begin{aligned} {{{\dot{V}}}_{1}}(t)&\le \frac{\beta }{{{\sigma }_{1}}}\int _{0}^{L}{{w_{t}^{2}}(x,t)dx+}\beta {{\sigma }_{1}}\int _{0}^{L}{{{Q}^{2}}(x,t)dx}\nonumber \\&+\frac{\beta }{2}\int _{0}^{L}{{P_{t}}(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}-\beta c_{1}P(L,t){w_{t}}\left( L,t\right) w\left( L,t\right) \nonumber \\&+\frac{\beta P(L,t)}{2}\left[ z_{2}^{2}-{{\left( {w_{x}}\left( L,t\right) \right) }^{2}}-{{\left( {w_{t}}\left( L,t\right) \right) }^{2}-c}_{1}^{2}w^{2}(L,t)\right] \nonumber \\&-\beta c_{1}P(L,t){w_{x}}\left( L,t\right) w\left( L,t\right) \end{aligned}$$

(6.40)

where ${{\sigma }_{1}}$ is a positive constant.

To go on, the term ${{\dot{V}}_{2}}(t)$ is rewritten as

$$\begin{aligned} {{\dot{V}}_{2}}(t)={{\dot{V}}_{21}}+{{\dot{V}}_{22}}+{{\dot{V}}_{23}}+{{\dot{V}}_{24}}\end{aligned}$$

(6.41)

where

$$\begin{aligned} {{\dot{V}}_{21}}&=\alpha \int _{0}^{L}\left( {x-L}\right) {{w_{x}}(x,t){P_{x}}(x,t){w_{x}}\left( x,t\right) dx}\end{aligned}$$

(6.42)

$$\begin{aligned} {{\dot{V}}_{22}}&=\alpha \int _{0}^{L}\left( {x-L}\right) {{w_{x}}(x,t)P(x,t){w_{xx}}\left( x,t\right) dx}\end{aligned}$$

(6.43)

$$\begin{aligned} {{\dot{V}}_{23}}&=\alpha \int _{0}^{L}\left( {x-L}\right) {{w_{x}}(x,t)Q(x,t)dx}\end{aligned}$$

(6.44)

$$\begin{aligned} {{\dot{V}}_{24}}&=\alpha \int _{0}^{L}{\rho \left( {x-L}\right) {w_{t}}(x,t){{w_{xt}}}(x,t)dx}\end{aligned}$$

(6.45)

Using the boundary conditions and integrating Eq. (6.43) by parts, we get

$$\begin{aligned} {{{\dot{V}}}_{22}}&=-\alpha \int _{0}^{L}{{w_{x}}(x,t)P(x,t){w_{x}}\left( x,t\right) dx}\nonumber \\&-\alpha \int _{0}^{L}\left( {x-L}\right) {{w_{x}}(x,t){P_{x}}(x,t){w_{x}}\left( x,t\right) dx}\nonumber \\&-\alpha \int _{0}^{L}\left( {x-L}\right) {{w_{xx}}(x,t)P(x,t){w_{x}}\left( x,t\right) dx}\end{aligned}$$

(6.46)

Combining (6.43) and (6.46), we have

$$\begin{aligned} {{{\dot{V}}}_{22}}&=-\frac{\alpha }{2}\int _{0}^{L}{P(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\end{aligned}$$

(6.47)

$$\begin{aligned}&-\frac{\alpha }{2}\int _{0}^{L}\left( {x-L}\right) {{P_{x}}(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\end{aligned}$$

(6.48)

According to Lemma 2.4, we obtain

$$\begin{aligned} {{\dot{V}}_{23}}\le \frac{\alpha L}{{{\sigma }_{2}}}\int _{0}^{L}{{{Q}^{2}}(x,t)dx}+\alpha L{{\sigma }_{2}}\int _{0}^{L}{{{\left[ {w_{x}}(x,t)\right] }^{2}}dx}\end{aligned}$$

