I want to solve my differential equation and plot velocity vectors but I am having some trouble with that. I tried this:
syms y(x);
ode = (1+exp(x))*y*diff(y,x)-2*exp(x) == 0;
ySol = dsolve(ode)
[X,Y] = meshgrid(-2:.2:2);
Z = 2*exp(X)/((1+exp(X)).*Y);
[DX,DY] = gradient(Z,.2,.2);
figure
contour(X,Y,Z)
hold on
quiver(X,Y,DX,DY)
hold off
and I get this error:
Warning: Matrix is singular to working precision.
Warning: Contour not rendered for non-finite ZData
It is probably something simple that I do not see but I am just starting using Matlab and I cold not find a right way to do my task. Please help me...
EDIT
As bconrad suggested, I changed my Z function like this:
Z = 2*exp(X)/((1+exp(X)).*Y);
and the previous errors are fixed. However, my prime goal, to plot velocity vectors is not accomplished yet because I get a graph like this:
Don’t have the ability to check at the moment, but I reckon you want an element by element division in that line. You’re missing a dot on the division, try
Z = 2*exp(X)./((1+exp(X)).*Y);
I took a closer look once at my station. The zero-division mentioned by Pablo forces inf's in Z, so quiver get's confused when scaling the vectors (understandably) and just doesn't show them. Try this (with the ode part removed):
[X,Y] = meshgrid(-2 : .2 : 2);
Z = 2 * exp(X) ./ ((1 + exp(X)) .* Y);
Z(isinf(Z)) = nan; % To avoid 0-division problems
[DX, DY] = gradient(Z, .2, .2);
figure
contour(X, Y, Z, 30, 'k')
hold on
quiver(X, Y, DX, DY, 6)
hold off
I've done 3 things here:
Added the line Z(isinf(Z)) = nan; forcing infinite values to be essentially ignored by quiver
Added the arguments 30, 'k' to the contour function to show 30 lines, and make them black (a bit more visible)
Added the argument 6 to the quiver function. This overrides the automatic length-scaling of the vectors.
You'll want to play with the arguments in the contour and quiver functions to make your figure appear as you'd like.
PS: There is a handy arrow function on the file exchange that I find gives better control when creating vector field plots. See https://www.mathworks.com/matlabcentral/fileexchange/278-arrow - the ratings do it justice.
Related
I need to plot a parabola and ellipse. However the ellipse is giving me trouble. Can anyone help? The equations are: y = -5*x^2 + 2 and (x^2/16) + (y^2/2) = 4
I've tried this code but obviously I feel like like it isn't right.
x = linspace(-5, 5);
y1 = (x.^2/16) + (y.^2/2) - 1;
y2 = -5*x.^2 +2;
figure(1)
plot(x, y1)
hold on
plot(x, y2)
hold off
Firstly, you did not define a range variable x. Secondly, the ellipse won't pass the vertical line test and can't be plotted like a regular function f(x). Thirdly, your equation y1 = (x.^2/16) + (y.^2/2) - 1; is non-sensical because you have y on each side.
You could correct your method by defining a range variable x1 and x2 that each have appropriate ranges for the functions your plotting. What I mean by this is that you probably don't want the same range for each function, because the ellipse is undefined over most of the range that the parabola is defined. To plot the ellipse using f(x) you could observe that there are + and - values that are identical, using this fact you could plot your ellipse by two functions one to represent the top half and one to represent the bottom half, each of these would pass the vertical line test.
OR
You could utilize ezplot and have a nice time with it because it makes your life easier. Here is a solution.
ezplot('x^2/16+y^2/2-4'); axis equal; hold on
ezplot('-5*x^2+2-y')
There are multiple ways to plot an ellipse, e.g. you could also use a parametric representation of the equation.
