Q5.3-26E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution to the initial value problem.

$$\begin{gathered} \mathrm{x}\text{'}=\mathrm{yz};\mathrm{x}\left(0\right)=0, \\ \mathrm{y}\text{'}=-\mathrm{xz};\mathrm{y}\left(0\right)=1, \\ \mathrm{z}\text{'}=\frac{-\mathrm{xy}}{2};\mathrm{z}\left(0\right)=1, \end{gathered}$$

At t=1.

Step-by-Step Solution

Answer

The required result is: $\mathrm{X}\left(1\right)=0.803,\mathrm{Y}\left(1\right)=0.59598,\mathrm{Z}\left(1\right)=0.82316$

Given conditions

Given initial value problem is:

$$\begin{gathered} x\text{'}=yz \\ y\text{'}=-xz \\ z\text{'}=\frac{-xy}{2} \end{gathered}$$

And the initial conditions are:

$$\begin{gathered} x\left(0\right)=0 \\ y\left(0\right)=1 \\ z\left(0\right)=1 \end{gathered}$$

Apply the Runge-Kutta method

To get the solution apply the Runge-Kutta method in mat lab for t=1.

 

Function n[t,x] = Runge_Kutta(f,t0,t_end,init_cond,h)
 %we begin at time t0 and end when we reach t_end
 %init_cond(i) contains the initial value of x_i
 %f contains functions such that x_i'=f_i(t,x1,x2,...)
 
 t(:,1)=t0; % t0 is the initial value of t
 x(:,1)=init_cond; % initial conditions are set
 
 i=1;
 while t(:,i) <t_end
 
     k1=f(t(i),x(:,i));
     k2=f(t(i)+0.5*h,x(:,i)+0.5*h*k1);
     k3=f(t(i)+0.5*h,x(:,i)+0.5*h*k2);
     k4=f(t(i)+h,x(:,i)+h*k3);
 
      x(:,i+1)=x(:,i)+(h/6)*(k1+2*k2+2*k3+k4);
 
     t(:,i+1)=t(:,i)+ h;
 i=i+1;
 
 end

clear all

 

init_cond=[0;1;1];

f=@(t,X) [X(2)*X(3);-X(1)*X(3);-X(1)*X(2)/2];

[t,x] = Runge_Kutta(f,0,1,init_cond,0.1);

 

table(t(:,end),x(1,end),x(2,end),x(3,end),'VariableNames',{'t','x','y','z'})

 

Then the results are;

T	X	Y	z
1	0.803	0.59598	0.82316

Therefore, $\mathrm{X}\left(1\right)=0.803,\mathrm{Y}\left(1\right)=0.59598,\mathrm{Z}\left(1\right)=0.82316$.