Question

Repeat Problem 17 for the initial-value problem ${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{y}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. The analytic solution is$\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{l}\mathit{n}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}$. Approximate y(0.5). See Problem 5 in Exercises 2.6.

Short answer

1. the formula for the local truncation error in the nth step is$y^{\text{'}\text{'}}\left(c\right)\frac{h^2}{2}=\frac{1}{{(c+1)}^2}\left(\frac{h^2}{2}\right)$
2. The upper bound of the local truncation error for the initial value problem $y^{\text{'}}=e^{-y},y\left(0\right)=0$is 0.005.
3. The value y(0.5) for step size h = 0.1 and h = 0.05 is obtained as 0.4197 and 0.4124 respectively.
4. The actual error when h = 0.05 is approximately half 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:

role="math" localid="1664283602826" $$\begin{gathered} {\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}^{\mathbf{*}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\mathit{h}\mathit{f}\left(x_n,y_n\right)\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ {\mathit{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathit{y}}_{\mathbf{n}}\mathbf{+}\frac{\mathbf{h}}{\mathbf{2}}\left(f\left(x_n,y_n\right)+f\left(x_{n+1},y_{n+1}^{*}\right)\right)\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

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 $y^{\text{'}}=e^{-y}$such that the value of y(0) is 0. The analytic solution is $y\left(x\right)=\ln (x+1)$.

Consider the analytic solution as follows:

$y\left(x\right)=\ln (x+1)$------------- (1)

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

$y^{\text{'}}\left(x\right)=\frac{1}{x+1}$

Further, differentiate the equation$y^{\text{'}}\left(x\right)=\frac{1}{x+1}$with respect to x as follows:

$y^{\text{'}\text{'}}\left(x\right)=-\frac{1}{{(x+1)}^2}$

The local truncation error of the function $y^{\text{'}\text{'}}\left(x\right)=-\frac{1}{{(x+1)}^2}$is as follows:

$y^{\text{'}\text{'}}\left(c\right)\frac{h^2}{2}=\frac{1}{{(c+1)}^2}\left(\frac{h^2}{2}\right)$

Thus, the formula for the local truncation error in nth step is$y^{\text{'}\text{'}}\left(c\right)\frac{h^2}{2}=\frac{1}{{(c+1)}^2}\left(\frac{h^2}{2}\right)$

b) Finding the upper bound value for local truncation error

From subpart (a) the obtained formula for the local truncation error is as follows:

$y^{\text{'}\text{'}}\left(c\right)=\frac{1}{{(c+1)}^2}\left(\frac{h^2}{2}\right)$

$y^{\text{'}\text{'}}\left(c\right)\frac{h^2}{2}$-------- (2)

The value of c is defined as follows:

$0<c<0.5$

Substitute 0.5 for c and 0.1 for h in equation (2) to obtain the upper bound for the local truncation error.

$y^{\text{'}\text{'}}\left(c\right)\frac{h^2}{2}=\frac{1}{{(0.5+1)}^2}\left(\frac{{(0.1)}^2}{2}\right)$

$\left(\frac{1}{{(1+x)}^2}\right.$is decreasing function) =0.005

Thus, the upper bound of the local truncation error for the initial value problem $y^{\text{'}}=e^{-y},y\left(0\right)=0$is 0.005.

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:

$y^{\text{'}}=e^{-y}$

The initial value is given as$y\left(0\right)=0$. This implies that$x_0=0$and$y_0=0$.

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

$f(x,y)=e^{-y}$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:

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

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 0 for$x_0,0$for$y_0$and 0.1 for h in the above equation.

$$\begin{gathered} y_1^{*}=0+(0.1)f(0,0) \\ {y}_1^{*}=0+(0.1)·\left(e^{-0}\right) \\ {y}_1^{*}=0+0.1 \\ {y}_1^{*}=0.1 \end{gathered}$$

Now for step size h = 0.05, calculate for value of y(0.5) 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 0 for $x_0,0$for $y_0$and $0.05$ for h in the above equation.

$$\begin{gathered} y_1=0+(0.05)f(0,0) \\ {y}_1=0+(0.05)·\left(e^0\right) \\ {y}_1=0+0.05 \\ {y}_1=0.05 \end{gathered}$$

Similarly, the value of $y(0.05),y(0.1),y(0.15),y(0.2),y(0.25),y(0.3),y(0.35)$, are obtained as shown in Table 2. $y(0.4),y(0.45)$and $y\left(0.5\right)$.

Thus, the value y(0.5) for step size h = 0.1 and h = 0.05 is obtained as 0.4197and 0.4124 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(0.5) for step size h = 0.1 is 0.4197, and for step size h = 0.05 the value of y(0.5) is 0.4124.

Consider the analytic solution as follows:

$y\left(x\right)=\ln (x+1)$

Now substitute 0.5 for x in equation (4) and solve.

$$\begin{gathered} y(0.5)=\ln (0.5+1) \\ y(0.5)=\ln (1.5) \\ y(0.5)=0.4054 \end{gathered}$$

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

localid="1664285870777" role="math" $Actualerror=y(0.5)-{\left.y(0.5)\right|}_{h=0.1}$------------(5)

Substitute 0.4054 for y(0.5) and 0.4197 for${\left.y(0.5)\right|}_{h=0.1}$in equation (5).

$$\begin{gathered} Actualerror=0.4054-0.4197 \\ Actualerror=0.0143 \end{gathered}$$

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

$Actualerror=y(0.5)-{\left.y(0.5)\right|}_{h=0.05}$------------(6)

Substitute 0.4054 for y(0.5) and 0.4124 for${\left.y(0.5)\right|}_{h=0.05}$in equation (6).

$Actualerror=0.4054-0.4124$

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

$Actualerror=y(0.5)-y(0.5)I_{h=0.05}^{\left(6\right)}$

Substitute 0.4054 for y(0.5) and 0.4124 for $y(0.5{)}_{h=0.05}$in equation .

$$\begin{gathered} Actualerror=0.4054-0.4124 \\ Actualerror=0.007 \end{gathered}$$

Thus, the actual error in subpart (d) when step size is halved (h = 0.05) is 0.007. The actual error when the step size (h = 0.1) is 0.0143. The actual error when h = 0.05 is approximately half of the actual error when h = 0.1.