Mechanical Engineering Assignment Help

Mechanical Engineering Assignment Help For any help regarding Mechanical Engineering Assignment Help visit :- https://www.matlabassignmentexperts.com/, Email :- info@matlabassignmentexperts.com or call us at :- +1 678 648 4277 matlabassignmentexperts.com

Problems Problem 1 : Nonlinear parametric pendulum The equation of motion governing an undamped simple pendulum is given by: d 2  02 sin  0 dt 2 g where  0 is the natural frequency of the pendulum, g the acceleration due to gravity, L the length of the pendulum and  the angle made with the vertical. Now, we can add a damping force to the equation by L including a term that is proportional to the velocity of the pendulum. The equation of motion now becomes: 2  0 d 22  2d   sin 0 dt dt where  is the damping coefficient.

In this problem, we are interested in studying the nonlinear parametric pendulum, whose equation of motion is given by: d 2  2 d   2 1 h cos 2 t sin   0 0    0  dt 2 dt

The above is a simple model for a child playing on a swing. The forcing term models the periodic pumping of the child's legs at approximately twice the natural frequency of the swing. h is the amplitude of forcing. Let us see if we can answer the following question in this problem: starting near the equilibrium point   ˙  0 , can the child get the swing going by pumping his/her legs this way, or does he/she need a push ?

matlabassignmentexperts.com

For this problem, choose 0  1 . Unless otherwise specified, use   0.1. Also, remember that  physically varies between 0 and 2 , and any value greater than 2 in the solution should be brought back to the physical range using the function mod.

i)

For the unforced and undamped pendulum ( h    0 ), find the time series  t  vs. t

with the initial conditions: a)

˙ 0   0,  0   0.01

b) ˙ 0   0,  0   3

Show that the trajectory in a) is sinusoidal, whereas the trajectory in b) is not. This happens because, in part b),  is so large thatsin   is not a good approximation, and hence nonlinear effects kick in. Also, show that the time period of oscillations is dependent on the initial conditions for large  (and ˙ ), and is independent of the initial conditions for small  (and ˙ ). ii) The rest 2 state   ˙  0 of the pendulum is known to be for unstable when h necessary sustained 2   h  / 16 . Verify the prediction of the critical value of oscillations of the pendulum. (Get the time series for two values of h , one that is just below and the other that is just 0 above the critical value of h given by the instability condition.) iii) Plot the time series  t 

(for any initial condition) for h =1.50, 1.80, 2.00, 2.05, 2.062.

Calculate the time periods and comment on the pattern you observe. (Remember to let the initial transients die away before you estimate the time period.)

matlabassignmentexperts.com

iv) The motion of the pendulum for h =2.2 is on a "strange attractor." An attractor (a set of points to which solutions get attracted) is informally described as strange if the dynamics on the attractor are chaotic. When a system gets into the chaotic regime, the behavior of the solution will seem random even though we have a deterministic system that governs the motion of the pendulum. For h =2.2, plot the trajectory on a  ˙ plane, and observe the chaotic (random) nature of the "strange attractor"! (Use any non-zero initial condition; you can also plot the time series of  or ˙ and observe that the behavior is random and there is no clear time period in the signal.) Again, remember to let the initial transients die away before you make your plots. Unpredictability is an essential feature of chaos. If a system is in the chaotic regime, it is impossible to predict the exact position and velocity of the pendulum at a future time even if we are given the current position and velocity with maximum accuracy. This property of unpredictability arises from what we call "sensitive dependence on initial conditions." We may start the system at (0.1,0.1) and observe a specific trajectory; but, starting at (0.1,0.10000001) may result in a completely different trajectory if the behavior is chaotic. Show that there is sensitive dependence on initial conditions for h =2.2. (For showing sensitive dependence on initial conditions, consider two initial conditions that are different only by a very small quantity (something like  = 0.001 radians should do the job), and show that the solutions are very different after some time when plotted on top of each other.) v) Finally, make a quick comparison of the time series of  for h =1.8, which is well below the critical value for chaos, and h =2.2 when the system is chaotic. Use the same initial conditions for the two cases.

Problem 2 : The growth/decay of population of animal species A popular model for the growth/decay of population of animal species is given by the logistic map. It takes into account the two basic factors that contribute to the birth or death of any animal species: a) Reproduction means the population will increase at a rate proportional to the current population. This can be stated as "the xn1isproportional xn population at a future time to the population at the current time," i.e. matlabassignmentexperts.com

Where is the population at time n 1 given that the population at time n is xn . xn1 The proportionality constant signifies the growth rate because of reproduction. b) Starvation means the population will decrease at a rate proportional to the value obtained by taking the theoretical "carrying capacity" of the environment less the current population. In a given environment, the presence of more number of animals will result in more rapid consumption of the available resources, and this will result in more likelihood of starvation. This can be written as Since xn and

xn 1  1  xn 

xn1

are normalized populations (actual population divided by the

maximum population possible), they cannot be greater than 1.

1 x  n

is a measure of

how close the system is to the maximum carrying capacity of the environment. The proportionality constant signifies the inverse of the death rate because of starvation.

Mathematically, the complete model that includes both the birth and the death processes can be written as: xn1  rxn 1  xn 

where xn is a number between zero and one, and represents the population at year n ; hence x0 represents the initial population (at year 0); r is a positive number, and represents a combined rate for reproduction and starvation. i) For values of r less than 1, show that, irrespective of the initial population, the population will eventually die. ii) With r between 1 and 2, show that the population quickly stabilizes to the value independent of the initial population. iii) What happens for values of r between 2 and 3 ? iv) Play around with the value of r , and see if you can find a critical value above which the n r

solution no more converges to r 1 . Plotting x vs. n should reveal that the solution never settles to a single value even for large n . Instead of converging to a single value,

matlabassignmentexperts.com

r 1 r ,

the solution converges to a 2-period oscillation or a 4-period oscillation, or some period oscillation as you increase r . At the first critical value of r , the solution will switch from settling to a single value to settling to an oscillation between two values; similarly, at the second critical value of r , the solution will switch from settling to an oscillation between two values to settling to an oscillation between four values; and so on. This is known as perioddoubling. As we can guess, for large enough r , the solution will k

converge to an oscillation between values, 2where k is large, and effectively, there is no finite-period oscillation to which the system settles, and this is known as the "period-doubling route to chaos".

