Fluid Flow. In the study of the no isothermal flow of a Newtonian fluid between parallel plates, the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{x}}^{\bf{2}}}}}{\bf{ + }}{{\bf{x}}^{\bf{2}}}{{\bf{e}}^{\bf{y}}}{\bf{ = 0,x > 0}}\) , was encountered. By a series of substitutions, this equation can be transformed into the first-order equation\(\frac{{{\bf{dv}}}}{{{\bf{du}}}}{\bf{ = u}}\left( {\frac{{\bf{u}}}{{\bf{2}}}{\bf{ + 1}}} \right){{\bf{v}}^{\bf{3}}}{\bf{ + }}\left( {{\bf{u + }}\frac{{\bf{5}}}{{\bf{2}}}} \right){{\bf{v}}^{\bf{2}}}\). Use the fourth-order Runge–Kutta algorithm to approximate \({\bf{v(3)}}\) if \({\bf{v(u)}}\) satisfies\({\bf{v(}}2{\bf{)}} = 0.1\). For a tolerance of, \({\bf{\varepsilon = 0}}{\bf{.0001}}\) use a stopping procedure based on the relative error.

Using the improved 4^th order Runge-Kutta subroutine with tolerance \(\xi {\bf{ = 0}}{\bf{.0001}}\).

Since \({\bf{f(x,y) = u}}\left( {\frac{{\bf{u}}}{{\bf{2}}}{\bf{ + 1}}} \right){{\bf{v}}^{\bf{3}}}{\bf{ + }}\left( {{\bf{u + }}\frac{{\bf{5}}}{{\bf{2}}}} \right){{\bf{v}}^{\bf{2}}}\) and \({\bf{u = }}{{\bf{u}}_{\bf{0}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}{\bf{.1}}\) and h = 1

\(\begin{array}{l}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(u,v) = h}}\left[ {{\bf{u}}\left( {\frac{{\bf{u}}}{{\bf{2}}}{\bf{ + 1}}} \right){{\bf{v}}^{\bf{3}}}{\bf{ + }}\left( {{\bf{u + }}\frac{{\bf{5}}}{{\bf{2}}}} \right){{\bf{v}}^{\bf{2}}}} \right]\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = h}}\left[ {\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}} \right)\left( {\frac{{{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}}}{{\bf{2}}}{\bf{ + 1}}} \right){{\left( {{\bf{v + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right)}^{\bf{3}}}{\bf{ + }}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{ + }}\frac{{\bf{5}}}{{\bf{2}}}} \right){{\left( {{\bf{v + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right)}^{\bf{2}}}} \right]\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = h}}\left[ {\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}} \right)\left( {\frac{{{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}}}{{\bf{2}}}{\bf{ + 1}}} \right){{\left( {{\bf{v + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right)}^{\bf{3}}}{\bf{ + }}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{ + }}\frac{{\bf{5}}}{{\bf{2}}}} \right){{\left( {{\bf{v + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right)}^{\bf{2}}}} \right]\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{u + h,v + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = h}}\left[ {\left( {{\bf{u + h}}} \right)\left( {\frac{{{\bf{u + h}}}}{{\bf{2}}}{\bf{ + 1}}} \right){{\left( {{\bf{v + }}{{\bf{k}}_{\bf{3}}}} \right)}^{\bf{3}}}{\bf{ + }}\left( {{\bf{u + h + }}\frac{{\bf{5}}}{{\bf{2}}}} \right){{\left( {{\bf{v + }}{{\bf{k}}_{{{\bf{3}}_{}}}}} \right)}^{\bf{2}}}} \right]\\{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(u,v) = 0}}{\bf{.049}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.088356}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.120795}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{u + h,v + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.348857}}\end{array}\)

\(\begin{array}{c}{\bf{u = }}{{\bf{u}}_{\bf{o}}}{\bf{ + h = 3}}\\{\bf{v = }}{{\bf{v}}_{\bf{o}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 0}}{\bf{.236}}\end{array}\)

Therefore \({\rm{v}}\left( 3 \right) = 0.236\).

This is the solution of IVP.

