I have a problem with some differential equations of first-order. I'm trying to solve them with ode23 and ode23s. The differential equations are:
y’ – z – u = 0
z’ – u – y = 0
u’ – y – z = 0
with the initial values: y(0) = –1, z(0) = 1, u(0) = 0.
I also want to compare it with the ANALYTICAL solution:
y = – exp(x)
z = exp(x)
u = 0
I want to do it this way because I need to do the comparison in order to choose the better solver: ode23 or ode23s, whichever one is closer to the analytical solution.
My code is:
%y'=z+u
%z'=u+y
%u'=y+z
%y(0)=-1,z(0)=1, u(0)=0 inital values
%y=y(0) z=y(1) u=y(2)
function dy = part1(t,k)
y=k(1);
z=k(2);
u=k(3);
dy=[z+u;u+y;y+z];
and:
%comparıson of numerıcal and analytıcal equatıons
%Fırst comparıson functıon
[t1,y1]=ode23('part1',[0 16],[0 1 -1]);
%Second comparıson functıon
[t2,y2]=ode23s('part1',[0 16],[0 1 -1]);
%ya,za,ua are analytıcal equtıons
ya =(-exp(t1));
za=exp(t1);
ua=0;
%ı have to plot ya,ua and za depends on ode23 and ode23s
%then ı wıll decıde the best solutıon depends on my shape
%either ode23 or ode23s ıs the best?
plot(t1,ya,t1,za,t1,ua);title('ya,za,ua')
figure
plot(t1,y1(:,1),t1,y1(:,2),t1,y1(:,3));title('ode23')
figure
plot(t2,y2(:,1),t2,y2(:,2),t2,y2(:,3));title('ode23s')
But it doesn't work. Could someone help me?
Related
I want to use MATLABs fmincon function to solve a non-linear problem of which I know that it can be solved in a different way very easily but I want to use fmincon (you might not need the following detailed information about the problem but I provided them in case you need):
Function f(x) is a quadratic function with its vertex at point (5|1).
f(x)=0.1(x-5)^2+1 for 0<=x<=5
Function g(x) is a polynom of order 4 with its vertex at Point (c|0).
g(x)=(x-c)^4 for 0<=x<=c
Function h is just a line on the x axis.
h=0 for c<=x<=5
I want to minimize the Area between the function f(x) and the two connected functions g(x) and h, in the interval [0,5]
minimize A=2*(int(f,[0,5])-int(g,[0,c]))=55/3 - (2*c^5)/5
Also I have the constraint that f(x) must always be 1 unit above the functions g(x) and h.
From the graph I know that the variable c must be between 0 and 2 (just a range for the fmincon function).
This is my .m file:
clc
clear
format long;
options = optimoptions(#fmincon, 'Display', 'iter', 'Algorithm', 'interior-point');
fun=#(x)55/3 - (2*(x(1))^5)/5;
lb = [0];
ub = [2];
[x,fval] = fmincon(fun,[0.1],[],[],[],[],lb,ub,#cons_Q6,options)
The constraints file looks like this (I inserted a lot of values for x with an increment of 0.1):
function [c,ceq]=cons_Q6(x)
c=[(0.0-x(1))^4-0.1*(0.0-5)^2
(0.1-x(1))^4-0.1*(0.1-5)^2
(0.2-x(1))^4-0.1*(0.2-5)^2
(0.3-x(1))^4-0.1*(0.3-5)^2
(0.4-x(1))^4-0.1*(0.4-5)^2
(0.5-x(1))^4-0.1*(0.5-5)^2
(0.6-x(1))^4-0.1*(0.6-5)^2
(0.7-x(1))^4-0.1*(0.7-5)^2
