Strong Local Optimality for a Bang–Bang–Singular Extremal: General Constraints

Abstract

The paper provides second-order sufficient conditions for the strong local optimality of bang–bang–singular extremals in a Mayer problem with general end point constraints. The sufficient conditions are expressed as a strengthening of the necessary ones plus the coerciveness of a suitable quadratic form related to a sub-problem of the given one. The sufficiency of the given conditions is proven via Hamiltonian methods.

Appendices

Appendices

Computation of the Extended Second Variation

Problem (15) can be written as

$$\begin{aligned}&\text {minimize } \varDelta C(y, \delta \upsilon _0, \delta \upsilon ) := \left( \widehat{\alpha }(y) + \beta (\eta (T))\right) - \left( \widehat{\alpha }({\widehat{x}}_T) + \beta ({\widehat{x}}_T)\right) \ \text { subject to} \nonumber \\&{\dot{\eta }}(t) = \varphi _t(\eta (t)) := {\left\{ \begin{array}{ll} \delta \upsilon _0(t) k_1(\eta (t)), \qquad &{} t \in ]0, {\widehat{\tau }}_{1}[ , \\ \delta \upsilon _0(t) k_2(\eta (t)), \quad &{} t \in ]{\widehat{\tau }}_{1}, {\widehat{\tau }}_{2}[ , \\ \delta \upsilon (t) g_{t}(\eta (t)), \quad &{} t \in ]{\widehat{\tau }}_{2}, T[ , \end{array}\right. } \nonumber \\&\delta \upsilon _0(t) > -1, \ \int _0^{{\widehat{\tau }}_{2}} \delta \upsilon _0(t) {\text {d}}t = 0, \;\; \left| \delta \upsilon (t)\right| < \delta , \quad \eta (0) = y \in \widehat{N}_0 , \;\; \eta (T) = x \in N_T. \end{aligned}$$

(41)

We can allow for the controls $\delta \upsilon _0$, $\delta \upsilon$ to be in $L^2 := L^2([0, T], {\mathbb {R}})$ since equation (41) is linear with respect to the controls. Defining

$$\begin{aligned}&\gamma :y \in {{\mathbb {R}}}^n\mapsto \left( \widehat{\alpha }+ \beta \right) (y) - \left( \widehat{\alpha }+ \beta \right) ({\widehat{x}}_T) \in {\mathbb {R}}, \\&{\mathcal {L}}:(t, y, \delta \upsilon _0, \delta \upsilon ) \in [0,T ]\times {{\mathbb {R}}}^n\times L^2 \times L^2 \mapsto {\mathcal {L}}_t( y, \delta \upsilon _0, \delta \upsilon ) := \langle {{\text {d}}\beta (y)} \, , \, {\varphi _t(y)} \rangle \in {\mathbb {R}}\end{aligned}$$

we get

$$\begin{aligned} \varDelta C(y, \delta \upsilon _0, \delta \upsilon ) := \gamma (y) + \int _0^T {\mathcal {L}}_t(\eta (t), \delta \upsilon _0(t), \delta \upsilon (t)){\text {d}}t . \end{aligned}$$

We aim at computing the second-order approximation $C^{\prime \prime }$ of $\varDelta C$.

By the properties of $\gamma$, the constraint on $\delta \upsilon _0$, and PMP it is not difficult to see that

Thus, the first-order approximation is null and the second-order approximation is intrinsically well defined. Obviously,

$$\begin{aligned} \partial ^2_{yy} C({\widehat{x}}_T, 0,0) = {\text {D}}^2\gamma ({\widehat{x}}_T). \end{aligned}$$

Denote as ${\mathcal {L}}^{\prime \prime }_t$ the second-order derivative of ${\mathcal {L}}_t$ at $({\widehat{x}}_T, 0,0)$ and let $\delta \eta$ be the linearization of $\eta$, i.e. $\delta \eta$ solves the problem

$$\begin{aligned} \dot{\delta \eta }(t) = \varphi _t({\widehat{x}}_T), \quad \delta \eta (0) = \delta y\in T_{{\widehat{x}}_T}\widehat{N}_0, \quad \delta \eta (T) = {\delta x}\in T_{{\widehat{x}}_T} N_T. \end{aligned}$$

