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A choice of Lyapunov function for this 2D system?
Clash Royale CLAN TAG#URR8PPP
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I am thinking of a choice of a suitable Lyapunov function$V(x_1,x_2)$ which can make the system stable around the fixed point $x_1=1,x_2=1$
$dotx_1 = x_1x_2 - x_1^2 $
$dotx_2 = x_2 - x_1x_2 + 2 -2x_2^2$
I thought of using $V(x_1,x_2) = a(x_1-1)^2 + b(x_2-1)^2$, but then I am unable to show the $fracdVdt <0$ but still we have $V(1,1) =0$, which I asked here - Determining $a$ and $b$ such that the expression is negative? and it seems I cannot take such a $V$? any help with the expression of $V(x_1,x_2)$?
dynamical-systems lyapunov-functions
asked 3 hours ago
BAYMAX
2,68621122
add a comment |Â
up vote
2
down vote
favorite
I am thinking of a choice of a suitable Lyapunov function$V(x_1,x_2)$ which can make the system stable around the fixed point $x_1=1,x_2=1$
$dotx_1 = x_1x_2 - x_1^2 $
$dotx_2 = x_2 - x_1x_2 + 2 -2x_2^2$
I thought of using $V(x_1,x_2) = a(x_1-1)^2 + b(x_2-1)^2$, but then I am unable to show the $fracdVdt <0$ but still we have $V(1,1) =0$, which I asked here - Determining $a$ and $b$ such that the expression is negative? and it seems I cannot take such a $V$? any help with the expression of $V(x_1,x_2)$?
dynamical-systems lyapunov-functions
asked 3 hours ago
BAYMAX
2,68621122
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am thinking of a choice of a suitable Lyapunov function$V(x_1,x_2)$ which can make the system stable around the fixed point $x_1=1,x_2=1$
$dotx_1 = x_1x_2 - x_1^2 $
$dotx_2 = x_2 - x_1x_2 + 2 -2x_2^2$
I thought of using $V(x_1,x_2) = a(x_1-1)^2 + b(x_2-1)^2$, but then I am unable to show the $fracdVdt <0$ but still we have $V(1,1) =0$, which I asked here - Determining $a$ and $b$ such that the expression is negative? and it seems I cannot take such a $V$? any help with the expression of $V(x_1,x_2)$?
dynamical-systems lyapunov-functions
asked 3 hours ago
BAYMAX
2,68621122
I am thinking of a choice of a suitable Lyapunov function$V(x_1,x_2)$ which can make the system stable around the fixed point $x_1=1,x_2=1$
$dotx_1 = x_1x_2 - x_1^2 $
$dotx_2 = x_2 - x_1x_2 + 2 -2x_2^2$
I thought of using $V(x_1,x_2) = a(x_1-1)^2 + b(x_2-1)^2$, but then I am unable to show the $fracdVdt <0$ but still we have $V(1,1) =0$, which I asked here - Determining $a$ and $b$ such that the expression is negative? and it seems I cannot take such a $V$? any help with the expression of $V(x_1,x_2)$?
dynamical-systems lyapunov-functions
dynamical-systems lyapunov-functions
asked 3 hours ago
BAYMAX
2,68621122
asked 3 hours ago
BAYMAX
2,68621122
edited 3 hours ago
asked 3 hours ago
BAYMAX
2,68621122
asked 3 hours ago
BAYMAX
2,68621122
asked 3 hours ago
BAYMAX
2,68621122
2,68621122
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
You only need that $dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
beginalign
dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
endalign
Thus for $a=b=1$ you get for $(x,y)approx(1,1)$ that $$dot Vapprox-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^3/2$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+8V^3/2
endalign
which is negative for $4V^1/2<1$ or $V<frac116$. Using better estimates that combine some of the third order terms one might get larger bounds,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+2Bigl[(x-1)^2+(y-1)^2Bigr]bigl(|x-1|+|y-1|bigr)\
&le-2V+2sqrt2V^3/2,
endalign
which means that the admissible region is $V<frac12$.
answered 2 hours ago
LutzL
50.5k31849
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
add a comment |Â
up vote
2
down vote
Using $V(x,y) = (x-1)^2+(y-1)^2$
First, a Lyapunov function, doesn't make a system stable. It serves to certify stability.
Second, the dynamical system has four equilibrium points as can be depicted in the attached stream plot.