(0.8-x(1))^4-0.1*(0.8-5)^2
(0.9-x(1))^4-0.1*(0.9-5)^2
(1.0-x(1))^4-0.1*(1.0-5)^2
(1.1-x(1))^4-0.1*(1.1-5)^2
(1.2-x(1))^4-0.1*(1.2-5)^2
(1.3-x(1))^4-0.1*(1.3-5)^2
(1.4-x(1))^4-0.1*(1.4-5)^2
(1.5-x(1))^4-0.1*(1.5-5)^2
(1.6-x(1))^4-0.1*(1.6-5)^2
(1.7-x(1))^4-0.1*(1.7-5)^2
(1.8-x(1))^4-0.1*(1.8-5)^2
(1.9-x(1))^4-0.1*(1.9-5)^2
(2.0-x(1))^4-0.1*(2.0-5)^2
(2.1-x(1))^4-0.1*(2.1-5)^2
(2.2-x(1))^4-0.1*(2.2-5)^2
(2.3-x(1))^4-0.1*(2.3-5)^2
(2.4-x(1))^4-0.1*(2.4-5)^2
(2.5-x(1))^4-0.1*(2.5-5)^2
(2.6-x(1))^4-0.1*(2.6-5)^2
(2.7-x(1))^4-0.1*(2.7-5)^2
(2.8-x(1))^4-0.1*(2.8-5)^2
(2.9-x(1))^4-0.1*(2.9-5)^2
(3.0-x(1))^4-0.1*(3.0-5)^2
(3.1-x(1))^4-0.1*(3.1-5)^2
(3.2-x(1))^4-0.1*(3.2-5)^2
(3.3-x(1))^4-0.1*(3.3-5)^2
(3.4-x(1))^4-0.1*(3.4-5)^2
(3.5-x(1))^4-0.1*(3.5-5)^2
(3.6-x(1))^4-0.1*(3.6-5)^2
(3.7-x(1))^4-0.1*(3.7-5)^2
(3.8-x(1))^4-0.1*(3.8-5)^2
(3.9-x(1))^4-0.1*(3.9-5)^2
(4.0-x(1))^4-0.1*(4.0-5)^2
(4.1-x(1))^4-0.1*(4.1-5)^2
(4.2-x(1))^4-0.1*(4.2-5)^2
(4.3-x(1))^4-0.1*(4.3-5)^2
(4.4-x(1))^4-0.1*(4.4-5)^2
(4.5-x(1))^4-0.1*(4.5-5)^2
(4.6-x(1))^4-0.1*(4.6-5)^2
(4.7-x(1))^4-0.1*(4.7-5)^2
(4.8-x(1))^4-0.1*(4.8-5)^2
(4.9-x(1))^4-0.1*(4.9-5)^2
(5.0-x(1))^4-0.1*(5.0-5)^2
];
ceq=[];
As you can see, I've set the bounds for the unknown variable so that x(1)=[0,2] and I have set the constraints in the range [0,5] although I would only need them in the range of [0,2] because of the bounds for x(1).
Now, when I solve it like this I get a solution that doesn't fit all the constraints. But when I delete the unneccessary constraints in the range ]2;5]
function [c,ceq]=cons_Q6(x)
c=[(0.0-x(1))^4-0.1*(0.0-5)^2
(0.1-x(1))^4-0.1*(0.1-5)^2
(0.2-x(1))^4-0.1*(0.2-5)^2
(0.3-x(1))^4-0.1*(0.3-5)^2
(0.4-x(1))^4-0.1*(0.4-5)^2
(0.5-x(1))^4-0.1*(0.5-5)^2
(0.6-x(1))^4-0.1*(0.6-5)^2
(0.7-x(1))^4-0.1*(0.7-5)^2
(0.8-x(1))^4-0.1*(0.8-5)^2
(0.9-x(1))^4-0.1*(0.9-5)^2
(1.0-x(1))^4-0.1*(1.0-5)^2
(1.1-x(1))^4-0.1*(1.1-5)^2
(1.2-x(1))^4-0.1*(1.2-5)^2
(1.3-x(1))^4-0.1*(1.3-5)^2
(1.4-x(1))^4-0.1*(1.4-5)^2
(1.5-x(1))^4-0.1*(1.5-5)^2
(1.6-x(1))^4-0.1*(1.6-5)^2
(1.7-x(1))^4-0.1*(1.7-5)^2
(1.8-x(1))^4-0.1*(1.8-5)^2
(1.9-x(1))^4-0.1*(1.9-5)^2
(2.0-x(1))^4-0.1*(2.0-5)^2
];
ceq=[];
then I get the right result. Does anyone know why this happens and why MATLAB does not respect the constraints when I put them up for the whole range [0,5]?
-Your problem is more related to calculus than matlab tool
constraints like
function [c]=cons_Q6(x)
c=[x < 0; x > 0]; are just ignored by fmincon, because they are not logical
Technically you need to know the optimum c before solving
this optimization problem
- Another issue A = int(f,[0,5])-int(g,[0,c]) = 55/6 - c^5/5 instead of
A = 2*(int(f,[0,5])-int(g,[0,c])) = 55/3 - (2*c^5)/5
Factor 2 is used whether for even whether for odd function (like cosine or since).
