## Topic 14.6: Stiff Differential Equations (Examples)

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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.

hApproximation
of y(0.5 + h)
Error
1.1.40.102
0.52.0.127
0.252.3157894740.0528
0.1252.4347826090.0160
0.06252.4754716980.00434
0.031252.4899135460.00112
0.0156252.4955194950.000285
0.00781252.4979026080.0000719
0.003906252.4989872830.0000181
0.0019531252.4995026700.00000452
0.00097656252.4997535950.00000113

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