Rapid Smooth Entry Trajectory Planning for High Lift/Drag Hypersonic Glide Vehicles

Abstract

This paper presents how to apply second-order cone programming, a subclass of convex optimization, to rapidly solve a highly nonlinear optimal control problem arisen from smooth entry trajectory planning of hypersonic glide vehicles with high lift/drag ratios. The common phugoid oscillations are eliminated by designing a smooth flight path angle profile. The nonconvexity terms of the optimal control problem, which include the nonlinear dynamics and nonconvex control constraints, are convexified via techniques of successive linearization, successive approximation, and relaxation. Lossless relaxation is also proved using optimal control theory. After discretization, the original nonconvex optimal control problem is converted into a sequence of second-order cone programming problems each of which can be solved in polynomial time using existing primal–dual interior-point algorithms whenever a feasible solution exists. Numerical examples are provided to show that rather smooth entry trajectory can be obtained in about 1 s on a desktop computer.

Appendix: Continued Proof of Theorem 3.1

Case 2

($\eta ^*>0$): Assume that there exists a finite interval $[e_m,e_n]\subseteq [e_i,e_j]$ where $[u_1^*(e)]^2 + [u_2^*(e)]^2<1$. As discussed in Case 1, we can also obtain $p_{\psi }=p_{\theta }=p_{\phi }=0$ by using the same method.

When $\eta ^*=|\gamma ^*-\hat{\gamma }|>0$, it implies that one of the constraints $\gamma ^*-\hat{\gamma }\ge -\eta ^*$ and $\gamma ^*-\hat{\gamma }\le \eta ^*$ is inactive but the other one is active. Without loss of generality, suppose $\gamma ^*-\hat{\gamma }\ge -\eta ^*$ is inactive and $\gamma ^*-\hat{\gamma }\le \eta ^*$ is active. Then, we have $\mu _2^-=0$ by Eq. (51). The condition $\partial _{\eta }L=0$ in Eq. (53) becomes $p_0c_{\gamma }+\mu _2^+=0$, that is

$$\begin{aligned} \mu _2^+ = -c_{\gamma }p_0 \end{aligned}$$

(80)

When $\gamma ^*-\hat{\gamma }=\eta ^*$, an optimal control $u_1^*$ can be found to generate the corresponding optimal flight path angle profile $\gamma ^*$, and $r^*$ can also be determined. The value of $u_1^*$ could make the constraints in Eq. (37) either active or inactive. In the following, we will consider two scenarios on whether the control constraints in Eq. (37) are active or not and prove $p_0=p_r=p_{\gamma }=\mu _1^-=\mu _1^+=\mu _2^+=\nu _1=\nu _2=0$.

• (1) Neither of the constraints $u_1^*\ge \varsigma _l$ and $u_1^*\le \varsigma _h$ is active. Then, we have $\mu _1^-=\mu _1^+=0$ based on Eq. (50). The condition $\partial _{u_1}L=0$ can be used to get the relation between $p_{\gamma }$ and $p_0$ as previously seen in Eq. (64), and the relation between $\dot{p}_{\gamma }$ and $p_0$ in Eq. (65) can also be obtained here.

For the constraints $r^*\ge l_1$ and $r^*\le l_2$ in Eq. (39), at most one of them is active. Let us suppose that at least the constraint $r^*\le l_2$ is inactive without loss of generality. Then, we have $\nu _2=0$ by using Eq. (52), while $\nu _1$ can be either zero or a positive value. Since $p_{\psi }=p_{\theta }=p_{\phi }=0$, $\nu _2=0$, $\mu _2^-=0$, and $\mu _2^+ = -c_{\gamma }p_0$, the costate differential equations for $p_{r}$ and $p_{\gamma }$ in Eqs. (44) and (47) can be rewritten as

$$\begin{aligned}&\dot{p}_r=-a_{11}p_r - a_{41}p_{\gamma } - \nu _1 \end{aligned}$$

