Mixing
Bifurcation points of differential equation (example)
$begingroup$
Assume the differential equation:
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R}.
$$
The critical points are the solutions to the equation:
$$
x'=0 iff 2x^2-8alambda x +lambda^2=0tag{1}
$$
which admits solutions:
$$
x=lambda cdot frac{8a pm sqrt{64a^2-8}}{2}
$$
Now, having in mind that by bifurcation point, we mean a point where change of the number of equilibrium points occur, it seems to me that the number of $(1)$'s solutions depends only on $a$ and not on $lambda$. Is this the case or did I get something wrong?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 7 at 19:29
LoneBoneLoneBone
888
$endgroup$
add a comment |
$begingroup$
Assume the differential equation:
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R}.
$$
The critical points are the solutions to the equation:
$$
x'=0 iff 2x^2-8alambda x +lambda^2=0tag{1}
$$
which admits solutions:
$$
x=lambda cdot frac{8a pm sqrt{64a^2-8}}{2}
$$
Now, having in mind that by bifurcation point, we mean a point where change of the number of equilibrium points occur, it seems to me that the number of $(1)$'s solutions depends only on $a$ and not on $lambda$. Is this the case or did I get something wrong?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 7 at 19:29
LoneBoneLoneBone
888
$endgroup$
add a comment |
$begingroup$
Assume the differential equation:
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R}.
$$
The critical points are the solutions to the equation:
$$
x'=0 iff 2x^2-8alambda x +lambda^2=0tag{1}
$$
which admits solutions:
$$
x=lambda cdot frac{8a pm sqrt{64a^2-8}}{2}
$$
Now, having in mind that by bifurcation point, we mean a point where change of the number of equilibrium points occur, it seems to me that the number of $(1)$'s solutions depends only on $a$ and not on $lambda$. Is this the case or did I get something wrong?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 7 at 19:29
LoneBoneLoneBone
888
$endgroup$
Assume the differential equation:
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R}.
$$
The critical points are the solutions to the equation:
$$
x'=0 iff 2x^2-8alambda x +lambda^2=0tag{1}
$$
which admits solutions:
$$
x=lambda cdot frac{8a pm sqrt{64a^2-8}}{2}
$$
Now, having in mind that by bifurcation point, we mean a point where change of the number of equilibrium points occur, it seems to me that the number of $(1)$'s solutions depends only on $a$ and not on $lambda$. Is this the case or did I get something wrong?
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
ordinary-differential-equations dynamical-systems stability-in-odes bifurcation
asked Jan 7 at 19:29
LoneBoneLoneBone
888
asked Jan 7 at 19:29
LoneBoneLoneBone
888
asked Jan 7 at 19:29
LoneBoneLoneBone
888
asked Jan 7 at 19:29
LoneBoneLoneBone
888
asked Jan 7 at 19:29
LoneBoneLoneBone
888
888
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To study bifurcation points for
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R},
$$
we see that $x'=0$ implies that
$$
x = frac{lambda}{2}left(4a pm sqrt{2(8a^2-1)}right).
$$
$textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $xin mathbb{R}$ that will give $x'=0$.
$textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2alambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2alambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2alambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point.
$textbf{Case 3}$. If $8a^2 > 1$, then there are two subcases to consider.
$color{green}{textbf{Subcase 1}}$. If $lambdanot=0$, then there are two fixed points:
$$
x_1^* = frac{lambda}{2}left(4a - sqrt{2(8a^2-1)}right),
quad
x_2^* = frac{lambda}{2}left(4a + sqrt{2(8a^2-1)}right).
$$
We can use a similar analysis as in $textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point.
$color{green}{textbf{Subcase 2}}$. If $lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $textbf{Case 2}$.
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
$endgroup$
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
add a comment |
$begingroup$
It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
$endgroup$
1
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
1
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
add a comment |
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2 Answers
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active
oldest
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2 Answers
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$begingroup$
To study bifurcation points for
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R},
$$
we see that $x'=0$ implies that
$$
x = frac{lambda}{2}left(4a pm sqrt{2(8a^2-1)}right).
$$
$textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $xin mathbb{R}$ that will give $x'=0$.
$textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2alambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2alambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2alambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point.
$textbf{Case 3}$. If $8a^2 > 1$, then there are two subcases to consider.
