Is sliding mode control complete in mathematics?
$begingroup$
Consider a double integrator control system
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u,end{cases}$$
where $dot{x} := mathrm{d}x/mathrm{d}t$. Then we apply the sliding mode control $u = -beta(x)mathrm{sgn}(s)$ with $s = a x_1+x_2$ ($a > 0$), in which $beta(x) geq a x_2 + beta_0$ ($beta_0 > 0$), and
$$mathrm{sgn}(s)=begin{cases}1, & s > 0,\ 0, & s = 0,\-1,& s<0.end{cases}$$
By setting the Lyapunov function as $V = s^2/2$, any trajectory initially with $s neq 0$ will reach $s = 0$ in finite time. This is because
$dot{V} = s dot{s} = s(a dot{x}_1 + dot{x}_2) = sa x_2 + su leq |s|ax_2 + s [-beta(x)mathrm{sgn}(s)] = |s|(ax_2 - beta(x)) leq -beta_0|s|.$
The above inequality also implies that: when reaching the manifold $s = 0$, the trajectory cannot leave it. On the manifold $s=0$, state $x_1$ follows $0 = s = ax_1 + x_2$, i.e., $dot{x}_1 = -a x_1$, which means $(x_1, x_2)$ will finally converge to $(0,0)$ (along $s = 0$, since the trajectory cannot leave $s = 0$) as time goes to infinity.
However, here is the interesting thing. Do you really think $(x_1,x_2)$ will move along $s = 0$? Let's check the vector field at $(x_1, x_2)$, and it will gives
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u=-beta(x)mathrm{sgn}(s)=0,end{cases}$$
which never points along $s = ax_1 + x_2$ (recall that $a > 0$), and it will leave $s = 0$ instantly. This result contradicts "when reaching the manifold $s = 0$, the trajectory cannot leave it" implied by the Lyapunov function. What's happened on $s = 0$?
ordinary-differential-equations control-theory discontinuous-functions
asked Dec 17 '18 at 14:15
RyanRyan
240118
$endgroup$
add a comment |
$begingroup$
Consider a double integrator control system
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u,end{cases}$$
where $dot{x} := mathrm{d}x/mathrm{d}t$. Then we apply the sliding mode control $u = -beta(x)mathrm{sgn}(s)$ with $s = a x_1+x_2$ ($a > 0$), in which $beta(x) geq a x_2 + beta_0$ ($beta_0 > 0$), and
$$mathrm{sgn}(s)=begin{cases}1, & s > 0,\ 0, & s = 0,\-1,& s<0.end{cases}$$
By setting the Lyapunov function as $V = s^2/2$, any trajectory initially with $s neq 0$ will reach $s = 0$ in finite time. This is because
$dot{V} = s dot{s} = s(a dot{x}_1 + dot{x}_2) = sa x_2 + su leq |s|ax_2 + s [-beta(x)mathrm{sgn}(s)] = |s|(ax_2 - beta(x)) leq -beta_0|s|.$
The above inequality also implies that: when reaching the manifold $s = 0$, the trajectory cannot leave it. On the manifold $s=0$, state $x_1$ follows $0 = s = ax_1 + x_2$, i.e., $dot{x}_1 = -a x_1$, which means $(x_1, x_2)$ will finally converge to $(0,0)$ (along $s = 0$, since the trajectory cannot leave $s = 0$) as time goes to infinity.
However, here is the interesting thing. Do you really think $(x_1,x_2)$ will move along $s = 0$? Let's check the vector field at $(x_1, x_2)$, and it will gives
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u=-beta(x)mathrm{sgn}(s)=0,end{cases}$$
which never points along $s = ax_1 + x_2$ (recall that $a > 0$), and it will leave $s = 0$ instantly. This result contradicts "when reaching the manifold $s = 0$, the trajectory cannot leave it" implied by the Lyapunov function. What's happened on $s = 0$?
ordinary-differential-equations control-theory discontinuous-functions
asked Dec 17 '18 at 14:15
RyanRyan
240118
$endgroup$
1
$begingroup$
The conventional solutions cease to exist at $s=0$, at least for $x_2>0$, as these are only defined for domains with continuous vector fields. What generalized solution definition are you using to extend these conventional solutions?
