Question

Q. Apply Runge Kutta Method to slove an I.V.P. comlicated problem.

Answer 1

Answer on Question #76736 – Math – Quantitative Methods

Question

Apply Runge Kutta Method to solve an I.V.P. complicated problem.

Solution

The motion of a material point of mass $m$ under the action of external forces is described by Newton's second law. Let the point move along the $x$ axis, then the function $x(t)$ is the position of the point at time $t$, it satisfies the ordinary differential equation of the second order $mx'' = F(t, x, x')$. For example, balance of forces for damped harmonic oscillators:

$$x'' + 2x' + 10x = 0, \quad x(0) = 2, \quad x'(0) = 5$$

The exact solution:

$$x(t) = e^{-t} \left( \frac{7}{3} \sin(3t) + 2 \cos(3t) \right)$$

For a numerical solution, we transform the second-order differential equation into a system of first-order differential equations by means of the change of variables:

$$y0(t) = x(t), \quad y1(t) = x'(t)$$ $$\left\{ \begin{array}{c} y0'(t) = x'(t) = y1(t) \\ y1'(t) = x''(t) = -2 \cdot y1(t) - 10 \cdot y0(t) \end{array} \right., \quad x(0) = y0(0) = 2, \quad x'(0) = y1(0) = 5$$

Approximations calculated using the Runge-Kutta method of order 4. For this method, each step requires the use of the following four values (h-step size):

$$k_1 = h \cdot f(t_n, y_n)$$ $$k_2 = h \cdot f\left(t_n + \frac{1}{2} h, y_n + \frac{1}{2} k_1\right)$$ $$k_3 = h \cdot f\left(t_n + \frac{1}{2} h, y_n + \frac{1}{2} k_2\right)$$ $$k_4 = h \cdot f(t_n + h, y_n + k_3)$$

Then we calculate the approximation to the solution:

$$y_{n+1} = y_n + \frac{1}{6} \left( k_1 + 2k_2 + 2k_3 + k_4 \right)$$

For the system of equations:

$$\left\{ \begin{array}{l} y_{n+1}^0 = y_n^0 + \frac{1}{6} \left( k_1^0 + 2k_2^0 + 2k_3^0 + k_4^0 \right) \\ y_{n+1}^1 = y_n^1 + \frac{1}{6} \left( k_1^1 + 2k_2^1 + 2k_3^1 + k_4^1 \right) \end{array} \right.$$

The difference of solutions:

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