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Recently I've started a few programming projects which involve simulating physical systems. However, more than most involve a differential equation which needs to be solved. Some examples are:

$$\frac{d^2\theta}{dt^2} = -\frac{g}{l}\sin \theta$$ $$\left(\frac{da}{dt}\times\frac{1}{a}\right)^2 = \frac{8\pi G\rho}{3} - \frac{kc^2}{a^2}$$

The technique I have been doing so far is using Wolfram Alpha to integrate the equation and then plot the result. For some of the equations I have found, WA has struggled, leaving me clueless.

I don't know a huge amount about differential equations beyond simple equations such as $\frac{dy}{dx}=2xy$, so I am unable to integrate the equations by hand.

That was why I was wondering if there is a way to find the solution on the fly, so to speak. For example, I would like to generate a graph of $\theta(t)$ in the first equation without first integrating it.

I mostly use JavaScript and Python, but am willing to use another, free, language which has the tools which will help me complete my projects.

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  • $\begingroup$ Have you looked to a free CAS like SAGE or a Matlab clone Octave or many others and using the built-in numerical DEQ solvers? $\endgroup$
    – Moo
    Aug 21, 2017 at 18:01
  • $\begingroup$ @Moo Not yet. I didn't know Octave had anything like that $\endgroup$
    – Beta Decay
    Aug 21, 2017 at 18:03
  • $\begingroup$ It has numerical solvers like Matlab and is free. $\endgroup$
    – Moo
    Aug 21, 2017 at 18:04
  • $\begingroup$ @Moo Nice, I'll have a look $\endgroup$
    – Beta Decay
    Aug 21, 2017 at 18:05
  • $\begingroup$ solving your first equation i got something with Jacobi Amplitude $\endgroup$ Aug 21, 2017 at 18:05

3 Answers 3

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ODE is your friend, available in:

In MATLAB/Octave, Others solvers ode23, ode15 are intended for specific cases of ODEs, when the default choice of ode45 is not working well, or when the problems include nonlinearities, discontinuities, etc.

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There is no difficulty writing a Runge-Kutta solver (RK4), for equations $y'=f(y,t)$ where $f$ is arbitrary. https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods

The scalar version is suited to first order ODEs.

For higher order, you turn the equation in a vector form:

$$y'''=f(y'',y',y,t)$$ becomes

$$(y_2,y_1,y_0)'=(f(y_2,y_1,y_0,t),y_2,y_1)$$ which is of the type

$$(\vec y)'=\vec g(\vec y,t).$$

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  • $\begingroup$ So I presume $y'' = f(y', y, t)$? $\endgroup$
    – Beta Decay
    Aug 21, 2017 at 21:24
  • $\begingroup$ @BetaDecay: what do you mean ? $\endgroup$
    – user65203
    Aug 21, 2017 at 21:26
  • $\begingroup$ You only show how to use the method on a first order differential equation and a third order. How about a second order? $\endgroup$
    – Beta Decay
    Aug 21, 2017 at 22:43
  • $\begingroup$ @BetaDecay: come on, the generalization to any order is obvious. $\endgroup$
    – user65203
    Aug 21, 2017 at 23:18
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With the first one $$\frac{d^{2}\theta(t)}{dt^{2}}=-g\sin{\theta(t)}$$ Multiply by $\frac{d\theta(t)}{dt}$ $$\frac{d\theta(t)}{dt}\frac{d^{2}\theta(t)}{dt^{2}}=-g\frac{d\theta(t)}{dt}\sin{\theta(t)}$$ Using the chain rule $$\frac{d}{dt}\frac{1}{2}\Big(\frac{d\theta(t)}{dt}\Big)^{2}=g\frac{d}{dt}\cos{\theta(t)}$$ Thus $$\Big(\frac{d\theta(t)}{dt}\Big)^{2}=2g\cos{\theta(t)}+c_{1}$$ Hence $$t+c_{2}=\frac{1}{\sqrt{2g}}\int^{\theta}\frac{d\theta'}{\pm\sqrt{\frac{c_{1}}+\cos{\theta'}}}=\frac{2\sqrt{\frac{c_{1}+2g\cos{\theta}}{c_{1}+2g}}F\Big(\frac{\theta}{2}\Big{|}\frac{4g}{c_{1}+1}\Big)}{\pm\sqrt{c_{1}+2g\cos{\theta}}}$$ Where $F(z|k^{2})$ is the incomplete elliptic integral of the first kind. With the second one $$\frac{da}{dt}=\pm\sqrt{\frac{8\pi{G}\rho}{3}a-\frac{kc^{2}}{a}}$$ $$t+c=\int^{a}\frac{da'}{\pm\sqrt{\frac{8\pi{G}\rho}{3}a'-\frac{kc^{2}}{a'}}}=\int^{a}\frac{\sqrt{a'}da'}{\pm\sqrt{\frac{8\pi{G}\rho}{3}(a')^{2}-kc^{2}}}$$ Let $a'=\sqrt{\frac{3kc^{2}}{8\pi{G}\rho}}\cos{\alpha}$ then $$t+c=\Big(\frac{3kc^{2}}{8\pi{G}\rho}\Big)^{3/4}\frac{1}{\mp{i}\sqrt{kc^{2}}}\int^{\sqrt{8\pi{G}\rho/2kc^{2}}\arccos{a}}\sqrt{\cos{\alpha}}d\alpha=\Big(\frac{3kc^{2}}{8\pi{G}\rho}\Big)^{3/4}\frac{1}{\mp{i}\sqrt{kc^{2}}}E\Big(\frac{1}{2}\sqrt{8\pi{G}\rho/2kc^{2}}\arccos{a}\Big{|}2\Big)$$ Where $E(z, k^{2})$ is the incomplete elliptic integral of the second kind.

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