Show sample plots of xn Vs n for various values of r to confirm that period- doubling happens. Find the first four values of r where period doubling occurs. v) Create a rough map of the asymptotic values of xn for varying r . (What we are looking for is the following: On the x-axis, you should plot r . On the y-axis should be the asymptotic values to which the population converges for the corresponding value of r . The asymptotic values to which the population converges will be k

either 1 or 22or 4 or 8 or some

in number depending on the value of r . You can do something like getting the

solution for a large number of iterations (say 50000) and plot the last 10000 values. If the solution converges to a single value, we can expect the last 10000 values to be the same; if the solution converges to a 2-period oscillation, we can expect the last 10000 values to be switching between the two values of the 2-period oscillation, and so plotting the last 10000 values is equivalent to plotting just the two values in the oscillation cycle; See the below figure which your rough map should look like!).

matlabassignmentexperts.com

vi) At r = 3.57 (approximately) is the onset of chaos, at the end of the period-doubling cascade. We can no longer see any oscillations. Take r =3.7, and show that there is now sensitive dependence on initial conditions. (Slight variations in the initial population yield dramatically different results over time, a prime characteristic of chaos. For more details on unpredictability and sensitive dependence on initial conditions, see part iv) of the previous question.)

Problem 3 : Double-Well Potential System

g

External Force

F t   F0 cos t

x x0 The above figure shows a double well potential system. The double well is attached to a rigid framework and the particle moves along the double well. As seen in the figure, the two wells are separated by a hump at x  0 has two stable states where their potentials are minimal. So, it is called bistable system. The potential energy is proportional to the height of the particle. Thedouble well potential can be expressed as V x    14 x 4  12 x 2. Suppose the double well is shaken periodically by the external force from side to side. What happens to the motion of the damped particle? If the shaking is very weak, the particle will go into one of the wells and stay near the bottom of the well, jiggling slightly. With strong shaking, the particle trajectory will become larger. At this moment, we can guess that two types of particle motion exist: a small-amplitude, low-energy oscillation about the bottom of a well, and a large-amplitude, high-energy oscillation where the particle goes back and forth over the hump without settling down in one of the wells. Your task is to observe the particle motion under Stokes type damping with coefficient  as a matlabassignmentexperts.com

function of initial conditions and the magnitude of external force. The motion equation of the particle along the double well is given as follows: 3

˙x˙   x˙  x  x  F 0cos t

In this problem, the damping coefficient and the frequency of the external force are given:   0.25 &   1. i) Suppose that small external force F  0.18 is applied. Show that the motion of the particle depends on the initial conditions. (You can choose different initial conditions, and calculate their trajectory, and compare together. 100-200 sec is enough to simulate the motion.) Confirm that all the trajectories of the particle are oscillatory and confined to single well at the steady state. Is there any change of the period for different initial conditions? ii) Suppose that large external force F  0.4 is applied. Compare the particle trajectories for different initial conditions. Are they relevant? Irrelevant? Or chaotic? Describe what you observe (for the definition of chaos, see problem 1). iii) Now, we introduce the phase plane analysis. Phase plane is sometimes very useful to analyze the nonlinear dynamic motions. Phase plane is defined as the plot of x t  (x axis) vs. x˙ t (y axis), and you can observe dynamic behavior, independent of time.   Generate phase plane plot for different initial conditions you used in ii) (separate plots), and compare them together. (Use simulation time of 1000-2000 sec to observe the behavior well.) iv) From iii), you may recognize phase plane analysis is hard to interpret. At this time, another nonlinear dynamic analysis tool is suggested, Poincaré section. It is obtained by plotting x t x˙ t Vs for the same phase.  obtain time values when they have same phase. For example, the phase of the external force at t=0, 2π/ω, 4π/ω,  are same.). From a physical view point, we take “snapshot” of the system at the same phase in each driver cycle. Plot Poincaré section for the initial conditions used in ii). (Plot the points, not line between two points in the figures. 10000-20000 sec simulation time is enough to generate Poincaré section.) Is it better than phase plane analysis in terms of visualization?

Describe what you observe for different initial conditions.

matlabassignmentexperts.com

v) In general, chaos still happens even when time goes to the infinity, but this system has different type of chaos: with two initial conditions which are close together, their trajectories aren’t similar even in the transient response. However, either trajectory eventually goes to a steady state, depending on particle’s transient behavior. It is called transient chaos, and chaos happens only in the transient response. To find suitable initial conditions, trial and error is used. Plot the trajectories of particle separately for following initial conditions. a. x 0   0.177

& 0   0.1 x˙

b. x 0   0.176

& 0   0.1 x˙

Use F  0.25 as the magnitude of external force and 500 sec as the simulation time. Describe what you observe by comparing both trajectories. vi) Now, you understand it can be very hard to predict the final state of this system depending on the initial condition.

Therefore, the sensitivity of initial condition is conveyed more vividly by the following graphical method. a. Initial conditions for position and velocity should be

2.5  x 0  , x˙ 0   2.5

b. Make a 101 x 101 color grids with each grid point representing the particle’s final state with the given initial condition. c. Calculate trajectory up to 500 second for each initial condition. d. A red dot is placed in a corresponding grid point if the trajectory ends up in the right well. e. Blue dot is placed in a corresponding grid point if the trajectory ends up in the left well.

Plot the color-coded grids for different initial conditions. Is it beautiful? (Hint: you plot the map with ‘imagesc’ function. This function requires matrix (grid) to be imaged in the figure. You define empty matrix for color coding. If color is red in corresponding grid, matrix element has 1. Otherwise, it has -1.)

matlabassignmentexperts.com

Solutions Problem 1 : i)

Nonlinear parametric pendulum

Nonlinear pendulum without damping and force can be simplified to

d 2   2 sin   0 . dt 2 0

Simulation results with given two initial conditions (small angle motion and large angle motion) are compared (See the below codes and figure). Runge-Kutta method is used, and simulation time is up to 50 second. For the increase of simulation accuracy, maximum time step is set to 0.01 second. In the figure, for a), small angular one is definitely sinusoidal, and its period (6.28 second) corresponds to the reciprocal of the natural frequency (1 rad/sec=1/2π Hz) of the pendulum, and it confirms the period is independent of the initial conditions, and sin   is a good approximation since natural frequency is closely related to the period of the response with no force in linear system. However, the period increases as the initial angle becomes larger. For b), the shape in the figure is not sinusoidal any more though it’s still periodic. Therefore, the period is related to the initial condition for the initial condition with large angle. function P81_i % Problem 8.1 : Nonlinear parametric pendulum % Part i. Unforced and undamped pendulum

% Define some constants omega0=1;

% Natural frequency of the pendulum

% Define some parameters for using Runge-Kutta metod matlabassignmentexperts.com

tspan=[0 50]; % Time span to be obtained % Choose time step smaller than 0.01 sec

optn=odeset('InitialStep',1e-

2,'MaxStep',1e-2);

% Simulation with small angle (theta(0)=0.01 radian)

[T1,Y1]=ode45(@(t,x)

Npp(t,x,omega0),tspan,[0.01 0],optn);

% Simulation with large angle (theta(0)=3 radian)

[T2,Y2]=ode45(@(t,x)

Npp(t,x,omega0),tspan,[3 0],optn);

