Consider the initial value problem,
f(x,y) = y(18.06 - y), y(0) = 12.
The exact solution of the problem increases from y(0) =12 to y = 18.06 as x increases without limit.
Determine the minimum upper bound of h for the classical 4th-order Runge Kutta method to be absolutely stable for this problem. Give your answer to 3 decimal places.
Note : You can make use of table (3.1) of Study Unit 2 Numerical Methods for Differential Equations.
In order to determine the minimum upper bound of h for the classical 4th-order Runge Kutta method to be absolutely stable for this problem, we need to refer to Table 3.1 of Study Unit 2 Numerical Methods for Differential Equations.
Let's first look at the Taylor series expansion of the exact solution of the initial value problem around the point (x0, y0):
y(x) = y(x0) + (x - x0)y'(x0) + (x - x0)^2/2! y''(x0) + (x - x0)^3/3! y'''(x0) + ...
The 4th-order Runge Kutta method has the form:
k1 = h f(x0, y0)
k2 = h f(x0 + 1/2 h, y0 + 1/2 k1)
k3 = h f(x0 + 1/2 h, y0 + 1/2 k2)
k4 = h f(x0 + h, y0 + k3)
y(x0 + h) = y0 + 1/6(k1 + 2k2 + 2k3 + k4)
For stability, we need to find the condition for which the error term remains bounded as h approaches zero. In Table 3.1, we can find the maximum value of Δy (the error term) for different values of Δt (the step size).
Given that the exact solution of the problem increases from y(0) = 12 to y = 18.06 as x increases without limit, we can approximate the upper bound of the error term as (18.06 - 12) = 6.06.
From Table 3.1, we can find the maximum value of Δy for 4th-order Runge Kutta method corresponding to Δt = h. To ensure the stability, we need to choose h such that the maximum value of Δy is less than or equal to 6.06.
By looking at Table 3.1, find the largest value of h for which the maximum value of Δy is less than or equal to 6.06. Round the answer to 3 decimal places.
Once you've found the value of h, check the corresponding value of Δy. If it is less than or equal to 6.06, then that is the minimum upper bound of h for the classical 4th-order Runge Kutta method to be absolutely stable for this problem.