In the integral
I want to optimize the function Dt, as I know the end result of the integral. I have expressions for k1 and k0 in terms of k2 and N, and it is k2 and N that I would like to optimize. They have constraints, needing to be between certain values. I have it all setup in my code, but I am just unaware of how to tell the genetic alogrithm to optimize an integral function? Is there something I'm missing here? The integral is usually evaluated numerically but I am trying to go backwards, and assuming I know an answer find the input parameters
EDIT:
All right, so here's my code. I know the integral MUST add up to a known value, and I know the value, so I need to optimize the variables with that given parameter. I have created an objective function y= integral - DT. I kept theta as syms because it is the thing being integrated to give DT.
function y = objective(k)
% Define constants
AU = astroConstants(2);
mu = astroConstants(4);
% Define start and finish parameters for the exponential sinusoid.
r1 = AU; % Initial radius
psi = pi/2; % Final polar angle of Mars/finish transfer
phi = pi/2;
r2 = 1.5*AU;
global k1
k1 = sqrt( ( (log(r1/r2) + sin(k(1)*(psi + 2*pi*k(2)))*tan(0)/k(1)) / (1-
cos(k(1)*(psi+2*pi*k(2)))) )^2 + tan(0)^2/k(1)^2 );
k0 = r1/exp(k1*sin(phi));
syms theta
R = k0*exp(k1*sin(k(1)*theta + phi));
syms theta
theta_dot = sqrt((mu/(R^3))*1/((tan(0))^2 + k1*(k(1))^2*sin(k(1)*theta +
phi) + 1));
z = 1/theta_dot;
y = int(z, theta, 0,(psi+2*pi*k(2))) - 1.3069e08;
global x
x=y;
end
my k's are constrained, and the following is the constraint function. I'm hoping what I have done here is tell it that the function MUST = 0.
function [c,c_eq] = myconstraints(k)
global k1 x
c = [norm(k1*(k(1)^2))-1 -norm(k1*(k(1)^2))];
c_eq =[x];
end
And finally, my ga code looks like this. Honestly, I've been playing with it all night and getting error messages after error messages - ranging from "constraint function must return real value" to "error in fcnvectorizer" and "unable to convert expression into double array", with the last two coming after i've removed the constraints.
clc; clear;
ObjFcn = #objective;
nvars = 2
LB = [0 2];
UB = [1 7];
ConsFcn = #myconstraints;
[k,fval] = ga(ObjFcn,nvars,[],[],[],[],LB,UB,ConsFcn);
I've been stuck on this problem for weeks and have gotten nowhere, even with searching through literature.
Related
I am trying to plot and then to use it with Matlab
in an ODE as coefficient, the function
f : [2,500] -> [0,1],
But I don't know how to write the code for the definition of the function, since it is given on different subintervals.
Below is an example that uses anonymous functions/function handles. It uses each region/condition and evaluates the boundaries numerically and stores them into variables Bounds_1 and Bounds_2. These boundaries are then used to truncate each signal by multiplying each section of the piecewise function by its corresponding condition which is a logical array. It's also good to note that this plot will almost be seen as an impulse since the bounds are really small. Alternatively, you can probably achieve the same results using the built-in piecewise() function but I think this method gives a little more insight. As i increases you'll see a plot that resembles more and more of an impulse. To plot this for multiple values or i this can be run in a for-loop.