% Plot and compare simulation results

figure(1);

subplot(2,1,1); plot(T1,Y1(:,1),'r'); grid on; axis tight; (Sec)'); ylabel('\bfAngle (Rad)'); \theta(0)=0.01 &

xlabel('\bfTime

title('\bfTrajectory of the Pendulum with

d\theta/dt(0)=0');

subplot(2,1,2); plot(T2,Y2(:,1),'b--'); grid on; axis tight; (Sec)'); ylabel('\bfAngle (Rad)'); \theta(0)=3 &

title('\bfTrajectory of the Pendulum with

d\theta/dt(0)=0');

end

function dx=Npp(t,x,omega0) % Describe nonlinear parametric pendulum motion dx(2,1)=-omega0^2*sin(x(1));

matlabassignmentexperts.com

end

xlabel('\bfTime

dx(1,1)=x(2);

x 10

-3

Angle (Rad)

10 5

Trajectory of the Pendulum with (0)=0.01 & d/dt(0)=0

0 -5

0

5

-2 0

5

3

10

15

20

25 30 35 Time (Sec) Trajectory of the Pendulum with (0)=3 & d/dt(0)=0

40

45

50

45

50

Angle (Rad)

2 1 0

-1

ii)

10

15

20

30 25 Time (Sec)

35

40

From  2  h 2 / 16 , the pendulum become stable when h  0.4 with given 0 parameters in this problem. To verify the above prediction. Two h values are chosen: the ones slightly smaller and larger the critical value h  0.4 ( h  0.39 and h  0.41 ) From the below figure generated by the following m-codes, it is shown that the pendulum with h  0.39 becomes stable since the angular position approaches to zero as time goes on (top plot), but the one with h  0.41 turns out to be unstable since the angular motion increases (bottom plot).

function P81_ii % Problem 8.1 : Nonlinear parametric pendulum % Part ii. Stability omega0=1; % Define some constants % Natural frequency of the pendulum gamma=0.1; % Damping coefficient

matlabassignmentexperts.com

h=[0.39 0.41]; % Two amplitudes of forcing

% Define some parameters for using Runge-Kutta metod Time span to be obtained

theta0=[0.01,0];

tspan=[0 500];

%

% Initial conditions

% Choose time step smaller than 0.01 sec optn=odeset('InitialStep',1e-2,'MaxStep',1e-2);

% Simulation with the amplitude of forcing smaller than critical value

[T1,Y1]=ode45(@(t,x)

Npp(t,x,gamma,omega0,h(1)),tspan,theta0,optn);

% Simulation with the amplitude of forcing larger than critical value

[T2,Y2]=ode45(@(t,x)

Npp(t,x,gamma,omega0,h(2)),tspan,theta0);

% Plot and compare simulation results

figure(1);

subplot(2,1,1); plot(T1,Y1(:,1),'r'); grid on; axis tight; ylabel('\bfAngle (Rad)'); \gamma>(h\omega_0)^2/16

xlabel('\bfTime (Sec)');

title('\bfTrajectory of the Pendulum with

(h=0.39)');

subplot(2,1,2); plot(T2,Y2(:,1),'b--'); grid on; axis tight; ylabel('\bfAngle (Rad)'); \gamma<(h\omega_0)^2/16

xlabel('\bfTime (Sec)');

title('\bfTrajectory of the Pendulum with

(h=0.41)');

end

function dx=Npp(t,x,gamma,omega0,h) % Describe nonlinear parametric pendulum motion

dx(1,1)=x(2);

dx(2,1)=-2*gamma*x(2)-omega0^2*(1+h*cos(2*omega0*t))*sin(x(1));

matlabassignmentexperts.com

end

Angle (Rad)

10

x 10

-3

Trajectory of the Pendulum with  >(h0)2/16 (h=0.39)

5 0 -5 0

50

100

150

200

250

300 350 Time (Sec)

400

450

500

Trajectory of the Pendulum with  <(h0)2/16 (h=0.41)

0.02

Angle (Rad)

0.01 0 -0.01 0 -0.02

50

100

150

200

250 300 350 Time (Sec)

400

450

500

iii) With the below m-codes, 5 plots are represented with different h (But, same initial conditions chosen arbitrarily). For plotting time series, all of them look so similar; the absolute values of slopes are almost same. However, their differences are apparent when they are plotted in the phase space. At h  1.50 , a given angular position has only one corresponding angular velocity (single curve), which means the period of the response is 2 second since 0  1 rad/sec . At h  1.80 , a given angular position has two corresponding angular velocities (two curves). It means that the pendulum follows one of two trajectories first, and the other next time. So, the pendulum rotates along the same trajectory with every other rotation. It confirms that the period is now doubled, compared with the one at h  1.50 . (If you plot the response along the angular position from 2 to 2 , the trajectory is just single curve.) It is called ‘period doubling’, and its period is 4 second. Similar phenomenon (period doubling) happens again at h  2.00 . (Its period is 8 second.) For h  2.05 , its period is close to 128 second or more since so many period doubling happen between h  2.00 and h  2.05 . Between h  2.05 and

matlabassignmentexperts.com

h  2.62 , it period decreases down to about 32 second. It’s a general behavior of period change in the nonlinear parametric pendulum. function P81_iii % Problem 8.1 : Nonlinear parametric pendulum % Part iii. Period doubling

% Define some constants % Natural frequency of the pendulum omega0=1; % Damping coefficient gamma=0.1; h=[ 1.50 1.80 2.00 2.05 2.062]; % Five amplitudes of forcing

% Define some parameters for using Runge-Kutta metod span to be obtained

theta0=[3,0];

tspan=[0 500];

% Time

% Initial conditions

% Choose time step smaller than 0.01 sec

optn=odeset('InitialStep',1e-

2,'MaxStep',1e-2);

% Define plot color set col={'r' ; 'b' ; 'g' ; 'm' ; 'c'};

for i=1:length(h) % Simulation with different h [T,Y]=ode45(@(t,x) Npp(t,x,gamma,omega0,h(i)),tspan,theta0,optn); % Remove transient response (remove data smaller than 100 second)

Y=Y(find(T>=100),:);

T=T(find(T>=100),:); % Generate figure

figure(i);

