Boundary energy control of a system governed by the nonlinear Klein–Gordon equation

Abstract

The boundary energy control problem for the sine-Gordon and the nonlinear Klein–Gordon equations is posed. Two control laws solving this problem based on the speed-gradient method with smooth and nonsmooth goal functions are proposed. The control law obtained via a nonsmooth goal function is proved to steer the system to any required nonzero energy level in finite time. The results of the numerical evaluation of the proposed algorithm for an undamped nonlinear elastic string demonstrate attainability of the control goal for the cases of both decreasing and increasing systems’ energy and show high rate of vanishing of the control error.

Appendices

Appendix A: Proof of Theorem 1

For any $\varepsilon > 0$ introduce the Lyapunov-like function

$$\begin{aligned} V(t) = H(z(t)) + \varepsilon {{\mathrm{sign}}}(H(z(t)) - H^*) g(t), \end{aligned}$$

where ${{\mathrm{sign}}}(0) = 0$ and

$$\begin{aligned} g(t) = \int _0^1 x z_t z_x \, \mathrm{d}x. \end{aligned}$$

Applying the inequalities

$$\begin{aligned} \Big | \int _0^1 x z_t z_x \, \mathrm{d}x \Big | \le \int _0^1 |z_t z_x| \, \mathrm{d}x \le \frac{1}{2} \int _0^1 z_t^2 \, \mathrm{d}x + \frac{1}{2} \int _0^1 z_x^2 \, \mathrm{d}x \le \max \left\{ 1, \frac{1}{k} \right\} H(z(t)), \end{aligned}$$

one obtains that

$$\begin{aligned} (1 - \varepsilon k_0) H(z(t)) \le V(t) \le (1 + \varepsilon k_0) H(z(t)) \end{aligned}$$

(12)

for all $t \ge 0$ and $\varepsilon \in (0, 1 / k_0)$, where $k_0 = \max \{ 1, 1 / k \}$. Thus, in particular, the function V(t) is nonnegative for any $\varepsilon \in (0, 1 / k_0)$.

For any $t \ge 0$ one has

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} g(t) = \int _0^1 \big ( x z_{tt} z_x + x z_t z_{tx} \big ) \, \mathrm{d}x = \int _0^1 \big ( x z_t z_{tx} + x (k z_{xx} - \beta \sin z) z_x \big ) \, \mathrm{d}x. \end{aligned}$$

Note that

$$\begin{aligned} \int _0^1 x z_t z_{tx} \, \mathrm{d}x = \frac{1}{2} \int _0^1 x (z_t^2)_x \, \mathrm{d}x = \frac{1}{2} z_t(t, 1)^2 - \frac{1}{2} \int _0^1 z_t^2 \, \mathrm{d}x \end{aligned}$$

and

$$\begin{aligned} \int _0^1 x z_{xx} z_x \, \mathrm{d}x = \frac{1}{2} \int _0^1 x (z_x^2)_x \, \mathrm{d}x = \frac{1}{2} z_x(t, 1)^2 - \frac{1}{2} \int _0^1 z_x^2 \, \mathrm{d}x. \end{aligned}$$

Observe also that

$$\begin{aligned}&- \int _0^1 x \sin z z_x \, \mathrm{d}x = \int _0^1 x (\cos z)_x \, \mathrm{d}x = \cos z(t, 1) - \int _0^1 \cos z \, \mathrm{d}x \\&\quad \le \int _0^1 (1 - \cos z) \, \mathrm{d}x \le \frac{1}{2} \int _0^1 z^2 \, \mathrm{d}x \le \frac{2}{\pi ^2} \int _0^1 z_x^2 \, \mathrm{d}x, \end{aligned}$$

where the last inequality is a consequence of Poincaré’s (or Wirtinger’s) inequality (see [16]). Hence, for any $\sigma > 0$ one has

