I want to solve 100 differential equation by ode45. My entire 100 ode equation has same pattern i.e. in the form of xdot= Ax + By + Cu where xdot, A, B and C have n1 order. So the first equation is xdot(1)=A(1)*x(1) + B(1)*y(1) +C(1)*u(1) and nth equation is xdot(n)= A(n)*x(n) + B(n)*y(n) +C(n)*u(n). It is very hectic to write this 100 equation in function script file of ode MATLAB. My question is that how to write this hundred equation using for loop or any other methods in MATLAB function script and solve the ode equation ?
You can define your system using matrices instead of a list of ODE's.
Given your vectors a, b and c
A = diag(a); B = diag(b); C = diag(c);
xdot = #(x, y, u) A*x + B*y + C*u;
then solve xdot using some solver like ode45.
Edit
Of course you can also use dot-notation. I initially wrote my answer in matrix form due to that being the standard in linear algebra.
xdot = #(x, y, u) a.*x + b.*y + c.*u;
Related
I would like to create a multivariate functional handle which the number of variables is changeable according to the input.
First, create n symbolic variables, and note that n can be changed according to your input.
n=3;
syms theta [1 n];
Now I create a function g. Via For loop, create the summation of g on all theta. As seen in the code, f is a symbolic expression.
g = #(x)(x^2);
f = 0;
for i = 1:n
f = f + g(sym(sprintfc('theta%d',i)))
end
Now I want to create a functional handle F according to f.
One potential way to do this F = #(theta1,theta2,theta3)(f). However, since n is user-specified, changeable variable, this approach is not doable.
Could someone give my hint? Many thanks!
Is this what you are looking for?
g = #(x)x.^2
fn = #(varargin) sum( cellfun(g,varargin) )
Now we have an anonymous function with a variable number of inputs. Example use below
fn(1) % = 1
fn(1,5,3) % = 35 = (1^2+5^2+3^2)
fn(1,2,3,4,5,6) % = 91 = (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2)
I have been working on a machine learning course and currently on Classification. I implemented the classification algorithm and obtained the parameters as well as the cost. The assignment already has a function for plotting the decision boundary and it worked but I was trying to read their code and cannot understand these lines.
plot_x = [min(X(:,2))-2, max(X(:,2))+2];
% Calculate the decision boundary line
plot_y = (-1./theta(3)).*(theta(2).*plot_x + theta(1));
Anyone explain?
I'm also taking the same course as you. I guess what the code does is to generate two points on the decision line.
As you know you have the function:
theta0 + theta1 * x1 + theta2 * x2 = 0
Which it can be rewritten as:
c + mx + ky = 0
where x and y are the axis corresponding to x1 and x2, c is theta(0) or the y-intercept, m is the slope or theta(1), and k is theta(2).
This equation (c + mx + ky = 0) corresponds to the decision boundary, so the code is finding two values for x (or x1) which cover the whole dataset (-2 and +2 in plot_x min and max functions) and then uses the equation to find the corresponding y (or x2) values. Finally, a decision boundary can be plotted -- plot(plot_x, plot_y).
In other words, what it does is to use the the equation to generate two points to plot the line on graph, the reason of doing this is that Octave cannot plot the line given an equation to it.
Hope this can help you, sorry for any mistake in grammar or unclear explanation ^.^
Rearranging equations helped me, so adding those here:
plot_y = -1/theta2 (theta1*plot_x + theta0)
note that index in Octave starts at 1, not at 0, so theta(3) = theta2, theta(2) = theta1 and theta(1) = theta0.
This plot_y equation is equivalent to:
c + mx + ky = 0 <=>
-ky = mx + c <=>
y = -1/k (mx + c)
How do I solve the differential equation y'+y=t with y(0)=24?
Do I need to defined the differential equation with a file in the form .m?
To solve ordinary differential equations you've got the function lsode (run lsode for help).
f = #(y,t) t-y;
t = linspace(0,5,50)';
y=lsode(f, 24, t);
plot(t,y);
print -djpg figure-lsnode.jpg
dy/dt + y = t
Multiply integrating factor e^t on both sides. Then,
d/dt [(e^t)y] = te^t
y = t-1+ce^(-t)
Because, y(0)=24 then c=25
y = t-1+25e^(-t)
I am attempting to use Octave to solve for a differential equation using Euler's method.
