I'm trying to see a laplace transform applied to a symbolic function, but it doesn't look like I would want to.
; file.m
syms t
f(t) = (3*t - 2) * cos(t);
laplace(f)
outputs
ans = (sym)
2 / 2 \
3*s - 2*s*\s + 1/ - 3
-----------------------
2
/ 2 \
\s + 1/
and I want
ans = (sym) (3*2^2 - 2*s*(s^2 + 1) - 3)/(s^2 + 1)^2
sympref display flat
See here: https://octave.sourceforge.io/symbolic/function/sympref.html
Also possibly useful: https://octave.sourceforge.io/symbolic/function/#sym/pretty.html
Good afternoon!
First things first, I looked for similar questions for a while, but (probably because of my inexperience) I've found nothing similar to what I'm going to ask.
I'm using matlab for the first time to solve this kind of problems, so I'm not sure of what to do. A brief explenation:
I'm doing a project for my Optimal Control course: I have to replicate the results of a paper about employment, and I'm stuck with the plots. I have the following data:
five variable functions (U(t), T(t), R(t), V1(t) and V2(t))
four control functions(u1(t), u2(t), u3(t), u4(t))
constraints on the control variables (each u must be between 0 and 1)
initial values for U, T, R, V1 and V2 (in t=0, in particular V1 and V2 are constant over time)
final values for the λ coefficients in the hamiltonian
(note: for the controls, I've already found the optimal expression, which is in this form: ui = min{1, max{0,"expression"}}. If needed, I can give also the four expressions, neglected to
synthesize a little)
Under professor's suggestions, I've tried to use fmincon, that theoretically should give me directly the information that I need to plot some result using only the cost function of the problem. But in this case I have some issues involving time in the calculations. Below, the code that I used for fmincon:
syms u
%note: u(5) corresponds to U(t), but this is the only way I've found to get
%a result, the other u(i) are in ascending order (u(1) = u1 and so on...)
g = #(u) 30*u(5) + (20/2)*(u(1))^2 + (20/2)*(u(2))^2 + (10/2)*(u(3))^2 + (40/2)*(u(4))^2;
%initial guesses
u0 = [0 0 0 0 100000]; %
A = [];
b = [];
Aeq = [];
beq = [];
lb = 0.0 * ones(1,2,3,4);
ub = 1.0 * ones(1,2,3,4);
[x,fval,output,lambda] = fmincon(g, u0, A, b, Aeq, beq, lb, ub);
Whit this code, i get (obviously) only one value for each variable as result, and since I've not found any method to involve time, as I said before, I start looking for other solving strategies.
I found that ode45 is a differential equation solver that has the "time iteration" already included in the algorithm, so I tried to write the code to get it work with my problem.
I took all the equations from the paper and put them in a vector as shown in the mathworks examples, and this is my matlab file:
syms u1(t) u2(t) u3(t) u4(t)
syms U(t) T(t) R(t) V1(t) V2(t)
syms lambda_u lambda_t lambda_r lambda_v1 lambda_v2
%all the parameters provided by the paper
delta = 500;
alpha1 = 0.004;
alpha2 = 0.005;
alpha3 = 0.006;
gamma1 = 0.001;
gamma2 = 0.002;
phi1 = 0.22;
phi2 = 0.20;
delta1 = 0.09;
delta2 = 0.05;
k1 = 0.000003;
k2 = 0.000002;
k3 = 0.0000045;
%these two variable are set constant
V1 = 200;
V2 = 100;
%weight values for the cost function (only A1 is used in this case, but I left them all since the unused ones are irrelevant)
A1 = 30;
A2 = 20;
A3 = 20;
A4 = 10;
A5 = 40;
%ordering the unknowns in an array
x = [U T R u1 u2 u3 u4];
%initial conditions, ordered as the x vector (for the ui are guesses)
y0 = [100000 2000 1000 0 0 0 0];
%system set up
f = #(t,x) [delta - (1 + x(4))*k1*x(1)*V1 - (1 + x(5))*k2*x(1)*V2 - alpha1*x(1) + gamma1*x(2) + gamma2*x(3);...
(1 + x(4))*k1*x(1)*V1 - k3*x(2)*V2 - alpha2*x(2) - gamma1*x(2);...
(1 + x(5))*k2*x(1)*V2 - alpha3*x(3) - gamma2*x(3) + k3*x(2)*V2;...
alpha2*x(2) + gamma1*x(2) + (1 + x(6))*phi1*x(1) + k3*x(2)*V2 - delta1*V1;...
alpha3*x(3) + gamma2*x(3) + (1 + x(7))*phi2*x(1) - delta2*V2;...