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} g(t) \le - \int _0^1 \left( \frac{z_t^2}{2} + k \frac{z_x^2}{2} \right) \, \mathrm{d}x + \frac{1}{2} z_t^2(t, 1) + \frac{k}{2} z_x^2(t, 1) \\&\quad + \beta \int _0^1 (1 - \cos z) \, \mathrm{d}x \le - \frac{1}{2} \int _0^1 z_t^2 \, \mathrm{d}x - \left( \frac{k}{2} - \frac{2 (1 + \sigma ) \beta }{\pi ^2} \right) \int _0^1 z_x^2 \, \mathrm{d}x\\&\quad - \sigma \beta \int _0^1 (1 - \cos z) \, \mathrm{d}x + \frac{1}{2} z_t^2(t, 1) + \frac{k}{2} z_x^2(t, 1). \end{aligned}$$

Since by our assumption $0 \le \beta < k \pi ^2 / 4$, there exists $\sigma > 0$ such that

$$\begin{aligned} C_0 := \min \left\{ \sigma , 1 - \frac{4(1 + \sigma ) \beta }{k \pi ^2} \right\} > 0. \end{aligned}$$

Consequently, one gets that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} g(t) \le - C_0 H(t) + \frac{1}{2} z_t^2(t, 1) + \frac{k}{2} z_x^2(t, 1) \end{aligned}$$

(13)

for any $t \ge 0$. Note also that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) = - \gamma k \psi (H(z(t)) - H^*) z_t^2(t, 1) + \varepsilon {{\mathrm{sign}}}(H(z(t)) - H^*) \frac{\mathrm{d}}{\mathrm{d}t} g(t). \end{aligned}$$

(14)

for any $t \ge 0$ such that $H(z(t)) \ne H^*$.

Suppose that $H(z(0)) > H^*$, and fix an arbitrary $\varDelta \in (0, H(z(0)) - H^*)$. Clearly, there exists $T_{\varDelta } \in [0, +\infty ]$ such that $H(z(t)) > H^* + \varDelta$ for any $t \in [0, T_{\varDelta })$, and $H(z(t)) \le H^* + \varDelta$ for any $t \in [T_{\varDelta }, +\infty )$ (see (7)). Our aim is to show that $T_{\varDelta }$ is finite.

Arguing by reductio ad absurdum, suppose that $T_{\varDelta } = + \infty$. Denote

$$\begin{aligned} \psi _{\varDelta } := \min \Big \{ \psi (s) \Bigm | s \in \big [ \varDelta , H(z(0)) - H^* \big ] \Big \} > 0 \end{aligned}$$

and

$$\begin{aligned} \varPsi _{\varDelta } := \max \Big \{ \psi (s) \Bigm | s \in \big [ \varDelta , H(z(0)) - H^* \big ] \Big \} > 0 \end{aligned}$$

Then for any $t \ge 0$ one has $\varPsi _{\varDelta } \ge \psi (H(z(t)) - H^*) \ge \psi _{\varDelta }$. Therefore, taking into account (14), (13) and (12) one obtains that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) \le - \gamma k \psi _{\varDelta } z_t^2(t, 1) - \frac{\varepsilon C_0}{1 + \varepsilon k_0} V(t) + \frac{\varepsilon }{2} z_t^2(t, 1) + \frac{k \varepsilon }{2} \gamma ^2 \varPsi _{\varDelta }^2 z_t^2(t, 1) \end{aligned}$$

for any $t \ge 0$ and $\varepsilon \in (0, 1 / k_0)$. Hence, for any sufficiently small $\varepsilon > 0$ one has

