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Time-dependent dynamic problem with nonlinear stiffness using ode45

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I am solving a time-dependent finite element problem with nonlinearities in the stiffness matrix which takes the following form:
where the dot symbolizes derivative with respect to time, U is the displacement vector, M is the mass matrix, C is the damping matrix, K(U) is the stiffness matrix which depends on U, and F is an external force. I would like to use ode45 to solve the dynamic problem. The general flow for solving this problem would be the following:
  1. Calculate the stiffness matrix K using the latest displacement U (Initial condition if it is the first time step).
  2. Formulate the state-space equations using K matrix, M matrix, C matrix, and F vector.
  3. Integrate the state-space equations with ode45 to caclulate the new displacement U.
  4. Recalculate the K matrix with the new displacement U. Check if the residual error meets the preselected tolerance.
  5. If the error is met, update velocities and accelerations, and go to the next time step. If the error is not met, then go back to step 1 using this latest K matrix.
I am familiar with solving this problem without the nonlinearity in K. However, I am unsure how to update the K matrix at every time step within the ode45 framework.
Thank you for the guidance!

Answers (1)

Torsten
Torsten on 8 Jul 2024 at 19:50
Moved: Torsten on 8 Jul 2024 at 19:50
You solve for U and Udot - thus you get values for U and Udot from ode45. Simply compute the matrix K(U) with these values and proceed as if K was constant.
  4 Comments
Andres Bellei
Andres Bellei on 9 Jul 2024 at 16:01
Thank you for your responses, Torsten. As Sam pointed out, I need help in how to update the K matrix for each time step, as I have only used ode45 with constant parameters for all time steps. In this case, at each time step the stiffness K matrix will change.
Torsten
Torsten on 9 Jul 2024 at 17:04
Edited: Torsten on 9 Jul 2024 at 17:28
Example:
M = [1 0;0 4];
invM = inv(M);
C = [4 -2;6 3];
K = @(u)[u(1) sin(u(2));exp(-u(2)) 1/u(1)];
F = [0 ;0];
tspan = [0 1];
y0 = [1; 2 ;3; 4]; % vector of initial values for (u(1),u(2),udot(1),udot(2))
[T,Y] = ode45(@(t,y)fun(t,y,invM,C,K,F),tspan,y0);
plot(T,[Y(:,1),Y(:,2)])
grid on
function dy = fun(t,y,invM,C,K,F)
n = numel(y)/2;
u = y(1:n);
udot = y(n+1:2*n);
dy = zeros(2*n,1);
dy(1:n) = udot;
dy(n+1:2*n) = invM*(F-C*udot-K(u)*u);
end

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