Now, the relative error is \(\xi {\bf{ = }}\left| {\frac{{{\bf{0}}{\bf{.236027 - 0}}{\bf{.1}}}}{{{\bf{0}}{\bf{.236027}}}}} \right|{\bf{ = 0}}{\bf{.57632 > 0}}{\bf{.0001}}\)

Apply the same procedure for h = 0.5, \({\bf{u = }}2{\bf{,v = }}0.1\).

\(\begin{array}{c}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(u,v) = 0}}{\bf{.245}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.033306}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.036114}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{u + h,v + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.053410}}\\{\bf{u = 2 + 0}}{\bf{.5 = 2}}{\bf{.5}}\\{\bf{v = 0}}{\bf{.1 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 0}}{\bf{.136125}}\end{array}\)

And

\(\begin{array}{c}{{\bf{k}}_{\bf{1}}}{\bf{ = h}}{\rm{f}}{\bf{(u,v) = 0}}{\bf{.053419}}\\{{\bf{k}}_{\bf{2}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{1}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.083702}}\\{{\bf{k}}_{\bf{3}}}{\bf{ = hf}}\left( {{\bf{u + }}\frac{{\bf{h}}}{{\bf{2}}}{\bf{,v + }}\frac{{{{\bf{k}}_{\bf{2}}}}}{{\bf{2}}}} \right){\bf{ = 0}}{\bf{.101558}}\\{{\bf{k}}_{\bf{4}}}{\bf{ = hf}}\left( {{\bf{u + h,v + }}{{\bf{k}}_{\bf{3}}}} \right){\bf{ = 0}}{\bf{.205709}}\\{\bf{u = 2}}{\bf{.5 + 0}}{\bf{.5 = 3}}\\{\bf{v = 0}}{\bf{.1 + }}\frac{{\bf{1}}}{{\bf{6}}}\left( {{{\bf{k}}_{\bf{1}}}{\bf{ + 2}}{{\bf{k}}_{\bf{2}}}{\bf{ + 2}}{{\bf{k}}_{\bf{3}}}{\bf{ + }}{{\bf{k}}_{\bf{4}}}} \right)\\{\bf{ = 0}}{\bf{.241066}}\end{array}\)

Thus, \({\rm{v}}\left( 3 \right) = 0.241066\)

The relative error is \(\xi {\bf{ = }}\left| {\frac{{{\bf{0}}{\bf{.2}}41066{\bf{ - 0}}.236027}}{{{\bf{0}}{\bf{.2}}41066}}} \right|{\bf{ = 0}}.020903{\bf{ > 0}}{\bf{.0001}}\)

Apply the same procedure for other values. The values are

\(\begin{array}{c}{\rm{v}}{\bf{(3) = v(3;0}}{\bf{.25) = 0}}{\bf{.241854}}\\\xi {\bf{ = }}\left| {\frac{{{\bf{0}}{\bf{.241854 - 0}}{\bf{.241066}}}}{{{\bf{0}}{\bf{.241854}}}}} \right|{\bf{ = 0}}{\bf{.003258 > 0}}{\bf{.0001}}\\{\rm{v}}{\bf{(3) = v(3;0}}{\bf{.125) = 0}}{\bf{.241924}}\\\xi {\bf{ = }}\left| {\frac{{{\bf{0}}{\bf{.241924 - 0}}{\bf{.241854}}}}{{{\bf{0}}{\bf{.241924}}}}} \right|{\bf{ = 0}}{\bf{.00029 > 0}}{\bf{.0001}}\\{\rm{v}}{\bf{(3) = v(3;0}}{\bf{.0625) = 0}}{\bf{.241929}}\\\xi {\bf{ = }}\left| {\frac{{{\bf{0}}{\bf{.241929 - 0}}{\bf{.241924}}}}{{{\bf{0}}{\bf{.241929}}}}} \right|{\bf{ = 0}}{\bf{.00002 > 0}}{\bf{.0001}}\end{array}\)

Hence,\({\rm{v}}\left( 3 \right) = 0.241929\)

Hence the solution is \({\rm{v}}\left( 3 \right) = 0.241929\)with h = 0.0625