A system of differential equations not having a analytical solution

[marellasunny]

I am going to quote an article below and there is a part I would like clarification.I do not understand why chaotic differential systems do not have an analytical solutions. In the simplest form,an equation such as this one [sin(x) + x - 0.5 = 0] does not have an analytical solution i.e it can be only solved numerically.
1.What would be its equivalent for a differential system not having a analytical solution?
2.When the differential equation/system gives a different response for different initial conditions,how does it imply "not having a analytical solution"? I'm a novice.

Not all mechanical problems can be solved analytically as many physicist thought before. They thought that if there were no analytical solution then it was only a matter of intelligence, it would need a more clever mathematical approach to solve these problems.
In fact there are only some special problems that can be solved analytically. This doesn’t mean that Newton was wrong, he predicted chaotic motion.
Henri Poincaré was the first to see this. In a contest, by King Oscar II of Sweden and Norway, a few problems was to be solved, one of them was to prove that the solar system was stable (i.e. analytical solution exist). Poincaré found that there could not be an analytical solution, even the simpler problem of a three-body system. He found that a small difference in the initial conditions could blow up to a totally different answer.

 

[lurflurf]

What equations do not have analytical solution depends on what functions are available.
usual examples would be
x''(t)+sin(x(t))=0
h x''(t)+t x'(t)-x(t)=0
h x''(t)+x'(t)-t x(t)=0

different response for different initial conditions is a different thing, but they often occur together
If you have the analytical solution it is (hopefully) easier to deal with the different responses
If we have an approximate solution we hope it is not to far from the actual solution
If we have very different solution we cannot distinguish we have a problem

We might want to use the differential equation to make a prediction. If tiny changes in the conditions make a huge difference in the response we cannot predict very well.

 

[pasmith]

I am going to quote an article below and there is a part I would like clarification.I do not understand why chaotic differential systems do not have an analytical solutions. In the simplest form,an equation such as this one [sin(x) + x - 0.5 = 0] does not have an analytical solution i.e it can be only solved numerically.
1.What would be its equivalent for a differential system not having a analytical solution?

Consider the one-dimensional system <br /> \frac{dx}{dt} = f(x)<br /> with continuous f. This has implicit solution <br /> \int_{x(0)}^{x(t)} \frac{1}{f(u)}\,du = t.<br /> There is then a question as to whether the integral can actually be done analytically (which it almost always can't) and whether the resulting equation can be solved analytically for x(t) (which it almost always can't). The only case we can always solve is the case where x(0) is such that f(x(0)) = 0, and the solution is then x(t) = x(0).

2.When the differential equation/system gives a different response for different initial conditions,how does it imply "not having a analytical solution"? I'm a novice.

"Sensitive dependence on initial conditions" is a phenomenon which marks a system out as being chaotic.

However even with non-chaotic systems the behaviour of a system may be drastically different depending on the initial conditions: consider <br /> \frac{dx}{dt} = x(1-x),<br /> which falls into the vanishingly small category of systems for which a solution can be found analytically:
<br /> x(t) = \begin{cases}<br /> \frac{x_0 e^t}{1 + x_0(e^t - 1)} &amp; 0 \leq t &lt; \log \left(1 + \frac{1}{|x_0|}\right), x_0 &lt; 0 \\<br /> 0 &amp; 0 \leq t &lt; \infty, x_0 = 0 \\<br /> \frac{x_0 e^t}{1 + x_0(e^t - 1)} &amp; 0 \leq t &lt; \infty, 0 &lt; x_0 &lt; 1 \\<br /> 1 &amp; 0 \leq t &lt; \infty, x_0 = 1 \\<br /> \frac{x_0 e^t}{1 + x_0(e^t - 1)} &amp; 0 \leq t &lt; \infty, x_0 &gt; 1<br /> \end{cases}<br />
The salient points are firstly that if x_0 = x(0) &lt; 0 then the trajectory diverges to -\infty in finite time, secondly that if x_0 = 0 then x_0 = 0 for all time, and thirdly that if x_0 &gt; 0 then x_0 \to 1 as t \to \infty. Thus if all we know is that -\epsilon &lt; x_0 &lt; \epsilon for some \epsilon &gt; 0, then we cannot predict the future behaviour of the system.

 

[marellasunny]

<br /> x(t) = \begin{cases}<br /> \frac{x_0 e^t}{1 + x_0(e^t - 1)} &amp; 0 \leq t &lt; \log \left(1 + \frac{1}{|x_0|}\right), x_0 &lt; 0 \\<br /> 0 &amp; 0 \leq t &lt; \infty, x_0 = 0 \\<br /> \frac{x_0 e^t}{1 + x_0(e^t - 1)} &amp; 0 \leq t &lt; \infty, 0 &lt; x_0 &lt; 1 \\<br /> 1 &amp; 0 \leq t &lt; \infty, x_0 = 1 \\<br /> \frac{x_0 e^t}{1 + x_0(e^t - 1)} &amp; 0 \leq t &lt; \infty, x_0 &gt; 1<br /> \end{cases}<br />
The salient points are firstly that if x_0 = x(0) &lt; 0 then the trajectory diverges to -\infty in finite time, secondly that if x_0 = 0 then x_0 = 0 for all time, and thirdly that if x_0 &gt; 0 then x_0 \to 1 as t \to \infty. Thus if all we know is that -\epsilon &lt; x_0 &lt; \epsilon for some \epsilon &gt; 0, then we cannot predict the future behaviour of the system.

The above solution of x(t) does not work for x(0)=0. Is there an inherent assumption that I am missing? I wrote the solution I got for the D.E x'=x(1-x) below. After substituting for x(0)=0,I get a fraction that tends to 1.
<br /> x(t)=\frac{1}{1+C_1*e^{-t}} \\<br /> x(0)=\frac{1}{1+C_1} \\<br /> C_1=x(0)-1 \\<br /> \Rightarrow x(t)=\frac{1}{1+(x(0)-1)e^{-t}} \\<br />

 

[pasmith]

The above solution of x(t) does not work for x(0)=0.

The solution in the case x(0) = 0 is x(t) = 0.

Is there an inherent assumption that I am missing? I wrote the solution I got for the D.E x'=x(1-x) below. After substituting for x(0)=0,I get a fraction that tends to 1.
<br /> x(t)=\frac{1}{1+C_1*e^{-t}} \\<br />

Separation of variables should give you <br /> \log \left| \frac{x}{1-x}\right| = t + C so that <br /> \left| \frac{x}{1-x}\right| = Ae^{t} from which it follows that <br /> A = \left| \frac{x_0}{1 - x_0}\right|. If at this point you set x_0 = 0 then you obtain A = 0, and hence <br /> \frac{x}{1-x} = 0.