A Convex Programming Approach to Mid-course Trajectory Optimization for Air-to-Ground Missiles

Abstract

This paper addresses trajectory optimization in the mid-course phase of an air-to-ground missile, when the main objectives are (a) to ensure that the target is locked on in the center of the missile’s field-of-view at a specified flight path angle and (b) to attain maximum possible speed to allow for sufficient maneuverability in the terminal phase. The method presents as a second-order cone program (SOCP) formulation for this trajectory optimization, taking advantage of partial linearization and lossless convexification techniques that effectively handle underlying non-convex characteristics of the problem. A well-established SOCP solver can then be readily used to obtain the optimal solution to this convex program. The proposed approach is validated by (a) proving the losslessness of the convexification, and (b) numerically comparing the results with an existing pseudo-spectral method.

Appendix: Proof of Lossless Convexification

The proof of lossless convexification is carried out using the proof by contradiction. Let us assume that there exists a certain interval $\left[ {x_{1} ,x_{2} } \right] \subset \left[ {x_{0} ,x_{\text{f}} } \right]$ that satisfies $u_{1}^{2} < u_{2}$. There should exist a constant $p_{0} \le 0$ for an optimal solution $\left( {q^{*} ,u^{*} } \right)$ satisfying the conditions in Eqs. (19)–(33) in Theorem 1.

• (a) In the interval $\left[ {x_{1} ,x_{2} } \right] \subset \left[ {x_{0} ,x_{\text{f}} } \right]$, the following relationship is obtained:

  $$ u_{1}^{2} < u_{2} ,\;u_{2} > 0. $$

  (55)

  The values of $\lambda_{1} ,\lambda_{2}$ can be determined from complementary slackness conditions (42) and (43):

  $$ \lambda_{0} = \lambda_{1} = 0. $$

  (56)

• (b) From the stationary condition (40), the following equation is derived:

  $$ \begin{aligned} \frac{\partial L}{{\partial u_{1} }} = p_{\gamma } b_{31} = p_{\gamma } \frac{{\left( {T + qS_{\text{ref}} C_{{{\text{L}}_{\alpha } }} } \right)}}{{mV^{2} \cos \gamma }} = 0, \hfill \\ \to p_{\gamma } = 0\;\quad \because T \ge 0,\;q > 0,\;S_{\text{ref}} > 0,\;C_{{{\text{L}}_{\alpha } }} > 0. \hfill \\ \end{aligned} $$

  (57)

• (c) From the stationary condition (41), the following equation is derived:

  $$ \begin{aligned} \frac{\partial L}{{\partial u_{2} }} = p_{V} b_{22} + \lambda_{2} = 0, \hfill \\ \to p_{V} b_{22} = - \lambda_{2} \ge 0. \hfill \\ \end{aligned} $$

  (58)

  Hamiltonian in Eq. (34) can be re-formulated as follows:

  $$ H = p_{y} \left( {a_{13} \gamma + c_{y} } \right) + p_{V} \left( {a_{22} V + a_{23} \gamma + c_{V} } \right) + p_{\gamma } \left( {a_{32} V + c_{\gamma } } \right) + p_{\gamma } b_{31} u_{1} + p_{V} b_{22} u_{2} . $$

  (59)

  Based on the pointwise maximum condition (20), $u_{2}$ is determined according to the switching function $p_{V} b_{22}$:

  $$ {\text{if}}\; p_{V} b_{22} = 0 \to u_{2} \in \left[ {0,u_{2\rm{max} } } \right], $$

  (60)
  $$ {\text{if }} p_{V} b_{22} > 0 \to u_{2} = 0. $$

  (61)

  Since $u_{2}$ should be positive as in Eq. (55), $p_{\text{V}}$ is determined as follows:

  $$ p_{V} = 0\quad \;\because b_{22} < 0. $$

  (62)

• (d) Since $p_{\gamma } = p_{V} = 0$ from (57) and (62), the differential equations (37)–(39) are described as follows:

  $$ p_{y}^{'} = - \frac{\partial L}{\partial y} = \nu_{y1} - \nu_{y2} , $$

  (63)
  $$ p_{V}^{'} = - \frac{\partial L}{\partial V} = - \left( {p_{V} a_{22} + p_{\gamma } a_{32} } \right) = 0, $$

  (64)
  $$ p_{\gamma }^{'} = - \frac{\partial L}{\partial \gamma } = - \left( {p_{y} a_{13} + p_{V} a_{23} } \right) = - p_{y} a_{13} = 0. $$

  (65)

  Since the downrange is monotonically increasing and $a_{13}$ cannot be 0, then

  $$ p_{y} = 0. $$

  (66)

