I've already searched it here but I couldn't found it the way I was looking for.
I kind managed to do it using Symbolic Math but I don't understand it quite well. For exemple, after doing that
syms y x
ezplot(-y + x + 1 == 0)
i get a nice graph, but can I use this expression later to calculate its value? like, first I want to plot -y + x + 1 == 0 and at another moment I want to solve f(3) for exemple, where f(x) = x + 1 (same equation).
I know I can write a function to do that, but as a function I don't know how to plot it. In the other way, I know how to plot using symbolic math, but I don't know how to calculate it after.
I'm writing a PLA algorithm and them I need to generate the 'a', 'b' 'c' for the equation, that why I need to know how to plot and solve in a "systematic code" way, and not typing one by one.
Thanks in advance!
The equation you gave us is a straight line, so a polynomial. The coefficients are y= -b/a*x -c/a.
% ay + bx + c = 0 reads y = -b/a*x - c/a*1
a = -1;
b = 1;
c = 1;
p = [-b/a, -c/a]; % polynomial representing your equation
% plot like this
x = linspace(-2,2, 50);
figure
plot(x, polyval(p,x)) % evaluate polynomial p at the positions x
% find the solution
roots(p) # -1
If you need or want to use ezplot, you can put the polyval-expression in an inline function and you can call ezplot with that handle:
f = #(x) polyval(p, x); % the function
ezplot(f)
Just define f to be a function of the symbolic variable x:
>> syms x
>> f = x+1;
Then you can use f as the input to ezplot:
>> ezplot(f)
which produces the graph
On the other hand, to solve the equation f(x)=0 use solve as follows:
>> solve(f)
ans =
-1
ezplot and solve can be used with string inputs as well, but the string has to be different in either case. To plot the graph:
>> ezplot('x+1');
To solve the equation:
>> solve('x+1=0')
ans =
-1
Related
I'm having trouble writing this code. So, I'm trying to make a code for
func = y*x(n) + z * x(n)
All the values are arbitrary and x(n) is the value at the position n. I need to plot a graph at each nth position. So if x(1) = 5 I plot a point at when x=1 and y=5. The issue is that I can't figure out how to make an arbitrary array and don't know how to get the answer for func when I add x(n) value at the nth position. I also am having trouble plotting a graph, but think this is because I can't figure out to use the array yet.
I'm new to MatLab.
So if I am following y & z are just constants? I think the confusion is typically this would be written something like "y = ax +bx"
Like Cris Luengo mentioned in the comments above, you should should go over some basic Matlab tutorials as this is very basic.
% y and Z are constants
y = 1;
z = 2;
%this makes x = [0,1,2,...10];
x = 0:1:10;
func = y.*x + z.*x;
plot(func)
This should do the trick:
% Define X as a range between -10 to 10 (+1 on every step)...
x = -10:10;
% Define your constants...
y = 3;
z = -1;
% Define the function...
fun = #(x) (y .* x) + (z .* x);
% Plot X on the x-axis and fun(x) on the y-axis...
% fun(x) numerically evaluates fun for the given x
plot(x,fun(x));
Refer to this page for more information about anonymous functions.
In order to obtain a graphical representation of the behaviour of a fluid it is common practice to plot its streamlines.
For a given two-dimensional fluid with speed components u = Kx and v = -Ky (where K is a constant, for example: K = 5), the streamline equation can be obtained integrating the flow velocity field components as follows:
Streamline equation: ∫dx/u = ∫dy/v
The solved equation looks like this: A = B + C (where A is the solution of the first integral, B is the solution of the second integral and C is an integration constant).
Once we have achieved this, we can start plotting a streamline by simply assigning a value to C, for example: C = 1, and plotting the resulting equation. That would generate a single streamline, so in order to get more of them you need to iterate this last step assigning a different value of C each time.
I have successfully plotted the streamlines of this particular flow by letting matlab integrate the equation symbolically and using ezplot to produce a graphic as follows:
syms x y
K = 5; %Constant.
u = K*x; %Velocity component in x direction.
v = -K*y; %Velocity component in y direction.
