If you visit the page I have just created, you may find the simulation of a mechanical system.

It is currently $$\mathcal H=p_1^2+p_2^2-\cos q_1-\cos q_2- \cos\!\left(q_1-q_2\right)$$ depicting two pendulum coupled with a spring (with $l_1=l_2$ and $m_1=m_2$, and the original length of the spring is zero, and the two hanging points overlap),

which is a classical example of non-linearly coupled system.

The pattern of the oscillation can be analyzed using discrete Fourier transformation, whose result can be found by clicking the buttons in the up-left corner (after the simulator has detected a period).

Hitting the space bar can make the simulation pause.

If you want to use it to simulate other mechanical systems, you can study the codes I wrote and write your own codes in the console.

By the way, the OpenRGSS version of the simulator is open-source here. Please star the repo if you like it.

To utilize the canonical equations $$\frac{\mathrm d\mathbf q}{\mathrm dt}= \frac{\partial\mathcal H}{\partial\mathbf p},\quad \frac{\mathrm d\mathbf p}{\mathrm dt}= -\frac{\partial\mathcal H}{\partial\mathbf p},$$ $$(1)$$ we need to calculate the partial derivatives of $\mathcal H$. Here is a simple code to calculate partial derivatives.

          
            def div x0, dx, f

            	f0 = f.(x0)

            	n = x0.size

            	Array.new n do |i|

            		(f.(x0 + Vector.basis(n, i) * dx) - f0) / dx

            	end

            end

          
        

(RGSS do not have matrix.rb, you can copy one from the attached file below.) Here x0 is a Vector, f is a call-able object as a function of vectors, dx is a small scalar which we are going to take 1e-6.

Let x = Vector[*q, *p], and then Formula 1 has the form $$\frac{\mathrm d\mathbf x}{\mathrm dt}=f\!\left(\mathbf x\right).$$ To solve this equation numerically, we need to use a famous method called the (explicit) Runge–Kutta method.

          
            def runge_kutta initial, max_t, dt, (*pyramid, coefs), func

            	(0..max_t).step(dt).reduce initial do |ret, t|

            		$canvas.trace t, ret if $canvas

            		coefs.zip(pyramid).each_with_object([]).sum do |(coef, row), ary|

            			coef * ary.push(func.(t, row.inner(ary) * dt + ret)).last

            		end * dt + ret#p(ret)

            	end

            end

          
        

Note that Runge–Kutta is a family of methods. The argument (*pyramid, coefs) takes one of the following, each of which is a single Runge–Kutta method.

          
            FORWARD_EULER = [[],[1]]

            EXPLICIT_MIDPOINT = [[],[1/2.0],[0,1]]

            HEUN = [[],[1],[1/2.0,1/2.0]]

            RALSTON = [[],[2/3.0],[1/4.0,3/4.0]]

            KUTTA_3RD = [[],[1/2.0],[-1,2],[1/6.0,2/3.0,1/6.0]]

            HEUN_3RD = [[],[1/3.0],[0,2/3.0],[1/4.0,0,3/4.0]]

            RALSTON_3RD = [[],[1/2.0],[0,3/4.0],[2/9.0,1/3.0,4/9.0]]

            SSPRK3 = [[],[1],[1/4.0,1/4.0],[1/6.0,1/6.0,2/3.0]]

            CLASSIC_4TH = [[],[1/2.0],[0,1/2.0],[0,0,1],[1/6.0,1/3.0,1/3.0,1/6.0]]

            RALSTON_4TH = [[],[0.4],[0.29697761,0.15875964],[0.21810040,-3.05096516,

            		3.83286476],[0.17476028, -0.55148066, 1.20553560, 0.17118478]]

            THREE_EIGHTH_4TH = [[],[1/3.0],[-1/3.0,1],[1,-1,1],[1/8.0,3/8.0,3/8.0,1/8.0]]

          
        

Here we are going to take CLASSIC_4TH.

The $canvas appearing here is an object that is going to draw the result onto the screen.

Here we also need to have some patches to get it work.

          
            class Float

            	alias ulysses20200426121236_add +

            	def + other

            		if zero? && [Vector, Matrix].any? { |c| other.is_a? c }

            			other

            		else

            			ulysses20200426121236_add other

            		end

            	end

            end

            module Enumerable

            	def sum init = 0, &block

            		(block ? map(&block) : self).reduce init, :+

            	end

            end

            class Array

            	def inner other

            		zip(other).sum { |a, b| a * b }

            	end

            end

          
        

(Again, note that this is Ruby 1.9.2.)

