Chapter 26, Problem 6P

Solve the following initial-value problem from $t=4\text{ to 5}$:

$\frac{dy}{dt}=-\frac{2y}{t}$

Use a step size of 0.5 and initial values of $y\left(2.5\right)=0.48,y\left(3\right)=0.333333,$ $y\left(3.5\right)=0.244898,\text{ and }y\left(4\right)=0.1875$. Obtain your solutions using the following techniques: (a) the non-self-starting Heun method $\left({\epsilon }_S=1\%\right)$, and (b) the fourth-order Adams method $\left({\epsilon }_S=0.01\%\right)$. [Note: The exact answers obtained analytically are $y\left(4.5\right)=0.148148\text{ and }y\left(5\right)=0.12$.] Compute the true percent relative errors ${\epsilon }_t$ for your results.

To determine

To calculate: The solution of differential equation, $\frac{dy}{dt}=-\frac{2y}{t}$ from $t=4\text{to}t=5$ using the non-self-starting Heun method with step size of 0.5 and initial values $y\left(2.5\right)=0.48,y\left(3\right)=0.333333,y\left(3.5\right)=0.244898\text{and}y\left(4\right)=0.1875$. Also, ${\epsilon }_s=1\%$. Compute the true percent relative as well.

Answer to Problem 6P

Solution:

First step:

The true error is ${\epsilon }_t=0.3028\%$.

Corrector yields:

$\begin{array}{ccc}j& y_{j+1}^j& {\epsilon }_a\\ 1& 0.1472683& 2.63\%\\ 2& 0.1476994& 0.292\%\end{array}$

Second step:

The true error is ${\epsilon }_t=0.39\%$.

Corrector yields:

$\begin{array}{ccc}j& y_{j+1}^j& {\epsilon }_a\\ 1& 0.11953193& 0.236\%\end{array}$

Explanation of Solution

Given Information:

Differential equation, $\frac{dy}{dt}=-\frac{2y}{t}$, $t=4\text{to}t=5$ with step size of 0.5 and initial values $y\left(2.5\right)=0.48,y\left(3\right)=0.333333,y\left(3.5\right)=0.244898\text{and}y\left(4\right)=0.1875$. Also, ${\epsilon }_s=1\%$.

Formula used:

(1) Predictor for non-self-starting Heun method is given by,

$y_{i+1}^0=y_{i-1}^m+f\left(t_{i,}{y}_i^m\right)2h$

(2) Corrector for non-self-starting Heun method is given by,

$y_{i+1}^j=y_i^m+\frac{f\left(t_i,y_i^m\right)+f\left(t_{i+1},y_{i+1}^{j-1}\right)}{2}h$

(3) Predictor modifier is given by:

$y_{i+1}^0=y_{i+1,u}^0+\frac{4}{5}\left(y_{i,u}^m-y_{i,u}^0\right)$

where, the subscript u designates that the variable is unmodified.

Calculation:

Consider the given differential equation, $\frac{dy}{dt}=-\frac{2y}{t}$. Here, $f\left(t,y\right)=-\frac{2y}{t}$.

Predictor for non-self-starting Heun method is given by,

$y_{i+1}^0=y_{i-1}^m+f\left(t_{i,}{y}_i^m\right)2h$.. .. .. (1)

For the given differential equation, it can be written as

$y_{i+1}^0=y_{i-1}^m+\left(\frac{-2y_i{}^m}{t_i}\right)2h$

Substitute, $i=0,m=0\text{and}h=0.5$ in above equation,

$\begin{array}{c}{y}_1^0=y_{-1}^0+\left(\frac{-2y_0{}^0}{t_0}\right)2\left(0.5\right)\\ =y_{-1}^0+\left(\frac{-2y_0{}^0}{t_0}\right)\end{array}$

The provided initials conditions are $y\left(2.5\right)=0.48,y\left(3\right)=0.333333,y\left(3.5\right)=0.244898\text{and}y\left(4\right)=0.1875$.

Now, substitute $y_{-1}^0=0.244898\text{ , }{y}_0^0=0.1875\text{ and }{t}_0=4$.

