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- Feb 29, 2012

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Consider the system of differential equations

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

$$y' = - y.$$

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

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- Thread starter
- #1

- Feb 29, 2012

- 342

Consider the system of differential equations

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

$$y' = - y.$$

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

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- Mar 5, 2012

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What are the fixed points?

Consider the system of differential equations

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

$$y' = - y.$$

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

Are they sources or sinks?

Perhaps describe the possible solution curves by graphical inspection?

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- Feb 29, 2012

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

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Right.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.

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

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.

- Jan 29, 2012

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

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- Feb 13, 2012

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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...

Consider the system of differential equations

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

$$y' = - y.$$

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

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

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

$\chi$ $\sigma$

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If $\mu>0$ we have 2 more invariant lines: $x=\pm \sqrt\mu$.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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- Feb 29, 2012

- 342

Are you sure I don't have any more fixed points? What if we somehow translate the complex number to two real solutions?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, 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.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.

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.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

$\chi$ $\sigma$

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?

- Jan 29, 2012

- 661

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 $$\begin{bmatrix}x'\\y'\end{bmatrix}=A \begin{bmatrix}x\\y\end{bmatrix},\quad A=J_v(x_0,y_0)= \begin{bmatrix}\frac{\partial v_1}{\partial x}(x_0,y_0)&\frac{\partial v_1}{\partial y}(x_0,y_0)\\\frac{\partial v_2}{\partial x}(x_0,y_0)&\frac{\partial v_1}{\partial x}(x_0,y_0)\end{bmatrix}=\begin{bmatrix}-2x_0&0\\0&-1\end{bmatrix}$$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.

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

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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- Mar 5, 2012

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

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- #11

- Mar 5, 2012

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According to its eigenvalues, $(\sqrt{\mu},0)$ is anasymptotically stable node. (Have a look here for $\mu=1$) and $(-\sqrt{\mu},0)$ is asaddle 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:

$\mu = 0$ with 1 saddle point:

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