I tried writing an Ordinary Differential Equation in MATLAB.
I wrote this code:
function [y] = odefun(t,y)
t = [0:0.01:10];
y = [0 0]';
y(1) = y(2);
y(2) = sin(2*t)-2*y(2)-2*y(1); % I get an error here
end
I get an error in the last line in this code. MATLAB doesn't show me what the error is. It just tells me I have an error in that line.
Why do I get this error and how to resolve it?
What you want is to carefully read the documentation of the diverse ode solvers and the examples there and then correct your code to something like
% Solve ODE y''(t)+2*y'(t)+2*y(t) = sin(2*t), y(0)=y'(0)=0
function ydot = odefun(t,y)
ydot = zeros_like(y)
ydot(1) = y(2);
ydot(2) = sin(2*t)-2*y(2)-2*y(1);
end
% or
% odefun = #(y,t) [ y(2); sin(2*t)-2*y(2)-2*y(1) ]
% define sample points
tspan = [0:0.01:10];
% define initial value to t=tspan(1)
y0 = [0 0]';
[ t, y ] = ode45(odefunc, tspan, y0)
% t,y now contain the times and values that
% the solution was actually computed for.
You try to assign to y(2) a vector of 1001 elements:
>> size(sin(2*t)-2*y(2)-2*y(1))
ans =
1 1001
The error message is pretty clear:
In an assignment A(:) = B, the number of elements in A and B must be the same.
Also, y and t are never use since you redefined them in the function.
Related
This is my code for Euler's explicit Method on MATLAB
function [t,x] = meuler(f, intervalo, x0, N)
h = (intervalo(2)-intervalo(1))/N;
t = intervalo(1):h:intervalo(2);
x(:,1) = x0(:);
for i = 1:N
x(i+1,:) = x(i,:) + h*(f(t(i),x(i,:)));
end
It takes a function f defined in another file. When I try to run the code with the following function
function f = funccorazon(t,x)
f1 = x(2);
f2 = 16*x(1)+4*sin(2*t);
f=[f1;f2];
I get this error
>> meuler(#funccorazon, [0 2*pi], [0 2], 1000)
Attempted to access y(2); index out of bounds because numel(y)=1.
Error in funccorazon (line 2)
f1 = y(2);
and I don't know why. Apparently, when I solve the differential equation using ode45, nothing seems to go wrong. Any help with that would be appreciated. Thank you!
The reason for the error is that you need to work with row vectors instead of column vectors.
first, in your function call, define the input x0 as row vector:
meuler(#funccorazon, [0 2*pi], [0,2], 1000)
second, in funccorazon function, define the output f similarly:
f=[f1,f2];
lastly, in meuler function do the following changes:
x = zeros(N,2);
x(1,:) = x0;
I need some help with finding solution to Cauchy problem in Matlab.
The problem:
y''+10xy = 0, y(0) = 7, y '(0) = 3
Also I need to plot the graph.
I wrote some code but, I'm not sure whether it's correct or not. Particularly in function section.
Can somebody check it? If it's not correct, where I made a mistake?
Here is separate function in other .m file:
function dydx = funpr12(x,y)
dydx = y(2)+10*x*y
end
Main:
%% Cauchy problem
clear all, clc
xint = [0,5]; % interval
y0 = [7;3]; % initial conditions
% numerical solution using ode45
sol = ode45(#funpr12,xint,y0);
xx = [0:0.01:5]; % vector of x values
y = deval(sol,xx); % vector of y values
plot(xx,y(1,:),'r', 'LineWidth',3)
legend('y1(x)')
xlabel('x')
ylabel('y(x)')
I get this graph:
ode45 and its related ilk are only designed to solve first-order differential equations which are of the form y' = .... You need to do a bit of work if you want to solve second-order differential questions.
Specifically, you'll have to represent your problem as a system of first-order differential equations. You currently have the following ODE:
y'' + 10xy = 0, y(0) = 7, y'(0) = 3
If we rearrange this to solve for y'', we get:
y'' = -10xy, y(0) = 7, y'(0) = 3
Next, you'll want to use two variables... call it y1 and y2, such that:
y1 = y
y2 = y'
The way you have built your code for ode45, the initial conditions that you specified are exactly this - the guess using y and its first-order guess y'.
Taking the derivative of each side gives:
y1' = y'
y2' = y''
Now, doing some final substitutions we get this final system of first-order differential equations:
y1' = y2
y2' = -10*x*y1
If you're having trouble seeing this, simply remember that y1 = y, y2 = y' and finally y2' = y'' = -10*x*y = -10*x*y1. Therefore, you now need to build your function so that it looks like this:
function dydx = funpr12(x,y)
y1 = y(2);
y2 = -10*x*y(1);
dydx = [y1 y2];
end
Remember that the vector y is a two element vector which represents the value of y and the value of y' respectively at each time point specified at x. I would also argue that making this an anonymous function is cleaner. It requires less code:
funpr12 = #(x,y) [y(2); -10*x*y(1)];
Now go ahead and solve it (using your code):
%%// Cauchy problem
clear all, clc
funpr12 = #(x,y) [y(2); -10*x*y(1)]; %// Change
xint = [0,5]; % interval
y0 = [7;3]; % initial conditions
% numerical solution using ode45
sol = ode45(funpr12,xint,y0); %// Change - already a handle
xx = [0:0.01:5]; % vector of x values
y = deval(sol,xx); % vector of y values
plot(xx,y(1,:),'r', 'LineWidth',3)
legend('y1(x)')
xlabel('x')
ylabel('y(x)')
Take note that the output when simulating the solution to the differential equation by deval will be a two column matrix. The first column is the solution to the system while the second column is the derivative of the solution. As such, you'll want to plot the first column, which is what the plot syntax is doing.