Thus,

$\begin{array}{c}{y}_1^0=0.244898+\left(\frac{-2\times 0.1875}{4}\right)\\ =0.151148\end{array}$

Corrector for non-self-starting Heun method is given by,

$y_{i+1}^j=y_i^m+\frac{f\left(t_i,y_i^m\right)+f\left(t_{i+1},y_{i+1}^{j-1}\right)}{2}h$

For the given differential equation, it can be written as,

$y_{i+1}^j=y_i^m+\left[\frac{\left(\frac{-2y_i{}^m}{t_i}\right)+\left(\frac{-2y_{i+1}{}^{j-1}}{t_{i+1}}\right)}{2}\right]h$

Substitute $i=0\text{,}m=0\text{ and }j=1$ in above equation,

$y_1^1=y_0^0+\left[\frac{\left(\frac{-2y_0{}^0}{t_0}\right)+\left(\frac{-2y_1{}^0}{t_1}\right)}{2}\right]h$

Put $y_0^0=0.1875,y_1^0=0.151148,t_0=4,t_1=4.5\text{ and }h=0.5$

$\begin{array}{c}{y}_1^1=0.1875+\left[\frac{\left(\frac{-2\times 0.1875}{4}\right)+\left(\frac{-2\times 0.151148}{4.5}\right)}{2}\right]0.5\\ =0.147268277\end{array}$

Similarly, for $i=0\text{ and }j=2$ corrector gives $y_1^2=0.1476994$.

Error is calculated as,

$\begin{array}{c}{\epsilon }_a=\left|\frac{y_1^2-y_1^1}{y_1^2}\right|\times 100\\ =\left|\frac{0.1476994-0.1472683}{0.1476994}\right|\times 100\\ =0.2918766\end{array}$

Also, the true error can be computed as,

$\begin{array}{c}{\epsilon }_t=\left|\frac{0.148148-0.1476994}{0.148148}\right|\times 100\\ =0.3028\end{array}$

Hence, the true error is ${\epsilon }_t=0.3028\%$.

$\begin{array}{ccc}j& y_{j+1}^j& {\epsilon }_a\\ 1& 0.1472683& 2.63\%\\ 2& 0.1476994& 0.292\%\end{array}$

Second step:

Predictor

Substitute $i=1$ and $m=0$ in equation (1),

$y_2^0=y_0^0+\left(\frac{-2y_1{}^0}{t_1}\right)2h$

Substitute $y_0^0=4.762673,y_1^0=3.913253\text{and}{t}_1=2.5$ in above equation,

$\begin{array}{c}{y}_2^0=0.1875+\left(\frac{-2\times 0.151148}{4.5}\right)1\\ =0.12032311\end{array}$

Predictor modifier is given by:

$y_{i+1}^0=y_{i+1,u}^0+\frac{4}{5}\left(y_{i,u}^m-y_{i,u}^0\right)$

where, the subscript u designates that the variable is unmodified.

Now, substitute $i=1\text{and}m=2$ in above equation,

$y_2^0=y_{2,u}^0+\frac{4}{5}\left(y_{1,u}^2-y_{1,u}^0\right)$

Substitute, $y_{2,u}^0=0.12032311,y_{1,u}^2=0.1476994\text{ and }{y}_{1,u}^0=0.151148$ in above equation,

$\begin{array}{c}{y}_2^0=0.12032311+\frac{4}{5}\left(0.1476994-0.151148\right)\\ =0.11756423\end{array}$

Corrector is given by:

$y_{i+1}^j=y_i^m+\frac{\left(\frac{-2y_i{}^m}{t_i}\right)+\left(\frac{-2y_{i+1}{}^{j-1}}{t_{i+1}}\right)}{2}h$

Substitute $i=1j=1,m=2\text{and}h=0.5$, m= 2 in above formula as shown below,

$y_2^1=y_1^2+\left[\frac{\left(\frac{-2y_1{}^2}{t_1}\right)+\left(\frac{-2y_2{}^0}{t_2}\right)}{2}\right]h$

Now, substitute $y_1^2=0.1476994$, $y_2^0=0.11756423$, $t_1=4.5$ and $t_2=5$ as shown below,

$\begin{array}{c}{y}_2^1=0.1476994+\left[\frac{\left(\frac{-2\times 0.1476994}{4.5}\right)+\left(\frac{-2\times 0.11756423}{5}\right)}{2}\right]0.5\\ =0.11953193\end{array}$

Error can be computed as,

$\begin{array}{c}{\epsilon }_a=\left|\frac{y_2^1-y_2^0}{y_2^0}\right|\times 100\\ =\left|\frac{0.11953193-0.11756423}{0.11953193}\right|\times 100\\ =1.646\%\end{array}$

The true error can be computed as,

$\begin{array}{c}{\epsilon }_t=\left|\frac{0.12-0.11953193}{0.12}\right|\times 100\\ =0.39\%\end{array}$

Hence, the true error is ${\epsilon }_t=0.39\%$.

