Question

In Problems \(5 - 8\)use the Adams-Bashforth-Moulton method to approximate \(y(1.0)\), where \(y(x)\) is the solution of the given initial-value problem. First use \(h = 0.2\)and then use \(h = 0.1\).Use the \(RK4\) method to compute \({y_1},{y_2}\)and \({y_3}\).

\(y' = {(x - y)^2},y(0) = 0\).

Short answer

The value of\(y(1.0)\) calculated using Adams-Bashforth-Moulton method with \(h = 0.2\) is\({y_5} = 0.24.\)

The value of\(y(1.0)\) calculated using Adams-Bashforth-Moulton method with \(h = 0.1\) is\({y_{10}} = 0.23873.\)

Step-by-step solution

Define Adams-Bashforth-Moulton Method.

• The Adams–Bashforth-Moulton methods allow us to calculate the estimated solution at a given instant from prior instants' solutions explicitly.
• Each phase of the Adams–Moulton approach produces an algebraic matrix Riccati equation (AMRE), which is then solved using Newton's method.
• The predictive formula for this method is,
• \({y_{n + 1,p}} = {y_n} + \frac{h}{{24}}\left( {55y_n^\cent - 59y_{n - 1}^\cent + 37y_{n - 2}^\cent - 9y_{n - 3}^\cent} \right)\).

Find the value of the constants with \(h = 0.2\).

We have the initial-value problem,

\(y' = {(x - y)^2},\quad y(0) = 0\)

and we have to approximate the value \(y(1.0)\) with \(h = 0.2\)using the following technique.

First we have to obtain three approximate values \({y_1},{y_2},{y_3}\)using the formula

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

Before that we have to find the constants \({k_1},{k_2},{k_3},{k_4}\)at \(n = 0\)and \(h = 0.2\)as the following.

Since we have \({x_0} = 0,{y_0} = 0\), then we have

\(\begin{aligned}{*{20}{c}}{{k_1} = f\left( {{x_0},{y_0}} \right)}\\{ = {{\left( {{x_0} - {y_0}} \right)}^2}}\\{ = 0}\\{{k_2} = f\left( {{x_0} + \frac{1}{2}h,{y_0} + \frac{1}{2}h{k_1}} \right)}\\{ = {{\left[ {\left( {{x_0} + \frac{1}{2}h} \right) - \left( {{y_0} + \frac{1}{2}h{k_1}} \right)} \right]}^2}}\\{ = {{\left[ {\left( {0 + \frac{1}{2} \times 0.2} \right) - \left( {0 + \frac{1}{2} \times 0.2 \times 0} \right)} \right]}^2}}\\{ = 0.01}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_3} = f\left( {{x_0} + \frac{1}{2}h,{y_0} + \frac{1}{2}h{k_2}} \right)}\\{ = {{\left[ {\left( {{x_0} + \frac{1}{2}h} \right) - \left( {{y_0} + \frac{1}{2}h{k_2}} \right)} \right]}^2}}\\{ = {{\left[ {\left( {0 + \frac{1}{2} \times 0.2} \right) - \left( {0 + \frac{1}{2} \times 0.2 \times 0.01} \right)} \right]}^2}}\\{ = {{(0.1 - 0.001)}^2}}\\{ = 0.009801}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_4} = f\left( {{x_0} + h,{y_0} + h{k_3}} \right)}\\{ = {{\left[ {\left( {{x_0} + h} \right) - \left( {{y_0} + h{k_2}} \right)} \right]}^2}}\\{ = {{[(0 + 0.2) - (0 + 0.2 \times 0.009081)]}^2}}\\{ = {{(0.2 - 0.0019602)}^2}}\\{ = 0.03922}\end{aligned}\)\(\)

Hence the values of the constants with \(h = 0.2\) are\({k_1} = 0,{k_2} = 0.01,{k_3} = 0.009801,{k_4} = 0.03922\)

Find the value of\({y_1}\)with\(h = 0.2\).

We need to find the value of \({y_1}\)by using \((1)\).

