Q5.3-17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem

$$\begin{gathered} \frac{\mathbf{du}}{\mathbf{dx}}\mathbf{=}\mathbf{3}\mathbf{u}\mathbf{-}\mathbf{4}\mathbf{v}\mathbf{;}\mathbf{u}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{\text{'}} \\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{3}\mathbf{v}\mathbf{;}\mathbf{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

at x = 1. Starting with h=1, continue halving the step size until two successive approximations of u(1)and v(1) differ by at most 0.001.

Step-by-Step Solution

Answer

The solution is $u\left(1\right)=0.36789$ and $v\left(1\right)=0.36789$.

Transform the equation

Write the equation as $u\text{'}=3u-4v$ and $v\text{'}=2u-3v$.

The transformation of the equation is:

$$\begin{gathered} \mathrm{u}\text{'}\left(\mathrm{t}\right)=\mathrm{x}{\text{'}}_2\left(\mathrm{t}\right) \\ \mathrm{v}\text{'}\left(\mathrm{t}\right)={\mathrm{x}}_3\left(\mathrm{t}\right) \\ {\mathrm{x}}_2\left(\mathrm{t}\right)=\mathrm{v}\left(\mathrm{t}\right) \\ {\mathrm{x}}_1=\mathrm{u}\left(\mathrm{t}\right) \\ \mathrm{x}{\text{'}}_2\left(\mathrm{t}\right)=3{\mathrm{x}}_1-4{\mathrm{x}}_2 \\ \mathrm{x}{\text{'}}_3\left(\mathrm{t}\right)=2{\mathrm{x}}_1-3{\mathrm{x}}_2 \end{gathered}$$

The initial conditions are

$$\begin{gathered} u\left(0\right)=1 \\ v\left(0\right)=1 \end{gathered}$$

Apply Runge –Kutta method

For the solution apply the Runge-Kutta method in MATLAB for h=0.25, and the solution is u(1)=0.36789 and v(1)=0.36789.

Find that u 1 and v ( 1 ) differ by at most 0.001.

Subtracting the values of $u\left(1\right)$ and $v\left(1\right)$ then 

 

$$\begin{gathered} u\left(1\right)-v\left(1\right)=0.36789-0.36678 \\ =0.001101 \end{gathered}$$

 

Therefore, $u\left(1\right)$ and $v\left(1\right)$ differ by at most 0.001.