Boundary stabilization of first-order hyperbolic equations with input delay

Abstract

This paper is concerned with the boundary stabilization problem of first-order hyperbolic equations with input delay. In our previous work, the same problem for parabolic equations with input delay was addressed. For the hyperbolic equation, the element of time lag is similarly replaced by a transport equation and a backstepping method is employed. However, unlike in the case of the parabolic equation, it is difficult to obtain the eigenvalues and eigenfunctions of the system operator for the hyperbolic equation. Hence, we cannot construct the target system and the controller by using the finite-dimensional control theory together. In this paper, using $C_0$-semigroups for hyperbolic equations with nonlocal boundary condition, we show that the proposed controller is expressed by an abstract form in a Hilbert space so-called a predictor, and that the predictor makes sense under a condition. Further, for the inverse integral transformation, we obtain an interesting result on its continuity under the same condition. A numerical algorithm is also proposed for solving a non-standard hyperbolic equation appearing in our controller design.

Appendices

Well-posedness of (6), (7)

Set $k(x,y)=e^{\int _y^x \gamma (\xi )d\xi }\overline{k}(x,y)$. Then, $\overline{k}$ satisfies

$$\begin{aligned}&\overline{k}_x(x,y)+\overline{k}_y(x,y)=\int _y^x \overline{k}(x,z)\overline{h}(z,y)-\overline{h}(x,y), \end{aligned}$$

(89)

$$\begin{aligned}&\overline{k}(x,0)=\int _0^x \overline{k}(x,y)\overline{g}(y)dy-\overline{g}(x), \end{aligned}$$

(90)

where $\overline{h}(x,y):=e^{-\int _y^x \gamma (\xi )d\xi }h(x,y)$, $\overline{g}(x):=e^{-\int _0^x \gamma (\xi )d\xi }g(x)$. Therefore, it follows from [11, Theorem 1] that (89), (90), i.e. (6), (7) is well-posed.

Existence of the solution of (16)

Let $Q=\{(x,y)\in \mathbf{R}^2 ; 0\le y\le x\le L\}$. We consider the following integral equation instead of (16):

$$\begin{aligned} \int _0^x k(x,y)\varphi (y)dy+1=\varphi (x), \quad x\in [0,L], \end{aligned}$$

(91)

where $k\in C^1(Q)$ is a unique solution to (6), (7). By using the method of successive approximations, we see that (91) has a unique solution $\varphi \in C^1[0,L]$ with bound $|\varphi (x)|\le e^{Mx}$, where $M:=\max _{(x,y)\in Q}|k(x,y)|$. It is clear that the $\varphi$ gives the solution to (16).

Left invertibility of backstepping transformation

In Sect. 3, we introduced the target system (12) and the integral transformation (13). We rewrite them equivalently as follows:

$\bullet$ Target system

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _t(t,x)=\varphi _x(t,x)-\gamma (x)\varphi (t,x)+g(x)\varphi (t,0) \\ \quad \quad \quad \quad \quad {+\int _0^x h(x,y)\varphi (t,y)dy, \quad (t,x)\in (0,\infty )\times (0,L),} \\ \displaystyle {\varphi (t,L)=\int _0^L k(L,y)\varphi (t,y)dy+w(t,0), \quad t>0,} \\ \varphi (0,x)=u_0(x), \quad x\in [0,L], \\ w_t(t,x)=w_x(t,x), \quad (t,x)\in (0,\infty )\times (0,\tau ), \\ w(t,\tau )=0, \quad t>0, \\ w(0,x)=w_0(x), \quad x\in [0,\tau ]. \end{array}\right. \end{aligned}$$

(92)

$\bullet$ Integral transformation

$$\begin{aligned} \left\{ \begin{array}{l} \varphi (t,x)=u(t,x), \\ w(t,x)=\mathcal{P}v(t,x)+\mathcal{Q}u(t,x), \end{array}\right. \end{aligned}$$