Even for those kind of function the integration interval is reduced by half
I updated your optimization function and the solution c is as follow
x = [0, c], constraint is g(x)-f(x)-1<= 0--> (x-c)^4 -0.1(x-5)^2 <=0
x = [c, 5], constraint is h(x)-f(x)-1<= 0--> -0.1(x-5)^2 <=0
c must be predefined or guessed in advance, here I supposed c = 2
because your upper bound ub = 2
As a result
x = [0, 2], --> (x-c)^4 -0.1(x-5)^2 <=0
x = [2, 5], --> -0.1(x-5)^2 <=0
cons_Q6(x) is as follow
function [c,ceq]=cons_Q6(x)
c=[(0.0-x)^4-0.1*(0.0-5)^2;
(0.1-x)^4-0.1*(0.1-5)^2;
(0.2-x)^4-0.1*(0.2-5)^2;
(0.3-x)^4-0.1*(0.3-5)^2;
(0.4-x)^4-0.1*(0.4-5)^2;
(0.5-x)^4-0.1*(0.5-5)^2;
(0.6-x)^4-0.1*(0.6-5)^2;
(0.7-x)^4-0.1*(0.7-5)^2;
(0.8-x)^4-0.1*(0.8-5)^2;
(0.9-x)^4-0.1*(0.9-5)^2;
(1.0-x)^4-0.1*(1.0-5)^2;
(1.1-x)^4-0.1*(1.1-5)^2;
(1.2-x)^4-0.1*(1.2-5)^2;
(1.3-x)^4-0.1*(1.3-5)^2;
(1.4-x)^4-0.1*(1.4-5)^2;
(1.5-x)^4-0.1*(1.5-5)^2;
(1.6-x)^4-0.1*(1.6-5)^2;
(1.7-x)^4-0.1*(1.7-5)^2;
(1.8-x)^4-0.1*(1.8-5)^2;
(1.9-x)^4-0.1*(1.9-5)^2;
(2.0-x)^4-0.1*(2.0-5)^2;
-0.1*(2.1-5)^2;
-0.1*(2.2-5)^2;
-0.1*(2.3-5)^2;
-0.1*(2.4-5)^2;
-0.1*(2.5-5)^2;
-0.1*(2.6-5)^2;
-0.1*(2.7-5)^2;
-0.1*(2.8-5)^2;
-0.1*(2.9-5)^2;
-0.1*(3.0-5)^2;
-0.1*(3.1-5)^2;
-0.1*(3.2-5)^2;
-0.1*(3.3-5)^2;
-0.1*(3.4-5)^2;
-0.1*(3.5-5)^2;
-0.1*(3.6-5)^2;
-0.1*(3.7-5)^2;
-0.1*(3.8-5)^2;
-0.1*(3.9-5)^2;
-0.1*(4.0-5)^2;
-0.1*(4.1-5)^2;
-0.1*(4.2-5)^2;
-0.1*(4.3-5)^2;
-0.1*(4.4-5)^2;
-0.1*(4.5-5)^2;
-0.1*(4.6-5)^2;
-0.1*(4.7-5)^2;
-0.1*(4.8-5)^2;
-0.1*(4.9-5)^2;
-0.1*(5.0-5)^2;
];
ceq=[];
The constraints in the range ]2;5] are very necessary keep them
clc
clear
format long;
options = optimoptions(#fmincon, 'Display', 'iter', 'Algorithm',...
'interior-point');
fun=#(x)55/6 - (x^5)/5;
lb = [0];
ub = [2];
[c, A] = fmincon(fun,[0.1],[],[],[],[],lb,ub,#cons_Q6,options)
solution :
c = 1.257432726024430
A = 8.537951710969493
Before anything, I can't use ANY built-in ODE solvers for this. I coded this system of ODEs with explicit euler's method but I need to instead rewrite it with implicit euler's. If I just switch the "i" to "i+1" (ex "y(1,i)" to "y(1,i+1)") then the answer comes out obscenely wrong.