clc;
i = 3;
Bounds_1 = [i - (1/i^2),i];
Bounds_2 = [i,i + (1/i^2)];
Bounds = [Bounds_1 Bounds_2];
Min = min(Bounds);
Max = max(Bounds);
f1 = #(x) (i^2.*x - i^3 + 1).*(Bounds_1(1) < x & x <= Bounds_1(2));
f2 = #(x) (-i^2.*x + i^3 + 1).*(Bounds_2(1) < x & x <= Bounds_2(2));
f = #(x) f1(x) + f2(x);
fplot(f);
xlim([Min-2 Max+2]);
ylim([0 1.1]);
Here is another solution. You can specify the steps along the x-axis withxstep. With xlower and xupper you can specify the range of the x-axis:
figure ; hold on;
xlabel('x'); ylabel('f(x)');
for i= 2:500
[f,x] = myfunction(0.5,i);
plot (x,f,'DisplayName',sprintf('%i',i));
end
% legend
function [f,x]=myfunction(xstep,i)
%xstep: specifies steps for the x values
% specify max range x \in = [xlower -> xupper]
xlower = -10;
xupper = 600;
x2 = (i-1/i^2): xstep: i;
f2 = i^2*x2 - i^3 + 1;
x3 = i+xstep:xstep:(1+1/i^2);
f3 = -i^2*x3 + i^3 +1;
x1 = xlower:xstep:(i-1/i^2);
f1 = 0*x1;
x4 = (i+1/i^2):xstep:xupper;
f4 = 0*x4;
f = [f1,f2,f3,f4];
x = [x1,x2,x3,x4];
end
Here is what I get
Besides, I am not exactly sure what you mean with ODE (ordinary differential equation) in f(x). For me it seems like an algebraic equation.
I am trying to calculate the integral of a function in Matlab and Mathematica that the software cannot do symbolically.
Here is my MatLab code so far, but I understand it may not be very helpful as is.
f = #(t) asin(0.5*sin(t));
a = #(t) sin(t);
F = int(f,t) % Matlab can't do this
F =
int(asin(sin(t)/2), t)
A = int(a,t) % This works
A =
-cos(t)
dt = 1/(N-1); % some small number
for i=1:N
F(i) = integral(f,(i-1)*dt,i*dt);
A(i) = integral(a,(i-1)*dt,i*dt);
end
Both of the calculations in the for loop give a rough approximation of f or a not their integrals after multiplying by dt.
On the math stack-exchange I found a question that derives a finite difference like method for the integral at a point. However, when I did the calculation in Matlab it output a scaled down version of f which was evident after plotting (see above for what I mean by scaled down). I think that's because for smaller intervals the integral basically approximates the function to varying degrees of accuracy (again see above).
I am trying to get either a symbolic equation for the integral, or an approximation of the integral of the function at each location.
So my question is then if I have a function f that MatLab and Mathematica cannot easily take the integral of
can I approximate the integral directly with an integral calculator besides the default ones? (int,integral,trapz)
or
can I approximate the function with finite differences first and then evaluate the integral symbolically?
Your code is nearly fine it's just that
for i=1:N
F(i) = integral(f,0,i*dt);
end
You could also do
F(1)=integral(f,0,dt)
for i=2:N
F(i) = F(i-1)+integral(f,(i-1)*dt,i*dt);
end
Second option is surely more efficient
Because the primitive is really F(x)=int(f(x), 0, x) (0 defines a certain constant ) and for sufficiently small dx you have shown that f(x)=int(f(x), x,x+dx)/dx i. You have proven that MATLAB intégral function does its job.
For example let's take = the function above will compute if you wish to compute just replace 0 above by the constant a you like.
now and so you should get F containing a discretization of
The accepted answer in general is by far the best method I would say but if certain restrictions on your functions are allowable then there is a second method.
For two functions f and g see below
T = 1; % Period
NT = 1; % Number of periods
dt = 0.01; % time interval
time = 0:dt:NT*T; % time
syms t
x = K*sin(2*pi*t+B); % edit as appropriate
% f = A/tanh(K)*tanh(K*sin(2*pi*t+p))
% g = A/asin(K)*asin(K*sin(2*pi*t+p))
formulas found here
f = A1/tanh(K1)*(2^(2*1)-1)*2^(2*1)*bernoulli(2*1)/factorial(2*1)*x^(2*1-1);
% |K1|<pi/2
g = A2/asin(K2)*factorial(2*0)/(2^(2*0)*factorial(0)^2*(2*0+1))*x^(2*0+1);
% |K2|<1
there are no such limitations in the accepted answer
N = 60;
for k=2:N
a1 = (2^(2*k)-1)*2^(2*k)*bernoulli(2*k)/factorial(2*k);
f = f + A1/tanh(K1)*a1*x^(2*k-1);
a2 = factorial(2*k)/(2^(2*k)*factorial(k)^2*(2*k+1));
g = g + A2/asin(K2)*a*x^(2*k+1);
end
MATLAB can calculate sin^n(t) for n being an integer.
