Appendix A Backstepping control—proof
System dynamics of the arc-shaped SMA actuator is defined in Eqs. (17), (18) and (20). Based on this system dynamics, we can define
$$\begin{aligned}&e_1 = x_1 - x_{1_d} \end{aligned}$$
(36)
$$\begin{aligned}&\dot{e}_1 = x_2 - \dot{x}_{1_d} \end{aligned}$$
(37)
where \(x_{1_d}\) is the desired trajectory. To achieve convergence for \(e_1\), we can define Lyapunov function as follows
$$\begin{aligned}&V_1 = \frac{1}{2}e_1^2 \end{aligned}$$
(38)
$$\begin{aligned}&\dot{V}_1 =e_1\dot{e}_1 \end{aligned}$$
(39)
$$\begin{aligned}&= e_1[x_2- \dot{x}_{1_d}] \end{aligned}$$
(40)
To realize \(\dot{V}_1<0\), we choose virtual control as
$$\begin{aligned} x_{2_d} = \dot{x}_{1_d}-k_1e_1. \end{aligned}$$
(41)
To achieve \(x_2\rightarrow x_{2_d}\), we define new error
$$\begin{aligned}&e_2 = x_2 - x_{2_d} \end{aligned}$$
(42)
so,
$$\begin{aligned}&x_2 = e_2+x_{2_d}. \end{aligned}$$
(43)
Solving Eqs. (40), (41) and (43) simultaneously, we get
$$\begin{aligned}&\dot{V}_1 = e_1[e_2+x_{2_d}- \dot{x}_{1_d}] \end{aligned}$$
(44)
$$\begin{aligned}&\dot{V}_1 = e_1[e_2+\dot{x}_{1_d}-k_1e_1- \dot{x}_{1_d}] \end{aligned}$$
(45)
$$\begin{aligned}&\dot{V}_1 = -k_1e_1^2+e_1e_2. \end{aligned}$$
(46)
Next, we obtain
$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 x_3 - \dot{x}_{2_d} \end{aligned}$$
(47)
Now, to achieve convergence of \(e_2\), we define Lyapunov function as
$$\begin{aligned}&{V}_2 = V_1 + \frac{1}{2}e_2^2 \end{aligned}$$
(48)
$$\begin{aligned}&\dot{V}_2 = \dot{V}_1+e_2\dot{e}_2 \end{aligned}$$
(49)
Simplifying Eq. (49) using (46) and (47), we get
$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2+e_1e_2+e_2[f_1(x)+g_1 x_3 - \dot{x}_{2_d} ] \end{aligned}$$
(50)
To realize \(\dot{V}_2 < 0\), we define virtual law as
$$\begin{aligned}&x_{3d} =\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-e_1-k_2e_2] \end{aligned}$$
(51)
where \(\vert f_1(x)-\hat{f}_1(x) \vert = \vert \tilde{f}_1(x) \vert \le F_1\) and \(g_1 = [1+\Delta _1]\hat{g}_1\), \(\vert \Delta _1\vert \le D_1\), \(0<D_1<1\). Next, to achieve \(x_3\rightarrow x_{3_d}\), we define new error
$$\begin{aligned}&e_3 = x_3 - x_{3d} \end{aligned}$$
(52)
so,
$$\begin{aligned}&x_3 = e_3+x_{3d}. \end{aligned}$$
(53)
Using Eqs. (53) and (51), we can simplify (47)
$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 [e_3+x_{3d}] - \dot{x}_{2_d} \nonumber \\&= f_1(x)+g_1 \Big [e_3+\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-e_1-k_2e_2]\Big ] - \dot{x}_{2_d} \end{aligned}$$
(54)
as \(g_1 = [1+\Delta _1]\hat{g}_1\), \(\vert \Delta _1\vert \le D_1\), \(0<D_1<1\) and \(\vert f_1(x)-\hat{f}_1(x) \vert = \vert \tilde{f}_1(x) \vert \le F_1\), therefore, we can write Eq. (54) as
