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deval

Evaluate solution of differential equation problem

Syntax

sxint = deval(sol,xint)
sxint = deval(xint,sol)
sxint = deval(sol,xint,idx)
sxint = deval(xint,sol,idx)
[sxint, spxint] = deval(...)

Description

sxint = deval(sol,xint) and sxint = deval(xint,sol) evaluate the solution of a differential equation problem. sol is a structure returned by one of these solvers:

• An initial value problem solver (ode45, ode23, ode113, ode15s, ode23s, ode23t, ode23tb, ode15i)

• A delay differential equations solver (dde23, ddesd, or ddensd),

• A boundary value problem solver (bvp4c or bvp5c).

xint is a point or a vector of points at which you want the solution. The elements of xint must be in the interval [sol.x(1),sol.x(end)]. For each i, sxint(:,i) is the solution at xint(i).

sxint = deval(sol,xint,idx) and sxint = deval(xint,sol,idx) evaluate as above but return only the solution components with indices listed in the vector idx.

[sxint, spxint] = deval(...) also returns spxint, the value of the first derivative of the polynomial interpolating the solution.

 Note   For multipoint boundary value problems, the solution obtained by bvp4c or bvp5c might be discontinuous at the interfaces. For an interface point xc, deval returns the average of the limits from the left and right of xc. To get the limit values, set the xint argument of deval to be slightly smaller or slightly larger than xc.

Examples

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Evaluate van der Pol Equation

This example solves the system y' = vdp1(t,y) using ode45 and plots the first component of the solution.

Solve the system using ode45.

```sol = ode45(@vdp1,[0 20],[2 0]);
```

Evaluate the first component of the solution at 100 points in the interval

```x = linspace(0,20,100);
y = deval(sol,x,1);
```

Plot the solution.

```plot(x,y)
```