(6.49)

where ${{\sigma }_{2}}$ is a positive constant. Integrating (6.45) by parts, we obtain

$$ {{{\dot{V}}}_{24}}=-\alpha \rho \int _{0}^{L}{{w_{t}^{2}}(x,t)dx}-\alpha \rho \int _{0}^{L}\left( {x-L}\right) {{w_{t}}(x,t){{w_{xt}}}(x,t)dx}$$

Considering (6.45), we then get

$$\begin{aligned} {{\dot{V}}_{24}}=-\frac{\alpha \rho }{2}\int _{0}^{L}{{w_{t}^{2}}(x,t)dx}\end{aligned}$$

(6.50)

Substituting (6.42), (6.48), (6.49) and (6.50) into (6.41), we obtain

$$\begin{aligned} {{{\dot{V}}}_{2}}(t)&\le \alpha \int _{0}^{L}\left( {x-L}\right) {{P_{x}}(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx} \nonumber \\&-\frac{\alpha }{2}\int _{0}^{L}{P(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}+\alpha L{{\sigma }_{2}}\int _{0}^{L}{{{\left[ {w_{x}}(x,t)\right] }^{2}}dx}\nonumber \\&-\frac{\alpha }{2}\int _{0}^{L}\left( {x-L}\right) {{P_{x}}(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\nonumber \\&+\frac{\alpha L}{{{\sigma }_{2}}}\int _{0}^{L}{{{Q}^{2}}(x,t)dx} -\frac{\alpha \rho }{2}\int _{0}^{L}{{{w}_{t}^{2}}(x,t)dx}\end{aligned}$$

(6.51)

Substituting (6.30), (6.40) and (6.51) into (6.35), we obtain

$$\begin{aligned} \dot{V}&={{{\dot{V}}}_{1}}+{{{\dot{V}}}_{2}}+{{{\dot{V}}}_{b}}\nonumber \\&\le -\frac{1}{2}\int _{0}^{L}{\left[ \alpha P(x,t)-\alpha (x-L){P_{x}}(x,t)\right] {{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\nonumber \\&-\frac{1}{2}\int _{0}^{L}{\left[ -2\alpha L{{\sigma }_{2}}-\beta {P_{t}}(x,t)\right] {{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\nonumber \\&-\left( \frac{\alpha \rho }{2}-\frac{\beta }{{{\sigma }_{1}}}\right) \int _{0}^{L}{{w_{t}^{2}}(x,t)dx}\nonumber \\&-\left( {{c}_{1}}+\frac{\beta P(L,t)c_{1}^{2}}{2}-\beta P(L,t)c_{1}^{2}-\frac{1}{{{\sigma }_{3}}}\right) z_{1}^{2}\nonumber \\&-\left( {{c}_{2}}-\frac{\beta P(L,t)}{2}\right) z_{2}^{2}-{{c}_{3}}z_{3}^{2}-\beta P(L,t)\frac{{w_{t}^{2}}(L,t)}{2}\nonumber \\&-\left( \frac{\beta P(L,t)}{2}-{{\sigma }_{3}}\right) {{\left[ {w_{x}}(L,t)\right] }^{2}}\nonumber \\&+\frac{1}{{{\gamma }_{\chi }}}\left( \xi N\left( \chi \right) -1\right) \dot{\chi }+\frac{1}{2l}\bar{d}^{2}\nonumber \\&+\left( \frac{\alpha L}{{{\sigma }_{2}}}+{{\sigma }_{1}}\beta \right) \int _{0}^{L}{{{Q}^{2}}(x,t)dx} \end{aligned}$$

(6.52)

We design parameters $\alpha$ and $\beta$ to satisfy the following inequality:

$$ \alpha {{P}_{\min }}-\beta {{{P}}_{t\max }}-\alpha L{{{P}}_{x\max }}-2\alpha L{{\sigma }_{2}}\ge \delta $$

