My function is:
function [tout, yout] = Lorenz (rho, x0)
%define constants
sigma = 10;
beta = 8/3;
%initial conditions
y0 = [x0; 1; 0];
f = #(t, y) [sigma*(y(2)-y(1)); (y(1)*(rho-y(3)))-y(2); (y(1)*y(2)) - (beta*y(3))];
[tout, yout] = ode45(f, [0 100], y0, 'RelTol', 1e-6, 'AbsTol', 1e-8);
end
When I run the function in the command window with
Lorenz(14,0)
I return
Error using Lorenz>#(t,y)[sigma*(y(2)-y(1));(y(1)*(rho-y(3)))-y(2);(y(1)*y(2))-(beta*y(3))]
Too many input arguments.
Any help would be much appreciated.
Although the syntax is not officially documented by MathWorks, the ODE suite does accept the syntax:
[t,y] = ode45(odefun,tspan,y0,options,extra1,extra2,...);
where options should be a struct created by odeset (not falling inline with the name-value system of other functions) and extra1,extra2,... is any number of extra, solve-constant parameters to be passed to odefun. I imagine it's a hold over from before anonymous functions possessed their own workspace to allow creation-time function parametrization.
So, since the options you're passing are not part of a struct, ode45 takes the syntax to be
[t,y] = ode45(odefun,tspan,y0,extra1,extra2,extra3,extra4);
and, via feval, will make the call odefun(t,y,extra1,extra2,extra3,extra4). A minor rewrite using odeset should do the job nicely:
options = odeset('RelTol', 1e-6, 'AbsTol', 1e-8);
[tout, yout] = ode45(f, [0 100], y0, options);
Related
I have written the following Matlab function:
function EulerMethod(t_min,t_max,h,f,Y,yzero)
tlist = t_min:h:t_max;
N = (t_max - t_min)/h;
ylist = transpose(zeros(N+1,1));
ylist(1) = yzero;
for i=1:N
term = f(tlist(i),ylist(i))*h;
ylist(i+1) = ylist(i) + term;
end
yrange = Y(tlist);
% modified to generate a new figure window each time.
figure;
plot(tlist,yrange,'red','LineWidth', 2);
hold;
plot(tlist,ylist,'blue','LineWidth', 2);
plot(tlist, abs(yrange - ylist),'magenta','LineWidth', 2)
% modified to wrap the title.
title({'Graphs of the True Solution, Euler Solution,', 'and the Absolute Value of the Global Error (GE)'})
xlabel('t')
ylabel('Y(t)')
legend({'True Solution','Euler Solution', 'Absolute Value of the GE'},'Location','southwest')
end
I now try and call this function in another script. The variables f and Y in the function are symbolic functions; so, in the other file, I first declare these functions before calling this function. Here's the code:
clc
syms f(t,y)
syms Y(t)
f(t,y) = -y + 2.0*cos(t); %This the derivative of the function whose solution we're trying to computed
Y(t) = sin(t) + cos(t); %This is the true solution;
% Calling function
EulerMethod(0.0, 6.0, 0.2, f, Y, 1.0);
I, however, get errors when I run the second script. Can anyone help me figure out what's going wrong? I suspect this may be because of the way I have both input and used the symbolic functions f and Y but I am not sure.
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'm new in Matlab, so sorry for asking a banal and probably already answered question, but I wasn't able to solve it.
I have this ODE to solve
g''(x)+f(x)g'(x)
with g(0)=0 and g'(0)=0.5
f(x) is a known vector, so I tried
xspan = linspace(0,10,100);
y0 = [0,0.5];
%f=f(xspan), known function, for example f=1./xspan;
[x,y] = ode45('odefun',xspan,y0);
and for odefun
function dy=odefun(x,y)
global f
dy(1) = y(2);
dy(2) = -f.*y(2);
dy = dy(:);
end
Of course it doesn't work. How can I pass the values of vector f to the solver?
EDIT
Thanks for help, but this is not working for me. Maybe I have an old Matlab version (7.11.0 (R2010B))? I tried to do as you told, but this is what I get in command window
??? In an assignment A(I) = B, the number of elements in B and
I must be the same.
Error in ==> odefun at 3
dy(2) = -f.*y(2);
Error in ==> #(x,y)odefun(x,y,f)
Error in ==> odearguments at 109
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ==> ode45 at 173
[neq, tspan, ntspan, next, t0, tfinal, tdir, y0, f0, odeArgs, odeFcn, ...
Error in ==> trying at 7
[x,y] = ode45(fun,xspan,y0);
This is the code trying.m
xspan=linspace(0,10,100);
y0=[0 0.5];
f=1./xspan;
[x,y] = ode45(#(x,y)odefun(x,y,f),xspan,y0);
and the odefun
function dy=odefun(x,y,f)
dy(1) = y(2);
dy(2) = -f.*y(2);
dy = dy(:);
end
It looks like you're using global variables incorrectly (f needs to be declared global in your main function as well), but you shouldn't be using globals to begin with. You should rarely, if ever, use globals. You should also specify your integration function, odefun, as an anonymous function rather than a string in modern Matlab. Here is how you can modify your code to fix these issues:
...
f = ...
[x,y] = ode45(#(x,y)odefun(x,y,f),xspan,y0);
And your integration function, which can be a sub-function or in a separate file, just needs to take an extra argument for the parameter f:
function dy=odefun(x,y,f)
dy(1) = y(2);
dy(2) = -f.*y(2);
dy = dy(:);
The methods should also be faster than using strings and globals. The code #(x,y)odefun(x,y,f) creates an anonymous function of x and y (ode45 expects a function that takes exactly two inputs) that captures the value of f from the current scope. This is also referred to as a closure in computer science. You can equivalently create a function handle and pass that in:
...
f = ...
fun = #(x,y)odefun(x,y,f)
[x,y] = ode45(fun,xspan,y0);
See also this article on parametrizing functions from The MathWorks for more information on this general technique.
I solved it, it was not so difficult, I just needed to interpolate the f at each value of the independent variable x. In this way the code works
function dy=odefun(x,y,f,xspan)
f=interp1(xspan,f,x);
dy(1) = y(2);
dy(2) = -f.*y(2);
dy = dy(:);
end
I would like to use odeplot so stepwise get the result instead of plotting the result afterwards. I've tried to write it like this but I can't get it to work so I would appreciate some help.
%Parameters
s = 1;
q = 1;
w = 0.1610;
y0 = [30 1 30]; % Initial values
tspan = [0 10]; % Time 0<t<10
plot=odeset('OutputFcn','odeplot');
[t, y] = ode45(#(t,y) concentration(t, y, s, q, w), plot, tspan, y0);
You need to specify your output function via an ODE Options argument:
options = odeset('OutputFcn', #odeplot);
[t, y] = ode45(#(t,y)concentration(t, y, s, q, w), tspan, y0, options);
Of course you can make your own custom output function. Type edit odeplot to see what is required (much simpler functions are possible). Also check out odephas2 and odephas3.