(42)

Hence

$$\begin{aligned} 2 \, C^{\prime \prime }[\delta y, \delta \upsilon _0, \delta \upsilon ]^2 = {\text {D}}^2\gamma ({\widehat{x}}_T)[\delta y]^2 + \int _0^T {\mathcal {L}}^{\prime \prime }_t[\delta \eta (t), \delta \upsilon _0(t), \delta \upsilon (t)]^2 {\text {d}}t . \end{aligned}$$

For the sake of computations, let us define

$$\begin{aligned} I_1(\delta y, \delta \upsilon _0)&:= \int _0^{{\widehat{\tau }}_{1}}{\mathcal {L}}^{\prime \prime }_t[\delta \eta (t), \delta \upsilon _0(t),0]^2{\text {d}}t = 2\int _0^{{\widehat{\tau }}_{1}} \delta \upsilon _0(t) L_{\delta \eta (t)}L_{k_1}{\beta }\left( {{\widehat{x}}_T}\right) {\text {d}}t, \\ I_2(\delta y, \delta \upsilon _0)&:= \int _{{\widehat{\tau }}_{1}}^{{\widehat{\tau }}_{2}} {\mathcal {L}}^{\prime \prime }_t[\delta \eta (t), \delta \upsilon _0(t),0]^2{\text {d}}t = 2\int _{{\widehat{\tau }}_{1}}^{{\widehat{\tau }}_{2}} \delta \upsilon _0(t) L_{\delta \eta (t)}L_{k_2}{\beta }\left( {{\widehat{x}}_T}\right) {\text {d}}t, \\ I_3(\delta y, \delta \upsilon )&:= \int _{{\widehat{\tau }}_{2}}^T {\mathcal {L}}^{\prime \prime }_t[\delta \eta (t), 0, \delta \upsilon (t)]^2{\text {d}}t = 2\int _{{\widehat{\tau }}_{2}}^T \delta \upsilon (t) L_{\delta \eta (t)}L_{g_t}{\beta }\left( {{\widehat{x}}_T}\right) {\text {d}}t . \end{aligned}$$

Let $\varepsilon _0 := \displaystyle \int _0^{{\widehat{\tau }}_{1}}\delta \upsilon _0(s) {\text {d}}s$. Then

$$\begin{aligned}&\delta \eta (t) = \delta y+ \int _0^t \delta \upsilon _0(s) {\text {d}}s\, k_1({\widehat{x}}_T), \qquad&t \in [0, {\widehat{\tau }}_{1}], \\&\delta \eta (t) = \delta y+ \varepsilon _0 k_1({\widehat{x}}_T) + \int _{{\widehat{\tau }}_{1}}^t \delta \upsilon _0(s) {\text {d}}s\, k_2({\widehat{x}}_T), \qquad&t \in [{\widehat{\tau }}_{1}, {\widehat{\tau }}_{2}], \\&\delta \eta (t) = \delta y+ \varepsilon _0 k({\widehat{x}}_T) + \int _{{\widehat{\tau }}_{2}}^t \delta \upsilon (s)g_s({\widehat{x}}_T) {\text {d}}s\, ,&t \in [{\widehat{\tau }}_{2}, T]. \end{aligned}$$

In particular

$$\begin{aligned} I_1(\delta y, \delta \upsilon _0)&= 2\varepsilon _0 L_{\delta y}L_{k_1}{\beta }\left( {{\widehat{x}}_T}\right) {\text {d}}t + \varepsilon _0^2 L^2_{k_1}{\beta } \left( {{\widehat{x}}_T}\right) , \\ I_2(\delta y, \delta \upsilon _0)&= - 2 \varepsilon _0 L_{\delta y+ \varepsilon _0 k_1}L_{k_2}{\beta }\left( {{\widehat{x}}_T}\right) + \varepsilon _0^2 L^2_{k_2}{\beta } \left( {{\widehat{x}}_T}\right) . \end{aligned}$$