(93)

where

$$\begin{aligned} \mathcal{P}v(t,x):=v(t,x)-\int _0^x q(x,y)v(t,y)dy, \quad \mathcal{Q}u(t,x):=-\int _0^L \beta (x,y)u(t,y)dy. \end{aligned}$$

Similarly, the inverse integral transformation (65) is rewritten as

$\bullet$ Inverse integral transformation

$$\begin{aligned} \left\{ \begin{array}{l} u(t,x)=\varphi (t,x), \\ v(t,x)=\mathcal{R}w(t,x)+\mathcal{S}\varphi (t,x), \end{array}\right. \end{aligned}$$

(94)

where

$$\begin{aligned} \mathcal{R}w(t,x):=w(t,x)+\int _0^x p(x,y)w(t,y)dy, \quad \mathcal{S}\varphi (t,x):=\int _0^L \alpha (x,y)\varphi (t,y)dy. \end{aligned}$$

Now, let us express the two transformations (93) and (94) as

$$\begin{aligned}&\left[ \begin{array}{c} \varphi (t,x) \\ w(t,x) \end{array}\right] =\mathcal{T}_\mathcal{P,Q}\left[ \begin{array}{c} u(t,x) \\ v(t,x) \end{array}\right] , \quad \mathcal{T}_\mathcal{P,Q}:=\left[ \begin{array}{cc} I &{} 0 \\ \mathcal{Q} &{} \mathcal{P} \end{array}\right] , \end{aligned}$$

(95)

$$\begin{aligned}&\left[ \begin{array}{c} u(t,x) \\ v(t,x) \end{array}\right] =\mathcal{T}_\mathcal{R,S}\left[ \begin{array}{c} \varphi (t,x) \\ w(t,x) \end{array}\right] , \quad \mathcal{T}_\mathcal{R,S}:=\left[ \begin{array}{cc} I &{} 0 \\ \mathcal{S} &{} \mathcal{R} \end{array}\right] . \end{aligned}$$

(96)

First, we calculate $\mathcal{T}_\mathcal{R,S}\mathcal{T}_\mathcal{P,Q}$ as follows:

$$\begin{aligned} \mathcal{T}_\mathcal{R,S}\mathcal{T}_\mathcal{P,Q} =\left[ \begin{array}{cc} I &{} 0 \\ \mathcal{S+RQ} &{} \mathcal{RP} \end{array}\right] , \end{aligned}$$

(97)

where

$$\begin{aligned} (\mathcal{S+RQ})u(t,x)= & {} \int _0^L \biggl \{\alpha (x,y)-\beta (x,y)-\int _0^x p(x,z)\beta (z,y)dz\biggr \}u(t,y)dy, \\ (\mathcal{RP})v(t,x)= & {} v(t,x)+\int _0^x \biggl \{p(x,y)-q(x,y)-\int _y^xp(x,z)q(z,y)dz\biggr \}v(t,y)dy. \end{aligned}$$

From this expression, if the kernels p, q, $\alpha$, and $\beta$ satisfy the equations

$$\begin{aligned}&\alpha (x,y)-\beta (x,y)=\int _0^x p(x,z)\beta (z,y)dz, \end{aligned}$$

(98)

$$\begin{aligned}&p(x,y)-q(x,y)=\int _y^x p(x,z)q(z,y)dz, \end{aligned}$$

(99)

it follows that

$$\begin{aligned} \mathcal{T}_\mathcal{R,S}\mathcal{T}_\mathcal{P,Q} =\left[ \begin{array}{cc} I &{} 0 \\ 0 &{} I \end{array}\right] . \end{aligned}$$

(100)

Indeed, we can verify that the kernels p, q, $\alpha$, and $\beta$ derived in Sect. 3 (see (60), (63), (78), and (81)) satisfy the above (98) and (99). It is done as follows: Introducing the variable $\varepsilon (x,y):=\alpha (x,y)-\beta (x,y)$, it follows from (23)–(25) and (66)–(68) that

$$\begin{aligned} \varepsilon _x(x,y)= & {} -\varepsilon _y(x,y)-\gamma (y)\varepsilon (x,y)+k(L,y)\alpha (x,L) \nonumber \\&+\int _y^L \varepsilon (x,z)h(z,y)dz, \end{aligned}$$

