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DDE of Neutral Type

This example shows how to use ddensd to solve the neutral DDE presented by Paul [1] for 0 ≤ tπ.

Click ddex4.m or type edit ddex4.m in a command window to view the code for this example in an editor.

The equation is

y '(t) = 1 + y(t) – 2y(t/2)2y '(tπ)

with history:

y(t) = cos (t) for t ≤ 0.

  1. Create a new program file in the editor. This file will contain a main function and four local functions.

  2. Define the first-order DDE as a local function.

    function yp = ddefun(t,y,ydel,ypdel) 
        yp = 1 + y - 2*ydel^2 - ypdel;
  3. Define the solution delay as a local function.

    function dy = dely(t,y) 
        dy = t/2;
  4. Define the derivative delay as a local function.

    function dyp = delyp(t,y) 
        dyp = t-pi;
  5. Define the solution history as a local function.

    function y = history(t)
        y = cos(t);
  6. Define the interval of integration and solve the DDE using the ddensd function. Add this code to the main function.

    tspan = [0 pi];
    sol = ddensd(@ddefun,@dely,@delyp,@history,tspan);
  7. Evaluate the solution at 100 equally spaced points between 0 and π. Add this code to the main function.

    tn = linspace(0,pi);
    yn = deval(sol,tn);
  8. Plot the results. Add this code to the main function.

    xlim([0 pi]);
    ylim([-1.2 1.2])
    xlabel('time t');
    ylabel('solution y');
    title('Example of Paul with 1 equation and 2 delay functions')
  9. Run your program to calculate the solution and display the plot.


[1] Paul, C.A.H. “A Test Set of Functional Differential Equations.” Numerical Analysis Reports. No. 243. Manchester, UK: Math Department, University of Manchester, 1994.

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