A = int(1/u,x); %First integral.
B = int(1/v,y); %Second integral.
for C = -10:0.1:10; %Loop. C is assigned a different value in each iteration.
eqn = A == B + C; %Solved streamline equation.
ezplot(eqn,[-1,1]); %Plot streamline.
hold on;
end
axis equal;
axis([-1 1 -1 1]);
This is the result:
The problem is that for some particular regions of the flow ezplot is not accurate enough and doesn't handle singularities very well (asymptotes, etc.). That's why a standard "numeric" plot seems desirable, in order to obtain a better visual output.
The challenge here is to convert the symbolic streamline solution into an explicit expression that would be compatible with the standard plot function.
I have tried to do it like this, using subs and solve with no success at all (Matlab throws an error).
syms x y
K = 5; %Constant.
u = K*x; %Velocity component in x direction.
v = -K*y; %Velocity component in y direction.
A = int(1/u,x); %First integral.
B = int(1/v,y); %Second integral.
X = -1:0.1:1; %Array of x values for plotting.
for C = -10:0.1:10; %Loop. C is assigned a different value in each iteration.
eqn = A == B + C; %Solved streamline equation.
Y = subs(solve(eqn,y),x,X); %Explicit streamline expression for Y.
plot(X,Y); %Standard plot call.
hold on;
end
This is the error that is displayed on the command window:
Error using mupadmex
Error in MuPAD command: Division by zero.
[_power]
Evaluating: symobj::trysubs
Error in sym/subs>mupadsubs (line 139)
G =
mupadmex('symobj::fullsubs',F.s,X2,Y2);
Error in sym/subs (line 124)
G = mupadsubs(F,X,Y);
Error in Flow_Streamlines (line 18)
Y = subs(solve(eqn,y),x,X); %Explicit
streamline expression for Y.
So, how should this be done?
Since you are using subs many times, matlabFunction is more efficient. You can use C as a parameter, and solve for y in terms of both x and C. Then the for loop is very much faster:
syms x y
K = 5; %Constant.
u = K*x; %Velocity component in x direction.
v = -K*y; %Velocity component in y direction.
A = int(1/u,x); %First integral.
B = int(1/v,y); %Second integral.
X = -1:0.1:1; %Array of x values for plotting.
syms C % C is treated as a parameter
eqn = A == B + C; %Solved streamline equation.
% Now solve the eqn for y, and make it into a function of `x` and `C`
Y=matlabFunction(solve(eqn,y),'vars',{'x','C'})
for C = -10:0.1:10; %Loop. C is assigned a different value in each iteration.
plot(X,Y(X,C)); %Standard plot call, but using the function for `Y`
hold on;
end
How can I use matlab to solve the following Ordinary Differential Equations?
x''/y = y''/x = -( x''y + 2x'y' + xy'')
with two known points, such as t=0: x(0)= x0, y(0) = y0; t=1: x(1) = x1, y(1) = y1 ?
It doesn't need to be a complete formula if it is difficult. A numerical solution is ok, which means, given a specific t, I can get the value of x(t) and y(t).
If matlab is hard to do this, mathematica is also OK. But as I am not familiar with mathematica, so I would prefer matlab if possible.
Looking forward to help, thanks!
I asked the same question on stackexchange, but haven't get good answer yet.
https://math.stackexchange.com/questions/812985/matlab-or-mathematica-solve-ordinary-differential-equations
Hope I can get problem solved here!
What I have tried is:
---------MATLAB
syms t
>> [x, y] = dsolve('(D2x)/y = -(y*D2x + 2Dx*Dy + x*D2y)', '(D2y)/x = -(y*D2x + 2Dx*Dy + x*D2y)','t')
Error using sym>convertExpression (line 2246)
Conversion to 'sym' returned the MuPAD error: Error: Unexpected 'identifier'.
