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- Thread starter dwsmith
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- Jan 26, 2012

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The first point you mentioned is not a steady state point, as it does not satisfy $\dot{x}=0$. Solve $\dot{y}=0$ to obtain $b=ay+x^{2}y$, and hence\begin{align}

\dot{x} =& -x + ay + x^2y\notag\\

\dot{y} =& b - ay - x^2y\notag

\end{align}

The only steady states are $(-b,0)$ and $\left(0,\frac{b}{2a}\right)$, correct?

I feel like there should be more but I don't think there any.

$$y=\frac{b}{a+x^{2}}.$$

Plug that into the first. Question: do you know the sign of $a$?

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I don't know the sign of a. I don't know why I said that was a steady state. I think I concatenated part -x with the 2nd equation in my head.The first point you mentioned is not a steady state point, as it does not satisfy $\dot{x}=0$. Solve $\dot{y}=0$ to obtain $b=ay+x^{2}y$, and hence

$$y=\frac{b}{a+x^{2}}.$$

Plug that into the first. Question: do you know the sign of $a$?

$$

-x + \frac{ab}{a+x^2} + \frac{x^2b}{a+x^2} = 0

$$

If x = 0, then b = 0 or a = 0.

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- Jan 26, 2012

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Right.I don't know the sign of a. I don't know why I said that was a steady state. I think I concatenated part -x with the 2nd equation in my head.

$$

-x + \frac{ab}{a+x^2} + \frac{x^2b}{a+x^2} = 0

$$

Hmm. Not sure I would go there. Try getting a common denominator on the LHS.If x = 0, then b = 0 or a = 0.

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- #5

Then it isRight.

Hmm. Not sure I would go there. Try getting a common denominator on the LHS.

$$

\frac{-xa-x^3+ab+x^2b}{a+x^2}=0

$$

How does this help?

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- #6

- Jan 26, 2012

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Well, your denominator is irrelevant now, right? (Unless $a$ can be negative!). You have a cubic in $x$. Solve that.Then it is

$$

\frac{-xa-x^3+ab+x^2b}{a+x^2}=0

$$

How does this help?

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- #8

- Jan 26, 2012

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I'd agree with $x=b$, but you need to check the two imaginary solutions, although that may be a moot point. If $x$ and $y$ represent physical quantities, they cannot be imaginary. You could rule out those solutions on physical grounds.If a is negative, we would have 3 real solutions of x since $-x^3+x^2b-xa+ab=(x-b)(-bx^2-a)=0$.

So $x = b, \pm i\sqrt{a/b}$

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- #9

If a is negative, we would have 3 real solutions. Why couldn't a be negative? This is about glycolysis so imaginary solutions would be ruled out.I'd agree with $x=b$, but you need to check the two imaginary solutions, although that may be a moot point. If $x$ and $y$ represent physical quantities, they cannot be imaginary. You could rule out those solutions on physical grounds.

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- #10

- Jan 26, 2012

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Well, unless you know something in advance about $a$, which you've said you don't, then you can't rule out that possibility. You'd have to state your solutions in a case-by-case fashion, like so:If a is negative, we would have 3 real solutions. Why couldn't a be negative? This is about glycolysis so imaginary solutions would be ruled out.

If $a<0$, then the steady-state points are the following: (list the three points).

If $a=0$, then the steady-state points are the following: (follow this logic from the original equations).

If $a>0$, then the steady-state points are the following: (list the point).

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- #11

What is a closed orbit?Well, unless you know something in advance about $a$, which you've said you don't, then you can't rule out that possibility. You'd have to state your solutions in a case-by-case fashion, like so:

If $a<0$, then the steady-state points are the following: (list the three points).

If $a=0$, then the steady-state points are the following: (follow this logic from the original equations).

If $a>0$, then the steady-state points are the following: (list the point).

I need to find the nullclines and prove that a closed orbit exist in a and b satisfying an appropriate condition (I don't know what that condition is unless is it a condition on what a and b are?).

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- Jan 26, 2012

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I'm thinking a closed orbit is a periodic path that $\langle x(t),y(t)\rangle$ makes in the $x,y$ plane. See the fifth paragraph down at this link. How to find conditions on $a$ and $b$ such that that is the case, I don't know. Maybe Danny could help out on that one.What is a closed orbit?