Corrector yields:

$\begin{array}{ccc}j& y_{j+1}^j& {\epsilon }_a\\ 1& 0.11953193& 0.236\%\end{array}$

To determine

To calculate: The solution of differential equation, $\frac{dy}{dt}=-\frac{2y}{t}$ from $t=4\text{to}t=5$ using the fourth-order Adams method step size of 0.5 and initial values $y\left(2.5\right)=0.48,y\left(3\right)=0.333333,y\left(3.5\right)=0.244898\text{and}y\left(4\right)=0.1875$. Also, the exact answers obtained analytically are $y\left(4.5\right)=0.148148\text{and}y\left(5\right)=0.12$ and ${\epsilon }_s=0.01\%$. Compute the true percent relative as well.

Answer to Problem 6P

Solution:

The solution obtained for the given differential equation are shown below in tabular form:

$\text{Predictor}=\text{ 0}\text{.1527937}$

Corrector iteration:

$\begin{array}{cccc}t& y& {\epsilon }_a& {\epsilon }_t\\ 4.5& 0.14760545& 3.5149\%& 0.3322\%\\ 4.5& 0.14803781& 0.292\%& 0.074\%\\ 4.5& 0.148001773& 0.02435& 0.099\%\end{array}$

$\text{Predictor = 0}\text{.12165973}$

Corrector iteration:

$\begin{array}{cccc}t& y& {\epsilon }_a& {\epsilon }_t\\ 5& 0.1196901& 1.6456\%& 0.258\%\\ 5& 0.1198378& 0.123\%& 0.133\%\\ 5& 0.1198267& 0.0092\%& 0.142\%\end{array}$

Explanation of Solution

Given Information:

Differential equation, $\frac{dy}{dt}=-\frac{2y}{t}$, $t=4\text{to}t=5$ with step size of 0.5 and initial values $y\left(2.5\right)=0.48,y\left(3\right)=0.333333,y\left(3.5\right)=0.244898\text{and}y\left(4\right)=0.1875$. Values obtained analytically $y\left(4.5\right)=0.148148\text{and}y\left(5\right)=0.12$. Also, ${\epsilon }_s=0.01\%$.

Formula used:

(1) The fourth order Adams-Bashforth formula as predictor is given by,

$y_{i+1}^0=y_i^m+\frac{h}{24}\left(55f_i^m-59f_{i-1}^m+37f_{i-2}^m-9f_{i-3}^m\right)$

(2) The fourth order Adams-Bashforth formula as corrector is given by,

$y_{i+1}^j=y_i^m+\frac{h}{24}\left(9f_{i+1}^{j-1}+19f_i^m-5f_{i-1}^m+f_{i-2}^m\right)$

(3) Error is given as,

${\epsilon }_a=\left|\frac{y_{i+1}^j-y_{i+1}^{j-1}}{y_{i+1}^j}\right|\times 100$

Calculation:

Consider the given differential equation, $\frac{dy}{dt}=-\frac{2y}{t}$.

Here, $f\left(t\right)=\frac{-2y}{t}$.. .. .. (1)

Starting values of AdamsBash fourth method are,

At $t=2.5$, $y=0.48$. So, $y_{-3}=0.48$. Substitute this value in equation (1)

Thus,

$\begin{array}{c}{f}_{-3}=\frac{-2\times 0.48}{2.5}\\ =-0.384\end{array}$

At $t=3$, $y\left(3\right)=0.333333$. So, $y_{-2}=0.333333$. Substitute this value in equation (1)

Thus,

$\begin{array}{c}{f}_{-2}=\frac{-2\times 0.333333}{3}\\ =-0.222222\end{array}$

At $t=3.5$, $y=0.244898$. So, $y_{-1}=0.244898$. Substitute this value in equation (1)

Thus,

$\begin{array}{c}{f}_{-1}=\frac{-2\times 0.244898}{3.5}\\ =-0.1399417\end{array}$

At $t=4.0$, $y=0.1875$. So, $y_0=0.1875$. Substitute this value in equation (1)