(81)

$$\begin{aligned}&\dot{p}_{\gamma } =-a_{14}p_r - a_{44}p_{\gamma } - c_{\gamma }p_0 \end{aligned}$$

(82)

The value of the Hamiltonian H is a constant, i.e.,

$$\begin{aligned} H=p_r \dot{r} + p_{\gamma }\dot{\gamma }+p_0c_{\gamma }\eta + p_0c_u u_1 = c \end{aligned}$$

(83)

where c is a constant and the superscript ($^*$) is suppressed again for simplicity. Taking the derivative of H results in

$$\begin{aligned} \dot{H}=\dot{p}_r\dot{r}+p_r \ddot{r}+\dot{p}_{\gamma }\dot{\gamma }+p_{\gamma }\ddot{\gamma } +p_0c_{\gamma }\dot{\eta }+p_0 c_u \dot{u}_1=0 \end{aligned}$$

(84)

Now we have five Eqs. (64)–(65), (81)–(82) and (84), which will be used to prove $p_r=p_{\gamma }=p_0=\nu _1=0$. First, by making use of Eqs. (64)–(65) and (81)–(82), we can obtain both $p_r$ and $\nu _1$ as a function of $p_0$ as follows

$$\begin{aligned} p_r=k_rp_0,\quad \nu _1=k_{\nu }p_0 \end{aligned}$$

(85)

where $k_r=(-\dot{k}_{\gamma }-a_{44}k_{\gamma }-c_{\gamma } )/a_{14}$ and $k_{\nu }=-\dot{k}_r-a_{11}k_r-a_{41}k_{\gamma }$ with $\dot{k}_r:=\partial k_r/\partial e$. Then, substituting Eqs. (64)–(65) and (85) into Eq. (84) yields

$$\begin{aligned} (\dot{k}_r\dot{r} + k_r\ddot{r} + \dot{k}_{\gamma }\dot{\gamma }+k_{\gamma }\ddot{\gamma } + c_{\gamma }\dot{\eta } + c_u\dot{u}_1 )p_0 = 0 \end{aligned}$$

(86)

Similarly, the above equation implies $p_0=0$ because its coefficient is time-varying and not a constant zero. With $p_0=0$, we can directly get $p_{\gamma }=p_r=\nu _1=0$ by using Eqs. (64) and (85) and get $\mu _2^+=0$ by using Eq. (80).

• (2) Either $u_1^*\ge \varsigma _l$ or $u_1^*\le \varsigma _h$ is active. Without loss of generality, suppose that $u_1^*\ge \varsigma _l$ is active. Then, we have $\nu _1=\nu _2=0$ based on Assumption 3.3 and the condition in Eq. (52). In addition, since the constraint $u_1^*\le \varsigma _h$ is inactive, we have $\mu _1^+=0$ by using Eq. (50). The condition $\partial _{u_1}L=0$ becomes

  $$\begin{aligned} \mu _1^- = -b_{41}p_{\gamma } - c_up_0 \end{aligned}$$

  (87)

  The above equation indicates that we no longer have a specific relation between $p_{\gamma }$ and $p_0$ as in Case 1 and the first scenario of this case. Nevertheless, we will still be able to show $p_r=p_{\gamma }=p_0=\mu _1^-=\mu _2^+=0$ in the following by making use of several equations.

First, the costate differential equations for $p_r$ and $p_{\gamma }$ in Eqs. (44) and (47) become

$$\begin{aligned}&\dot{p}_r=-a_{11}p_r - a_{41}p_{\gamma } \end{aligned}$$

(88)

$$\begin{aligned}&\dot{p}_{\gamma } =-a_{14}p_r - a_{44}p_{\gamma } - c_{\gamma }p_0 \end{aligned}$$

(89)

which are obtained by using $p_{\psi }=p_{\theta }=p_{\phi }=\mu _2^-=0$ and Eq. (80). As before, H is a constant along the optimal solution, i.e.,

$$\begin{aligned} H=p_r \dot{r} + p_{\gamma }\dot{\gamma }+p_0c_{\gamma }\eta + p_0c_u u_1 = c \end{aligned}$$