-A1 + (1 + x(4))*k1*V1*(lambda_u - lambda_t) + (1 + x(5))*k2*V2*(lambda_u - lambda_r) + lambda_u*alpha1 - lambda_v1*(1 + x(6))*phi1 - lambda_v2*(1 + x(7))*phi2;...
-lambda_u*gamma1 + (alpha2 + gamma1)*(lambda_t - lambda_v1) + k3*V2*(lambda_t - lambda_r - lambda_v1);...
-lambda_u*gamma2 + (alpha3 + gamma2)*(lambda_r - lambda_v2);...
(1 + x(4))*k1*x(1)*(lambda_u - lambda_t) + lambda_v1*delta1;...
(1 + x(5))*k2*x(1)*(lambda_u -lambda_r) + k3*x(2)*(lambda_t - lambda_r - lambda_v1) + lambda_v2*delta2];
%using ode45 to solve over the chosen time interval
[t,xa] = ode45(f,[0 10],y0);
With this code, I get the following error:
Error using odearguments (line 95)
#(T,X)[DELTA-(1+X(4))*K1*X(1)*V1-(1+X(5))*K2*X(1)*V2-ALPHA1*X(1)+GAMMA1*X(2)+GAMMA2*X(3);(1+X(4))*K1*X(1)*V1-K3*X(2)*V2-ALPHA2*X(2)-GAMMA1*X(2);(1+X(5))*K2*X(1)*V2-ALPHA3*X(3)-GAMMA2*X(3)+K3*X(2)*V2;ALPHA2*X(2)+GAMMA1*X(2)+(1+X(6))*PHI1*X(1)+K3*X(2)*V2-DELTA1*V1;ALPHA3*X(3)+GAMMA2*X(3)+(1+X(7))*PHI2*X(1)-DELTA2*V2;-A1+(1+X(4))*K1*V1*(LAMBDA_U-LAMBDA_T)+(1+X(5))*K2*V2*(LAMBDA_U-LAMBDA_R)+LAMBDA_U*ALPHA1-LAMBDA_V1*(1+X(6))*PHI1-LAMBDA_V2*(1+X(7))*PHI2;-LAMBDA_U*GAMMA1+(ALPHA2+GAMMA1)*(LAMBDA_T-LAMBDA_V1)+K3*V2*(LAMBDA_T-LAMBDA_R-LAMBDA_V1);-LAMBDA_U*GAMMA2+(ALPHA3+GAMMA2)*(LAMBDA_R-LAMBDA_V2);(1+X(4))*K1*X(1)*(LAMBDA_U-LAMBDA_T)+LAMBDA_V1*DELTA1;(1+X(5))*K2*X(1)*(LAMBDA_U-LAMBDA_R)+K3*X(2)*(LAMBDA_T-LAMBDA_R-LAMBDA_V1)+LAMBDA_V2*DELTA2]
returns a vector of length 10, but the length of initial conditions vector is 7. The vector returned by
#(T,X)[DELTA-(1+X(4))*K1*X(1)*V1-(1+X(5))*K2*X(1)*V2-ALPHA1*X(1)+GAMMA1*X(2)+GAMMA2*X(3);(1+X(4))*K1*X(1)*V1-K3*X(2)*V2-ALPHA2*X(2)-GAMMA1*X(2);(1+X(5))*K2*X(1)*V2-ALPHA3*X(3)-GAMMA2*X(3)+K3*X(2)*V2;ALPHA2*X(2)+GAMMA1*X(2)+(1+X(6))*PHI1*X(1)+K3*X(2)*V2-DELTA1*V1;ALPHA3*X(3)+GAMMA2*X(3)+(1+X(7))*PHI2*X(1)-DELTA2*V2;-A1+(1+X(4))*K1*V1*(LAMBDA_U-LAMBDA_T)+(1+X(5))*K2*V2*(LAMBDA_U-LAMBDA_R)+LAMBDA_U*ALPHA1-LAMBDA_V1*(1+X(6))*PHI1-LAMBDA_V2*(1+X(7))*PHI2;-LAMBDA_U*GAMMA1+(ALPHA2+GAMMA1)*(LAMBDA_T-LAMBDA_V1)+K3*V2*(LAMBDA_T-LAMBDA_R-LAMBDA_V1);-LAMBDA_U*GAMMA2+(ALPHA3+GAMMA2)*(LAMBDA_R-LAMBDA_V2);(1+X(4))*K1*X(1)*(LAMBDA_U-LAMBDA_T)+LAMBDA_V1*DELTA1;(1+X(5))*K2*X(1)*(LAMBDA_U-LAMBDA_R)+K3*X(2)*(LAMBDA_T-LAMBDA_R-LAMBDA_V1)+LAMBDA_V2*DELTA2]
and the initial conditions vector must have the same number of elements.