In your approach though, when plotting functions using plot(x,y) command, you need to express your dependent variable (y) through independent variable (x). You defined the range for x, which is what you substitute into your equations in order to find y's. While for the parabola, the dependency of y from x is obvious, you forgot to derive such a relationship for the ellipse. In this case it will be +-sqrt((1 - x^2/16)*2). So in your approach, you'll have to take into account both negative and positive y's for the same value of x. Also there's discrepancy in your written equation for the ellipse (=4) and the one in Matlab code (=1).
x = linspace(-5, 5);
y1 = sqrt((1 - x.^2/16)*2);
y2 = -5*x.^2 +2;
figure(1)
plot(x, real(y1), 'r', x, -real(y1), 'r')
hold on
plot(x, y2)
hold off
Since the ellipse has real y's not on the whole x domain, if you want to plot only real parts, specify real(y1) or abs(y1) (even though Matlab does it for you, too). You can also dismiss complex numbers for certain x when computing y1, but you'll need a for-loop for that.
In order to make things simpler, you can check the function fimplicit, ezplot is not recommended according to Matlab's documentation. Or if you want to plot the ellipse in a parametric way, fplot will work, too.
Another (more classic) approach for parametric plotting is given here already, then you don't need any other functions than what you already use. I think it is the simplest and most elegant way to plot an ellipse.
You will not be able to generate points for the ellipse using a function f(x) from a Cartesian linspace range. Instead, you can still use linspace but for the angle in a polar notation, from 0 to 2*pi. You should also be able to easily adjust radius and offset on both axis on the cos and sin expressions.
x = linspace(-5, 5);
y2 = -5*x.^2 +2;
figure(1)
clf;
plot(x, y2)
hold on
a = linspace(0,2*pi);
x2 = 4*cos(a);
y2 = sqrt(2)*sin(a);
plot(x2, y2)
xlim([-5,5]);
ylim([-5,5]);
hold off
I wish to create a limit cycle in Matlab. A limit cycle looks something like this:
I have no idea how to do it though, I've never done anything like this in Matlab.
The equations to describe the limit cycle are the following:
x_1d=x_2
x_2d=-x_1+x_2-2*(x_1+2*x_2)x_2^2
It is to be centered around the equilibrium which is (0,0)
Can any of you help me?
If you use the partial derivatives of your function to make a vector field, you can then use streamlines to visualize the cycle that you are describing.
For example, the function f = x^2+y^2
Gives me partial derivatives dx = 2x, dy=2y For the visualization, I sample from the partial derivatives over a grid.
[x,y] = meshgrid(0:0.1:1,0:0.1:1);
dx = 2*x;
dy = 2*y;
To visualize the vector field, I use quiver;
figure;
quiver(x, y, dx, dy);
Using streamline, I can visualize the path a particle injected into the vector field would take. In my example, I inject the particle at (0.1, 0.1)
streamline(x,y, dx, dy, 0.1, 0.1);
This produces the following visualization
In your case, you can omit the quiver step to remove the hedgehog arrows at every grid point.
Here's another example that shows the particle converging to an orbit.
Edit: Your function specifically.
So as knedlsepp points out, the function you are interested in is a bit ambiguously stated. In Matlab, * represents the matrix product while .* represents the element-wise multiplication between matrices. Similarly, '^2' represents MM for a matrix M, while .^2 represents taking the element-wise power.
So,
[x_1,x_2] = meshgrid(-4:0.1:4,-4:0.1:4);
dx_1 = x_2;
dx_2 = -x_1+x_2-2*(x_1+2*x_2)*(x_2)^2;
figure; streamline(x_1,x_2, dx_1, dx_2, 0:0.1:4, 0:0.1:4);
Looks like
This function will not show convergence because it doesn't converge.
knedlsepp suggests that the function you are actually interested in is
dx_1 = -1 * x_2;
dx_2 = -1 * -x_1+x_2-2*(x_1+2*x_2).*(x_2).^2;
His post has a nice description of the rest.
This post shows the code to produce the integral lines of your vector field defined by:
dx/dt = y
dy/dt = -x+y-2*(x+2*y)*y^2.