(101)

$$\begin{aligned} \varepsilon (x,0)= & {} \int _0^L \varepsilon (x,y)g(y)dy, \end{aligned}$$

(102)

$$\begin{aligned} \varepsilon (0,y)= & {} 0. \end{aligned}$$

(103)

By using the operators A and H defined by (28), (29), we can formulate (101)–(103) as

$$\begin{aligned} \varepsilon '(x,\cdot )=(A+H)\varepsilon (x,\cdot )+k(L,\cdot )\alpha (x,L), \quad \varepsilon (0,\cdot )=0, \end{aligned}$$

(104)

from which we have the solution

$$\begin{aligned} \varepsilon (x,\cdot )=\int _0^x T_{A+H}(x-z)k(L,\cdot )\alpha (z,L)dz. \end{aligned}$$

(105)

The solution (105) is expressed as

$$\begin{aligned} \varepsilon (x,y)= & {} \int _0^x (T_{A+H}(x-z)k(L,\cdot ))(y)\alpha (z,L)dz \nonumber \\= & {} \int _0^x \beta (x-z,y)\alpha (z,L)dz \nonumber \\= & {} \int _0^x \beta (z,y)\alpha (x-z,L)dz \nonumber \\= & {} \int _0^x \beta (z,y)T_{B+H}(x-z)k(L,\cdot )(L)dz \nonumber \\= & {} \int _0^x p(x,z)\beta (z,y)dz. \end{aligned}$$

(106)

This means that p, $\alpha$, and $\beta$ of Sect. 3 actually satisfy (98).

Next, introducing the variable $r(x,y):=p(x,y)-q(x,y)$, from (26), (27) and (69), (70), we have

$$\begin{aligned}&r_x(x,y)+r_y(x,y)=0, \end{aligned}$$

(107)

$$\begin{aligned}&r(x,0)=\varepsilon (x,L). \end{aligned}$$

(108)

It is easy to see that eqs. (107), (108) has the solution

$$\begin{aligned} r(x,y)=\varepsilon (x-y,L), \end{aligned}$$

(109)

which leads to

$$\begin{aligned} r(x,y)= & {} \int _0^{x-y} (T_{A+H}(x-y-z)k(L,\cdot ))(L)\alpha (z,L)dz \nonumber \\= & {} \int _0^{x-y} (T_{A+H}(x-y-z)k(L,\cdot ))(L)(T_{B+H}(z)k(L,\cdot ))(L)dz \nonumber \\= & {} \int _0^{x-y} q(x-z,y)p(x,x-z)dz \nonumber \\= & {} \int _y^x p(x,z)q(z,y)dz. \end{aligned}$$

(110)

Therefore, we see that p and q of Sect. 3 also satisfy (99). In this way, the two transformations (95) and (96) with the kernels (60), (63), (78), and (81) satisfy (100). However, as for $\mathcal{T}_\mathcal{P,Q}\mathcal{T}_\mathcal{R,S} =\left[ \begin{array}{cc} I &{} 0 \\ \mathcal{Q+PS} &{} \mathcal{PR} \end{array}\right]$, we can verify that they do not satisfy the relation

$$\begin{aligned} \mathcal{T}_\mathcal{P,Q}\mathcal{T}_\mathcal{R,S} =\left[ \begin{array}{cc} I &{} 0 \\ 0 &{} I \end{array}\right] , \end{aligned}$$

(111)

through a discussion similar to the above. In fact, $\mathcal{PR}=I$ is satisfied, but $\mathcal{Q+PS}=0$ is not. Hence, the backstepping transformation (95) with the kernels (60) and (63) is left invertible, but not right invertible.