\(\begin{aligned}{*{20}{c}}{{y_1} = {y_0} + \frac{h}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)}\\{ = 0 + \frac{{0.2}}{6}[0 + 2(0.01) + 2(0.009801) + 0.03922]}\\{ = 0 + \frac{{0.2}}{6} \times (0.078822)}\\{ = 0.006274}\end{aligned}\)

After this we can obtain the value of \(y\)for each value of \(x\)until \(x = 0.6\)and \({y_3} = 0.065699\)using the below table.

\(h = 0.2\)
\({x_n}\)	\({k_1} + 2{k_2} + 2{k_3} + {k_4}\)	\({y_n}\)
\(0.00\)	\(6.02008\)	\(0.00\)
\(0.20\)	\(6.17432\)	\(0.006274\)
\(0.40\)	\(6.39923\)	\(0.023349\)
\(0.60\)		\(0.065699\)

Hence the value of \({y_1}\)with \(h = 0.2\) is\(0.006274\).

Find the value of the predictor \({y_4}*.\)

We have to find the value of \(y(1.0)\)using Adams-Bashforth-Moulton method. Since we have, \({x_0} = 0,{x_1} = 0.2,{x_2} = 0.4{\rm{,\;}}{x_3} = 0.6\)and the value of \(y\)from the table we can find,

\(\begin{aligned}{*{20}{c}}{y_0^\cent = f\left( {{x_0},{y_0}} \right)}\\{ = {{\left( {{x_0} - {y_0}} \right)}^2}}\\{ = {{(0 - 0)}^2}}\\{ = 0}\\{y_1^\cent = f\left( {{x_1},{y_1}} \right)}\\{ = {{\left( {{x_1} - {y_1}} \right)}^2}}\\{ = {{(0.2 - 0.006274)}^2}}\\{ = 0.038956}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{y_2^\cent = f\left( {{x_2},{y_2}} \right)}\\{ = {{\left( {{x_2} - {y_2}} \right)}^2}}\\{ = {{(0.4 - 0.023349)}^2}}\\{ = 0.141866}\\{y_3^\cent = f\left( {{x_3},{y_3}} \right)}\\{ = {{\left( {{x_3} - {y_3}} \right)}^2}}\\{ = {{(0.6 - 0.065699)}^2}}\\{ = 0.285478}\end{aligned}\)

After these we can get the predictor \({y_4}*\)using,

\(\begin{aligned}{*{20}{c}}{y_{n + 1}^ * = y_4^ * = {y_n} + \frac{h}{{24}}\left( {55y_n^\cent - 59y_{n - 1}^\cent + 37y_{n - 2}^\cent - 9y_{n - 3}^\cent} \right)}\\{ = {y_3} + \frac{h}{{24}}\left( {55y_3^\cent - 59y_2^\cent + 37y_1^\cent - 9y_0^\cent} \right)}\\{ = 0.065699 + \frac{{0.2}}{{24}}[55 \times 0.285478 - 59 \times 0.141866 + 37 \times 0.038956 - 0]}\\{ = 0.065699 + \frac{{0.2}}{{24}} \times (8.772568)}\\{ = 0.138804}\end{aligned}\)

where \(n = 3\).

Hence the value of the predictor \({y_4}*\)is \(0.138804\).

Find the value of \(y(1.0).\)

After that using the predictor and \({x_4} = 0.8\)we can find,

\(\begin{aligned}{*{20}{c}}{y_4^\cent = f\left( {{x_4},y_4^ * } \right)}\\{ = {{\left( {{x_4} - y_4^ * } \right)}^2}}\\{ = {{(0.8 - 0.138804)}^2}}\\{ = 0.437181}\end{aligned}\)

Now we can find the approximate value of \(y(0.8)\)as

\(\begin{aligned}{*{20}{c}}{{y_{n + 1}} = {y_4} = {y_n} + \frac{h}{{24}}\left( {9y_{n + 1}^\cent + 19y_n^\cent - 5y_{n - 1}^\cent + y_{n - 2}^\cent} \right)}\\{ = {y_3} + \frac{h}{{24}}\left( {9y_4^\cent + 19y_3^\cent - 5y_2^\cent + y_1^\cent} \right)}\\{ = 0.065699 + \frac{{0.2}}{{24}}[9 \times 0.437181 + 19 \times 0.285478 - 5 \times 0.141866 + 0.038956]}\\{ = 0.065699 + \frac{{0.2}}{{24}}(8.688333)}\\{ = 0.138102}\end{aligned}\)

where \(n = 3\).

