Question

Repeat Problem 17 using the improved Euler's method, which has a global truncation error$\mathit{O}\left(h^2\right)$. See Problem 1. You might need to keep more than four decimal places to see the effect of reducing the order of the error.

Short answer

1. The formula for the local truncation error in the nth step is $\mathrm{y}\text{'}\text{'}\text{'}\left(\mathrm{c}\right)\frac{{\mathrm{h}}^3}{6}=19{\mathrm{e}}^{-3(\mathrm{c}-1)}{\mathrm{h}}^3$
2. The upper bound of the local truncation error for the initial value problem $y^{\text{'}}=2x-3y+1,y\left(1\right)=5$is 0.019.
3. The value y(1.5) for step size h = 0.1 and h = 0.05 is obtained as 2.080108 and 2.059165 respectively.
4. The actual error when h = 0.05 is approximately half of the actual error when h = 0.1.

Step-by-step solution

Improved Euler's method

As per the Improved Euler's method, the solution of a linear differential equation of the form $\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$is given as follows:

$$\begin{gathered} {\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}^{\mathbf{*}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hf}\left(x_n,y_n\right)\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ {\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\frac{\mathbf{h}}{\mathbf{2}}\left(f\left(x_n,y_n\right)+f\left(x_{n+1},y_{n+1}^{*}\right)\right)\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Euler's method:

As per Euler's method, the solution of the equation of the form $\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$is given as follows:

${\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\mathit{h}\mathit{f}\left(x_n,y_n\right)$

a) Formula for the local truncation error.

The initial value problem is $y\text{'}=2x-3y+1$such that the value of y(1) is 5.n The analytic solution is $y=\frac{1}{9}+\frac{2}{3}x+\frac{38}{9}{e}^{-3(x-1)}$.

Consider the analytic solution as follows:

$y=\frac{1}{9}+\frac{2}{3}x+\frac{38}{9}{e}^{-3(x-1)}$(1)

Differentiate the equation (1) with respect to x as follows:

$$\begin{gathered} y\text{'}=\frac{2}{3}+\frac{38}{9}(-3)e^{-3(x-1)} \\ y\text{'}=\frac{2}{3}-\frac{38}{3}{e}^{-3(x-1)} \end{gathered}$$

Further, the second and third derivatives of the equation $y=\frac{1}{9}+\frac{2}{3}x+\frac{38}{9}{e}^{-3(x-1)}$with respect to x is obtained as follows:

$$\begin{gathered} y"=-\frac{38}{3}(-3)e^{-3(x-1)}y" \\ y"=38e^{-3(x-1)}y\text{'}\text{'}\text{'} \\ y"=38(-3)e^{-3(x-1)}y\text{'}\text{'}\text{'} \\ y"=-114e^{-3(x-1)} \end{gathered}$$

The local truncation error of the function $y\text{'}\text{'}\text{'}=-114e^{-3(x-1)}$is as follows:

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=114e^{-3(c-1)}\left(\frac{h^3}{6}\right) \\ {y}^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=19e^{-3(c-1)}{h}^3 \end{gathered}$$

Thus, the formula for the local truncation error in nth step is$y^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=19e^{-3(c-1)}{h}^3$

b) Finding the upper bound value for local truncation error

From subpart (a), the obtained formula for the local truncation error is as follows:$y^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=19e^{-3(c-1)}{h}^3$(2)

The value of c is defined as follows:

Substitute 1.5 for c and 0.1 for h in equation (2) to obtain the upper bound for the local truncation error.

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=19e^{-3(c-1)}{h}^3 \\ {y}^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=19\left(1\right)(0.001) \end{gathered}$$

$\left(e^{-3(x-1)}\right.$ is a decreasing function)
$$\begin{gathered} y^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=19(0.001) \\ {y}^{\text{'}\text{'}\text{'}}\left(c\right)\frac{h^3}{6}=0.019 \end{gathered}$$

Thus, the upper bound of the local truncation error for the initial value problem $y^{\text{'}}=2x-3y+1,y\left(1\right)=5$is 0.019.

c) Approximation value for y(1.5) using h = 0.1 and h = 0.05 with Euler's method

The differential equation is given as follows:

$y^{\text{'}}=2x-3y+1$

The initial value is given as y(1) = 5. This implies that$x_0=1$and$y_0=5$.

The given differential equation is of the form$y^{\text{'}}=f\left(x\right)$.

$f(x,y)=2x-3y+1$

Obtain the solution of the given differential equation by Euler's method for h = 0.1.