Define $w(t) := \displaystyle \int _{{\widehat{\tau }}_{2}}^t - \delta \upsilon (s) {\text {d}}s$, $\varepsilon _1 := w(T)$ and let $\zeta :[{\widehat{\tau }}_{2}, T] \rightarrow {{\mathbb {R}}}^n$ solve the Cauchy problem

$$\begin{aligned} {\dot{\zeta }}(t) = w(t)\dot{g}_t({\widehat{x}}_T), \qquad \zeta ({\widehat{\tau }}_{2}) = \delta \eta ({\widehat{\tau }}_{2}). \end{aligned}$$

By (42), $\zeta (T) = {\delta x}+ \varepsilon _1 f_{\mathrm{d}}({\widehat{x}}_T)$. Moreover, applying an intrinsic version of Goh transformation as in [9] we obtain

$$\begin{aligned} I_3 (\delta y, \delta \upsilon )= & {} L_{ \delta y+ \varepsilon _0 k }\int _{{\widehat{\tau }}_{2}}^T -\dot{w}(t) L_{g_t}{\beta } \left( {{\widehat{x}}_T}\right) {\text {d}}t\\&+ \int _{{\widehat{\tau }}_{2}}^T \dot{w}(t) \int _{{\widehat{\tau }}_{2}}^t \dot{w}(s) L_{g_s }L_{g_t}{\beta }\left( {{\widehat{x}}_T}\right) {\text {d}}s{\text {d}}t \\= & {} - \varepsilon _1^2 L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) - 2\varepsilon _1 L_{{\delta x}}L_{f_{\mathrm{d}}}{\beta }\left( {{\widehat{x}}_T}\right) \\&+ \int _{{\widehat{\tau }}_{2}}^T \left( 2\, w(t) L_{\zeta (t)}L_{\dot{g}_{t}}{\beta }\left( {{\widehat{x}}_T}\right) + w(t)^2 R(t) \right) \, {\text {d}}t . \end{aligned}$$

Thus

$$\begin{aligned}&2 C^{\prime \prime }[\delta y, \delta \upsilon _0, \delta \upsilon ]^2 \\&\quad = {\text {D}}^2\gamma ({\widehat{x}}_T)[\delta y]^2 + I_1 + I_2 + I_3 \\&\quad = {\text {D}}^2\gamma ({\widehat{x}}_T)[\delta y]^2 + 2\varepsilon _0 L_{\delta y}L_{k}{\beta }\left( {{\widehat{x}}_T}\right) {\text {d}}t + \varepsilon _0^2\left( L^2_{k}{\beta } \left( {{\widehat{x}}_T}\right) + L_{ \left[ {k_2}, {k_1} \right] }{\beta } \left( {{\widehat{x}}_T}\right) \right) \\&\qquad - \varepsilon _1^2 L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) - 2\varepsilon _1 L_{{\delta x}}L_{f_{\mathrm{d}}}{\beta }\left( {{\widehat{x}}_T}\right) + \int _{{\widehat{\tau }}_{2}}^T \left( 2\, w(t) L_{\zeta (t)}L_{\dot{g}_{t}}{\beta }\left( {{\widehat{x}}_T}\right) + w(t)^2 R(t) \right) \, {\text {d}}t \end{aligned}$$

subject to

$$\begin{aligned} {\dot{\zeta }}(t) = w(t)\dot{g}_t({\widehat{x}}_T), \quad \zeta ({\widehat{\tau }}_{2}) = \delta y+ \varepsilon _0 k ({\widehat{x}}_T), \quad {\delta x}= \zeta (T) - \varepsilon _1 f_{\mathrm{d}}({\widehat{x}}_T) \in T_{{\widehat{x}}_T}N_T. \end{aligned}$$

Notice that $\delta \upsilon _0$ appears only through $\varepsilon _0$, while the immersion

$$\begin{aligned} \delta \upsilon \in L^2([{\widehat{\tau }}_{2}, T], {\mathbb {R}}) \mapsto \left( w(t), w(T) \right) \in L^2([{\widehat{\tau }}_{2}, T], {\mathbb {R}}) \times {\mathbb {R}}\end{aligned}$$

is continuous and dense. Thus we can extend $C^{\prime \prime }$ to variations ${\delta e}{:=} \left( {\delta x}, \delta y, \varepsilon _0, \varepsilon _1, w \right) \in {\mathcal {W}}_\mathrm{ext}$ as defined in Sect. 5, and the extension coincides with $J_\mathrm{ext}$.