After that we can obtain the value of \(y(1.0)\)which equals \(0.24\)as shown in the table below.

\(h = 0.2\)
\({x_n}\)	\(y{*_{n + 1}}\)	\({y_{n + 1}}\)
\(0.80\)	\(0.138804\)	\(0.138102\)
\(1.00\)	\(0.238940\)	\(0.240051\)

Hence the value of \(y(1.0)\)is\(0.24.\)

Find the value of the constants with \(h = 0.1\).

We have the initial-value problem,

\(y' = {(x - y)^2},\quad y(0) = 0\)

and we have to approximate the value \(y(1.0)\) with \(h = 0.1\)using the following technique.

First we have to obtain three approximate values \({y_1},{y_2},{y_3}\)using the formula

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

Before that we have to find the constants \({k_1},{k_2},{k_3},{k_4}\)at \(n = 0\)and \(h = 0.1\)as the following.

Since we have \({x_0} = 0,{y_0} = 0\), then we have

\(\begin{aligned}{*{20}{c}}{{k_1} = f\left( {{x_0},{y_0}} \right)}\\{ = {{\left( {{x_0} - {y_0}} \right)}^2}}\\{ = 0}\\{{k_2} = f\left( {{x_0} + \frac{1}{2}h,{y_0} + \frac{1}{2}h{k_1}} \right)}\\{ = {{\left[ {\left( {{x_0} + \frac{1}{2}h} \right) - \left( {{y_0} + \frac{1}{2}h{k_1}} \right)} \right]}^2}}\\{ = {{\left[ {\left( {0 + \frac{1}{2} \times 0.1} \right) - \left( {0 + \frac{1}{2} \times 0.1 \times 0} \right)} \right]}^2}}\\{ = 0.0025}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_3} = f\left( {{x_0} + \frac{1}{2}h,{y_0} + \frac{1}{2}h{k_2}} \right)}\\{ = {{\left[ {\left( {{x_0} + \frac{1}{2}h} \right) - \left( {{y_0} + \frac{1}{2}h{k_1}} \right)} \right]}^2}}\\{ = {{\left[ {\left( {0 + \frac{1}{2} \times 0.1} \right) - \left( {0 + \frac{1}{2} \times 0.1 \times 0.0025} \right)} \right]}^2}}\\{ = {{(0.05 - 0.000125)}^2}}\\{ = 0.002488}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{{k_4} = f\left( {{x_0} + h,{y_0} + h{k_3}} \right)}\\{ = {{\left[ {\left( {{x_0} + h} \right) - \left( {{y_0} + h{k_2}} \right)} \right]}^2}}\\{ = {{[(0 + 0.1) - (0 + 0.1 \times 0.002488)]}^2}}\\{ = {{(0.1 - 0.0002488)}^2}}\\{ = 0.009950}\end{aligned}\)

Hence the values of the constants are \({k_1} = 0,{k_2} = 0.0025,{k_3} = 0.002488,{k_4} = 0.009950\).

Find the value of\({y_1}\).

We need to find the value of \({y_1}\)by using \((1)\).

\(\begin{aligned}{*{20}{c}}{{y_1} = {y_0} + \frac{h}{6}\left( {{k_1} + 2{k_2} + 2{k_3} + {k_4}} \right)}\\{ = 0 + \frac{{0.1}}{6}[0 + 2(0.0025) + 2(0.002488) + 0.009950]}\\{ = 0 + \frac{{0.1}}{6} \times (0.0647)}\\{ = 0.001078}\end{aligned}\)

After this we can obtain the value of \(y\)for each value of \(x\)with another four constants as shown in the table below.

\(h = 0.1\)
\({x_n}\)	\({k_1} + 2{k_2} + 2{k_3} + {k_4}\)	\({y_n}\)
\(0.00\)	\(0.604700\)	\(0.00\)
\(0.10\)	\(0.136256\)	\(0.001078\)
\(0.20\)	\(0.361699\)	\(0.003349\)
\(0.30\)		\(0.009377\)

Hence the value of \({y_1}\)is\(0.001078\).