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) \le - C_{\varepsilon } V(t) \quad \forall t \ge 0, \end{aligned}$$

where $C_{\varepsilon } := \varepsilon C_0 / (1 + \varepsilon k_0) > 0$. Therefore,

$$\begin{aligned} V(t) \le V(0) e^{- C_{\varepsilon } t} \quad \forall t \ge 0, \end{aligned}$$

which implies that $V(t) \rightarrow 0$ as $t \rightarrow \infty$. Consequently, with the use of (12) one obtains that $H(t) \rightarrow 0$ as $t \rightarrow +\infty$, which contradicts the definition of $T_{\varDelta }$ and the fact that $H^* + \varDelta > 0$. Thus, $T_{\varDelta } < + \infty$ and $H(z(t)) < H^* + \varDelta$ for any $t > T_{\varDelta }$. Hence, $H(z(t)) \rightarrow H^*$ as $t \rightarrow + \infty$ due to the fact that $\varDelta > 0$ was chosen arbitrarily.

Suppose, now, that $H(z(0)) < H^*$ and $H(z(t)) \le H^* - \varDelta$ for any $t \in [0, + \infty )$, where $\varDelta \in (0, H^* - H(z(0)))$ is arbitrary. Then arguing in a similar way to the case $H(z(0)) > H^*$, one can show that for any sufficiently small $\varepsilon > 0$ the following inequality holds true

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) := \frac{\mathrm{d}}{\mathrm{d}t} H(z(t)) - \varepsilon \frac{\mathrm{d}}{\mathrm{d}t} g(t) \ge C_{\varepsilon } V(t) \quad \forall t \ge 0. \end{aligned}$$

Therefore,

$$\begin{aligned} V(t) \ge V(0) e^{C_{\varepsilon } t} \quad \forall t \ge 0, \end{aligned}$$

which implies that $V(t) \rightarrow + \infty$ as $t \rightarrow \infty$. Then applying (12) one obtains that $H(t) \rightarrow + \infty$, which contradicts the assumption that $H(z(t) \le H^* - \varDelta$ for any $t \in [0, + \infty )$. Hence, $H(z(t)) \rightarrow H^*$ as $t \rightarrow +\infty$ due to the fact that $\varDelta > 0$ is arbitrary. $\square$

Appendix B: Proof of Theorem 2

For any $\varepsilon > 0$ introduce the same Lyapunov-like function

$$\begin{aligned} V(t) = H(z(t)) + \varepsilon {{\mathrm{sign}}}(H(z(t)) - H^*) \int _0^1 x z_t z_x \, \mathrm{d}x, \end{aligned}$$

as in the smooth case. Suppose that $H(z(0)) > H^*$. Then from the fact that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} H(z(t)) {\left\{ \begin{array}{ll} \le 0, &{} \text {if} \quad H(z(t)) > H^*, \\ \ge 0, &{} \text {if} \quad H(z(t)) < H^* \end{array}\right. } \end{aligned}$$

(see (7)) it follows that there exists $T \in (0, + \infty ]$ such that $H(z(t)) > H^*$ for any $t \in [0,T)$ and $H(z(t)) = H^*$ for $t\in [T, +\infty )$. Our aim is to prove that T is finite.

Observe that for any $t \in [0, T)$ one has

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) = - \gamma k {{\mathrm{sign}}}(H(z(t)) - H^*) z_t^2(t, 1) + \varepsilon {{\mathrm{sign}}}(H(t) - H^*) \frac{\mathrm{d}}{\mathrm{d}t} \int _0^1 x z_t z_x \, \mathrm{d}x. \end{aligned}$$

Then applying inequalities (13) and (12) one obtains that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) \le - \gamma k z_t^2(t, 1) - \frac{\varepsilon C_0}{1 + \varepsilon k_0} V(t) + \frac{\varepsilon }{2} z_t^2(t, 1) + \frac{k \varepsilon }{2} \gamma ^2 z_t^2(t, 1) \end{aligned}$$

for any $t \in [0, T)$ and $\varepsilon \in (0, 1)$. Hence, for any sufficiently small $\varepsilon > 0$ one has

