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Autonomous differential equations

  • #1
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Consider the autonomous differential equation that follows

[tex]\frac{dy}{dt} = e^y - 1, \qquad -\infty < y_0 < \infty \mbox{.}[/tex]

I'm supposed to plot [tex]f(y)[/tex] versus [tex]y[/tex], and determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. Anyhow, you can find my plot at

http://mygraph.cjb.net/

which gives

[tex]y_0 < 0 \Longrightarrow \left\{ \begin{array}{ll} y^{\prime} < 0 & \mbox{(y is decreasing),} \\ y^{\prime \prime} < 0 & \mbox{(y is concave down),} \end{array} \right.[/tex]

and

[tex]y_0 > 0 \Longrightarrow \left\{ \begin{array}{ll} y^{\prime} > 0 & \mbox{(y is increasing),} \\ y^{\prime \prime} > 0 & \mbox{(y is concave up).} \end{array} \right.[/tex]

However, I picture the critical point [tex]y = \phi (t) = 0[/tex] to be semistable. The book says it is unstable, but I really can't find my mistake.

Any help is highly appreciated.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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thiago_j said:
Consider the autonomous differential equation that follows

[tex]\frac{dy}{dt} = e^y - 1, \qquad -\infty < y_0 < \infty \mbox{.}[/tex]



[tex]y_0 < 0 \Longrightarrow \left\{ \begin{array}{ll} y^{\prime} < 0 & \mbox{(y is decreasing),} \\ y^{\prime \prime} < 0 & \mbox{(y is concave down),} \end{array} \right.[/tex]

and

[tex]y_0 > 0 \Longrightarrow \left\{ \begin{array}{ll} y^{\prime} > 0 & \mbox{(y is increasing),} \\ y^{\prime \prime} > 0 & \mbox{(y is concave up).} \end{array} \right.[/tex]

However, I picture the critical point [tex]y = \phi (t) = 0[/tex] to be semistable. The book says it is unstable, but I really can't find my mistake.
When a negative number decreases its absolute value increases so it goes away further from the origin.

ehild
 
  • #3
372
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Absolutely! I see it now. Thanks
 

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