You can find the serial also in the yearbook.
We are sorry, this serial has not been translated.
$$x(t) = x_0 \cos\left[\omega(x_0) t\right]\,, \quad \omega(x_0) = 2\pi \left(1 - \frac{x_0^2}{l_0^2}\right)\,,$$
where $l_{0}is$ some length scale. We think that are letting go of the pendulum from $x_{0}=l_{0}⁄2$ but actually it is from $x_{0}=l_{0}(1+ε)⁄2$. B By how much does the argument of the cosine differ from 2π after one predicted period? How many periods will it take for the pendulum to displaced to the other side than which we expect? Tip Argument of the cosine will in that moment differ from the expected one by more than π ⁄ 2.
function xidot = f(t,xi)
…
xdot=…;
ydot=…;
zdot= …;
xidot = [xdot;ydot;zdot];
endfunction
config = odeset('InitialStep', 0.01,'MaxStep',0.1);
initialCondition=[0.2,0.3,0.4];
solution=ode45(@f,[0,300],initialCondition,config);
plot3(solution.y(:,1),solution.y(:,2),solution.y(:,3)); </pre> Just instead of three dots fill in the rest of the code (just as in the second part of the series) and use $σ=9,5$, $b=8⁄3.Then$ figure out with a precision of at least units for what positive $r$ the system goes from asymptomatic stopping to chaotic oscillation(it is independent of the initial conditions).
pkg load odepkg
function xidot = f(t,xi)
alfa=0.1;
vx=xi(3);
vy=xi(4);
r=sqrt(xi(1)^2+xi(2)^2);
ax=-xi(1)/r^3;
ay=-xi(2)/r^3;
xidot = [vx;vy;ax;ay];
endfunction
config = odeset('InitialStep', 0.01,'MaxStep',0.1);
x0=0;
y0=1;
vx0=…;
vy0=0;
initialCondition=[x0,y0,vx0,vy0];
solution=ode45(@f,[0,100],initialCondition,config)
plot(solution.y(:,1),solution.y(:,2));
pause()</pre>
Y01=2;
Z01=5;
X02=…;
Y02=…;
Z02=…;
nastaveni = odeset('InitialStep', 0.01,'MaxStep',0.1);
pocPodminka1=[X01,Y01,Z01];
reseni1=ode45(@f,[0,45],pocPodminka1,nastaveni);
pocPodminka2=[X02,Y02,Z02];
reseni2=ode45(@f,[0,45],pocPodminka2,nastaveni);
plot(reseni1.x,reseni1.y(:,1),reseni2.x,reseni2.y(:,1));
pause()
</pre> Instead of three dots $X02,Y02,Z02you$ have to give the initial conditions for the second trajectory. Run the code for at least five different orders of magnitude that are all still small and note the time, in which the second trajectory shall differ qualitatively from the first(ie will go in the opposite direaction). Don't decrease the deviation under cca 10^{$-8}$, because then the imprecision's of numerical integration start to show. Chart the dependency of the ungluing time on the order of magnitude of the deviation.
Bonus: Attempt to use the gained dependency of the ungluing time on the size of the deviation estimate Ljapun's exponent. You will need more than five runs and you can assume that at the moment of ungluing it will always overcome some constant $Δ_{c}$.
$$x_{n} = x_{n-1} y_{n-1},$$
$$\\ y_n = y_{n-1} K \sin(x),$$
where $x$, y$ are somehow scaled d$φ⁄dt,φ$. Show that the physical parameter $K$, x, y$$.
Copy the function $iterace_stanMap$ from the series and using the following commands choose ten very close initial conditions for some $K$.
K=…;
X01=…;
Y01=…;
Iter1 = iterace_stanMap(X01,Y01,1000,K);
…
X10=…;
Y10=…;
Iter10 = iterace_stanMap(X10,Y10,1000,K);
</pre> Between $Iter1$ and $Iter10$ there are hidden a thousand iterations of given initial conditions using the Standard map. As to see how the ten points look after the $nth$ iteration, you have to write
n=…;
plot(Iterace1(n,1),Iterace1(n,2),„o“,…,Iterace10(n,1),Iterace10(n,2),„o“)
xlabel („x“);
ylabel („y“);
axis([0,2*pi,-pi,pi],„square“);
refresh;
</pre> we write $"o"$ into $plot$ so that the points will draw themselves as circles. The rest of the commands is then included so that the graph will include the whole square and that it would have the correct labels.