Finally, just combine them up, and we can solve a Hamiltonian numerically.

          
            def solve_hamiltonian n, qp0, max_t, dt, hamiltonian

            	runge_kutta qp0, max_t, dt, CLASSIC_4TH, ->t, qp do

            		dqpdt = div qp, 1e-6, ->x { hamiltonian.(t, x) }

            		Vector[*dqpdt[n...n*2], *dqpdt[0...n].map(&:-@)]

            	end

            end

          
        

For example, let’s simulate a double pendulum.

          
            solve_hamiltonian 2,Vector[PI/2,0.0,0.0,0.0],Float::INFINITY,1e-3,

            		->t,(q1,q2,p1,p2){p1**2+p2**2/2+cos(q1-q2)*p1*p2-cos(q1)-cos(q2)}

          
        

The codes are not complete in this post. See the attached file for details. You can open the project using RPG Maker VX Ace. The Game.exe file is not the official Game.exe executable but the third-party improved version of it called RGD (of version 1.3.2, while the latest till now is 1.5.1).

In Hamiltonian mechanics, if we let $$\mathbf c\coloneqq\alpha\mathbf q+\mathrm i\beta\mathbf p,$$ $$(1)$$ where $\alpha$ and $\beta$ are non-zero real numbers, then two real vectors $\mathbf q$ and $\mathbf p$ becomes a complex vector $\mathbf c$. In other words, we use $s$ complex numbers instead of $2s$ real numbers to represent the status of a system, where $s$ is the DOF.

We are going to find out the form of some theorems in Hamiltonian mechanics with respect to the complex variable that we have just defined.

Note that if you do not care about the units, it is recommended to let $\alpha=\beta=\frac1{\sqrt2}$ due to the convenience.

Hamiltonian

In this way, the Hamiltonian is a function of real value with respect to a complex vector. To be clear, $$\mathcal H\!\left(\mathbf c,t\right)= \mathcal H\!\left(\mathbf q,\mathbf p,t\right).$$ The Hamiltonian is not an analytical function, so we need to redefine how a function can be differentiated in order that the Hamiltonian is “differentiable.” The approach to this is to use the average of the limit along the real axis and that along the imaginary axis, which means $$\frac{\mathrm d}{\mathrm d\left(x+\mathrm iy\right)}\coloneqq \frac12\left(\frac\partial{\partial x}- \mathrm i\frac\partial{\partial y}\right).$$ Thus, $$\frac\partial{\partial\mathbf c}= \frac1{2\alpha}\frac\partial{\partial\mathbf q}- \frac{\mathrm i}{2\beta}\frac\partial{\partial\mathbf p}.$$ $$(2)$$ There is also an obvious property that for any function $f:\mathbb C\rightarrow\mathbb R$, we have $$\frac{\partial f}{\partial z^*}= \left(\frac{\partial f}{\partial z}\right)^*.$$ Furthermore, $$\frac\partial{\partial\mathbf q}= \alpha\left(\frac\partial{\partial\mathbf c}+ \frac\partial{\partial\mathbf c^*}\right),\quad \frac\partial{\partial\mathbf p}= \mathrm i\beta\left(\frac\partial{\partial\mathbf c}- \frac\partial{\partial\mathbf c^*}\right).$$ $$(3)$$

Canonical equations

Now we may be curious about what will the canonical equations $$\frac{\mathrm d\mathbf q}{\mathrm dt}= \frac{\partial\mathcal H}{\partial\mathbf p},\quad \frac{\mathrm d\mathbf p}{\mathrm dt}= -\frac{\partial\mathcal H}{\partial\mathbf q}$$ $$(4)$$ change into after $\mathbf c$ is introduced.

Apply Formula 2 to $2\mathrm i\alpha\beta\mathcal H$, and we can derive that $$2\mathrm i\alpha\beta\frac{\partial\mathcal H}{\partial\mathbf c}= \alpha\frac{\partial\mathcal H}{\partial\mathbf p}+ \mathrm i\beta\frac{\partial\mathcal H}{\partial\mathbf q}.$$ $$(5)$$ On the other hand, take the derivative of both sides of Formula 1, and substitute Formula 4 into it, and then we can derive that $$\frac{\mathrm d\mathbf c}{\mathrm dt}= \alpha\frac{\partial\mathcal H}{\partial\mathbf p}- \mathrm i\beta\frac{\partial\mathcal H}{\partial\mathbf q}.$$ $$(6)$$ Compare Formula 5 and 6, we get the useful formula $$\frac{\mathrm d\mathbf c}{\mathrm dt}= -2\mathrm i\alpha\beta \frac{\partial\mathcal H}{\partial\mathbf c^*}.$$ $$(7)$$ The new canonical equations are a set of $s$ ODEs of $1$st degree, and there should be only $s$ (instead of $2s$) arbitrary constants in the solution.

