Editing Autonomous system (mathematics) (section)

=== Second order ===

The second-order autonomous equation
<math display="block">\frac{d^2x}{dt^2} = f(x, x')</math>
is more difficult, but it can be solved<ref>{{cite book |last=Boyce |first=William E. |author2=Richard C. DiPrima | title=Elementary Differential Equations and Boundary Volume Problems |edition=8th |year=2005 |publisher=John Wiley & Sons | isbn=0-471-43338-1 |page=133}}</ref> by introducing the new variable
<math display="block">v = \frac{dx}{dt}</math>
and expressing the [[second derivative]] of <math>x</math> via the [[chain rule]] as
<math display="block">\frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dx}{dt}\frac{dv}{dx} = v\frac{dv}{dx}</math>
so that the original equation becomes
<math display="block">v\frac{dv}{dx} = f(x, v)</math>
which is a first order equation containing no reference to the independent variable <math>t</math>. Solving provides <math>v</math> as a function of <math>x</math>. Then, recalling the definition of <math>v</math>:

<math display="block">\frac{dx}{dt} = v(x)  \quad \Rightarrow \quad t + C = \int \frac{d x}{v(x)} </math>

which is an implicit solution.

====Special case: {{math|1=''x''&Prime; = ''f''(''x'')}}====

The special case where <math>f</math> is independent of <math>x'</math>

<math display="block">\frac{d^2 x}{d t^2} = f(x)</math>

benefits from separate treatment.<ref>{{cite web |title=Second order autonomous equation |url=https://eqworld.ipmnet.ru/en/solutions/ode/ode0301.pdf |website=[[Eqworld]] |access-date=28 February 2021 |language=en }}</ref> These types of equations are very common in [[classical mechanics]] because they are always [[Hamiltonian system]]s.

The idea is to make use of the identity

<math display="block">\frac{d x}{d t} = \left(\frac{d t}{d x}\right)^{-1}</math>

which follows from the [[chain rule]], barring any issues due to [[division by zero]].

By inverting both sides of a first order autonomous system, one can immediately integrate with respect to <math>x</math>:

<math display="block">\frac{d x}{d t} = f(x) \quad \Rightarrow \quad \frac{d t}{d x} = \frac{1}{f(x)} \quad \Rightarrow \quad t + C = \int \frac{dx}{f(x)}</math>

which is another way to view the separation of variables technique. The second derivative must be expressed as a derivative with respect to <math>x</math> instead of <math>t</math>:

<math display="block">\begin{align}
\frac{d^2 x}{d t^2} &= \frac{d}{d t}\left(\frac{d x}{d t}\right)
= \frac{d}{d x}\left(\frac{d x}{d t}\right) \frac{d x}{d t} \\[4pt]
&= \frac{d}{d x}\left(\left(\frac{d t}{d x}\right)^{-1}\right) \left(\frac{d t}{d x}\right)^{-1} \\[4pt]
&= - \left(\frac{d t}{d x}\right)^{-2} \frac{d^2 t}{d x^2} \left(\frac{d t}{d x}\right)^{-1}
= - \left(\frac{d t}{d x}\right)^{-3} \frac{d^2 t}{d x^2} \\[4pt]
&= \frac{d}{d x}\left(\frac{1}{2}\left(\frac{d t}{d x}\right)^{-2}\right)
\end{align}</math>

To reemphasize: what's been accomplished is that the second derivative with respect to <math>t</math> has been expressed as a derivative of <math>x</math>. The original second order equation can now be integrated:

<math display="block">\begin{align}
\frac{d^2 x}{d t^2} &= f(x) \\
\frac{d}{d x}\left(\frac{1}{2}\left(\frac{d t}{d x}\right)^{-2}\right) &= f(x) \\
\left(\frac{d t}{d x}\right)^{-2} &= 2 \int f(x) dx + C_1 \\
\frac{d t}{d x} &= \pm \frac{1}{\sqrt{2 \int f(x) dx + C_1}} \\
t + C_2 &= \pm \int \frac{dx}{\sqrt{2 \int f(x) dx + C_1}}
\end{align}</math>

This is an implicit solution. The greatest potential problem is inability to simplify the integrals, which implies difficulty or impossibility in evaluating the integration constants.

====Special case: {{math|1=''x''&Prime; = ''x''&prime;<sup>''n''</sup> ''f''(''x'')}}====

Using the above approach, the technique can extend to the more general equation

<math display="block">\frac{d^2 x}{d t^2} = \left(\frac{d x}{d t}\right)^n f(x)</math>

where <math>n</math> is some parameter not equal to two. This will work since the second derivative can be written in a form involving a power of <math>x'</math>. Rewriting the second derivative, rearranging, and expressing the left side as a derivative:

<math display="block">\begin{align}
&- \left(\frac{d t}{d x}\right)^{-3} \frac{d^2 t}{d x^2} = \left(\frac{d t}{d x}\right)^{-n} f(x) \\[4pt]
&- \left(\frac{d t}{d x}\right)^{n - 3} \frac{d^2 t}{d x^2} = f(x) \\[4pt]
&\frac{d}{d x}\left(\frac{1}{2 - n}\left(\frac{d t}{d x}\right)^{n - 2}\right) = f(x) \\[4pt]
&\left(\frac{d t}{d x}\right)^{n - 2} = (2 - n) \int f(x) dx + C_1 \\[2pt]
&t + C_2 = \int \left((2 - n) \int f(x) dx + C_1\right)^{\frac{1}{n - 2}} dx
\end{align}</math>

The right will carry +/− if <math>n</math> is even. The treatment must be different if <math>n = 2</math>:

<math display="block">\begin{align}
- \left(\frac{d t}{d x}\right)^{-1} \frac{d^2 t}{d x^2}    &= f(x) \\
-\frac{d}{d x}\left(\ln\left(\frac{d t}{d x}\right)\right) &= f(x) \\
\frac{d t}{d x} &= C_1 e^{-\int f(x) dx} \\
t + C_2 &= C_1 \int e^{-\int f(x) dx} dx
\end{align}</math>