Question

Repeat Problem 18 for the initial-value problem ${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{y}}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}$The analytic solution is $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{l}\mathit{n}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{.}$ Approximate y(0.5) See Problem 7

(a) Find a formula involving cand h for the local truncation error in the nth step if the RK4 method is used.

(b) Find a bound for the local truncation error in each step if h = 0.1is used to approximate

(c) Approximate y(1.5)using the RK4 method with h = 0 and h = 0.05 See Problem 3. You will need to carry more than six decimal places to see the effect of reducing the step size.

Short answer

(a)The truncation error in the nth step if the RK4 method is used is $y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}$.

(b)The approximatebound for the local truncation error is 2 x 10^-6.

(c)The approximate values is for x = 0.5, h = 0.06 is y₅ = 0.406007 and for x = 0.5, h = 0.06 is y₁₀= 0.405608

Step-by-step solution

Obtaining the local truncation error

The difference between a truncated value and the actual value is referred to as a truncation error. A truncated quantity is represented by a numeral with a set number of allowed digits, with any extra digits "chopped off" (hence the expression "truncated").

$y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}$

Obtaining the truncation error if RK4 method is used

We have the initial value problem

$y^{\text{'}}=f(x,y)=e^{-y},y\left(0\right)=0$

With the analytical solution

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

and we have to obtain the required points as the following technique :

(a)

We can obtain the local truncation erroe in the nth if the RK4 method from the formula

$y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}$ (a)

$Wherec=valueofx.$

After that, we have to find the fifth derivative of the given analytical solution as

$$\begin{gathered} y^{\text{'}}\left(x\right)=\frac{1}{x+1}{ \\ }^{\text{'}\text{'}}\left(x\right)=-\frac{1}{{(x+1)}^2} \\ {y}^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{2}{{(x+1)}^3} \\ {y}^{\left(4\right)}\left(x\right)=-\frac{6}{{(x+1)}^4} \\ {y}^{\left(5\right)}\left(x\right)=-\frac{24}{{(x+1)}^5} \end{gathered}$$

Then replace x by c, then we have

$y^{\left(5\right)}\left(c\right)=-\frac{24}{{(c+1)}^5}$

After that, substitute with y⁽⁵⁾(c) into (a) then we can have the genera local truncation error as

$y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}=\frac{24}{{(c+1)}^5}\frac{h^5}{5!}$ (b)

Therefore, the truncation error in the nth step if the RK4 method is used is $y^{\left(5\right)}\left(c\right)\frac{h^5}{5!}$.

Determining the bound for the local truncation error

(b)

Now, with $c=x_0=1,$we can obtain the truncation error in each step if h = 0.1 using equation (b) as

$Error⩽\frac{24}{{(0+1)}^5}\frac{{(0.1)}^5}{120}$

$$\begin{gathered} =\frac{{(0.1)}^5}{120} \\ =2\times {10}^{-6} \end{gathered}$$

Therefore,the approximatebound for the local truncation error is 2 x 10^-6.

Obtaining the approximate values

(c)

We can obtain the approximate values for ywith one step using the formula

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$

But before that, we have to find the constant k and at and as the following

Since we have x₀= 0 and y₀ = 0 then we have

localid="1663962607588" $$\begin{gathered} k_1=f\left(x_0,y_0\right) \\ =e^{-y} \\ =e^0 \\ =1 \end{gathered}$$

$$\begin{gathered} k_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ =e^{-\left(y0+\frac{1}{2}h{k}_1\right)} \\ =e^{-\left(0+\frac{1}{2}\times 0.1\times 1\right)} \\ =e^{-0.05} \\ =0.951229 \\ {k}_3=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_2\right) \\ =e^{-\left(y0+\frac{1}{2}h{k}_1\right)} \\ =e^{-\left(0+\frac{1}{2}\times 0.1\times 0.95123\right)} \\ =e^{-0.0475615} \\ =0.953552 \end{gathered}$$

$$\begin{gathered} k_4=f\left(x_0+h,y_0+h{k}_3\right) \\ =e^{-\left(y_0+h{k}_1\right)} \\ =e^{-(0+0.1\times 0.953552)} \\ =e^{-0.09533552} \\ =0.90905 \end{gathered}$$

After that when n = 0 since we have y₀ = 0 then by substituting with the four constants we can obtain from equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =0+\frac{0.1}{6}[1+2(0.951229)+2(0.953552)+0.90905] \\ =0+\frac{0.1}{6}\times (5.718612) \\ =0.09531 \end{gathered}$$

After that, we can obtain the value of y for each value of x with another four constants in the table shown below until x = 0.5 we can have y₅= 0.406007

After that, we can obtain the approximate values with two steps using the formula

$y_{n+1}=y_n+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right)$

But before that, we have to find the constant k₁, k₂ and at and as the following

Since we have x₀ = 1 and y₀ = 0 then we have

$$\begin{gathered} k_1=f\left(x_0,y_0\right) \\ =e^{-y} \\ =e^0 \\ =1 \end{gathered}$$

$$\begin{gathered} k_2=f\left(x_0+\frac{1}{2}h,y_0+\frac{1}{2}h{k}_1\right) \\ =e^{-\left(y_0+\frac{1}{2}h{k}_1\right)} \\ =e^{-\left(0+\frac{1}{2}\times 0.05\times 1\right)} \\ =e^{-0.025} \\ =0.975310 \end{gathered}$$

$$\begin{gathered} k_4=f\left(x_0+h,y_0+h{k}_3\right) \\ =\left[2\left(x_0+h\right)-3\left(y_0+h{k}_1\right)+1\right] \\ =\left[2\right(1+0.05)-3(5+0.05\times (-11.12125))+1] \\ =2.1-13.3318125+1 \\ =-10.2318125 \end{gathered}$$

After that when n = 0 since we have y₀ = 5 then by substituting with the four constants we can obtain y₁ from equation (1) as

$$\begin{gathered} y_1=y_0+\frac{h}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =0+\frac{0.05}{6}[1+2(0.975310)+2(0.975912)+0.952375] \\ =0+\frac{0.05}{6}\times (5.85482) \\ =0.04879 \end{gathered}$$

After that, we can obtain the value of y for each value of x with another four constants in the table shown below until x = 0.5 we can have y = 0.405608

Therefore the approximate values is for x = 0.5, h = 0.06 is y₄= 0.406007 and for x = 0.5, h = 0.06 is y₁₀= 0.405608