Thus,

$\begin{array}{c}{f}_0=\frac{-2\times 0.1875}{4}\\ =-0.09375\end{array}$

The fourth order Adams-Bashforth formula as predictor is given by,

$y_{i+1}^0=y_i^m+\frac{h}{24}\left(55f_i^m-59f_{i-1}^m+37f_{i-2}^m-9f_{i-3}^m\right)$.. .. .. (2)

The fourth order Adams formula as the corrector is,

$y_{i+1}^j=y_i^m+\frac{h}{24}\left(9f_{i+1}^{j-1}+19f_i^m-5f_{i-1}^m+f_{i-2}^m\right)$.. .. .. (3)

Use predictor to compute a value at $t=4.5$, substitute $i=0$ and $m=0$ in equation (2),

$y_1^0=y_0^0+\frac{h}{24}\left(55f_0^{}-59f_{-1}^{}+37f_{-2}^{}-9f_{-3}^{}\right)$

Substitute values $f_0=-0.09375$, $f_{-1}=-0.1399417$, $f_{-2}=-0.222222$, $f_{-3}=-0.384$ and $h=0.5$.

$\begin{array}{c}{y}_1^0=0.1875+\frac{0.5}{24}\times \left(\begin{array}{l}55\left(-0.09375\right)-59\left(-0.1399417\right)+37\left(-0.222222\right)\\ -9\left(-0.384\right)\end{array}\right)\\ =0.1527937\end{array}$

Hence, at $t=4.5,y_1^0=0.1527937$

$\begin{array}{c}{f}_1=\frac{-2\times 0.1527937}{4.5}\\ =-0.0679083\end{array}$

Thus, $f_1=-0.0679083$.

Now, use corrector,

Put $i=0\text{ , }j=1$ in equation (3),

$y_1^1=y_0^0+\frac{h}{24}\left(9f_1+19f_0-5f_{-1}^{}+f_{-2}^{}\right)$

Substitute values $f_0=-0.09375$, $f_{-1}=-0.1399417$, $f_{-2}=-0.222222$, $f_1=-0.0679083$ and $h=0.5$.

$\begin{array}{c}{y}_1^1=0.1875+\frac{0.5}{24}\left(\begin{array}{l}9\left(-0.0679083\right)+19\left(-0.09375\right)-5\left(-0.1399417\right)\\ +\left(-0.222222\right)\end{array}\right)\\ =0.14760545\end{array}$

At $t=4.5\text{and}{y}_1^1=0.14760545$

$\begin{array}{c}{f}_1=\frac{-2\times 0.14760545}{4.5}\\ =-0.0656024\end{array}$

Thus, $f_1=-0.0656024$.

This result can be substituted back into equation (3) to iteratively correct the estimate,

$\begin{array}{c}{y}_1^2=0.1875+\frac{0.5}{24}\left(\begin{array}{l}9\left(-0.0656024\right)+19\left(-0.09375\right)-5\left(-0.1399417\right)\\ +\left(-0.222222\right)\end{array}\right)\\ =0.14803781\end{array}$

Again,

At $t=4.5\text{and}{y}_1^2=0.14803781$

$\begin{array}{c}{f}_1=\frac{-2\times 0.14803781}{4.5}\\ =-0.0657946\end{array}$

Thus, $f_1=-0.0657946$.

This result can be substituted back into equation (3) to iteratively correct the estimate,

$\begin{array}{c}{y}_1^3=0.1875+\frac{0.5}{24}\left(\begin{array}{l}9\left(-0.0657946\right)+19\left(-0.09375\right)-5\left(-0.1399417\right)\\ +\left(-0.222222\right)\end{array}\right)\\ =0.148001773\end{array}$

At $t=4.5\text{and}{y}_1^3=0.148001773$

$\begin{array}{c}{f}_1=\frac{-2\times 0.148001773}{4.5}\\ =-0.0657786\end{array}$

Thus, $f_1=-0.0657786$.

Now, use the predictor to find the value at $t=5$.

Substitute $i=1$ in equation (2),

$y_2^0=y_1^{}+\frac{h}{24}\left(55f_1^{}-59f_0^{}+37f_{-1}^{}-9f_{-2}^{}\right)$

Substitute values, $f_1=-0.0657786$ $f_0=-0.09375$, $f_{-1}=-0.1399417$ and $f_{-2}=-0.222222$ in above equation, to get

$\begin{array}{c}{y}_2^0=0.148001773+\frac{0.5}{24}\times \left(\begin{array}{l}55\left(-0.0657786\right)-59\left(-0.09375\right)+37\left(-0.1399417\right)\\ -9\left(-0.222222\right)\end{array}\right)\\ =0.12165973\end{array}$

At $t=5\text{and}{y}_2^0=0.12165973$.