[line 1, col 31]
Error in sym>convertChar (line 2157)
s = convertExpression(x);
Error in sym>convertCharWithOption (line 2140)
s = convertChar(x);
Error in sym>tomupad (line 1871)
S = convertCharWithOption(x,a);
Error in sym (line 104)
S.s = tomupad(x,'');
Error in dsolve>mupadDsolve (line 324)
sys = [sys_sym sym(sys_str)];
Error in dsolve (line 186)
sol = mupadDsolve(args, options);
--------MATLAB
Also, I tried to add conditions, such as x(0) = 2, y(0)=8, x(1) = 7, y(1) = 18, and the errors are still similar. So what I think is that this cannot be solve by dsolve function.
So, again, the key problem is, given two known points, such as when t=0: x(0)= x0, y(0) = y0; t=1: x(1) = x1, y(1) = y1 , how I get the value of x(t) and y(t)?
Update:
I tried ode45 functions. First, in order to turn the 2-order equations into 1-order, I set x1 = x, x2=y, x3=x', x4=y'. After some calculation, the equation becomes:
x(1)' = x(3) (1)
x(2)' = x(4) (2)
x(3)' = x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)) (3)
x(4)' = -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2) (4)
So the matlab code I wrote is:
myOdes.m
function xdot = myOdes(t,x)
xdot = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)]
end
main.m
t0 = 0;
tf = 1;
x0 = [2 3 5 7]';
[t,x] = ode45('myOdes',[t0,tf],x0);
plot(t,x)
It can work. However, actually this is not right. Because, what I know is that when t=0, the value of x and y, which is x(1) and x(2); and when t=1, the value of x and y. But the ode functions need the initial value: x0, I just wrote the condition x0 = [2 3 5 7]' randomly to help this code work. So how to solve this problem?
UPDATE:
I tried to use the function bvp4c after I realized that it is a boundary value problem and the following is my code (Suppose the two boundry value conditions are: when t=0: x=1, y=3; when t=1, x=6, y=9. x is x(1), y is x(2) ):
1. bc.m
function res = bc(ya,yb)
res = [ ya(1)-1; ya(2)-3; yb(1) - 6; yb(2)-9];
end
2. ode.m
function dydx = ode(t,x)
dydx = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)];
end
3. mainBVP.m
solinit = bvpinit(linspace(0,6,10),[1 0 -1 0]);
sol = bvp4c(#ode,#bc,solinit);
t = linspace(0,6);
x = deval(sol,t);
plot(t,x(1,:));
hold on
plot(t,x(2,:));
plot(t,x(3,:));
plot(t,x(4,:));
x(1,:)
x(2,:)
It can work, but I don't know whether it is right. I will check it again to make sure it is the right code.
As mentioned, this isn't a math site, so try to give code or something showing some effort.
However, the first step you need to do is turn the DE into normal form (i.e., no 2nd derivatives). You do this by making a separate variable equal to the derivative. Then, you use
syms x y % or any variable instead of x or y
to define variables as symbolic. Use matlabfunction to create a symbolic function based on these variables. Finally, you can use the ode45 function to solve the symbolic function while passing variable values. I recommend you look up the full documentation in matlab in order to understand it better, but here is a very basic syntax:
MyFun= matlabFunction(eq,'vars',{x,y});
[xout,yout]=ode45(#(x,Y) MyFun(variables),[variable values],Options);
Hopefully this puts you in the right direction, so try messing around with it and provide code if you need more help.
EDIT:
This is how I would solve the problem. Note: I don't really like the matlabFunction creator but this is simply a personal preference for various reasons I won't go into.
% Seperate function of the first order state equations
function dz = firstOrderEqns(t,z)
dz(4,1) = 0;
dz(1) = -2.*z(3).*z(1).*z(4)./(1 + z(4).^2 + z(2).^2);
dz(2) = z(1);
dz(3) = -2.*z(2).*z(3).*z(1)./(1 + z(4).^2 + z(2).^2);
dz(4) = z(3);
end
% runfirstOrderEqns
%% Initial conditions i.e. # t=0
z1 = 5; % dy/dt = 5 (you didn't specify these initial conditions,
% these will depend on the system which you didn't really specify
z2 = 0; % y = 0
z3 = 5; % dx/dt = 5 (The same as for z1)
z4 = 0; % x = 0
IC = [z1, z2, z3, z4];
%% Run simulation
% Time vector: i.e closed interval [0,20]
t = [0,20]; % This is where you have to know about your system
% i.e what is it's time domain.