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) \le - C_{\varepsilon } V(t) \quad \forall t \in [0, T), \end{aligned}$$

where $C_{\varepsilon } = \varepsilon C_0 / (1 + \varepsilon k_0)$, which yields

$$\begin{aligned} V(t) \le V(0) e^{- C_{\varepsilon } t} \quad \forall t \in [0, T). \end{aligned}$$

Then applying (12) one obtains that

$$\begin{aligned} H(t) \le \frac{V(0)}{1 - \varepsilon k_0} e^{- C_{\varepsilon } t} \quad \forall t \in [0, T). \end{aligned}$$

Hence, with the use of the definition of T one obtains that $T < + \infty$ and $H(z(t)) \rightarrow H^*$ as $t \rightarrow T$ in the case $H^* > 0$.

Suppose, now, that $H(z(0)) < H^*$. Then applying inequalities (13) and (12) one obtains that

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} V(t) \ge \gamma k z_t^2(t, 1) + \frac{\varepsilon C_0}{1 + \varepsilon k_0} V(t) \\&\quad - \frac{\varepsilon }{2} z_t^2(t, 1) - \frac{\varepsilon }{2} \gamma ^2 z_t^2(t, 1) \end{aligned}$$

for any $t \ge 0$ such that $H(z(t)) < H^*$. Consequently, for any sufficiently small $\varepsilon$ and for any $t \ge 0$ such that $H(z(t)) < H^*$ one has

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t) \ge C_{\varepsilon } V(t), \end{aligned}$$

which implies that

$$\begin{aligned} V(t) \ge V(0) e^{C_{\varepsilon } t} \quad \forall t \ge 0 :H(z(t)) < H^*. \end{aligned}$$

Hence, with the use of (12) one obtains the desired result. $\square$

Appendix C: Proof of Theorem 3

Let us prove the theorem in the case of the smooth control law (5). The case of the nonsmooth control law (6) is proved in a similar way.

For any $\varepsilon > 0$ and $c > 0$ introduce the Lyapunov-like function

$$\begin{aligned} V(t) = H(z(t)) + \varepsilon {{\mathrm{sign}}}(H(z(t)) - H^*) g(t), \end{aligned}$$

where z is a solution of the problem (1)–(3), ${{\mathrm{sign}}}(0) = 0$, and

$$\begin{aligned} g(t) = \int _0^1 x z_t z_x \, \mathrm{d}x + c \int _0^1 z z_t \, \mathrm{d}x. \end{aligned}$$

Let us obtain some estimates of the function V(t).

Since $z(t, 0) = 0$ for any $t \ge 0$, for all $x \in [0, 1]$ one has

$$\begin{aligned} |z(t, x)| = \left| \int _0^x z_x \, \mathrm{d}x \right| \le \int _0^1 |z_x| \, \mathrm{d}x \le \sqrt{\int _0^1 z_x^2 \, \mathrm{d}x} \end{aligned}$$

(15)

and

$$\begin{aligned} \int _0^1 z^2 \, \mathrm{d}x \le \int _0^1 z_x^2 \, \mathrm{d}x. \end{aligned}$$

(16)

Therefore,

$$\begin{aligned} c \int _0^1 z z_t \, \mathrm{d}x\le & {} \frac{c}{2} \int _0^1 z^2 \, \mathrm{d}x + \frac{c}{2} \int _0^1 z_t^2 \, \mathrm{d}x \le \frac{c}{2} \int _0^1 \big ( z_t^2 + z_x^2 \big ) \, \mathrm{d}x \\\le & {} \max \left\{ c, \frac{c}{k} \right\} H(z(t)). \end{aligned}$$

Note also that

$$\begin{aligned} \Big | \int _0^1 x z_t z_x \, \mathrm{d}x \Big | \le \int _0^1 |z_t z_x| \, \mathrm{d}x \le \frac{1}{2} \int _0^1 z_t^2 \, \mathrm{d}x + \frac{1}{2} \int _0^1 z_x^2 \, \mathrm{d}x \le \max \left\{ 1, \frac{1}{k} \right\} H(z(t)). \end{aligned}$$

Define

$$\begin{aligned} K_0 = \max \left\{ 1, \frac{1}{k} \right\} + \max \left\{ c, \frac{c}{k} \right\} . \end{aligned}$$

Then

$$\begin{aligned} 0 \le (1 - \varepsilon K_0) H(z(t)) \le V(t) \le (1 + \varepsilon K_0) H(z(t)), \end{aligned}$$

(17)

for all $t \ge 0$ and for any sufficiently small $\varepsilon > 0$.