$$\begin{aligned}&\dot{e}_2= f_1(x)+g_1 e_3+\Big [[1+\Delta _1][-\hat{f}_1(x)+\dot{x}_{2d}-e_1-k_2e_2]\Big ] - \dot{x}_{2_d} \nonumber \\&\dot{e}_2 \le f_1(x)+g_1 e_3+\Big [[1+D_1][-\hat{f}_1(x)+\dot{x}_{2d} \nonumber \\ & \quad-e_1-k_2e_2]\Big ] - \dot{x}_{2_d} \dot{e}_2 \le \tilde{f}_1(x)-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}+\Big [[1+D_1][-e_1-k_2e_2]\Big ] \end{aligned}$$
(55)
$$\begin{aligned}&\dot{e}_2 \le F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}+\Big [[1+D_1] \nonumber \\ & \quad [-e_1-k_2e_2]\Big ] \end{aligned}$$
(56)
Using Eq. (56), we can simplify (50)
$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2+e_1e_2+e_2\bigg [F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d} \nonumber \\ &\quad +\Big [[1+D_1][-e_1-k_2e_2]\Big ]\bigg ]\nonumber \\&\quad \dot{V}_2 = -k_1e_1^2 - [1+D_1]k_2e_2^2-D_1e_1e_2 \nonumber \\ &\quad +e_2\bigg [F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}\bigg ] \nonumber \\&\quad \dot{V}_2 = -k_1e_1^2 - [1+D_1]k_2e_2^2-D_1e_1e_2 \nonumber \\ &\quad +e_2\big [g_1 e_3 + V_{2_e}\big ] \end{aligned}$$
(57)
where
$$\begin{aligned} V_{2_e} = \bigg [F_1-D_1\hat{f}_1+D_1\dot{x}_{2d}\bigg ]. \end{aligned}$$
(58)
Next, to achieve convergence of \(e_3\), we define Lyapunov function as
$$\begin{aligned}&V_3 = V_2+\frac{1}{2}e_3^2 \end{aligned}$$
(59)
$$\begin{aligned}&\dot{V}_3 = \dot{V}_2 + e_3\dot{e}_3 \end{aligned}$$
(60)
where
$$\begin{aligned}&\dot{e}_3 = \dot{x}_3 - \dot{x}_{3d} \end{aligned}$$
(61)
$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)u-\dot{x}_{3d} \end{aligned}$$
(62)
To achieve \(\dot{V}_3 < 0\), we define control law as
$$\begin{aligned}&u = \frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3-\hat{g}_1e_2] \end{aligned}$$
(63)
Using Eq. (63), we can simplify (62) as
$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)\Big [\frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3 \nonumber \\ &\quad -\hat{g}_1e_2] \Big ]-\dot{x}_{3d} \end{aligned}$$
(64)
As \(g_2 = [1+\Delta _2]\hat{g}_2\), \(\vert \Delta _2\vert \le D_2\), \(0<D_2<1\) and \(\vert f_2(x)-\hat{f}_2(x) \vert = \vert \tilde{f}_2(x) \vert \le F_2\), therefore, we can write Eq. (64)
$$\begin{aligned} \dot{e}_3&= f_2(x)+[1+\Delta _2][-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3-\hat{g}_1e_2] -\dot{x}_{3d} \nonumber \\ \dot{e}_3&\le f_2(x)+[1+D_2][-\hat{f}_2(x)+\dot{x}_{3d}-k_3e_3-\hat{g}_1e_2] -\dot{x}_{3d} \nonumber \\ \dot{e}_3&\le \tilde{f}_2(x)-D_2\hat{f}_2+D_2 \dot{x}_{3d} +[1+D_2][-k_3e_3-\hat{g}_1e_2] \end{aligned}$$
(65)
$$\begin{aligned} \dot{e}_3&\le F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} +[1+D_2][-k_3e_3-\hat{g}_1e_2] \end{aligned}$$
(66)
Now, using Eq. (66), we can simplify (60) as
