System of differential equations with constant (as variables) coefficients

Asked by Marc Rosales

Marc Rosales (view profile)

on 28 Oct 2016
Latest activity Commented on by Marc Rosales

Marc Rosales (view profile)

on 29 Oct 2016
Accepted Answer by Torsten

Torsten (view profile)

Hi
I have 4 differential equations that I need to solve, but the coefficients for each term in each equation are different unknown constants. I read about solving it using a matrix and came up with:
syms A(t) B(t) C(t) D(t) k1 kn1 k2 kn2 k3 kn3 k4 kn4 k5 kn5 k6 kn6
Z = [A;B;C;D];
X = [-(k1+k3+k5), kn1, kn3, kn5; k1, -(kn1+k2+k6), kn6, kn2; k3, k6, -(kn3+k4+kn6), kn4; k5, k2, k4, -(kn2+kn4+kn5)];
Y = zeros(4,1);
eqn = diff(Z) == X*Z + Y;
[ASol(t), BSol(t), CSol(t), DSol(t),] = dsolve(eqn);
however that does not seem to work. Any help would be appreciated

Torsten (view profile)

on 28 Oct 2016

Your system of ODEs is too complicated to be solved symbolically.
Specify the constants and use a numerical solver (e.g. ODE15s).
Best wishes
Torsten.

Marc Rosales

Marc Rosales (view profile)

on 28 Oct 2016
I see. :(
Unfortunately this came from a theoretical chemical reaction, thus I cannot provide any values for the 12 constants I have. :'( Any suggestions on what application can handle these types of systems symbolically? Thanks!
Torsten

Torsten (view profile)

on 28 Oct 2016
I wonder what you want to do with a symbolic solution if you don't know reasonable values for the constants ?
Best wishes
Torsten.
Marc Rosales

Marc Rosales (view profile)

on 28 Oct 2016
^Exactly my question too.
To be completely honest, this is a problem a chemistry professor gave me, and I too am at a complete loss why I need a symbolic solution hahaha. Anyway, the constants may be any non-negative value afaik.
If I get a system of ODEs of 4 equations, around how many constants(as variables) can I use to still be solvable by Matlab? or do I need to eliminate the number of equations too? :/

Answer by Teja Muppirala

Teja Muppirala (view profile)

on 28 Oct 2016

As has been mentioned, with the k's explicitly accounted for, it's very complicated.
But if you consider the matrix X simply as some constant matrix X, regardless of what's in it, then the analytical solution is
Z(t) = expm(X*t)*x0
where x0 is the initial condition.
For example
% Make some random stable X matrix
X = randn(3);
X = X*X';
X = -X;
x0 = [1;2;3]; % Some initial condition
ode45(@(t,Z)X*Z,[0 3],x0) % Using ODE45
Z = [];
tList = 0:0.01:3;
for t = tList;
Z(:,end+1) = expm(X*t)*x0; % Using matrix exponential solution
end
hold on;
plot(tList, Z','k','linewidth',1);
title('Same answer with ode45 and expm')

Torsten

Torsten (view profile)

on 28 Oct 2016
And does expm work for the symbolic (4x4)-matrix from above ?
I doubt it - and if it works, the solution will contain roots of a fourth-order symbolic polynomial, I guess.
Best wishes
Torsten.
Marc Rosales

Marc Rosales (view profile)

on 29 Oct 2016
I've looked at the results after tweaking the initial conditions ([1,0,0]), but the graph always slopes downwards, whereas I'm expecting a downwards slope for only one graph, and upwards for the rest until they stabilize.