It is important to properly vectorize this function. (i.e. Introducing dots at all the important places)
dxdt = #(x,y) y;
dydt = #(x,y) -x+y-2*(x+2*y).*y.^2;
[X,Y] = meshgrid(linspace(-4,4,100));
[sx,sy] = meshgrid(linspace(-3,3,20));
streamline(stream2(X, Y, ... % Points
dxdt(X,Y), dydt(X,Y),... % Derivatives
sx, sy)); % Starting points
axis equal tight
To get a picture more similar to yours, change the grid size and starting points:
[X,Y] = meshgrid(linspace(-1,1,100));
[sx,sy] = meshgrid(linspace(0,0.75,20),0.2);
I want to plot relations like y^2=x^2(x+3) in MATLAB without using ezplot or doing algebra to find each branch of the function.
Does anyone know how I can do this? I usually create a linspace and then create a function over the linspace. For example
x=linspace(-pi,pi,1001);
f=sin(x);
plot(x,f)
Can I do something similar for the relation I have provided?
What you could do is use solve and allow MATLAB's symbolic solver to symbolically solve for an expression of y in terms of x. Once you do this, you can use subs to substitute values of x into the expression found from solve and plot all of these together. Bear in mind that you will need to cast the result of subs with double because you want the numerical result of the substitution. Not doing this will still leave the answer in MATLAB's symbolic format, and it is incompatible for use when you want to plot the final points on your graph.
Also, what you'll need to do is that given equations like what you have posted above, you may have to loop over each solution, substitute your values of x into each, then add them to the plot.
Something like the following. Here, you also have control over the domain as you have desired:
syms x y;
eqn = solve('y^2 == x^2*(x+3)', 'y'); %// Solve for y, as an expression of x
xval = linspace(-1, 1, 1000);
%// Spawn a blank figure and remember stuff as we throw it in
figure;
hold on;
%// For as many solutions as we have...
for idx = 1 : numel(eqn)
%// Substitute our values of x into each solution
yval = double(subs(eqn(idx), xval));
%// Plot the points
plot(xval, yval);
end
%// Add a grid
grid;
Take special care of how I used solve. I specified y because I want to solve for y, which will give me an expression in terms of x. x is our independent variable, and so this is important. I then specify a grid of x points from -1 to 1 - exactly 1000 points actually. I spawn a blank figure, then for as many solutions to the equation that we have, we determine the output y values for each solution we have given the x values that I made earlier. I then plot these on a graph of these points. Note that I used hold on to add more points with each invocation to plot. If I didn't do this, the figure would refresh itself and only remember the most recent call to plot. You want to put all of the points on here generated from all of the solution. For some neatness, I threw a grid in.
This is what I get:
Ok I was about to write my answer and I just saw that #rayryeng proposed a similar idea (Good job Ray!) but here it goes. The idea is also to use solve to get an expression for y, then convert the symbolic function to an anonymous function and then plot it. The code is general for any number of solutions you get from solve:
clear
clc
close all
syms x y
FunXY = y^2 == x^2*(x+3);
%//Use solve to solve for y.
Y = solve(FunXY,y);
%// Create anonymous functions, stored in a cell array.
NumSol = numel(Y); %// Number of solutions.
G = cell(1,NumSol);
for k = 1:NumSol
G{k} = matlabFunction(Y(k))
end
%// Plot the functions...
figure
hold on
for PlotCounter = 1:NumSol
fplot(G{PlotCounter},[-pi,pi])
end
hold off
The result is the following:
n = 1000;
[x y] = meshgrid(linspace(-3,3,n),linspace(-3,3,n));
z = nan(n,n);
z = (y .^ 2 <= x .^2 .* (x + 3) + .1);
z = z & (y .^ 2 >= x .^2 .* (x + 3) - .1);
contour(x,y,z)
It's probably not what you want, but I it's pretty cool!