$\forall (x,t)\in \left[ 0,L \right] \times \left[ 0,\infty \right)$, for a positive constant $\delta$, and the following conditions:

$$\begin{aligned}&\frac{\alpha \rho }{2}-\frac{\beta }{{{\sigma }_{1}}}>0\\&{{c}_{1}}+\frac{\beta {{P}_{\min }}c_{1}^{2}}{2}-\beta {{P}_{\max }}c_{1}^{2}-\frac{1}{{{\sigma }_{3}}}>0\\&{{c}_{2}}-\frac{\beta {{P}_{\max }}}{2}\ge 0\\&\frac{\beta {{P}_{\min }}}{2}-{{\sigma }_{3}}\ge 0 \end{aligned}$$

Equation (6.52) can be rewritten as

$$\begin{aligned} \dot{V}(t)&\le -{{\gamma }_{1}}\frac{\beta }{2}\int _{0}^{L}{P(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}-{{\gamma }_{2}}\frac{\beta \rho }{2}\int _{0}^{L}{{w_{t}^{2}}(x,t)dx}\nonumber \\&-{{\gamma }_{3}}\frac{1}{2}z_{1}^{2}-{{\gamma }_{4}}\frac{m}{2}z_{2}^{2}-{{\gamma }_{5}}\frac{m}{2}z_{3}^{2}\nonumber \\&+\frac{1}{{{\gamma }_{\chi }}}\left( \xi N\left( \chi \right) -1\right) \dot{\chi }+\varepsilon \end{aligned}$$

(6.53)

where

$$\begin{aligned} {{\gamma }_{1}}&=\frac{\delta }{\beta {{P}_{\max }}}\\ {{\gamma }_{2}}&=\left( \frac{\alpha }{\beta }-\frac{2}{{{\sigma }_{1}}\rho } \right) \\ {{\gamma }_{3}}&=2\left( {{c}_{1}}+\frac{\beta {{P}_{\min }}c_{1}^{2}}{2}-\beta {{P}_{\max }}c_{1}^{2}-\frac{1}{{{\sigma }_{3}}} \right) \\ {{\gamma }_{4}}&=\frac{2}{m}\left( {{c}_{2}}-\frac{\beta {{P}_{\max }}}{2} \right) \\ {{\gamma }_{5}}&=\frac{2}{m}{{c}_{3}}\\ \varepsilon&=\frac{1}{2l}\bar{d}^{2}+\left( \frac{\alpha L}{{{\sigma }_{2}}}+{{\sigma }_{1}}\beta \right) LQ_{\max }^{2} \end{aligned}$$

We further obtain

$$\begin{aligned} \dot{V}(t)\le -{{\lambda }_{1}}\left[ {{V}_{1}}(t)+{{V}_{b}}(t)\right] +\frac{1}{{{\gamma }_{\chi }}}\left( \xi N\left( \chi \right) -1\right) \dot{\chi }+\varepsilon \end{aligned}$$

(6.54)

where ${{\lambda }_{1}}=\min \left( {{\gamma }_{1}},{{\gamma }_{2}},{{\gamma }_{3}},{{\gamma }_{4}},{{\gamma }_{5}}\right)$.

Combining (6.34) and (6.54), we have

$$\begin{aligned} \dot{V}(t)\le -\lambda V(t)+\frac{1}{{{\gamma }_{\chi }}}\left( \xi N\left( \chi \right) -1\right) \dot{\chi }+\varepsilon \end{aligned}$$

(6.55)

where $\lambda ={{{\lambda }_{1}}}/{{{\beta }_{3}}}\;>0$.

Then multiplying Eq. (6.55) by ${{e}^{\lambda t}}$, we obtain

$$\begin{aligned} \frac{\partial }{\partial t}\left( \left( V(t){{e}^{\lambda t}}\right) \right) \le \varepsilon {{e}^{\lambda t}}+\frac{1}{{{\gamma }_{\chi }}}\left( \xi N\left( \chi \right) -1\right) \dot{\chi }{{e}^{\lambda t}} \end{aligned}$$

(6.56)

