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Qualitative theory of two-dimensional system

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I am having trouble to work this question. It was asked in my exam in ordinary differential equations and I have no idea how to solve it.

Consider the system of differential equations

$$x' = \mu - x^2,$$
$$y' = - y.$$

Describe the qualitative behavior according to the value of $\mu$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,905
I am having trouble to work this question. It was asked in my exam in ordinary differential equations and I have no idea how to solve it.

Consider the system of differential equations

$$x' = \mu - x^2,$$
$$y' = - y.$$

Describe the qualitative behavior according to the value of $\mu$.
What are the fixed points?
Are they sources or sinks?
Perhaps describe the possible solution curves by graphical inspection?
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Fixed points happen at $(\pm \sqrt{\mu}, 0)$, but here is when the problems arise. If $\mu >0$ then we have one source and one sink, depending on whether it has positive and negative sign, respectively.

On the other hand, if $\mu \leq 0$ then I have complex eigenvalues with zero real part. How to go on about it?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,905
Fixed points happen at $(\pm \sqrt{\mu}, 0)$, but here is when the problems arise. If $\mu >0$ then we have one source and one sink, depending on whether it has positive and negative sign, respectively.
Right.

On the other hand, if $\mu \leq 0$ then I have complex eigenvalues with zero real part. How to go on about it?
Then you don't have fixed points.


For a qualitative analysis I would tend to draw the solution curves for both $\mu > 0$ and $\mu < 0$, which would complete the qualitative analysis as far as I'm concerned.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
On the other hand, if $\mu \leq 0$ then I have complex eigenvalues with zero real part. How to go on about it?
If $\mu<0$ we can easily identify the vector field, $v(x,y)=(-,-)$ if $y>0$, $v(x,y)=(-,+)$ if $y<0$. Besides $v(x,0)=(-,0)$ for all $x\in\mathbb{R}$, so the $x$-axis is an invariant line. With this, you'll easily plot the phases plane.
 
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chisigma

Well-known member
Feb 13, 2012
1,704
I am having trouble to work this question. It was asked in my exam in ordinary differential equations and I have no idea how to solve it.

Consider the system of differential equations

$$x' = \mu - x^2,$$
$$y' = - y.$$

Describe the qualitative behavior according to the value of $\mu$.
The two ODE are independent [one in x and one in y...] and, because $\mu$ is in the first, we focalize our attention on the ODE...

$\displaystyle x^{\ '} = \mu - x^{2}\ (1)$

It is a symple separate variables ODE, so that standard approach my be used. Supposing $\mu \ne 0$ [the case $\mu=0$ is to be treated separately...] You have...

$\displaystyle \frac{dx}{\mu - x^{2}} = d t \implies \frac{\tanh^{-1} (\frac{x}{\sqrt{\mu}})}{\sqrt{\mu}} = t + c \implies x = \sqrt{\mu}\ \tanh(\sqrt{\mu}\ t + c)\ (2)$

Very well!... now Your next step are...

a) find, using standard formulas, what is $\tanh(\sqrt{\mu}\ t + c)$...

b) analyse separately the cases $\mu>0$ and $\mu < 0$ [in the last case remember that is $\tanh i\ t = i\ \tan t$]...

c) analyse the case $\mu=0$...

Kind regards

$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,905
If $\mu<0$ we can easily identify the vector field, $v(x,y)=(-,-)$ if $y>0$, $v(x,y)=(-,+)$ if $y<0$. Besides $v(x,0)=(-,0)$ for all $x\in\mathbb{R}$, so the $x$-axis is an invariant line. With this, you'll easily plot the phases plane.
If $\mu>0$ we have 2 more invariant lines: $x=\pm \sqrt\mu$.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Then you don't have fixed points.

For a qualitative analysis I would tend to draw the solution curves for both $\mu > 0$ and $\mu < 0$, which would complete the qualitative analysis as far as I'm concerned.
Are you sure I don't have any more fixed points? What if we somehow translate the complex number to two real solutions?

If $\mu<0$ we can easily identify the vector field, $v(x,y)=(-,-)$ if $y>0$, $v(x,y)=(-,+)$ if $y<0$. Besides $v(x,0)=(-,0)$ for all $x\in\mathbb{R}$, so the $x$-axis is an invariant line. With this, you'll easily plot the phases plane.
Fernando, by $v(x,y) = (-,-)$ do you mean that it is a sink? Sorry, I don't understand what you want to convey with this notation. :confused:

The two ODE are independent [one in x and one in y...] and, because $\mu$ is in the first, we focalize our attention on the ODE...