(90)

where c is a constant and the superscript ($^*$) is suppressed for simplicity. Taking the derivative of H results in

$$\begin{aligned} \dot{H}=\dot{p}_r\dot{r}+p_r \ddot{r}+\dot{p}_{\gamma }\dot{\gamma }+p_{\gamma }\ddot{\gamma } +p_0c_{\gamma }\dot{\eta }=0 \end{aligned}$$

(91)

where $\dot{u}_1=\dot{\varsigma }_l=0$ is used. Next, Eq. (90) can be used to get

$$\begin{aligned} p_r=-(\dot{\gamma }/\dot{r})p_{\gamma }-[(c_{\gamma }\eta +c_u u_1)/\dot{r}]p_0+c/\dot{r} \end{aligned}$$

(92)

which, together with Eqs. (88)–(89), are then substituted into Eq. (91) to get

$$\begin{aligned} p_{\gamma }=k_{\gamma }p_0 + \alpha _{\gamma }c \end{aligned}$$

(93)

where $k_{\gamma }=[c_{\gamma }\dot{\gamma }-c_{\gamma }\dot{\eta }-(c_{\gamma }\eta +c_u u_1)(a_{11}\dot{r}+a_{14}\dot{\gamma }-\ddot{r} )/\dot{r} ] /\beta$ and $\alpha _{\gamma }=(a_{11}\dot{r}+a_{14}\dot{\gamma }-\ddot{r} )/(\dot{r}\beta )$ with $\beta =(a_{11}\dot{r}+a_{14}\dot{\gamma }-\ddot{r} )\dot{\gamma }/\dot{r}-a_{41}\dot{r}-a_{44}\dot{\gamma }+\ddot{\gamma }$. When the above equation is substituted into Eq. (92), we have

$$\begin{aligned} p_r=k_rp_0 + \alpha _r c \end{aligned}$$

(94)

where $k_r=-(\dot{\gamma }/\dot{r})k_{\gamma }-(c_{\gamma }\eta +c_u u_1)/\dot{r}$ and $\alpha _r=-(\dot{\gamma }/\dot{r})\alpha _{\gamma }+1/\dot{r}$. Now we have got both $p_r$ and $p_{\gamma }$ as a function of $p_0$ and c as given in Eqs. (93)–(94). Next, they are substituted into Eq. (88) to get a relation between $p_0$ and c as follows

$$\begin{aligned} (\dot{k}_r + a_{11}k_r + a_{41}k_{\gamma } )p_0 = (-\dot{\alpha }_r - a_{11}\alpha _r - a_{41}\alpha _{\gamma } )c \end{aligned}$$

(95)

In the above equation, $p_0$ has a varying coefficient. When it is not zero, the above equation can be rewritten as

$$\begin{aligned} p_0 = k_p c \end{aligned}$$

(96)

where $k_p=(-\dot{\alpha }_r - a_{11}\alpha _r - a_{41}\alpha _{\gamma } )/(\dot{k}_r + a_{11}k_r + a_{41}k_{\gamma } )$. Note that the value of $k_p$ is also time-varying. If c is a nonzero constant, $k_p c$ must not be a constant, which will violate Eq. (96) since $p_0$ is a constant. Therefore, c must be zero, and as a result $p_0=0$. When $p_0=c=0$, it becomes straightforward to obtain $p_r=p_{\gamma }=0$ by using Eqs. (93) and (94) and $\mu _2^+=0$ by using Eq. (80). Substituting $p_{\gamma }=p_0=0$ into Eq. (87), we have $\mu _1^-=0$.

In this case, first we have proven $p_{\psi }=p_{\theta }=p_{\phi }=\mu _2^-=0$ and then have proven $p_r=p_{\gamma }=p_0=\mu _1^-=\mu _1^+=\mu _2^+=\nu _1=\nu _2=0$ in the above two scenarios, which obviously violate the condition in Eq. (41). $\square$