$$\begin{aligned} \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2+e_2\big [g_1e_3+V_{2_e}\big ] \nonumber \\&+ e_3\Big [ F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} +[1+D_2][-k_3e_3-\hat{g}_1e_2]\Big ] \nonumber \\ \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2\nonumber \\&+e_2\big [g_1e_3+V_{2_e}\big ] + e_3\Big [-\hat{g}_1e_2 +V_{3_e}\Big ] \end{aligned}$$
(67)
where
$$\begin{aligned} V_{3_e} = F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} -D_2\hat{g}_1e_2 \end{aligned}$$
(68)
As, \(g_1 \le (1+D_1)\hat{g}_1\), we can write Eq. (67) as
$$\begin{aligned} \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ & \quad +(1+D_1)\hat{g}_1e_2e_3 - \hat{g}_1e_2e_3 + V_{t_e} \end{aligned}$$
(69)
where
$$\begin{aligned} V_{t_e} = e_2 V_{2_e} + e_3V_{3_e}. \end{aligned}$$
(70)
Therefore, we can write Eq. (69) as
$$\begin{aligned} \dot{V}_3 \le -k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ \quad +D_1\hat{g}_1e_2e_3+ V_{t_e} \end{aligned}$$
(71)
So, to achieve \(\dot{V}_3 < 0\), we have to
$$\begin{aligned} \big [D_1\hat{g}_1e_2e_3+ V_{t_e}\big ] < \big [k_1e_1^2+[1+D_2]k_2e_2^2 \nonumber \\ &\quad +[1+D_2]k_3e_3^2 \big ] \end{aligned}$$
(72)
Therefore, if we shall choose \(k_1, k_2\) and \(k_3\) greater than zero and large enough to satisfy Eq. (72), we shall always have \(\dot{V}_3 < 0\).
Appendix B Integral backstepping control—proof
Just like backstepping controller, procedure to derive integral backstepping controller is also same. It starts with system dynamics of the arc-shaped SMA actuator as defined in Eqs. (17), (18) and (20). Based on the dynamics, we define
$$\begin{aligned}&e_1 = x_1 - x_{1_d} \end{aligned}$$
(73)
$$\begin{aligned}&\dot{e}_1 = x_2 - \dot{x}_{1_d} \end{aligned}$$
(74)
where \(x_{1_d}\) is the desired trajectory. To achieve convergence for \(e_1\), we can define Lyapunov function as follows
$$\begin{aligned}&V_1 = \frac{1}{2}e_1^2 \end{aligned}$$
(75)
$$\begin{aligned}&\dot{V}_1 =e_1\dot{e}_1 \end{aligned}$$
(76)
$$\begin{aligned}&= e_1[x_2- \dot{x}_{1_d}] \end{aligned}$$
(77)
To realize \(\dot{V}_1<0\), we choose virtual control as
$$\begin{aligned} x_{2_d} = \dot{x}_{1_d}-k_1e_1. \end{aligned}$$
(78)
To achieve \(x_2\rightarrow x_{2_d}\), we define new error
$$\begin{aligned}&e_2 = x_2 - x_{2_d} \end{aligned}$$
(79)
so,
$$\begin{aligned}&x_2 = e_2+x_{2_d}. \end{aligned}$$
(80)
Solving Eqs. (77), (78) and (80) simultaneously, we get
$$\begin{aligned}&\dot{V}_1 = e_1[e_2+x_{2_d}- \dot{x}_{1_d}]\end{aligned}$$
(81)
$$\begin{aligned}&\dot{V}_1 = e_1[e_2+\dot{x}_{1_d}-k_1e_1- \dot{x}_{1_d}]\end{aligned}$$
(82)
$$\begin{aligned}&\dot{V}_1 = -k_1e_1^2+e_1e_2. \end{aligned}$$
(83)
Next, we obtain
$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 x_3 - \dot{x}_{2_d} \end{aligned}$$
(84)
Now, to achieve convergence of \(e_2\), we define Lyapunov function as