  Therefore, we obtain $\nu_{y1} = \nu_{y2}$ from Eq. (63). If the altitude is larger than 0, then $\nu_{y1} = 0$ from Eq. (45) and if the altitude is less than $y_{\hbox{max} }$, then $\nu_{y2} = 0$ from Eq. (46). Thus, the multipliers $\nu_{y1}$ and $\nu_{y2}$ should be 0 regardless of the altitude condition:

  $$ \nu_{y1} = \nu_{y2} = 0. $$

  (67)

• (e) From Assumption 2, Eqs. (47), (48), (49) give

  $$ \zeta_{y1} = \zeta_{y2} = p_{y} \left( {x_{\text{f}} } \right) = 0. $$

  (68)

• (f) The altitude constraint is described as follows:

  $$ S\left( q \right) = y - y_{\rm{max} } \le 0. $$

  The constraint $S\left( q \right)$ is second-order constraint with respect to the control and then the derivatives of the constraint is calculated

  $$ \begin{aligned} \frac{S\left( q \right)}{\partial q} = \left( {\begin{array}{*{20}c} 1 \\ 0 \\ 0 \\ \end{array} } \right), \;\frac{{S^{\prime}\left( q \right)}}{\partial q} = \left( {\begin{array}{*{20}c} 0 \\ 0 \\ {\sec^{2} \gamma \left( {1 + 2\gamma \tan \gamma } \right)} \\ \end{array} } \right) , \hfill \\ {\text{where }}S^{\prime}\left( q \right) = y^{\prime} = \gamma \sec^{2} \gamma . \hfill \\ \end{aligned} $$

  (69)

  The jump condition (29) for the costate is

  $$ \left( {\begin{array}{*{20}c} {p_{y} \left( {\tau_{1}^{ - } } \right)} \\ {p_{V} \left( {\tau_{1}^{ - } } \right)} \\ {p_{\gamma } \left( {\tau_{1}^{ - } } \right)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {p_{y} \left( {\tau_{1}^{ + } } \right)} \\ {p_{V} \left( {\tau_{1}^{ + } } \right)} \\ {p_{\gamma } \left( {\tau_{1}^{ + } } \right)} \\ \end{array} } \right) + \eta_{1} \left( {\tau_{1} } \right)\left( {\begin{array}{*{20}c} 1 \\ 0 \\ 0 \\ \end{array} } \right) + \eta_{2} \left( {\tau_{1} } \right)\left( {\begin{array}{*{20}c} 0 \\ 0 \\ {\sec^{2} \gamma \left( {1 + 2\gamma \tan \gamma } \right)} \\ \end{array} } \right). $$

  (70)

  If the altitude constraint is active, the flight path angle should be 0. Therefore,

  $$ \left( {\begin{array}{*{20}c} {p_{y} \left( {\tau_{1}^{ - } } \right)} \\ {p_{V} \left( {\tau_{1}^{ - } } \right)} \\ {p_{\gamma } \left( {\tau_{1}^{ - } } \right)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {p_{y} \left( {\tau_{1}^{ + } } \right) + \eta_{1} \left( {\tau_{1} } \right)} \\ {p_{V} \left( {\tau_{1}^{ + } } \right)} \\ {p_{\gamma } \left( {\tau_{1}^{ + } } \right) + \eta_{2} \left( {\tau_{1} } \right)} \\ \end{array} } \right). $$

  (71)

  Since $p_{\gamma } ,p_{V} ,p_{y}$ are 0 from (57), (62), (66), then $\eta_{1} \left( {\tau_{1} } \right),\eta_{2} \left( {\tau_{1} } \right)$ should be 0:

  $$ \eta_{1} \left( {\tau_{1} } \right) = \eta_{2} \left( {\tau_{1} } \right) = 0. $$

  (72)

• (g) Because the Hamiltonian, endpoint, and state constraint functions are autonomous, then

  $$ H\left( t \right) = 0 \forall t. $$

  (73)

From Eqs. (51) and (68), $p_{\gamma } \left( {x_{\text{f}} } \right)$ and $p_{y} \left( {x_{\text{f}} } \right)$ are 0 and then:

$$ p_{V} \left( {x_{\text{f}} } \right) = - p_{0} = 0. $$

(74)

From the above results, we conclude

$$ \left( {p_{0} ,p\left( t \right),\lambda \left( t \right),\nu \left( t \right),\xi ,\zeta ,\eta_{1} ,\eta_{2} , \ldots } \right) = 0. $$

(75)

This contradicts the non-triviality condition (19). Therefore, the following equation holds:

$$ u_{1}^{2} \left( x \right) = u_{2} \left( x \right)\;{\text{a}} . {\text{e}} . x \in \left[ {x_{0} ,x_{\text{f}} } \right]. $$

(76)