I get this plot now:
I am trying to implement the Taylor method for ODEs in MatLab:
My code (so far) looks like this...
function [x,y] = TaylorEDO(f, a, b, n, y0)
% syms t
% x = sym('x(t)'); % x(t)
% f = (t^2)*x+x*(1-x);
h = (b - a)/n;
fprime = diff(f);
f2prime = diff(fprime);
y(0) = y0,
for i=1:n
T((i-1)*h, y(i-1), n) = double(f((i-1)*h, y(i-1)))+(h/2)*fprime((i-1)*h, y(i-1))
y(i+1) = w(i) + h*T(t(i), y(i), n);
I was trying to use symbolic variables, but I don´t know if/when I have to use double.
I also tried this other code, which is from a Matlab function, but I do not understand how f should enter the code and how this df is calculated.
http://www.mathworks.com/matlabcentral/fileexchange/2181-numerical-methods-using-matlab-2e/content/edition2/matlab/chap_9/taylor.m
As error using the function from this link, I got:
>> taylor('f',0,2,0,20)
Error using feval
Undefined function 'df' for input arguments of type 'double'.
Error in TaylorEDO (line 28)
D = feval('df',tj,yj)
The f I used here was
syms t
x = sym('x(t)'); % x(t)
f = (t^2)*x+x*(1-x);
This is a numerical method, so it needs numerical functions. However, some of them are computed from the derivatives of the function f. For that, you need symbolic differentiation.
Relevant Matlab commands are symfun (create a symbolic function) and matlabFunction (convert a symbolic function to numerical).
The code you have so far doesn't seem salvageable. You need to start somewhere closer to basics, e.g., "Matlab indices begin at 1". So I'll fill the gap (computation of df) in the code you linked to. The comments should explain what is going on.
function [T,Y] = taylor(f,a,b,ya,m)
syms t y
dfs(1) = symfun(f, [t y]); % make sure that the function has 2 arguments, even if the user passes in something like 2*y
for k=1:3
dfs(k+1) = diff(dfs(k),t)+f*diff(dfs(k),y); % the idea of Taylor method: calculate the higher derivatives of solution from the ODE
end
df = matlabFunction(symfun(dfs,[t y])); % convert to numerical function; again, make sure it has two variables
h = (b - a)/m; % the rest is unchanged except one line
T = zeros(1,m+1);
Y = zeros(1,m+1);
T(1) = a;
Y(1) = ya;
for j=1:m
tj = T(j);
yj = Y(j);
D = df(tj,yj); % syntax change here; feval is unnecessary with the above approach to df
Y(j+1) = yj + h*(D(1)+h*(D(2)/2+h*(D(3)/6+h*D(4)/24)));
T(j+1) = a + h*j;
end
end
Example of usage:
syms t y
[T, Y] = taylor(t*y, 0, 1, 2, 100);
plot(T,Y)
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 was wondering if it's possible to combine for-loops/while-loops and ODE45?, please look at the example below:
I have a function (ode) which I want to solve at different ic (initial conditions):
ser = #(x) x.^(-0.3) - x.^(1.8); % the function
tspan = 1:0.02:2;
x0 = 0.5;
% x0 = 0.8;
% x0 = 1.2;
% x0 = 1.8;
% x0 = 2;
% x0 = 2.5;
[~, x_t] = ode45(#(t,x) ser(x), tspan, x0);
plot(tspan,x_t,'r-')
And plot the solution curves at the end.
Is there a way to pass the other icwithout doing it manually, like a loop? or any other way to optimize this step? If I could receive some help with this, it will be very appreciated because I have to compute many ODEs (more complex ones) at 15 to 25 different ic.
Thanks in advance!
PS. If the code has to change (e.g. different names for the x0s or solutions x_t) it will be OK!
If you have a predefined no. of ic you can use this for loop:
for i=1:NoOfic
test_mat(i,:)=test_ode_45(x(i),tspan);
end
You should predefine your matrix (test_mat) in a sufficient dimension e.g.
test_mat = zeros(NoOfic, SizeOftspan)
Obviously I have defined the variables NoOficand SizeOftspanto the specific numbers
My whole test-code is:
tspan = 1:0.02:2;
x= [0.4 0.8 1.2 1.5 2.6];
sizeOftspan = size(tspan);
sizeOfFamily= size(x);
test_mat = zeros(sizeOfFamily(2),sizeOftspan(2));
test_tes= test_ode_45(x(1),tspan);
for i=1:sizeOfFamily(2)
test_mat(i,:)=test_ode_45(x(i),tspan);
end
plot(tspan, test_mat)
Where my test_ode_45 function is a simple x*sin(t) function