% Plot steady state response for given h plot(T,Y(:,1),col{i});

grid on; axis tight;

xlabel('\bfTime (second)');

matlabassignmentexperts.com

subplot(2,1,1);

ylabel('\bfAngular Position \theta (rad)');

title(['\bfSteady State Response (t,\theta) for given h=' num2str(h(i))]); % Plot phase space for given h % Use mod for the angular position (-pi<=theta<=+pi) mod(Y(:,1),2*pi),Y(:,2),[col{i} '.']);

subplot(2,1,2); p=plot(pi-

set(p,'MarkerSize',2);

grid on; axis tight; xlabel('\bfAngular Position \theta (rad)'); d\theta/dt(rad/sec)'); num2str(h(i))]);

ylabel('\bfAngular Velocity

title(['\bfPhase Space (\theta,d\theta/dt) for given h='

end

end

function dx=Npp(t,x,gamma,omega0,h) % Describe nonlinear parametric pendulum motion

dx(1,1)=x(2);

dx(2,1)=-2*gamma*x(2)-omega0^2*(1+h*cos(2*omega0*t))*sin(x(1));

matlabassignmentexperts.com

end

iv) First figure contains the time responses of angular position and angular velocity of the pendulum, and second one represents phase space of the pendulum’s response. (They are generated with following m-codes.) If the system is deterministic, the small change of initial conditions makes the system response change a little. However, the system responses are very different even with very similar initial conditions if system is chaotic. In the case of our nonlinear parametric pendulum case, the system is either deterministic or chaotic, depending on its some parameters. In this problem, the amplitude of forcing decides characteristics of pendulum’s response. Therefore, the pendulum with h  2.20 is chaotic and it shows sensitive dependence on initial conditions, based on figures below. function P81_iv % Problem 8.1 : Nonlinear parametric pendulum % Part iv. Sensitive dependence on initial conditions

matlabassignmentexperts.com

% Define some constants omega0=1; % Natural frequency of the pendulum gamma=0.1; % Damping coefficient h=2.20; % Amplitude of forcing % Define some parameters for using Runge-Kutta metod tspan=[0 500]; % Time span to be obtained theta0=[1e-1,0]; % Initial conditions dtheta0=[0,1e-8]; % Change of initial conditions % Choose time step smaller than 0.01 sec optn=odeset('InitialStep',1e-2,'MaxStep',1e-2);

% Simulation with given initial condition [T,Y]=ode45(@(t,x) Npp(t,x,gamma,omega0,h),tspan,theta0,optn); % Simulation with given initial condition + delta [Td,Yd]=ode45(@(t,x) Npp(t,x,gamma,omega0,h),tspan,theta0+dtheta0,optn); % Generate figure

figure(1);

% Plot steady state response for given h

subplot(2,1,1);

plot(T,Y(:,1),'r-',Td,Yd(:,1),'b--'); grid on; axis tight;

xlabel('\bfTime

(second)'); ylabel('\bfAngular Position \theta (rad)');

title('\bfSteady State

Response (t,\theta)'); legend('\bf\theta_0','\bf\theta_0+\delta\theta', ...

'Location','NorthWest'); subplot(2,1,2); plot(T,Y(:,2),'r-',Td,Yd(:,2),'b--'); grid on; axis tight;

xlabel('\bfTime

(second)'); ylabel('\bfAngular Velocity d\theta/dt (rad/sec)'); title('\bfResponse (t,d\theta/dt) with different initial conditions'); legend('\bf\theta_0','\bf\theta_0+\delta\theta', ...

'Location','SouthWest');

matlabassignmentexperts.com

% Plot phase space % Use mod for the angular position (-pi<=theta<=+pi)

figure(2);

p=plot(pi-mod(Y(:,1),2*pi),Y(:,2),'ro', ... pi-mod(Yd(:,1),2*pi),Yd(:,2),'bs');

set(p,'MarkerSize',1);

grid on; axis tight; xlabel('\bfAngular Position \theta (rad)'); d\theta/dt(rad/sec)');

ylabel('\bfAngular Velocity

title('\bfPhase Space (\theta,d\theta/dt)');

legend('\bf\theta_0','\bf\theta_0+\delta\theta', ...

'Location','SouthWest'); end

function dx=Npp(t,x,gamma,omega0,h) % Describe nonlinear parametric pendulum motion

dx(1,1)=x(2);

dx(2,1)=-2*gamma*x(2)-omega0^2*(1+h*cos(2*omega0*t))*sin(x(1));

matlabassignmentexperts.com

end

v) With the below m-codes, the comparisons with different h are shown. As you see plots, they seem to have similar response at the beginning.

However, the responses are quite different as time goes on. For h  1.80 , its response becomes steady and predictable. (the pendulum seems to rotate clockwise constantly) However, the response with h  2.20 is neither stabilized nor predictable. The pendulum rotates counterclockwise constantly, and it suddenly changes the rotating direction. Therefore, h  2.20 makes pendulum’s response chaotic.

function P81_v % Problem 8.1 : Nonlinear parametric pendulum % Part v. Sensitive dependence on initial conditions

% Define some constants omega0=1; gamma=0.1;

% Natural frequency of the pendulum % Damping coefficient

matlabassignmentexperts.com

h=[1.80 2.20];

% Two amplitudes of forcing

% Define some parameters for using Runge-Kutta metod % Time span to be obtained

tspan=[0 500];

theta0=[1e-1,0];

% Initial conditions

% Choose time step smaller than 0.01 sec optn=odeset('InitialStep',1e-2,'MaxStep',1e-2);

% Simulation with h=1.8 [T1,Y1]=ode45(@(t,x) Npp(t,x,gamma,omega0,h(1)),tspan,theta0,optn); % Simulation with h=2.2 [T2,Y2]=ode45(@(t,x) Npp(t,x,gamma,omega0,h(2)),tspan,theta0,optn); % Generate figure

figure(1);

% Plot steady state response for given h plot(T1,Y1(:,1),'r-',T2,Y2(:,1),'b--'); grid on; axis tight;

xlabel('\bfTime

(second)'); ylabel('\bfAngular Position \theta (rad)'); different h');

title('\bfResponse (t,\theta) with

legend(['\bfh=' num2str(h(1))],['\bfh=' num2str(h(2))], ...

'Location','NorthWest'); end

function dx=Npp(t,x,gamma,omega0,h) % Describe nonlinear parametric pendulum motion

dx(1,1)=x(2);

dx(2,1)=-2*gamma*x(2)-omega0^2*(1+h*cos(2*omega0*t))*sin(x(1));

matlabassignmentexperts.com

end

Response (t, ) with different h

h=1.8

400

h=2.2

Angular Position 

200

0

(rad)

-200

-400

-600 0

50

100

150

200 250 300 Time (second)

350 400

450

500

-800

Problem 2 :

The growth/decay of population of animal species

i) The following figures with m-codes described as below represent the normalized population with respect to year n with different initial populations. Each subplot has different r (0.1, 0.5, 0.7, and 0.9). As seen in the plots, for arbitrary r between 0 and 1, all the populations approach to zero as time goes on regardless of their initial conditions. function P82_i % Problem 8.2 : The growth/decay of population of animal species % i. Show x_n converges to zero when 0<r<1.