$$\begin{aligned}&{V}_2 = V_1 + \frac{1}{2}e_2^2 \end{aligned}$$
(85)
$$\begin{aligned}&\dot{V}_2 = \dot{V}_1+e_2\dot{e}_2 \end{aligned}$$
(86)
Simplifying Eq. (86) using (83) and (84), we get
$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2+e_1e_2+e_2[f_1(x)+g_1 x_3 - \dot{x}_{2_d} ] \end{aligned}$$
(87)
To realize \(\dot{V}_2 < 0\), we define virtual law as
$$\begin{aligned}&x_{3d} =\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-k_i\int e_1\;\text {d}t-e_1-k_2e_2]. \end{aligned}$$
(88)
This Eq. (88) is the brings major difference between backstepping controller and integral backstepping controller. We can compare Eq. (51) for backstepping controller with Eq. (88). Next, we define error
$$\begin{aligned}&e_3 = x_3 - x_{3d} - k_i \int e_1\;\text {d}t \end{aligned}$$
(89)
$$\begin{aligned}&x_3 = e_3+x_{3d}+k_i \int e_1\;\text {d}t \end{aligned}$$
(90)
$$\begin{aligned}&\dot{e}_2 = \dot{x}_2 -\dot{x}_{2_d} = f_1(x)+g_1 [e_3+x_{3d}+k_i \int e_1\;\text {d}t] - \dot{x}_{2_d} \end{aligned}$$
(91)
$$\begin{aligned}&\dot{e}_2 = f_1(x)+g_1 \Big [e_3+\frac{1}{\hat{g}_1}[-\hat{f}_1(x)+\dot{x}_{2d}-k_i\int e_1\;\text {d}t \nonumber \\ & \quad -e_1-k_2e_2] +k_i \int e_1\;\text {d}t\Big ] - \dot{x}_{2_d} \end{aligned}$$
(92)
as \(g_1 = [1+\Delta _1]\hat{g}_1\), \(\vert \Delta _1\vert \le D_1\), \(0<D_1<1\) and \(\vert f_1(x)-\hat{f}_1(x) \vert = \vert \tilde{f}_1(x) \vert \le F_1\), therefore, we can write Eq. (92) as
$$\begin{aligned}&\dot{e}_2 = f_1(x)+g_1 e_3+\Big [[1+\Delta _1][-\hat{f}_1(x)+\dot{x}_{2d}-k_i \nonumber \\ & \quad \int e_1\;\text {d}t-e_1-k_2e_2] +k_i \int e_1\;\text {d}t\Big ] - \dot{x}_{2_d} \end{aligned}$$
(93)
$$\begin{aligned}&\dot{e}_2 \le f_1(x)+g_1 e_3+\Big [[1+D_1][-\hat{f}_1(x)+\dot{x}_{2d}-k_i \nonumber \\ & \quad \int e_1\;\text {d}t-e_1-k_2e_2] +k_i \int e_1\;\text {d}t\Big ] - \dot{x}_{2_d} \end{aligned}$$
(94)
$$\begin{aligned}&\dot{e}_2 \le \tilde{f}_1(x)-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t \nonumber \\ & \quad +\Big [[1+D_1][-e_1-k_2e_2]\Big ] \dot{e}_2 \le F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t +\Big [[1+D_1][-e_1-k_2e_2]\Big ] \end{aligned}$$
(95)
Using Eq. (95), we can simplify (87) as
$$\begin{aligned}&\dot{V}_2 \le -k_1e_1^2+e_1e_2\nonumber \\&\quad +e_2\bigg [F_1-D_1\hat{f}_1+g_1 e_3+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t \nonumber \\ & \quad +\Big [[1+D_1][-e_1-k_2e_2]\Big ]\bigg ] \end{aligned}$$
(96)
$$\begin{aligned}&\dot{V}_2 = -k_1e_1^2 - [1+D_1]k_2e_2^2-D_1e_1e_2 +e_2\bigg [g_1 e_3+V_{2_e}\bigg ] \end{aligned}$$
(97)
where
$$\begin{aligned} V_{2_e} = F_1-D_1\hat{f}_1+D_1\dot{x}_{2d}-D_1k_i \int e_1\;\text {d}t. \end{aligned}$$
(98)
Next, to achieve convergence of \(e_3\), we define Lyapunov function as