Integrating of the inequality (6.56), we have

$$\begin{aligned} V(t)&\le V(0){{e}^{-\lambda t}}+\frac{\varepsilon }{\lambda }\left( 1-{{e}^{-\lambda t}}\right) \nonumber \\&+\frac{{{e}^{-\lambda t}}}{{{\gamma }_{\chi }}}\int _{0}^{t}{\left( \xi N\left( \chi \right) -1\right) \dot{\chi }{{e}^{\lambda \tau }}d\tau }\nonumber \\&\le V(0){{e}^{-\lambda t}}+\frac{\varepsilon _{0}}{\lambda }\end{aligned}$$

(6.57)

where $\varepsilon _{0}=\varepsilon +\frac{\lambda }{{{\gamma }_{\chi }}}\int _{0}^{t}{\left( \xi N\left( \chi \right) -1\right) \dot{\chi }{{e}^{-\lambda \left( t-\tau \right) }}d\tau }$.

Applying Lemma 2.8, we can conclude that V(t), $\chi$ and $\int _{0}^{t}{\left( \xi N\left( \chi \right) -1 \right) \dot{\chi }d\tau }$ are bounded on $\left[ 0,t \right)$.

This completes the proof.

Appendix 2: Proof of Theorem 6.1

Proof

According to Lemma 6.1, we can conclude that ${{z}_{1}}$, ${{z}_{2}}$, ${{z}_{3}}$, w(x, t), ${w_{t}}(x,t)$ and ${w_{x}}(x,t)$ are all bounded.

Note that

$$\begin{aligned} \left| u_{g}\left( {{u}_{0}}\right) \right|&={{u}_{M}}\left| \tanh \left( \frac{{{u}_{0}}}{{{u}_{M}}}\right) \right| \le {{u}_{M}}\end{aligned}$$

(6.58)

$$\begin{aligned} \left| \frac{\partial u_{g}\left( {{u}_{0}}\right) }{\partial {{u}_{0}}}\right|&=\left| \frac{4}{{{\left( {{e}^{{{u}_{0}}/{{u}_{M}}}}+{{e}^{-{{u}_{0}}/{{u}_{M}}}}\right) }^{2}}}\right| \le 1\end{aligned}$$

(6.59)

$$\begin{aligned} \left| \frac{\partial u_{g}\left( {{u}_{0}}\right) }{\partial {{u}_{0}}}{{u}_{0}}\right|&=\left| \frac{4{{u}_{0}}}{{{\left( {{e}^{{{u}_{0}}/{{u}_{M}}}}+{{e}^{-{{u}_{0}}/{{u}_{M}}}}\right) }^{2}}}\right| \le \frac{{{u}_{M}}}{2}\end{aligned}$$

(6.60)

Then we can obtain that $\bar{\omega }$ is bounded from (6.58)–(6.60) and (6.26). This further implies that $\omega$ and $u_{0}(t)$ are bounded.

Combining with (6.32) and according to Lemma 2.5, we have

$$\begin{aligned}&\frac{\beta {{P}_{\min }}}{2L}{{w}^{2}}(x,t)\le \frac{\beta }{2}\int _{0}^{L}{P(x,t){{\left[ {w_{x}}\left( x,t\right) \right] }^{2}}dx}\le {{V}_{1}}(t)\nonumber \\&\le {{V}_{1}}(t)+{{V}_{b}}(t)\le \frac{V(t)}{{{\beta }_{2}}}\end{aligned}$$

(6.61)

Then we get

$$\begin{aligned} w(x,t)\le \sqrt{\frac{2l(t)}{\beta {{P}_{\min }}{{\beta }_{2}}}\left( V(0){{e}^{-\lambda t}}+\frac{{{\varepsilon }_{0}}}{\lambda }\right) }\end{aligned}$$

(6.62)

It follows that, $\underset{t\rightarrow \infty }{\mathop {\lim }}\,\left| w(x,t)\right| \le \sqrt{\frac{2L\varepsilon _{0}}{\beta {{P}_{\min }}{{\beta }_{2}}\lambda }},\forall (x,t)\in [0,L]\times [0,\infty ),$ so w(x, t) is uniformly ultimate bounded.

This completes the proof.