$endgroup$
– LutzL
Dec 17 '18 at 16:08
$begingroup$
@LutzL Thanks very much! Yes, the vector fields are not continuous (PS: Can we say that the vector fields do not exist on the manifold $s = 0$?). For the generalized solution of an ODE, so sorry I am not familiar wtih its definition. What kind of generalized solution can describe the actual trajectory of the system? Thanks a lot!
$endgroup$
– Ryan
Dec 18 '18 at 1:05
1
$begingroup$
One way is to approximate the right side, like with $sgn(x)approx 2frac{x}{epsilon+|x|}$ and the same shifted a little to the side. Then look if anything is common quantivatively or only qualitatively among the solutions to these approximations. This essentially is the distributional approach of Fillipov(?).
$endgroup$
– LutzL
Dec 18 '18 at 1:44
$begingroup$
@LutzL Thanks so much for your answer. I think I need to read some textbooks and get some background on the generalized solutions in Filippov sense.
$endgroup$
– Ryan
Dec 19 '18 at 2:28
add a comment |
$begingroup$
Consider a double integrator control system
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u,end{cases}$$
where $dot{x} := mathrm{d}x/mathrm{d}t$. Then we apply the sliding mode control $u = -beta(x)mathrm{sgn}(s)$ with $s = a x_1+x_2$ ($a > 0$), in which $beta(x) geq a x_2 + beta_0$ ($beta_0 > 0$), and
$$mathrm{sgn}(s)=begin{cases}1, & s > 0,\ 0, & s = 0,\-1,& s<0.end{cases}$$
By setting the Lyapunov function as $V = s^2/2$, any trajectory initially with $s neq 0$ will reach $s = 0$ in finite time. This is because
$dot{V} = s dot{s} = s(a dot{x}_1 + dot{x}_2) = sa x_2 + su leq |s|ax_2 + s [-beta(x)mathrm{sgn}(s)] = |s|(ax_2 - beta(x)) leq -beta_0|s|.$
The above inequality also implies that: when reaching the manifold $s = 0$, the trajectory cannot leave it. On the manifold $s=0$, state $x_1$ follows $0 = s = ax_1 + x_2$, i.e., $dot{x}_1 = -a x_1$, which means $(x_1, x_2)$ will finally converge to $(0,0)$ (along $s = 0$, since the trajectory cannot leave $s = 0$) as time goes to infinity.
However, here is the interesting thing. Do you really think $(x_1,x_2)$ will move along $s = 0$? Let's check the vector field at $(x_1, x_2)$, and it will gives
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u=-beta(x)mathrm{sgn}(s)=0,end{cases}$$
which never points along $s = ax_1 + x_2$ (recall that $a > 0$), and it will leave $s = 0$ instantly. This result contradicts "when reaching the manifold $s = 0$, the trajectory cannot leave it" implied by the Lyapunov function. What's happened on $s = 0$?
ordinary-differential-equations control-theory discontinuous-functions
asked Dec 17 '18 at 14:15
RyanRyan
240118
$endgroup$
Consider a double integrator control system
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u,end{cases}$$
where $dot{x} := mathrm{d}x/mathrm{d}t$. Then we apply the sliding mode control $u = -beta(x)mathrm{sgn}(s)$ with $s = a x_1+x_2$ ($a > 0$), in which $beta(x) geq a x_2 + beta_0$ ($beta_0 > 0$), and
$$mathrm{sgn}(s)=begin{cases}1, & s > 0,\ 0, & s = 0,\-1,& s<0.end{cases}$$
By setting the Lyapunov function as $V = s^2/2$, any trajectory initially with $s neq 0$ will reach $s = 0$ in finite time. This is because
$dot{V} = s dot{s} = s(a dot{x}_1 + dot{x}_2) = sa x_2 + su leq |s|ax_2 + s [-beta(x)mathrm{sgn}(s)] = |s|(ax_2 - beta(x)) leq -beta_0|s|.$
The above inequality also implies that: when reaching the manifold $s = 0$, the trajectory cannot leave it. On the manifold $s=0$, state $x_1$ follows $0 = s = ax_1 + x_2$, i.e., $dot{x}_1 = -a x_1$, which means $(x_1, x_2)$ will finally converge to $(0,0)$ (along $s = 0$, since the trajectory cannot leave $s = 0$) as time goes to infinity.
