Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Transmission Lines. In the study of the electric field that is induced by two nearby transmission lines, an equation of the form $\frac{\mathit{d}\mathit{z}}{\mathit{d}\mathit{x}}\mathbf{+}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ arises. Let $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{2}$ and $\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$. If z(0)=1, use the fourth-order Runge–Kutta algorithm to approximate z(1). For a tolerance of $\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0001}$, use a stopping procedure based on the absolute error.

Step-by-Step Solution

Answer

$z\left(1\right)=2.870833$ with  $h=0.03125$

Step 1: Find the values of k i . i = 1 , 2 , 3 , 4

Using the improved 4^th order Runge-Kutta subroutine with tolerance $\xi =0.0001$.

 

Since $f(x,z)=5x+2-x^2{z}^2$ and $x=x_0=0,z=z_o=1$ and $h=1$

$$\begin{gathered} k_1=hf(x,z)=h(5x+2-x^2{z}^2) \\ {k}_2=hf\left(x+\frac{h}{2},z+\frac{k_1}{2}\right)=h\left[5\left(x+\frac{h}{2}\right)+2-{\left(x+\frac{h}{2}\right)}^2{\left(z+\frac{k_1}{2}\right)}^2\right] \\ {k}_3=hf\left(x+\frac{h}{2},z+\frac{k_2}{2}\right)=h\left[5\left(x+\frac{h}{2}\right)+2-{\left(x+\frac{h}{2}\right)}^2{\left(z+\frac{k_2}{2}\right)}^2\right] \\ {k}_4=hf\left(x+h,z+k_3\right)=h\left[5(x+h)+2(x+h{)}^2{(z+k_3)}^2\right] \\ {k}_1=h(x,z)=2 \\ {k}_2=hf\left(x+\frac{h}{2},z+\frac{k_1}{2}\right)=3.5 \\ {k}_3=hf\left(x+\frac{h}{2},z+\frac{k_2}{2}\right)=2.60938 \\ {k}_4=hf\left(x+h,z+k_3\right)=0.-6.02759 \end{gathered}$$ 

Step 2: Find the values of x and z

$$\begin{gathered} x=x_o+h=1 \\ z=z_o+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =2.365194 \end{gathered}$$

Therefore $z\left(1\right)=2.465194$.

 

This is the solution of IVP. 

Now,

$$\begin{gathered} \xi =\left|\frac{1-2.365194}{1}\right| \\ \xi =1.365194>0.0001 \end{gathered}$$  

Step 3: Determine the other values

Apply the same procedure for $h=0.5$, $x=0,z=1$.

$$\begin{gathered} k_1=hf(x,z)=1 \\ {k}_2=hf\left(x+\frac{h}{2},z+\frac{k_1}{2}\right)=1.55469 \\ {k}_3=hf\left(x+\frac{h}{2},z+\frac{k_2}{2}\right)=1.52628 \\ {k}_4=hf\left(x+h,z+k_3\right)=1.45224 \\ x=0+0.5=0.5 \\ z=1+\frac{1}{6}\left(k_1+2k_2+2k_3+k_4\right) \\ =2.4357 \end{gathered}$$ 

Thus, $z\left(1\right)=2.573986$

 

Now  

$\xi =\left|2.365194-2.573986\right|=0.208792>0.0001$

Step 4: Evaluate the other values

Apply the same procedure for other values. The values are

$$\begin{gathered} f\left(1\right)=v(1;0.25)=2.854533 \\ \xi =\left|2.573986-2.854553\right|=0.280574>0.0001 \\ f\left(1\right)=v(1;0.125)=2.870202 \\ \xi =\left|2.870202-2.854553\right|=0.015649>0.0001 \\ f\left(1\right)=v(1;0.0625)=2.870805 \\ \xi =\left|2.870805-2.870202\right|=0.000603>0.0001 \\ f\left(1\right)=v(1;0.03125)=2.870833 \\ \xi =\left|2.870833-2.870805\right|=0.000028<0.0001 \end{gathered}$$ 

 

Hence, $z\left(1\right)=2.870833$.

 

Hence the solution is $z\left(1\right)=2.870833$ with $h=0.03125$