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If a particle of mass $m$ moves in the $x$-$y$ plane, then its equations of motion are

$$m\frac{d^2x}{dt^2}=f(t,x,y)\space \space \text{and} \space \space m\frac{d^2y}{dt^2}=g(t,x,y).$$

Here $f$ and $g$ represent the $x$ and $y$ components, respectively, of the force acting on the particle. Replace this system of two second-order equations by an equivalent system of four first order equations of the form:

$$y_1'=f_1(x,y_1,...,y_n)$$

$$y_2'=f_2(x,y_1,...,y_n)$$

$$y_n'=f_n(x,y_1,...,y_n)$$


I understand how replace a differential equation by an equivalent system of first order equations when the differential looks something like

$$xy''-x^2y'-x^3y=0$$

An equivalent system is:

$$y_0'=y_1$$ $$y_1'=x^2y_0+xy_1$$


Therefore, I need someone to point me in the right direction for my question stated at the top.

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2 Answers 2

It's the same thing, you just state $x'(t) = r(t)$, and $y'(t) = s(t)$ and you'll have the system \begin{align} x'(t) &= r(t)\\ y'(t) &= s(t)\\ r'(t) &= \frac{1}{m}f(t,x,y)\\ s'(t) &= \frac{1}{m}g(t,x,y) \end{align}

Note that if $f=f(t,x,y,x',y')$ then the substitution would lead to $r' = \frac{1}{m}f(t,x,y,r,s)$; the same thing for $g$.


On a side note, if you take $\dot{x_i}(t) = \frac{1}{m_i} \dot{p_i}(t)$, $1 \le i \le n$, you'll have the problem formulated clearly in the language of classical mechanics. In that case, for a $n$ particle system \begin{align} \dot{x_1}(t) &= \frac{1}{m_1}p_1(t)\\ \\ \dot{x_2}(t) &= \frac{1}{m_2}p_2(t)\\ &\vdots \\ \dot{x_n}(t) &= \frac{1}{m_n}p_n(t)\\ \\ \\ \dot{p_1}(t) &= f_1(t,x_1,\ldots,x_n,p_1,\ldots,p_n)\\ \\ \dot{p_2}(t) &= f_2(t,x_1,\ldots,x_n,p_1,\ldots,p_n)\\ &\vdots \\ \dot{p_n}(t) &= f_n(t,x_1,\ldots,x_n,p_1,\ldots,p_n)\\ \end{align} This are Newton's equations rewritten in terms of momenta. The space $\{x_1, \ldots, x_n, p_1,\ldots, p_n\}$ is called the Configuration Space, and it's very important for the study of the behavior of the system.

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Given the physical flavour of the problem, I'll explain as follws: If you know the Lagrangian of the system, say $L\equiv L(q,\dot{q})=T-V$, the equations of motion are (2nd order)

$$ \frac{d}{dt}\frac{\partial L}{\partial{\dot{q}}}=\frac{\partial L}{\partial q} $$

Now, state the problem in Hamiltonian Mechanics. Define the momenta as

$$ p=\frac{\partial L}{\partial \dot{q}} $$

and solve for $\dot{q}=\dot{q}(p)$. Write the Hamiltonian in that new variables:

$$ H\equiv H(p,q)=\displaystyle\sum_{i}p_i\dot{q}_i -L$$

Then the equations of motion are (1st order)

$$ \dot{p}=-\frac{\partial H}{\partial q}, \dot{q}=\frac{\partial H}{\partial p} $$

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