I need to find the nullclines and prove that a closed orbit exist in a and b satisfying an appropriate condition (I don't know what that condition is unless is it a condition on what a and b are?).

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- #13

Suppose a < 0, then $x = \pm\sqrt{a/b}$.

$0 = -\sqrt{a/b} + ay + \frac{a}{b}y\Rightarrow y =\frac{\sqrt{a/b}}{(a+a/b)}$

$0 = b - ay -\frac{a}{b}y\Rightarrow y =\frac{b}{a+a/b}$

Since they don't agree, \sqrt{a/b} isn't a correct x for a steady state then?

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- #14

- Jan 26, 2012

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I think you should revisit post # 8.

Suppose a < 0, then $x = \pm\sqrt{a/b}$.

$0 = -\sqrt{a/b} + ay + \frac{a}{b}y\Rightarrow y =\frac{\sqrt{a/b}}{(a+a/b)}$

$0 = b - ay -\frac{a}{b}y\Rightarrow y =\frac{b}{a+a/b}$

Since they don't agree, \sqrt{a/b} isn't a correct x for a steady state then?

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- #15

But the question still stand because then I have $y = \sqrt{a}/2a$ and $y = b/2a$I think you should revisit post # 8.

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- #16

- Jan 26, 2012

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I think we really have to rule out those two "complex" solutions. If you look at the displayed equation in post # 2, you'll see why. The only thing that could relieve the situation is if $b=0$.But the question still stand because then I have $y = \sqrt{a}/2a$ and $y = b/2a$

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- #17

I did rule out the complex solutions.I think we really have to rule out those two "complex" solutions. If you look at the displayed equation in post # 2, you'll see why. The only thing that could relieve the situation is if $b=0$.

If $a > 0$, then $x = b, \pm i\sqrt{a}$.

If $a < 0$, then $x = b, \pm\sqrt{a}$.

If $a = 0$, then $x = b, 0$ where $0$ is a double root.

Since we are dealing with a glycolysis model, the complex solutions wouldn't play any part since $x$ is ADP which has to be real.

If $a > 0$, then our only steady state is $\left(b,\frac{b}{a + b^2}\right)$.

If $a < 0$, then our steady states are $\left(b,\frac{b}{a + b^2}\right)$ and $\left(\sqrt{a}, \frac{\pm\sqrt{a}}{2a}\right)$ but this is only a solution if $\pm\sqrt{a} = b$ or $b^2 = a$.

I guess this could be said then correct?

---------- Post added at 10:37 PM ---------- Previous post was at 10:27 PM ----------

If $a = 0$, then our steady states are $\left(b,\frac{b}{a + b^2}\right)$ and $\left(0,\frac{b}{a}\right)$ but this is only a steady state is $b = 0$.I did rule out the complex solutions.

If $a > 0$, then $x = b, \pm i\sqrt{a}$.

If $a < 0$, then $x = b, \pm\sqrt{a}$.

If $a = 0$, then $x = b, 0$ where $0$ is a double root.

Since we are dealing with a glycolysis model, the complex solutions wouldn't play any part since $x$ is ADP which has to be real.

If $a > 0$, then our only steady state is $\left(b,\frac{b}{a + b^2}\right)$.

If $a < 0$, then our steady states are $\left(b,\frac{b}{a + b^2}\right)$ and $\left(\sqrt{a}, \frac{\pm\sqrt{a}}{2a}\right)$ but this is only a solution if $\pm\sqrt{a} = b$ or $b^2 = a$.

I guess this could be said then correct?

I suppose then the only steady state for all cases is $\left(b,\frac{b}{a + b^2}\right)$ since the other restrict a and b to work.

If it were true that kinetic parameters could never be equal, it would eliminate the trouble with the steady states. However, I am not sure if that is an always be the case though.

---------- Post added at 10:55 PM ---------- Previous post was at 10:37 PM ----------

I guess we could evaluate two cases then. Case 1 kinetic parameters are never equal leading to only $\left(b,\frac{b}{a+b^2}\right)$.