% Choose arbitrary r between 0 and 1 % Set initial population matrix % Set the population year % Define

color

setr

figure(1); for j=1:length(r)

matlabassignmentexperts.com

r=[0.1 0.5 0.7 0.9];

x0=[0.1:0.1:0.9]';

nf=15;

col={'r' ; 'b' ; 'g' ; 'm'};

% Set the normalized population matrix % row: different initial populatipon % column: population year n xn=zeros(length(x0),nf); % Calculate the quadratic recurrence equation x=x0; for i=1:nf xn(:,i)=r(j)*x.*(1-x); x=xn(:,i); end % Plot the normalized population for n subplot(2,2,j); h=plot(0:nf,[x0 xn],[col{j} 's-']); set(h,'MarkerSize',6); set(gca,'FontSize',18); grid on; axis tight; xlabel('\bfThe population at year (a.u.)'); ylabel('\bfThe normalized population x_n (a.u.)'); title(['\bfThe Normalized Population for r=' num2str(r(j))]); end end

The normalized

The normalized population x (a.u.) population x (a.u.) n n

The Normalized Population for r=0.1 0.8 0.6 0.4 0.2

The normalized

The Normalized Population for r=0.5 0.8 0.6 0.4 0.2

The normalized population x (a.u.) population x (a.u.) n n

0

5

10

15

The population at year (a.u.)

The Normalized Population for r=0.7

0

5

10

15

The population at year (a.u.)

The Normalized Population for r=0.9

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

5

10

15

The population at year (a.u.)

0

5

10

15

The population at year (a.u.)

matlabassignmentexperts.com

ii) In the same way as i), each subplot has different r (1.3, 1.5, 1.7, and 1.9). with these r ’s, below plots show the normalized population as time goes on. Unlike i), the normalized population don’t approach to zero any more. Following output lists the final values which the populations approach to.

1. r=1.300000 the population approaches to 0.230754 2. r=1.500000 the population approaches to 0.333333 3. r=1.700000 the population approaches to 0.411765 4. r=1.900000 the population approaches to 0.473684

With given

r , each final values are same to

r 1

r . Therefore, the final values for

r between 1 and 2 converge to function P82_ii

r 1

.

r

% Problem 8.2 : The growth/decay of population of animal species % ii. Show x_n converges to (r-1)/r when 1<r<2.

% Choose arbitrary r between 1 and 2 % Set initial population matrix % Set the population year % Define

color

matlabassignmentexperts.com

setr

r=[1.3 1.5 1.7 1.9];

x0=[0.1:0.1:0.9]';

nf=30;

col={'r' ; 'b' ; 'g' ; 'm'};

figure(1); for j=1:length(r) % Set the normalized population matrix % row: different initial populatipon % column: population year n xn=zeros(length(x0),nf); % Calculate the quadratic recurrence equation

x=x0;

for i=1:nf xn(:,i)=r(j)*x.*(1-x);

x=xn(:,i);

end % Plot the normalized population for n

subplot(2,2,j);

figure(1); h=plot(0:nf,[x0 xn],[col{j} 's-']); set(h,'MarkerSize',6); set(gca,'FontSize',18);

grid on; axis

tight; xlabel('\bfThe population at year (a.u.)');

ylabel('\bfThe normalized

population x_n (a.u.)'); title(['\bfThe Normalized Population for r=' num2str(r(j))]); % Display final values which the population approaches to

fprintf('%d. r=

%f\n',j,r(j)); fprintf('the population approaches to %f\n',mean(xn(:,end))); end

end

matlabassignmentexperts.com

The normalized

The normalized population x (a.u.) population x (a.u.) n n

The normalized

The Normalized Population for r=1.3 0.8

The Normalized Population for r=1.5 0.8

0.6

0.6

0.4

0.4

The normalized population x (a.u.) population x (a.u.) n n

0.2

0.2

0

5

10 15 20 25

30

The population at year (a.u.)

The Normalized Population for r=1.7

0

5

10 15 20 25

30

The population at year (a.u.)

The Normalized Population for r=1.9

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

5

10 15 20 25

30

The population at year (a.u.)

0

5

10 15 20 25

30

The population at year (a.u.)

iii) What happens for values of r between 2 and 3 ? function P82_iii % Problem 8.2 : The growth/decay of population of animal species % iii. What happens to x_n when 2<r<3.

% make series of r between 2 and 3 r=2.10:0.10:2.90;

% Set initial population matrix x0=[0.1:0.1:0.9]';

% Set the population year nf=100; for j=1:length(r) % Set the normalized population matrix % row: different initial populatipon

% column: population year n

xn=zeros(length(x0),nf);

% Calculate the quadratic recurrence equation for i=1:nf xn(:,i)=r(j)*x.*(1-x);

x=x0;

matlabassignmentexperts.com

x=xn(:,i); end

xf(j)=mean(xn(:,end));

end % Plot the normalized population for n

% Generate series

figure(1);

r1=[2:0.01:3];

xf1=(r1-1)./r1; % h=plot(r,xf,'rs',r1,xf1,'b-'); set(h,'MarkerSize',6,'LineWidth',2); set(gca,'Fontsize',15); legend('\bfFinal Value of Population','\bf(r-1)/r', ...

'Location','SouthEast'); grid on; axis tight; xlabel('\bfThe population at year (a.u.)');

ylabel('\bfThe normalized

population x_n (a.u.)'); title(['\bfThe Normalized Population for r=' num2str(r(j))]); end The normalized population n x (a.u.)

The Normalized Population for 2<r<3 0.66 0.64 0.62 0.6 0.58 0.56 0.54 0.5 0.52

matlabassignmentexperts.com

Final Value of Population (r-1)/r 2

2.2 2.4 2.6 2.8 The population at year (a.u.)

3

iv) Below m-codes generate four plots whenever period-doubling happens as r increases from 2.90 to 3.60. As seen in the figure, each normalized population converges to an 2k oscillation between different values. Another figure shows how many values exists in single oscillation with respect to r . In addition to figures, the following output for m- code below is shown. at r=2.900000, number of population =1 at r=3.000000, number of population =2 at r=3.450000, number of population =4 at r=3.545000, number of population =8 at r=3.565000, number of population =16 at r=3.569000, number of population =28 at r=3.570000, number of population =69 at r=3.571000, number of population =129 at r=3.572000, number of population =150 at r=3.573000, number of population =174 …… ……

The number of different final populations is displayed doubled as r increases (2-period oscillation at r =3.000, 4-period oscillation at r =3.450, 8-period oscillation at r =3.455, and 16-period oscillation at r =3.565). When r is close to 3.57, period-doubling happens any more since the number of different final populations increases randomly (look like chaotic), and no finite-period oscillation doesn’t seem to exist. Therefore, 3.000, 3.450, 3.455, and 3.565 are the first four values of r where period doubling occurs. function P82_iv % Problem 8.2 : The growth/decay of population of animal species % iv. period-doubling

matlabassignmentexperts.com

% make series of r between 2.90 and 3.60 % with step size of 0.001 % Set initial population

r=2.90:0.001:3.60; x0=0.5;