% Note: when a system has unstable poles at
% certain places the solver can crash you need
% to understand these.
% using default settings (See documentation ode45 for 'options')
[T,Y] = ode45(#firstOrderEqns,t,IC);
%% Plot function
plot(T,Y(:,1),'-',T,Y(:,2),'-.',T,Y(:,3),':',T,Y(:,4),'.');
legend('dy/dt','y','dx/dt','x')
As in my comments I have made a lot of assumtions that you need to fix for example, you didn't specify what the initial conditions for the first derivatives of the states are i.e. (z1, z3) which is important for the response of the system. Also you didn't specify the time interval your interested for the simulation etc.
Note: The second m file can be used with any state function in the correct format
The following is the answer we finally get #Chriso: use matlab bvp4c function to solve this boundary value problem (Suppose the two boundry value conditions are: when t=0: x=1, y=3; when t=1, x=6, y=9. x is x(1), y is x(2) ):
1. bc.m
function res = bc(ya,yb)
res = [ ya(1)-1; ya(2)-3; yb(1) - 6; yb(2)-9];
end
2. ode.m
function dydx = ode(t,x)
dydx = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)];
end
3. mainBVP.m
solinit = bvpinit(linspace(0,6,10),[1 0 -1 0]);
sol = bvp4c(#ode,#bc,solinit);
t = linspace(0,6);
x = deval(sol,t);
plot(t,x(1,:));
hold on
plot(t,x(2,:));
plot(t,x(3,:));
plot(t,x(4,:));
x(1,:)
x(2,:)
I am trying to write a function that implements Newton's method in two dimensions and whilst I have done this, I have to now adjust my script so that the input parameters of my function must be f(x) in a column vector, the Jacobian matrix of f(x), the initial guess x0 and the tolerance where the function f(x) and its Jacobian matrix are in separate .m files.
As an example of a script I wrote that implements Newton's method, I have:
n=0; %initialize iteration counter
eps=1; %initialize error
x=[1;1]; %set starting value
%Computation loop
while eps>1e-10&n<100
g=[x(1)^2+x(2)^3-1;x(1)^4-x(2)^4+x(1)*x(2)]; %g(x)
eps=abs(g(1))+abs(g(2)); %error
Jg=[2*x(1),3*x(2)^2;4*x(1)^3+x(2),-4*x(2)^3+x(1)]; %Jacobian
y=x-Jg\g; %iterate
x=y; %update x
n=n+1; %counter+1
end
n,x,eps %display end values
So with this script, I had implemented the function and the Jacobian matrix into the actual script and I am struggling to work out how I can actually create a script with the input parameters required.
Thanks!
If you don't mind, I'd like to restructure your code so that it is more dynamic and more user friendly to read.
Let's start with some preliminaries. If you want to make your script truly dynamic, then I would recommend that you use the Symbolic Math Toolbox. This way, you can use MATLAB to tackle derivatives of functions for you. You first need to use the syms command, followed by any variable you want. This tells MATLAB that you are now going to treat this variable as "symbolic" (i.e. not a constant). Let's start with some basics:
syms x;
y = 2*x^2 + 6*x + 3;
dy = diff(y); % Derivative with respect to x. Should give 4*x + 6;
out = subs(y, 3); % The subs command will substitute all x's in y with the value 3
% This should give 2*(3^2) + 6*3 + 3 = 39
Because this is 2D, we're going to need 2D functions... so let's define x and y as variables. The way you call the subs command will be slightly different:
syms x, y; % Two variables now
z = 2*x*y^2 + 6*y + x;
dzx = diff(z, 'x'); % Differentiate with respect to x - Should give 2*y^2 + 1
dzy = diff(z, 'y'); % Differentiate with respect to y - Should give 4*x*y + 6
out = subs(z, {x, y}, [2, 3]); % For z, with variables x,y, substitute x = 2, y = 3
% Should give 56
One more thing... we can place equations into vectors or matrices and use subs to simultaneously substitute all values of x and y into each equation.
