MATH 308 Lecture 7

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Lecture Notes

Autonomous Equations

Of the form

$y^{\prime }=f(y)$

nonlinear, always seperable

if $y(t)=k$ is a solution, find $k$ (Constant solutions)

Example

$y^{\prime }(t)=y^2(y^2-4)$

$\begin{array}{rlr}f(k)=0& ⟹k^2(k^2-4)=0k& \in \{-2,0,2\}\end{array}$

Example 2

Prove that $Y(t)$ is a solution to $y(t)=y(t-t_0)$

$\begin{array}{rl}\frac{\mathrm{d}Y(t)}{\mathrm{d}t}& =\frac{\mathrm{d}}{\mathrm{d}t}(y(t-t_0))\\ & =\frac{\mathrm{d}y}{\mathrm{d}t}(t-t_0)\cdot \frac{\mathrm{d}(t-t_0)}{\mathrm{d}t}\\ & =(y^2(t-t_0))(y^2(t-t_0)-4)\\ & =Y(t{)}^2(y^2(t)-4)\\ \end{array}$

Note that $\frac{\mathrm{d}y(T)}{\mathrm{d}t}=y^2(T)(y^2(T)-4)$ for any $T$

The graph of $y(t-t_0)$ is a horizontal shift of the graph of $y(t)$ to the right (if $t_0$ is positive).

From yesterday's theorem, any shifted solution graph cannot intersect with any other shifted solution graph.

Can we find where solutions are increasing/decreasing?

$y(t)$ is increasing when $y^{\prime }(t)>0$. Because $y^{\prime }(t)=y^2(y^2-4)$ is a solution, $y(t)$ is increasing when $\left|y\right|>2$ and decreasing when $\left|y\right|<2$ except when $y=0$

If $y\in (0,2)$, we know that $y(t)$ is decreasing, and since it cannot cross the equilibrium solution $y(t)=0$, the limit is finite.