MATH 308 Lecture 7

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Lecture Notes

Autonomous Equations

Of the form

$y'=f(y)\,\!$

nonlinear, always seperable

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

Example

$y'(t)=y^{2}(y^{2}-4)$

${\begin{aligned}f(k)=0&\implies k^{2}(k^{2}-4)=0k&\in \{-2,0,2\}\end{aligned}}$

Example 2

Prove that $Y(t)$ is a solution to $y(t)=y(t-t_{0})$

${\begin{aligned}{\frac {\mathrm {d} Y(t)}{\mathrm {d} t}}&={\frac {\mathrm {d} }{\mathrm {d} t}}\left(y(t-t_{0})\right)\\&={\frac {\mathrm {d} y}{\mathrm {d} t}}\left(t-t_{0}\right)\cdot {\frac {\mathrm {d} (t-t_{0})}{\mathrm {d} t}}\\&=\left(y^{2}(t-t_{0})\right)\left(y^{2}(t-t_{0})-4\right)\\&=Y(t)^{2}\left(y^{2}(t)-4\right)\\\end{aligned}}$

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

The graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t-t_0)} is a horizontal shift of the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t)} to the right (if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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?

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t)} is increasing when $y'(t)>0$. Because Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'(t) = y^2(y^2-4)} is a solution, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t)} is increasing when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| y \right| > 2} and decreasing when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| y \right| < 2 } except when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = 0}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.