System of differential equations with constant (as variables) coefficients

Question

Marc Rosales 2016-10-28

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/309621-system-of-differential-equations-with-constant-as-variables-coefficients

评论： Marc Rosales 2016-10-29

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

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Torsten 2016-10-28

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/309621-system-of-differential-equations-with-constant-as-variables-coefficients#answer_241128

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.

3 个评论
显示 1更早的评论隐藏 1更早的评论

Torsten 2016-10-28

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 2016-10-28

^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? :/

请先登录，再进行评论。

Answer 2

Teja Muppirala 2016-10-28

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/309621-system-of-differential-equations-with-constant-as-variables-coefficients#answer_241145

在 MATLAB Online 中打开

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')

2 个评论
显示无隐藏无

Torsten 2016-10-28

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 2016-10-29

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.

请先登录，再进行评论。