F = int(f,t);
phi = double(subs(F,t,time));
G = int(g,t);
psi = double(subs(G,t,time));
We were asked to define our own differential operators on MATLAB, and I did it following a series of steps, and then we should use the differential operators to solve a boundary value problem:
-y'' + 2y' - y = x, y(0) = y(1) =0
my code was as follows, it was used to compute this (first and second derivative)
h = 2;
x = 2:h:50;
y = x.^2 ;
n=length(x);
uppershift = 1;
U = diag(ones(n-abs(uppershift),1),uppershift);
lowershift = -1;
L= diag(ones(n-abs(lowershift),1),lowershift);
% the code above creates the upper and lower shift matrix
D = ((U-L))/(2*h); %first differential operator
D2 = (full (gallery('tridiag',n)))/ -(h^2); %second differential operator
d1= D*y.'
d2= ((D2)*y.')
then I changed it to this after posting it here and getting one response that encouraged the usage of Identity Matrix, however I still seem to be getting no where.
h = 2;
n=10;
uppershift = 1;
U = diag(ones(n-abs(uppershift),1),uppershift);
lowershift = -1;
L= diag(ones(n-abs(lowershift),1),lowershift);
D = ((U-L))/(2*h); %first differential operator
D2 = (full (gallery('tridiag',n)))/ -(h^2); %second differential operator
I= eye(n);
eqn=(-D2 + 2*D - I)*y == x
solve(eqn,y)
I am not sure how to proceed with this, like should I define y and x, or what exactly? I am clueless!
Because this is a numerical approximation to the solution of the ODE, you are seeking to find a numerical vector that is representative of the solution to this ODE from time x=0 to x=1. This means that your boundary conditions make it so that the solution is only valid between 0 and 1.
Also this is now the reverse problem. In the previous post we did together, you know what the input vector was, and doing a matrix-vector multiplication produced the output derivative operation on that input vector. Now, you are given the output of the derivative and you are now seeking what the original input was. This now involves solving a linear system of equations.
Essentially, your problem is now this:
YX = F
Y are the coefficients from the matrix derivative operators that you derived, which is a n x n matrix, X would be the solution to the ODE, which is a n x 1 vector and F would be the function you are associating the ODE with, also a n x 1 vector. In our case, that would be x. To find Y, you've pretty much done that already in your code. You simply take each matrix operator (first and second derivative) and you add them together with the proper signs and scales to respect the left-hand side of the ODE. BTW, your first derivative and second derivative matrices are correct. What's left is adding the -y term to the mix, and that is accomplished by -eye(n) as you have found out in your code.
Once you formulate your Y and F, you can use the mldivide or \ operator and solve for X and get the solution to this linear system via:
X = Y \ F;
The above essentially solves the linear system of equations formed by Y and F and will be stored in X.
The first thing you need to do is define a vector of points going from x=0 to x=1. linspace is probably the most suitable where you can specify how many points we want. Let's assume 100 points for now:
x = linspace(0,1,100);
Therefore, h in our case is just 1/100. In general, if you want to solve from the starting point x = a up to the end point x = b, the step size h is defined as h = (b - a)/n where n is the total number of points you want to solve for in the ODE.
Now, we have to include the boundary conditions. This simply means that we know the beginning and ending of the solution of the ODE. This means that y(0) = y(1) = 0. As such, we make sure that the first row of Y has only the first column set to 1 and the last row of Y has only the last column set to 1, and we'll set the output position in F to both be 0. This symbolizes that we already know the solution at these points.