% Define number of different final populations n=zeros(1,length(r)); % Counter for number of period doubling happening % Define

color

setr

j=1;

col={'r' ; 'b' ; 'g' ;

'm'}; % define figure

figure(1);

% Simulation with different r

for i=1:length(r)

% get 10000 values of final population % 1. Simulation up to 40000 secs % 2. Collect populations for next 10000 secs %

as final population.

xn=LogiEqn(r(i),x0);

% Discard numbers below 0.001 % Extract unique populations

xf=fix(xn*1000); xf=unique(xf);

% Count number of unique populations

n(i)=length(xf);

if i>1 % if period-doubling happens, display r

if n(i)~=n(i-1)

fprintf('at r=%f, number of population =%d\n', ... % if the first 4 period-dobling, plot n vs. xn subplot(2,2,j);

matlabassignmentexperts.com

r(i),n(i));

if j<=4

plot([40001:50000],xn,[col{j} '.']); axis tight; grid on; xlabel('\bfYear (n)');

set(gca,'FontSize',16); ylabel('\bfNormalized Population

(x_n)'); title(['Number of period-doubling =' num2str(n(i))]);

j=j+1;

end end % For starting point,

else

fprintf('at r=%f, number of population =%d\n', ...

r(i),n(i));

end end % Plot r vs. number of unique populations

figure(2);

plot(r,n,'r-'); grid on; axis tight; xlabel('\bfThe combined rate for reproduction and starvation (a.u.)'); Unique normalized populations'); title('\bfNumber of Uniqie normalized population for 2.9<=r<=3.6');

function y=LogiEqn(r,x) % Calculate quadratic reccureance relations % x_n+1=r*x_n

% Calculate population for first 40000 years xn=r*x.*(1-x);

for i=1:40000

x=xn;

end % Define variable to store population for next 10000 years y=zeros(length(x),10000); matlabassignmentexperts.com

end

ylabel('\bfNumber of

% Calculate population for next 10000 years for i=1:10000 xn=r*x.*(1-x); x=xn; y(:,i)=xn; end end

matlabassignmentexperts.com

Number of Unique normalized populations

Number of Uniqie normalized population for 2.9<=r<=3.6

350 300 250 200 150 1002.9 50

3

3.1

3.2

3.3

3.4

3.5

3.6

The combined rate for reproduction and starvation (a.u.)

v) Following m-codes are for creating a rough map, and the figure represent a rough map of the asymptotic values of xn for varying r . function P82_v % Problem 8.2 : The growth/decay of population of animal species % v.Plot r vs. x_infinity for logistic equation

% define the series of r from 0.01 to 4 with time step of 0.01 % Calculate x_n when n approach to the infinity. % Obtain x_n for different very large n % Prepare the plot

r=0.01:0.01:4;

for i=1:length(r)

y=LogiEqn(r(i),0.1);

x=r(i)*ones(length(y),1);

% Generate the plot x_infinity distribution for the given r

plot(x,y,'r.'); hold

on; end hold off; axis tight; grid on; set(gca,'FontSize',16); xlabel('\bfThe combined rate for reproduction and starvation (a.u.)'); normalized population x_n for n\rightarrow\infty'); end

function y=LogiEqn(r,x0) % % Calculate quadratic reccureance relations % x_n+1=r*x_n %

for i=1:40000 matlabassignmentexperts.com

xn=r*x0.*(1-x0);

ylabel('\bfThe

x0=xn; end

y=zeros(length(x0),10000); xn=r*x0.*(1-x0);

end

x0=xn;

for i=1:10000

y(:,i)=xn;

end

vi) Even if there is no oscillation, the sensitive dependence on initial conditions doesn’t 0 0 at r  3.57 . However, happen when initial populations x  x  0.5 105 are chosen

the output of population for very very small amount of difference between two initial 0 ) at 0r  3.7 is quite different, and it confirms that there populations ( x  x  0.5 105

is sensitive dependence on initial conditions. The m-codes and graphs for those outputs are shown as below. function P82_vi % Problem 8.2 : The growth/decay of population of animal species

matlabassignmentexperts.com

% vi. sensitive dependence on initial conditions

% Set two r's given in problem % Set initial population

r=[3.57 3.7];

x0=0.5;

% set small difference of initial population

dx0=0.00001;

for i=1:length(r) % calcuate population with x0-dx0 xn1=LogiEqn(r(i),x0-dx0); % calcuate population with x0-dx0 xn2=LogiEqn(r(i),x0+dx0); % plot and compare the results with above initial populations plot(0:length(xn1),[x0;xn1],'r*',0:length(xn2),[x0;xn2],'bo'); 1]; axis(d); grid on;

figure(i); d=axis(); d(3:4)=[0

set(gca,'FontSize',24);

xlabel('\bfThe population at year (a.u.)');

ylabel('\bfThe normalized

population x_n (a.u.)'); title(['\bfThe Normalized Population for r=' num2str(r(i)) '& x_0='

num2str(x0)]);

legend('\bfx_0-Deltax_0','\bfx_0+\Deltax_0','Location','SouthEast'); end

end

function xn=LogiEqn(r,x0) % % Calculate quadratic reccureance relations % x_n+1=r*x_n %

nf=500;

xn=zeros(nf,1); for i=1:nf

matlabassignmentexperts.com

x=x0;

xn(i)=r*x.*(1-x); x=xn(i); end end

n

The normalized population

x (a.u.)

1 0.9 0.8 0.7 0.6 0.5 0.4 00.1.3 0.020 1 0.9 0.8 0.7 0.6 0.5 0.4 00.1.3 0.020

The Normalized Population for r=3.57& x0=0.5 50 100 150 200 250 300 350 400 450 500 The population at year (a.u.)

The Normalized Population for r=3.7& x0=0.5 50 100 150 200 250 300 350 400 450 500 The population at year (a.u.)

x0-x0 x0+x0 x0-x0 x0+x0

n

The normalized population

x (a.u.)

Problem 8.3 : Double-Well Potential System i) Two different initial conditions are used to observe their response. All the m-codes and figures are shown as below. First of all, there responses are quite different, depending on

matlabassignmentexperts.com

initial conditions. Second, their steady state responses are oscillatory. Finally, even though particles with different initial conditions stay in one of two wells, their periods are didn’t change. function P83_i % Problem 8.3 : Double-Well Potential System % i. Initial Condition Sensitivity for Small External Force

% Define some parameters x01=[-0.05 0];

% First initial condition

x02=[-0.05 -0.05];

% Second initial condition

F=0.18; % Amplitude

of external force % Solve equation of motion with the first initial conditions [T1,X1]=ode45(@(t,x)DoubleWell(t,x,F),[0,100],x01); % Solve equation of motion with the second initial conditions [T2,X2]=ode45(@(t,x)DoubleWell(t,x,F),[0,100],x02); % Plot and compare above two time response plot(T1,X1(:,1),'r-',T2,X2(:,1),'b--'); grid on; axis tight;

set(gca,'FontSize',14);

xlabel('\bfTime (second)'); ylabel('\bfLateral Trajectory of Particle (a.u.)'); Sensitivity for Small External Force

title('\bfInitial Condition

F=0.18');

legend('\bfx_0=-0.05&v_0=0','\bfx_0=-0.05&v_0=-0.05');

function dx=DoubleWell(t,x,F) % Describe euation of motion % in the double-well potetial system

dx(1,1)=x(2);

dx(2,1)=F*cos(t)-0.25*x(2)+x(1)-x(1)^3;

matlabassignmentexperts.com

end

end

Initial Condition Sensitivity for Small External Force F=0.18

Lateral Trajectory of Particle (a.u.)