$$\begin{aligned}&V_3 = V_2+\frac{1}{2}e_3^2 \end{aligned}$$
(99)
$$\begin{aligned}&\dot{V}_3 = \dot{V}_2 + e_3\dot{e}_3. \end{aligned}$$
(100)
Here,
$$\begin{aligned}&\dot{e}_3 = \dot{x}_3 - \dot{x}_{3d} - k_i e_1 \end{aligned}$$
(101)
$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)u-\dot{x}_{3d} - k_i e_1 \end{aligned}$$
(102)
Then, we define
$$\begin{aligned}&u = \frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}+k_ie_1-k_3e_3-\hat{g}_1e_2] \end{aligned}$$
(103)
Using Eq. (103), we can simplify (102) as
$$\begin{aligned}&\dot{e}_3 = f_2(x)+g_2(x)\bigg [\frac{1}{\hat{g}_2(x)}[-\hat{f}_2(x)+\dot{x}_{3d}+k_ie_1 \nonumber \\ & \quad -k_3e_3-\hat{g}_1e_2] \bigg ]-\dot{x}_{3d} - k_i e_1 \end{aligned}$$
(104)
as \(g_2 = [1+\Delta _2]\hat{g}_2\), \(\vert \Delta _2\vert \le D_2\), \(0<D_2<1\) and \(\vert f_2(x)-\hat{f}_2(x) \vert = \vert \tilde{f}_2(x) \vert \le F_2\), therefore, we can write Eq. (104) as
$$\begin{aligned}&\dot{e}_3 \le \bigg [F_2-D_2\hat{f}_2(x)+[1+D_2]\Big [\dot{x}_{3d}+k_1e_1 \nonumber \\ & \quad -k_3e_3-\hat{g}_1e_2\Big ]-\dot{x}_{3d} -k_ie_1\bigg ] \end{aligned}$$
(105)
$$\begin{aligned}&\dot{e}_3 \le \bigg [F_2-D_2\hat{f}_2(x)+D_2\dot{x}_{3d}+D_2k_1e_1+ [1+D_2] \nonumber \\ & \quad \Big [-k_3e_3-\hat{g}_1e_2\Big ]\bigg ] \end{aligned}$$
(106)
Using Eq. (105), we can simplify (100) as
$$\begin{aligned}&\dot{V}_3 \le -k_1e_1^2-[1+D_1]k_2e_2^2+e_2\Big [g_1e_3+V_{2_e}\Big ] \nonumber \\&\quad + e_3\bigg [F_2-D_2\hat{f}_2(x)+[1+D_2]\Big [\dot{x}_{3d}+k_1e_1-k_3e_3-\hat{g}_1e_2\Big ] \nonumber \\ & \quad -\dot{x}_{3d} -k_ie_1\bigg ] \end{aligned}$$
(107)
$$\begin{aligned}&\dot{V}_3 \le -k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 +e_2\big [g_1e_3+V_{2_e}\big ] \nonumber \\ & \quad + e_3\Big [-\hat{g}_1e_2 +V_{3_e}\Big ] \end{aligned}$$
(108)
where
$$\begin{aligned} V_{3_e} = F_2-D_2\hat{f}_2+D_2 \dot{x}_{3d} -D_2\hat{g}_1e_2 \end{aligned}$$
(109)
SAs, \(g_1 \le (1+D_1)\hat{g}_1\), we can write Eq. (108) as
$$\begin{aligned} \dot{V}_3 \le&-k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ & \quad +(1+D_1)\hat{g}_1e_2e_3 - \hat{g}_1e_2e_3 + V_{t_e} \end{aligned}$$
(110)
where
$$\begin{aligned} V_{t_e} = e_2 V_{2_e} + e_3V_{3_e}. \end{aligned}$$
(111)
Therefore, we can write Eq. (110) as
$$\begin{aligned} \dot{V}_3 \le -k_1e_1^2-[1+D_2]k_2e_2^2-[1+D_2]k_3e_3^2 \nonumber \\ \quad +D_1\hat{g}_1e_2e_3+ V_{t_e} \end{aligned}$$
(112)
So, to achieve \(\dot{V}_3 < 0\), we have to
$$\begin{aligned} \big [D_1\hat{g}_1e_2e_3+ V_{t_e}\big ] < \big [k_1e_1^2+[1+D_2]k_2e_2^2 \nonumber \\ \quad +[1+D_2]k_3e_3^2 \big ] \end{aligned}$$
(113)
Therefore, if we shall choose \(k_1, k_2\) and \(k_3\) greater than zero and large enough to satisfy Eq. (113), we shall always have \(\dot{V}_3 < 0\).