Q5.3-27E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Generalized Blasius Equation. H. Blasius, in his study of the laminar flow of a fluid, encountered an equation of the form $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{yy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{(}\mathbf{y}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$. Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution that satisfies the initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{32824}$. Sketch this solution on the interval [0, 2].

Step-by-Step Solution

Answer

The result can get by the Runge-Kutta method, and the result is y(2)=1.6001.

Transform the equation

Here the equation is $\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{yy}\text{'}\text{'}=(\mathrm{y}\text{'}{)}^2-1$.

 

The system can be written as:

$$\begin{gathered} {\mathrm{x}}_1=\mathrm{y} \\ {\mathrm{x}}_2=\mathrm{y}\text{'}=\mathrm{x}{\text{'}}_1 \\ {\mathrm{x}}_3=\mathrm{y}\text{'}\text{'}=\mathrm{x}{\text{'}}_2 \end{gathered}$$

The transform equation is:

 

$$\begin{gathered} \mathrm{x}{\text{'}}_1={\mathrm{x}}_2 \\ \mathrm{x}{\text{'}}_2={\mathrm{x}}_3 \\ \mathrm{x}{\text{'}}_3=-{\mathrm{x}}_1{\mathrm{x}}_3+{{\mathrm{x}}^2}_2-1 \end{gathered}$$

 

The initial conditions are:

 $$\begin{gathered} {\mathrm{x}}_1\left(0\right)=\mathrm{y}\left(0\right)=0 \\ {\mathrm{x}}_2\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0 \\ {\mathrm{x}}_3\left(0\right)=\mathrm{y}\text{'}\text{'}\left(0\right)=1.32824 \end{gathered}$$

Apply the Runge-Kutta method

For h=0.1

t	Y	T	Y
0	0	1.1	0.599
0.1	0.00647	1.2	0.69515
0.2	0.0252	1.3	0.79515
0.3	0.0553	1.4	0.89926
0.4	0.0957	1.5	1.0072
0.5	0.1456	1.6	1.1189
0.6	0.20407	1.7	1.234
0.7	0.27032	1.8	1.3526
0.8	0.34363	1.9	1.4747
0.9	0.4233	2	1.6001
1	0.50882

Graph

Therefore, the value of y(2)=1.6001.

 

Thus, this is the required result.