1 0.5 0

0 0

-0.5 0

20 -1

40

x =-0.05&v =0 0

x =-0.05&v =-0.05

60

0

80

100

Time (second)

ii) In the same way as i), we compare particle trajectories for different initial conditions. Their responses are quite different finally, even though their transient responses are a little similar. They are not periodic any more, they are irrelevant, and seem to be chaotic. function P83_ii % Problem 8.3 : Double-Well Potential System % i. Initial Condition Sensitivity for Large External Force

% Define some parameters x01=[-0.05 0];

% First initial condition

x02=[-0.05 -0.05];

% Second initial condition

F=0.40; % Amplitude

of external force % Solve equation of motion with the first initial conditions [T1,X1]=ode45(@(t,x)DoubleWell(t,x,F),[0,100],x01); % Solve equation of motion with the second initial conditions [T2,X2]=ode45(@(t,x)DoubleWell(t,x,F),[0,100],x02); % Plot and compare above two time response plot(T1,X1(:,1),'r-',T2,X2(:,1),'b--'); grid on; axis tight; matlabassignmentexperts.com set(gca,'FontSize',14);

xlabel('\bfTime (second)'); ylabel('\bfLateral Trajectory of Particle (a.u.)'); Sensitivity for Small External Force

title('\bfInitial Condition

F=0.40');

legend('\bfx_0=-0.05&v_0=0','\bfx_0=-0.05&v_0=-0.05');

end

function dx=DoubleWell(t,x,F) % Describe euation of motion % in the double-well potetial system

dx(1,1)=x(2);

dx(2,1)=F*cos(t)-0.25*x(2)+x(1)-x(1)^3;

end

Initial Condition Sensitivity for Small External Force F=0.40

Lateral Trajectory of Particle (a.u.)

1.5 1 0

x =-0.05&v =0 0 0 x =-0.05&v =-0.05

0.5

0

-0.5

0

-1 0 -1.5

20

40

60

80

100

Time (second)

iii) Time responses obtained in i) and ii) are converted into position and velocity of particle trajectories. Phase planes for different F and initial conditions are shown, which are generated by following m-codes. For F  0.18 , they shows periodicity in steady state, but phase planes with F  0.40 don’t seems to have periodicity, even if the responses are independent of time. (See below Figures)

matlabassignmentexperts.com

function P83_iii % Problem 8.3 : Double-Well Potential System % iii. Phase Plane for different initial conditions %

and the amplitude of external force

x01=[-0.05 0]; % Define some parameters x02=[-0.05 -0.05];

% First initial condition

F=[0.18 0.40];

% Amplitude of external force

% Second initial condition

% Create figure figure(1); % Solve equation of motion with the first initial conditions for i=1:length(F) [T1,X1]=ode45(@(t,x)DoubleWell(t,x,F(i)),[0,200],x01); % Plot phase plane for the first initial conditions plot(X1(:,1),X1(:,2),'r-'); set(gca,'FontSize',14); (a.u.)');

subplot(2,2,i*2-1);

grid on; axis tight;

xlabel('\bfLateral Position

ylabel('\bfLateral Velocity (a.u.)');

title(['\bfPhase Plane with x_0=-0.05&v_0=0 for F=' num2str(F(i))]); % Solve equation of motion with the second initial conditions [T2,X2]=ode45(@(t,x)DoubleWell(t,x,F(i)),[0,200],x02); % Plot phase plane for the second initial conditions

subplot(2,2,i*2);

plot(X2(:,1),X2(:,2),'b--'); grid on; axis tight;

set(gca,'FontSize',14);

xlabel('\bfLateral Position (a.u.)');

ylabel('\bfLateral

Velocity (a.u.)'); title(['\bfPhase Plane with x_0=-0.05&v_0=-0.05 for F=' end

end

function dx=DoubleWell(t,x,F) matlabassignmentexperts.com

num2str(F(i))]);

% Describe euation of motion % in the double-well potetial system dx(1,1)=x(2); dx(2,1)=F*cos(t)-0.25*x(2)+x(1)-x(1)^3; end

Lateral Velocity (a.u.)

0.4 0.3 00..12 0 -0.1 -0.2 -0.3 0 1

=-0.05&v =0 for F=0.18

Phase Plane with x0

0

0.2

0.4 0.6 0.8

1

Lateral Position (a.u.)

=-0.05&v =0 for F=0.4

Phase Plane with x0

0

0.5

0

-0.5

-1 -1.5

-1

-0.5

0

0.5

Lateral Position (a.u.)

1.2

1

1.5

Lateral Velocity (a.u.)

Lateral Velocity (a.u.)

0.4 0.2 0 -0.2 -0.4 -0.6 1 0.5

=-0.05&v =-0.05 for F=0.18

Phase Plane with x0

0

-1.2

-1 -0.8 -0.6 -0.4 Lateral Position (a.u.)

-0.2 0

=-0.05&v =-0.05 for F=0.4

Phase Plane with x0

0

0

-0.5

-1 -1.5

-1

-0.5

0

0.5

Lateral Position (a.u.)

