nonlinear system of differential algebraic equations

2011 4 12
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Hi,
I have a system of coupled nonlinear equations ( 10 differential equations and three algebraics). I used ode15s for integration.I want to study the behavaior of the system to step change in one parameter defined in my code. So, I considered that the system at t=0 is in the state before step change (vector of initial condition) and at exactly t=0 the parameter changes to the new value. ODE15s does the integration with no problem, but the values at t=0 is not the one defined in the code( initial condition vector). The values for the three variables defined by algeraic equations will change.
So, I thought that I should apply this step change at sometime after t=0 and using If statement in my code. like, if t<100 param=5; else; param=6; But, this time I got this error Warning: Failure at t=1.000000e+002. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.273737e-013) at time t.

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Hooman
Hooman 2012-6-14
Hi, I was just had a general question. I am trying to solve a system of coupled nonlinear equations. Did u need to decouple the equations to be able to use the ode solvers? Also did u need to linearize the terms. I didn't do any of that, just wrote my equations in the form the ode accepts and solved them and got the same error you have.
Thanks!

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Andrew Newell
Andrew Newell 2011-4-12

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I think ode15s is trying to estimate the derivative at the step change. You should run the integration with the first parameter to the time of your step change and then use the output as the initial condition for a new run using the second parameter.
EDIT: Suppose your function is f(x,param) and x has length 2. You could do something like this:
param = 5;
odefun = @(x) f(x,param);
tspan = [0 100];
x0 = [0; 0];
[t1,x1] = ode15s(odefun,tspan,x0);
param = 6;
odefun = @(x) f(x,param);
tspan = [100 200];
x0 = x1(:,end);
[t2,x2] = ode15s(odefun,tspan,x0);
t = [t1 t2];
x = [x1 x2];

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Mona
Mona 2011-4-13
Andrew,
In this line x0 = x2(:,end);
Shouln't it be x0 = x1(:,end)?
Andrew Newell
Andrew Newell 2011-4-13
Oops! This just a sketch of what needs doing and is not tested.
Mona
Mona 2011-4-13
Thanks Andrew. For some parameter values, this method works, but still for some parameters I get this error
Unable to meet integration tolerances without reducing the step size below the smallest value allowed
Do you know how can I fix that?
Andrew Newell
Andrew Newell 2011-4-13
Maybe this solution will help: http://www.mathworks.com/support/solutions/en/data/1-192Z0/index.html?product=ML&solution=1-192Z0

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Mona
Mona 2011-4-12

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Hi Andrew,
Thanks for your response. I think I am doing the same thing. But, still the variables corresponding to algebraic equations are changing. The only thing I thought is just adding the vector of initial conditions to the first row of the matrix of state variables (final result of integrarion) to study the system behavior. I am curious to know why if I use "if" statement( to implement the step change), the integarator stops?
Is there anyway to fix that?

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Andrew Newell
Andrew Newell 2011-4-12
My guess is that if you're using an if statement, you're not doing what I suggest. I have edited the above answer.

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