Third, the sought local lyapunov function characterizes a limited region around the equilibrium point $(1,1)$ which is an approximation to the attraction basin for $(1,1)$
Attached the MATHEMATICA script
V0 = (x - 1)^2 + (y - 1)^2;
gr0 = StreamPlot[x y - x^2, y - x y + 2 - 2 y^2, x, -3, 3, y, -3,
3, Mesh -> 20];
sols = Solve[x y - x^2 == 0, y - x y + 2 - 2 y^2 == 0, x, y]
gr1 = ContourPlot[V0, x, 1 - 1, 1 + 1, y, 1 - 1, 1 + 1,
ContourShading -> None, Contours -> 30,
RegionFunction -> Function[x, y, (x - 1)^2 + (y - 1)^2 < 1/2]];
gr = Table[Graphics[Red, Disk[(x, y /. sols[[k]]), 0.1]], k, Length[sols]];
Show[gr0, gr1, gr]
answered 1 hour ago
Cesareo
6,4892413
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You only need that $dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
beginalign
dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
endalign
Thus for $a=b=1$ you get for $(x,y)approx(1,1)$ that $$dot Vapprox-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^3/2$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+8V^3/2
endalign
which is negative for $4V^1/2<1$ or $V<frac116$. Using better estimates that combine some of the third order terms one might get larger bounds,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+2Bigl[(x-1)^2+(y-1)^2Bigr]bigl(|x-1|+|y-1|bigr)\
&le-2V+2sqrt2V^3/2,
endalign
which means that the admissible region is $V<frac12$.
answered 2 hours ago
LutzL
50.5k31849
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
You only need that $dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
beginalign
dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
endalign
Thus for $a=b=1$ you get for $(x,y)approx(1,1)$ that $$dot Vapprox-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^3/2$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+8V^3/2
endalign
which is negative for $4V^1/2<1$ or $V<frac116$. Using better estimates that combine some of the third order terms one might get larger bounds,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+2Bigl[(x-1)^2+(y-1)^2Bigr]bigl(|x-1|+|y-1|bigr)\
&le-2V+2sqrt2V^3/2,
endalign
which means that the admissible region is $V<frac12$.
answered 2 hours ago
LutzL
50.5k31849
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You only need that $dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
beginalign
dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
endalign
Thus for $a=b=1$ you get for $(x,y)approx(1,1)$ that $$dot Vapprox-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^3/2$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+8V^3/2
endalign
which is negative for $4V^1/2<1$ or $V<frac116$. Using better estimates that combine some of the third order terms one might get larger bounds,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+2Bigl[(x-1)^2+(y-1)^2Bigr]bigl(|x-1|+|y-1|bigr)\
&le-2V+2sqrt2V^3/2,
endalign
which means that the admissible region is $V<frac12$.
answered 2 hours ago
LutzL
50.5k31849
You only need that $dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
beginalign
dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
endalign
Thus for $a=b=1$ you get for $(x,y)approx(1,1)$ that $$dot Vapprox-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^3/2$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+8V^3/2
endalign
which is negative for $4V^1/2<1$ or $V<frac116$. Using better estimates that combine some of the third order terms one might get larger bounds,
beginalign
dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\
&le-2V+2Bigl[(x-1)^2+(y-1)^2Bigr]bigl(|x-1|+|y-1|bigr)\
&le-2V+2sqrt2V^3/2,
endalign
which means that the admissible region is $V<frac12$.
answered 2 hours ago
LutzL
50.5k31849
edited 1 hour ago
answered 2 hours ago
LutzL
50.5k31849
answered 2 hours ago
LutzL
50.5k31849
answered 2 hours ago
LutzL
50.5k31849
50.5k31849
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
add a comment |Â
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
Great!!, but how could I find the region of asymptotically stable? like it is now known that near (1,1) the system is Asymptotically stable but how can we find the region precisely? any help LutzL??
– BAYMAX
1 hour ago
add a comment |Â
up vote
2
down vote
Using $V(x,y) = (x-1)^2+(y-1)^2$
First, a Lyapunov function, doesn't make a system stable. It serves to certify stability.
Second, the dynamical system has four equilibrium points as can be depicted in the attached stream plot.
