Question

Question: A count of the number of evaluations of the function used in solving the initial-value problem$\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{f}\left(\mathrm{xy}\right)\mathbf{;}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}{\mathit{x}}_{\mathbf{0}}\mathbf{\right)}\mathbf{=}{\mathit{y}}_{\mathbf{0}}$ is used as a measure of the computational complexity of a numerical method. Determine the number of evaluations of required for each step of Euler's, the improved Euler's, and the RK4 methods. By considering some specific examples, compare the accuracy of these methods when used with comparable computational complexities.

Short answer

The computational complexities for Euler's method, improved Euler's method and RK4 method, are 1,2 and 5 respectively. By using the example $y=\left(e^x-y\right)/xy\left(1\right)=2,y\left(2\right)$, using the same number of evaluations for the three methods, it shows that RK4 is the most accurate method.

Step-by-step solution

Comparison of Euler’s,Improved Euler’s and RK4 Method

• The solution curve is approximated using a straight line (as in the original Euler's method), but the slope of this line is considered as the average of the slopes at points 0 and 1, where points 0 and 1 are the end points for each sub - interval.
• That is, slope = [(dy/dx) at point 0 + (dy/dx) at point 1] / 2.
• However, in the original Euler method, each yn+1 approximation is based on the slope at the previous point (dy/dx) at point n. So y1 is based on (dy/dx) at point 0.
• The comparison shows that the Euler method produces more accurate approximate results than the Runge-Kutta method.

Derivations of Euler’s, Improved Euler’s and RK4 Method

Euler’s method

$y_{n+1}=y_n+\mathrm{hf}\left(x_n{y}_n\right)$

Improved Euler’s method

$$\begin{gathered} y_{n+1}=y_n+\mathrm{hf}\left(x_n{y}_n\right) \\ {y}_{n+1}=y_n+h/2\left[f(x_n+1,y_n)+f(x_n+1,y_n*)\right] \end{gathered}$$

RK4 Method

$$\begin{gathered} k_1=f\left(x_n,y_n\right) \\ {k}_2=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_1}{2}\right)\right) \\ {k}_3=f\left(\left(x_n+\frac{h}{2}\right),\left(y_n+\frac{h{k}_2}{2}\right)\right) \\ {k}_4=f\left(\left(x_n+h\right),\left(y_n+h{k}_3\right)\right) \\ {y}_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \end{gathered}$$

From the previous, we have 1 evaluation for Euler's method, 2 evaluations for improved Euler's method and 5 evaluations for RK4 method, hence the complexities are 1, 2 and 5 respectively.

Comparison using a specific example

Assume we have the initial value problem

$y=\left(e^x-y\right)/x,y\left(1\right)=2$

and we want to determine $y\left(2\right)$.Note that this problem is of number $\left(25\right)$in section $2.3$and its exact solution$y=\left(e^x+2-e\right)/x$ is hence $y\left(2\right)=3.3358714$

Note that we want to make all methods make the same number of evaluations.

Now we will use RK4 method to determine in only 20 steps $\left(h=0.05\right)$hence we need 100 evaluations.

Thus, we will make improved Euler's method use 50 steps, hence $h=0.02$And we will make Euler's method use 100 steps, hence $h=0.01$

By using the mentioned conditions, we've got $y_{Ealce}\left(2\right)=3.3308y_{lampowodEuler}\left(2\right)=3.33546y_{RKA}\left(2\right)=3.33539$

And by comparing the results by the exact result, method gives the most accurate result.

The computational complexities for Euler's method, improved Euler's method and method, are 1,2 and 5 respectively.

By using the example $y=\left(e^x-y\right)/x,y\left(1\right)=2$to determine $y\left(2\right)$using the same number of evaluations for the three methods, we have seen that RK4 is the most accurate method.

Therefore RK4 method is the most accurate of the three methods.