Error in ode45 (line 115)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in test (line 62)
[t,xa] = ode45(f,[0 10],y0);
For which I can't find a solution, since I have used all the initial values given in the paper. The only values that I have left are the final values for the lambda coefficients, since they are final values, and I am not sure if they can be used.
In this case, I can't also understand where I should put the bounds on the control variable.
For completeness, I will provide also the link to the paper in question:
https://www.ripublication.com/ijss17/ijssv12n3_13.pdf
Can you help me figure out what I can do to solve my problems?
P.S: I know this is a pretty bad code, but I'm basing on the basics tutorials on mathworks; for sure this should need to be refactored and ordered in various file (one for the cost function and one for the constraints for example) but firstly I would like to understand where the problem is and then I will put all in a pretty form.
Thank you so much!
Generally you confused something with Vectors. In initial conditions you declared 7 values:
%initial conditions, ordered as the x vector (for the ui are guesses)
y0 = [100000 2000 1000 0 0 0 0];
But you declared 10 ODE's:
%system set up
f = #(t,x) [delta - (1 + x(4))*k1*x(1)*V1 - (1 + x(5))*k2*x(1)*V2 - alpha1*x(1) + gamma1*x(2) + gamma2*x(3);...
(1 + x(4))*k1*x(1)*V1 - k3*x(2)*V2 - alpha2*x(2) - gamma1*x(2);...
(1 + x(5))*k2*x(1)*V2 - alpha3*x(3) - gamma2*x(3) + k3*x(2)*V2;...
alpha2*x(2) + gamma1*x(2) + (1 + x(6))*phi1*x(1) + k3*x(2)*V2 - delta1*V1;...
alpha3*x(3) + gamma2*x(3) + (1 + x(7))*phi2*x(1) - delta2*V2;...
-A1 + (1 + x(4))*k1*V1*(lambda_u - lambda_t) + (1 + x(5))*k2*V2*(lambda_u - lambda_r) + lambda_u*alpha1 - lambda_v1*(1 + x(6))*phi1 - lambda_v2*(1 + x(7))*phi2;...
-lambda_u*gamma1 + (alpha2 + gamma1)*(lambda_t - lambda_v1) + k3*V2*(lambda_t - lambda_r - lambda_v1);...
-lambda_u*gamma2 + (alpha3 + gamma2)*(lambda_r - lambda_v2);...
(1 + x(4))*k1*x(1)*(lambda_u - lambda_t) + lambda_v1*delta1;...
(1 + x(5))*k2*x(1)*(lambda_u -lambda_r) + k3*x(2)*(lambda_t - lambda_r - lambda_v1) + lambda_v2*delta2];
Every line in above code is recognized as one ODE.
But that's not all. The second problem is with your construction. You mixed symbolic math (lambda declared as syms) with numerical solving, which will be tricky. I'm not familiar with the exact scientific problem you are trying to solve, but if you can't avoid symbolic math, maybe you should try dsolve from Symbolic Math Toolbox?
How to convert a symbolic expression to a Octave function from the Symbolic Package?
After installing the symbolic package on octave with pkg install -forge symbolic. Using the symbolic package on octave I may write this:
octave> pkg load symbolic;
octave> a = sym( "a" );
octave> int ( a^2 + csc(a) )
which will result in:
ans = (sym)
3
a log(cos(a) - 1) log(cos(a) + 1)
-- + --------------- - ---------------
3 2 2
But how to do this Integral (int(1)) symbolic result just above to became a valuable function like this below?
function x = f( x )
x = x^3/3 + log( cos(x) - 1 )/2 - log( cos(x) + 1 )/2
end
f(3)
# Which evaluates to: 11.6463 + 1.5708i
I want to take the symbolic result from int ( a^2 + csc(a) ), and call result(3), to compute it at 3, i.e., return the numeric value 11.6463 + 1.5708i, from the symbolic expression integral a^2 + csc(a). Basically, how to use the symbolic expression as numerically evaluable expressions? It is the as this other question for Matlab.