Third, the sought local lyapunov function characterizes a limited region around the equilibrium point $(1,1)$ which is an approximation to the attraction basin for $(1,1)$
Attached the MATHEMATICA script
V0 = (x - 1)^2 + (y - 1)^2;
gr0 = StreamPlot[x y - x^2, y - x y + 2 - 2 y^2, x, -3, 3, y, -3,
3, Mesh -> 20];
sols = Solve[x y - x^2 == 0, y - x y + 2 - 2 y^2 == 0, x, y]
gr1 = ContourPlot[V0, x, 1 - 1, 1 + 1, y, 1 - 1, 1 + 1,
ContourShading -> None, Contours -> 30,
RegionFunction -> Function[x, y, (x - 1)^2 + (y - 1)^2 < 1/2]];
gr = Table[Graphics[Red, Disk[(x, y /. sols[[k]]), 0.1]], k, Length[sols]];
Show[gr0, gr1, gr]
answered 1 hour ago
Cesareo
6,4892413
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
add a comment |Â
up vote
2
down vote
Using $V(x,y) = (x-1)^2+(y-1)^2$
First, a Lyapunov function, doesn't make a system stable. It serves to certify stability.
Second, the dynamical system has four equilibrium points as can be depicted in the attached stream plot.
Third, the sought local lyapunov function characterizes a limited region around the equilibrium point $(1,1)$ which is an approximation to the attraction basin for $(1,1)$
Attached the MATHEMATICA script
V0 = (x - 1)^2 + (y - 1)^2;
gr0 = StreamPlot[x y - x^2, y - x y + 2 - 2 y^2, x, -3, 3, y, -3,
3, Mesh -> 20];
sols = Solve[x y - x^2 == 0, y - x y + 2 - 2 y^2 == 0, x, y]
gr1 = ContourPlot[V0, x, 1 - 1, 1 + 1, y, 1 - 1, 1 + 1,
ContourShading -> None, Contours -> 30,
RegionFunction -> Function[x, y, (x - 1)^2 + (y - 1)^2 < 1/2]];
gr = Table[Graphics[Red, Disk[(x, y /. sols[[k]]), 0.1]], k, Length[sols]];
Show[gr0, gr1, gr]
answered 1 hour ago
Cesareo
6,4892413
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using $V(x,y) = (x-1)^2+(y-1)^2$
First, a Lyapunov function, doesn't make a system stable. It serves to certify stability.
Second, the dynamical system has four equilibrium points as can be depicted in the attached stream plot.
Third, the sought local lyapunov function characterizes a limited region around the equilibrium point $(1,1)$ which is an approximation to the attraction basin for $(1,1)$
Attached the MATHEMATICA script
V0 = (x - 1)^2 + (y - 1)^2;
gr0 = StreamPlot[x y - x^2, y - x y + 2 - 2 y^2, x, -3, 3, y, -3,
3, Mesh -> 20];
sols = Solve[x y - x^2 == 0, y - x y + 2 - 2 y^2 == 0, x, y]
gr1 = ContourPlot[V0, x, 1 - 1, 1 + 1, y, 1 - 1, 1 + 1,
ContourShading -> None, Contours -> 30,
RegionFunction -> Function[x, y, (x - 1)^2 + (y - 1)^2 < 1/2]];
gr = Table[Graphics[Red, Disk[(x, y /. sols[[k]]), 0.1]], k, Length[sols]];
Show[gr0, gr1, gr]
answered 1 hour ago
Cesareo
6,4892413
Using $V(x,y) = (x-1)^2+(y-1)^2$
First, a Lyapunov function, doesn't make a system stable. It serves to certify stability.
Second, the dynamical system has four equilibrium points as can be depicted in the attached stream plot.
Third, the sought local lyapunov function characterizes a limited region around the equilibrium point $(1,1)$ which is an approximation to the attraction basin for $(1,1)$
Attached the MATHEMATICA script
V0 = (x - 1)^2 + (y - 1)^2;
gr0 = StreamPlot[x y - x^2, y - x y + 2 - 2 y^2, x, -3, 3, y, -3,
3, Mesh -> 20];
sols = Solve[x y - x^2 == 0, y - x y + 2 - 2 y^2 == 0, x, y]
gr1 = ContourPlot[V0, x, 1 - 1, 1 + 1, y, 1 - 1, 1 + 1,
ContourShading -> None, Contours -> 30,
RegionFunction -> Function[x, y, (x - 1)^2 + (y - 1)^2 < 1/2]];
gr = Table[Graphics[Red, Disk[(x, y /. sols[[k]]), 0.1]], k, Length[sols]];
Show[gr0, gr1, gr]
answered 1 hour ago
Cesareo
6,4892413
edited 1 hour ago
answered 1 hour ago
Cesareo
6,4892413
answered 1 hour ago
Cesareo
6,4892413
answered 1 hour ago
Cesareo
6,4892413
6,4892413
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
add a comment |Â
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
Nice, visualization, which tool u used to plot the graph , vector fields?
– BAYMAX
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
@BAYMAX The corresponding MATHEMATICA script is now included.
– Cesareo
1 hour ago
add a comment |Â
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