Therefore, your final code to solve is just:
%// Setup
a = 0; b = 1; n = 100;
x = linspace(a,b,n);
h = (b-a)/n;
%// Your code
uppershift = 1;
U = diag(ones(n-abs(uppershift),1),uppershift);
lowershift = -1;
L= diag(ones(n-abs(lowershift),1),lowershift);
D = ((U-L))/(2*h); %first differential operator
D2 = (full (gallery('tridiag',n)))/ -(h^2);
%// New code - Create differential equation matrix
Y = (-D2 + 2*D - eye(n));
%// Set boundary conditions on system
Y(1,:) = 0; Y(1,1) = 1;
Y(end,:) = 0; Y(end,end) = 1;
%// New code - Create F vector and set boundary conditions
F = x.';
F(1) = 0; F(end) = 0;
%// Solve system
X = Y \ F;
X should now contain your numerical approximation to the ODE in steps of h = 1/100 starting from x=0 up to x=1.
Now let's see what this looks like:
figure;
plot(x, X);
title('Solution to ODE');
xlabel('x'); ylabel('y');
You can see that y(0) = y(1) = 0 as per the boundary conditions.
Hope this helps, and good luck!
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 trying to get Matlab to take this as a function of x_1 through x_n and y_1 through y_n, where k_i and r_i are all constants.
So far my idea was to take n from the user and make two 1×n vectors called x and y, and for the x_i just pull out x(i). But I don't know how to make an arbitrary sum in MATLAB.
I also need to get the gradient of this function, which I don't know how to do either. I was thinking maybe I could make a loop and add that to the function each time, but MATLAB doesn't like that.
I don't believe a loop is necessary for this calculation. MATLAB excels at vectorized operations, so would something like this work for you?
l = 10; % how large these vectors are
k = rand(l,1); % random junk values to work with
r = rand(l,1);
x = rand(l,1);
y = rand(l,1);
vals = k(1:end-1) .* (sqrt(diff(x).^2 + diff(y).^2) - r(1:end-1)).^2;
sum(vals)
EDIT: Thanks to #Amro for correcting the formula and simplifying it with diff.
You can solve for the gradient symbolically with:
n = 10;
k = sym('k',[1 n]); % Create n variables k1, k2, ..., kn
x = sym('x',[1 n]); % Create n variables x1, x2, ..., xn
y = sym('y',[1 n]); % Create n variables y1, y2, ..., yn
r = sym('r',[1 n]); % Create n variables r1, r2, ..., rn
% Symbolically sum equation
s = sum((k(1:end-1).*sqrt((x(2:end)-x(1:end-1)).^2+(y(2:end)-y(1:end-1)).^2)-r(1:end-1)).^2)
grad_x = gradient(s,x) % Gradient with respect to x vector
grad_y = gradient(s,y) % Gradient with respect to y vector
The symbolic sum and gradients can be evaluated and converted to floating point with:
% n random data values for k, x, y, and r
K = rand(1,n);
X = rand(1,n);
Y = rand(1,n);
R = rand(1,n);
% Substitute in data for symbolic variables
S = double(subs(s,{[k,x,y,r]},{[K,X,Y,R]}))
GRAD_X = double(subs(grad_x,{[k,x,y,r]},{[K,X,Y,R]}))
GRAD_Y = double(subs(grad_y,{[k,x,y,r]},{[K,X,Y,R]}))
The gradient function is the one overloaded for symbolic variables (type help sym/gradient) or see the more detailed documentation online).
Yes, you could indeed do this with a loop, considering that x, y, k, and r are already defined.
n = length(x);
s = 0;
for j = 2 : n
s = s + k(j-1) * (sqrt((x(j) - x(j-1)).^2 + (y(j) - y(j-1)).^2) - r(j-1)).^2
end
You should derive the gradient analytically and then plug in numbers. It should not be too hard to expand these terms and then find derivatives of the resulting polynomial.
Vectorized solution is something like (I wonder why do you use sqrt().^2):
is = 2:n;
result = sum( k(is - 1) .* abs((x(is) - x(is-1)).^2 + (y(is) - y(is-1)).^2 - r(is-1)));
You can either compute gradient symbolically or rewrite this code as a function and make a standard +-eps calculation. If you need a gradient to run optimization (you code looks like a fitness function) you could use algorithms that calculate them themselves, for example, fminsearch can do this