References:
http://octave.sourceforge.net/symbolic/index.html
How do I declare a symbolic matrix in Octave?
Octave symbolic expression
Julia: how do I convert a symbolic expression to a function?
What is symbolic computation?
You can use pretty.
syms x;
x = x^3/3 + log( cos(x) - 1 )/2 - log( cos(x) + 1 )/2;
pretty(x)
which gives this:
3
log(cos(x) - 1) log(cos(x) + 1) x
--------------- - --------------- + --
2 2 3
Update (Since the question is edited):
Make this function:
function x = f(y)
syms a;
f(a) = int ( a^2 + csc(a) );
x = double(f(y));
end
Now when you call it using f(3), it gives:
ans =
11.6463 + 1.5708i
it seems like you answered your own question by linking to the other question about Matlab.
Octave has an implementation of matlabFunction which is a wrapper for function_handle in the symbolic toolbox.
>> pkg load symbolic;
>> syms x;
>> y = x^3/3 + log( cos(x) - 1 )/2 - log( cos(x) + 1 )/2
y = (sym)
3
x log(cos(x) - 1) log(cos(x) + 1)
-- + --------------- - ---------------
3 2 2
>> testfun = matlabFunction(y)
testfun =
#(x) x .^ 3 / 3 + log (cos (x) - 1) / 2 - log (cos (x) + 1) / 2
testfun(3)
>> testfun(3)
ans = 11.6463 + 1.5708i
>> testfun([3:1:5]')
ans =
11.646 + 1.571i
22.115 + 1.571i
41.375 + 1.571i
>> testfun2 = matlabFunction(int ( x^2 + csc(x) ))
testfun2 =
#(x) x .^ 3 / 3 + log (cos (x) - 1) / 2 - log (cos (x) + 1) / 2
>> testfun2(3)
ans = 11.6463 + 1.5708i
>> testfun2([3:1:5]')
ans =
11.646 + 1.571i
22.115 + 1.571i
41.375 + 1.571i
I'm sure there are other ways you could implement this, but it may allow you to avoid hardcoding your equation in a function.
I have this cubic spline function:
s(x) = 4 + k1*x + 2x^2 - (1/6)*x^3 for x in [0,1]
s(x) = 1 - (4/3)*(x-1) + k2*(x-1)^2 - (1/6) * (x-1)^3 for x in [1,2]
s(x) = 1 + k3*(x-2) + (x-2)^2 - (1/6) * (x-2)^3 for x in [2,3]
I would like to implement a simple function, that giving this function, it determines the coefficients k1, k2 and k3, but I couldn't do that..
Has anyone an idea?
Denoting the 3 functions as s0(x), s1(x) and s2(x), you can expand s1(x) and s2(x) and collect the coefficients for x^3, x^2, x and constant terms. Then, you get
s1(x)= 4 +k1*x+2*x^2-(1/6)*x^3 s2(x)=
(5/2+k2)+(-11/6-2k2)*x+(k2+1/2)*x^2-(1/6)*x^3 s3(x)= ..... (left our
for your own practice)
For these 3 functions to be all cubic polynomials, they should be identical. So, you can have
4 = 5/2 + k2
k1 = -11/6 - 2k2
2 = k2+1/2
from which you can get k1=-29/6, k2=3/2.
Compare the equations s0(x) and s2(x), you can also get k3 = 7/6.
I passed the following nonlinear system to Matlab:
2(x−p1)+2(xy−p3)y = 0
2(y−p2)+2(xy−p3)x = 0
and used syms to find solution for x and y symbolically but I got:
sol.x
ans =
(p1^3 + p3*p1^2*z1 + p1*z1^4 - 1.0*p2*p1*z1^3 + p1*z1^2 - 1.0*p2*p1*z1 + p3*z1^3 - 1.0*p2*p3*z1^2 + p3*z1 - 1.0*p2*p3)/(p1^2 + p3^2)
and
sol.y
ans =
z1
where z1 = RootOf(z^5 - p2*z^4 + 2*z^3 - z^2*(2*p2 - p1*p3) + z*(p1^2 - p3^2 + 1) - p1*p3 - p2, z)
I dont understand where z come from? what is z?
Your y solution is expressed in terms of roots of a polynomial in z that depends on your equation's parameters also.
To show why is difficult to answer your question in the present form, please allow me to rephrase it: the numbers I'm looking for are the roots of an equation f(z) = 0; now, where z comes from? :-)