Splitting of the Second Variation

Lemma B.1

Assume $f_{\mathrm{d}}({\widehat{x}}_T) \in T_{{\widehat{x}}_T}N_T$. Then the coerciveness of $J_{\mathrm{ext}}$ on ${\mathcal {W}}_{\mathrm{ext}}$ splits into $L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) > 0$ plus the coerciveness of J on ${\mathcal {W}}$.

Proof

We decompose ${\delta x}\in T_{{\widehat{x}}_T}N_T$ as ${\delta x}= \delta z+ rf_{\mathrm{d}}({\widehat{x}}_T)$, $\delta z\in T_{{\widehat{x}}_T}\widetilde{N}_T$, where $\widetilde{N}_T$ is the manifold defined in Sect. 5.1. We can compute

$$\begin{aligned} 2 \, J_\mathrm{ext}[{\delta e}]^2&= {\text {D}}^2({\widehat{\alpha }}+ \beta \pm \widetilde{\beta })({\widehat{x}}_T)[\delta y]^2 +\varepsilon _0^2 \left( L^2_{k}{(\beta \pm \widetilde{\beta })} \left( {{\widehat{x}}_T}\right) + H_{12}({{\widehat{\ell }}_1}) \right) \\&\quad + 2\varepsilon _0 L_{\delta y}L_{k}{(\beta \pm \widetilde{\beta })}\left( {{\widehat{x}}_T}\right) - \varepsilon _1^2 L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) - 2 \varepsilon _1 L_{\delta z+ rf_{\mathrm{d}}}L_{f_{\mathrm{d}}}{\beta }\left( {{\widehat{x}}_T}\right) \\&\quad + \int _{{\widehat{\tau }}_{2}}^T \left( 2\, w(t) L_{\zeta (t)}L_{\dot{g}_{t}}{(\beta \pm \widetilde{\beta })}\left( {{\widehat{x}}_T}\right) + w(t)^2 R(t) \right) \, {\text {d}}t \\&= \varGamma [\delta y]^2 + \varepsilon _0^2 J_0 + 2 \varepsilon _0 L_{\delta y}L_{k}{\widetilde{\beta }}\left( {{\widehat{x}}_T}\right) + {\text {D}}^2(\beta - \widetilde{\beta })({\widehat{x}}_T)[\delta y+ \varepsilon _0 k]^2 \\&\quad - \left( \varepsilon _1^2 + 2 \varepsilon _1 r \right) L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) + \left. {\text {D}}^2(\beta - \widetilde{\beta })({\widehat{x}}_T)[\zeta (t)]^2 \right| _{t = {\widehat{\tau }}_{2}}^{t= T} \\&\quad + \int _{{\widehat{\tau }}_{2}}^T \left( 2\, w(t) L_{\zeta (t)}L_{\dot{g}_{t}}{\widetilde{\beta }}\left( {{\widehat{x}}_T}\right) + w(t)^2 R(t) \right) \, {\text {d}}t \\&= 2 J[(\delta z+ (r + \varepsilon _1) f_{\mathrm{d}}({\widehat{x}}_T), \delta y, \varepsilon _0, w)]^2 \\&\quad +(r + \varepsilon _1) ^2 {\text {D}}^2(\beta - \widetilde{\beta })({\widehat{x}}_T)[f_{\mathrm{d}}({\widehat{x}}_T)]^2 - \left( \varepsilon _1^2 + 2 \varepsilon _1 r \right) L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) \\&= 2 J[(\delta z+ (r + \varepsilon _1) f_{\mathrm{d}}({\widehat{x}}_T), \delta y, \varepsilon _0, w)]^2 + r^2 L^2_{f_{\mathrm{d}}}{\beta } \left( {{\widehat{x}}_T}\right) . \end{aligned}$$

The above computation shows that the real variable r is decoupled and $\delta z+ (r + \varepsilon _1) f_{\mathrm{d}}({\widehat{x}}_T)$ is a generic vector ${\delta x}\in T_{{\widehat{x}}_T}N_T$. This proves the claim. $\square$