$\displaystyle x^{\ '} = \mu - x^{2}\ (1)$

It is a symple separate variables ODE, so that standard approach my be used. Supposing $\mu \ne 0$ [the case $\mu=0$ is to be treated separately...] You have...

$\displaystyle \frac{dx}{\mu - x^{2}} = d t \implies \frac{\tanh^{-1} (\frac{x}{\sqrt{\mu}})}{\sqrt{\mu}} = t + c \implies x = \sqrt{\mu}\ \tanh(\sqrt{\mu}\ t + c)\ (2)$

Very well!... now Your next step are...

a) find, using standard formulas, what is $\tanh(\sqrt{\mu}\ t + c)$...

b) analyse separately the cases $\mu>0$ and $\mu < 0$ [in the last case remember that is $\tanh i\ t = i\ \tan t$]...

c) analyse the case $\mu=0$...

Kind regards

$\chi$ $\sigma$
Thank you $\chi$ $\sigma$, but it was written at the exam that we weren't required to solve for the equations, he said there was a simple way to do this. Furthermore, the original question said $x' = \mu + (-1)^{\text{academic number}} x^2$, which for some meant $x' = \mu + x^2$ and for others $x' = \mu - x^2$. He also explicitly stated that it did not matter much the sign, as long as the analysis was correct, hence why I chose the minus sign.

At the exam we didn't have access to the Hartman-Grobman theorem, but now we do. How could be such analysis carried out then?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Fernando, by $v(x,y) = (-,-)$ do you mean that it is a sink? Sorry, I don't understand what you want to convey with this notation.
By $v(x,y)=(-\mu-x^2,-y)$ we denote the vector field associated to the system. The linearized system associated to an equilibrium point $(x_0,y_0)$ is $$\begin{bmatrix}x'\\y'\end{bmatrix}=A \begin{bmatrix}x\\y\end{bmatrix},\quad A=J_v(x_0,y_0)= \begin{bmatrix}\frac{\partial v_1}{\partial x}(x_0,y_0)&\frac{\partial v_1}{\partial y}(x_0,y_0)\\\frac{\partial v_2}{\partial x}(x_0,y_0)&\frac{\partial v_1}{\partial x}(x_0,y_0)\end{bmatrix}=\begin{bmatrix}-2x_0&0\\0&-1\end{bmatrix}$$
For $\mu >0$ we have two equilibrium points $(\pm\sqrt{\mu},0)$ so,
$$J_v(\sqrt{\mu},0)=\begin{bmatrix}-2\sqrt{\mu}&0\\0&-1\end{bmatrix},\quad J_v(\sqrt{\mu},0)=\begin{bmatrix}2\sqrt{\mu}&0\\0&-1\end{bmatrix}$$
According to its eigenvalues, $(\sqrt{\mu},0)$ is an asymptotically stable node. (Have a look here for $\mu=1$) and $(-\sqrt{\mu},0)$ is a saddle point. (Have a look here for $\mu=1$). Also, have a look here to see the the complete phases plane

For $\mu=0$ we have only the equilibrium point $(0,0)$ and according to the eigenvalues of $J_v(0,0)$ we have an inconlusive case. But drawing the vector field we get its phases plane.

For $\mu<0$ we have no equilibrium points, but again, drawing the vector field we get its phases plane.
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,905
Are you sure I don't have any more fixed points? What if we somehow translate the complex number to two real solutions?
If x and y are real valued functions (which is what I assumed), an imaginary fixed point means that there is no fixed point (in real coordinates).

If x and y are complex valued functions, for $\mu < 0$ you get 2 fixed points at the complex coordinates $(\pm i \sqrt{-\mu},0)$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,905
According to its eigenvalues, $(\sqrt{\mu},0)$ is an asymptotically stable node. (Have a look here for $\mu=1$) and $(-\sqrt{\mu},0)$ is a saddle point. (Have a look here for $\mu=1$). Also, have a look here to see the the complete phases plane

For $\mu=0$ we have only the equilibrium point $(0,0)$ and according to the eigenvalues of $J_v(0,0)$ we have an inconlusive case. But drawing the vector field we get its phases plane.

For $\mu<0$ we have no equilibrium points, but again, drawing the vector field we get its phases plane.

Thanks for that!
I was trying to figure out how to make vector plots with W|A... and there they are! ;)

Allow me to present them as graphs.

$\mu < 0$ with no fixed points:
ode_mu_negative.png

$\mu = 0$ with 1 saddle point:
ode_mu_zero.png

$\mu > 0$ with 1 saddle point and 1 sink:
ode_mu_positive.png