Basically, I have a function f(X,Y) that would return one value for each X,Y that I give. Is there any function in matlab where I can pass the function f, the ranges for X,Y so that it plots a 3d graph showing the magnitude of f (along the z axis) for all values within the given range.
ezplot3, does this kind of, but it takes only one parameter 't'. I am very new to matlab and am trying my best to learn it fast, but I couldnt find much regarding this. Any help would be appreciated
Keep in mind, that with matlab, you're never really plotting "functions"; You're plotting arrays/vectors. So instead of trying to plot g = f(X,Y), you'll actually by plotting the vectors X, Y, and g, where X and Y are your original inputs and g is a vector containing your outputs.
I'm having a hard time visualizing what exactly you're trying to plot but basically, you can follow any standard matlab plotting example such as: http://web.cecs.pdx.edu/~gerry/MATLAB/plotting/plotting.html
It does not produce a 3D plot, but I have found the 2D scatter plot useful for this kind of task before:
scatter(x, y, 5, z)
Where z is the value of the function at the point (x, y) will produce something similar to what you want. Its perhaps not quite as pretty as a full 3D plot but can be used to good effect.
See:
http://www.mathworks.com/matlabcentral/fileexchange/35287-matlab-plot-gallery-scatter-plot-2d/content/html/Scatter_Plot_2D.html
Here is some (very ugly) code I put together to demonstrate the difference:
j=1;
y = -100:1:100;
for i = -100:1:100
y = [y -100:1:100];
count = 0;
while count < 202;
x(j) = i;
j = j+1;
count = count + 1;
end
end
z = (abs(x) + abs(y));
figure(1)
scatter(x, y, 10, z)
h=colorbar;
figure(2)
ezsurf('(abs(x) + abs(y))')
Well, this is what I was going for : http://www.mathworks.com/help/matlab/ref/ezsurf.html
if i do this
ezsurf('f(x,y)');
I get the 3d graph I wanted.
Thanks anyways!
I need to draw an elliptic curve over the finite field F17(in other words, I want to draw some specific dots on the curve), but somehow I don't get it right.
The curve is defined by the equation:
y^2 = x^3 +x + 1 (mod 17)
I tried the way below, but it can't work.
for x = 0:16, plot(x, mod(sqrt(x^3+x+1), 16),'r')', end
Can someone help ?
[Update]
According to Nathan and Bill's suggestions, here is a slightly modified version.
x = 0:18
plot(mod(x,16), mod(sqrt(x.^3+x+1), 16),'ro')
However, I feel the figure is WRONG , e.g.,y is not an integer when x=4 .
You have to test all points that fulfill the equation y^2 = x^3 +x + 1 (mod 17). Since it is a finite field, you cannot simply take the square root on the right side.
This is how I would go about it:
a=0:16 %all points of your finite field
left_side = mod(a.^2,17) %left side of the equation
right_side = mod(a.^3+a+1,17) %right side of the equation
points = [];
%testing if left and right side are the same
%(you could probably do something nicer here)
for i = 1:length(right_side)
I = find(left_side == right_side(i));
for j=1:length(I)
points = [points;a(i),a(I(j))];
end
end
plot(points(:,1),points(:,2),'ro')
set(gca,'XTick',0:1:16)
set(gca,'YTick',0:1:16)
grid on;
Matlab works with vectors natively.
your syntax was close, but needs to be vectorized:
x = 0:16
plot(x, mod(sqrt(x.^3+x+1), 16),'r')
Note the . in x.^3. This tells Matlab to cube each element of x individually, as opposed to raising the vector x to the 3rd power, which doesn't mean anything.
You can use this code if you want to plot on Real numbers:
syms x y;
v=y^2-x^3-x-1;
ezplot(v, [-1,3,-5,5]);
But, for plot in modulo, at first you can write below code;
X=[]; for x=[0:16], z=[x; mod(x^3+x+1,17)]; X=[X, z]; end, X,
Then, you can plot X with a coordinate matrix.