Question 7.6: CLASSICAL RK4 METHOD Consider the initial-value problem in E...

CLASSICAL RK4 METHOD

Consider the initial-value problem in Examples 7.4 and 7.5:

y′ – x²y = 2x²,  y(0) = 1, 0 ≤ x ≤ 1, h = 0.1

1. Using hand calculations, compute the estimated value of y_1 = y(0.1) by the classical RK4 method.

2. Solve the initial-value problem by executing the user-defined function RK4.

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1. Noting that f(x, y) = x²(2 + y), the calculations are carried out as follows:

k_1=f(x_0,y_0)=f(0,1)=0

k_2=f(x_0+\frac{1}{2} h,y_0+\frac{1}{2}k_1h) =f(0.05,1)=0.0075

k_3=f(x_0+\frac{1}{2}h,y_0+\frac{1}{2}k_2h )=f(0.05,1.0004)=0.0075

k_4=f(x_0+ h,y_0+k_3h) =f(0.1,1.0008)=0.0300

y_1=y_0+\frac{1}{6}h(k_1+2k_2+2k_3+k_4)=\boxed{1.0010}

2.

>> f = @(x,y)((x^2)*(2+y));
>> x = 0:0.1:1;
>> y0 = 1;
>> y = RK4(f,x,y0); y = y'
y =
1.0000
1.0010
1.0080
1.0271
1.0647
1.1276
1.2240
1.3634
1.5583
1.8252
2.1868

A summary of all calculations is provided in Table 7.3 where it is easily seen that the global % relative error for the classical RK4 method is significantly lower than all previous methods used up to this point. As expected, starting with Euler’s method, which is indeed a first-order Runge–Kutta method, the accuracy  improves with the order of the RK method.

TABLE 7.3
Summary of Calculations in Example 7.6
x y_{\text{RK4}} RK4

e_{\text{RK4}}

RK3

e_{\text{RK3}}

RK2

e_{\text{Heun}}

RK1

e_{\text{Euler}}

0.0 1.000000 0.000000 0.0000 0.00 0.00
0.1 1.001000 0.000001 -0.0000 -0.05 0.10
0.2 1.008011 0.000002 -0.0001 -0.10 0.50
0.3 1.027122 0.000003 -0.0004 -0.15 1.18
0.4 1.064688 0.000004 -0.0010 -0.19 2.12
0.5 1.127641 0.000005 -0.0018 -0.23 3.27
0.6 1.223966 0.000006 -0.0030 -0.25 4.56
0.7 1.363377 0.000007 -0.0044 -0.27 5.96
0.8 1.558286 0.000010 -0.0059 -0.28 7.40
0.9 1.825206 0.000016 -0.0074 -0.27 8.87
1.0 2.186837 0.000028 -0.0087 -0.25 10.37

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