The Euler method was given to me (and is correct), which works for the given Initial Value Problem,
y*y'' + (y')^2 + 1 = 0; y(1) = 1;
That initial value problem is defined in the following Octave function:
function [YDOT] = f(t, Y)
YDOT(1) = Y(2);
YDOT(2) = -(1 + Y(2)^2)/Y(1);
The question I have is about this function definition. Why is YDOT(1) != 1? What is Y(2)?
I have not found any documentation on the definition of a function using function [YDOT] instead of simply function YDOT, and I would appreciate any clarification on what the Octave code is doing.
First things first: You have a (non linear) differential equation of order two which will require you to have two initial conditions. Thus the given information from above is not enough.
The following is defined for further explanations: A==B means A is identical to B; A=>B means B follows from A.
It seems you are mixing a few things. The guy who gave you the files rewrote the equation in the following way:
y*y'' + (y')^2 + 1 = 0; y(1) = 1; | (I) y := y1 & (II) y' := y2
(I) & (II)=>(III): y' = y2 = y1' | y2==Y(2) & y1'==YDOT(1)
Ocatve is "matrix/vector oriented" so we are writing everything in vectors or matrices. Rather writing y1=alpha and y2=beta we are writing y=[alpha; beta] where y(1)==y1=alpha and y(2)==y2=beta. You will soon realize the tremendous advantage of using especially this mathematical formalization for ALL of your problems.
(III) & f(t,Y)=>(IV): y2' == YDOT(2) = y'' = (-1 -(y')^2) / y
Now recall what is y' and y from the definitions in (I) and (II)!
y' = y2 == Y(2) & y = y1 == Y(1)
So we can rewrite equation (IV)
(IV): y2' == YDOT(2) = (-1 -(y')^2) / y == -(1 + Y(2)^2)/Y(1)
So from equation (III) and (IV) we can derive what you already know:
YDOT(1) = Y(2)
YDOT(2) = -(1 + Y(2)^2)/Y(1)
These equations are passed to the solver. Differential equations of all types are solved numerically by retrieving the "next" value in a near neighborhood to some "previously known" value. (The step size inside this neighborhood is one of the key questions when writing solvers!) So your solver uses your initial condition Y(1)==y(1)=1 to make the next step and calculate the "next" value. So right at the start YDOT(1)=Y(2)==y(2) but you didn't tell us this value! But from then on YDOT(1) is varied by the solver in dependency to the function shape to solve your problem and give you ONE unique y(t) solution.
It seems you are using Octave for the first time so let's make a last comment on function [YDOT] = f(t, Y). In general a function is defined in this way:
function[retVal1, retVal2, ...] = myArbitraryName(arg1, arg2, ...)
Where retVal is the return value or output and arg is the argument or input.
I am a first time MATLAB user. I have a 2d array of t vs f in MATLAB. This 2d array correponds to a function, say f(t). I am using ode45 to solve a set of differential equations and f(t) is one of the coefficients i.e. I have a set of equations of the form x'(t)=f(t)x(t) or variations thereof. How do I proceed?
I need a way to convert my array of t vs f into a function that can be used in the ode45 method. Presumably, the conversion will do some linear interpolation and give me the equation of the best fit curve.
Or if there is another approach, I'm all ears!
Thank you!
This is a simple example with passing function as a parameter to right hand side of the ODE. Create files in the same directory and run main
main.m
t = 0:0.2:10; f = sin(t);
fun = #(xc) interp1(t, f, xc);
x0=0.5
% solve diff(x, t)=sin(t)
% pass function as parameter
[tsol, xsol] = ode45(#(t, x) diff_rhs(t, x, fun),[0.0 8.0], 0.5);
% we solve equation dx/dt = sin(x), x(0)=x0
% exact solution is x(t) = - cos(t) + x(0) + 1
plot(tsol, xsol, 'x');
hold on
plot(tsol, -cos(tsol) + x0 + 1, '-');
legend('numerical', 'exact');
diff_rhs.m
function dx = diff_rhs(t, x, fun)
dx = fun(t);
end
References in the documentation: interp1 and anonymous Functions.