I have a question about plotting x(t), the solution to the following differential equation knowing that dx/dt equals the expression below. The value of x is 0 at t = 0.
syms x
dxdt = -(1.0*(6.84e+45*x^2 + 5.24e+32*x - 2.49e+42))/(2.47e+39*x + 7.12e+37)
I want to plot the solution of this first-order nonlinear differential equation. The analytical solution involves complex numbers so that's not relevant because this equation models a real-life process, but Matlab can solve the equation using numerical methods and plot it. Can someone please suggest how to do this?
in matlab try this
tspan = [0 10];
x0 = 0;
[t,x] = ode45(#(t,x) -(1.0*(6.84e+45*x^2 + 5.24e+32*x - 2.49e+42))/(2.47e+39*x + 7.12e+37), tspan, x0);
plot(t,x,'b')
i try it and i got this
hope that help you.
I have written an example for how to use Python with SymPy and matplotlib. SymPy can be used to calculate both definite and indefinite integrals. By calculating the indefinite integral and adding a constant to set it to evaluate to 0 at t = 0. Now you have the integral, so just a matter of plotting. I would define an array from a starting point to an endpoint with 1000 points between (could likely be less). You can then calculate the value of the integral with the constant at each time point, which can then be plotted with matplotlib. There are plenty of other questions on how to customize plots with matplotlib.
This displays a basic plot of the indefinite integral of the function dxdt with assumption of x(t) = 0. Variation of the tuple when running Plotting() will set what range of x values to plot. This is set to plot 1000 data points between the minimum and maximum values set when calling the function.
For more information on customizing the plot, I recommend matplotlib documentation. Documentation on the integral can be found in SymPy documentation.
import pandas as pd
from sympy import *
from sympy.abc import x
import matplotlib as mpl
import matplotlib.pyplot as plt
import numpy as np
def Plotting(xValues, dxdt):
# Calculate integral
xt = integrate(dxdt,x)
# Convert to function
f = lambdify(x, xt)
C = -f(0)
# Define x values, last number in linspace corresponding to number of points to plot
xValues = np.linspace(xValues[0],xValues[1],500)
yValues = [f(x)+C for x in xValues]
# Initialize figure
fig = plt.figure(figsize = (4,3))
ax = fig.add_axes([0, 0, 1, 1])
# Plot Data
ax.plot(xValues, yValues)
plt.show()
plt.close("all")
# Define Function
dxdt = -(1.0*(6.84e45*x**2 + 5.24e32*x - 2.49e42))/(2.47e39*x + 7.12e37)
# Run Plotting function, with left and right most points defined as tuple, and function as second argument
Plotting((-0.025, 0.05),dxdt)
Related
I am trying to use MATLAB to solve the ODE involving step function
Here is my code to solve the above ODE:
syms t s Y y(t) Dy(t) u(t) f(t) k
Dy = diff(y,t)
D2y = diff(Dy,t)
f = heaviside(t) + 2*symsum((-1)^k*heaviside(t-k*pi()),k,1,Inf)
eqn = D2y +0.1*Dy + y == f
leqn = laplace(eqn,t,s)
LT_Y=subs (leqn, laplace (y,t,s),Y);
LT_Y=subs (LT_Y, y(0), 0);
LT_Y=subs(LT_Y, subs(diff(y(t), t), t, 0), 0);
Y=solve(LT_Y,Y);
y=ilaplace(Y,s,t)
However, the result seems strange and I cannot plot the graph of the solution. Could anyone tell me what should I do to my code so that I can plot the solution? Thank you so much
The decay rate of the friction term is 0.1/2=0.05, so you need about t=40 for a decay of 0.1. In a standard plot features down to 1% of the plot size are visible, which would require a time span of t=80, or longer for some visual confirmation of the stabilization of the amplitude at that level.
The right side is about in resonance with the left one, so the steady-state solution will roughly look like -10*cos(t) plus higher frequency components.
Just for plotting a numerical solution is sufficient
opts=odeset("AbsTol",1e-6,"RelTol",1e-9);
[T,Y] = ode45(#(t,y)[y(2); f(t)-0.1*y(2)-y(1)], [0, 100], [0;0], opts);
plot(T,Y(:,1),T,-13*cos(T));
function z=f(t)
z = sign(sin(t));
end
This gives the plot below, blue for the solution and green for the reference (guessed amplitude)
Computing the Fourier series for the right side, the amplitude of the cosine component of the solution is 40/pi=12.732.
I'm a completely new user to Matlab. I want to calculate PDEs like this one: PDE But I have no idea how to code it.
I have tried g(x)=diff(x,y)^2/(diff(x)*diff(y)); but it doesn't work.
By using the symbolic toolbox and diff() function the partial derivatives can be simplified as follows. I'm not sure if you'd like to solve any specific values or plots but going based on the image posted here are some techniques that may help.
Example Partial Derivatives of Function f Taken with Respect to X and Y:
%Creating symbolic variables/representations%
syms x y
%Function dependent on variables x and y%
f = 2*y*sin(5*x);
%Taking the partial derivative with respect to x%
g = diff(f,x);
%Taking the partial derivative with respect to y%
g = diff(g,y)
Example Equation:
Extension: Graphing
Plotting the original function and its derivative can be done by using the fsurf() function which plot symbolic functions.