A_init= 4;
B_init= 1.1;
C_init= 4;
y0 = [A_init; B_init; C_init];
h = 20/100;
n = 100;
t0 = 0;
tf = 20;
t = linspace(t0,tf,n);
y = zeros(numel(y0) , n);
y(:, 1) = y0(:);
dAdt= #(a,b) 7.27*(b-a*b+ a-(8.375*10^-5)*a^3);
dBdt= #(b,a,c)(-b-a*b+c)/77.27;
dCdt= #(c,a) 0.4*(a-c);
for i = 1:n-1
y(1,i+1) = y(1,i) +h*feval(dAdt,y(1,i),y(2,i));
y(3,i+1) = y(3,i) +h*feval(dCdt,y(3,i),y(1,i));
y(2,i+1) = y(2,i) +h*feval(dBdt,y(2,i),y(1,i),y(3,i));
end
Your idea is correct, but remember that if you change the i on the right-hand side to i+1, you will have y(1,i+1), y(2,i+1), y(3,i+1) appearing on both sides of the equation. This means that you actually have to solve for y(1,i+1), y(2,i+1) and y(3,i+1) at each step. Since your three equations are coupled, you have to solve a nonlinear system of equations at each time-step, using fsolve or fzero.
Read the answer to this question, which shows how to do this for the single equation case.
this is my first time posting, so if you need anymore info or I havn't done something correct please let me know!
I want to plot a system of ODES, which would be no problem as I would do the following;
F=#(t,x) [
.the RHS of my first order ODES.
];
[t x]=ode45(F,[Range], [Initial conditions]);
Where a vector would be created containing x'(1), x'(2),... where x(1), x(2) would be the time dependant variable.
However my system is a little more complicated. I have 6 differential equations which are made up of equations that include differential equations and the time dependant variables. For example, the RHS of my ODES are something like
2*x(2)*x'(1)*f(a)
Where f(a) could be another function based on constants, a.
I have set up my code in the following order;
Constants, then equations of the form f(a), then my differential equations followed by the ode45 solver and the plot commands. However I am getting several errors "Undefined function or variable" as early equations depend on variables/equations that are not defined until later.
Thank you ever so much for you help :)
As suggested, here an example of the type of my code;
`%Constants
a=34
b=31
c=20
%Equations
A=b*cos(2*pi)
B=a*EQ1
C=c*x(2)
%DifferentialEquations
EQ1=x(1)*A
EQ2=(EQ3-EQ1)*(B-C)
EQ3=x(2)*x(3)
F=#(t,x) [EQ1;EQ2;EQ3;];[t x]=ode45(F,[0 10], [0 0 0 ]);
Provides the error.
Produces the undefined function or variable 'DEQ1A'.
Based on the comment you are trying to use a variable before it is defined. In MATLAB you need to define your variables and functions before you use them. Because the equations are functions of each other you can use anonymous functions to define F.
Example:
a=34;
b=31;
c=20;
A = b*cos(2*pi);
EQ1 = #(x) x(1)*A;
B = #(x) a*EQ1(x);
C = #(x) c*x(2);
EQ3 = #(x) x(2)*x(3);
EQ2 = #(x) (EQ3(x) - EQ1(x))*(B(x)-C(x));
F = #(t,x) [EQ1(x);EQ2(x);EQ3(x)];
[t,x] = ode45(F,[0 10], [0 0 0]);
I have a large underdetermined equation system for which I search an unique solution in respect of any given constraints. I simplified my problem into the following one:
x²-4=0,
y²-9=0,
x*y=myMin,
x+y=myMin.
What is the best way to implement this in Matlab symbolically, so that it returns
x=2
y=-3
I'm searching something like
syms x y
S=solve(...
x²-4==0,...
y²-9==0,...
x*y==myMin,...
x+y==myMin);
I do not know how specify the min as a function command to solve. But here's an approach that solves the equations and then post-processes the result according to your constraints:
syms x y
S=solve(x^2-4==0,y^2-9==0);
[~,idx] = min(double(S.x .* S.y)+double(S.x + S.y));
X = double(S.x(idx))
Y = double(S.y(idx))
This gives:
X =
2
Y =
-3
The symbolic results have to be converted using the double command to allow processing with the min function.
The problem you seem to run into is that there is no solution, not even matlab can deal with that.
Try it like this:
myMin = -6;
syms x y
S=solve(...
x²-4==0,...
y²-9==0,...
x*y==myMin,...
x+y==myMin + 5); %Note the +5 to make it feasible
Cannot try myself, but a quick calculation tells me that this one is at least solvable.