Find the value of the predictor \({y_4}*.\)

We have to find the value of \(y(1.0)\)using Adams-Bashforth-Moulton method. Since we have, \({x_0} = 0,{x_1} = 0.1{\rm{,\;}}{x_2} = 0.2,{x_3} = 0.3\)and the value of \(y\)from the table we can find,

\(\begin{aligned}{*{20}{c}}{y_0^\cent = f\left( {{x_0},{y_0}} \right)}\\{ = {{\left( {{x_0} - {y_0}} \right)}^2}}\\{ = {{(0 - 0)}^2}}\\{ = 0}\\{y_1^\cent = f\left( {{x_1},{y_1}} \right)}\\{ = {{\left( {{x_1} - {y_1}} \right)}^2}}\\{ = {{(0.1 - 0.001078)}^2}}\\{ = 0.009786}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}{y_2^\cent = f\left( {{x_2},{y_2}} \right)}\\{ = {{\left( {{x_2} - {y_2}} \right)}^2}}\\{ = {{(0.2 - 0.003349)}^2}}\\{ = 0.038672}\\{y_3^\cent = f\left( {{x_3},{y_3}} \right)}\\{ = {{\left( {{x_3} - {y_3}} \right)}^2}}\\{ = {{(0.3 - 0.009377)}^2}}\\{ = 0.084462}\end{aligned}\)

After these we can get the predictor \({y_4}*\)using,

\(\begin{aligned}{*{20}{c}}{y_{n + 1}^ * = y_4^ * = {y_n} + \frac{h}{{24}}\left( {55y_n^\cent - 59y_{n - 1}^\cent + 37y_{n - 2}^\cent - 9y_{n - 3}^\cent} \right)}\\{ = {y_3} + \frac{h}{{24}}\left( {55y_3^\cent - 59y_2^\cent + 37y_1^\cent - 9y_0^\cent} \right)}\\{ = 0.009377 + \frac{{0.1}}{{24}}[55 \times 0.084462 - 59 \times 0.038672 + 37 \times 0.009786 - 0]}\\{ = 0.009377 + \frac{{0.1}}{{24}} \times (2.72583)}\\{ = 0.020735}\end{aligned}\)

where \(n = 3\).

Hence the value of the predictor \({y_4}*\)is\(0.020735\).

Find the value of \(y(1.0).\)

After that using the predictor and \({x_4} = 0.4\)we can find,

\(\begin{aligned}{*{20}{c}}{y_4^\cent = f\left( {{x_4},y_4^ * } \right)}\\{ = {{\left( {{x_4} - y_4^ * } \right)}^2}}\\{ = {{(0.4 - 0.020735)}^2}}\\{ = 0.143842}\end{aligned}\)

Now we can find the approximate value of \(y(1.0)\)as

\(\begin{aligned}{*{20}{c}}{{y_{n + 1}} = {y_4} = {y_n} + \frac{h}{{24}}\left( {9y_{n + 1}^\cent + 19y_n^\cent - 5y_{n - 1}^\cent + y_{n - 2}^\cent} \right)}\\{ = {y_3} + \frac{h}{{24}}\left( {9y_4^\cent + 19y_3^\cent - 5y_2^\cent + y_1^\cent} \right)}\\{ = 0.009377 + \frac{{0.1}}{{24}}[9 \times 0.143842 + 19 \times 0.084462 - 5 \times 0.038672 + 0.009786]}\\{ = 0.009377 + \frac{{0.1}}{{24}}(2.715784)}\\{ = 0.020693}\end{aligned}\)

where \(n = 3\).

After that we can obtain the value of \(y(1.0)\)which equals \(0.238732\)as shown in the table below.

\(h = 0.1\)
\({x_n}\)	\(y{*_{n + 1}}\)	\({y_{n + 1}}\)
\(0.40\)	\(0.020735\)	\(0.020693\)
\(0.50\)	\(0.038488\)	\(0.038469\)
\(0.60\)	\(0.063490\)	\(0.063480\)
\(0.70\)	\(0.096092\)	\(0.096106\)
\(0.80\)	\(0.136365\)	\(0.136385\)
\(0.90\)	\(0.184044\)	\(0.184073\)
\(1.00\)	\(0.238705\)	\(0.238732\)

Hence the value of \(y(1.0)\)is\(0.238732\).