$\begin{array}{c}{f}_2=\frac{-2\times 0.12165973}{5}\\ =-0.0486639\end{array}$

Thus, $f_2=-0.0486639$

Now error is given as,

${\epsilon }_a=\left|\frac{y_{i+1}^j-y_{i+1}^{j-1}}{y_{i+1}^j}\right|\times 100$

Calculate ${\epsilon }_a$ as shown below,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.14760545-0.1527937}{0.14760545}\right|\times 100\\ =3.5149\%\end{array}$

The true percent relative errors can be calculated as,

$\begin{array}{c}{\epsilon }_t=\left|\frac{0.148148-0.14760545}{0.148148}\right|\times 100\\ =0.36622\%\end{array}$

Corrector formula is given as,

$y_{i+1}^j=y_i^m+\frac{h}{24}\left(9f_{i+1}^{j-1}+19f_i^m-5f_{i-1}^m+f_{i-2}^m\right)$

Substitute $i=1\text{and}j=1$ in above formula,

$y_2^1=y_1^{}+\frac{h}{24}\left(9f_2^0+19f_1^{}-5f_0^{}+f_{-1}^{}\right)$

Substitute values, $f_1=-0.0657786$ $f_0=-0.09375$, $f_{-1}=-0.1399417$ and $f_2=-0.0486639$ in above equation, to get

$\begin{array}{c}{y}_2^1=0.148001773+\frac{0.5}{24}\left(\begin{array}{l}9\left(-0.0486639\right)+19\left(-0.0657786\right)-5\left(-0.09375\right)\\ +\left(-0.1399417\right)\end{array}\right)\\ =0.1196901\end{array}$

At $t=5\text{and}{y}_2^1=0.1196901$

$\begin{array}{c}{f}_2=\frac{-2\times 0.1196901}{5}\\ =-0.0478760\end{array}$

Thus, $f_2=-0.0478760$.

Hence,

$\begin{array}{c}{y}_2^2=0.148001773+\frac{0.5}{24}\left(\begin{array}{l}9\left(-0.0478760\right)+19\left(-0.0657786\right)-5\left(-0.09375\right)\\ +\left(-0.1399417\right)\end{array}\right)\\ =0.11983783\end{array}$

At $t=5\text{and}{y}_2^2=0.11983783$

$\begin{array}{c}{f}_2=\frac{-2\times 0.11983783}{5}\\ =-0.0479351\end{array}$

Thus, $f_2=-0.0479351$.

Hence,

$\begin{array}{c}{y}_2^3=0.148001773+\frac{0.5}{24}\left(\begin{array}{l}9\left(-0.0479351\right)+19\left(-0.0657786\right)-5\left(-0.09375\right)\\ +\left(-0.1399417\right)\end{array}\right)\\ =0.11982675\end{array}$

Now error is given as,

${\epsilon }_a=\left|\frac{y_{i+1}^j-y_{i+1}^{j-1}}{y_{i+1}^j}\right|\times 100$

Calculate ${\epsilon }_a$ as shown below,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.1196901-0.12165973}{0.1196901}\right|\times 100\\ =1.6456\%\end{array}$

The true percent relative errors can be calculated as,

$\begin{array}{c}{\epsilon }_t=\left|\frac{0.12-0.1196901}{0.12}\right|\times 100\\ =0.25825\%\end{array}$

Results are shown below in tabular form:

$\text{Predictor}=\text{ 0}\text{.1527937}$

$\begin{array}{cccc}t& y& {\epsilon }_a& {\epsilon }_t\\ 4.5& 0.14760545& 3.5149\%& 0.3322\%\\ 4.5& 0.14803781& 0.292\%& 0.074\%\\ 4.5& 0.148001773& 0.02435& 0.099\%\end{array}$

Corrector iteration:

$\text{Predictor = 0}\text{.12165973}$

Corrector iteration:

$\begin{array}{cccc}t& y& {\epsilon }_a& {\epsilon }_t\\ 5& 0.1196901& 1.6456\%& 0.258\%\\ 5& 0.1198378& 0.123\%& 0.133\%\\ 5& 0.1198267& 0.0092\%& 0.142\%\end{array}$