However, here is the interesting thing. Do you really think $(x_1,x_2)$ will move along $s = 0$? Let's check the vector field at $(x_1, x_2)$, and it will gives
$$begin{cases}dot{x}_1 = x_2,\dot{x}_2=u=-beta(x)mathrm{sgn}(s)=0,end{cases}$$
which never points along $s = ax_1 + x_2$ (recall that $a > 0$), and it will leave $s = 0$ instantly. This result contradicts "when reaching the manifold $s = 0$, the trajectory cannot leave it" implied by the Lyapunov function. What's happened on $s = 0$?
ordinary-differential-equations control-theory discontinuous-functions
ordinary-differential-equations control-theory discontinuous-functions
asked Dec 17 '18 at 14:15
RyanRyan
240118
asked Dec 17 '18 at 14:15
RyanRyan
240118
asked Dec 17 '18 at 14:15
RyanRyan
240118
asked Dec 17 '18 at 14:15
RyanRyan
240118
asked Dec 17 '18 at 14:15
RyanRyan
240118
240118
1
$begingroup$
The conventional solutions cease to exist at $s=0$, at least for $x_2>0$, as these are only defined for domains with continuous vector fields. What generalized solution definition are you using to extend these conventional solutions?
$endgroup$
– LutzL
Dec 17 '18 at 16:08
$begingroup$
@LutzL Thanks very much! Yes, the vector fields are not continuous (PS: Can we say that the vector fields do not exist on the manifold $s = 0$?). For the generalized solution of an ODE, so sorry I am not familiar wtih its definition. What kind of generalized solution can describe the actual trajectory of the system? Thanks a lot!
$endgroup$
– Ryan
Dec 18 '18 at 1:05
1
$begingroup$
One way is to approximate the right side, like with $sgn(x)approx 2frac{x}{epsilon+|x|}$ and the same shifted a little to the side. Then look if anything is common quantivatively or only qualitatively among the solutions to these approximations. This essentially is the distributional approach of Fillipov(?).
$endgroup$
– LutzL
Dec 18 '18 at 1:44
$begingroup$
@LutzL Thanks so much for your answer. I think I need to read some textbooks and get some background on the generalized solutions in Filippov sense.
$endgroup$
– Ryan
Dec 19 '18 at 2:28
add a comment |
1
$begingroup$
The conventional solutions cease to exist at $s=0$, at least for $x_2>0$, as these are only defined for domains with continuous vector fields. What generalized solution definition are you using to extend these conventional solutions?
$endgroup$
– LutzL
Dec 17 '18 at 16:08
$begingroup$
@LutzL Thanks very much! Yes, the vector fields are not continuous (PS: Can we say that the vector fields do not exist on the manifold $s = 0$?). For the generalized solution of an ODE, so sorry I am not familiar wtih its definition. What kind of generalized solution can describe the actual trajectory of the system? Thanks a lot!
$endgroup$
– Ryan
Dec 18 '18 at 1:05
1
$begingroup$
One way is to approximate the right side, like with $sgn(x)approx 2frac{x}{epsilon+|x|}$ and the same shifted a little to the side. Then look if anything is common quantivatively or only qualitatively among the solutions to these approximations. This essentially is the distributional approach of Fillipov(?).
