kixEffinsoj

Answered

2022-06-25

Solve non-linear ode system as a function of $t$.

$\{\begin{array}{l}\dot{x}=y\\ \dot{y}=-x+{x}^{2}=x(x-1)\end{array}$

To solve it as a function $x(y)$ or $y(x)$ is trivial, but I need the solution as a function of time: $(x(t),y(t))$.

The system has a first integral: $H(x,y)=\frac{1}{2}({x}^{2}+{y}^{2})-\frac{1}{3}{x}^{3}$. In particular, if no general solution is possible, I'm searching for a solution for $H=\frac{1}{6}$.

$\{\begin{array}{l}\dot{x}=y\\ \dot{y}=-x+{x}^{2}=x(x-1)\end{array}$

To solve it as a function $x(y)$ or $y(x)$ is trivial, but I need the solution as a function of time: $(x(t),y(t))$.

The system has a first integral: $H(x,y)=\frac{1}{2}({x}^{2}+{y}^{2})-\frac{1}{3}{x}^{3}$. 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 $\ddot{x}=x(x-1)$ by $\dot{x}$ on both sides and integrating with respect to time, we get a nonlinear 1st order equation for $x$:

$\ddot{x}=x(x-1)\phantom{\rule{0ex}{0ex}}\ddot{x}\dot{x}=x(x-1)\dot{x}\phantom{\rule{0ex}{0ex}}{\int}_{0}^{t}\dot{x}(\tau )\ddot{x}(\tau )d\tau ={\int}_{0}^{t}x(\tau )(x(\tau )-1)\dot{x}(\tau )d\tau \phantom{\rule{0ex}{0ex}}{\int}_{{\dot{x}}_{0}}^{\dot{x}(t)}\dot{x}\phantom{\rule{thinmathspace}{0ex}}d\dot{x}={\int}_{{x}_{0}}^{x(t)}x(x-1)\phantom{\rule{thinmathspace}{0ex}}dx$

$\frac{1}{2}({\dot{x}}^{2}-{\dot{x}}_{0}^{2})=\frac{1}{3}(x-1{)}^{3}+\frac{1}{2}(x-1{)}^{2}-\frac{1}{3}({x}_{0}-1{)}^{3}-\frac{1}{2}({x}_{0}-1{)}^{2}\phantom{\rule{0ex}{0ex}}{\dot{x}}^{2}={\dot{x}}_{0}^{2}+\frac{2}{3}(x-1{)}^{3}+(x-1{)}^{2}-\frac{2}{3}({x}_{0}-1{)}^{3}-({x}_{0}-1{)}^{2}=f(x;{x}_{0},{\dot{x}}_{0})$

$\frac{1}{2}({\dot{x}}^{2}-{\dot{x}}_{0}^{2})=\frac{1}{3}(x-1{)}^{3}+\frac{1}{2}(x-1{)}^{2}-\frac{1}{3}({x}_{0}-1{)}^{3}-\frac{1}{2}({x}_{0}-1{)}^{2}\phantom{\rule{0ex}{0ex}}{\dot{x}}^{2}={\dot{x}}_{0}^{2}+\frac{2}{3}(x-1{)}^{3}+(x-1{)}^{2}-\frac{2}{3}({x}_{0}-1{)}^{3}-({x}_{0}-1{)}^{2}=f(x;{x}_{0},{\dot{x}}_{0})$

Once we have ${\dot{x}}^{2}=f(x;{x}_{0},{\dot{x}}_{0})$, we can at least find an implicit equation for $x$:

$\frac{dx}{dt}=\sqrt{f(x;{x}_{0},{\dot{x}}_{0})}\phantom{\rule{0ex}{0ex}}{\int}_{{x}_{0}}^{x}\frac{d\chi}{\sqrt{f(\chi ;{x}_{0},{\dot{x}}_{0})}}={\int}_{0}^{t}d\tau =t.$

The final integral will in general ony be computable in terms of elliptic integrals.

$\ddot{x}=x(x-1)\phantom{\rule{0ex}{0ex}}\ddot{x}\dot{x}=x(x-1)\dot{x}\phantom{\rule{0ex}{0ex}}{\int}_{0}^{t}\dot{x}(\tau )\ddot{x}(\tau )d\tau ={\int}_{0}^{t}x(\tau )(x(\tau )-1)\dot{x}(\tau )d\tau \phantom{\rule{0ex}{0ex}}{\int}_{{\dot{x}}_{0}}^{\dot{x}(t)}\dot{x}\phantom{\rule{thinmathspace}{0ex}}d\dot{x}={\int}_{{x}_{0}}^{x(t)}x(x-1)\phantom{\rule{thinmathspace}{0ex}}dx$

$\frac{1}{2}({\dot{x}}^{2}-{\dot{x}}_{0}^{2})=\frac{1}{3}(x-1{)}^{3}+\frac{1}{2}(x-1{)}^{2}-\frac{1}{3}({x}_{0}-1{)}^{3}-\frac{1}{2}({x}_{0}-1{)}^{2}\phantom{\rule{0ex}{0ex}}{\dot{x}}^{2}={\dot{x}}_{0}^{2}+\frac{2}{3}(x-1{)}^{3}+(x-1{)}^{2}-\frac{2}{3}({x}_{0}-1{)}^{3}-({x}_{0}-1{)}^{2}=f(x;{x}_{0},{\dot{x}}_{0})$

$\frac{1}{2}({\dot{x}}^{2}-{\dot{x}}_{0}^{2})=\frac{1}{3}(x-1{)}^{3}+\frac{1}{2}(x-1{)}^{2}-\frac{1}{3}({x}_{0}-1{)}^{3}-\frac{1}{2}({x}_{0}-1{)}^{2}\phantom{\rule{0ex}{0ex}}{\dot{x}}^{2}={\dot{x}}_{0}^{2}+\frac{2}{3}(x-1{)}^{3}+(x-1{)}^{2}-\frac{2}{3}({x}_{0}-1{)}^{3}-({x}_{0}-1{)}^{2}=f(x;{x}_{0},{\dot{x}}_{0})$

Once we have ${\dot{x}}^{2}=f(x;{x}_{0},{\dot{x}}_{0})$, we can at least find an implicit equation for $x$:

$\frac{dx}{dt}=\sqrt{f(x;{x}_{0},{\dot{x}}_{0})}\phantom{\rule{0ex}{0ex}}{\int}_{{x}_{0}}^{x}\frac{d\chi}{\sqrt{f(\chi ;{x}_{0},{\dot{x}}_{0})}}={\int}_{0}^{t}d\tau =t.$

The final integral will in general ony be computable in terms of elliptic integrals.

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