1

1.5

Lateral Velocity (a.u.)

iv) Sometimes, Poincare section is quite useful in nonlinear dynamics analysis. We choose phase which is identical to the multiple of 2 in the phase of oscillating external force. For small external force ( F  0.18 ), Poincare section have several points for both initial conditions, but the position of points in Poincare section are completely different, depending on initial conditions. Therefore, the phases of responses are synchronized with those of external forces, and they are also periodic, but the position and velocity of particles during the oscillation depend on initial conditions. For large external force( F  0.40 ), Their shapes for different initial conditions in Poincare section are almost identical, independent of initial condition, even though they didn’t show any similarities in both time response and phase plane analysis. It means that the position and velocity pairs of the particle at the multiple of period for external force are same, independent of their pair’s sequence. Therefore, it confirms that time responses in ii) are not chaotic, and they are relevant each other.

matlabassignmentexperts.com

function P83_iv % Problem 8.3 : Double-Well Potential System % iv. Poincare section for different initial conditions %

and the amplitude of external force

x01=[-0.05 0]; % Define some parameters x02=[-0.05 -0.05];

% First initial condition

F=[0.18 0.40];

% Amplitude of external force

% Second initial condition

% Create figure figure(1); % Solve equation of motion with the first initial conditions for i=1:length(F) [T1,X1]=ode45(@(t,x)DoubleWell(t,x,F(i)),[0:2*pi:20000],x01); % Plot phase plane for the first initial conditions 1);

plot(X1(:,1),X1(:,2),'r.');

set(gca,'FontSize',14); (a.u.)');

subplot(2,2,i*2-

grid on; axis tight;

xlabel('\bfLateral Position

ylabel('\bfLateral Velocity (a.u.)');

title(['\bfPoincare Section with x_0=-0.05&v_0=0 for F='

num2str(F(i))]);

% Solve equation of motion with the second initial conditions [T2,X2]=ode45(@(t,x)DoubleWell(t,x,F(i)),[0:2*pi:20000],x02); % Plot phase plane for the second initial conditions subplot(2,2,i*2); plot(X2(:,1),X2(:,2),'b.'); grid on; axis tight;

set(gca,'FontSize',14);

xlabel('\bfLateral Position (a.u.)'); ylabel('\bfLateral Velocity (a.u.)'); title(['\bfPoincare Section with x_0=-0.05&v_0=-0.05 for F=' end

end

matlabassignmentexperts.com

num2str(F(i))]);

function dx=DoubleWell(t,x,F) dx(1,1)=x(2); dx(2,1)=F*cos(t)-0.25*x(2)+x(1)-x(1)^3; end

Lateral Velocity (a.u.)

=-0.05&v =0 for F=0.18

0.1

Poincare Section with x0

0

0.05

0

-0.05

-0.1

0

0.2 0.4 0.6 0.8

1

Lateral Position (a.u.)

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4

=-0.05&v =0 for F=0.4

Poincare Section with x0

0

-1

-0.5

0

0.5

Lateral Position (a.u.)

1.2 1

Lateral Velocity (a.u.)

Lateral Velocity (a.u.)

0.3 0.25 0.2 0.15 0.1 0.05 0

=-0.05&v =-0.05 for F=0.18

Poincare Section with x0

0

-0.05 -1.2

-1

-0.8 -0.6 -0.4

Lateral Position (a.u.)

-0.2

=-0.05&v =-0.05 for F=0.4

1 Poincare Section with x0

0

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-1

-0.5

0

0.5

1

Lateral Position (a.u.)

Lateral Velocity (a.u.)

v) Very small different initial condition (position difference is only 0.001, whereas velocity is same) give completely different response and final well where the particle stays at the end. At the beginning, their transient responses are almost same since they have very similar initial condition, but one of their response ( x0  0.178 ) goes to another well suddenly due to some nonlinearity. (See following figure.) m-codes is also provided as below.

function P83_v

% Problem 8.3 : Double-Well Potential System

% v. Large Initial Condition Sensitivity % Define some parameters

x01=[0.177 0.1]; x02=[0.178 0.1];

% Initial condition for i) % Initial condition for ii)

matlabassignmentexperts.com

F=0.25;

% Amplitude of external force

% Solve equation of motion with initial conditions for i) [T1,X1]=ode45(@(t,x)DoubleWell(t,x,F),[0,500],x01); % Solve equation of motion with initial conditions for ii) [T2,X2]=ode45(@(t,x)DoubleWell(t,x,F),[0,500],x02); % Plot and compare above two time response plot(T1,X1(:,1),'r-',T2,X2(:,1),'b--'); grid on; axis tight;

set(gca,'FontSize',14);

xlabel('\bfTime (second)'); ylabel('\bfLateral Trajectory of Particle (a.u.)'); Sensitivity for Small External Force

title('\bfInitial Condition

F=0.25');

legend('\bfx_0=0.177&v_0=0.1','\bfx_0=0.178&v_0=0.1');

end

function dx=DoubleWell(t,x,F) % Describe euation of motion % in the double-well potetial system

dx(1,1)=x(2);

dx(2,1)=F*cos(t)-0.25*x(2)+x(1)-x(1)^3;

end

Initial Condition Sensitivity for Small External Force F=0.25

Lateral Trajectory of Particle (a.u.)

1 x =0.177&v =0.1 x0=0.178&v =0.1

0.5

0

0

-0.5

0

-1 0

100

200

300

Time (second)

matlabassignmentexperts.com

0

400

500

vi) Following m-codes let you know how to generate color-coded grid, and display it as an image. See the below figure shows the result of initial condition sensitivity for F  0.25 . As expected, the results of which well where the particle stays eventually quite depend on the initial conditions in some range of initial condition. function P83_vi % Problem 8.3 : Double-Well Potential System % vi. Plot the color-coded grids %

for different initial conditions.

% Define number of steps

nx=50; ny=50;

% Define range of initial conditions

mx=2.5; my=2.5;

% Generate full range of initial conditions

x=[-nx:nx]/nx*mx;

y=[-ny:ny]/ny*my; % Define Grid (101 X 101) o=zeros(length(x),length(y)); % Calculate the final position o particle, % depending on the initial conditions

for i=1:length(x)

for j=1:length(y) % Set initial condition

x0=[x(i);y(j)];

% Calculate trajectory of particle [T,X]=ode45(@DoubleWell,[0,500],x0); % Determine which well the particle stays finally o(i,j)=sign(X(end,1)); end end % Plot grid as an image with color code % Swap row and column (Matrix transpose)

matlabassignmentexperts.com

o=o';

imagesc(o); axis image;

set(gca,'YDir','normal')

set(gca,'XTick',[1:10:101]) set(gca,'XTickLabel',[-2.5:0.5:2.5]) set(gca,'YTick',[1:10:101]) set(gca,'YTickLabel',[-2.5:0.5:2.5]) set(gca,'FontSize',20); set(gca,'FontWeight','bold'); (x_0)');

xlabel('\bfInitial Position

ylabel('\bfInitial Velocity (v_0)');

title('\bfInitial Condition Sensitivity in the Color-Coded Grid');

end

function dx=DoubleWell(t,x) % Describe euation of motion % in the double-well potetial system

dx(1,1)=x(2);

dx(2,1)=0.25*cos(t)-0.25*x(2)+x(1)-x(1)^3; end Initial Condition Sensitivity in the Color-Coded Grid 2.5 2

Initial Velocity (v 0)

1.5 1 0.5 0 -0.5 -1 -2.5

-2

-1.5 -1 -1.5 -2 -2.5

matlabassignmentexperts.com

-0.5

0 0.5 Initial Position (x0)

1

1.5

2

2.5

Note: Below figures shows initial condition sensitivity with same range of initial conditions as iv), but use different steps (201 X 201 for left one, and 901 X 901 for right one.)

matlabassignmentexperts.com