$color{green}{textbf{Subcase 1}}$. If $lambdanot=0$, then there are two fixed points:
$$
x_1^* = frac{lambda}{2}left(4a - sqrt{2(8a^2-1)}right),
quad
x_2^* = frac{lambda}{2}left(4a + sqrt{2(8a^2-1)}right).
$$
We can use a similar analysis as in $textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point.
$color{green}{textbf{Subcase 2}}$. If $lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $textbf{Case 2}$.
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
$endgroup$
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
add a comment |
$begingroup$
To study bifurcation points for
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R},
$$
we see that $x'=0$ implies that
$$
x = frac{lambda}{2}left(4a pm sqrt{2(8a^2-1)}right).
$$
$textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $xin mathbb{R}$ that will give $x'=0$.
$textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2alambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2alambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2alambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point.
$textbf{Case 3}$. If $8a^2 > 1$, then there are two subcases to consider.
$color{green}{textbf{Subcase 1}}$. If $lambdanot=0$, then there are two fixed points:
$$
x_1^* = frac{lambda}{2}left(4a - sqrt{2(8a^2-1)}right),
quad
x_2^* = frac{lambda}{2}left(4a + sqrt{2(8a^2-1)}right).
$$
We can use a similar analysis as in $textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point.
$color{green}{textbf{Subcase 2}}$. If $lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $textbf{Case 2}$.
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
$endgroup$
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
add a comment |
$begingroup$
To study bifurcation points for
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R},
$$
we see that $x'=0$ implies that
$$
x = frac{lambda}{2}left(4a pm sqrt{2(8a^2-1)}right).
$$
$textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $xin mathbb{R}$ that will give $x'=0$.
$textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2alambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2alambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2alambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point.
$textbf{Case 3}$. If $8a^2 > 1$, then there are two subcases to consider.
$color{green}{textbf{Subcase 1}}$. If $lambdanot=0$, then there are two fixed points:
$$
x_1^* = frac{lambda}{2}left(4a - sqrt{2(8a^2-1)}right),
quad
x_2^* = frac{lambda}{2}left(4a + sqrt{2(8a^2-1)}right).
$$
We can use a similar analysis as in $textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point.
$color{green}{textbf{Subcase 2}}$. If $lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $textbf{Case 2}$.
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
$endgroup$
To study bifurcation points for
$$
x'=lambda^2-8alambda x+2x^2, quad ain mathbb{R},
$$
we see that $x'=0$ implies that
$$
x = frac{lambda}{2}left(4a pm sqrt{2(8a^2-1)}right).
$$
$textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $xin mathbb{R}$ that will give $x'=0$.
$textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2alambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2alambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2alambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point.
$textbf{Case 3}$. If $8a^2 > 1$, then there are two subcases to consider.
$color{green}{textbf{Subcase 1}}$. If $lambdanot=0$, then there are two fixed points:
$$
x_1^* = frac{lambda}{2}left(4a - sqrt{2(8a^2-1)}right),
quad
x_2^* = frac{lambda}{2}left(4a + sqrt{2(8a^2-1)}right).
$$
We can use a similar analysis as in $textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point.
$color{green}{textbf{Subcase 2}}$. If $lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $textbf{Case 2}$.
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
edited Jan 8 at 6:09
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
answered Jan 8 at 5:55
Mee Seong ImMee Seong Im
2,8051617
2,8051617
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
add a comment |
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
But isn't $a$ a fixed constant? I thought only $lambda$ was a bifurcation parameter and it seems that it doesn't affect how many equilibrium points there are.
$endgroup$
– LoneBone
Jan 8 at 14:27
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
$begingroup$
@LoneBone Yes, $a$ is a fixed constant. But your solution set depends on the value of $a$ as you can see from above. I'm considering all possible values of $a$ and giving a complete set of solutions.
$endgroup$
– Mee Seong Im
Jan 8 at 15:32
add a comment |
$begingroup$
It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
$endgroup$
1
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
1
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
add a comment |
$begingroup$
It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
$endgroup$
1
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
1
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
add a comment |
$begingroup$
It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
$endgroup$
It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
answered Jan 7 at 19:48
Robert LewisRobert Lewis
45.4k23065
45.4k23065
1
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
1
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
add a comment |
1
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
1
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
1
1
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
$begingroup$
Does this mean that there are no bifurcation points?
$endgroup$
– LoneBone
Jan 7 at 19:50
1
1
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
$begingroup$
Also will it make sense to consider cases for $a=const.$?
$endgroup$
– LoneBone
Jan 7 at 19:51
add a comment |
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