kixEffinsoj

Answered

2022-06-25

Solve non-linear ode system as a function of $t$.
$\left\{\begin{array}{l}\stackrel{˙}{x}=y\\ \stackrel{˙}{y}=-x+{x}^{2}=x\left(x-1\right)\end{array}$
To solve it as a function $x\left(y\right)$ or $y\left(x\right)$ is trivial, but I need the solution as a function of time: $\left(x\left(t\right),y\left(t\right)\right)$.
The system has a first integral: $H\left(x,y\right)=\frac{1}{2}\left({x}^{2}+{y}^{2}\right)-\frac{1}{3}{x}^{3}$. In particular, if no general solution is possible, I'm searching for a solution for $H=\frac{1}{6}$.

Answer & Explanation

luisjoseblash2

Expert

2022-06-26Added 16 answers

Multiplying the 2nd order equation $\stackrel{¨}{x}=x\left(x-1\right)$ by $\stackrel{˙}{x}$ on both sides and integrating with respect to time, we get a nonlinear 1st order equation for $x$:
$\stackrel{¨}{x}=x\left(x-1\right)\phantom{\rule{0ex}{0ex}}\stackrel{¨}{x}\stackrel{˙}{x}=x\left(x-1\right)\stackrel{˙}{x}\phantom{\rule{0ex}{0ex}}{\int }_{0}^{t}\stackrel{˙}{x}\left(\tau \right)\stackrel{¨}{x}\left(\tau \right)d\tau ={\int }_{0}^{t}x\left(\tau \right)\left(x\left(\tau \right)-1\right)\stackrel{˙}{x}\left(\tau \right)d\tau \phantom{\rule{0ex}{0ex}}{\int }_{{\stackrel{˙}{x}}_{0}}^{\stackrel{˙}{x}\left(t\right)}\stackrel{˙}{x}\phantom{\rule{thinmathspace}{0ex}}d\stackrel{˙}{x}={\int }_{{x}_{0}}^{x\left(t\right)}x\left(x-1\right)\phantom{\rule{thinmathspace}{0ex}}dx$
$\frac{1}{2}\left({\stackrel{˙}{x}}^{2}-{\stackrel{˙}{x}}_{0}^{2}\right)=\frac{1}{3}\left(x-1{\right)}^{3}+\frac{1}{2}\left(x-1{\right)}^{2}-\frac{1}{3}\left({x}_{0}-1{\right)}^{3}-\frac{1}{2}\left({x}_{0}-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\stackrel{˙}{x}}^{2}={\stackrel{˙}{x}}_{0}^{2}+\frac{2}{3}\left(x-1{\right)}^{3}+\left(x-1{\right)}^{2}-\frac{2}{3}\left({x}_{0}-1{\right)}^{3}-\left({x}_{0}-1{\right)}^{2}=f\left(x;{x}_{0},{\stackrel{˙}{x}}_{0}\right)$
$\frac{1}{2}\left({\stackrel{˙}{x}}^{2}-{\stackrel{˙}{x}}_{0}^{2}\right)=\frac{1}{3}\left(x-1{\right)}^{3}+\frac{1}{2}\left(x-1{\right)}^{2}-\frac{1}{3}\left({x}_{0}-1{\right)}^{3}-\frac{1}{2}\left({x}_{0}-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\stackrel{˙}{x}}^{2}={\stackrel{˙}{x}}_{0}^{2}+\frac{2}{3}\left(x-1{\right)}^{3}+\left(x-1{\right)}^{2}-\frac{2}{3}\left({x}_{0}-1{\right)}^{3}-\left({x}_{0}-1{\right)}^{2}=f\left(x;{x}_{0},{\stackrel{˙}{x}}_{0}\right)$
Once we have ${\stackrel{˙}{x}}^{2}=f\left(x;{x}_{0},{\stackrel{˙}{x}}_{0}\right)$, we can at least find an implicit equation for $x$:
$\frac{dx}{dt}=\sqrt{f\left(x;{x}_{0},{\stackrel{˙}{x}}_{0}\right)}\phantom{\rule{0ex}{0ex}}{\int }_{{x}_{0}}^{x}\frac{d\chi }{\sqrt{f\left(\chi ;{x}_{0},{\stackrel{˙}{x}}_{0}\right)}}={\int }_{0}^{t}d\tau =t.$
The final integral will in general ony be computable in terms of elliptic integrals.

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