Q5.3-21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Fluid Ejection. In the design of a sewage treatment plant, the following equation arises: $\mathbf{60}\mathbf{-}\mathbf{H}\mathbf{=}\mathbf{(}\mathbf{77}\mathbf{.}\mathbf{7}\mathbf{)}\mathbf{H}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{19}\mathbf{.}\mathbf{42}\mathbf{\left)}\mathbf{\right(}\mathbf{H}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{2}}\mathbf{;}\mathbf{H}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{H}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ where H is the level of the fluid in an ejection chamber, and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h = 0.5 to approximate $\mathbf{H}\left(t\right)$ over the interval [0, 5].

Step-by-Step Solution

Answer

The result can get by the Runge-Kutta method.

Transform the equation

Here the equation is $60-\mathrm{H}=(77.7)\mathrm{H}\text{'}\text{'}+(19.42\left)\right(\mathrm{H}\text{'}{)}^2;\mathrm{H}\left(0\right)=\mathrm{H}\text{'}\left(0\right)=0.$

 

The system can be written as:

 $$\begin{gathered} {\mathrm{x}}_1=\mathrm{H}\left(\mathrm{t}\right) \\ {\mathrm{x}}_2=\mathrm{H}\text{'}=\mathrm{x}{\text{'}}_1 \\ \mathrm{H}\text{'}\text{'}=\mathrm{x}{\text{'}}_2 \end{gathered}$$

 

The transform equation:

 

$$\begin{gathered} \mathrm{x}{\text{'}}_1={\mathrm{x}}_2 \\ \mathrm{x}{\text{'}}_2=\frac{[60-{\mathrm{x}}_1-(19.42){{\mathrm{x}}^2}_2]}{77.7} \end{gathered}$$

 

The initial conditions are:

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

Apply the Runge-Kutta method.

Here h=0.5, N=10 steps, ${\mathrm{x}}_{1,0}=0,{\mathrm{x}}_{2,0}=0$ then;

$$\begin{gathered} x_{1,0}=0,x_{2,0}=0 \\ {k}_{1,1}=h{x}_{2,0}=0.5\left(0\right)=0 \\ {k}_{2,1}=\frac{h\left[60-x_{1,0}-(19.42){x^2}_{2,0}\right]}{77.7}=0.38610 \\ {k}_{1,2}=0.09653 \\ {k}_{2,2}=0.38144 \\ {k}_{1,3}=0.09536 \\ {k}_{2,3}=0.38124 \\ {k}_{1,4}=0.19062 \\ {k}_{2,4}=0.36732 \end{gathered}$$

Apply the recursive formula

$$\begin{gathered} {\mathrm{t}}_1={\mathrm{t}}_0+\mathrm{h} \\ 0+0.5=0.5 \\ {\mathrm{x}}_1(0.5)={\mathrm{x}}_{1,1}=0.09573 \\ {\mathrm{x}}_2(0.5)={\mathrm{x}}_{2,1}=0.37980 \end{gathered}$$

Apply the same procedure for n=1, 2,…., 9.

N	$t_n$	${\mathrm{x}}_{1,\mathrm{n}}=\mathrm{H}({\mathrm{t}}_{\mathrm{n}})$
0	0	0
1	0.5	0.09573
2	1.0	0.37389
3	1.5	0.81045
4	2.0	1.37361
5	2.5	2.03111
6	3.0	2.7549
7	3.5	3.52322
8	4.0	4.319
9	4.5	5.13307
10	5.0	5.95554

This is the required result.