Why is my model not converging with ode45 solver ?

I am trying to simulate an electronical device that can be modeled by a mass-spring-damper system with an additional non-linear force. The equation at the equilibrium for the system is the following :

The goal of my MATLAB code is to solve this equation for $x$ and find the equilibrium. For this purpose, I use the `ode45` function like this (all coefficient are defined in my code but not shown here) :

x0 = 0;

[t, x] = ode45(@(t, x) odefun(t, x, eps0, a, V0, B, d0, K), tspan, x0);

function dxdt = odefun(t, x, eps, a, V, B, d0, K)

dxdt = ((eps * a * V^2) ./ (2 * B * (d0 - x).^2)) + ((K / B) .* x);

end

The equation is good according to my teachers and several papers but the solver never converges and I can't see why. All coefficient are defined according to the dimensions of a capacitive micromachined ultrasound transducer. The solution of ode45 is the following :

This is obviously wrong because the dimensions of a cmut are bellow millimeter.

Can you see any mistakes in the way I use the ode45 solver ?

Full code :

% DIMENSIONS DE LA MEMBRANE
e = 500e-9;                 % epaisseur [m]
r = 20e-6;                  % rayon [m]
d0 = 550e-9;                % epaisseur de cavite [m]
a = pi * r^2;               % surface de la membrane [m^2]

% PARAMETRES MECANIQUES
E = 200e9;                  % module d'Young du SiN [Pa]
nu = 0.25;                  % coefficient de poisson du SiN
eta = 18.5e-6;              % viscosité dynamique de l'air [Pa.s]
p0 = 1e5;                   % pression exterieure [Pa]
rhoSiN = 3170;              % densite du SiN [Kg/m^3]
m = rhoSiN * a * e;         % masse membrane [Kg]
K = (16 * E * e^3) / (3 * (1 - nu^2) * r^2);   % raideur [N/m]
B = (eta * pi * r^2) / e;                      % amortissement [N.s/m]

% PARAMETRES ELECTRIQUES
eps0 = 8.85e-12;            % permittivité diélectrique du vide [F/m]
V0 = 10;                    % tension de polarisation [V]

% Conditions initiales et plage de temps
x0 = 0;
ti = 0; tf = 1e-3;
dt = 1e-6;
tspan = linspace(ti, tf, 1/dt);

% Résolution par RK4
[t, x] = ode45(@(t, x) odefun(t, x, eps0, a, V0, B, d0, K), tspan, x0);

plot(t,x,'-')

function dxdt = odefun(t, x, eps, a, V, B, d0, K)
dxdt = ((eps * a * V^2) ./ (2 * B * (d0 - x).^2)) + ((K / B) .* x);
end

References :

- Y. Wang, L. -M. He, Z. Li, W. Xu and J. Ren, "A computationally efficient nonlinear dynamic model for cMUT based on COMSOL and MATLAB/Simulink"

- T. Merrien, A. Boulmé and D. Certon, "Lumped-Parameter Equivalent Circuit Modeling of CMUT Array Elements"

I tried several solver from MATLAB and implemented my own Runge-Kutta algorithm based on the Wikipedia example. I also verified the coefficients according to my references.

Answer 1

[Sam Chak]

Hi @ALOYS

My guess is that if the coefficient is a positive value, then the system will be unstable.

[Sam Chak]

Hi @ALOYS

I changed the sign of the coefficient to test if my guess is correct.

% DIMENSIONS DE LA MEMBRANE
e   = 500e-9;                   % epaisseur [m]
r   = 20e-6;                    % rayon [m]
d0  = 550e-9;                   % epaisseur de cavite [m]
a   = pi * r^2;                 % surface de la membrane [m^2]

% PARAMETRES MECANIQUES
E   = 200e9;                    % module d'Young du SiN [Pa]
nu  = 0.25;                     % coefficient de poisson du SiN
eta = 18.5e-6;                  % viscosité dynamique de l'air [Pa.s]
p0  = 1e5;                      % pression exterieure [Pa]
rhoSiN = 3170;                  % densite du SiN [Kg/m^3]
m   = rhoSiN * a * e;           % masse membrane [Kg]
K   = (16 * E * e^3) / (3 * (1 - nu^2) * r^2);   % raideur [N/m]
B   = (eta * pi * r^2) / e;     % amortissement [N.s/m]

% PARAMETRES ELECTRIQUES
eps0= 8.85e-12;                 % permittivité diélectrique du vide [F/m]
V0  = 10;                       % tension de polarisation [V]

% Conditions initiales et plage de temps
x0  = 0;
ti  = 0; 
% tf  = 1e-3;
tf  = 2e-9;
dt  = 1e-6;
% tspan = linspace(ti, tf, 1/dt);
tspan = [ti, tf];

% Résolution par RK4
opts   = odeset('RelTol', 1e-9, 'AbsTol', 1e-11);
[t, x] = ode15s(@(t, x) odefun(t, x, eps0, a, V0, B, d0, K), tspan, x0, opts);
plot(t, x), grid on
xlabel('t'), ylabel('x(t)')

function dxdt = odefun(t, x, eps, a, V, B, d0, K)
    dxdt = ((eps * a * V^2) ./ (2 * B * (d0 - x).^2)) - ((K / B) .* x);
end

[Sam Chak]

The sign looks "positive" to me. But you placed a 'minus' sign at .

[William Rose]

THat is a very nice and insightful answer from @Sam Chak.

[ALOYS]

You are right ! The sign on the K/B term was wrong. Now my model gives accurate predictions according to my tests on the real device. This will be the first part of my internship report.

[Sam Chak]

Hi @ALOYS, I'm glad it worked out. If you find the solution helpful, please consider clicking 'Accept' ✔ on the answer and voting 👍 for it. Your support is greatly appreciated!

Thank you, @William Rose.