MATLAB Mathematics - SERC - Index of
MATLAB Mathematics - SERC - Index of MATLAB Mathematics - SERC - Index of
5 Differential Equations 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 0 2 4 6 8 10 12 Example: Stiff Problem (van der Pol Equation) vdpode illustrates the solution of the van der Pol problem described in “Example: The van der Pol Equation, m = 1000 (Stiff)” on page 5-12. The differential equations y′ 1 = y 2 2 y′ 2 = µ(1 – y 1 )y 2 – y 1 involve a constant parameter µ . As µ increases, the problem becomes more stiff, and the period of oscillation becomes larger. When µ is 1000 the equation is in relaxation oscillation and the problem is very stiff. The limit cycle has portions where the solution components change slowly and the problem is quite stiff, alternating with regions of very sharp change where it is not stiff (quasi-discontinuities). By default, the solvers in the ODE suite that are intended for stiff problems approximate Jacobian matrices numerically. However, this example provides a nested function J(t,y) to evaluate the Jacobian matrix ∂f ⁄ ∂y analytically at 5-20
Initial Value Problems for ODEs and DAEs (t,y) for µ = MU. The use of an analytic Jacobian can improve the reliability and efficiency of integration. To run this example, click on the example name, or type vdpode at the command line. From the command line, you can specify a value of µ as an argument to vdpode. The default is µ = 1000. function vdpode(MU) %VDPODE Parameterizable van der Pol equation (stiff for large MU) if nargin < 1 MU = 1000; % default end tspan = [0; max(20,3*MU)]; y0 = [2; 0]; options = odeset('Jacobian',@J); % Several periods [t,y] = ode15s(@f,tspan,y0,options); plot(t,y(:,1)); title(['Solution of van der Pol Equation, \mu = ' num2str(MU)]); xlabel('time t'); ylabel('solution y_1'); axis([tspan(1) tspan(end) -2.5 2.5]); --------------------------------------------------------------- function dydt = f(t,y) dydt = [ y(2) MU*(1-y(1)^2)*y(2)-y(1) ]; end % End nested function f --------------------------------------------------------------- function dfdy = J(t,y) dfdy = [ 0 1 -2*MU*y(1)*y(2)-1 MU*(1-y(1)^2) ]; end % End nested function J end 5-21
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5 Differential Equations<br />
1<br />
0.8<br />
0.6<br />
0.4<br />
0.2<br />
0<br />
−0.2<br />
−0.4<br />
−0.6<br />
−0.8<br />
−1<br />
0 2 4 6 8 10 12<br />
Example: Stiff Problem (van der Pol Equation)<br />
vdpode illustrates the solution <strong>of</strong> the van der Pol problem described in<br />
“Example: The van der Pol Equation, m = 1000 (Stiff)” on page 5-12. The<br />
differential equations<br />
y′ 1 = y 2<br />
2<br />
y′ 2 = µ(1 – y 1 )y 2 – y 1<br />
involve a constant parameter µ .<br />
As µ increases, the problem becomes more stiff, and the period <strong>of</strong> oscillation<br />
becomes larger. When µ is 1000 the equation is in relaxation oscillation and<br />
the problem is very stiff. The limit cycle has portions where the solution<br />
components change slowly and the problem is quite stiff, alternating with<br />
regions <strong>of</strong> very sharp change where it is not stiff (quasi-discontinuities).<br />
By default, the solvers in the ODE suite that are intended for stiff problems<br />
approximate Jacobian matrices numerically. However, this example provides<br />
a nested function J(t,y) to evaluate the Jacobian matrix ∂f ⁄ ∂y analytically at<br />
5-20
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