%Plotting the symbolic result of the partial derivatives%
subplot(1,2,1); fsurf(f);
title('Original Function');
xlabel('X-Axis'); ylabel('Y-Label');
subplot(1,2,2); fsurf(g);
title('Derivative of Function');
xlabel('X-Axis'); ylabel('Y-Label');
Extension: Evaluating at Specific Values
%Evaluating partial derivative at specific values%
x = 1;
y = 1;
Answer = subs(g);
Answer_As_A_Double = double(Answer);
Using MATLAB version: R2019b
I want to sample R random numbers from a custom probability density function in Matlab.
This is the expression of the probability density function evaluated at x.
I thought about using slicesample
R=10^6;
f = #(x) 1/(2*pi^(1/2))*(1/(x^(3/2)))*exp(-1/(4*x));
epsilon= slicesample(0.3,R,'pdf',f,'thin',1,'burnin',1000);
However, it does not work because I get the error
Error using slicesample (line 175)
The step-out procedure failed.
I tried to change starting value and values of thin and burning parameters but it does not seem to work. Could you advise, either on how to make slicesample work or on alternative solutions to sample random numbers from a custom probability density function in Matlab?
Let X be a random variable distributed according to your target pdf. Applying the change of variable y = 1/x and using the well-known theorem for a function of a random variable, the distribution of Y = 1/X is recognized to be a Gamma distribution with parameters α = 1/2, β = 1/4.
Therefore, it suffices to generate a Gamma random variable (using gamrnd) with those parameters and take the inverse. Note that Matlab's definition of the Gamma distribution uses parameters A = α, B = 1/β.
R = 1e5; % desired sample size
x = 1./gamrnd(1/2, 4, [1 R]); % result
Check:
histogram(x, 'Normalization', 'pdf', 'BinEdges', 0:.1:10)
hold on
f = #(x) 1/2/sqrt(pi)./x.^(3/2).*exp(-1/4./x); % target pdf
fplot(f, 'linewidth', .75)
My intention is to plot the original mathematical function values from the differential equation of the second order below:
I(thetadbldot)+md(g-o^2asin(ot))sin(theta)=0
where thetadbldot is the second derivative of theta with respect to t and m,d,I,g,a,o are given constants. Initial conditions are theta(0)=pi/2 and thetadot(0)=0.
My issue is that my knowledge and tutoring is limited to storing the values of the derivatives and returning them, not values from the original mathematical function in the equation. Below you can see a code that calculates the differential in Cauchy-form and gives me the derivatives. Does anyone have suggestions what to do? Thanks!
function xdot = penduluma(t,x)
% The function penduluma(t,x) calculates the differential
% I(thetadbldot)+md(g-o^2asin(ot))sin(theta)=0 where thetadbldot is the second
% derivative of theta with respect to t and m,d,I,g,a,o are given constants.
% For the state-variable form, x1=theta and x2=thetadot. x is a 2x1 vector on the form
% [theta,thetadot].
m=1;d=0.2;I=0.1;g=9.81;a=0.1;o=4;
xdot = [x(2);m*d*(o^2*a*sin(o*t)-g)*sin(x(1))/I];
end
options=odeset('RelTol', 1e-6);
[t,xa]=ode45(#penduluma,[0,20],[pi/2,0],options);
% Then the desired vector from xa is plotted to t. As it looks now the desired
% values are not found in xa however.
Once you have the angle, you can calculate the angular velocity and acceleration using diff:
options=odeset('RelTol', 1e-6);
[t,xa]=ode45(#penduluma,[0,20],[pi/2,0],options);
x_ddot = zeros(size(t));
x_ddot(2:end) = diff(xa(:,2))./diff(t);
plot(t,xa,t,x_ddot)
legend('angle','angular velocity','angular acceleration')
which gives the following plot in Octave (should be the same in MATLAB):
Alternatively, you can work it out using your original differential equation:
x_ddot = -m*d*(o^2*a*sin(o*t)-g).*sin(xa(:,1))/I;
which gives a similar result:
My overall goal is to use the MATLAB symbolic toolbox to simplify the process of formulating and solving for the sensitivities of solutions to ordinary differential equations with respect to the parameters in the equations. In my case I have an ODE with 2 states and 10 parameters. A smaller, but representative, example would look like
X = sym('X', [2 1]) % Vector representing state variables
p = sym('p', [3 1]) % Vector representing parameters
% Fitzhugh Nagumo Equations
rhs_1 = symfun(p(3)*(X(1) - X(1)^3/3 + X(2)), [X; p])
rhs_2 = symfun(-(X(1) - p(1) + p(2)*X(2))/p(3), [X; p])
I can then get the partial derivatives, which are used to solve for the sensitivities, of the RHS of the ODE wrt to the parameters using a command like 'gradient(rhs_1, p)'. But then I would like to convert this gradient to a matlab function that is a function of the vectors X and p, not a function of the elements of these vectors. I need these functions to be of this form because otherwise I cannot use the CVODES solver in the sundialsTB toolbox. Is this possible? Is there an easier way to accomplish what I am trying to do?
Recognizing that a comma-separated list of function inputs is really just a cell array, you can do this by converting your vector inputs to a cell arrays of scalar using mat2cell:
x=1:2;
p=1:3;
v = mat2cell([x(:);p(:)],ones(numel(x)+numel(p),1),1);
y1 = rhs_1(v{:})
y2 = rhs_2(v{:})