So if I wanted to graph the null clines for case 1, I would simply plot x = b and $y = \frac{b}{a+b^2}$, correct?

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- Jan 26, 2012

- 890

For a steady state \( \dot x = \dot y = 0 \) simultaneously, so:\begin{align}

\dot{x} =& -x + ay + x^2y\notag\\

\dot{y} =& b - ay - x^2y\notag

\end{align}

The only steady states are $(-b,0)$ and $\left(0,\frac{b}{2a}\right)$, correct?

I feel like there should be more but I don't think there any.

\[ \begin{aligned} y(a+x^2)=x \\ y(a+x^2)=b \end{aligned} \]

So \( x=b \), leaving you with a linear equation for \(y\) in terms of \(a\) and \(b\)

CB

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- #19

I have these as my steady states:For a steady state \( \dot x = \dot y = 0 \) simultaneously, so:

\[ \begin{aligned} y(a+x^2)=x \\ y(a+x^2)=b \end{aligned} \]

So \( x=b \), leaving you with a linear equation for \(y\) in terms of \(a\) and \(b\)

CB

If $a > 0$, then $x = b, \pm i\sqrt{a}$.

If $a < 0$, then $x = b, \pm\sqrt{a}$.

If $a = 0$, then $x = b, 0$ where $0$ is a double root.

Since we are dealing with a glycolysis model, the complex solutions wouldn't play any part since $x$ is ADP which has to be real.

If $a > 0$, then our only steady state is $\left(b,\frac{b}{a + b^2}\right)$.

If $a < 0$, then our steady states are $\left(b,\frac{b}{a + b^2}\right)$ and $\left(\sqrt{a}, \frac{\pm\sqrt{a}}{2a}\right)$ but this is only a solution if $\pm\sqrt{a} = b$ or $b^2 = a$.

Are kinetic parameters always positive? If so, we would only have case one with one steady state.

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- Jan 26, 2012

- 890

Did you read the post you quoted?I have these as my steady states:

If $a > 0$, then $x = b, \pm i\sqrt{a}$.

If $a < 0$, then $x = b, \pm\sqrt{a}$.

If $a = 0$, then $x = b, 0$ where $0$ is a double root.

Since we are dealing with a glycolysis model, the complex solutions wouldn't play any part since $x$ is ADP which has to be real.

If $a > 0$, then our only steady state is $\left(b,\frac{b}{a + b^2}\right)$.

If $a < 0$, then our steady states are $\left(b,\frac{b}{a + b^2}\right)$ and $\left(\sqrt{a}, \frac{\pm\sqrt{a}}{2a}\right)$ but this is only a solution if $\pm\sqrt{a} = b$ or $b^2 = a$.

Are kinetic parameters always positive? If so, we would only have case one with one steady state.

CB

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- #21

Yes.Did you read the post you quoted?

CB

- Jan 26, 2012

- 890

Then you know that for this problem there are no complex solutions, and in fact the work to find the steady state is trivial once the key observation is made.Yes.

(Unless there is a typo and you are trying to solve something other than what is in the first post in this thread)

CB

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- #23

So it is only the one steady state $\left(b, \frac{b}{a+b^2}\right)$ but is it also true that kinetic parameters must be non-negative?Then you know that for this problem there are no complex solutions, and in fact the work to find the steady state is trivial once the key observation is made.

(Unless there is a typo and you are trying to solve something other than what is in the first post in this thread)

CB

- Jan 26, 2012

- 890

You need to look at the definitions of \(a\) and \(b\). Since \(a\) looks like it is the backwards rate when there is only \(y\) present, then it must be positive.So it is only the one steady state $\left(b, \frac{b}{a+b^2}\right)$ but is it also true that kinetic parameters must be non-negative?

Also, check the first post the rate equations do not look right to me. Assuming one molecule of precursor results in one molecule of product they violate the mass balance constraint that I expect to see.

CB

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- #25

Those are the correct equations.You need to look at the definitions of \(a\) and \(b\). Since \(a\) looks like it is the backwards rate when there is only \(y\) present, then it must be positive.

Also, check the first post the rate equations do not look right to me. Assuming one molecule of precursor results in one molecule of product they violate the mass balance constraint that I expect to see.

CB