This example shows how to use `ddensd`

to solve a system of initial value DDEs (delay differential equations) with time-dependent delays. The example was originally presented by Jackiewicz [1].

The equation is

$${\mathit{y}}^{\prime}\left(\mathit{t}\right)=2\text{\hspace{0.17em}}\mathrm{cos}\left(2\mathit{t}\right){\mathit{y}\left(\frac{\mathit{t}}{2}\right)}^{2\text{\hspace{0.17em}}\mathrm{cos}\left(\mathit{t}\right)}+\mathrm{log}\left({\mathit{y}}^{\prime}\left(\frac{\mathit{t}}{2}\right)\right)-\mathrm{log}\left(2\text{\hspace{0.17em}}\mathrm{cos}\left(\mathit{t}\right)\right)-\mathrm{sin}\left(\mathit{t}\right).$$

This equation is an *initial* *value* DDE because the time delays are zero at ${\mathit{t}}_{0}$. Therefore, a solution history is unnecessary to calculate a solution, only the initial values are needed:

$$\mathit{y}\left(0\right)=1,$$

$${\mathit{y}}^{\prime}\left(0\right)=\mathit{s}.$$

$\mathit{s}$ is the solution of $2+\mathrm{log}\left(\mathit{s}\right)-\mathrm{log}\left(2\right)=0$. The values of $\mathit{s}$ that satisfy this equation are ${\mathit{s}}_{1}=2$ and ${\mathit{s}}_{2}=0.4063757399599599$.

Since the time delays in the equations are present in a ${\mathit{y}}^{\prime}$ term, this equation is called a *neutral* *DDE*.

To solve this equation in MATLAB, you need to code the equation and delays before calling the delay differential equation solver `ddensd`

, which is the solver for neutral equations. You either can include the required functions as local functions at the end of a file (as done here), or save them as separate files in a directory on the MATLAB path.

First, write an anonymous function to define the delays in the equation. Since both $\mathit{y}$ and ${\mathit{y}}^{\prime}$ have delays of the form $\frac{\mathit{t}}{2}$, only one function definition is required. This delay function is later passed to the solver twice, once to indicate the delay for $\mathit{y}$ and once for ${\mathit{y}}^{\prime}$.

delay = @(t,y) t/2;

Now, create a function to code the equation. This function should have the signature `yp = ddefun(t,y,ydel,ypdel)`

, where:

`t`

is time (independent variable).`y`

is the solution (dependent variable).`ydel`

contains the delays for*y*.`ypdel`

contains the delays for ${\mathit{y}}^{\prime}=\frac{\mathrm{dy}}{\mathrm{dt}}$.

These inputs are automatically passed to the function by the solver, but the variable names determine how you code the equation. In this case:

`ydel`

$\to \mathit{y}\left(\frac{\mathit{t}}{2}\right)$`ypdel`

$\to {\mathit{y}}^{\prime}\left(\frac{\mathit{t}}{2}\right)$

function yp = ddefun(t,y,ydel,ypdel) yp = 2*cos(2*t)*ydel^(2*cos(t)) + log(ypdel) - log(2*cos(t)) - sin(t); end

*Note: All functions are included as local functions at the end of the example.*

Finally, define the interval of integration $\left[{\mathit{t}}_{0}\text{\hspace{0.17em}\hspace{0.17em}}{\mathit{t}}_{\mathit{f}}\right]$ and the initial values, and then solve the DDE using the `ddensd`

solver. Pass the initial values to the solver by specifying them in a cell array in the fourth input argument.

tspan = [0 0.1]; y0 = 1; s1 = 2; sol1 = ddensd(@ddefun, delay, delay, {y0,s1}, tspan);

Solve the equation a second time, this time using the alternate value of $\mathit{s}$ for the initial condition.

s2 = 0.4063757399599599; sol2 = ddensd(@ddefun, delay, delay, {y0,s2}, tspan);

The solution structures `sol1`

and `sol2`

have the fields `x`

and `y`

that contain the internal time steps taken by the solver and corresponding solutions at those times. However, you can use `deval`

to evaluate the solution at the specific points.

Plot the two solutions to compare results.

plot(sol1.x,sol1.y,sol2.x,sol2.y); legend('y''(0) = 2','y''(0) = .40637..','Location','NorthWest'); xlabel('Time t'); ylabel('Solution y'); title('Two Solutions of Jackiewicz''s Initial-Value NDDE');

Listed here are the local helper functions that the DDE solver `ddensd`

calls to calculate the solution. Alternatively, you can save these functions as their own files in a directory on the MATLAB path.

function yp = ddefun(t,y,ydel,ypdel) yp = 2*cos(2*t)*ydel^(2*cos(t)) + log(ypdel) - log(2*cos(t)) - sin(t); end

[1] Jackiewicz, Z. “One step Methods of any Order for Neutral Functional Differential Equations.” *SIAM Journal on Numerical Analysis.* Vol. 21, Number 3. 1984. pp. 486–511.

`dde23`

| `ddensd`

| `ddesd`

| `deval`