syms x, y;
z1 = 3*x + 6*y + 3;
z2 = 3*y + 4*y + 4;
f = [z1; z2];
out = subs(f, {x,y}, [2, 3]); % Produces a 2 x 1 vector with [27; 25]
We can do the same thing for matrices, but for brevity I won't show you how to do that. I will defer to the code and you can see it then.
Now that we have that established, let's tackle your code one piece at a time to truly make this dynamic. Your function requires the initial guess x0, the function f(x) as a column vector, the Jacobian matrix as a 2 x 2 matrix and the tolerance tol.
Before you run your script, you will need to generate your parameters:
syms x y; % Make x,y symbolic
f1 = x^2 + y^3 - 1; % Make your two equations (from your example)
f2 = x^4 - y^4 + x*y;
f = [f1; f2]; % f(x) vector
% Jacobian matrix
J = [diff(f1, 'x') diff(f1, 'y'); diff(f2, 'x') diff(f2, 'y')];
% Initial vector
x0 = [1; 1];
% Tolerance:
tol = 1e-10;
Now, make your script into a function:
% To run in MATLAB, do:
% [n, xout, tol] = Jacobian2D(f, J, x0, tol);
% disp('n = '); disp(n); disp('x = '); disp(xout); disp('tol = '); disp(tol);
function [n, xout, tol] = Jacobian2D(f, J, x0, tol)
% Just to be sure...
syms x, y;
% Initialize error
ep = 1; % Note: eps is a reserved keyword in MATLAB
% Initialize counter
n = 0;
% For the beginning of the loop
% Must transpose into a row vector as this is required by subs
xout = x0';
% Computation loop
while ep > tol && n < 100
g = subs(f, {x,y}, xout); %g(x)
ep = abs(g(1)) + abs(g(2)); %error
Jg = subs(J, {x,y}, xout); %Jacobian
yout = xout - Jg\g; %iterate
xout = yout; %update x
n = n + 1; %counter+1
end
% Transpose and convert back to number representation
xout = double(xout');
I should probably tell you that when you're doing computation using the Symbolic Math Toolbox, the data type of the numbers as you're calculating them are a sym object. You probably want to convert these back into real numbers and so you can use double to cast them back. However, if you leave them in the sym format, it displays your numbers as neat fractions if that's what you're looking for. Cast to double if you want the decimal point representation.
Now when you run this function, it should give you what you're looking for. I have not tested this code, but I'm pretty sure this will work.
Happy to answer any more questions you may have. Hope this helps.
Cheers!
I'm new to Matlab and would greatly appreciate if anyone could help.
I have double integral which i calculated using quad2d() function:
>> syms x y
>> ymin=#(x)x.^2
>> ymax=#(x)2*x
>> fun=#(x,y)x+y
>> quad2d(fun,0,2,ymin,ymax)
ans = 3.4667
How can i plot the graph of the integral?
Thanks in advance!
Instead of using quad2d, why not define fun as a symbolic function, perform symbolic integration and then use ezplot to plot the result?
Something along these lines:
syms x y t
fun = x + y;
I = int(fun, 0, t, x ^ 2, 2 * x);
ezplot(I)
If you want to plot the integral for different values of x limits (in the above example it is between 0 and 2) you can do something like this:
counter = 1;
for xmin = 0:10
xminv(counter) = xmin;
xmax = xmin+2;
xmaxv(counter) = xmax;
z(counter) = quad2d(fun,xmin,xmax,ymin,ymax);
counter = counter +1;
end
plot(xminv,z);
Of course you can change the range that you do the integral instead of 2 to any value you want. For example if you want your x changes from over a range of 3 instead of 2 change the xmax from xmax=xmin+2 to xmax=xmin+3.