Taking into account the fact that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} H(z(t)) {\left\{ \begin{array}{ll} \le 0, &{} \text {if} \quad H(z(t)) > H^*, \\ \ge 0, &{} \text {if} \quad H(z(t)) < H^* \end{array}\right. } \end{aligned}$$

(see (7)) one obtains that

$$\begin{aligned} \int _0^1 z_x^2 \, \mathrm{d}x \le \frac{2}{k} \max \{ H^*, H(z(0)) \}. \end{aligned}$$

Hence, and from (15) it follows that the function z(t, x) is uniformly bounded for all $t \ge 0$ and $x \in [0, 1]$. Recall that $\varPi \in C^2(\mathbb {R})$ and $\varPi '(0) = 0$. Therefore, there exists $L > 0$ such that

$$\begin{aligned} | \varPi '(z(t, x)) | \le L |z(t, x)| \quad \forall t \ge 0 \quad \forall x \in [0, 1]. \end{aligned}$$

(18)

For any $t \ge 0$ such that $H(z(t)) \ne H^*$ one has

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} \int _0^1 x z_x z_t \, \mathrm{d}x = \int _0^1 x z_{tx} z_t \, \mathrm{d}x + \int _0^1 x z_x (k z_{xx} - \varPi '(z)) \, \mathrm{d}x \nonumber \\&\quad = - \frac{1}{2} \int _0^1 \big ( z_t^2 + k z_x^2 \big ) \, \mathrm{d}x + \frac{1}{2} \big ( z_t^2(t, 1) + k z_x^2(t, 1) \big )\nonumber \\&\qquad \,\, + \int _0^1 \left( - \frac{d}{\mathrm{d}x}\Big ( x \varPi (z) \Big ) + \varPi (z) \right) \, \mathrm{d}x \nonumber \\&\quad = - \frac{1}{2} \int _0^1 \big ( z_t^2 + k z_x^2 \big ) \, \mathrm{d}x + \frac{1}{2} \big ( z_t^2(t, 1) + k z_x^2(t, 1) \big ) + \int _0^1 \varPi (z) \, \mathrm{d}x - \varPi (z(t, 1)) \nonumber \\&\quad \le - \frac{1}{2} \int _0^1 \big ( z_t^2 + k z_x^2 \big ) \, \mathrm{d}x + \int _0^1 \varPi (z) \, \mathrm{d}x + \frac{1}{2} \big ( z_t^2(t, 1) + k z_x^2(t, 1) \big ), \end{aligned}$$

(19)

and

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} \int _0^1 z z_t \, \mathrm{d}x = \int _0^1 z_t^2 \, \mathrm{d}x + \int _0^1 z (k z_{xx} - \varPi '(z)) \, \mathrm{d}x \\&\quad = \int _0^1 z_t^2 \, \mathrm{d}x + k z(t, 1) z_x(t, 1) - k \int _0^1 z_x^2 \, \mathrm{d}x - \left( 1 + \frac{k}{2 L} \right) \int _0^1 z \varPi '(z) \, \mathrm{d}x \\&\qquad + \frac{k}{2 L} \int _0^1 z \varPi '(z) \, \mathrm{d}x \le \int _0^1 z_t^2 \, \mathrm{d}x + k z(t, 1) z_x(t, 1) - k \int _0^1 z_x^2 \, \mathrm{d}x \\&\qquad - \eta \left( 1 + \frac{k}{2 L} \right) \int _0^1 \varPi (z) \, \mathrm{d}x + \frac{k}{2 L} \int _0^1 z \varPi '(z) \, \mathrm{d}x. \end{aligned}$$

