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Autonomous system (mathematics)
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== Solution techniques == The following techniques apply to one-dimensional autonomous differential equations. Any one-dimensional equation of order <math>n</math> is equivalent to an <math>n</math>-dimensional first-order system (as described in [[Ordinary differential equation#Reduction to a first-order system|reduction to a first-order system]]), but not necessarily vice versa. === First order === The first-order autonomous equation <math display="block">\frac{dx}{dt} = f(x)</math> is [[Separation of variables|separable]], so it can be solved by rearranging it into the integral form <math display="block">t + C = \int \frac{dx}{f(x)}</math> === 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''″ = ''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''″ = ''x''′<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 +/โ 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> === Higher orders === There is no analogous method for solving third- or higher-order autonomous equations. Such equations can only be solved exactly if they happen to have some other simplifying property, for instance [[linear differential equation|linearity]] or dependence of the right side of the equation on the dependent variable only<ref>[http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf Third order autonomous equation] at [[eqworld]].</ref><ref>[http://eqworld.ipmnet.ru/en/solutions/ode/ode0506.pdf Fourth order autonomous equation] at [[eqworld]].</ref> (i.e., not its derivatives). This should not be surprising, considering that nonlinear autonomous systems in three dimensions can produce truly [[chaos theory|chaotic]] behavior such as the [[Lorenz attractor]] and the [[Rรถssler attractor]]. Likewise, general non-autonomous equations of second order are unsolvable explicitly, since these can also be chaotic, as in a periodically forced pendulum.<ref>{{cite book | author = Blanchard | author2 = Devaney | author2-link = Robert L. Devaney | author3 = Hall | title = Differential Equations | publisher = Brooks/Cole Publishing Co | year = 2005 | pages = 540โ543 | isbn = 0-495-01265-3}}</ref> ===Multivariate case=== {{main|Matrix differential equation}} In <math>\mathbf x'(t) = A \mathbf x(t)</math>, where <math>\mathbf x(t)</math> is an <math>n</math>-dimensional column vector dependent on <math>t</math>. The solution is <math>\mathbf x(t) = e^{A t} \mathbf c</math> where <math>\mathbf c</math> is an <math>n \times 1</math> constant vector.<ref>{{cite web |title=Method of Matrix Exponential |url=https://www.math24.net/method-matrix-exponential/ |website=Math24 |access-date=28 February 2021}}</ref> === Finite durations === For non-linear autonomous ODEs it is possible under some conditions to develop solutions of finite duration,<ref>{{cite book |author = Vardia T. Haimo |title = 1985 24th IEEE Conference on Decision and Control |chapter = Finite Time Differential Equations |year = 1985 |pages = 1729โ1733 |doi = 10.1109/CDC.1985.268832 |s2cid = 45426376 |chapter-url=https://ieeexplore.ieee.org/document/4048613}}</ref> meaning here that from its own dynamics, the system will reach the value zero at an ending time and stay there in zero forever after. These finite-duration solutions cannot be [[analytical function]]s on the whole real line, and because they will be non-[[Lipschitz function]]s at the ending time, they don't stand{{clarification needed|date=October 2024}} uniqueness of solutions of Lipschitz differential equations. As example, the equation: :<math>y'= -\text{sgn}(y)\sqrt{|y|},\,\,y(0)=1</math> Admits the finite duration solution: :<math>y(x)=\frac{1}{4}\left(1-\frac{x}{2}+\left|1-\frac{x}{2}\right|\right)^2</math>
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