I have all the data and an ODE system of three equations which has 9 unknown coefficients (a1, a2,..., a9).
dS/dt = a1*S+a2*D+a3*F
dD/dt = a4*S+a5*D+a6*F
dF/dt = a7*S+a8*D+a9*F
t = [1 2 3 4 5]
S = [17710 18445 20298 22369 24221]
D = [1357.33 1431.92 1448.94 1388.33 1468.95]
F = [104188 104792 112097 123492 140051]
How to find these coefficients (a1,..., a9) of an ODE using Matlab?
I can't spend too much time on this, but basically you need to use math to reduce the equation to something more meaningful:
your equation is of the order
dx/dt = A*x
ergo the solution is
x(t-t0) = exp(A*(t-t0)) * x(t0)
Thus
exp(A*(t-t0)) = x(t-t0) * Pseudo(x(t0))
Pseudo is the Moore-Penrose Pseudo-Inverse.
EDIT: Had a second look at my solution, and I didn't calculate the pseudo-inverse properly.
Basically, Pseudo(x(t0)) = x(t0)'*inv(x(t0)*x(t0)'), as x(t0) * Pseudo(x(t0)) equals the identity matrix
Now what you need to do is assume each time step (1 to 2, 2 to 3, 3 to 4) is an experiment (therefore t-t0=1), so the solution would be to:
1- Build your pseudo inverse:
xt = [S;D;F];
xt0 = xt(:,1:4);
xInv = xt0'*inv(xt0*xt0');
2- Get exponential result
xt1 = xt(:,2:5);
expA = xt1 * xInv;
3- Get the logarithm of the matrix:
A = logm(expA);
And since t-t0= 1, A is our solution.
And a simple proof to check
[t, y] = ode45(#(t,x) A*x,[1 5], xt(1:3,1));
plot (t,y,1:5, xt,'x')
You have a linear, coupled system of ordinary differential equations,
y' = Ay with y = [S(t); D(t); F(t)]
and you're trying to solve the inverse problem,
A = unknown
Interesting!
First line of attack
For given A, it is possible to solve such systems analytically (read the wiki for example).
The general solution for 3x3 design matrices A take the form
[S(t) D(t) T(t)].' = c1*V1*exp(r1*t) + c2*V2*exp(r2*t) + c3*V3*exp(r3*t)
with V and r the eigenvectors and eigenvalues of A, respectively, and c scalars that are usually determined by the problem's initial values.
Therefore, there would seem to be two steps to solve this problem:
Find vectors c*V and scalars r that best-fit your data
reconstruct A from the eigenvalues and eigenvectors.
However, going down this road is treaturous. You'd have to solve the non-linear least-squares problem for the sum-of-exponentials equation you have (using lsqcurvefit, for example). That would give you vectors c*V and scalars r. You'd then have to unravel the constants c somehow, and reconstruct the matrix A with V and r.
So, you'd have to solve for c (3 values), V (9 values), and r (3 values) to build the 3x3 matrix A (9 values) -- that seems too complicated to me.
Simpler method
There is a simpler way; use brute-force:
function test
% find
[A, fval] = fminsearch(#objFcn, 10*randn(3))
end
function objVal = objFcn(A)
% time span to be integrated over
tspan = [1 2 3 4 5];
% your desired data
S = [17710 18445 20298 22369 24221 ];
D = [1357.33 1431.92 1448.94 1388.33 1468.95 ];
F = [104188 104792 112097 123492 140051 ];
y_desired = [S; D; F].';
% solve the ODE
y0 = y_desired(1,:);
[~,y_real] = ode45(#(~,y) A*y, tspan, y0);
% objective function value: sum of squared quotients
objVal = sum((1 - y_real(:)./y_desired(:)).^2);
end
So far so good.
However, I tried both the complicated way and the brute-force approach above, but I found it very difficult to get the squared error anywhere near satisfyingly small.
The best solution I could find, after numerous attempts:
A =
1.216731997197118e+000 2.298119167536851e-001 -2.050312097914556e-001
-1.357306715497143e-001 -1.395572220988427e-001 2.607184719979916e-002
5.837808840775175e+000 -2.885686207763313e+001 -6.048741083713445e-001
fval =
3.868360951628554e-004
Which isn't bad at all :) But I would've liked a solution that was less difficult to find...