Observe that

$$\begin{aligned} | k z(t, 1) z_x(t, 1) | \le \frac{k}{2} z^2(t, 1) + \frac{k}{2} z_x^2(t, 1) \le \frac{k}{2} \int _0^1 z_x^2 \, \mathrm{d}x + \frac{k}{2} z_x^2(t, 1). \end{aligned}$$

(see (15)). Hence, one obtains that

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} \int _0^1 z z_t \, \mathrm{d}x \le \int _0^1 z_t^2 \, \mathrm{d}x - \frac{k}{2} \int _0^1 z_x^2 \, \mathrm{d}x \\&\quad - \eta \left( 1 + \frac{k}{2 L} \right) \int _0^1 \varPi (z) \, \mathrm{d}x + \frac{k}{2 L} \int _0^1 z \varPi '(z) \, \mathrm{d}x + \frac{k}{2} z_x^2(t, 1). \end{aligned}$$

Then applying (18) and (16) one gets that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \int _0^1 z z_t \, \mathrm{d}x \le \int _0^1 z_t^2 \, \mathrm{d}x - \eta \left( 1 + \frac{k}{2 L} \right) \int _0^1 \varPi (z) \, \mathrm{d}x + \frac{k}{2} z_x^2(t, 1). \end{aligned}$$

(20)

Recall that $\eta \ge 2$. Therefore, $c \eta ( 1 + k / 2 L ) > 1$ for some $c \in (0, 1/2)$. Hence, taking into account (19) and (20) one obtains that

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t} g(t) \le - \left( \frac{1}{2} - c \right) \int _0^1 z_t^2 \, \mathrm{d}x - \frac{k}{2} \int _0^1 z_x^2 \, \mathrm{d}x - \left( c \eta \Big ( 1 + \frac{k}{2 L} \Big ) - 1 \right) \int _0^1 \varPi (z) \, \mathrm{d}x \\&\quad + \frac{k}{2}(1 + c) z_x^2(t, 1) + \frac{1}{2} z_t^2(t, 1) \le - C_0 H(z(t)) + \frac{1}{2} z_t^2(t, 1) + \frac{k}{2}(1 + c) z_x^2(t, 1), \end{aligned}$$

for any $t \ge 0$ such that $H(z(t)) \ne H^*$, where

$$\begin{aligned} C_0 = \min \left\{ 1 - 2c, c \eta \Big ( 1 + \frac{k}{2 L} \Big ) - 1 \right\} > 0. \end{aligned}$$

Therefore, applying (17) one obtains that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t)\le & {} - C_{\varepsilon } V(t) - \gamma k \psi (H(z(t)) - H^*) z_t^2(t, 1) \\&+ \frac{\varepsilon }{2} z_t^2(t, 1) + \frac{\varepsilon }{2}(1 + c) k \gamma ^2 \psi (H(z(t)) - H^*)^2 z_t^2(t, 1) \end{aligned}$$

for any $t \ge 0$ such that $H(z(t)) > H^*$, and

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(t)\ge & {} C_{\varepsilon } V(t) + \gamma k |\psi (H(z(t)) - H^*)| z_t^2(t, 1) \\&- \frac{\varepsilon }{2} z_t^2(t, 1) - \frac{\varepsilon }{2}(1 + c) k \gamma ^2 \psi (H(z(t)) - H^*)^2 z_t^2(t, 1) \end{aligned}$$

for any $t \ge 0$ such that $H(z(t)) < H^*$, where $C_{\varepsilon } = \varepsilon C_0 / (1 + \varepsilon K_0)$. Then arguing in the same way as in the proof of Theorem 1, one can easily obtain the desired result. $\square$