$endgroup$
– LutzL
Dec 18 '18 at 1:44
$begingroup$
@LutzL Thanks so much for your answer. I think I need to read some textbooks and get some background on the generalized solutions in Filippov sense.
$endgroup$
– Ryan
Dec 19 '18 at 2:28
1
1
$begingroup$
The conventional solutions cease to exist at $s=0$, at least for $x_2>0$, as these are only defined for domains with continuous vector fields. What generalized solution definition are you using to extend these conventional solutions?
$endgroup$
– LutzL
Dec 17 '18 at 16:08
$begingroup$
The conventional solutions cease to exist at $s=0$, at least for $x_2>0$, as these are only defined for domains with continuous vector fields. What generalized solution definition are you using to extend these conventional solutions?
$endgroup$
– LutzL
Dec 17 '18 at 16:08
$begingroup$
@LutzL Thanks very much! Yes, the vector fields are not continuous (PS: Can we say that the vector fields do not exist on the manifold $s = 0$?). For the generalized solution of an ODE, so sorry I am not familiar wtih its definition. What kind of generalized solution can describe the actual trajectory of the system? Thanks a lot!
$endgroup$
– Ryan
Dec 18 '18 at 1:05
$begingroup$
@LutzL Thanks very much! Yes, the vector fields are not continuous (PS: Can we say that the vector fields do not exist on the manifold $s = 0$?). For the generalized solution of an ODE, so sorry I am not familiar wtih its definition. What kind of generalized solution can describe the actual trajectory of the system? Thanks a lot!
$endgroup$
– Ryan
Dec 18 '18 at 1:05
1
1
$begingroup$
One way is to approximate the right side, like with $sgn(x)approx 2frac{x}{epsilon+|x|}$ and the same shifted a little to the side. Then look if anything is common quantivatively or only qualitatively among the solutions to these approximations. This essentially is the distributional approach of Fillipov(?).
$endgroup$
– LutzL
Dec 18 '18 at 1:44
$begingroup$
One way is to approximate the right side, like with $sgn(x)approx 2frac{x}{epsilon+|x|}$ and the same shifted a little to the side. Then look if anything is common quantivatively or only qualitatively among the solutions to these approximations. This essentially is the distributional approach of Fillipov(?).
$endgroup$
– LutzL
Dec 18 '18 at 1:44
$begingroup$
@LutzL Thanks so much for your answer. I think I need to read some textbooks and get some background on the generalized solutions in Filippov sense.
$endgroup$
– Ryan
Dec 19 '18 at 2:28
$begingroup$
@LutzL Thanks so much for your answer. I think I need to read some textbooks and get some background on the generalized solutions in Filippov sense.
$endgroup$
– Ryan
Dec 19 '18 at 2:28
add a comment |
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$begingroup$
The conventional solutions cease to exist at $s=0$, at least for $x_2>0$, as these are only defined for domains with continuous vector fields. What generalized solution definition are you using to extend these conventional solutions?
$endgroup$
– LutzL
Dec 17 '18 at 16:08
$begingroup$
@LutzL Thanks very much! Yes, the vector fields are not continuous (PS: Can we say that the vector fields do not exist on the manifold $s = 0$?). For the generalized solution of an ODE, so sorry I am not familiar wtih its definition. What kind of generalized solution can describe the actual trajectory of the system? Thanks a lot!
$endgroup$
– Ryan
Dec 18 '18 at 1:05
1
$begingroup$
One way is to approximate the right side, like with $sgn(x)approx 2frac{x}{epsilon+|x|}$ and the same shifted a little to the side. Then look if anything is common quantivatively or only qualitatively among the solutions to these approximations. This essentially is the distributional approach of Fillipov(?).
$endgroup$
– LutzL
Dec 18 '18 at 1:44
$begingroup$
@LutzL Thanks so much for your answer. I think I need to read some textbooks and get some background on the generalized solutions in Filippov sense.
$endgroup$
– Ryan
Dec 19 '18 at 2:28