The solution of a linear differential equation of the form $y^{\text{'}}=f(x,y)$by the Improved Euler's method is given as follows:

$$\begin{gathered} y_{n+1}^{*}=y_n+hf\left(x_n,y_n\right)-----\left(3\right) \\ {y}_{n+1}=y_n+\frac{h}{2}\left(f\left(x_n,y_n\right)+f\left(x_{n+1},y_{n+1}^{*}\right)\right)----\left(4\right) \end{gathered}$$

Substitute 0 for n in equation (3) to obtain the equation oflocalid="1664281692686" $y_1^{*}$as shown below.

localid="1664281685938" $y_1^{*}=y_0+hf\left(x_0,y_0\right)$

Substitute 1 forlocalid="1664281697729" $x_0,5$forlocalid="1664281702236" $y_0$and 0.1 for h in the above equation.

localid="1664281705767" $$\begin{gathered} y_1^{*}=5+(0.1)f(1,5) \\ {y}_1^{*}=5+(0.1)·\left(2\right(1)-3(5)+1) \\ {y}_1^{*}=5-1.2 \\ {y}_1^{*}=3.8 \end{gathered}$$

Substitute 0 for n in equation (4).

localid="1664281709590" $y_1=y_0+\frac{h}{2}\left(f\left(x_0,y_0\right)+f\left(x_1,y_1^{*}\right)\right)$

Substitute 1 for localid="1664281714688" $x_0,5$for localid="1664281718822" $y_0,1.1$for localid="1664281723026" $x_1,3.8$for localid="1664281727525" ${y_1}^{*}$and 0.1 for h in the above equation.

localid="1664281731399" $$\begin{gathered} y_1=5+\frac{0.1}{2}\left(f\right(1,5)+f(1.1,3.8\left)\right) \\ {y}_1=5+(0.05)\left(2\right(1)-3(5)+1+2(1.1)-3(3.8)+1) \\ {y}_1=5-1.01 \\ {y}_1=3.99 \end{gathered}$$

Similarly, the value of localid="1664281735305" $y(1.1),y(1.2),y(1.3),y(1.4)$and localid="1664281740436" $y\left(1.5\right)$are obtained as shown in Table 1.

Now for step size h = 0.05, calculate for value of y(1.5) as shown below.

Substitute 0 for n in equation (3) to obtain the equation of$y_1^{*}$as shown below.

$y_1^{*}=y_0+hf\left(x_0,y_0\right)$

Substitute 1 for$x_0,5$for$y_0$and 0.05 for h in the above equation.

$$\begin{gathered} y_1^{*}=5+(0.05)f(1,5) \\ {y}_1^{*}=5+(0.05)·\left(2\right(1)-3(5)+1) \\ {y}_1^{*}=5-0.6 \\ {y}_1^{*}=4.4 \end{gathered}$$

Substitute 0 for n in equation (4).

$y_1=y_0+\frac{h}{2}\left(f\left(x_0,y_0\right)+f\left(x_1,y_1^{*}\right)\right)$

Substitute 1 for $x_0,5$for $y_0,1.05$for $x_1,4.4$for ${y_1}^{*}$and 0.05 for h in the above equation.

$$\begin{gathered} y_1=5+\frac{0.05}{2}\left(f\right(1,5)+f(1.05,4.4\left)\right) \\ {y}_1=5+(0.025)\left(2\right(1)-3(5)+1+2(1.05)-3(4.4)+1) \\ {y}_1=5-0.5525 \\ {y}_1=4.4475 \end{gathered}$$

Similarly, the value of $y(1.05),y(1.1),y(1.15),y(1.2),y(1.25),y(1.3),y(1.35)$, are obtained as shown in Table 2. $y(1.4),y(1.45)$and $y(1.5)$

Thus, the value y(1.5) for step size h = 0.1 and h = 0.05 is obtained as 2.080108 and 2.059165 respectively.

d) Calculate the errors in part (c) and verify that the global truncation error of Euler's method

From subpart (c) the obtained value of y(1.5) for step size h = 0.1 is 2.080108, and for step size h = 0.05 the value ofis 2.059165.

Consider the analytic solution as follows:

$y=\frac{1}{9}+\frac{2}{3}x+\frac{38}{9}{e}^{-3(x-1)}$------(5)

Now substitute 1.5 for x in equation (5) and solve.

$$\begin{gathered} y(1.5)=\frac{1}{9}+\frac{2}{3}(1.5)+\frac{38}{9}{e}^{-3(1.5-1)} \\ y(1.5)=0.1111+1+0.9421 \\ y(1.5)=2.0532 \end{gathered}$$

For h = 0.1, the actual error is calculated by the equation as follows:

$Actualerror=y(1.5)-{\left.y(1.5)\right|}_{h=0.1}$

Substitute 2.0532 for y(1.5) and 2.080108 for ${\left.y(1.5)\right|}_{h=0.1}$in equation (6).

$$\begin{gathered} Actualerror=2.0532-2.080108 \\ Actualerror=0.026908 \end{gathered}$$

For h = 0.05, the actual error is calculated by the equation as follows:

$actualerror=y(1.5)-y(1.5)I_{h=0.05}$-------(7)

Substitute 2.0532 for y(1.5) and 2.059165$ for${\left.y(1.5)\right|}_{h=0.05}$in equation (7).

$$\begin{gathered} Actualerror=2.0532-2.59165 \\ Actualerror=0.005965 \end{gathered}$$

Thus, the actual error in subpart (d) when step size is halved (h = 0.05) is 0.026908. The actual error when the step size (h = 0.1) is 0.005965.

The actual error when h = 0.05 is approximately half of the actual error when h = 0.1.