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The following are not stiff differential equations, however, the techniques may still be applied.

Example 1

Given the IVP y⁽¹⁾(t) = 1 - t y(t) with y(0) = 1, approximate y(1) with one step.

First, let t₀ = 0, y₀ = 1, and h = 1. Thus, we write down the equation

-υ + y₀ + h*f(t₁, υ) = 0

and, after substituting the appropriate values, we get

-υ + 1 + 1*f(1, υ) = -2 υ + 2 = 0

Solving this equation yields υ = 1, and therefore we set y₁ = 1. The absolute error is 0.33.

Example 2

Given the same IVP shown in Example 1, approximate y(0.5).

First, let t₀ = 0, t₁ = 0.5, y₀ = 1, and h = 0.5. Thus, we write down the equation

-υ + y₀ + h*f(t₁, υ) = 0

and, after substituting the appropriate values, we get

-υ + 1 + 0.5*f(0.5, υ) = -1.25 υ + 1.5 = 0

Solving this equation yields υ = 1.2, and therefore we set y₁ = 1.2. The absolute error is 0.14 which is approximately the absolute error in Example 1.

Example 3

Repeat Examples 1 and 2 but with with the initial value y(0.5) = 2.5 and approximating y(1.5) and y(1.0).

To find y(1.5), let t₀ = 0.0, t₁ = 1.5, y₀ = 2.5, and h = 1. Thus, the equation

-υ + 2.5 + 1*f(1.5, υ) = -2.5 υ + 3.5 = 0

Solving this equation yields υ = 1.4, and therefore we set y₁ = 1.4. The actual value is y(1.5) = 1.502483616, and therefore the absolute error is 0.102.

To find y(1.0), let t₀ = 0.0, t₁ = 1.0, y₀ = 2.5, and h = 0.5. Thus, the equation is

-υ + 2.5 + 0.5*f(1.0, υ) = -1.5 υ + 3.0 = 0

Solving this equation yields υ = 2, and therefore we set y₁ = 2. The actual value is y(1) = 2.126611964 and therefore the absolute error is 0.127.

This absolute error is larger than it was when h = 1.0, and thus, to show that the error is O(h²), we must use smaller values of h. These are shown in Table 1.

Table 1. Errors when approximating y(t₀ + h) for decreasing values of h.

h	Approximation of y(0.5 + h)	Error
1.	1.4	0.102
0.5	2.	0.127
0.25	2.315789474	0.0528
0.125	2.434782609	0.0160
0.0625	2.475471698	0.00434
0.03125	2.489913546	0.00112
0.015625	2.495519495	0.000285
0.0078125	2.497902608	0.0000719
0.00390625	2.498987283	0.0000181
0.001953125	2.499502670	0.00000452
0.0009765625	2.499753595	0.00000113

The quadratic behaviour becomes obvious with the last step, being smaller by almost exactly 4.

Topic 14.6: Stiff Differential Equations (Examples)

Example 1

Example 2

Example 3