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#### dwsmith

##### Well-known member
\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.

#### Ackbach

##### Indicium Physicus
Staff member
\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.
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 $a$?

• dwsmith

#### dwsmith

##### Well-known member
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 $a$?
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.

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

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

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#### Ackbach

##### Indicium Physicus
Staff member
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.

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

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

#### dwsmith

##### Well-known member
Right.

Hmm. Not sure I would go there. Try getting a common denominator on the LHS.
Then it is
$$\frac{-xa-x^3+ab+x^2b}{a+x^2}=0$$

How does this help?

#### Ackbach

##### Indicium Physicus
Staff member
Then it is
$$\frac{-xa-x^3+ab+x^2b}{a+x^2}=0$$

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

• dwsmith

#### dwsmith

##### Well-known member
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 $x = b, \pm i\sqrt{a/b}$

#### Ackbach

##### Indicium Physicus
Staff member
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 $x = b, \pm i\sqrt{a/b}$
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.

#### dwsmith

##### Well-known member
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. Why couldn't a be negative? This is about glycolysis so imaginary solutions would be ruled out.

#### Ackbach

##### Indicium Physicus
Staff member
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.
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).

• dwsmith

#### dwsmith

##### Well-known member
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).
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?).

#### Ackbach

##### Indicium Physicus
Staff member
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?).
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.

• dwsmith

#### dwsmith

##### Well-known member
For the steady states, as I pick the x for each respective a cases, both equations would have to agree correct? As an example,

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?

#### Ackbach

##### Indicium Physicus
Staff member
For the steady states, as I pick the x for each respective a cases, both equations would have to agree correct? As an example,

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?
I think you should revisit post # 8.

• dwsmith

#### Ackbach

##### Indicium Physicus
Staff member
But the question still stand because then I have $y = \sqrt{a}/2a$ and $y = b/2a$
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 $b=0$.

#### dwsmith

##### Well-known member
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 $b=0$.
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?

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

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?
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 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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#### CaptainBlack

##### Well-known member
\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.
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

• dwsmith

#### dwsmith

##### Well-known member
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
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.

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#### CaptainBlack

##### Well-known member
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.
Did you read the post you quoted?

CB

#### CaptainBlack

##### Well-known member
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

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• dwsmith

#### dwsmith

##### Well-known member
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
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?

#### CaptainBlack

##### Well-known member
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?
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

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#### dwsmith

##### Well-known member
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
Those are the correct equations.