Poisson bracket

The Poisson bracket $\left\{\cdot,\cdot\right\}$ can be defined just as usual: $$\left\{f,g\right\}\coloneqq \frac{\partial f}{\partial\mathbf q}\cdot \frac{\partial g}{\partial\mathbf p}- \frac{\partial f}{\partial\mathbf p}\cdot \frac{\partial g}{\partial\mathbf q};$$ $$(8)$$ while something beautiful will occur if we substitute Formula 3 into 8: $$\left\{f,g\right\}=-2\mathrm i\alpha\beta \left(\frac{\partial f}{\partial\mathbf c}\cdot \frac{\partial g}{\partial\mathbf c^*}- \frac{\partial f}{\partial\mathbf c^*}\cdot \frac{\partial g}{\partial\mathbf c}\right).$$ $$(9)$$ With Formula 9, you can also verify that $$\frac{\mathrm d}{\mathrm dt}= \frac\partial{\partial t}-\left\{\mathcal H,\cdot\right\}.$$ $$(10)$$

Canonical transformation

Consider some kind of transformation $\mathbf c'=\mathbf c'\!\left(\mathbf c\right)$ that will preserve the form of the canonical equation, which means $$\frac{\mathrm d\mathbf c'}{\mathrm dt}= -2\mathrm i\alpha\beta \frac{\partial\mathcal H}{\partial\mathbf c'^*}.$$ $$(11)$$ (We do not consider those transformations that involves $t$. As we all know, if a canonical transformation involves $t$, an additional part should be added to $\mathcal H$.)

Apply Formula 10 to $\mathbf c'$ and make use of Formula 11, we can derive that $$\frac{\partial\mathcal H}{\partial\mathbf c'^*}= \left\{\mathbf c',\mathcal H\right\}.$$ The equation should be true for all $\mathcal H$. In other words, $$\frac\partial{\partial\mathbf c'^*}= \left\{\mathbf c',\cdot\right\},$$ so $$\frac{\partial\mathbf c^*}{\partial\mathbf c'^*} \frac\partial{\partial\mathbf c^*}+ \frac{\partial\mathbf c}{\partial\mathbf c'^*} \frac\partial{\partial\mathbf c}= \frac{\partial\mathbf c'}{\partial\mathbf c} \frac\partial{\partial\mathbf c^*}- \frac{\partial\mathbf c'}{\partial\mathbf c^*} \frac\partial{\partial\mathbf c}.$$ Note that usually $\frac\partial{\partial\mathbf c}$ and $\frac\partial{\partial\mathbf c^*}$ are linearly independent, so we can derive that $$\frac{\partial\mathbf c^*}{\partial\mathbf c'^*}= \frac{\partial\mathbf c'}{\partial\mathbf c},\quad \frac{\partial\mathbf c}{\partial\mathbf c'^*}= -\frac{\partial\mathbf c'}{\partial\mathbf c^*}.$$ Here it is a much more convenient way to judge whether a transformation is a canonical transformation than to find a generating function.

With the property we have just found, it is easy to find out that the Poisson brackets corresponding to different set of canonical variables have the same value, which is to say that $$\left\{f,g\right\}_{\mathbf c}=\left\{f,g\right\}_{\mathbf c'}.$$

Phase space

With the introduction of the complex variable $\mathbf c$, the phase space becomes the vector space of $\mathbf c$, which is $\mathbb C^s$.

I have not learned differential geometry about complex manifold, so maybe there is some awesome extensions that can be made in the phase space, while I will not be able to find them out…

Some examples

Free particle

The Hamiltonian of a free particle is $$\mathcal H=\frac{\left(\Im\mathbf c\right)^2}{2m},$$ where $\mathbf c$ is a $3$-dimensional complex vector, $\alpha=\beta=1$. Substitute it into 7, and then we can derive that $$\frac{\mathrm d\mathbf c}{\mathrm dt}= \frac{\Im\mathbf c}m.$$ By solving it, we can derive that $$\mathbf c=\frac{\Im\mathbf c_0}mt+\mathbf c_0,$$ where $\mathbf c_0$ is $3$ arbitrary complex constants.

Harmonic oscillator

The Hamiltonian of a harmonic oscillator is $$\mathcal H=\frac{\left|c\right|^2}2,$$ where $c$ is a complex number, $\alpha=\sqrt k$, $\beta=\frac1{\sqrt m}$. Substitute it into 7, and then we can derive that $$\frac{\mathrm dc}{\mathrm dt}=-\mathrm i\omega c,$$ where $\omega\coloneqq\alpha\beta=\sqrt{\frac km}$. By solving it, we can derive that $$c=c_0\mathrm e^{-\mathrm i\omega t},$$ where $c_0$ is an arbitrary complex constant.